honors topics in electrodynamics - high energy physics · physics 212, spring 2003. .1 introduction...

Physics 212, Spring 2003. .1 Introduction and syllabus

Physics 212

Honors topics in electrodynamics

Introduction and syllabus

Spring, 2003

George Gollin University of Illinois at Urbana-Champaign

2003



Physics 212

(...though it's officially PHYCS 199&HO, and it will be renumbered "PHYCS 222" next fall)

Instructor: George Gollin; Loomis 437D. 333-4451; [email protected] Teaching assistant: To be announced Course secretary: Donna Guzy; Loomis 441. 333-4452; [email protected] Prerequisites: Physics 111 and associated math courses. Corequisites: Concurrent registration in Physics 112. Description: In Physics 212 (called PHYCS 199&HO by the registrar for the time being), students will investigate some of the most interesting topics in electricity and magnetism at a more sophisticated level than would be appropriate for Physics 112. We welcome all interested students who have done well in Physics 111; it is likely that about half of Physics 212 students will be physics majors. Enrollment is limited (by fire safety considerations) to 30 students. Electrodynamics is a beautiful subject which has, at its heart, the fever-dream alienness of Special Relativity. The subject offers students their first opportunity to see how profoundly different physical reality is from our familiar (and grossly inaccurate) classical world. We will cover a number of subjects related to those presented in Physics 112, but with a different slant: our intention is to delve more deeply into a limited number of topics in order to supplement the material of Physics 112. We will call upon the superior mathematical skills of our Physics 212 students to allow the course to deal with electrodynamics in a more sophisticated manner which complements (but does not replace) Physics 112. The format of the course comprises weekly meetings in which students will work together in small groups (closely supervised by the instructor and teaching assistant) to solve a number of simple problems, then to discuss the conclusions one can draw from the results. In this way, students will derive for themselves some of the surprising features of our post-classical physical world. For example, the simple fact that the speed of light is constant gives rise to the existence of magnetic fields, the nature of special relativity, and the origins of electromagnetic radiation. We will review mathematical techniques as required. Students will be expected to become comfortable using calculus to solve problems as necessary. Completion of each of the weekly homework sets will require an hour or two. Grades will be based most strongly on participation in the weekly meetings and performance on problem sets.

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We want the course to be interesting, engaging, challenging, and lots of fun for both the students and the staff. Since Physics 212 is a "topics" course, we'll get to focus on the cool stuff, while relying on Physics 112 to provide broad coverage of the entire subject. It is likely that most students who are capable of working at the required level for this honors course will receive high course grades. In more detail, we will cover: 1. Special relativity: time dilation and Lorentz contraction 2. Special relativity: non-simultaneity; the Lorentz transformations 3. A microscopic version of Gauss’ law 4. Why superposition works 5. Pinwheels (curl and the vector potential) 6. The origin of the magnetic field as a consequence of special relativity 7. Operational amplifier circuits (meet in Loomis 336) 8. Electronic models of mechanical systems (meet in Loomis 336) 9. Analog computers (meet in Loomis 336) 10. Calculating magnetic fields using the vector potential 11. Maxwell's equations 12. Visualizing fields and potentials with Mathematica (meet in Loomis 257) 13. Studying electromagnetic radiation with Mathematica (meet in Loomis 257) 14. Oops! It doesn't really work like that! (Quantum mechanics enters the picture) 15. The electrodynamics of barbecuing We will review techniques involving differential equations and vector calculus as necessary. Do not be alarmed by the amount of calculus in some of the units! It’ll sink in as you work with it. Office hours: Professor Gollin: just call, or walk in. Email is also usually a reasonably quick way to reach me. Teaching assistant: TBA

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Physics 212, Spring 2003. Unit 1 1.1 Special relativity: time dilation and Lorentz contraction

Physics 212


Unit 1

Special relativity: time dilation and Lorentz contraction

Spring, 2003

George Gollin

University of Illinois at Urbana-Champaign 2003

5


Unit 1: Special relativity-- time dilation and Lorentz contraction Introduction: the speed of light is finite, and constant. You're going to love this: its consequences are the weirdest things you'll see in physics before you learn about quantum mechanics. All the strange features of special relativity come from the fact that the speed of light is exactly 2.9972458 × 108 meters per second. It doesn't matter whether the source of light moves with respect to the measuring apparatus: any device measuring c (the speed of light) will obtain, within experimental error, this value. Nothing else works this way: sound travels at fixed speed with respect to the air, for example. The value 2.9972458 × 108 meters per second really is exact: it serves as a definition, along with some time standard, of the length of one meter. The speed of light is very close to 1 foot per nanosecond, so we'll use that as a convenient approximation this week and next week. Here’s what you’ll discover today: • Events which are simultaneous in one frame of reference are not necessarily simultaneous in

another frame. • The rate at which time passes in a moving frame of reference is slowed down. • Moving objects become shorter along their direction of motion. Exercise 1.1: constant c; simultaneity. (a) Mr. Urkin, sociopathic co-proprietor of Urkin's Deli and Hardware Emporium, is capable of throwing a snowball at 50 miles per hour. While riding as a passenger in a delivery truck traveling 40 miles per hour, he hurls snowballs at stationary pedestrians ahead of, and behind,

the moving truck.

40 mph

foward victimrearward victim

Urkin's Deliand

Hardware Emporium

Calculate the speeds of the snowballs from the perspectives of the two pedestrians shown in the diagram.

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(b) The starship Nostromo, drifting in deep space, passes a large interstellar recreational facility which is also adrift, as shown in the figure. Observers on the recreational facility see the Nostromo as moving to the right, with speed 0.8c. While the ship is between the ends of the rec facility, it fires its communications lasers to send signal pulses in the forward and aft directions to receivers built into the ends of the rec facility.

burst oflaser light burst of

laser light

N

Nostromo v=0.8c

receiver receiver

Six Flags Over τ Ceti iv

According to technicians on the recreation facility, how fast are the two light pulses traveling as they pass through the receiver hardware built into each end of the rec facility? Note that (slowly moving) snowballs do not behave the same way as bursts of light!

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(c) The Sulaco, another fast ship, glides past the same drifting interstellar recreational facility. As before, observers on the recreational facility see the ship moving to the right with speed 0.8c. Unfortunately, Sulaco's X-band antenna strikes the rec facility's TV parabola, interrupting the final episode of Pride and Prejudice. Sulaco's wrecked antenna is midway along the ship's length; the rec facility's ruined parabola is also midway along the facility's length. Observers on the recreation facility see the Sulaco's length as 600 feet. The rec facility's length, according to the manufacturer, is 2000 feet.

receiver receiver


v=0.8c

boom!

SulacoS

600 feet

2000 feetAccording to technicians on the recreation facility, when does the flash of light from the collision reach the rec facility's right- and left-receivers (assume the flash occurs at t = 0)? According to technicians on the recreation facility, when does light from the collision reach Sulaco's aft periscope (located at the rear of the ship)? When does light reach the forward periscope?

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(d) From the perspective of the Sulaco, the collision happens because the incompetently-piloted recreational facility smashes into Sulaco’s antenna as it lumbers past at 0.8c, moving to the left, as shown in the figure.


v=0.8c

SulacoS

boom!

According to cosmonauts on the Sulaco, does the flash of light from the collision reach the Sulaco's forward and aft periscopes simultaneously? According to cosmonauts on the Sulaco, does the flash of light from the collision reach the rec facility's right- and left-receivers simultaneously? Discussion Imagine we use the receipt of the light flash as a way to synchronize clocks on the rec facility: we set a pair of stopped clocks, one at each end of the rec facility, to t = +1000 nsec. When the light flash arrives at the left end, we start the left clock ticking. When the light flash arrives at the right end, we start the right clock ticking. How do the clocks look from the two frames of reference (recreation facility's and Sulaco's)?

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Exercise 1.2: a light clock; time dilation (a) Mrs. Urkin, the irritable, but surprisingly fit co-proprietress of Urkin's D&HE, drifts in deep space aboard the gigantic starship Vulcan Flashrocket. Since she realizes that the speed of light is approximately one foot per nanosecond, Mrs. Urkin constructs a clock using two mirrors, separated by a distance of 500 million feet, and a bouncing light pulse. The pulse is of brief duration-- the laser that produced it fired for only a nanosecond-- and it bounces between the mirrors without attenuation.

Free delivery

Vulcan Flashrocket

BombsR Us

path ofbouncing lightpulse

500

mill

ion

feet

Urkin's Deliand

Hardware Emporium

As shown in the figure, she holds the lower mirror in one hand, moving it briefly into place to reflect the descending light beam. After each successful reflection she removes the mirror. Because Mrs. Urkin knows her resting pulse rate, she has constructed the device so that she will need to move the mirror into position immediately after each heart beat. It is fortunate that she does not suffer from cardiac arrhythmia since the bomb stored below the light clock will detonate if its optical sensor is struck by the light beam. What is Mrs. Urkin's pulse rate?

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(b) The Flashrocket is passed by that hazardous interstellar recreational facility while Mrs. Urkin is occupied with her light clock. Fortunately, no collision takes place. From the perspective (the "frame of reference") of technicians on the rec facility, the Flashrocket is seen to glide past the facility at speed v, as shown in the figure below.

v (=0.8c)


Urkin's Delia nd

Hard ware Emporium

Free delivery

Vulca nFlashro cke

As seen from the rec facility's perspective, does Mrs. Urkin's bomb detonate? (c) As seen from the recreational facility, Mrs. Urkin's light pulse travels along a diagonal path, as shown below. Keep in mind that the Flashrocket is seen to move with speed v, and that the light pulse is seen to travel one foot per nanosecond!

path oflight pulse

mirror assemblymoves with speed v

500

mill

ion

feet c

How long is the path followed by the light pulse during one lower-mirror-to-upper-mirror-to-lower-mirror trip? (Express your answer in terms of v, c, and so forth.) If v = 0.8c, how long does it take for the light pulse to make the bottom-to-top-to-bottom trip?

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(d) If we define ∆t to be the time required for one bottom-to-top-to-bottom light pulse cycle in Mrs. Urkin's frame of reference, and ∆t' to be the time required for one cycle according to observers on the recreational facility, what is the ratio ∆t' / ∆t? Express your result in terms of v and c. (e) What is Mrs. Urkin's pulse rate, according to observers on the recreational facility? Discussion • To Mrs. Urkin, everything seems fine. To observers in the other frame, Mrs. Urkin is doing

everything slowly. • This MUST be true if something (light) exists which has the same velocity in all frames

regardless of any source-observer relative motion. • All systems used to measure time are affected the same way (heart beat, light clocks,

mechanical clocks,...)

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Exercise 1.3: measuring the length of a moving object using a single clock. (a) Floating comfortably aboard the Vulcan Flashrocket, Mrs. Urkin measures the length of the Nostromo by timing how long it takes for the Nostromo to coast past her position, moving to the right at speed v. Here's how Mrs. Urkin's length measurement looks to the crew of the Nostromo: they see the Flashrocket gliding past with speed v and, naturally, see that Mrs. Urkin's clock is ticking slowly. The Nostromo crew can measure the length of their own ship (let's call it L) by seeing how long it takes the Flashrocket to fly the full length of the Nostromo. (Let's call the time required ∆t).

(As seen from the Nostromo's frame of reference)

v = 0.8cFlashrocket clock starts

when its clock lever strikesNostromo's forward probe

Flashrocket clock stopswhen its clock lever strikes

Nostromo's aft probe

Urkin's Deliand

Hardware Emporium

Free delivery

Flashrocket

N

NostromoNostromo

Flashrocket

L

∆t∆t

∆t′

What is L in terms of ∆t and so on? (L is the Nostromo's rest length. This is a ridiculously easy problem!)

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(b) In Mrs. Urkin’s rest frame, the measurement looks as shown in the figure. Her clock is started and stopped by the probes mounted near Nostromo’s nose and tail. Naturally, she sees that Nostromo’s clocks (not shown) are ticking slowly, and are not synchronized. (Note that Nostromo’s aft clock must still read ∆t, however, as it passes Mrs. Urkin.)

(As seen from the Flashrocket's frame of reference)

Urkin's Deliand

Hardware Emporium

Free delivery

Flashrocket

v = 0.8c

Flashrocket clock startswhen Nostromo's forwardprobe strikes clock lever

Flashrocket clock stopswhen Nostromo's aft probe

strikes clock lever

N

Nostromo

L´

∆t´

∆t ?

Define L′ to be the Nostromo length measured by Mrs. Urkin; the time interval which passed on the Flashrocket clock is ∆t′. In terms of v and ∆t′, what is the value she obtains for L′? (Clearly, we have another difficult problem here!) (c) What is the ratio L' / L, expressed only in terms of v and c? (Lorentz contraction!) Keep in mind that ∆t′ < ∆t. Discussion All methods of performing a length measurement (tape measure, timing how long it takes a moving object to coast past a fixed point,...) will yield consistent results in one frame of reference.

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Exercise 1.4: is there any “transverse” contraction? Lengths parallel to the direction of motion become shorter as v → c. What happens to lengths perpendicular to the direction of motion?

paint balls on moving stick

1

2

3

Consider a long stick which carries a pair of paint balls, spaced along the edge of the stick which is perpendicular to its velocity. The stick crashes into a stationary ruler, as shown in the figure. How far apart are the paint marks which are left on the ruler?

paint balls on stationary stick

1

2

3

Imagine that a second observer views the same process (the very same ruler-stick collision shown in the above diagram), but from a frame in which the stick is at rest and the ruler is in motion. How far apart will this observer think the paint marks are? What conclusion can you draw concerning contraction perpendicular to the direction of motion from this system? Summary and wrap-up discussion • Time is messed up:

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1. moving clocks take longer per tick by a factor of ( )2 2v c−1 1 .

2. simultaneity doesn't work: clocks "at the front" read earlier times than clocks "at the back"

• All time references are affected identically: light clocks, heartbeat,... • Conclusions are valid if anything (light, neutrinos, gravitons,...) is found to move with

constant velocity, independent of source-observer motion. • Some things don't change when one switches frames of reference (e.g. bomb did/didn't

explode). • Length is messed up: moving objects are shorter by a factor of ( )2 21 v c− , but only along

their direction of motion. Comments about homework assigned for next week

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Reading and homework should be completed by this Friday, 4pm. Please leave your finished problem set in the TA’s mailbox by that time. We are available for help and questions as they arise; please call (let the phone ring until a secretary picks it up if I am not in my office), drop in, or send us email. Required reading: 1. Introduction and syllabus handout. 2. Notes on Special Relativity, pp. 1 – 23 as necessary to clarify points that remain obscure after

class. Optional reading, just for fun: Einstein’s Dreams, Alan Lightman. Warner Books (paperback), 179 pages, 1994. Comments from Amazon.com: "The book takes flight when Einstein takes to his bed and we share his dreams, 30 little fables about places where time behaves quite differently. In one world, time is circular; in another a man is occasionally plucked from the present and deposited in the past: "He is agonized. For if he makes the slightest alteration in anything, he may destroy the future ... he is forced to witness events without being part of them ... an inert gas, a ghost ... an exile of time." Einstein's Bridge, John Cramer. Avon Books (paperback), 1998. A high energy physics experiment at the SSC lab in Texas opens a channel into another universe. There are good things and bad things lurking out there, and both find their way through the portal. I suppose this really is about a quantum version of General Relativity, rather than Special Relativity, but it was really fun to read. Problem 0 (5 points): Make your own hardcopies… Print (from the web) your own copy of the solutions to this week’s in-class exercises and write a (true!) statement on your problem set asserting that you have successfully printed your personal copy. Problem 1 (10 points): KS

0 meson decay Neutral K mesons (“kaons”) are unstable particles composed of a quark and an antiquark. They can be produced copiously in energetic collisions between stable particles. The neutral kaon mass is roughly 1/2 the proton mass; short-lived neutral kaons usually decay into pairs of (lighter) pi-mesons through the decays KS

0 → π+π- and KS0 → π0π0. The average KS

0 lifetime for a kaon at rest is 0.8935 × 10-10 seconds. A beam of fast KS

0 is produced at a national laboratory. The average distance traveled by a kaon before decaying in flight is found to be approximately 9.2 cm. How fast are the kaons moving? (They are not traveling faster than c!!) You should assume all kaons are moving at the same speed v and use the value c = 3 × 1010 cm/sec in your calculations.

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Problem 2 (10 points): The space telescope sees clearly Two photons from a pair of faint, distant objects simultaneously enter the open end of the Hubble Space Telescope’s barrel. One of the objects is moving at 0.5 c directly towards the telescope, while the other is moving at 0.5 c directly away from the telescope. Assuming that the distance from the telescope’s open end to its primary mirror is 20 feet, calculate how long it takes for each of the two photons to arrive at the primary mirror after entering the open end of the telescope barrel (as determined by observers in the rest frame of the telescope). Problem 3 (10 points): The Minoans synchronize their watches The identical sister ships Linear A and Linear B pass each other in deep space, as shown from the perspective of the Linear A’s crew in the figure below. In its rest frame, each ship is 2000 feet long; the crew of Linear A sees the other ship glide by with speed 0.8 c. As the midpoint of Linear B passes the midpoint of Linear A, parabolic antennas on the two ships collide. The brief flash of light from the collision travels outwards from the point of contact.

boom!

Linear AL-A

v=0.8c

2000 feet

Linear BL-B

? ?

(a) According to the crew of Linear A, how long is the ship Linear B? (b) According to the crew of Linear B, how long is the ship Linear A? (c) Both ships carry (initially stopped) clocks at their forward and aft ends. Crew members stationed next to each of the clocks start them ticking as soon as they see the light from the explosion. After all clocks have finally begun ticking, how do the clocks on Linear B look to the crew on Linear A? If you conclude that the Linear A crew believes them to be out-of-synch, which Linear B clock reads the later time according to Linear A observers?

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Physics 212, Spring 2003. Unit 2 2.1 Special relativity: Non-simultaneity; the Lorentz transformations

Physics 212


Unit 2

Special relativity: Non-simultaneity; the Lorentz transformations

Spring, 2003

George Gollin


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Unit 2: Special relativity-- Non-simultaneity; the Lorentz transformations Recap of last week Introduction: reference frames and measurements. The only quantitative features of special relativity we've seen so far are the rates at which moving clocks slow down and moving objects shrink. If we're clever, we can use this to bootstrap our way into drawing other quantitative conclusions concerning measurements such as the non-simultaneity of events which are simultaneous in a different reference frame. What do we mean by "reference frame?" The idea is simple: it's just a coordinate system in which all our clocks and rulers are at rest, and all our clocks are synchronized. A measurement of a spatial interval between two events amounts to determining the distance between the two clocks placed next to the events. The time interval is just the difference in the readings on the faces of the two clocks. For example, the space-time interval between waking up and entering my office involves a distance of about 2 miles and a time interval that depends on how sleepy I am. Synchronizing clocks in a reference frame is easy: we preset all of them so that a clock which is R feet from the origin reads +R nanoseconds. Each clock (initially paused) starts running when it senses the light from a strobe, placed at the origin, which fires at time zero. Keep in mind that most measurements involve both position and time: measuring the length of a moving object as it glides past our (stationary) ruler requires us to see where the two ends of the object are, in comparison to the scale printed on our ruler, at the same time. Timing how long a moving object needs to coast past our stopwatch requires that the stopwatch remain in the same place during the timed interval. We'll continue to use the approximation c = 1 foot per nanosecond this week. Here’s what you’ll find/work on today: • A quantitative expression for the non-simultaneity of events viewed from one frame when

the events occur simultaneously in another frame • Putting it all together: deriving the Lorentz transformations

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Exercise 2.1: quantitative description of non-simultaneity. We know from last week that clocks at the front of a moving object read earlier times than clocks at the back of the object. Let's work this up, quantitatively, now. Here's the setup: a flash is produced at the tail of the Nostromo as its rear clock reads zero. The light travels the length of the ship, arriving at a sensor at its nose. We can describe the production and reception of the light in terms of a pair of events (an event is something which happens at one point in space and lasts only an instant). The first "event" is the production of the spark near the Nostromo's aft clock. The second event is the arrival at the forward sensor of light from the spark.

rest length L

flash of light

N

Nostromot = L/c t = L/c

flash of light

N

Nostromot = 0 t = 0

event 1

event 2

Since Nostromo's rest length is L, and all its clocks are synchronized in its rest frame, we know that the light will arrive at the forward clock when that clock reads L/c. Perhaps the light from the arriving flash is used to expose some photographic film to record permanently the reading on the forward clock when the light arrives. There is a well-defined spatial and temporal interval between the two events in this reference frame (∆x = L and ∆t = L/c). From the perspective of observers who see Nostromo moving to the right, the ship's length is

Lorentz contracted to ( ) 21L v c− . Nostromo's forward clock is seen to show an earlier time than its aft clock. (Let's call this difference δ.) The two events look like this:

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flash of light

event 1N

Nostromo t = -δt = 0

event 2

t = L/ct = L/c + δ

flash of light

N

Nostromo

v


( )21L v c−

( )21L v c−

0 0

t' t'

vt'ct'

Event 1: spark is produced at the aft end of Nostromo. The Nostromo's tail clock reads 0; the nose clock -δ. (We already know that δ is a positive number, since the forward clock reads an earlier time than the aft clock.) Since we are viewing events from the rest frame of the observers on the rec facility, all clocks on the recreation facility read zero. Event 2: light from spark reaches forward end of Nostromo. From the photograph taken in the Nostromo rest frame, we know that Nostromo's forward clock will read L/c when the light arrives. The tail clock always reads δ later than the nose clock, according to rec facility observers. The rec facility clocks all read t'. In time t', the light from the spark has traveled a distance ct' to catch up with the nose of Nostromo, while the Nostromo itself has moved a distance vt'. There is a well-defined spatial and temporal interval between the two events in this reference frame. (∆x' = ct' and ∆t' = t' -- note that I'm using coordinates with primes to describe variables in the rec facility frame.).

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(a) Note that (in the rec facility rest frame) the light beam travels a distance ∆x' = c∆t' which is the same as the sum of the Lorentz-contracted Nostromo length and the distance traveled by the Nostromo. Write an equation which says the same thing as the previous sentence, and then use this to solve for ∆t' in terms of L, v, and c. (b) When event 2 occurs in the rec facility rest frame, Nostromo’s forward clock must read L/c and its aft clock must still be ahead of the forward clock by an amount δ. As a result, the reading on Nostromo’s slowly-ticking aft clock has advanced from 0 to L/c + δ during the time it took for a rec facility clock to advance (in the rec facilty rest frame) from 0 to ∆t'. Since the moving Nostromo clocks are ticking slowly in comparison to the rec facility’s clocks, we must have 2 21L c t v cδ ′+ = ∆ − . Use this fact to derive an expression for δ in terms of L, v, and c. Please simplify your expression as much as possible. Discussion How do the rec facility clocks look to the Nostromo crew?

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Exercise 2.2: Deriving the Lorentz transformations Observers in different frames of reference which are in relative motion will tend to disagree about lengths, distances, time intervals, and the synchronization of clocks. Here’s a summary of what we know so far:

1. A single, moving clock ticks slowly so that a time interval ∆t between two events measured by this one clock will be shorter than the time interval ∆t' measured by observers who see that this clock is in motion: 2 21t t v c′∆ = ∆ − .

2. The length L' of a moving object is shorter than the length L of the object as measured in the object’s rest frame: 2 21L v c ′L− = .

3. Clocks which are separated by a distance ∆x in their rest frame and are synchronized in their rest frame are seen to be out-of-synch in a frame in which the clocks are moving with speed v by an amount 2v x c− ∆ . (The negative sign means that the forward clock reads an earlier time than the aft clock.)

Imagine that a pair of events is seen by observers in the two different coordinate systems O and O'. Observers in O (the "unprimed" frame) measure the space-ime intervals between the events to be ∆x, ∆y, ∆z, ∆t while observers in O' measure the interval to be ∆x′, ∆y′, ∆z′, ∆t′. (I’m defining ∆x = x2 – x1, ∆t = t2 – t1, and so forth.) Let’s assume that the origins of O and O' coincide when clocks at the origins of both frames read zero. Frame O' (according to observers in O) is moving with velocity v along the x axis of O as shown in the figure. (Note that the sign of v matters.)

x

z

y

O

x'

z'

y'

O'

v

event 1(x1, y1, z1, t1) or (x1', y1', z1', t1')

event 2(x2, y2, z2, t2) or (x2', y2', z2', t2')

The Lorentz transformations allow us to calculate the space-time interval between the two events measured by observers in one frame if we know the interval between the events measured by observers in the other frame and also know the relative velocities of the two frames. They allow us to write ∆x′ as a function of ∆x, ∆t, v, and c and also ∆t′ as a (different) function of ∆x, ∆t, v, and c. (a) Let’s redraw the figure used in the last exercise in order to make the choice of primed and unprimed coordinates agree with the discussion in this exercise. Here’s the redrawn figure:

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flash of light

event 1N

Nostromot′ = 0

event 2

t′ = ∆t′

flash of light

N

Nostromo

v


t = 0 t = 0

t = ∆t t = ∆t

v∆t∆x = c∆t

∆x′ = rest length

In the rec facility frame the light beam travels a distance which is the same as the sum of the Lorentz-contracted Nostromo length and the distance traveled by the Nostromo. The equation which says the same thing as this is 2 21x x v c v′ t∆ = ∆ − + ∆ . Solve this equation for ∆x' as a function of ∆x and ∆t.

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(b) Since the two events are separated by ∆x′ and ∆t′ in the Nostromo rest frame, we know that =t x′ ′∆ ∆ c . We also know that =t x∆ ∆ c and that 2=t c x c∆ ∆ .

Use these facts to rewrite your answer to the previous exercise as an equation for ∆t', and simplify it as much as possible to express ∆t' as a function of ∆t and ∆x. Discussion You have derived the Lorentz transformations! The frames O and O' correspond to the rest frames of the rec facility and the Nostromo respectively; the space-time separation of any pair of events measured by observers in the unprimed frame can be used to derived the space-time interval between the events that would be measured by observer in the primed frame. Just to make sure we have them written down compactly: define: 2 21 1 v cγ = − so that

( ) ( )

( ) ( )2 2 .

x x v t x x v ty yz z

t t v x c t t v x c

γ γ

γ γ

′ ′∆ = ∆ − ∆ ∆ = ∆ + ∆

′∆ = ∆′∆ = ∆

′ ′∆ = ∆ − ∆ ∆ = ∆ + ∆

′

′

If we know the intervals in one frame, we can calculate what they'll be in the other frame. Keep in mind that positive v means that O' is moving to the right, as viewed from O. Note that we can extract all of the quantities we’ve calculated so far using the Lorentz transformations as long as we choose the right pair of events. Keep in mind that almost all measurements require you to know something about space and time. For example, a time interval measured using a single clock is done so that the spatial separation between the “starting” event and “stopping” event used to define the time interval is zero.

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Summary and wrap-up discussion • Time AND space are messed up.

1. rate of passage of time is slowed in a moving frame:

( )2 21t t v c∆ = ∆ −′

2. simultaneity doesn't work: clocks "at the front" read earlier by v∆x / c2 than clocks "at the back," where ∆x is the separation between the clocks as measured in their own rest frame.

3. moving objects are shorter: ( )2 21L L v c= −′ where L is

the rest length of the object. • The Lorentz transformations can be used to calculate the intervals ∆x, ∆y, ∆z, ∆t between

events observed from one reference frame using the intervals ∆x', ∆y', ∆z', ∆t' between the same events observed from another reference frame.

• v → c. Comments about homework assigned for next week Please do not ever ever ever search for solutions to assigned problems in texts, lecture notes or problem set solutions from other courses, or solutions distributed in previous offerings of this course! Scout’s honor!

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Reading and homework should be completed by 4pm Friday. Required reading: Notes on Special Relativity, pages as appropriate to our work. Optional reading, just for fun: The Feynman Lectures on Physics (volume I), chapter 15. The Forever War, Joe Haldeman. Avon books (paperback), 1974. This science fiction novel won both the Hugo and Nebula awards. Time dilation plays a role in the messes the characters make of their lives. Problem 0 (5 points): Make your own hardcopies… Print (from the web) your own copy of the solutions to this week’s in-class exercises as well as last week’s problem set and write a (true!) statement on your problem set asserting that you have successfully printed your personal copies. Problem 1 (10 points): Can anything go faster than c? In this problem you'll derive the relativistic velocity addition formula. To replenish the Nostromo's perilously low supply of jell-o, Mrs. Urkin's fast shuttle is launched on a resupply mission. As seen from Nostromo, the shuttle streaks away with constant velocity u in the positive x' direction. At position x1' and time t1' the shuttle strikes a dust mote, vaporizing it. Somewhat later, the shuttle hits a micrometeoroid.

event 1: Mrs. Urkinclobbers a dust mote

N

Nostromo U u

event 2: Mrs. Urkin strikesa micrometeoroid

N

Nostromo U u

∆x', ∆t'(Nostromo rest frame)

The light flashes from both events are seen by Nostromo's cook, who calculates (after correcting for the light's propagation time to the ship) that the space-time interval between the events was (∆x', ∆t') = (x2' - x1', t2' - t1'). Not surprisingly, the cook notices that u = ∆x' / ∆t'. The same events are viewed by Mrs. Urkin's personal trainer on board the interstellar recreation facility. He sees the Nostromo travelling with speed v in the positive x direction, and determines that the space-time interval between the two events was (∆x, ∆t) = (x2 - x1, t2 - t1). He concludes that the shuttle craft was moving in the positive x direction with speed ∆x / ∆t. (a) By making use of the Lorentz transformations, calculate ∆x / ∆t in terms of u (the shuttle's velocity as seen from Nostromo), v (Nostromo's velocity, as seen from the rec facility), and c.

28


(b) How does your expression for ∆x / ∆t behave in the limit that both u and v are close to c? Problem 2 (10 points): The train-tunnel "paradox" Here's how this one works. We have a tunnel, 800 feet long in its rest frame, with doors on each end which can be used to seal the tunnel. The train is 1,000 feet long in its own rest frame, as shown in the illustration:

Tunnel rest frame: The train travels at speed 0.8 c so that its Lorentz-contracted length allows it to fit entirely inside the tunnel. When (synchronized in their rest frame) tunnel clocks by both doors read zero, just as the train is neatly centered inside the tunnel, the doors slam shut, trapping the entire train inside the tunnel. The front of the train crashes through the right-side tunnel door 125 nanoseconds later, but the (closed) left-side door was able to close without interfering with the train.

29


Train rest frame: From the train's rest frame, things look rather different: the Lorentz-contracted tunnel rushes towards it as shown in the following figure.

The tunnel is much too short to be able to trap the train entirely inside itself, yet both doors will drop, and neither will touch the train (although the front of the train will crash through the already-closed right-side door). How can this be possible?

30


Problem 3 (10 points): An introduction to partial derivatives We will begin using calculus to address problems next unit. Here’s a reminder about the definition of an ordinary derivative, and an introduction to partial derivatives. (a) The Urkins are hiking in eastern France’s Jura mountains. They have been told by residents of a nearby village that the altitude above sea level is well-described by the function h(x,y) = 10-7 x3, where x and y are the distances due east and due north (in meters) of the center of town. A plot of the elevation near the town would therefore look something like this:

-500

0

500

1000.

-1000

-500

0

500

1000

-100

-50

0

50

100

-500

0

500x

y

altitude is a function only of x

h(x,y) = 10-7 x3

village

(i) The Urkins hike due east from the point (x, y) one morning, stopping for lunch at the point (x + a, y). Write an algebraic expression for their change in altitude, ( ) ( ), ,h h x a y h x y∆ = + − and for the average slope of their path, m h a= ∆ . Please simplify your result as much as possible, expressing it as a function of x and a. (ii) Evaluate the average slope in the limit that a , expressing your answer as a function of x. (Since the definition of the derivative of a function of one variable is

0→

( ) ( ) ( )0

limx

df x dx f x x f x x∆ →

= + ∆ − ∆ , your answer shouldn’t come as a surprise to you!)

31


(b) The next day the Urkins are hiking in a region where the altitude is well-described by the function h(x,y) = 10-10 x3y, where x and y are the distances due east and due north of the center of a different town. An elevation plot would therefore look something like the following:

-500

0

500

-500

0

500

-100

-50

0

50

100

-500

0

500x

y

altitude

altitude as a function of x,y is 10-10 x3y

(i) The Urkins hike due east from the point (x, y) to the point (x + a, y). Write an algebraic expression for their change in altitude, ( ) ( ),h h x a y h x y+ − ,∆ = and for the average slope of their path, m h= ∆ a . Simplify your result, expressing it as a function of x, y, and a. (ii) Evaluate the average slope in the limit that a as a function of x and y. 0→ You have just evaluated a partial derivative! The definition of a partial derivative of a function of several variables is quite similar to the definition of the derivative of a function of one variable:

( ) ( ) ( )0

constant

, ,limx

y

,f x y f x x y f x yx x∆ →

∂ + ∆=

∂ ∆−

and ( ) ( ) ( )0

constant

, ,limy

x

,f x y f x y y f x yy y∆ →

∂ + ∆=

∂ ∆−

.

The machinery is easy to use: just treat all “the other” variables as if they were constants and take the derivative of the function in the usual way.

32

Physics 212, Spring 2003. Unit 3 3.1 A microscopic version of Gauss’ law

Physics 212


Unit 3

A microscopic version of Gauss’ law

Spring, 2003


2003

33


Unit 3: A microscopic version of Gauss’ law Recap of last week Introduction: the connection between Coulomb’s and Gauss’ laws The Physics 112 discussion of the electric field and electrostatic potential describes them as quantities which depend on the global distribution of electric charge:

( ) ( ) 3

all charges 0 0all space

1 1 4 4

i

i

rQV r rr r r r

ρπε πε

′= ⇒ d ′

− − ′∑ ∫

( ) ( ) 32 2

all charges 0 0all space

1 1 4 4

i i

ii

rQ r r r rE r rr r r rr rr r

ρπε πε

′− − ′= ⇒ ′

− − ′− ′−∑ ∫ d .

By “global” I mean that you’ll need to know where all the charge is in order to calculate the sums or integrals. Don’t be alarmed by the volume integrals—an integral is just a gigantic sum, which is probably why an integral sign is meant to resemble a large letter “S.” It is possible to craft equations which relate characteristics of the fields (or potentials) to the charge distribution using only on local properties of the charge density. (You’ll do that today.) In particular, knowing the charge density at a single point in space lets us know about the spatial derivatives of the electric field at that same point in space. The electric field falls off like 1/r2 while the area of a surface surrounding a finite charge distribution grows like r2. One consequence of this is Gauss’ law: the net flux of electric field through any closed surface depends only on the amount of charge inside the surface. This is another global statement since knowledge of the field over an extended surface is necessary to calculate the flux in order to establish the connection between the electric field and the (extended) charge distribution. Again, it is possible to concoct a local-properties-only formulation of Gauss’ law. Naturally, the electric field is the force per Coulomb felt by a point test charge placed somewhere, while the electric potential is the test charge’s potential energy per Coulomb.

34


Here’s what you’ll find/work on today: • Verifying that Gauss’ law is true even when the electric field is not perpendicular to the

Gaussian surface enclosing a point charge • A microscopic (differential) version of Gauss’ law which depends only on local properties of

the electric field and charge distribution. • Another differential equation (Poisson’s equation) which relates the potential to the charge

distribution. Exercise 3.1: Coulomb ⇒ Gauss in cylindrical geometry Gauss’ law describes the connection between the flux of electric field lines passing through a closed surface and the net charge contained inside the surface. The idea is relatively simple: electric field lines can only begin on positive charges and can only end on negative charges. If there are no charges inside a closed surface, any field line entering the surface must also exit. If the surface does enclose a net positive charge, more field lines will leave it than will enter. Naturally, the situation is reversed if the surface encloses a net negative charge. The flux of electric field lines passing through a small area patch ∆A depends on how strong the field is, how large the area of the patch is, and how close the electric field lines are to being perpendicular to the surface of the patch. In symbols, cos ,E A E Ai θ∆Φ = ∆ = ∆ where the direction associated with the area patch is along the (outward) normal to the surface of the patch. Summing the flux of electric field over a closed surface involves a sum (an integral) over all area patches which comprise the surface:

A dAsurface surface surface surface

E A d E dAi i∆ →Φ = ∆Φ = ∆ → Φ =∑ ∑ ∫ ∫ .

The formal statement of Gauss’ law is

0

enclosed

surface

QE dAε

=∫ i .

It’s easy to evaluate this integral when the surface is a sphere of radius R which is centered on a point charge Q: the integral is just the product of the strength of the electric field at a distance R and the surface area of a sphere, since all the cosines in the dot product are unity:

2 22

0 0

4 44surface

Q QE dA E R RR

i π ππε ε

= × = × =∫ .

The dot product inside the integral is important: it is what makes Gauss’ law true regardless of the shape of the bounding surface used in the integral.

35


To convince yourself (or at least to make it seem more plausible) that the shape doesn’t matter, consider an infinitely long cylindrical surface of radius a surrounding a point charge Q as in the following figure.

a

xθ

dA E

Q z

y

φ

The charge is at the origin, so the distance to an area patch on the surface of the cylinder at (r, z, φ) = (a, z, φ) is (a2 + z2)1/2. In addition, the cosine of the angle between the electric field and the normal to the area patch is cos(θ) = a / (a2 + z2)1/2. An area patch of extent dφ, dz has area dA = a dφ dz. For this surface, evaluate the electric field flux through the cylinder, namely .

surface

E dA∫ i

Don’t get spooked by this: a surface integral involves taking a double integral, which can be written this way:

( )

( )

2 2

2 2 2 20

2

2 2 2 200

4

.4

surface surface

surface

aE dA E dAa z

Q a dAa z a z

Q a a d dza z a z

i

π

πε

φπε

+∞

−∞

= +

= + +

= + +

∫ ∫

∫

∫ ∫

.

cos θE dA (A correct calculation will yield the answer Q/ε0. You will probably find the trig substitution z = a⋅tan(α), in combination with the identity 1 + tan2(α) = sec2(α) = 1/cos2(α), to be helpful.) Discussion

36


Exercise 3.2: A microscopic version of Gauss’ law If a surface encloses a distribution of charge with density ( )rρ C/m3, Gauss’ law can be written

( )0 0

1enclosed

surface volume

QE dA r dVρε ε

= =∫ ∫i

since the amount of charge contained inside a small volume element dV is ρdV and the integral just sums up the charge contained by all volume elements inside the bounding surface. Let’s develop a microscopic version of this. Consider a small volume which is a parallelepiped centered at (x,y,z) with side lengths dx, dy, dz as shown in the figure. Assume that ρ and E vary slowly over the size of the volume so that the charge inside the volume is .

x

z

y

dx

dz

dy

If the volume is small enough, the surface integral can be rewritten as a sum over the faces of the box so that Gauss’ law becomes

( )6

1 0

, ,i

i ii

x y z dxdydzE dA

ρε

=

=

=∑ i .

(a) Write an explicit expression for each of the six area patch vectors dAi in terms of the unit vectors x , , and the volume’s side lengths dx, dy, dz. y z

37


(b) The electric field at the center of one particular face of the box can be written as

ˆ ˆ ˆ, , , , , , , ,2 2 2 2x y zdx dx dx dxE x y z E x y z x E x y z y E x y z z − = − + − + −

while the field at the center of the opposite face is

ˆ ˆ ˆ, , , , , , , ,2 2 2 2x y zdx dx dx dxE x y z E x y z x E x y z y E x y z z + = + + + + +

.

There are similar expressions for the electric field at the other four faces of the box. Use this fact, and the definition of a partial derivative, to rewrite Gauss’ law for a vanishingly small volume entirely in terms of ρ and partial derivatives of the components of the electric field.

(You should obtain ( ) ( ) ( ) ( )0

, ,, , , , , ,yx zE x y zE x y z E x y z x y zx y z

ρε

∂∂ ∂+ + =

∂ ∂ ∂.)

In Cartesian coordinates, the definition of the operator ∇ (“del”) is ˆ ˆ ˆx y zx y z

∂ ∂∇ ≡ + +

∂ ∂∂∂

so

that the divergence of any vector field G can be written

( ) ( ) ( ), ,, , , ,yx zG x y zG x y z G x y zG

x yi

∂∂ ∂∇ = + +

∂ ∂ ∂z.

As a result, the microscopic (differential) version of Gauss’ law can be written compactly as

0

E ρε

∇ =i .

38


Discussion: divergence (and the divergence theorem) • A net “outflow” of electric field lines automatically means that the electric field has positive divergence. A net “inflow” means the field has negative divergence.

• The divergence theorem (true for all vector fields): use 0

E ρε

∇ =i to rewrite Gauss’ law as

( ) ( )0surface volume volume

rE dA dV E dVi i

ρε

= = ∇∫ ∫ ∫ :

The flux of field through a closed surface is the same as the integral of the field’s divergence over the volume enclosed by the surface. In general, for any vector field G it is true that

( )surface volume

G dA G dVi i= ∇∫ ∫ .

Exercise 3.3: Poisson’s equation

The equation 0

E ρε

∇ = depends only on the i local properties of the charge density and the

derivatives of the electric field. We can work up a similar equation connecting the potential with the charge density with the help of the work-energy theorem. (This only applies to static systems, where all free charge stays put, and all currents are constant!) Recall that the (small) change in an object’s potential energy is

when it is pushed through a displacement U F r∆ = ∆i r∆ by a force F . If you are applying the force to overcome the effects of an electric field opposing your efforts, your force will satisfy

. Imagine that you move the test charge through a displacement in three steps, as illustrated below.

F QE= −ˆ ˆr x x∆ = ∆ + ˆy y z∆ + ∆z

x

z

y

1. dx

3. dz2. dy

Q starts here Q ends here

39


3

z z

The work you do will be with

. 1 2U U U U∆ = ∆ + ∆ + ∆

1 2ˆ ˆ , x yU F y y F y∆ = ∆ = ∆ ∆i i 3 ˆ, U F x x F x U F z z F∆ = ∆ = ∆ = ∆ = ∆i

Since 1x

U Fx

∆=

∆, with some thought it should be apparent that , , x y

U U UF Fx y z zF∂ ∂ ∂

= = =∂ ∂ ∂

so

that ˆ ˆ ˆU U UF x y z Ux y z

∂ ∂ ∂= + + = ∇

∂ ∂ ∂.

Because and U , we can conclude that F Q= − E QV=

(static systems only)E V= −∇ . Use this to construct Poisson’s equation, which relates various second derivatives of the potential to the charge distribution, from the differential form of Gauss’ law.

(You should obtain ( )2 2 22

2 2 20

, ,x y zV V VVx y z

ρε

∂ ∂ ∂∇ ≡ + + = −

∂ ∂ ∂. The operator is called the

“Laplacian.”)

2∇

Discussion Summary and wrap-up discussion • The 1/r2 nature of the electric field leads to Gauss’ law which relates the flux of electric field

through a closed surface to the net charge enclosed by the surface. • In combination with the connection between the electric field and the electrostatic potential,

the microscopic (differential) version of Gauss’ law can be used to produce Poisson’s equation, which relates various second derivatives of the potential with the local charge density.

Comments about homework assigned for next week

40


41


Reading and homework should be completed by 4pm Friday. Required reading: Purcell, chapter 1, sections 7 through 10. Chapter 2, sections 1 through 5 and 7 through 12. Note that Purcell uses CGS, instead of MKS (SI), units so there won’t be any 4πε0’s anywhere. Read over (but don't worry about understanding) the Unit 4 material we'll tackle in class next week. Optional reading: The Feynman Lectures on Physics (volume II), chapter 2, sections 1-5 and 7-8; chapter 3, sections 1-3. Comment on notation: In general, I’ll use ˆ ˆ ˆ, ,x y z to represent unit vectors parallel to the x, y, and z axes. Problem 0 (5 points): Make your own hardcopies… Print (from the web) your own copy of the solutions to this week’s in-class exercises as well as last week’s problem set and write a (true!) statement on your problem set asserting that you have successfully printed your personal copies. Problem 1 (10 points): This is not why the universe is still expanding Theoretical food chemists in the Urkin's D&HE Research Division are completing an analysis of a hypothetical universe filled with electrically charged mayonnaise. In this alternative reality the electrostatic potential created by the mayo is found to be

( ) ( )0, , sin cosV x y z V kx ky= +

where k is a constant. (Note that the potential depends only on x and y.)

(a) Calculate the electric field ( ), ,E x y z produced by mayonnaise.

(b) Calculate the net charge density in Coulombs per cubic meter ( ), ,x y zρ responsible for creating this potential.

42


Problem 2 (10 points): At the surface of the FlashOven The electric field inside, and at the external surface of Urkin’s Vulcan FlashOven is accurately described as The oven is a rectangular box whose sides are bounded by the planes

( ) 20ˆ, , .E x y z z E x yz=

0, , 0, , 0, x x a y y= = = b z z c= = = .

(a) Evaluate the integralsurface

E dA⋅∫ over the surface of the oven.

( points outwards; the six faces have dA ˆ ˆ ˆ, , and .dA x dydz y dzdx z dxdy= ± ± ± )

(b) Evaluate the integral over the interior volume of the oven. volume

E dV∇ ⋅∫

(c) How might the divergence theorem helped you in part (b)? Problem 3 (10 points): The point is… Anthrax, the Urkin’s pet cat, is resting comfortably at the origin after eating an electrically charged canary. He exudes the (radial) electric field

( )( )

32 2 2 2

ˆ ˆ ˆ, , x x y y z zE x y z

x y z

+ +=

+ +

where ˆ ˆ ˆ, ,x y z are unit vectors parallel to the x, y, and z axes. (Assume this expression is accurate for all points outside the cat.) Prove that the density of electric charge ( ), ,x y zρ is zero for all points (x, y, z) external to the cat.

43

Physics 212, Spring 2003. Unit 4 4.1 Why superposition works; Green’s functions

Physics 212


Unit 4

Why superposition works; Green’s functions

Spring, 2003

George Gollin


44


Unit 4: Why superposition works; Green’s functions Recap of last week Introduction: superposition One of the most important/useful/convenient/fundamental properties of the equations which characterize electric and magnetic fields is that all the fields and potentials obey the superposition principle. Imagine that we are able to solve for the electric field (call it ) in a universe which contains nothing but charge Q

1E

1 at position 1r and also that we can solve for 2E in a (different) universe which contains nothing but charge Q2 at 2r . Because of superposition we can be sure that the electric field in a universe which contains both charges will be the sum of the fi -charge-universe field calculations: elds we found doing our pair of lone

both charges present 1 2E E= + E . Superposition allows us to break a complicated problem up into a number of simple problems which we can solve, then to sum the simple problems’ solutions to generate the solution to the complicated problem. In the two-charge example mentioned above Q1 and Q2 are sources of the electric field. Naturally, they don’t contribute equally to the electric field everywhere: the closer an observer is to one of the charges, the more important that charge’s contribution to the measured electric field will be. If we placed unit charges at 1r and 2r we would need to weight their contributions to the

field at by r ( )20 ir rπε −1 4 . (For sources which are different from one Coulomb we will also

need to scale their contributions according to how much charge each particle carries.) The unit-charge weighting function ( )2

0 ir rπε −1 4 is a Green’s function (named for the British

miller George Green (1793-1841), whose hobby was mathematics). A Green’s function is a “helper” function which tells us how important each source’s contribution is when we superpose the effects (fields) of all the individual sources.

45


Discussion: superposition and Poisson’s equation Last week we cooked up a pair of powerful differential equations which can be used (at least in principle) to generate the electrostatic potential (and from it the electric field) if we know the charge distribution:

( )2

0

, ,x y zV

ρε

∇ = − (with solution ( ) ( ) 3

0all space

1 d4

rr

r rV r

ρπε

′= ′

− ′∫ )

and

E V= −∇ . The expression is Poisson’s equation, named for the French mathematician Simeon Denis Poisson (1781 - 1840). It is, in a sense, the master equation for the electrostatic potential: given ρ, there will always exist a solution to Poisson’s equation which we can construct (perhaps numerically, as opposed to analytically) by calculating an integral.

2V∇

An important property of Poisson’s equation is that the solutions superpose. Imagine that we have already calculated the potential V1 which corresponds to a charge distribution ρ1, and also the potential V2 which corresponds to a different charge distribution ρ2. If we want to know the potential Vsum which corresponds to a charge distribution ρsum = ρ1 + ρ2 all we have to do is add the potentials we previously found: Vsum = V1 + V2. It’s not necessary to crank through an integral over the combined charge distribution ρsum. From Physics 112 you already know about the simplest version of this: a pair of charges q1 and q2 at positions r1 and will yield a potential

2r

( ) 1 2

0 1

14

q qV rr r r rπε

= + − − 2

.

It is obvious that two identical charges (q1 = q2) would not, in general, contribute equally to the potential except at points which are equidistant from both charges. The weighting function which determines the size of the ith charge’s contribution to the overall potential is the (Coulomb Green’s) function ( )0 ir rπε1 4 − . Here’s what you’ll find/work on today: • Proving that potentials which satisfy Poisson’s equation really do obey the principle of

superposition. • Investigating superposition for some other systems. Exercise 4.1: Proving that Coulomb potentials really do obey the principle of superposition

46


Poisson’s equation says that ( )2

0

, ,x y zV

ρε

−∇ = with solution ( ) ( ) 3

0all space

1 d4

rr

r rV r

ρπε

′= ′

− ′∫ .

Imagine that you’ve solved for the potential V1 corresponding to a particular charge distribution ρ1, and also for the potential V2 which corresponds to a different charge distribution ρ2. Prove

that ( )2

0

, ,sumsum

x y zV

ρε

−∇ = where Vsum = V1 + V2 and ρsum = ρ1 + ρ2 .

discussion Exercise 4.2: Another superposition proof The net force on a forced, damped harmonic oscillator is given by . The damping, described by the –βv term to the right of the “=” sign, is velocity-dependent and tends to slow the oscillator down. The external driving force is f(t). We can rewrite the equation this way:

( )ma kx v f tβ= − − +

( )2

2

d x dxm kxdt dt

β+ + = f t .

Imagine that our oscillator is being driven by a particular driving force f1(t) and that we have already solved for its position as a function of time x1(t). A different oscillator with identical mass and damping is driven by a driving force f2(t), and we have already solved for this oscillator’s position as a function of time x2(t). Prove that when one of the oscillators is subjected to a force fsum(t) = f1(t)+ f2(t), it will move so that its position is xsum(t) = x1(t) + x2(t) . discussion

47


Exercise 4.3: Yet again another proof The “Karma” of central Illinois is described by the differential equation ( )K x

3 2

3 2

( ) ( ) ( ) ( ) ( )d K x d K x dK xA B C DK xdx dx dx

+ + + = U x .

The function U(x) (incorrectly described in the Bhagavad-Gita as “the source of all visible kinetic energy in the Universe”) depends on the local weather. Imagine that you have already solved for the Karma on rainy days (Krain(x) corresponding to Urain(x)) and on days with hail (Khail(x) corresponding to Uhail(x)). During an especially unpleasant day in which it rains and hails simultaneously, the central Illinois U(x) function is Urain+hail(x) = Urain(x) + Uhail(x). Does the mid-state Karma function obey the superposition principle? (Is it true that Krain+hail(x) = Krain(x) + Khail(x))? Prove your answer. discussion Exercise 4.4: Yet again another (and not, perhaps, the last) proof The “Karma” of Indiana behaves differently from that in Illinois, obeying the differential equation

( )K x

( )2

22

( ) ( ) ( )d K xA B K xdx

+ = U x .

As in Illinois, the function U(x) depends on the weather. Imagine that you have already solved for the Indiana Karma on rainy days (Krain(x) corresponding to Urain(x)) and on days with hail (Khail(x) corresponding to Uhail(x)). Once again, when it rains and hails simultaneously, the function U(x) is given by Urain+hail(x) = Urain(x) + Uhail(x). Does the Indiana Karma function superpose? (Is it true that Krain+hail(x) = Krain(x) + Khail(x))? Prove your answer. discussion Exercise 4.5: Probably the last superposition proof

48


The Iowa Karma function obeys the differential equation ( )K x

( )( ) ( ) ( )dK xAK x BK x U xdx

+ = .

As before, U(x) depends on the weather. Imagine that you know Krain(x), corresponding to Urain(x), and Khail(x), corresponding to Uhail(x). When it rains and hails simultaneously, the function U(x) is again given by Urain+hail(x) = Urain(x) + Uhail(x). Does the Iowa Karma function superpose? (Is it true that Krain+hail(x) = Krain(x) + Khail(x))? Prove your answer. discussion: linear vs. nonlinear differential equations Linear equations will obey superposition, while nonlinear ones (often, usually, always??) will not. discussion: cheap audio equipment “Total harmonic distortion:”

(kHz)1 2 3 4 51 2 3 4 5 (kHz)

In Out

Pure frequency in yields various frequencies out if amplifier is (slightly) nonlinear.

49


Summary and wrap-up discussion • The principle of superposition generally holds for linear differential equations. • The Green’s function for a linear system (which obeys superposition) is the “helper function”

which indicates how to sum the effects caused by various sources. For example, the Coulomb potential’s Green’s function is ( ) ( )0, 1 4i iG r r r rπε≡ − .


50


Reading and homework should be completed by 4pm Friday. Required reading: The Feynman Lectures on Physics (volume I), chapter 25, sections 1 and 2. Read over (but don't worry about understanding) the Unit 5 material we'll tackle in class next week. Problem 0 (5 points): Make your own hardcopies… Print (from the web) your own copy of the solutions to this week’s in-class exercises as well as last week’s problem set and write a (true!) statement on your problem set asserting that you have successfully printed your personal copies. Problem 1 (10 points): Anthrax’s personal MP3 player sounds funny Anthrax, the Urkins’ unpleasant feline, enjoys listening to music on a defective MP3 player. The relationship between the player’s input signal ( )inA t and its output signal is described by the equation

( )outA t

( ) ( )310 inout

dA tA t

dt−= .

Notice that this equation is linear: superposition works for it. (a) The closing note of Grunt Like a Chicken Thief is a sustained 500 Hz triangle wave whose amplitude vs. time is shown in the following figure:

1 ms

+0.5 V

-0.5 V

Ain is a 500 Hz triangle wave

V

t

Draw an accurate plot of the defective MP3 player’s output signal, Aout. (b) Any periodic function can be well-represented by a (possibly infinite) sum of sines and cosines. In particular, the triangle-wave final wail of Grunt Like a Chicken Thief can be represented this way:

( ) ( )32 2

1,3,5,...

4 1 cos 10k

ink

A t k tk

ππ

=∞

=

= − ⋅ ×∑ .

51


An expression of this sort is called a Fourier series. Using this, along with the knowledge gleaned from your answer to part (a), derive an expression for the Fourier series which represents the following 500 Hz square wave:

1 ms

+1 V

-1 V

500 Hz square wave

Vt

. Notice that superposition allows you to draw conclusions about the behavior of the input and output signals based on an understanding of how the individual Fourier components (the sines and cosines) behave when passed through the defective MP3 player. Problem 2 (10 points): Starburn In this problem you will find it convenient to use (struggle with??) a volume integral in spherical coordinates (r,θ,φ) instead of the more familiar Cartesian (x,y,z) coordinates. Please find Brian or me if you need help setting up the problem’s solution. Here’s how spherical coordinates work:

x

y

zr

θ

φ

dr

dφ

dθ

dV = dr × rdθ × r sinθ dφ

dVθ

φ

The lengths of the sides of an almost-rectangular volume element dV at (r,θ,φ) are dr, rdθ, and r sinθ dφ. Its volume is therefore dV = dr × rdθ × r sinθ dφ = r2

sinθ dr dθ dφ.

A volume integral of a function F can be written in the two coordinate systems as

52


( ), ,f f f

i i i

z y x

cartesianz y x

I F x y z dx dy dz =

∫ ∫ ∫ or ( ), , sinf f f

i i i

r

sphericalr

I F r r d rd drθ φ

θ φ

θ φ θ φ θ =

∫ ∫ ∫ .

You will derive an argument against a model of our universe as infinite and static, in which new stars appear periodically to replace old stars as they burn out. It is thought that there are roughly 1020 stars inside the 15 billion lightyear diameter observable universe. (This number could be wrong by a factor of 1000!) Since a lightyear is approximately 1016 meters, let’s assume that the average density of stars in our universe is 1 star per 1058 cubic meters, and that we can treat a distant star as a point source of light. If all stars shine with the same luminosity as our sun, each one is pouring 3.8×1026 watts of power into space. (Recall that a watt is the same as a joule per second.) Naturally, the intensity of power per square meter a distance r from a star is just the fraction of the area of a sphere of radius r subtended by one square meter, namely ( )2rπ1 4 . (a) Imagine that starlight falls on a sphere of radius 1 π floating in space. (The cross-sectional area of the sphere is a square meter.) Using the assumptions mentioned above, derive an expression for the power, in watts, striking the sphere from all stars within a distance R from the sphere. Since it will be necessary for you to sum the intensity from lots of stars, you will find it convenient to recall how to do an integral in spherical coordinates using the volume element

2 sindV r drd dθ θ φ= with (0, )θ π∈ but ( )0, 2φ π∈ . (Neglect light from the sun!) (b) The intensity of sunlight reaching the ground at noon in Champaign on a clear spring day is about 1000 watts per square meter. What value (if any) of R would make the starlight illuminating the sphere in part (a) be just as bright as this? (Express you answer in lightyears, please.) (c) Using the same assumptions, estimate the intensity of starlight at the Earth's surface.

53


Problem 3 (10 points): New technology ice cream Research scientists at Urkin’s D&HE contemplate upgrading their sauerkraut mill so that it can be used to prepare gallons of ever-popular brussels sprout ice cream. The complication is that the ice cream matrix needs to be agitated at frequency 4 Hz in order to set properly, while the brussels sprouts need to detect vibrations at 3 Hz in order to develop the correct flavor. Can the modified kraut mill handle the task? This is, the scientists, realize, a problem of superposition. Will the simultaneous installation of new 4 Hz and 3 Hz drivers in the kraut mill manage to deliver both frequencies to the ice cream under preparation? Sophisticated computer simulations of the sauerkraut mill show that the mixing amplitude ( )x t

delivered to the ice cream in the mill’s crucible by a single driver supplying force ( )f t obeys one of the following equations:

(a) ( )4 2

4 2

d x d xA B C fdt dt

+ + = t

(b) ( )2

22

d x dxA Bx Cx fdt dt

+ + = t

(c) ( )3 2

3 2

d x d x dxA B C Dx fdt dt dt

+ + + = t .

The coefficients A, B, C, and D are constant. The researchers realize that the question is this: when two drivers exert a combined force ( ) ( ) ( )4 Hz 3 Hzf t f t f t= + will the response be the same

as the sum of the single-driver responses (so that ( ) ( ) ( )3 Hz4 Hzx t x t= + x t )? If this is the case, the sauerkraut mill upgrade will work nicely. Which (if any) of the possible response functions (a), (b), and (c) will allow a successful upgrade? Please prove, or otherwise justify, your answers.

54

Physics 212, Spring 2003. Unit 5 5.1 Pinwheels (curl and the vector potential); le souvenir du temps passé

Physics 212


Unit 5

Pinwheels (curl and the vector potential); le souvenir du temps passé

Spring, 2003

George Gollin


55


Unit 5: Pinwheels (curl and the vector potential); le souvenir du temps passé Recap of last week Introduction: Does it spin? We’ve been looking at some of the characteristics of static electric potentials and fields with the help of a variety of tools such as Poisson’s equation. As you’ve seen in Physics 112, the connection between the electric potential and electric field is closely tied to the idea that the work done moving a charge from one place to another is independent of the path taken. If our charges are not static (in particular, if they are accelerating) all bets are off. We’re going to investigate another kind of differential operator today: the curl of a vector function. Curl will prove useful when we discuss systems in which currents are flowing, and when we investigate (derive/discover) the existence of electromagnetic radiation. We will also look at a vector analog of the scalar potential. This is less mysterious than it s unds: the scalar potential at created by a point charge q at position o r r′ moving with velocity

is v approximately

( )0

14

qr r

≈V r′πε −

while the vector potential is approximately A

( ) 002 2

0 0

1 1since4 4

q qA r v vc r r r r c

µ≈ = µ

′ ′πε − π − ε= .

. (There are some subtleties concerning what we need to do to make these equations exactly true.) Here’s what you’ll find/work on today: • properties of the electric field that are impossible to find in fields generated by static charges • curl and divergence of electric fields • Stokes’ Theorem • The vector potential • Time retardation’s influence on the vector potential Exercise 5.1: Pinwheel statics (or should we say dynamics?)

56


Imagine we make a tiny pinwheel with point charges q on each of its four blades as drawn in the following figure. Electric field lines in the vicinity of the pinwheel, caused by (static) charges placed elsewhere are shown in the diagram. The pinwheel’s center bearing is kept fixed. Let’s see if it is possible for the electric field to cause the pinwheel to spin faster and faster.

z

y

x

pinwheelcenter at (x,y) 2 2

x δ∆=

2 2y δ∆

=q

q

q q

(a) The charges are at positions ( , , 2xx y 0)∆

+ , ( , , 0)2yx y ∆

+ , ( , , 2xx y 0)∆

− , ( , , 0)2yx y ∆

− .

Recall that torque is . Assuming that the pinwheel’s arms are of equal lengths (so that ∆x = ∆y = δ), write an expression for the torque on the pinwheel in terms of the x and y components of the electric field. (Assume the z component of the electric field is zero.)

r F×

57


(b) If you haven’t already done so, transform your answer to part (a) into an expression involving partial derivatives of the fields by taking the limit that ∆x and ∆y go to zero. discussion

You should end up with something like this: 2

ˆ - 2

y xE Eq zx y

δτ∂ ∂

= ∂ ∂ .

If our pinwheel also had had a pair of arms parallel to the z axis, as drawn here, we could have calculated torque components along all three axes to find

( )

2

2

ˆ ˆˆ - + - + - 2

E .2

∂ ∂ ∂ ∂∂ ∂ = ∂ ∂ ∂ ∂ ∂ ∂

≡ ∇×

y yx xz zE EE EE Ez xx y y z z x

y δτ

δ z

y

x

E∇× is called the “curl of .” If the electric field has a nonzero curl, it’ll make the pinwheel

spin. E

How might such electric fields look? Here are two examples.

58


Exercise 5.2: Can the electric field from a charge distribution do this? If no charges are moving, we can use Poisson’s equation to calculate the potential, and then take its gradient to solve for the electric field:

( )2

0

, ,x y zV

ρε

∇ = − so that ( ) ( ) 3

0all space

1 d4

rr

r rV r

ρπε

′= ′

− ′∫ , and E V= −∇ .

When the electric field is the gradient of a scalar field is it possible for ( )E V∇ × = −∇ × ∇ to be

non-zero? (Prove your answer.) Exercise 5.3: What’s the divergence of a curl? Let’s say we find a different vector field (call it A ) whose curl is non-zero. Can the divergence of the curl of ( ∇ ∇ ) also be different from zero? (Prove your answer.) A (i A× ) discussion Path independence, work done along a closed path, line integrals, and curl...

59


Exercise 5.4: Line integrals around microscopic loops; curl Imagine that an electric field (with non-zero curl) does work moving a charged particle counterclockwise around a tiny rectangular loop, as shown in the diagram. The midpoints of the

four sides are at positions ( , y,2xx ∆

+ 0) , ( , y , 0)2yx ∆

+ , ( , y, 2xx 0)∆

− , and ( , ;

the work done by the electric field is

y , 0)2yx ∆

−

iloop

qE dl∫ .

z

y

x

Rectangular loop is centered at (x,y)

∆y

∆x

(a) Calculate the work done by the electric field, and then re-express it in terms of ∆x, ∆y, and various partial derivatives of the electric field. (You should assume that the electric field along each side of the loop is well-represented by its value at the midpoint of the side, and write the field explicitly in terms of its x and y components: ˆ ˆ( , , ) ( , , ) ( , , )x yE x y z E x y z x E x y z y= + .) (b) Re-express your answer in terms of E∇ × if you haven’t already done so. Discussion dA vs. ; (normal to the loop via the right hand rule). dA n

60


Exercise 5.5: Line integrals around lots of microscopic loops Now imagine that we tile part of the x-y plane with several small loops as shown below. Prove that the work done by the field when a charge moves around the outer perimeter of the tiled region is equal to the sum of the work done moving a charge around each of the small loops:

( )i ithi loop big loop

small loops

qE dl qE dl=∑ ∫ ∫ .

z

y

x

∆y

∆x

1 2

3 4

61


Discussion: Stokes’ theorem From an earlier exercise, you should have concluded that

( ) ( )ˆi itiny loop

E dl E n dx dy E dA= ∇ × = ∇ ×∫ i

where is a unit vector normal to the surface of the loop (its direction is given by the right hand rule) and dA . We can use this to rewrite

nn dx dy=

( )i ithi loop big loop

small loops

E dl E dl=∑ ∫ ∫

as

( ) ( ) .i i ii big loopsmall loops surface surrounded

by big loop

E dA E dA E dl∇ × ⇒ ∇ × =∑ ∫ ∫

This is Stokes’ Theorem, named for the Irish mathematical physicist George Gabriel Stokes (1819-1903); it is true for any vector field.

62


s f

Discussion: the Helmholtz theorem Suppose we are told what the divergence and curl are for some vector function , but

don’t know the explicit form of it el :

( ), ,F x y z

F( ) ( ), , , ,F x y z D x y z∇ =i

and ( ) ( ), , , ,F x y z C x y z∇× = .

(Naturally, ∇ will be zero since the divergence of a curl is always zero.) Ci Is there a general technique we can use to determine F if we know C and D? (Being told that

0E ρ ε∇ =i and ∇× for a static charge distribution is an example of this.) 0E = According to something called the Helmholtz theorem, we can always write any well-behaved vector function as the sum of a gradient and a curl as long as C and D go to zero sufficiently rapidly at large distances. The exact form is

( ) ( ) ( )3 3

all space all space

1 14 4

D r C rF r d r d r

r r r rπ π

′ ′′ ′= −∇ + ∇× ′ ′− −

∫ ∫ .

If you use this to calculate something, keep in mind that the partial derivatives in the gradient and curl are taken with respect to , ,x y z and not with respect to , ,x y z′ ′ ′ . As a result, terms which depend only on the primed variables act like constants when you evaluate the partial derivatives.

63


Discussion: The vector potential

The scalar potential for a stationary point charge at position ′r is ( )0

14

qr r

=′πε −

V r .

If we have a continuous distribution of static charge so that the amount of charge contained in a small volume dV is dVρ , we can calculate the potential at r by doing an integral ( 3d r′ is just a volume element):

( ) ( ) 3

0all space

1 4

rV r d r

r rρ

πε′

= ′− ′∫ .

Are there ever times when we might want to superpose something that is a vector instead of a scalar? Certainly! It is possible to determine the electric field by calculating the gradient of the potential; since the potential obeys the superposition principle when we have several charges in our universe, the electric field also obeys superposition. We know that the electric field at r caused by a small amount of stationary charge dVρ at ′r is

( ) 20

14

dV r rE rr rr r

ρπε

′−=

′−′−.

(The ( )r r r r′− − ′ piece is just a unit vector pointing away from the charge.) As a result, we can calculate the electric field (without referring explicitly to the potential) this way:

( ) ( ) 32

0all space

1 4

r r rE r d rr rr r

ρπε

′ − ′= ′

− ′− ′∫ .

( is just the volume element dV3d r′ ′ .) An integral is really just a big sum. The presence of the vector’s direction as part of the integrand may make the integral harder to evaluate, but that’s just a practical matter-- it doesn’t make the integral’s meaning (more??) obscure. Remember that we're only dealing with the case right now that the charge density is unchanging: the number of Coulombs per cubic meter at a particular point never changes, no matter how long we wait. We can have constant ρ when charges are moving: as long as the amount of charge dq = ρdV flowing into a small volume element equals the amount flowing out, we'll never have a change in the amount of charge contained by dV. Here’s another quantity that has a direction associated with it. Let's say our charge distribution is swirling around so that the charge dq = ( ), ,x y zρ dV at the point (x,y,z) has velocity

. By the way, is called the ( , ,v x y z ) ( ) ( )r v rρ current density ( )J r : it corresponds to the

64


number of amperes per second per square meter flowing through a surface at which is perpendicular to . I'm going to impose the constraint that

r( )J r ( )J r never changes with time,

even though its value can differ from place to place.

3 3d rr r 02

al

′ ′

ace

, A

r ′

), ,x y z

( , ,x xv x

) (xr vr r

′ ′− −′ ′

For grins, let’s define the vector potential associated with this sort of current flow as follows:

( ) ( ) ( )02

0 0l space all space

1 1 since4 4

J r J rA r d r

c r rµ µ

πε π′ ′

≡ =− −′ ′∫ ∫ . 1

cε=

Keep in mind that this equation is really shorthand for three distinct equations:

( ) ( ) ( ) ( ) ( ) ( )3 30 0

all space all space all sp

, 4 4

yx zx y z

J rJ r J rA r d r A r d r r d r

r r r r r r′′ ′µ µ′ ′= = =

′ ′π − π − π −∫ ∫ ∫µ ′

′. 30

4 Of what use could this possibly be? Think about the scalar potential: the reason that it is useful to work with the function

( ) ( ) 3

0all space

1 4

rV r d

r rρ

πε′

=− ′∫

is that it is the solution to Poisson’s equation

( )2

0

, ,x y zV

ρε

∇ = − .

Imagine that we encounter an equation which resembles Poisson’s equation, except that the function of x,y,z on the right side is ( ) ( ) (, , , , xf x y z x y z vρ= instead of ( , , )x y zρ , and the constant on the right side is -µ0 instead of -1/ε0. Our equation would be

( ) ( ) )20 0, , , ,A f x y z x y z y zµ µ ρ∇ = − = − .

with solution

( ) ( ) ( )3 30 0

all space all space

4 4x

f r rA r d r d r

r rρµ µ

π π= =′ ′∫ ∫ .

′

(This would correspond to a charge distribution ( ), ,x y zρ swirling around, with the velocity of the charge differing from place to place, so that the “local velocity” of charge at ( ), ,x y z is

. Neither function is time dependent.) ( , ,xv x y z)

65


If we encounter similar equations involving the y and z velocity components, we could multiply both sides of each equation by the appropriate unit vector, then add them:

( ) ( )( ) ( )( ) ( )

( ) ( )

20

2 20 0

20

ˆ ˆ, , , ,

ˆ ˆ, , , , , , , ,

ˆ ˆ, , , ,

x x

y y

z z

A x x y z v x y z x

A y x y z v x y z y A x y z v x y z

A z x y z v x y z z

µ ρ

µ ρ µ ρ

µ ρ

∇ = −∇ = − ∇ = −

∇ = −

⇒ .

Why might we ever bump into an equation of this sort? How could something as abstract as the vector potential possibly matter? Who knows? (Who cares?) In any event, if for some reason we do encounter this equation, we’ll already know how to solve it because of our experience with Poisson’s equation:

2 A∇

( ) ( ) ( ) 30

all space

4

r v rA r d r

r rρµ

π′ ′

= ′− ′∫ .

(There actually is a good reason for messing with the vector potential: the magnetic field is exactly the same as the curl of the vector potential: B A= ∇× . We’ll deal with this later.) We might want to discuss the vector potential for a point charge in motion, but this sort of charge distribution does not satisfy the constraint that ( ), ,x y zρ is constant. (As the point charge moves, the charge density changes a great deal at each point the charge occupies during its travels.) However, as long as the charge is moving slowly compared to the speed of light, it's not a bad approximation to use

( ) ( ) ( ) ( )( ) ( ) ( )( )

3 30 0

all space all space

4 4

q

q

q r r t v tr v r qv tA r d r d r

r r r r r r t0

4δρµ µ

π π−′′ ′

= ≈ =′ ′− −′ ′ −∫ ∫

µπ

where ( )qr t and represent the charge's position and velocity. ( )v t Exercise 5.6: Sketching the vector potential A slowly moving point charge is passing through the origin at time t = 0 with constant velocity

. Ignoring any effects associated with the difference between t and mentioned earlier, draw a reasonably good sketch of the vector potential associated with this charge’s behavior at t = 0.

0 ˆv v z= t′

66


Exercise 5.7: Sketching another vector potential Two slowly moving point charges are passing through the points ( ), , (0, 1,0)x y z = − and

with constant velocities ( ), , (0, 1,0)x y z = + 0 ˆv v z= . Ignoring any effects associated with the difference between t and t mentioned earlier, draw a reasonably good sketch of the vector potential at t = 0.

′

67


Exercise 5.8: Sketching yet again another vector potential Two slowly moving point charges are passing through the points ( ), , (0, 1,0)x y z = − and

with constant velocities ( ), , (0, 1,0)x y z = + 0 ˆv v z= and 0 ˆv v z= − respectively. Ignoring any effects associated with the difference between t and t′ mentioned earlier, draw a reasonably good sketch of the vector potential at t = 0.

68


Discussion: retardation The correct expressions for the scalar and vector potentials include the fact that an observer at a position sees the charge as it r was at an earlier time: the information about the position of the charge takes a finite amount of time to propagate from the charge to the observer. (We see the sun as it was eight minutes ago; we see Alpha Centauri as it was 4.35 years ago.) If the charge density and/or current density changes with time, the integrals used to calculate the potentials will need to take this into account. If the observer at position (at observer-time t) sees charge at position r r′ , the time at which the charge was really at that point was at a time earlier than t by t t t r r c′ ′∆ = − −= . We need to modify our equations like this:

( ) ( ) ( ) ( )3 3

0 0all space all space

, 1 1 , 4 4

r t r r crV r d r V r t d r

r r r rρρ

πε πε− −′ ′′

= ⇒ =′ ′− −′ ′∫ ∫

( ) ( ) ( ) ( )3 30 0

all space all space

, ,

4 4J r t r r cJ r

A r d r A r t d rr r r r

µ µπ π

− −′ ′′= ⇒ =′ ′

− −′ ′∫ ∫

What happens to the potentials from a finite charge distribution with total charge Q when we

include retardation? It's tempting to think that V and A will satisfy ( ) ( )0

1,4 Q

Qtr r t

V rπε

≈− ′

and ( ) ( )( )

0,4 Q

Qv tA r t

r r tµπ

′≈

− ′

( )

if we're far from the charge. (The retarded time satisfies t′

qt t r r t c= − −′ ′ ). However, this is wrong! The problem arises because ( ) 3

all space

, t r r cQ r d rρ≠ − −′ ′∫ ′ . As we integrate over the volume

containing the charge cloud, the integral passes through different regions at different times due to retardation. (It will pick up contributions to V from distant regions further in the past than from close regions.) Here's a concrete example. Imagine that a small charge, shaped in its rest frame like a tiny cube with side a, is moving towards an observer. The observer will evaluate the potential (scalar or vector) caused by this charge as a sum of contributions from small slices in space as the cube moves towards him, as shown in the figure.

69


3 5 7 92 4 6 8

a v

2930

2 21a v c−

distantobserver

2 211

a v cv c

−−

cube is here wheninformation fromfirst slice reaches

observer

cube is here wheninformation fromlast slice reaches

observer

1

The observer will take a "snapshot" of the amount of charge filling different regions of space and will calculate the net potential as the sum of the contributions from all the small regions of space in his snapshot. Because of retardation, the "local time" (the retarded time) in each region of space in the snapshot will be different: the further away a region is, the earlier in time he'll be seeing it. The slice of space labeled "1" in the observer's snapshot just happened to be holding the back of the Lorentz-contracted cube. Since the cube is moving, the front of the cube just happened to be occupying slice 30 at the correct time for it to appear in the snapshot: slice 30 is seen at a less retarded time than slice 1 since it is closer to the observer. As far as the observer is concerned, the potential must be the same as it would be with slices 1 through 30 all containing electric charge! (If the charge were moving away from the observer, there would seem to be less charge there.) With some algebra, it's easy to show that the length of the region in space which seems to hold charge is a factor of ( )1 1 longer than the side of the charge which is parallel to its line-of-site to the observer. If the charge were not moving directly towards/away from the observer, the correction would involve the component of velocity parallel to the observer's line-of-site:

v c−

( )ˆ1 1 v cε− ⋅ where ε is the unit vector from the object to the observer, as shown in the following figure.

70


v

observer

ε

line-of-sightto observer

As a result, our expressions for the potentials associated with a point charge need to be written

( ) ( ) ( ) ( )( )0

1 1,ˆ4 1Q

QV r tv t t cr r tπε ε

= ×− ⋅′ ′− ′

and

( ) ( )( ) ( ) ( )( )

0 1,ˆ4 1Q

Qv tA r t

v t t cr r tµπ ε

′= ×

− ⋅′ ′− ′.

These are know as the Liénard-Wiechert potentials. (A.Liénard, 1898; E.Wiechert, 1900.)

71


Exercise 5.9: Retardation Consider a point charge which oscillates along the z axis so that its position is ( ), , (0,0, sin( ))x y z a tω= . The amplitude and frequency of its motion are a=0.001 feet and ω=2π×109 radians per second (so that the period of its motion is one nanosecond.) Including effects associated with the finite speed of light (c ≈ 1 foot per nanosecond), sketch how the vector potential looks along the line y = z = 0 within a few feet of the origin. (Note that

ˆ 0v ε⋅ ≈ so you can ignore the ( )ˆv c1 1 ε− ⋅ correction.) Discussion Summary and wrap-up discussion • properties of the electric field that are impossible to find in fields generated by static charges • curl and divergence • Stokes’ Theorem • The vector potential • Time retardation’s influence on the vector potential • Why did we do this, anyway? Comments about homework assigned for next week

72


.

Reading and homework should be completed by 4pm Friday. Required reading: Purcell, chapter 2, sections 13 - 16; chapter 6, section 3. The Feynman Lectures on Physics (volume II), chapter 21, section 5. Read over the remaining Unit 5 material we didn’t get to in class this week. Also read over (but don't worry about understanding) the Unit 6 material we'll tackle next week. Optional reading: The Feynman Lectures on Physics (volume II), chapter 14. Problem 0 (5 points): Make your own hardcopies… Print (from the web) your own copy of the solutions to this week’s in-class exercises as well as last week’s problem set and write a (true!) statement on your problem set asserting that you have successfully printed your personal copies. Problem 1 (10 points): Anthrax’s litter box is exposed to an electric field Mrs. Urkin creates an E field across the top Anthrax’s cat box in order to lay claim to the invention of the field of electroscatology. (The top of the cat’s litter box is in the x-y plane at z = 0 with 0 1 ) She experiments with different field configurations as described below.

0 1,x≤ ≤ y≤ ≤

(a) The first electric field configuration is ( ) ( )ˆ ˆ ˆ, ,E x y z xyz x y z= + + at the top of the

aquarium. Is it possible for this particular electric field to be the gradient of a scalar potential ? Be sure to prove, or otherwise justify your answer. ( , ,V x y z )

(b) The next electric field configuration is ( ) ˆ ˆ ˆ, , +E x y z yz x zx y xy z= + at the top of the

aquarium. Can this electric field be the gradient of a scalar potential? Prove your answer. (c) Another configuration is . Can this electric field be the gradient of a

scalar potential? Prove your answer. ( ) 2 2ˆ, , -E x y z x x y y= ˆ

(d) The final field configuration is ( ) 2 2ˆ, , -E x y z y x x y= ˆ . Can this electric field be the gradient

of a scalar potential? Prove your answer.

73


74


Problem 2 (10 points): Being clever may help you avoid doing the line integrals

Mr. Urkin fabricates Christmas presents for his nephews from electrically charged marbles and polyvinyl chloride sewer pipes as illustrated in the figure. One of the “toys” comprises a square loop of (frictionless) sewer pipe in which a marble rolls. Another is a circular loop of pipe, also frictionless, in which a second marble rolls. Each marble carries charge q.

charged marble

plastic pipe

charged marble

1 meter 1 meter

1 meter xx

y y

plastic pipe

(a) Placed flat on the floor during tests with its center at ( ) ( ), 0,x y = 0 and its sides parallel to

the x and y axes, the square loop is exposed to an electric field ( ) ˆ, ,y z x y=E x . (The floor is in the plane z = 0.) Assuming the marble does not lose any energy as it rounds the corners, calculate the change in its kinetic energy after the field drives it once around the loop.

(b) The round loop is tested in a similar fashion: it is placed in the same electric field, with its

center at the origin. How much kinetic energy does the marble gain traveling once around the loop?

Problem 3 (10 points): Vector potential sketch Read through the in-class material about retardation, digesting it as best as you can (see me with questions) and then do Unit 5’s in-class exercise 5.9.

75

Physics 212, Spring 2003. Unit 6 6.1 The origin of the magnetic field as a consequence of special relativity

Physics 212


Unit 6

The origin of the magnetic field as a consequence of special relativity

Spring, 2003

George Gollin


76


Unit 6: The origin of the magnetic field as a consequence of special relativity Recap of last week Introduction: magnetic fields are just electric fields seen from a different perspective. Magnetic fields and forces exist because of Coulomb’s law and special relativity. Though observers in one frame may determine that a current-carrying wire is electrically neutral, observers in another frame might conclude that the wire does carry a net electric charge. We’ll derive this today, for the particular case of motion parallel (or antiparallel) to the current. You’ll slog through much less vector calculus today than last week! Here’s what you’ll find/work on today: • analyzing the charge density in a wire loop which suddenly finds itself carrying a current,

according to observers in two different frames of reference • describing how electric fields from a static charge transform when changing to a frame in

which the charge is in motion

77


Exercise 6.1: Whap! They’re moving! A rectangular loop is made from a thin filament of superconducting wire as drawn below. Initially no current flows in the wire; there are as many (positive) metal ions in the wire’s crystal lattice as there are conduction electrons. The spacing between adjacent electrons is a while the spacing between positive metal ions is b. (Naturally, since the wire carries no net charge, a = b.)

wire loop

b

a

metal ions

electrons

synchronizedclocks

before...

electron flow

...after

u

u

When all (synchronized) clocks in the lab frame read zero, all the electrons are suddenly accelerated from rest to a speed u so that the flow of electrons in the wire loop is in a counter-clockwise sense. (a) According to observers in the lab frame (the rest frame of the still-stationary metal ions), what is the spacing between adjacent (but now-moving) electrons? Discussion ...but what about Lorentz contraction?

78


(b) An observer in a different frame of reference sees the wire loop glide past with speed v, moving to the left as shown in the figure. When the observer’s clock reads zero, clocks close to the right-most edge of the (moving) wire loop are also seen to read zero.

v

Sketch the readings on the other (moving) clocks’ faces at the instant the right-side clocks read zero. discussion

79


(c) Keep in mind that an electron is accelerated when the lab-frame clock next to it reads zero. From the perspective of the observer who sees the wire loop moving to the left, the densities of electrons and metal ions are identical before any of the electrons begin circulating in the wire loop. Describe the relative densities of electrons and metal ions in each of the four sides of the loop after all the electrons have been set in motion. It may help to think in terms of which electrons begin moving when. (d) What would happen to a test charge, stationary in the frame of the observer who sees the wire loop moving to the left, if the test charge were placed close to the bottom side of the loop? How would this charge behave when seen from the rest frame of the wire loop? discussion

80


Exercise 6.2: Quantitatively, now... Let’s consider the bottom leg of the wire loop now, after the electrons have started moving. You should have concluded that the electron spacing, viewed from the lab frame (the rest frame of the wire), does not change after the electrons are accelerated. (How could it possibly change? In this frame all electrons started moving simultaneously, so there was never a time interval during which one electron was moving but its neighbor was not.)

a

a

metal ions

electronsu

(a) Calculate (in terms of a and u) the spacing between electrons in the electrons’ rest frame. (Even though u for realistic currents you’ll need to consider the effects of special relativity.)

c

(b) Recall that the binomial expansion for ( )1 n+ ε is ( ) ( ) 211 1

1! 2!n n nn −

+ ε = + ε + ε + …

(This works even when n isn’t an integer, and is useful for approximations.) Some examples for : 1ε 1 1+ ε ≈ + ε 2 ( ); ( )1+ ε ≈ − ε1 1 ; ( ) 1 21 1 2−1 1+ ε = + ε ≈ − ε . In these exercises you’ll

only need to calculate things to second order accuracy in any of the velocities present in your equations (terms proportional to can be ignored). 3 3, ,v u u 2...v Use the binomial expansion to calculate a version of the relativistic velocity addition formula which is accurate to second order in the velocities.

81


c(c) In another frame of reference the wire is seen to move to the left at speed . (By this I mean that the metal lattice is moving with speed v). How fast are the electrons moving in this frame? (Please use the second-order expression you just derived.)

v

(d) Calculate the spacing between metal ions in the frame in which the wire is moving. Do the same for the spacing between electrons in this frame. (e) Use the binomial expansion to generate an approximate expression for the spacing between adjacent metal ions in the frame with the wire moving. Please include all terms up to second order in the ratios of small velocities (u and v). Do the same thing for the electrons.

(f) Calculate the apparent charge density in the wire in terms of the fundamental charge (e), a, u, v, and c. Verify that your expression yields zero if v = 0 or u = 0.

82


Discussion

83


Exercise 6.3: Real numbers Let’s calculate what happens in a real wire. Imagine that we have a 1000 ampere current flowing in a heavy copper cable with cross-sectional area 1 cm2. The density of copper is about 9 grams per cm3 while the mass of a single copper atom is approximately 10-22 grams. As a result, every cubic centimeter of copper cable holds 9 × 1022 copper atoms. Each copper atom gives up two conduction electrons so that there are 1.8 × 1023 mobile electrons per cm3 of cable. The charge on each electron is 1.6 × 10-19 Coulombs. (a) How fast are the electrons moving, on average? (b) How large is the electric field generated by the (charged!!) wire at a distance of 10 centimeters according to a test charge which observes the wire moving to the left with speed 100 m/s? (Use Gauss’ law!)

84


(c) If the test charge is an electron (with mass 9.1 × 10-31 kg), how large is the acceleration it experiences due to the passage of the current-carrying wire? (d) In the wire loop’s rest frame the charge moves with initial velocity 100 m/s parallel to the wire. Its acceleration is nearly the same in this frame. What is the radius of curvature of its path at the instant that it is moving parallel to the wire? Discussion Exercise 6.4: How the fields transform (let’s call this one “optional” in case we don’t have enough time)

85


Relativistic effects will also deflect a moving test charge which heads directly towards the current-carrying wire. Let’s investigate this now, starting with the fact that Gauss’ law is always true, even for charges in motion. The flux of electric field through a closed surface is always

0enclosedsurfaceE dA Q=∫ i ε , even when the enclosed charge is in motion. (An experimental proof

of this: the electric field far from a hydrogen atom is nearly zero, even though its proton is nearly at rest, while its electron is moving with 137v c∼ .) We’ll need to determine how the electric field at a point in space transforms as we shift frames of reference. Let’s say that observers in two frames of reference, O (the "unprimed" frame) and O', measure the electric field at the origins of their respective coordinate systems just as the origins coincide. Frame O' (according to observers in O) is moving with velocity v along the x axis of O as shown in the figure. In the figure, the field measurements have already been performed.

x

z

y

O

x'

z'

y'

O'

v

electric fieldlines

86


(a) The field lines have been created by a pair of capacitor plates which are stationary in O as shown in the next figure. The surface charge densities are ±σ .

x

z

yO

x'

z'

y'O'

vcapacitor plates

+σ

−σ

What are the surface charge densities ′σ measured by observers in O'? (b) What is E E′ ?

Discussion Fields which are perpendicular to the direction of motion... (Note: in O there is no magnetic field.)

87


(c) In a different universe, field lines have been created by a pair of capacitor plates which are stationary in O as shown, but whose faces are perpendicular to the direction of motion of O'. As before, the surface charge densities are ±σ .

x

z

yO

x'

z'

y'O'

v

capacitorplates

+σ−σ

What are the surface charge densities ′σ measured by observers in O'? (d) What is E E′ ?

Discussion Fields which are parallel to the direction of motion... (Note: in O there is no magnetic field.)

88


Discussion

You should have concluded that E E′ = and ( )2 21E v c E E⊥ ⊥′ = − = ⊥γ where and E E⊥

represent electric field components parallel to and perpendicular to the relative velocities of the two frames. An important point: this is only true if there is no magnetic field in the frame in which the electric charge (the surface charge on the capacitors) is at rest. Naturally, this applies to electric fields created by any charge distribution, not just the sheets of charge on capacitor plates. In particular, the electric field around a moving point charge becomes stronger in the plane perpendicular to its direction of travel. If we use the density of field lines to represent the strength of the electric field, we’ll find our field lines seem to bunch up like this:

charge, at rest in O charge, as seen from O'.

v

To a test charge moving towards the current-carrying wire, the ions’ and electrons’ velocities will appear as follows (in the rest frame of the test charge):

metal ions

electrons

test charge

89


(e) On the following diagram (which shows one pair of conduction electrons and metal ions, placed symmetrically about the test charge), sketch the field lines at the test charge associated with the two electrons. (In the test charge’s rest frame, the wire is moving towards the test charge.)

test charge

Discussion Direction of the force, and comparison with qv B× ...

90


Summary and wrap-up discussion • Magnetic force is just a consequence of relativity: observers in different frames do not agree

about charge densities! Comments about homework assigned for next week

91


Reading and homework should be completed by 4pm Friday. Required reading: Purcell, chapter 5, sections 4 - 6; chapter 5, section 9. Read over (but don't worry about understanding) the Unit 7 material we'll tackle next week. Optional reading: The Feynman Lectures on Physics (volume II), chapter 26. The Feynman Lectures on Physics (volume II), chapter 42. (Some of these problems require you to know how to multiply matrices, and to remember that the transpose of a matrix is what you get when you swap its rows and columns. Please see me for a quick lesson if you've never learned about this.) Problem 0 (5 points): Make your own hardcopies… Print (from the web) your own copy of the solutions to this week’s in-class exercises as well as last week’s problem set and write a (true!) statement on your problem set asserting that you have successfully printed your personal copies. Problem 1: More on Lorentz transformations Imagine observers in the frame O see another frame O' moving in the positive x direction with velocity v so that the origins of the two coordinate systems overlap at t = t' = 0. We can use the Lorentz transformations to relate the space-time coordinates of an event in one frame with those in another frame. Defining v cβ ≡ and 21 1γ β≡ − , the Lorentz transformations become ( ) ( ), , , x x ct y y z z t t x cγ β γ β′ ′ ′ ′= − = = = −

( ) ( ), , , x x ct y y z z t t x cγ β γ β′ ′ ′ ′ ′ ′= + = = = + . Keep in mind that v is the velocity of the primed frame's coordinate axes as seen from the unprimed frame. We can write the Lorentz transformations in matrix form this way:

0 00 0

0 0 1 00 0 0 1

ct ctx xy yz z

γ γβγβ γ

′ − ′ − =

′ ′

and

0 00 0

0 0 1 00 0 0 1

ct ctx xy yz z

γ γβγβ γ

′+ ′+ =

′ ′

.

92


More compactly, defining

ctx

xyz

′ ′ ′ ≡

′ ′

,

ctx

xyz

≡

, and

0 00 0

0 0 1 00 0 0 1

γ γβγβ γ

− − Λ ≡

allows us to write the Lorentz transformations as

x x′ = Λ , 1 x x− ′= Λ where 1 1

1 0 0 00 1 0 00 0 1 00 0 0 1

I− −

ΛΛ = Λ Λ = ≡

.

Any quantity which transforms this way under changes of reference frame is called a 4-vector. Another four vector can be built from a particle's energy E and momentum p :

x

y

z

E cp

ppp

≡

.

since it transforms like this: p p′ = Λ . Naturally, this is just shorthand for the four equations

( ) ( ), , , x x y y z z xp p E c p p p p E c E c pγ β γ β′ ′ ′ ′= − = = = − .

(a) For Λ defined as ,

0 00 0

0 0 1 00 0 0 1

γ γβγβ γ

− −

prove that its inverse 1−Λ is

0 00 0

0 0 1 00 0 0 1

γ γβγβ γ

+ +

.

(b) A proton at rest has energy . (Its momentum, not surprisingly, is zero.) The protons which circulate inside the Fermilab Tevatron have energies close to 1,000 GeV. (1 GeV = 1,000 MeV.) What value of γ does a Tevatron proton have?

2 938 MeVproton pE m c= ≈

93


(c) Somehow a Tevatron proton captures an electron, becoming a fast-moving hydrogen atom without changing its speed. An electron at rest has . What is the electron's energy in the rest frame of Fermilab's sedentary buffalo herd?

2 0.511 MeVelectron eE m c= ≈

Problem 2: Field transformation laws As you've seen, the electric and magnetic fields are intimately connected by relativity. We can construct the "electromagnetic field strength tensor" out of the fields' components like this:

00

00

x y z

x z

y z

z y x

E c E c E cE c B B

FE c B BE c B B

− − − − ≡ − −

y

x.

What makes this a "second rank tensor" (a 4-vector is a "first rank tensor") is that the electric and magnetic fields transform when one changes frames this way:

0

0

0

0

x y z

x z y T

y z x

z y x

E c E c E c

E c B BF F

E c B B

E c B B

′ ′ ′− − − ′ ′ ′− ′ = = ′ ′ ′− ′ ′ ′−

FΛ Λ = Λ Λ

If an observer in one frame of reference measures E and B at the point (x,y,z) at time t, that observer can predict what an observer in a different frame would measure for and E′ B′ should their field-measuring devices glide past each other just as the measurements are being performed. (a) The electric and magnetic fields of a 1 Coulomb charge Q are measured by a pair of gizmos as shown in the figure. From the perspective of observers in frame O, the charge is at rest at the origin and one of the field-measuring devices is also at rest, with position (x,y,z) = (0,1,0). Observers in O see the second field-measuring gizmo moving at high speed, with velocity

ˆv cxβ= .

94


x

z

y

E B

v=βcE Bv=0

1 meter

Q

At time t = 0 this gizmo passed very close to the stationary device at (x,y,z) = (0,1,0). The readings on the dials of the moving device were photographed by observers in O as it was illuminated by the lights from the stationary gizmo's panel lights. • Calculate the components of the electric and magnetic field vectors measured by the

stationary device at the time the photograph was taken. • Calculate the components of the electric and magnetic field vectors measured by the moving

device. • According to observers in the rest frame of the moving device, how far away is the charge

Q? (b) The same gizmos are now placed on the x axis with the stationary device at position (x,y,z) = (1,0,0) and the moving device traveling along the x axis with velocity v ˆcxβ= . As the moving device brushes past the stationary device, photographs are taken. • Calculate the components of the electric and magnetic field vectors measured by the

stationary device at the time the photograph was taken. • Calculate the components of the electric and magnetic field vectors measured by the moving

device. • According to observers in the rest frame of the moving device, how far away is the charge

Q?

95


Problem 3: And now for something completely different The idealized “operational amplifier” shown in the figure has the following properties:

-

+

V-

V+

Vout

Vout = G(V+ - V-)

• No current enters its inputs, regardless of how large the input voltages V+ and V- become. • The output voltage is always exactly Vout = G(V+ - V-) where G is the (constant) gain of the

amplifier. • The gain G is “nearly” infinite. The inputs to which V+ and V- are connected are called the amplifier’s “non-inverting” and “inverting” inputs, respectively. Note that voltages can be bipolar: both positive and negative values are fine. (a) In the following circuit, what is Vout as a function of Vin?

-

+Vin

Vout

(b) What is Vout as a function of Vin for this circuit? Keep in mind that no current flows into the amplifier’s input terminals and that Ohm’s law requires the voltage drop across a resistor to be equal to the product of its resistance and the current flowing through it. .

96


-

+Vout = ?

Vin

R1 = 10kΩ

R2 = 20kΩ

97

Physics 212, Spring 2003. Unit 7 7.1 Operational amplifier circuits

Physics 212


Unit 7

Operational amplifier circuits

Spring, 2003


2003

98


Unit 7: Operational amplifier circuits Recap of last week Introduction: opamps and other circuit elements We’re going to build toys today. It’s a lot more fun to build electronic circuits with active components (amplifiers and the like) than it is to be stuck in the passive-only world of electrical circuits built only with resistors and capacitors. Our principal building block will be the operational amplifier (opamp), a differential amplifier whose “open loop” gain is so high, and input impedance so large, that it is always used with feedback to control its behavior. Let’s discuss opamps using a particular example, the venerable LF411. This opamp is built on a single sliver of silicon, and comprises twenty-two transistors, one diode, one capacitor, and eleven resistors. You can find a description of it at the Texas Instruments web site http://www.ti.com/: run a search on “LF411” to find the data sheets. One of the many ways the chip is packaged is in an 8-pin dual-inline package (DIP):

IN- and IN+ are the “inverting” and “non-inverting” inputs to the opamp. OUT is, of course, the opamp’s output pin. VCC+ and VCC- are the power supply pins, held somewhere between ±3.5 and ±18 volts. The pins BAL1 and BAL2 may be used to fine-tune the amplifier’s output when the inputs IN+ and IN- are held at exactly the same voltage. We will leave them disconnected. We’ll use a symbols like the following to represent the opamp, omitting the power and balance pins.

-

+or

99


(It’s OK to draw them with the non-inverting input at the top instead of the bottom.) The output voltage is V G , where G is the opamp’s “open loop” gain. According to Texas Instruments, the typical open loop gain for an LF411 is 200,000. (However, the output cannot go higher or lower than the power supply voltages!) The LF411’s input impedance is typically 10

( )out V V+ −= −

12Ω: increasing the input voltage by a volt will only change the current flowing into the input pin by a picoamp. The device can drive output currents as large as 20 milliamps, and output voltages within a few volts of VCC+ and VCC-, so it has enough “oomph” to drive a pair of headphones without additional amplification. (Headphones have input impedances of several dozen ohms and can handle input voltages of a couple of volts before melting.) We NEVER use an opamp without feedback between its output and its inputs: its gain, “input offset voltage,” and other parameters are not particularly stable against changes in power supply voltage, temperature, and so forth. The following circuit is a BAD BAD THING!

-

+

~

Vout = -GVin

Bad circuit!!!Vin Here’s how to think about an opamp:

1. An opamp’s gain is (almost) infinite; 2. An opamp’s input impedance is (almost) infinite.

This sounds crazy, but there are three very useful rules of thumb which follow from this:

1. In a properly designed circuit (which won’t “pin” its output), the opamp does whatever is necessary to make its inputs (almost) exactly equal;

2. (almost) Absolutely no current flows into the opamp’s input pins. (almost) As much current as needed flows from (or to) the output pin to maintain its output voltage.

3. There must always be a DC path between the opamp output and (at least) one input, even if it’s made with a huge resistor that does very little in setting the circuit’s behavior.

Here’s what you’ll find/work on today: • Analysis (and construction!) of a few simple opamp circuits

100


Exercise 7.1: Opamp voltage follower You analyzed this circuit as a homework problem. Be sure that you understand how the circuit works. (Do you see why Vout as a function of Vin behaves the way it does?)

-

+Vout = ?

Vin

Exercise 7.2: Mystery amplifier #1 This was another homework problem, and is the circuit that we have setup for you on the breadboard at your lab bench. Make sure you are clear about how the circuit works.

-

+Vout = ?

Vin

R1 = 10kΩ

R2 = 20kΩ

Discussion What happens to Vout if we load the above circuit’s output, as shown?

-

+

VoutVin

load resistance

101


Hands-on time: We’ve already assembled the simple “inverting amplifier” circuit in the previous exercise for you, provided power to the opamp chip, attached a signal generator to its input, and set up the oscilloscope to trigger sensibly. Note that we’ve put some capacitors on the board between ground and the power supply voltages to smooth out noise and glitches on the power “rails.” The capacitors are probably the kind that care about the polarity they see at their input leads: one particular lead must always be more positive than the other. (Reversing the polarity can cause the cap to explode, or catch fire, in a most dramatic fashion.) Investigate the following with your circuit: 1. How close is the gain to what you’d expect? 2. How close to being equal are the voltages on the two input pins? 3. Change one of the resistors. Does the gain change in the way you expect? 4. Change (increase) the frequency of the input signal. At what frequency do you begin to

notice the amplifier’s gain dropping? 5. Change the input signal frequency again, keeping an eye on the phase difference between the

input and output signals. When does the phase shift begin to change noticeably? Exercise 7.3: Mystery amplifier #2 What does the following (idealized) circuit do? (By this I mean what is Vout in terms of Vin?)

C = .01µF

-

+Vout = ?

Vin

R = 100kΩ

Note that this circuit violates our third rule of thumb and is NOT A GOOD CIRCUIT!

102


103


Discussion and hands-on time: An important design rule: due to imprecision in manufacturing, the point at which an opamp produces zero volts at its output will be when its inputs are held at slightly unequal voltages. This input offset voltage can be as large as several millivolts. Generally, it is necessary to have a DC feedback path from the opamp’s output to an input. As a result, a better version of the integrator circuit shown above would be the following:

.01µF-

+Vout

Vin

100kΩ

22MegΩ

The 22Meg resistor is small compared to the LF411’s input impedance, but large compared to the 100k input resistor. The circuit’s behavior as an integrator will change to that of an ordinary inverting amplifier for times which are not short compared to RC = 22Meg × .01µF = 0.22 seconds. Build this circuit and see how it responds to sinusoidal, square, and triangle wave inputs. Vary the signal generator’s frequency and observe how the circuit’s response to a sinusoidal input changes (the circuit’s gain, in particular). Exercise 7.4: Mystery amplifier #3 What does this circuit do?

C = .01µF

-

+Vout = ?

Vin

R = 100kΩ

104


Discussion The above circuit is vulnerable to high frequency noise (some of which is generated by the amplifier itself) since its gain increases with frequency. If you need to use one of these, a better version includes another (smaller) resistor and capacitor, as in the following circuit.

.01µF

-

+Vout

Vin100kΩ1kΩ

50pF

Build one of these and see if it behaves as you’d expect.

Exercise 7.5: Mystery amplifier #4 A diode’s current-vs.-voltage behavior is 40( ) V

SI V I e≈ . The parameter IS (the “reverse saturation current) depends on details of the diode’s construction and changes rapidly with temperature. Here are a couple of voltage vs. current curves for a 1N914 silicon junction diode. Since the current axis scale is logarithmic, the V vs. I curves are straight lines.

Let’s assume that IS is 10-15 amperes for the diode in the following circuit. As drawn, the diode likes to pass current from left to right.

105


In the following circuit what is Vout as a function of Vin? (Assume that Vin ≥ 0.)

diode-

+Vout = ?

Vin

R = 1kΩ

current flows in this direction Discussion Since the diode will only pass current in one direction, a better scheme might be to use a pair of diodes in parallel, with the second diode passing current from right to left. Build the circuit with the pair of diodes as described above, and see how it behaves in response to triangle waves. Exercise 7.6: Do-it-yourself summing amplifier How would you design an opamp circuit which sums two voltages so that Vout = -(Vin1 + Vin2)? (Hint: one version of the circuit requires three resistors and one opamp.) Discussion Build an example of your circuit and make it work. (Figure out a clever way to have your function generator provide a second input to your circuit.) Summary and wrap-up discussion

106



107


Reading and homework should be completed by 4pm Friday. Required reading: Read over (but don't worry about understanding) the Unit 8 material we'll tackle next week. Optional reading: Horowitz & Hill, chapter 4 through section 4.06. (The book is on reserve in the Physics Department library.) Problem 0 (5 points): Make your own hardcopies… Print (from the web) your own copy of the solutions to this week’s in-class exercises as well as last week’s problem set and write a (true!) statement on your problem set asserting that you have successfully printed your personal copies. Problem 1 (10 points): Diode circuit IS is 10-15 amperes for the diode in the following circuit. (As drawn, the diode likes to pass current from left to right.) What is Vout as a function of Vin? (Assume that Vin ≥ 0. There is a description of the current-voltage behavior of a diode in the in-class materials.)

diode

-

+Vout = ?

Vin

R = 1kΩ

Problem 2 (10 points): Multiplying voltages You now know about opamp circuits which incorporate diodes and resistors. (You will need the results of the previous problem in order to work this problem.) Keeping in mind that

( )1 2 1 2ln ln ln1 2

V V V VV V e e× +× = = , design a circuit whose output is

108


1 2out in inV V V= ×

where V and V . You should assume that the (ideal) diode behaves as described previously, with I

10in >

20in >

S = 10-15 amperes. (Your circuit will probably require several opamps.) Problem 3 (10 points): Mrs. Urkin commissions a prototype for something more terrible Anticipating a possible upturn in the market for horrible instruments of torture and punishment, scientists in the UD&HE Research Division are charged with the responsibility of developing a current-source poultry incinerator based on a high-voltage, high-power operational amplifier. The potentiometer (a variable resistor) at the opamp's input controls the amount of current passing through the dressed bird in a species-independent fashion. (For a given potentiometer setting the same current will flow through a chicken as a turkey.)

10 Ω

-

+

1000 V

potentiometer:Rpin 1 to pin 3 fixed at 1000 ΩRpin 2 to pin 3 varies from 0 to 1000 ΩRpin 1 to pin 2 = Rpin 1 to pin 3 - Rpin 2 to pin 3

pin 1

pin 2

pin 3

Determine the current I which flows through the bird as a function of the resistance Rpin 2 to pin 3 dialed into the potentiometer. (The utility of this circuit is that the current which flows through the "load" [the bird] is independent of the properties of the load.)

109

Physics 212, Spring 2003. Unit 8 8.1 Electronic models of mechanical systems; analog computers 1

Physics 212


Unit 8

Electronic models of mechanical systems; analog computers 1

Spring, 2003

George Gollin


110


Units 8 & 9: Electronic models of mechanical systems; analog computers Recap of last week Introduction: using opamp circuits to represent (differential) equations Last week you analyzed (and built) a number of circuits which can be used to add voltages, take logarithms, multiply voltages by constants, and so forth. I have installed a table of the circuits (some will probably be new to you) on the last page of the unit 8 material. As you may already know, it is possible to build circuits whose behavior mimics that of non-electronic systems. I think it’s easiest to show this with an example. Imagine we apply a driving force F(t) to a mass immersed in a viscous liquid so that the drag force it experiences is proportional to velocity. We’ll have ( )net dragF F t F= + or

( )ma F t vβ= − (β is a positive constant, and the drag force opposes the velocity) so that

( )mdv dt F t vβ= − . Rewrite this slightly so that the highest derivative is by itself on the left:

( )1dv F t vdt m m

β= − .

A circuit which “solves” ( )dv dt F t m v mβ= − can be arranged like this:

1. assume we have a voltage which correctly represents dv dt coming in from the left. 2. integrate it to produce a voltage which represents -v. 3. multiply the voltage corresponding to –v by the constant β/m. 4. receive (from a function generator) a voltage which corresponds to F(t)/m. 5. add the voltages to produce a voltage representing –(F(t)/m - vβ/m). 6. invert –(F(t)/m - vβ/m) to produce (F(t)/m - vβ/m). 7. bring this signal out from the right, inject it into the input of step 2’s integrator.

For example, this circuit produces an integral of its input signal:

-+ Vout

Vin

R C

22 MegΩ

1 ( 22 Megout inV V dt RRC

= − Ω∫ )

. Other functions are shown in the table at the end of this unit.

111


Here’s what we can do when β/m < 1:

Nillog

-+

-vdv/dt R C

22 MegΩ

Integrate with RC=1

-+

-vβ/mR1

R2Scale with

( )2 1 2R R R mβ+ =

Add

-+

R R

F(t)/m R

-vβ/m-F(t)/m + vβ/m

-+

R R

Invert

F(t)/

m -

v β/m

(= d

v/dt

)

We could have combined steps 2 and 3 ( reducing our component count) by picking the integrator’s RC value to be m/β instead of 1:

Nillog

-+

dv/dt R C

22 MegΩ

Integrate with RC=m/β

-vβ/m

Add

-+

R R

F(t)/m R

-vβ/m

-F(t)/m + vβ/m-+

R R

Invert

F(t)/

m -

v β/m

(= d

v/dt

)

112


If we are interested in seeing a representation of the object’s position we can add an extra integrator to calculate x v dt= ∫ :

Nillog

-+

dv/dt R C

22 MegΩ

Integrate with RC=m/β

-vβ/m

Add

-+

R R

F(t)/m R

-vβ/m

-F(t)/m + vβ/m-+

R R

Invert

F(t)/

m -

v β/m

(= d

v/dt

)

-+

R C

22 MegΩ

Integrate with RC=β/m

x

By watching the circuit’s voltages on an oscilloscope we’ll see a good representation of the behavior of the mechanical system’s position and velocity. To make scope work easier, let’s use one millisecond as the unit of time. This is easy to do: instead of building the circuit so that RC = 1, or β/m, or m/β, we’ll use RC = 10-3, or β/m × 10-3, or m/β × 10-3 instead. (For example, use R = 100 kΩ and C =.01 µF to obtain RC = 10-3.) In addition, let’s imagine we’ve chosen a system where β/m ≈ 0.2. Defining ( ) ( )f t F t≡ m , our equation becomes

( ) 0.2dv f tdt

= − v or ( )2

2 0.2d x dxf tdt dt

= − if you prefer. The circuit becomes

-+

dv/dt

500kΩ.01µF

22 MegΩ

Integrate-0.2v

Add and flip sign

-+

-F(t)/m + 0.2v-+

Invert

F(t)/m – 0.2v (= dv/dt)

Nillog

F(t)/m 100kΩ

100kΩ100kΩ

100kΩ100kΩ

.

113


Exercise 8.1: Analytic solution

Imagine that you drive your mass-in-viscous fluid with a time-dependent driving force as shown in the following figure.

t

f(t)

f0

The equation of motion for the velocity of the mass is 0.2dt vdv = − when the force is zero and

0 0.2dv dt f v= − when the force is f0. (a) Let’s call the time at which the force goes to zero t = 0. If the velocity at the instant the force drops from f0 to zero is v0, prove that is a solution to the equation of motion during the time when the force remains zero.

0.20( ) tv t v e−=

(b) Now let’s call the time at which the force changes from zero t = 0. If the velocity at the instant the force increases from zero to f0 is v1, prove that ( ) 0.2

1 0( ) 5 5tv f e f−= − + 0v t is a solution to the eqution of motion as long as the force remains f0. Exercise 8.2: Build it! Build the three-opamp-circuit drawn above. Try driving it with 1 volt square waves of various frequencies. How does your result for the voltage corresponding to -vβ/m correspond to the exact solution from the prevuous exercise? Discussion You’ve just built an analog computer! Why might one want to do (to have done) that?

114


kx v

Optional material: Modeling a damped, driven simple harmonic oscillator Introduction A damped, driven harmonic oscillator experiences a net force with contributions from a spring ( ), a velocity-dependent drag ( ), and a driving force (F ). We can write its equation of motion as . After a small amount of manipulation, the equation becomes

springF = − dragF β= −

drag drivF F= + +net spring eF Fdrive

( ) ( ) ( ) ( )2

2drived x t dx t F tk x t

dt m dt m mβ

= − − + .

You may remember that the “natural frequency” of this oscillator is radians/sec. Its natural frequency in Hz (cycles/sec) is .

k mω =

( )2ω π For homework you designed a circuit which models this oscillator using the values k/m = 1 and β/m = 0.2. Defining , your oscillator’s equation becomes ( ) ( )f t F t≡ m

( ) ( ) ( ) ( )2

2 0.2d x t dx t

x t fdt dt

= − − + t .

Build it! To make scope work easy, let’s use milliseconds as the unit of time so that your integrators will have RC values in the vicinity of 10 instead of 1. (R = 100 kΩ and C =.01 µF gives RC = 10 .) The oscillator’s natural frequency is 1 2 cycles per millisecond, or about 159 Hz.

-3 -3

( )π Build your circuit, and try driving it with 100 millivolt sine waves of various frequencies. Describe how the amplitudes of the voltages which correspond to x(t) and v(t) behave as you change the frequency of the driving “force.” Also take note of the relative phases of x(t) and f(t). Configure your oscilloscope to display an x-y plot of the voltages representing x(t) and v(t). What do you see? Discussion Summary and wrap-up discussion

115



116


Opamp circuit function blocks

add voltages and flip sign (R1=R2=R3=R4) subtract two voltages (R1=R2=R3=R4)

-+ Vout

V1R1

R3

V2R2

...

3 31 2

1 2out

R RV V VR R

= − +

-+ Vout

V1R1

R2

V2R3

R4

( )( )

1 2 4 22 1

3 4 1 1out

R R R RV VR R R R

+ = − +

V

Multiply by a positive constant < 1 multiply by a positive constant > 1 -+ VoutVin

R1

R2

2

1 2out in

RV VR R

= ++

-+ VoutVin

R1 R2

1 2

1out in

R RV VR+

= +

Multiply by a negative constant exponentiation

-+ VoutVin

R1R2

2

1out in

RV VR

= −

-+ Vout

Vin

Rdiodes

( ) 40 inVout in sV sign V I R e≈ − ⋅ ⋅

integrate differentiate

-+ Vout

Vin

R C

22 MegΩ

1 ( 22 Megout inV V dt R

RC= − Ω∫ )

-+ Vout

RC

Vin 1 kΩ

50 pF

( 1 k , 50 pF)inout

dVV RC R Cdt

= − Ω

take logarithm Useful facts

-+ Vout

Vin

R diodes

( ) ( ) 15ln ln ( 10 )

40in

out in s s

sign VV V I R I −−

≈ ⋅ − ∼

• RC has units of time. • If you’d like to have a millisecond act

as your unit quantity of time (this is convenient for oscilloscope work), chose RC to be 10-3. For example, R = 100 kΩ and C =.01 µF should work fine

117


118


Reading and homework should be completed by 4pm Friday. Required reading: Read over (but don't worry about understanding) the Unit 9 material we'll tackle next week. Optional reading: Horowitz & Hill, chapter 4 through section 4.06. (The book is on reserve in the Physics Department library.) Problem 0 (5 points): Make your own hardcopies… Print (from the web) your own copy of the solutions to this week’s in-class exercises as well as last week’s problem set and write a (true!) statement on your problem set asserting that you have successfully printed your personal copies. Problem 1 (10 points): No Venusians were found Research electricians at Urkin’s have shown the following circuit to be unable to detect signs of extraterrestrial life, though it has proved useful for other purposes. Calculate VA and VB in terms of Vin.

VA

-

+

Vin

-

+

R

VB

R

Problem 2 (20 points): Doing A/B electronically

119


Design a circuit whose output is Vout = Vin1 / Vin2 where V and V . You should assume that (ideal) diodes behave as described previously, with I

10in >

20in >

S = 10-15 amperes. (Your circuit will probably require several opamps.)

120


Physics 212


Unit 9

Electronic models of mechanical systems; analog computers 2

Spring, 2003

George Gollin


121


Unit 9 is now included with the unit 8 writeup which covers both weeks.

122


Reading and homework should be completed by 4pm Friday. Required reading: Read over (but don't worry about understanding) the Unit 10 material we'll tackle next week. Optional reading: Horowitz & Hill, chapter 4 through section 4.06. (The book is on reserve in the Physics Department library.) Problem 0 (5 points): Make your own hardcopies… Print (from the web) your own copy of the solutions to this week’s in-class exercises as well as last week’s problem set and write a (true!) statement on your problem set asserting that you have successfully printed your personal copies. Problem 1 (20 points): We are so clever, and so frugal Design an opamp circuit which “solves” the equation

( ) ( ) ( ) ( )2

2 0.2d x t dx t

x t fdt dt

= − − + t

using only four opamps. (f(t) [or -f(t) if it should prove convenient] is provided by a function generator.) Be sure to include explanatory comments on your circuit diagram so we can tell what you had in mind. Problem 2 (10 points): Yep, it subtracts them! Prove that the following circuit

-+ Vout

V1R1

R2

V2R3

R4 produces an output voltage

( )( )

1 2 4 22 1

3 4 1 1out

R R R RV VR R R R

+V

= − +

.

123

Physics 212, Spring 2003. Unit 10 10.1 Magnetic fields and the vector potential

Physics 212


Unit 10

Magnetic fields and the vector potential

Spring, 2003


2003

124


Unit 10: Magnetic fields and the vector potential Recap of last week Introduction: A and B Today we’ll investigate the connection between the vector potential and the magnetic field. It should be clear that there’s some sort of link between the two: if no currents are flowing, both the vector potential and the magnetic field are zero. In addition, both grow linearly with the magnitude of the current. (In addition, recall the problem set for unit 6 in which you used the Lorentz transformations to calculate the electric and magnetic fields of a moving charge by transforming the electric field of a stationary charge.) Since the vector potential is defined in terms of an integral over the current density , there’s also some sort of direct connection between the

JB and J . You’ll discover that it is the curl of

the magnetic field which is proportional to the current density. Here’s what you’ll do today: • Calculate the curl of the vector potential associated with a current-carrying wire • Compare with A∇× B • Calculate ∇× for a point charge moving with constant velocity A• Discover that 0B Jµ∇× =

• Derive the Biot-Savart law • Derive Ampère’s law Discussion: A definition of the magnetic field, and a recap of your previous results From Physics 112 you know that the (Lorentz) force on a moving charge is . F qE qv B= + × Let’s use this as the definition of the magnetic field: it is what is responsible for the velocity-dependent portion of the force acting on a charged particle. You can immediately see that the observers in different frames will disagree about the values of

and E B since all the force in the rest frame of a particle must come from . Both and E E B contribute to the force in a frame in which the particle is moving. A few units back you worked up a calculation of the apparent charge density of a current-carrying wire as seen by a moving test charge. Your calculation concerned a test charge moving

125


to the right with speed v parallel to a wire in which electrons flow to the right (as seen from the wire’s rest frame) with speed u . Your calculations should have revealed that the apparent net charge density on the wire (according to the test charge) was ( ) ( )2quv acλ = C/m. q is the charge on a metal ion:

+1.6×10-19 C. Further calculation would/should have shown that the force on the test charge Q was

20 02 2

Q quv QFr ac r

λπ ε π

= =ε

.

The magnitude of the current flowing in the wire is the amount of charge per second that flows past one point in the wire. Since the spacing between electrons (in the wire’s rest frame) is a while the average electron velocity is u, there will be u/a electrons per second flowing past one position in the wire. As a result, the magnitude of the current is I qu a= . We can use this to rewrite the force as

20

1 12

IF Qv Qv Bc rε π

= ≡ × .

From our definition of the magnetic field, we conclude that its magnitude a distance r from a straight current-carrying wire must be

02

0

1 12 2

I IBc r

µrε π π

= =

since ( )2

0 01 cµ ε= . This agrees with the Physics 112 result. (The direction also agrees if you work it all out correctly.)

126


Exercise 10.1: Line current vector potential A few weeks ago we found that the vector potential for a non-relativistic charge is

( ) 0

4qA r v

r rµ

≈′π −

. Use superposition for several charges: ( ) 0

all charges i4i i

i

q vA rr r

µ≈

π −∑ .

For a distribution of charge swirling around so that the distribution’s velocity depends on position, we can superpose the contributions from cells of volume dV containing charge dq = ρdV this way:

( )rρ ( )v r

( ) ( ) ( ) ( )3 3 30 0

all space all space

( is the same as )4 4

r v r J rA r d r d r d r dV

r r r rρµ µ

π π′ ′ ′

≈ =′ ′ ′− −′ ′∫ ∫ .

The current density is the number of Coulombs per second per square meter which flow through a surface perpendicular to . (The above equation is exact only if does not change with time.)

JJ J

If the current flow is confined to a wire, it is convenient to “do the integral” along the two coordinates perpendicular to the direction along the wire. For example, if we have a current I flowing in a thin wire along the z axis…

x

z

y

the vector potential created by the current is ( ) 0 ˆ4

z

z

I zA r dzr r

µπ

=+∞

=−∞

= ′− ′∫ .

127


(a) In Cartesian coordinates, the vector potential created by this current is

( )( )

022 2

, ,4

IdzA x y zx y z z

µπ

+∞

−∞

′=

+ + − ′∫ .

Note that the integral diverges: for large z′ it behaves like ( )lndz z z′ ′ ′=∫ . We encounter the same difficulty when we calculate the electrostatic potential from a line charge:

( ) ( )( )

322 20 0all space

1 1, , 4 4

rV x y z d r dz

r r x y z z

ρ λπε πε

+∞

−∞

′= =′ ′

− ′ + + − ′∫ ∫ .

One way around this would be to calculate the electric field instead (which behaves like 1/r), then to do the integral to work with the potential difference between two points which are a finite distance from the line charge. Since we’re interested in showing that the curl of the vector potential is the same as the magnetic field, we can circumvent the difficulty here by taking the curl of A , and moving the curl inside the integral. This will yield a convergent integral:

( )( ) ( )

0 02 22 2 2 2

1 ˆ, ,4 4

IIdzA x y z z dzx y z z x y z z

µ µπ π

+∞ +∞

−∞ −∞

′ ∇ × = ∇ × = ∇ × ′

+ + − + + −′ ′ ∫ ∫

.

Note that the partial derivatives of the curl act only on the unprimed variables , ,x y z and not on the primed variables , ,x y z′ ′ ′ .

Calculate ( ) 22 2

1 zx y z z

∇ × + + − ′

.

128


(b) The vector potential is symmetric under rotations about the z axis, and is independent of the value of z. As a result, we can work with points on the x axis (so that y = z = 0) without loss of generality in order to simplify things. (We could do the calculation in cylindrical coordinates instead.)

Calculate ( )( )

022 2

1 ˆ, ,4

IA x y z z dzx y z z

µπ

+∞

−∞

∇ × = ∇ × ′ + + − ′

∫ for points in the x axis.

(You may find that ( )3 2 2 2 22 2

dz zx x zx z

′ ′=

′+′+∫ is useful.)

129


(c) Compare your answer in part (b) with the magnetic field found using the relativity-based calculation described earlier. (The results should agree!) Discussion: is the magnetic field always the curl of the vector potential? It looks like the vector potential as we’ve defined it is a good candidate for what we want! (In a more advanced course you’ll probably encounter a proof that the definition we’re using for the vector potential is correct. I’ll just assert that it’s true and forget about the formalism.) Can we always write the magnetic field as the curl of the vector potential? There seem to be no isolated sources or sinks of magnetic field lines anywhere in the universe. (These would be the magnetic analogs of positive and negative electric charge.) Think of the field lines created by a solenoid: they don’t begin (or end) anywhere. Consequently, the magnetic version of Poisson’s equation, 0magneticBi ρ ε∇ = , becomes 0Bi∇ = . Remember Helmholtz’s theorem? We didn’t get to it during unit 5, but you may have read about it if you looked through the notes. Here’s how we can use it: For any vector field B which dies away very far from the origin (not just the netic field), we begin by defining the scalar and vector fields D and

magC so that ( ) (, , , ,D x y z B x y z≡ ∇i )

) and

. According to Helmholtz’s theorem we can then ( ), ,C x y z B y z≡ ∇× ( , ,x reconstruct B from

D and this way: C

( ) ( ) ( )3 3

all space all space

1 14 4

D r C rB r d r

r r r rπ π

′ ′d r′ ′= −∇ + ∇× ′ ′− −

∫ ∫ .

If (so that D = 0), this reduces to ( ), , 0B x y z∇i =

( ) ( ) ( )3 3

all space all space

1 14 4

C r B rB r d r d

r r r rπ π

′ ′ ′∇ ×′ ′= ∇× = ∇× ≡ ∇× ′ ′− −

∫ ∫ r A

.

The prime on the curl inside the integral indicates that its derivatives are with respect to primed, not unprimed, coordinates.

130


We can always write a divergenceless function as the curl of some ector f eld. As a result, we can

v ialways find a vector potential which satisfies the equation A B A= ∇× :

( ) ( ) 3

all space

14

B rA r d

r rπ′ ′∇ ×

r′=′−∫ .

One last point: since the curl of a gradient is always zero (you showed that and

a few weeks ago), we are always free to add the gradient of any old scalar

f gnetic field which results since

( ) 0V∇ × ∇ =

( ) 0Ai∇ ∇ × =

unction to the

( )A Vvector potential without changing the ma

B( )A V A∇ × + ∇ = ∇ × + ∇ × ∇ = ∇ × = .

It turns out that this allows us to impose the condition on the vector potential that it has zero divergence: ∇ . (This isn’t a requirement for 0Ai = A , but it is sometimes convenient to calculate a vector potential which satisfies this condition to simplify other calculations.) Exercise 10.2: Magnetic field caused by a point charge moving with constant velocity A point charge moves with constant velocity along the z axis so that its position is ( ) (, , 0,0, )x y z vt′ ′ ′ = . An observer positioned at ( ) ( ), , ,0,0x y z x= measures the magnetic field associated with the particle. (Assume that the particle is moving slowly so that retardation effects can be ignored: during the time it would take light to travel from the charge to the observer the charge’s position doesn’t change significantly.) Make use of the vector potential to calculate the magnetic field as a function of time at the observer’s position.

131


discussion Note the 1/r2 dependence to the field as the charge passes through the origin. discussion: cylindrical coordinates We could have calculated B A= ∇ × using a non-Cartesian coordinate system. In fact, since the problems we’ve done so far have all been symmetric under rotations about the z axis, it would have made sense to use cylindrical coordinates, illustrated in the figure. Naturally, we’d find the same values for B , but they would be parameterized in terms of ( ), ,r zθ instead of ( ), ,x y z .

The unit vectors r ˆ ˆ, ,ˆ zθ point in the direction of increasing r, θ, and z.

x

z

yθ

r

In cylindrical coordinates the curl of the vector potential is

( )1 1ˆˆ ˆz r z rAAA A AA rr z z r r r

θθ θθ θ

∂ ∂∂ ∂ ∂ ∇× = − + − + − ∂ ∂ ∂ ∂ ∂ ∂ rA z∂ .

132


discussion: retarded time We’ve concocted a couple of ways of calculating the magnetic field. The first uses relativity to describe how the effects of Lorentz contraction and non-simultaneity make observers in different frames disagree about charge density. The second involves a global calculation if we know the “current density” everywhere. It is common to use ( ) ( )r v rρ ( ) ( ) (, ,J r t r t v r t ),ρ≡ to represent the current density so that

( ) ( ) ( ) 30 ,, ,

4 all space

J r tB r t A r t d r

r rµπ

′ ′= ∇ × = ∇ × ′ − ′ ∫ .

Note that the curl only acts on unprimed coordinates. One subtle point is that the time t inside the integral is the ′ retarded time: at each value of r′ that the integral touches, t represents the present time t set back by the amount of time a light beam would have required to arrive at the observer’s position (

′r , no prime). The connection

between the two times is t t r′ ′= − −r c . Sometimes the need to use the retarded time inside the integral can make it very difficult to evaluate! Since the curl acts on unprimed coordinates, any time dependence in the current density will cause it to “react” to the curl’s partial derivatives due to the presence of unprimed spatial coordinates in the retardation term. If the current density depends on position but J not on time (the currents never change), things simplify somewhat. Exercise 10.3: a microscopic version of some of this Starting with the global version of Gauss’ law

( )0 0

1enclosed

surface volume

QE dA r dVρε ε

= =∫ ∫i ,

we were able to construct a microscopic (local) version

0

E ρε

∇ =i

several weeks ago.

133


Let’s do the same thing for the magnetic field now: we’ll establish a connection between its curl and the current density, assuming the current density is constant in time. There’s a hard way to do this, which proceeds as follows: • ( ) ( )( )B r∇ × = ∇ × ∇ ×

( )A r . Show that, for example, the x component of ( )A∇ ×∇ ×

satisfies ( ) 2xA A x

xAi∇ ∂ − ∇ ∇ × ∇ × = ∂ .

• Since we are able to work with vector potentials for which 0Ai∇ = , ( ) 2

xxB r A∇ × = −∇ . There are similar equations for the y and z components which can be written this way:

( )( )( )

2

2 2

2

ˆ ˆ

ˆ ˆ

ˆ ˆ

xx

yy

zz

B x A x

B y A y B

B z A z

A

∇ × = −∇ ∇ × = −∇ ∇ × = −∇∇ × = −∇

⇒ .

Rewrite 2B A= −∇

( )J r∇ × so that the right side if the equation is expressed in terms of the

current density . (You could find what you need in a previous unit.)

(a) You should demonstrate this the easy way instead: compare the expression for the vector potential in the earlier discussion (which refers to the Helmholtz formula) with our original definition of the vector potential. From Helmholtz:

( ) ( ) 3

all space

14

B rA r d

r rπ′ ′∇ ×

r′=′−∫ .

From our original definition of the vector potential:

( ) ( ) ( ) ( )03 30

all space all space

1 4 4

r v r J rA r d r d r

r r r rρ µµ

π π′ ′ ′

= =′ ′− −′ ′∫ ∫ .

By inspection, what’s B∇× in terms of the current density J ? Exercise 10.4: Biot-Savart law Let’s consider a current density which is confined to a thin wire. Earlier, when you calculated the vector potential for a wire running along the z axis, you found that

134


( ) ( ) 30 0

all space

, , 4 4

J r IdzA x y z d rr r r r

µ µπ π

+∞

−∞

′ ′= =′

− −′ ′∫ ∫ .

Define to be a short distance along the wire in the direction that current is flowing so we can rewrite the integral more generally, for arbitrary wires:

dl

( ) 0, ,4along

wire

IdlA x y zr r

µπ

=− ′∫ .

This suggests that we write

( ) 0so that4contributions

from entirewire

IdlA r dA dAr r

µπ

= =− ′∫

and

0so that4contributions

from entirewire

IdlB A dA dBr r

µπ

= ∇ × = ∇ × = ∇ × − ′ ∫ .

(a) Evaluate 0

4IdldB

r rµπ

= ∇ × − ′

ˆ

for a small segment of current at the origin flowing in the z

direction: Idl I z dz= . Assume that the observer is located at r as shown in the diagram (in the y-z plane), and that . (Keep in mind that the curl acts only on 0r′ = unprimed coordinates.) After grinding through the differentiation, reexpress your answer in terms of θ and r .

(b) Rewrite your expression for in terms of dB ˆdl r× , where is a unit vector which points from the small current element

rIdl (at the origin here) towards the observer’s location.

Idl

x

z

y

rθ

observer

135


discussion • Written this way (using a cross product), your expression is independent of the choice of

coordinate system: very convenient!

• You should have obtained 02

ˆ4

I dl rdBr

µπ

×= . This is the Biot-Savart law, named for Jean-

Baptiste Biot (1774-1862) and Felix Savart (1791-1841), two French mathematicians. This is the magnetic equivalent of Coulomb’s law: the contribution to B from a small amount of current somewhere follows an inverse-square law.

• If we have to contend with current densities instead of currents in thin wires, we’ll need to

replace I dl with in our expression for JdV dB .

136


Exercise 10.5: Ampère’s law A short time ago you found that 0B Jµ∇× . Using Stokes’ theorem, you’ll now derive the macroscopic version of this, known as Ampère’s law (named for the French mathematician and physicist André Ampère [1775-1836]). Imagine we have a current density g t ugh a loop, as shown in the diagram. We can integrate both sides of the equation

=

flowin hroJ

0B Jµ∇× = over the surface enclosed by the loop:

x

z

y

J

0.

surface surfaceenclosed enclosed

J dAB dA µ=∇×∫ ∫ ii

Use Stokes’ theorem to rewrite the left side as a line integral of the magnetic field around the loop. Note that the details of how the surface extends across the loop do not matter: it can be as lumpy as we want without spoiling the connection between the line integral and the surface integral guaranteed by Stokes’ theorem. Now rewrite the right side as an expression involving the total current I which flows through the loop. Ampère’s law! discussion Use of Ampere’s law with cylindrical symmetry

137


Summary and wrap-up discussion • Calculating the curl of the vector potential for a current-carrying wire • is the same as A∇× B • for a point charge moving with constant velocity A∇×

• The Biot-Savart law: 02

ˆ4

I dl rdBr

µπ

×=

• Ampère’s law: 0 throughsurfaceperimeter

IB dl µ⋅ =∫


138


Reading and homework should be completed by 4pm Friday. Required reading: Purcell, chapter 2, sections 13 - 16; chapter 6, section 3. Optional reading: The Feynman Lectures on Physics (volume II), chapter 14. Problem 0 (5 points): Make your own hardcopies… Print (from the web) your own copy of the solutions to this week’s in-class exercises as well as last week’s problem set and write a (true!) statement on your problem set asserting that you have successfully printed your personal copies. Problem 1 (10 points): Les belles choses qu’on peut dessiner You sketched a vector potential, including retardation effects, as the last problem of unit 5’s homework assignment. Reproduce your sketch, and then draw on the sketch the vectors corresponding to the magnetic field associated with (the curl of ) this vector potential. (As in the unit 5 problem, only indicate the field on the x axis.) Problem 2: Mrs. Urkin’s industrial jell-o dispenser To speed its distribution of electrically charged jell-o, Urkin’s D&HE builds a pipeline from the Emporium to the central New Jersey rail head at which tank cars are to be filled with the product. You may approximate the pipe as a 2 kilometer long straight cylinder of radius 0.1 meter. The jell-o flows due north through the pipe at 10 m/s and carries uniform charge 1 µC per cubic meter. (a) Calculate the current density inside and outside the pipe. J (b) Calculate the curl of the magnetic field B∇× inside and outside the pipe. (Note that you do not need to know B to calculate B∇× here! See exercise 9.3 in the class materials if you’re stuck.) (c) Calculate the value of the vector potential A along the centerline (r = 0) of the pipe midway between its ends. (It’s easiest to do this problem in cylindrical coordinates. Keep in mind that the volume element in your integral will be dV = rdrdθdz, and not drdθdz.) (continued next page...)

139


(d) Recall Stokes’ theorem, true for any vector field B (not just a magnetic field):

( )

surface perimeter

of surface

B dA B dl∇× ⋅ = ⋅∫ ∫ .

Use this to determine the magnetic field B as a function of distance from the center of the pipe.

140

Physics 212, Spring 2003. Unit 11 11.1 Maxwell’s equations

Physics 212


Unit 11

Maxwell’s equations

Spring, 2003


2003

141


Unit 11: Maxwell’s equations Recap of last week Introduction: Establishing a connection between E and B We’ve worked up a number of connections between electric fields and charge distributions, magnetic fields and current distributions, potentials and fields, and so forth. In many ways our equations are still incomplete, however: some of them are only true for static charge distributions, or for currents which never change. In addition, we don’t have anything which provides a direct link between the electric and magnetic fields, even though we know that special relativity requires a connection between them. Here’s a table which summarizes what we know so far.

Scalar and vector potentials in terms of charge and current distributions

( ) ( ) 3

0all space

1 d4

rV r r

r rρ

πε′

= ′− ′∫

static charges, currents

( ) ( ) 30

all space

4

J rA r d r

r rµπ

′= ′

− ′∫ static charges, currents

Electric and magnetic fields in terms of charge and current distributions

( ) ( ) 32

0all space

1 d4

r r rE rr rr r

ρπε

′ − ′= r ′

− ′− ′∫ (Coulomb’s law) static charges, currents

( ) 30 2 11 22

2 1

ˆ4space

I J rB rr

µπ

→

→

×= ∫ d r (Biot-Savart law)

static charges, currents

Electric and magnetic fields in terms of potentials E V= −∇ static charges, currents

B A= ∇ × always true Divergences (flux integrals) of electric and magnetic fields

0

E ρε

∇ =i or0

enclosed

surface

QE dAε

=∫ i (Gauss’ law) always true

0Bi∇ = (No magnetic charge) always true Curls (line integrals) of electric and magnetic fields

0E∇ × = static charges, currents

0B Jµ∇ × = or 0 throughsurfacesurface

IB dAi µ=∇ ×∫ static charges, currents

Today we’ll work on “correcting” the curl equations to make them accurate even when charges move and currents are time-dependent.

142


Here’s what you’ll be working on: • using Stokes’ theorem and Ampère’s law to investigate the connection between changing

electric fields and the existence of non-zero magnetic fields • developing the connection between changing magnetic fields and the existence of non-zero

electric fields. • Maxwell’s equations! Exercise 11.1: Stokes’ theorem, applied to two different surfaces A current I flows through a wire, as shown in the following figure. The wire passes through two different surfaces: the first is a flat circle, while the second resembles an already-opened can of tuna.

I

“tuna can” isopen herefirst surface

surface A

surface Bthis loop enclosesthe (entire) surface

this loop enclosesthe (entire) surface

The two loops which bound the surfaces are identical in size and shape. The part of the second surface labeled “surface A” is a cylinder, while “surface B” is a flat circle. (a) Evaluate

surfaceA

B dA∇×∫ i for surface A only in terms of I. (Divide it into two pieces to obtain a

pair of surfaces with simply-connected boundaries if you like. Stokes’ theorem will be helpful to you here.)

143


(b) Prove .first second

surface surface

B dA B dA=∇× ∇×∫ ∫i i (Making use of Stokes’ theorem should help.)

(c) Evaluate

firstloop

B dl∫ i in terms of I. Do the same for secondloop

B dl∫ i .

discussion The integrals are independent of details of shapes of surfaces and bounding curves.

144


Exercise 11.2: Another version of the last two surfaces A current I flows through a wire, charging a parallel-plate capacitor, as shown in the following figure. The surfaces shown are similar to those in the previous exercise: the first surface is a circular disk, perpendicular to the wire, seen edge-on. The second looks like a soup can with the top removed: the open end is on the left. The closed end of the soup can passes between the capacitor plates. As before, line integrals around the boundaries of the two surfaces involve traveling along circles of identical radii, indicated in the figure. You should assume that the capacitor plates are very close together so that the magnetic fields on the circles labeled “boundaries of surfaces” are identical.

first surface

I

parallel-platecapacitor

I

second surface

boundaries of surfaces

(a) Evaluate

boundaryof firstsurface

B dl∫ i in terms of I. (Since the magnetic field is parallel to the boundary

used in your integral, you could have used 2boundaryof firstsurface

B dl r Bπ=∫ i if you had known the value of

the magnetic field.)

145


(b) It is certainly true that boundary boundaryof first of secondsurface surface

B dl B dl=∫ ∫i

firstsurface

i since the magnetic fields are the same along

the two boundaries. Must it still be true that ?secondsurface

B dA B dA= ∇×∫ ∫∇× i i (Remember Stokes’

theorem!) discussion But I through the surface is zero! What’s going on?

146


Exercise 11.3: The electric field gets into the act There must be a magnetic field between the capacitor plates while it is charging up. There’s no way to have a non-zero line integral along the boundary of that soup can without having a non-zero curl to the magnetic field: that would violate Stokes’ theorem, which is true for any vector field. If the magnetic field were zero over a finite (non-pointlike) volume of space, its curl would also be zero. We are forced to conclude that something going on between the capacitor plates is creating a magnetic field so that Stokes’ theorem is satisfied. This is an especially nice example of how mathematics can lead us to a better understanding of physical reality: it is impossible for Stokes’ theorem to be wrong. As a result, it is impossible for the magnetic field between the capacitor plates to be zero, even though there is no current flowing across the gap between the plates. We know there’s no current between the plates of an already-charged capacitor: with no charges in motion there can be no relativistic effects giving rise to a magnetic field due to different Lorentz contraction effects on (stationary) positive and (stationary) negative charges. It must be something associated with the change in the electric field between the plates that is giving rise to a magnetic field. (a) Let’s assume that our capacitor plates each have area A and are uncharged at time t = 0. Assume the current flowing onto the left-hand plate is I, and is constant. Calculate the electric field between the plates as a function of time, E(t). (If you don’t remember how to do this from Physics 112, use Gauss’ law, and keep in mind that the electric field is zero outside the plates.)

(b) Now calculate Et

∂∂

.

147


(c) We know that:

02boundary secondof second surfacesurface

B dl B dA r B Iπ µ= ∇× = =∫ ∫i i

and also that

0 0

so thatsecondsurface

E I E dAt A t

Iε ε

∂ ∂= =

∂ ∂∫ i

and also that 12 7 7 2

0 0 08.854 10 , 4 10 1.257 10 , 1 cε µ π ε− − −= × = × ≈ × =0µ .

Fill in the missing terms in curly brackets:

second secondsurface surface

EB dA expression in terms of dAt

∂∇× = ∂

∫ ∫i i

so that

EB expression in terms oft

∂∇× = ∂

.

148


Discussion In the absence of currents, you’ve just proposed a connection between the curl of the magnetic field and the time derivative of the electric field:

0 0 2

1E EBt c t

µ ε ∂ ∂∇× = =

∂ ∂.

In integral form this can be written

2 2

1 1

surface boundary surface surfaceof surface

EB dA B dl dA E dAc t c t

∂ ∂∇× = = =

∂ ∂ ∫ ∫ ∫ ∫i i i i

so that

2

1 E

boundaryof surface

B dlc t

∂Φ=

∂∫ i

where ΦE is the flux of electric field through the surface. Imagine that we had the following situation, in which a current passes through part of our surface, while a changing electric field crosses another part of the surface, as shown in the figure.

parallel-plate

I2

I1

surfacecapacitor

In this case we’ll expect the curl to contain contributions from both the current and the changing electric field:

0 2

1 EB Jc t

µ ∂∇× = +

∂.

149


Exercise 11.4: Now to investigate the curl of E ... This one’s easier to get. We’ll start with a Physics 112 approach, then turn the integral expression into a differential one. (a) A rectangular conducting loop slowly glides with speed v into a region in which there is magnetic field B , pointing in the positive z direction as shown in the figure. The mobile charge carriers inside the conductor each have positive charge q.

magnetic fie ld B

no magnetic field here: 0B =

ab v

loopx

y

z

Calculate the force (magnitude and direction) on a charge q (due to its motion through the magnetic field) in the side of the loop which has just entered the magnetic field..

150


(b) Somewhat later, half the moving rectangular loop has entered the magnetic field as shown.

v

x

y

z

Calculate the direction of the magnetic force on charges q in the sides of the loop which are parallel to the loop’s direction of motion. (Assume that the charges are almost at rest with respect to the wire of the loop.) (c) In the rest frame of the loop, each mobile charge will ascribe the force it experiences to the presence of an electric field. Calculate

conductingloop

E dl∫ i in the loop’s rest frame when it has

partially entered the magnetic field. Note that the line integral involves traveling around the loop in the positive sense, as given by the right hand rule. (The path around the loop is counterclockwise, when seen from above.)

151


(d) Notice that the effects of the induced electric field disappear as soon as the entire loop is inside the field: when 0Bd dtΦ = , 0E dl⋅ =∫ . (The magnetic flux is B

loop

B dAΦ = ⋅∫ : positive

flux is in the positive z direction.)

loop

dA E

= − ∫

p loop

B dA Et

∂= ⋅ = −

∂∂Φ

E∇×

Prove that B

loop

E dlt

∂Φ= −

∂ ∫ i for this loop. (This is Faraday’s law, named for English chemist

Michael Faraday [1791 – 1867]. It is true in general, not just for our rectangular loop moving with constant velocity.)

(e) You know that B

loop

B dlt t

∂Φ ∂= ⋅

∂ ∂ ∫ i . This also works in the rest frame of the

loop, where the magnetic field sweeps across the (stationary) loop. In fact, it also holds if the magnetic field is spatially constant across the loop, but varies with time. As a result, we can pull

the time derivative inside the integral to write B

loop loo

B dA dlt t

∂= ⋅

∂ ∂ ∫ ∫ ∫ i .

Use Stokes’ theorem to rewrite the rightmost integral as a surface integral of the curl of the electric field:

loop

E dl something involving− =∫ i

152


(f) What should go between the curly brackets?

B something involving Et

∂= ∇

∂×

153


discussion We can correct the last several rows in the table at the start of this unit. The following two tables hold equivalent statements: the first are differential, the second integral equations which describe exactly the same physics. These are known as Maxwell’s equations, named for James Clerk Maxwell (1831 – 1879), Scottish physicist and mathematician. They appeared in his 1873 work Treatise on Electricity and Magnetism. They represent one of the most profound intellectual achievements that the human species has yet produced. Fully relativistic, these equations (and extensions of them) have served as the basis for much of modern physics.

Divergences of electric and magnetic fields

0

E ρε

∇ =i always true (Gauss’ law)

0Bi∇ = always true (No magnetic charge)

Curls of electric and magnetic fields BEt

∂∇ × = −

∂

always true (Faraday’s law)

0 2

1 EB Jc t

µ ∂∇ × = +

∂

always true (Ampère’s law)

Flux integrals of electric and magnetic fields

0

enclosed

surface

QE dAε

=∫ i always true (Gauss’ law)

0surface

B dA⋅ =∫ always true (No magnetic charge)

Line integrals of electric and magnetic fields B

perimeter

E dlt

∂Φ⋅ = −

∂∫ always true (Faraday’s law)

0 2

1 Ethroughsurfaceperimeter

IB dl c tµ ∂Φ

+⋅ = ∂∫ always true (Ampère’s law)

The differential versions of Maxwell’s equations relate the behavior of the electric and magnetic fields (and their derivatives) at all points in space. It does not matter what is done to produce the fields: once they are known to exist they MUST obey the Maxwell equations. Often one works with electric and magnetic fields external to the region where there are currents or free charges. (This way ρ = 0 and 0J = .) When this is the case, Maxwell’s equations simplify to

154


Maxwell’s equations in the absence of charges and currents

0Ei∇ = 0Bi∇ =

BEt

∂∇ × = −

∂ 2

1 EBc t

∂∇ × =

∂

Note the symmetry between the electric and magnetic fields in Maxwell’s equations. The presence of 1/c2 in only one of the curl equations is a consequence of our ill-advised choice of units for the magnetic field. If we had used, say, the “Nillog” (where 1 Nillog = c × 1 Tesla), the revised curl equations would each have a 1/c coefficient in front of the time derivative. The negative sign in the ∇× equation is important: it’s why Lenz’s law works. Without it, a change in magnetic flux through a wire loop would induce a current in the wire which would

E

increase the change in magnetic flux, inducing an even stronger current. The current would rise exponentially, vaporizing the wire. The positive sign in the B∇× equation is needed so that the changing electric field between capacitor plates will create a magnetic field whose curl matches that of the field made just upstream/downstream of the capacitor (where there is a current flowing in the circuit). As long as we include time retardation, we can still calculate the vector potential and use it to extract the magnetic field:

( ) 30

all space

,,

4

r rJ r t

cA r t d r

r rµπ

− ′−′

= ′− ′∫ and B A= ∇ × .

When there are free charges in motion we can calculate a scalar potential in a similar fashion:

( ) 3

0all space

,1,

4

r rr t

cV r t d r

r r

ρ

πε

− ′−′

= ′− ′∫ .

Note that the vector potential also lets us calculate the electric field when ρ = 0 (so that

) in our part of the universe: 0E∇⋅ =

155


BEt

∂∇ × = −

∂ so AE

t t∂∇ × ∂

∇ × = − = −∇ ×∂

A∂

, yielding AEt

∂= −

∂.

The more general case (ρ ≠ 0) requires the inclusion of the V−∇ term. To summarize:

Scalar and vector potentials in terms of charge and current distributions

( ) 3

0all space

,1,

4

r rr t

cV r t d r

r r

ρ

πε

− ′−′

= ′− ′∫ ( ) 30

all space

,,

4

r rJ r t

cA r t d r

r rµπ

− ′−′

= ′− ′∫

Electric and magnetic fields in terms of potentials AE Vt

∂= −∇ −

∂ B A= ∇ ×

Divergences of electric and magnetic fiel s d

0

E ρε

∇ =i 0Bi∇ =

Curls of electric and magnetic fields BEt

∂∇ × = −

∂ 0 2

1 EB Jc t

µ ∂∇ × = +

∂

Summary and wrap-up discussion Here’s what you were working on: • using Stokes’ theorem and Ampère’s law to investigate the connection between changing

electric fields and the existence of non-zero magnetic fields • developing the connection between changing magnetic fields and the existence of non-zero

electric fields. • Maxwell’s equations! Comments about homework assigned for next week

156


Reading and homework should be completed by 4pm Friday. Required reading: nothing this week. Optional reading: Purcell, chapter 9 (you’ll have to borrow it from the library). The Feynman Lectures on Physics (volume II), chapter 4, section 1. The Feynman Lectures on Physics (volume II), chapter 18. Recall that the Maxwell equations are

Sources (charges and currents) are present

0

E ρε

∇ =i BEt

∂∇ × = −

∂

0Bi∇ = 0 2

1 EB Jc t

µ ∂∇ × = +

∂

Sources are absent (ρ = 0, J = 0) 0Ei∇ = BE

t∂

∇ × = −∂

0Bi∇ = 2

1 EBc t

∂∇ × =

∂

They are always true for electric and magnetic fields. It is impossible to find a time-varying electric field which is not accompanied by a magnetic field with non-zero curl. Problem 0 (5 points): Make your own hardcopies… Print (from the web) your own copy of the solutions to this week’s in-class exercises as well as last week’s problem set and write a (true!) statement on your problem set asserting that you have successfully printed your personal copies. Problem 1 (15 points): How can we make these fields satisfy Maxwell’s equations? The electric and magnetic fields that exist in some universe which is empty of charges and currents are known to be given by

( ) (0 ˆ, , ,E x y z t E f kz t xω= − ) and ( ) ( )0 ˆ, , ,B x y z t B f kz t yω= − where E0, B0, k, and ω are constants, and f is a function. Since these fields do exist, they must satisfy all four of the Maxwell equations.

157


(a) Prove that these fields satisfy the divergence equations 0Ei∇ = and 0Bi∇ = .

(b) Show that it is also possible for the fields to satisfy the curl equations BEt

∂∇ × = −

∂ and

21 EBc t

∂∇ × =

∂ if the ratios 0E B0 and kω are chosen properly, and determine the required

values of 0E B0 and kω . (c) Imagine that the function f has a peak when its argument is zero (so that the peak is at z = 0 at time t = 0, but is located elsewhere when t 0≠ ). Calculate the speed with which this peak travels, making use of your answers to part b. In an entirely different universe, Mrs. Urkin realizes that she had only partially folded the egg whites into the chocolate mousse that she has just placed before her dinner guests. Fortunately, the circular traces of egg in the dark chocolate background remind everyone of a simple electric field with non-zero curl, namely

[ ]0 ˆ ˆ( , , )E x y z E yx xy= − . Her guests notice that the local magnetic field is zero at the instant that dessert is placed on the table and that, to their amazement, they are actually immersed in the time-independent electric field [ ]0 ˆ ˆ( , , )E x y z E yx xy= − . (There are no charges or currents in the vicinity of the dining room table.) (d) Sketch the electric field and calculate its curl E∇ × . (e) Solve for the magnetic field ( , , , )B x y z t in Mrs. Urkin’s dining room. Keep in mind that the Maxwell equations are always satisfied by real electric and magnetic fields. Problem 2 (15 points): Deriving the wave equation The Maxwell equations couple the electric and magnetic fields: a changing electric field creates a magnetic field with non-zero curl; a changing magnetic field creates an electric field with non-zero curl. In this problem you’ll manipulate the Maxwell equations to produce a pair of decoupled wave equations for the electric and magnetic fields. Imagine that an electric field is known to exist in space.

ˆ ˆ ˆ( , , , ) ( , , , ) ( , , , ) ( , , , )x y zE x y z t E x y z t x E x y z t y E x y z t z= + +

158


(a) With the help of the Maxwell equations, prove that ( ) ( ) 2

2 2

1B EEt c t

∂ ∇ × ∂∇ × ∇ × = − = −

∂ ∂.

(b) Show that the z component of ( )E∇ × satisfies the following: ∇ ×

( ) ( ) ( )2z

z

EE E

z

∂ ∇⋅ ∇ × ∇ × = −∇ + ∂

.

(The equations for the x and y components of ( )E∇ × ∇ × are similar to this, allowing us to

combine the three equations to write ( ) ( ) ( )2E E E∇ × ∇ + ∇ ∇⋅× = −∇ .)

(c) If the charge density ρ is zero we’ll have 0 0E ρ ε∇⋅ = = so that

( ) (2

22 2

1 zz

z

E Ec t

)E∂ − = ∇ × ∇ × = −∇ ∂, yielding the wave equation ( )

22

2 2

1 0zz

EEc t

∂∇ − =

∂

for Ez. (If you like, you can combine this with the wave equations for the x and y components

into the single equation ( )2

22 2

1 0EEc t

∂∇ − =

∂.)

Derive the relationship between k and ω necessary to make ( )0 ˆ( , , , ) cosE x y z t E kz t xω= − a solution to the wave equations. (It should look familiar to you if you’ve already done the first problem!) Note that you can derive the wave equation for the magnetic field in a similar fashion:

• ( ) ( ) 2

2 2

1 1E2

BBc t c t

∂ ∇ × ∂∇ × ∇ × = + = −

∂ ∂ ⇒

• ( ) ( ) ( ) ( )2 2B B B∇ × ∇ × = −∇ + ∇ ∇⋅ = −∇ ⇒ B

• ( )2

22 2

1 0BBc t

∂∇ − =

∂.

(d) Recall where that 1/c2 factor in front of the time derivative came from: 1/c2 is really just the product of µ0 and ε0. The constant ε0 comes from Coulomb’s law: it sets the scale for the strength of the Coulomb force between a pair charges. The constant µ0 comes from the connection between the amount of current flowing in a wire and the magnitude of the resulting magnetic field a fixed distance from the wire.

159


Imagine a different universe in which the speed of light is ten times greater, but in which the Coulomb force is the same as it is for us. How would magnetic fields created by current-carrying wires compare in the two universes?

160

Physics 212, Spring 2003. Unit 12 12.1 Visualizing fields and potentials with Mathematica

Physics 212


Unit 12

Visualizing fields and potentials with Mathematica

Spring, 2003

George Gollin


161


Unit 12: Visualizing fields and potentials with Mathematica Recap of last week Introduction: Visualization and Mathematica We’ve been working with fairly abstract quantities—electric and magnetic fields and their derivatives, as well as scalar and vector potentials (and their derivatives). Let’s look into using the software package Mathematica to generate plots of potentials and fields in order to better visualize them. This week will serve as an introduction to Mathematica. Next week we’ll use the program as a tool to investigate the nature of the fields associated with a moving charge. We’ll be working in Loomis 257 where a large number of personal computers are available to run Mathematica. It’s likely that the PC’s are running Windows NT, and are loaded with Mathematica version 4. If you’ve never used Mathematica, you’re likely to find that its interface is fairly intuitive, and that the software behaves quite sensibly once you’re used to it. It’s a good tool, and you may find it useful in a number of other courses. It is able to do symbolic manipulation of equations and all sorts of other good stuff. Most of your interactions with Mathematica are through a window displaying a notebook, which will contain your instructions to Mathematica and the results of its efforts. I’ve been noticing a certain tendency for Mathematica to spit out a dozen error message when it corrupts some of its internal settings files. According to Wolfram Research, sometimes this can be fixed by restarting the program while holding down the ctrl and shift keys simultaneously. (I find that this doesn’t work.) In addition, the program has a tendancy to insert non-displaying formatting characters into notebook files in a manner which corrupts the files occasionally. Try this: copy the suspect text from your notebook into a text-only editor (MicroSoft Notepad, for example), remove things which don’t belong, then paste it back into your notebook. Alternatively, retype the misbehaving lines, then delete the original text. This is an annoying problem, and should not be present in software judged ready for commercial release. Sometimes stopping the kernel (do this from the kernel menu), then restarting it will fix problems. Screen shots are taken from my PC’s, so what you see will be slightly different.

162


Discussion: Introduction to Mathematica Fire up Mathematica on your PC. The path probably is something similar to the following: Start menu → Programs → Mathematica → Mathematica 4 You’ll see something like this:

The large window labeled “Untitled-1” is a Mathematica notebook in which you can enter calculations and instructions to the Mathematica program. The palette to the right is useful for entering various quantities (Greek letters, integrals, and so forth) into your notebook. Mathematica’s online help facility is pretty good—you can get to it using the “Help browser...” item in the “Help” menu, as shown below. The Mathematica program is split into two parts: a “front end,” which manages the user interface, and a “kernel,” which actually performs various computations for you. You do not interact directly with the kernel.

163


Many of your interactions with Mathematica require you to type instructions into the notebook, then request Mathematica to evaluate them. To cause the program to evaluate your inputs, often you’ll type keypad-enter or shift-enter. Have Mathematica calculate 2 + 2 for you...

...then hit keypad-enter (or shift-enter) to tell it to evaluate what it sees. Notice that Mathematica has added the In[x]: and Out[x]: labels in the example below. In addition, the strange brackets at the right mark “cells,” subunits in the Mathematica notebook.

164


You can define constants like this:

They can then be used in other expressions:

The palette is useful for building a variety of functions. For example, the tile labeled “π” represents 3.141592653... Mathematica contains loads of functions, all of whose arguments must be placed between square brackets ( [ and ] ). Parentheses will not work! For example (the “π” came from clicking the appropriate tile in the palette):

165


You can define your own functions this way:

Note the use of the underscore at the end of the argument name ( x_ ), the use of the palette to enter x3 (use the tab key to move from one input box to the next), and the := assignment symbol in the definition. Here are some other useful Mathematica symbols. Since Mathematica also does symbolic manipulation of equations, it sometimes does surprising things: take note of the difference between 8/5 and 8./5. for example.

A few obscure things: • % means “the result of the previous calculation” • == means “logical test” • >= prints as ≥ I always seem to get burned by this: ALL (nearly all??) Mathematica built-in functions begin with an uppercase letter: Sin[π] works, but sin[π] doesn’t. The program is case-sensitive. It is easy to plot stuff in Mathematica but keep in mind that sometimes you’ll need to tell it to have the kernel load some “add-on” definitions.

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No add-on stuff was needed here. Pay attention to the format: x2 Exp[-x] is the function to be plotted, while x, 0, 10 is an argument list (lists are always enclosed in curly brackets). The arguments, as you can surely tell, inform Mathematica of the variable to plotted along the abscissa, as well as the minimum and maximum values to be plotted.

Mathematica is good at symbolic manipulations. For example, the palette symbol represents a partial derivative. The two boxes are for you to fill in: the hollow one is requested first, the solid one is requested second. Use the tab key to move from one box to the next. ( ( )x f x∂ means ( )f x∂ x∂ ; it’s a common, and convenient, notation used frequently in physics.) Here’s how this looks in your notebook (but note one way in which you must actually make use of a function defined this way):

This is very convenient! (Note the upper case first letters in Tan[x] and Sec[x].) Because a reference to f[x] asks Mathematica to take a derivative of Sin[x], it is necessary to let the program do the differentiation before replacing x with a numerical value: asking it to evaluate

f[Pi] is the same as asking it to calculate sin( )dd

ππ

instead of sin( )

x

d xdx π=

. (Thus the

f[Pi]/.x->Pi format. There are other ways to do this also...)

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Mathematica also does integrals, using the indefinite integral palette tile . Note the two ways I’ve used to enter x2: the first is with a “^” to indicate an exponent, while the second is done using the palette.

Not surprisingly, Mathematica notebooks can be stored so that you can resume/correct/extend your work in a convenient fashion. To find more information about Mathematica functions that you might find useful, use the Help menus:

There are lots of cool things available. A far-from-complete list: Plot[] Plot3D[] PlotVectorField[] PlotVectorField3D[] ContourPlot[] ContourPlot3D[] PlotGradientField[] PlotGradientField3D[] SphericalPlot3D[] Here are some common mistakes/problems: • Leaving out the underscore at the end of an argument in a function definition:

Wombat[x]:=x^1.5; instead of Wombat[x_]:=x^1.5; • Forgetting that built-in functions begin with upper case letters: sin[x] instead of

Sin[x] • Forgetting that arguments are surrounded by square brackets instead of parentheses:

Sin(x) instead of Sin[x]

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• Forgetting to load a required software package, such as Graphics`PlotField` or leaving off the back-apostrophe(s) in the package name.

• Having Mathematica insert non-printing characters in places where it shouldn't. Retype, then delete the offendin text.

We’ll work through some of the notebooks I’ve already set up so you can see some examples of Mathematica at work.

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Exercise 12.1: Plotting potentials with Mathematica Let’s start with potentials for linear charges which are oriented parallel to the z axis. (This way we’ll be dealing with a system which can be drawn nicely in two dimensions since everything we’ll calculate is independent of z.) To avoid problems with singularities, we’ll make each of our charges a hollow cylinder. Recall that the potential varies logarithmically with distance from an infinite line charge or cylinder. Make a local copy of the file p212_unit12_mathematica_1.nb for yourself, then open it in Mathematica. Have Mathematica evaluate the entire notebook, then read through it to make sense of what’s there. If there are headphones available, plug them into the PC to hear the sounds generated by the cells at the end of the notebook. When you think you understand what’s going on, change the charge and/or position of one of the cylinders and reevaluate the cell which generates a contour plot. Here is a partial list of what’s in (an earlier version??) of the notebook. (* Comments are placed between (* and *). They can span multiple lines and be nested. Notice the long brace near the right edge of this window. A brace indicates the extent of a "cell," a subunit of a notebook. *) (*Generally good programming practice is to use lots of comments and to define functions only once, then refer to them in your work *) (* To cause part, or all, of the notebook to be evaluated go to the Kernel -> Evaluation -> Evaluate Cells (or Evaluate notebook) menu path. Another way is to select a cell (or cells), then hit keypad - enter. *) (* RealTime3D` will let us use the mouse to rotate some of our plots. An important thing to note is the back - apostrophe as the last character in"RealTime3D`" I'll also load other packages here. *) Needs["RealTime3D`"] Needs["Graphics`PlotField`"] Needs["Calculus`VectorAnalysis`"] Off[General::spell1] (* Define some constants here. The semicolons suppress printing of the values. Note that quantities separated by blank spaces are multiplied together : 4 and Pi for example (see below). *)

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EpsilonNought = 8.854 10^-12; OneOver4PiEpsilonNought = 1. / (4 Pi EpsilonNought); (* Define the square of distance between x1, y1, z1 and x2, y2, z2. Note the use of := and also the syntax concerning underscores at the ends of argument names. *) DistanceSquared[x1_, y1_, z1_, x2_, y2_, z2_] := (x1 - x2)^2 + (y1 - y2)^2 + (z1 - z2)^2; (* Now define the distance between x1, y1, z1 and x2, y2, z2 *) Distance[x1_, y1_, z1_, x2_, y2_, z2_] := Sqrt[DistanceSquared[x1, y1, z1, x2, y2, z2]]; (* Now define the Green's function for the potential at (x, y, z) associated with a hollow line charge of radius rmin at (xprime, yprime, zprime). Recall that the Green's function is the same as the potential caused at position r by a unit whatever - it - is placed at rprime. *) GreensFunctionLinePotential[x_, y_, z_, xprime_, yprime_, zprime_, rmin_] := 2 * OneOver4PiEpsilonNought * Log[Max[rmin, Distance[x, y, z, xprime, yprime, zprime]]]; (* specify the coordinates, radius, and charge per meter for the first cylinder *) x1 = -1.; y1 = 0.; rmin1 = 0.05; lambda1 = -1./ OneOver4PiEpsilonNought; (* Now do the same for the second cylinder *) x2 = 1.; y2 = 0.; rmin2 = 0.05; lambda2 = +1./ OneOver4PiEpsilonNought; (* Here's the contribution to the potential at x, y due to the first cylinder ... *) v1[x_, y_] := lambda1 * GreensFunctionLinePotential[x, y, 0., x1, y1, 0., rmin1] ; (* ... and due to the second cylinder *) v2[x_, y_] := lambda2 * GreensFunctionLinePotential[x, y, 0., x2, y2, 0., rmin2] ; (* The potential due to both of them is just the sum (superpositon) of the individual contributions *) v[x_, y_] := v1[x, y] + v2[x, y];

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(* Let's put the requests for plots into separate cells. Since you can ask the kernel to reevaluate a single cell instead of the entire notebook this'll save some execution time when we want to redo only one of the plots. *) (* Make a contour plot. None of the options (PlotPoints, and so on) are required. *) ContourPlot[v[x, y], x, -5, 5, y, -5, 5, PlotPoints -> 50, ColorFunction -> Hue, Contours -> 20, PlotLabel -> "Contour Plot"];

-4

-2

0

2

4

Contour Plot

-4 -2 0 2 4 (* Now make a 3 - d plot. You can click on the plot, then resize it, and rotate it with the mouse to inspect it from different angles. Note that loading the RealTime3D package will suppress the axis labels. *) Plot3D[v[x, y], x, -2, 2, y, -2, 2, ColorFunction -> Hue, PlotPoints -> 50, PlotLabel -> "3D plot", AxesLabel -> "x", "y", "potential"];

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(* Now make a field vectors plot. Note that we need to have already loaded the package Graphics`PlotField`. Also, notice that we're not getting field lines--only a bunch of vectors whose lengths and directions tell us about the electric field strength and direction. They are placed on a grid rather than being drawn so that the density of them would indicate the field strength. *) PlotGradientField[v[x, y], x, -2., 2., y, -2., 2., Frame -> True, PlotLabel -> "Electric field vectors", AxesLabel -> "x", "y"];

-2 -1 0 1 2

-2

-1

0

1

2

Electric field vectors

(* Now redraw that field vectors plot, but also draw in the positions of the charged cylinders. The use of % indicates "the previous thing." *) Show[%, Graphics[Disk[x1, y1, rmin1]], Graphics[Disk[x2, y2, rmin2]]];

-2 -1 0 1 2

-2

-1

0

1

2

Electric field vectors

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(*** * Another way to do this would have been as follows-- MyPlot = PlotGradientField[v[x, y], x, -2., 2., y, -2., 2., Frame -> True]; Show[MyPlot, Graphics[Disk[x1, y1, rmin1]], Graphics[Disk[x2, y2, rmin2]]]; so that we wouldn't need to put this request to redisplay the plot immediately after the previous plot was generated. ****) (* Let's read a sound from a file. *) Needs["Miscellaneous`Audio`"] (* Construct the full file pathname for where the current notebook file lives. There has to be an easier way to do this but I haven't yet discovered it! *) thingone = Options[$FrontEnd, NotebookDirectory]; thingone = ToString[thingone[[1]][[2]][[1]]]; thingone = StringDrop[thingone, 3]; thingone = StringDrop[thingone, -1]; thingone = thingone <> ", "; NotebookDirectoryPath = StringReplace[thingone, ", " -> "\\"]; (* Now concatenate the directory path onto the sound file name. *) SoundFile = NotebookDirectoryPath <> "sound1.wav" noise = ReadSoundFile[ SoundFile, PrintHeader -> True]; ListPlay[noise, PlayRange -> -2^15, 2^15, SampleRate -> 11025]; (* Now unload the audio package since it may conflict with the music package if that is requested later. *) Remove[Audio] Out[37]= "H:\\Physics_education\\PHYCS_212\\p212_unit12\\sound1.wav" From In[37]:= "Format: "\[InvisibleSpace]"Microsoft PCM WAVE RIFF" From In[37]:= "Duration: "\[InvisibleSpace]2.61279\[InvisibleSpace]" seconds" From In[37]:= "Channels: "\[InvisibleSpace]1 From In[37]:= "Sampling rate: "\[InvisibleSpace]11025 From In[37]:= "Bits per sample: "\[InvisibleSpace]8 From In[37]:= "Data size: "\[InvisibleSpace]28806\[InvisibleSpace]" bytes" From In[37]:= "Number of samples: "\[InvisibleSpace]28806

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Needs["Miscellaneous`Music`"] (* Names and frequencies (in Hz) for notes are available : for example, A4 is 440, C4 is 523.251, Bflat4 is 466.164. Lowest note is A0 (27.5 Hz), highest is Gsharp7 (6644.88 Hz). *) (* Define a wacko time - varying frequency (frequency in radians/sec) *) WeirdFrequency1[t_] := 2 Pi A3 Sin[5. t]; WeirdFrequency2[t_] := 2 Pi Aflat3 Sin[5. t]; (* Define left and right channel signal strength vs time *) LeftChannel[t_] := Sin[2 t]*Sin[WeirdFrequency1[t]]; RightChannel[t_] := Cos[2 t]*Sin[WeirdFrequency2[t]]; (* Now play it! Also, assign it to a sound object so we can replay it quickly.*) WeirdSound = Play[LeftChannel[t], RightChannel[t], t, 0, 5, PlayRange -> 0, 4.]; (* Now let's unload the package for music routines *) Remove[Music]

discussion: grids of field vectors vs. display of field lines

175


Local vs. global concerns in graphical representations... If it’s not 1/r2, field lines don’t make sense (they fade out, disappear).

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Exercise 12.2: Potentials in 3D It’s harder to represent fields and potentials in three dimensions: there are problems with surfaces obscuring other surfaces, making it hard for you to see what you’re working with, and calculations running slowly. Make a local copy of the file p212_unit12_mathematica_2.nb for yourself, then open it in Mathematica. Take a look at what it does: there are a few 3D plot-generating cells, as well as a couple of animations. Play with code in the notebook to familiarize yourself with how it works. Here’s a listing of (a preliminary version of) the notebook: (* Load required packages here. *) Needs["RealTime3D`"] Needs["Graphics`ContourPlot3D`"] Needs["Graphics`PlotField3D`"] Needs["Calculus`VectorAnalysis`"] (* Define some constants. *) EpsilonNought = 8.854 10^-12; OneOver4PiEpsilonNought = 1. / (4 Pi EpsilonNought); (* Define the square of distance between x1, y1, z1 and x2, y2, z2. Note the use of := and also the syntax concerning underscores at the ends of argument names *) DistanceSquared[x1_, y1_, z1_, x2_, y2_, z2_] := (x1 - x2)^2 + (y1 - y2)^2 + (z1 - z2)^2; (* Now define the distance between x1, y1, z1 and x2, y2, z2 *) Distance[x1_, y1_, z1_, x2_, y2_, z2_] := Sqrt[DistanceSquared[x1, y1, z1, x2, y2, z2]]; (* Now define the Green's function for the potential at (x, y, z) associated with a hollow spherical charge of radius rmin at (xprime, yprime, zprime). Recall that the Green's function is the same as the potential caused at position r by a unit whatever - it - is placed at rprime. *) GreensFunctionPointPotential[x_, y_, z_, xprime_, yprime_, zprime_, rmin_] := OneOver4PiEpsilonNought / Max[rmin, Distance[x, y, z, xprime, yprime, zprime]];

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(* specify the coordinates, radius, and charge for the first charge *) x1 = -2.; y1 = 0.; z1 = 0.; rmin1 = 0.05; charge1 = -1. / OneOver4PiEpsilonNought; (* Now do the same for the second charge *) x2 = 2.; y2 = 0.; z2 = 0.; rmin2 = 0.05; charge2 = +1. / OneOver4PiEpsilonNought; (* Here's the contribution to the potential at x, y due to the first charge ... *) v1[x_, y_, z_] := charge1 * GreensFunctionPointPotential[x, y, z, x1, y1, z1, rmin1] ; (* ... and due to the second charge *) v2[x_, y_, z_] := charge2 * GreensFunctionPointPotential[x, y, z, x2, y2, z2, rmin2] ; (* The potential due to both of them is just the sum (superpositon) of the individual contributions *) v[x_, y_, z_] := v1[x, y, z] + v2[x, y, z]; (* Let's put the requests for plots into separate cells. Since you can ask the kernel to reevaluate a single cell instead of the entire notebook this'll save some execution time when we want to redo only one of the plots. *) (* Make a contour plot. None of the options (PlotPoints, and so on) are required. After the plot is finished, click on it, resize it, and rotate it around using the mouse. The list of values for the "Contours" option says for which values of the potential we'll draw our contours. *) ContourPlot3D[v[x, y, z], x, -5, 5, y, -5, 5, z, -5, 5, PlotPoints -> 5, 5, Contours -> -0.1, -0.5, 0., 0.5, 0.1, Axes -> True ];

178


(* Now make a field vectors plot. Note the z range I have chosen. You can resize the plot, also rotate it around.*) PlotGradientField3D[v[x, y, z], x, -5., 5., y, -5., 5., z, 4., 10., VectorHeads -> True];

(* Now do an animation. *) Needs["Graphics`Animation`"]; Plot3D[Sin[x*y], x, 0, 3, y, 0, 3, Axes -> None, Mesh -> False, ColorFunction -> GrayLevel];

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SpinShow[%, RotateLights -> True]; "Double-click on a plot to see something spin!"

(Lots of these get drawn) (* Now do an animation. *) dn = 0.1; nstart = 1; nstop = 3; Do[ Plot3D[Sin[n*x]*Sin[n*y], x, 0, 2 Pi, y, 0, 2 Pi], n, nstart, nstop, dn]; Do[ Plot3D[Sin[n*x]*Sin[n*y], x, 0, 2 Pi, y, 0, 2 Pi], n, nstop - dn, nstart + dn, -dn]; "Double-click on a plot to see something move!"

180


181


Exercise 12.3: Going solo with Mathematica Go play! I suggest you try doing something of this sort:

• animate a scalar potential contour plot, also a field vectors plot for a pair of charges whose separation oscillates

• cook up the strangest sound you possibly can using Mathematica’s sound-generating tools

Summary and wrap-up discussion Next week we’ll work on using Mathematica to plot vector potentials, and to investigate the associated magnetic field. Comments about homework assigned for next week

182


Reading and homework should be completed by 4pm Friday. Optional reading: The Mathematica Book, S. Wolfram, or any other Mathematica reference you might happen to find useful. Problem 0 (5 points): Make your own hardcopies… Print (from the web) your own copy of the solutions to last week’s problem set and write a (true!) statement on your problem set asserting that you have successfully printed your personal copies. Problem 1 (5 points): Some symbolic manipulations Have Mathematica evaluate the following integrals and derivatives for you. Please submit a hardcopy of an evaluated Mathematica notebook to show us your results for this problem. (I can give you a hand if you have problems generating the hardcopy or moving your files off the PC in Loomis 257.) To display useful palettes, follow this sort of menu path: File → Palettes → Complete Characters → Letter-like Forms → Technical Symbols (you’ll find ∞ there)

(a) [ ]2tan( )x dx∫ (remember that Tan[x]and not tan(x) is what Mathematica uses to represent a tangent.)

(b) 2 2xe d

+∞−

−∞∫ x

(c) ( )4lnd xdx

(d) ( )2 3 4 2x y z

x y∂

∂ ∂

Problem 2 (10 points): An animation For the x range 0 2x π≤ ≤ please use the Plot[…] function inside a Do loop to plot sin(kx) using k as the animation variable, with k stepping over values from 0 to 10 in increments of 0.1 unit. When you’ve done it correctly, Mathematica will generate a large number of plots for you; double-clicking on one of them will cause the program to play all the plots as an animation. In order to keep Mathematica from adjusting the minimum, maximum values of the y axis you can

183


include the expression PlotRange->0, 2*Pi, -1, 1 as a last argument to the Plot[…] function.

Problem 3 (5 points): Plotting an implicit function One can define a function (or relation) implicitly by specifying a rule that is satisfied by the all points (x, f(x)). For example, the rule x2 + y2 = 4 is an implicit definition of a circle of radius 2 centered at the origin. Mathematica is able to plot functions defined this way, although you’ll need to load one of its “packages” using the statement Needs["Graphics`ImplicitPlot`"] at the start of your notebook. (Be careful about the exact form for this: the single quotes are the back-apostrophes usually found near the upper-left corner of a standard keyboard.) Look up the ImplicitPlot[…] function in Mathematica’s online help facility and use it to generate plots of:

(a) the Folium of Descartes, defined implicitly through the equation 3 3 3x y x+ = y (a) a cardioid, defined implicitly through the equation

22 2 2 2x y x x y + − = + Your plots should look something like this:

-3 -2 -1 1 2 3

-4

-3

-2

-1

1

2

Folium of Descartes

0.5 1 1.5 2

-1

-0.5

0.5

1

Cardiod

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Problem 4 (10 points): Solving an algebraic equation Mathematica can solve a variety of algebraic equations, both through symbolic manipulation, and through numerical means. See the program’s online help documentation concerning the Solve[…] function. Here are a couple of examples:

Note the required pair of equal signs “==” used to define the equation. (a) Use the Solve[…] function to generate the roots of the most general form of a cubic equation in one variable, . 3 2 0ax bx cx d+ + + = (b) Some equations do not lend themselves to solution through symbolic techniques built into Mathematica. If you encounter one of these, try using the FindRoot[…] function instead to search for roots to the equation. (See the online help information for the format of FindRoot’s arguments.) Use FindRoot to determine numerical values of the solutions to sin 2θ θ= .

-4 -2 2 4

-2

-1

1

2plots of sinHtheta L and theta ê2

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Physics 212, Spring 2003. Unit 13 13.1 Studying electromagnetic radiation with Mathematica

Physics 212


Unit 13

Studying electromagnetic radiation with Mathematica

Spring, 2003

George Gollin


186


Unit 13: Studying electromagnetic radiation with Mathematica Recap of last week Introduction: Those Liénard-Wiechert potentials for a point charge When a point charge is in motion, the scalar potential it creates is modified in two fundamental ways:

( ) ( )( ) ( ) ( )( )0 0

1 1,ˆ4 4 1Q Q

Q QV r V r tv t t cr r r r tπε πε ε

1 = =

− ⋅′ ′− − ′ ⇒

1. The potential depends, in part, on the charge’s retarded position-- its location at the retarded

time t , defined through the relation ′ ( )Qr r tt t c′ ′= − − .

2. The 1/r form of the potential is modified by a velocity-dependent factor ( ) ( )( )ˆv t t cε− ⋅′ ′1 1

where is a unit vector pointing from the charge’s retarded position to the observer’s position as shown in the following figure.

( )ˆ tε ′

v

observer

ε


Q Q is really here nowobserver sees Q here

(at its retarded position)

The charge’s vector potential is modified in similar fashion so that

( ) ( )( ) ( ) ( )( )

( ) ( )2 20

1 1, ,ˆ4 1Q

v t v tQA r t V r tc cv t t cr r tπε ε

′ ′= × =

− ⋅′ ′− ′.

(Recall that ( )20 01 cµ ε= .)

This complicates things somewhat—the electric field is no longer guaranteed to be conservative, so it is no longer the case that we can always define it as the gradient of a potential. The scalar and vector potentials are still useful, however, since E V A t= −∇ − ∂ ∂ and B A= ∇ × , so we can use them to calculate the fields if we’re willing to do all the differentiation. Here’s what you’ll work on, with the help of Mathematica:

187


• generating contours of constant retarded time for observers watching a charged particle moving with constant velocity, and a separate particle which stops suddenly at the origin;

• graphing the behavior of the ( ) ( )( )ˆv t t c1 1 ε− ⋅′ ′ factor as a function of observer position for the constant-velocity and stop-at-the-origin cases just mentioned

• studying the scalar potential including retardation, but neglecting the ( ) ( )( )ˆv t t cε− ⋅′ ′1 1 correction factor (there’ll be some surprises here)

• graphing the vector potential • plotting the relative contributions to E of V−∇ and A t− ∂ ∂ (more surprises!) Exercise 13.1: Retarded time for a moving charge Since Alpha Centauri is 4.3 light years away, something we see happening there at the present time actually took place several years ago. The retarded time we would use in calculating the potentials at Earth induced by electric charges on Alpha Centauri is set back from the present time by 4.3 years. It works the same way for something in motion: imagine that a moving object emitted a flash of light nine nanoseconds ago. Since c 0.3≈ m/nsec, observers who are three meters from the charge’s position at the time it emitted the flash will see the flash now. (Since emitting the flash the charge may have moved elsewhere.) Here’s a graphical example, involving an observer watching a charge moving at half the speed of light up the y axis. The true position of the charge is 0.5 (it passes through the origin at t = 0) while the observer is stationary at x = 1.5 meters. At t = 5 nsec in the observer’s frame, the charge is really at y = 0.75 meters but the observer is seeing light that left it earlier, when it was at y = 0.

ˆct y

x

y

observer at x=1.5 m

charge’s true position: x=0.75 m when t=5 nsec

charge’s retarded position: x=0

present time is t=5 nsec In this particular case, the retarded time corresponding to this observer’s t = 5 nsec is

( ) 0obs Qt t r r t c′ ′= − − =

( )t t′

. It is a simple algebra problem to solve for the retarded time as a function of the observer’s time when the charge moves with constant velocity: to find the form of the function . When the charged particle’s velocity isn’t constant, though, it can be

nearly impossible to solve for . ( )t t′

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Consider one particle moving with constant velocity , and another moving with velocity

for t < 0, which stops at the origin at t = 0. Graphs of the particles’ y positions as functions of time are shown below. (Remember that

ˆ0.5ct y

0.3cˆ0.5ct y

≈ m/nsec.)

charge is movingwith v = 0.5c

t (nsec)

y (0.3 m per division)

charge is movingwith v = 0.5c

until t = 0

t

y

rvst1y rvst2y

(a) Imagine that the present time is now t = 10 nsec and that observers are watching the charge which moves with constant velocity. Sketch possible locations of observers who are seeing (at t = 10 nsec) light which had illuminated the moving particle at the earlier times t = 0 nsec and t = 5 nsec. (These observers would use retarded times 0 nsect′ = and t 5 nsec′ = in their calculations of the potentials associated with the charge.)

charge here now (t = 10 nsec)

charge is moving with v = 0.5cgrid cell (∆x, ∆y) is (0.3m, 0.3m)

charge was here at t = 0 nsec

charge was here at t = -10 nsec

present time ist = 10 nsec

x

y

(b) Imagine that the present time is now t = 5 nsec (not 10 nsec). Observers are watching the charge which moved with constant velocity for t < 0 nsec, but then stopped at the origin. Sketch possible locations of observers who are seeing (at t = 5 nsec) light which had illuminated the

189


moving particle at the earlier times t = 4, 2, 0, -5, -10 nsec. (These observers would use retarded times t in their calculations of the potentials associated with the charge.) 4, 2, 0, 5, 10 nsec′ = − −

charge here since t = 0 nsec

charge was here at t = -10 nsec

present time ist = 5 nsec

x

y

charge was moving with v = 0.5cbut stopped at (0,0) 5 nsec ago

(c) Let’s do this with Mathematica now. I’ve set up a file of useful functions for you in a utility file p212_mathematica.m in the Physics 212 area. (They are automatically loaded when you evaluate the first cell in the notebook file p212_unit13_mathematica_1.nb, which you should use as a template for your work.) Here is a list of the functions defined for you: Distance[x1,y1, x2,y2] distance between points 1 and 2 Norm[x,y] length of the vector extending from 0,0 to x,y DotProd[x1,y1, x2,y2] dot product between two (2D) vectors CrossProd[x1,y1, x2,y2] cross product between two (2D) vectors UnitVector[x,y] unit vector parallel to the line from 0,0 to x,y rvst1[t] x,y at time t for a particle moving with constant velocity up the y axis so that x=0, y=0.5ct

190


rvst2[t] like rvst1[t], but the particle stops in 0.1 nsec at the origin at t=0. rvst3[t] like rvst2[t], but the particle takes 10 nsec to stop at the origin. NOTE: you must select one of these, or define your own function so that rvst[t] is defined. For example, to select rvst1 do this: rvst[time_] := rvst1[time] being careful to include the underscore at the end of time_ to the left of the “:=” symbol, but NOT to the right of the “:=” vvst[t] time derivative of rvst[t]. (assumes user has defined rvst) avst[t] time derivative of vvst[t]. (assumes user has defined rvst) TRetard[x,y, t] returns retarded time at charge's position for observer at x,y, t (uses rvst function to calculate this) VDotETerm[x,y, t] value of 1/(1 - v dot E) evaluated using the retarded velocity and line-of-sight vector e (not the electric field!) ScalarPotential[x,y, t] scalar potential when charge is at rvst[t] and observer is at x,y at time t. Calculation includes retardation effects. ScalarPotentialNoVDotE[x,y, t] like ScalarPotential, but omitting the velocity-dependent correction factor. GradV[x,y, t] gradient (in 2 dimensions) of scalar potential when charge is at rvst[t] and observer is at x,y at time t. Includes retardation effects. VectorPotential[x,y, t] vector potential when charge is at rvst[t] and observer is at x,y at time t. Calculation includes retardation effects. dAdt[x,y, t] time derivative of vector potential when charge is at rvst[t] etc. ECoulomb[x,y, t] Coulomb (1/r^2) portion of the electric field ERadiation[x,y, t] Radiation (1/r) portion of the electric field ETotalx,y, t] Total electric field (Coulomb + radiation) NonRadialAngle[xobs,yobs, fx, fy, t] returns the angle (in degrees) between the vector fx, fy and the line from the instantaneous (not the retarded) position of the charge to the observer’s location.

Here's an example of how to use the predefined functions: (* As an example, assign rvst to be the same as rvst2, then plot its y component. *) (* define the (new) function rvst here *) rvst[timearg_] := N[rvst2[timearg]] (* print one value of it to remind everyone that rvst returns a vector, not a number *) Print["rvst at -20 nsec is ", rvst[-20*10^-9]]; (* Now plot its y component (thus the [[2]] index) from - 20 nsec to + 20 nsec *) Plot[rvst[t][[2]], t, -20*10^-9, 20*10^-9];

191


rvst at -20 nsec is 0., -2.99792

-2×10-8 -1×10-8 1×10-8 2×10-8

-3

-2.5

-2

-1.5

-1

-0.5

Graphs of the y components of rvst1 and rvst2 appeared in the writeup a few pages ago. Do the following: 1. Assign the constant velocity function rvst1 to rvst. 2. Assume the present time is 30 nsec after the charge has reached (or passed through) the

origin in our frame of reference. 3. Make a contour plot of the retarded time in the region . (The

contour closest to the charge’s t = 30 nsec position corresponds to observers who see the charge as it is at t = 30 nsec. Contours further away correspond to observers who see the charge as it was at progressively earlier times.)

20 20, 20 20x y− ≤ ≤ + − ≤ ≤ +

4. Assign the stop-at-the-origin velocity function rvst2 to rvst. 5. Make a contour plot with the same x,y limits. 6. Comment sensibly on what you see, in particular about the differences in the two plots. Your plots should look something like these:

192


-20 -10 0 10 20

20

10

0

10

20

d time vs. position , constant velocity 0.5c, time now 3

-20 -10 0 10 20-20

-10

0

10

20

Retarded time vs. position , stop -at-origin , time now 30 nsec

The concentric region in the second plot extending 9 meters ( ) out from the origin expands as time passes: observers at greater distance become aware at later times that the charge is stopped.

30 nsecc= ×

193


′Exercise 13.2: That v t correction factor ( ) ( )ˆ tε′ ⋅ Let’s investigate the effects of retardation and that 1 1 factor on the scalar potential now. We’ll work with two different cases, both involving a point charge. The first charge is moving up the y axis at half the speed of light so that its position is . The second is also moving up the y axis, but decelerates rapidly with constant acceleration after it passes through the origin, coming to rest in 0.1 nanosecond at . (These are the two cases rvst1, rvst2 used in the previous exercise.)

( ) ( )( )ˆv t t cε− ⋅′ ′

Qr

( ) ˆ0.5Qr t ct y=

y0.0075=

While moving, our charge has v = 0.5c so the correction factor is 2 in forward direction, 0.5 in the backwards direction, and 1 at 90 degrees. Use Plot3D to make graphs of the 1 1 factor using the same time (30 nsec) and x,y ranges for the plots. (The function is defined in the utility file as VDotETerm[x,y, t].) Do it for both velocity cases we’ve been using. Your plots should look something like the two which follow.

( ) ( )( )ˆv t t cε− ⋅′ ′

Bear in mind that the boundary of the flat region within 9 meters ( ) of the origin in the second plot expands outwards at c as more distant observer become aware that the charge has stopped. This means that the 1 1 correction factor included in the scalar potential will change suddenly for observers as they realize that the charge is no longer moving.

30 nsecc= ×

( ) ( )( )ˆv t t cε− ⋅′ ′

V dot E correction term vs. position , uniform velocity , t=30 nsec

-20

-10

0

10

20 -20

-10

0

10

20

1

1.5

2

-20

-10

0

10

20

194


V dot E correction term vs. position , stop -at-origin , t=30 nsec

-20

-10

0

10

20 -20

-10

0

10

20

1

1.5

2

-20

-10

0

10

20

195


Exercise 13.3: Scalar potential with retarded time, but no v dot e correction When retardation is included, but the velocity-dependent correction factor is omitted, the scalar caused by a moving charge at an observer's position is the same as that caused by a stationary charge at the moving charge's retarded position. The equipotential contours for a charge moving with constant velocity are circles centered on the retarded positions of the charge. Use ContourPlot to make graphs of the scalar potential the 1 1 factor using the same time (30 nsec) and x,y ranges for the plots as before. (The function is defined in the utility file as ScalarPotentialNoVDotE[x,y, t].) Do it for both velocity cases we’ve been using. Your plots should look something like the two which follow.

( ) ( )( )ˆv t t cε− ⋅′ ′without

Note the bunching of field lines towards the tops of the plots: this means the potential would change more rapidly in front of the moving charges if this were the correct expression for the scalar potential. (It's not, since we've left off the necessary 1 1 factor.) ( ) ( )( )ˆv t t cε− ⋅′ ′ As before, take note in the second plot of the change that occurs at the boundary 9 meters from the origin.

-20 -10 0 10 20

-20

-10

0

10

20

calar potential NOT including V dot E correction term , constant v, t=30 nsec

196


-20 -10 0 10 20

-20

-10

0

10

20

calar potential NOT including V dot E correction term , stop −at−0, t=30 nsec

Exercise 13.4: Scalar (and vector) potential with retarded time and v dot e correction Now use ContourPlot to make graphs of the scalar potential including the 1 1 factor using the same time (30 nsec) and x,y ranges for the plots as before. (The function is defined in the utility file as ScalarPotential[x,y, t].) Do it for both velocity cases we’ve been using. Your plots should look something like the two which follow.

( ) ( )( )ˆv t t cε− ⋅′ ′

Note the surprising fact that the equipotential contours in the first plot are concentric, and are centered on the instantaneous position of the charge! Distant observers cannot know that the charge is really 4.5 meters up the y axis, since it might have stopped recently. This should be clear from the second plot-- observers far from the (now stopped) charge still see the same equipotentials as are seen by distant observers in the constant velocity plot shown first. The effects at the 9 meter boundary are especially dramatic in the second plot in the forward direction where the velocity correction factor changes from 2 to 1 suddenly. The rapid change in the potential at this point causes an impressive bunching of the equipotential contours. We haven't done anything with the vector potential yet. The constant velocity case is easy, since

( ) ( ) (2,v t

A r t V r tc

′= ), . Except for the direction (along y) associated with the vector potential, it

behaves just like the scalar potential. The second case is also a snap-- the vector potential is zero close to the (stopped) charge where all observers realize that it is stationary.

197


198


-20 -10 0 10 20

20

10

0

10

20

potential including V dot E correction term , uniform velocity , t=30

-20 -10 0 10 20

-20

-10

0

10

20

Scalar potential including V dot E correction term , stop −at−0, t=30 nsec

199


Exercise 13.5: Vector potential for the stop-at-0 charge Use PlotVectorField to make a graph of the vector potential using the same time (30 nsec) and x,y ranges for the plots as before. (The function is defined in the utility file as VectorPotential[x,y, t].) Do it only for the stop-at-zero velocity case. Your plot should look something like the one which follows. Note that vector potential is always along y, and that the forward-backward symmetry has been broken by the velocity-denominator term. The boundary at 9 meters moves outwards at the speed of flight.

-20 -10 0 10 20

-20

-10

0

10

20

Vector potential , stop −at−0, t = 30 ns

200


Discussion: Liénard-Wiechert fields It’s always true that and that . Calculating and is “just” a matter of taking some derivatives!

E V A= −∇ − ∂ ∂ B A= ∇ × E Bt

After a fair amount of messy calculation it can be shown that:

( ) ( ) ( )( )

( )( )

2 2

2 30

320

ˆ ˆ1 11,4 ˆ1

ˆ ˆ1 14 ˆ1

Q

Q

v c v c v cQV r tv cr r

aQc v cr r

ε επε ε

ε επε ε

− − − ⋅′ ′ ′ ′ ′ −∇ = − ⋅′ ′′ −

⋅′ ′ ′ +

′ − ⋅′ ′−

Primed quantities are evaluated at the appropriate retarded time; is the (retarded) acceleration. (The acceleration enters because of the subtleties involving functions of t being differentiated with respect to t.)

a′′

Note that the first term in the gradient is independent of acceleration, and is proportional to 1/r . (This is the “Coulomb term.”) However, the second term is proportional to acceleration, and falls off as 1/r.

2

It is possible to pound out the derivatives to calculate too: the result is A t− ∂ ∂

( ) ( ) ( )( )

( )( )2 3 2

0 0

ˆ ˆ, 1 1 14 4ˆ ˆ1 1

QQ

v c v cA r t a a v cv Q Qt c cv c v cr rr r

ε επε πεε ε

⋅ −′ ′ ′∂ − ′ ′ ′ ′− ′ − = +∂ ′ − ⋅ − ⋅′ ′ ′ ′′ − −

3

+ × ×

.

As with the equation, the first (Coulomb) term in is independent of acceleration, and is proportional to 1/r . The second term, proportional to acceleration, falls as 1/r.

V−∇ A t− ∂ ∂2

Let’s study the “Coulomb terms” of and first by looking at a charge moving with constant velocity.

V−∇ A t− ∂ ∂

201


Exercise 13.6: -Grad V: constant velocity (a) Use PlotVectorField to make a graph of the negative of the gradient of the scalar potential using the same time (30 nsec) and x,y ranges for the plots as before. (The function is defined in the utility file as GradV[x,y, t]; do it only for the constant velocity case.) To make it easier to see the directions of the vectors, also use the utility function UnitVector to replot . V−∇ Your plots should look something like the first two which follow. It looks like the vectors point radially away from the instantaneous position of the charge, without retardation. How close is this to the truth?

202


Vectors...

-20 -10 0 10 20

-20

-10

0

10

20

−grad V, constant v, t = 30 nsec

Unit vectors...

-20 -10 0 10 20

-20

-10

0

10

20

Unit vetors of −grad V, constant v, t = 30 nsec

203


(b) Is really radial? (Does it really point directly away from the instantaneous position of the charge?) Use Plot3D and the utility function NonRadialAngle to investigate, using the same observer time and plot boundaries as before.

V−∇

Your plot (which may take two or three minutes to appear) will look something like this:

non −radialness angle Hdegrees L of −grad V, uniform velocity , t = 30 nsec

-20

-10

0

10

20-20

-10

0

10

20

-5

0

5

-20

-10

0

10

20 What do you conclude?

204


Exercise 13.7: -dA/dt for constant velocity Use PlotVectorField to make a graph of the negative of the time derivative of the vector potential using the same time (30 nsec) and x,y ranges for the plots as before. (The function is defined in the utility file as dAdt[x,y, t]; do it only for the constant velocity case.) To make it easier to see the directions of the vectors, also use the utility function UnitVector to replot

. A t−∂ ∂ Your plots should look something like the first two which follow. Not surprisingly, can only point along y. A t−∂ ∂ Vectors... Unit vectors...

-20 -10 0 10 20

-20

-10

0

10

20

−dAêdt, constant v, t=0

-20 -10 0 10 20

-20

-10

0

10

20

unit vactors along −dAêdt, constant v, t=0

Note the sign flip at instantaneous charge position

205


Exercise 13.8: E = -Grad V - dA/dt for constant velocity Since , we can investigate the electric field by combining the last few functions we've been using. Do it-- generate vector, and unit vector plots of , then plot the degree to which the electric field points radially away form the instantaneous position of the charge.

E V A= −∇ − ∂ ∂

E V A t= −∇ − ∂ ∂

t

Here are my versions of the plots. Don't be mislead by the round-off errors which fool Mathematica's arrow-drawing routine when the fields are too weak. Vectors...

-20 -10 0 10 20

-20

-10

0

10

20

−grad V −dAêdt, constant v, t = 30 nsec

206


Unit vectors...

-20 -10 0 10 20

-20

-10

0

10

20

unit vector along −grad V −dAêdt, constant v, t = 30 nsec

"Radialness..."

non −radialness angle Hdegrees L of −grad V −dAêdt, constant velocity , t = 30

-20

-10

0

10

20-20

-10

0

10

20

-0.0002

-0.0001

0

0.0001

0.0002

-20

-10

0

10

20

207


Well, look at that! What can we conclude?

208


Discussion: Electromagnetic radiation! When do the terms in proportional to acceleration dominate ? Since the Coulomb terms are proportional to 1/r , while the acceleration terms are proportional to 1/r, it's clear that at large distances the acceleration terms must dominate. But what constitutes a "large" distance? Very roughly, the ratio of the "radiation" (1/r) and Coulomb terms for is

, where r represents the distance to the observer and a is the charge's

acceleration. As a result, when the observer's distance is large compared to the acceleration-based term will dominate.

E V A= −∇ − ∂ ∂ E

E( )2 2ra c r c a=

2c a

t2

Exercise 13.9: Plotting the radiation terms Including only the terms proportional to 1/r gives

( )( ) ( ) 321 only

0

1 ˆ ˆ ˆˆ4 1r

Q

QE a ac v cr r

a v cε ε επε ε

= − − ⋅ − × ×′ ′ ′ ′ ′ ′ ′′ − ⋅′ ′−

.

To simplify matters, let’s only consider acceleration (anti)parallel to v . (Clearly this will not work for circular motion!). In this case, ( so we can ignore the term with the cross products.

a)a v c 0× =′ ′

Let’s look at the remaining terms in the curly brackets: . Here’s a diagram. ( )ˆ â aε ε− ⋅′ ′ ′ ′

( )ˆ â a ε ε′ ′ ′ ′− ⋅

observer

( )ˆ â ε ε′ ′ ′⋅

ε ′a′

Q With some thought you should be able to see that is just the component of the (retarded) acceleration which is perpendicular to the observer’s line of site towards the (retarded) position of the charge. Let’s give this component the name a (“a prime perp”). (Even if we

( )ˆ â aε ε− ⋅′ ′ ′ ′

⊥′

209


hadn’t asked that the acceleration be along the direction of the velocity we’d still have the field perpendicular to since the ignored term is perpendicular to due to the cross products.)

ε ′ ( )ˆ a v cε × ×′ ′ ′ ε ′

)3

With these simplifications, we can write . (21 only

0

1ˆ4 1r

Q

QaEc vr r cπε ε

⊥′= −′ − ⋅′ ′−

Keep in mind that most everything in this expression is a retarded variable, and that the velocity-correction denominator is cubed. Things to notice: 1. The electric field is perpendicular to the line-of-sight vector . ε ′2. The electric field is proportional to the charge’s (retarded) acceleration. 3. The field falls off like 1/r. 4. There’s a negative sign in front: the radiation electric field of a positive charge with upwards

acceleration points . down Use ContourPlot to make graphs of the of the radiation electric field using the same time (30 nsec) and x,y ranges for the plots as before. (The function is defined in the utility file as ERadiation[x,y, t].) Do it for the as-yet unused position-vs.-time function rvst3 which describes a particle which begins decelerating from at t = -10 nsec, coming to a complete stop at t = 0.

magnitide

ˆ0.5v c y=

Your plot should look something like the sample plot which follows. The spikes are an artifact of the granularity of the plot I’ve chosen. Notice the following: • Since the radiation fields are blowing past an observer only while that observer sees the

particle’s (retarded) acceleration as different from zero, the region in space containing a non-zero radiation field is only three meters thick.

• The fields would be strongest in directions perpendicular to the particle’s acceleration if the velocity-correction factor 1 1 weren’t present. Because of it, the fields are somewhat stronger in the forward direction. (“Forward” means parallel to the particle’s velocity during acceleration.)

( )3ˆ v cε− ⋅′ ′

210


Radiation E field , stop -at-zero , 10 ns deceleration , t=30 ns

-20

-10

0

10

20-20

-10

0

10

20

0

2×108

4×108

6×108

-20

-10

0

10

20 Now do the same thing, but using the position-vs.-time function rvst4 which describes a particle which begins decelerating from at t = -10 nsec, coming to a complete stop at t = 0.

ˆ0.01v c= y

Your plot should look something like the sample plot which follows. Because the particle is nonrelativistic, the radiation electric fields are nearly forward-backward symmetric, and are strongest perpendicular to the particle’s acceleration. In addition, the fields are considerably weaker due to the reduced acceleration.

211


Radiation E field , non-relativistic , 10 ns deceleration , t=30 ns

-20

-10

0

10

20-20

-10

0

10

20

0

2×106

4×106

6×106

-20

-10

0

10

20 Discussion: What about the magnetic field? We haven’t said anything about the magnetic field! It’s just a matter of calculation to take the curl of the vector potential; what one finds is that the magnetic and electric fields obey

for electromagnetic radiation. The magnetic field is perpendicular to the electric field, and also perpendicular to the line-of-sight direction. B E c=

Summary and wrap-up discussion We’ve only made plots for particles which move with constant velocity, or stop rapidly. However, everything we’ve done would work for a particle moving with arbitrary acceleration. (Mathematica might gag, but that doesn’t mean that there’s anything wrong with our equations.)

212


Here are the main points. • When a point charge is in motion, the scalar and vector potentials it creates are modified in

two fundamental ways:

1. The potential depends, in part, on the charge’s position-- its location at the retarded time t , defined through the relation

.

retarded′

( )Qt t r r t′ ′= − − c2. The 1/r form of the potential is modified by a velocity-

dependent factor 1 1 where is a unit vector pointing from the charge’s retarded position to the observer’s position.

( ) ( )( )ˆv t t cε− ⋅′ ′ ( )ˆ tε ′

• The electric field depends on both the gradient of the scalar potential the time derivative

of the vector potential. These two contributions conspire in a remarkable fashion so that the electric field of a charge moving with constant velocity is radial, always pointing exxactly away from the position of the charge.

and

instantaneous • Observers far from a charge which has stopped cannot know instantaneously that it has

stopped moving. As a result, they will continue to see the fields associated with a charge in motion until enough time has passed so that they become aware the charge has stopped.

• The sudden change in the 1 1 term in the potentials (which corresponds to

the 1 1 term in the fields) induces dramatic changes in the derivatives of the potentials, giving rise to electromagnetic radiation: electric fields which are proportional to (retarded) acceleration, and propagate away from the charge, falling off like 1/r instead of 1/r .

( ) ( )( )ˆv t t cε− ⋅′ ′

( ) ( )( )3ˆv t t cε− ⋅′ ′

2


213


Reading and homework should be completed by 4pm Friday. Required reading: Read all of the in-class exercise material for this unit. Purcell, chapter 5, section 7. Optional reading: Purcell, chapter 9 and appendix B. The Feynman Lectures on Physics (volume II), chapters 20 and 21. Radiation review/reminder: When a point charge is in motion, the scalar and vector potentials it creates are modified in two fundamental ways:

3. The potentials depend, in part, on the charge’s position-- its location at the retarded

time t , defined through the relation t t . retarded

′ ( )Qr r t′ ′= − − c4. The 1/r forms of the potentials are modified by a velocity-dependent factor

where is a unit vector pointing from the charge’s retarded position to the observer’s position as shown in the following figure. (Sometimes I’ll use as an abbreviation for .)

( ) ( )( )ˆ1 1 v t t cε− ⋅′ ′ ( )ˆ tε ′

ε ′(ˆ tε ′)

v

observer

ε


Q Q is really here nowobserver sees Q here

(at its retarded position)

The changes to the potentials work like this:

( ) ( )( ) ( ) ( )( )0 0

1 1,ˆ4 4 1Q Q

Q QV r V r tv t t cr r r r tπε πε ε

1 = =

− ⋅′ ′− − ′ ⇒

( ) ( )( ) ( ) ( )( )

( ) (2 20

1 1, ,ˆ4 1Q

v t v tQA r t V r tc cv t t cr r tπε ε

′ ′= × =

− ⋅′ ′− ′( )2

0 01 cµ ε=) . (Recall that .)

It is still true that and so we can use the potentials to calculate the fields. The in and will be proportional to 1/r. If you crank out the derivatives you’ll find that the radiation term in the expression for the electric field is this:

E V A= −∇ − ∂ ∂ B A= ∇ ×

E Bt

radiation terms

214


( )( ) ( ) 321 only

0

1 ˆ ˆ ˆˆ4 1r

Q

QE a ac v cr r

a v cε ε επε ε

= − − ⋅ − × ×′ ′ ′ ′ ′ ′ ′′ − ⋅′ ′−

.

The particle’s (retarded) acceleration is a . Don’t be intimidated by all the cross products! Asdescribed in the in-class exercise material, if we only consider acceleration (anti)parallel to we can write

′a v

( )321 only0

1ˆ4 1r

Q

QaEc vr r cπε ε

⊥′= −′ − ⋅′ ′−

where (“a prime perp”) is the component of (retarded) acceleration perpendicular to the line-of-sight to the observer. Keep in mind that most everything in this expression is a retarded variable, and that the velocity-correction denominator is cubed.

a⊥′

Calculating the magnetic field is “just” a matter of taking the curl of the vector potential; what one finds is that the radiation (1/r) magnetic and electric fields obey and that the magnetic field is perpendicular to the electric field, and also perpendicular to the line-of-sight direction.

B E c=

Things to keep in mind for the radiation fields: 5. The electric and magnetic fields are perpendicular to the line-of-sight vector . ε ′6. The electric and magnetic fields are proportional to the charge’s (retarded) acceleration. 7. T all off like 1/r. he fields f8. .

9. There’s a negative sign in front of the expression for : the radiation electric field of a positive charge with upwards (retarded) acceleration points .

B E c=

Edown

Problem 0 (5 points): Make your own hardcopies… Print (from the web) your own copy of the solutions to last week’s problem set and write a (true!) statement on your problem set asserting that you have successfully printed your personal copies.

215


t z

Problem 1 (15 points): Non-relativistic dipole radiation A particle with charge q oscillates about the origin with position . Since the particle is always moving non-relativistically, the electric fi ld it generates at r , the location of a

distant observer, is well-approximated by . Note that as

shown in the following figure, and that the retarded time is .

3 7 ˆ10 sin(10 )qr−=

( ) 20

1,4

qaE r tr cπε

⊥′≈ − sina a θ⊥ =′ ′

t t′ r c≈ −

e

a′

a′

a⊥′

observerat r

ε ′

q

z

θ

You learned in Physics 112 that the power flux (energy per second per square meter) transported by an electromagnetic wave has magnitude and that . 0S E B µ= × 2

0 0 1 cµ ε =

(a) Calculate the RMS power flux (in watts per square meter) arriving at an observer located at the point (r, θ), as shown in the figure. (b) What is the total RMS power radiated by the oscillating charge? (You’ll need to calculate to answer this question. The integral is easiest to do in spherical

coordinates (r, θ, φ) where , , and is a unit vector pointing

away from the origin.)

alldirections

S dA⋅∫

ˆS r S= 2ˆ sindA r r d dθ θ φ= r

(c) Assuming q = 1 Coulomb, what is the numerical value (in watts) for your answer to part (b)?

216


z

Problem 2 (15 points): Mrs. Urkin builds her own x-ray machine After seeing Marathon Man for the 97 time, Mrs. Urkin decides to teach herself dentistry. Knowing that x-ray machines employ bremsstrahlung (German for “braking radiation”) to produce short-wavelength electromagnetic radiation, she borrows the UD&HE electron linear accelerator to use as the electron gun for her device. When a fast-moving electron slams into a dense target its rapid deceleration will cause it to radiate x-rays.

th

Assume that an electron from the “linac” initially travels upwards along the z axis (so that its initial velocity is ), then decelerates with constant acceleration to stop near the origin at t = 0. The teeth of an unfortunate patient are located at (r, θ ), as shown. Please assume that is very large compared to the distance through which the electron moves while coming to

rest so that always.

0 ˆv v z= ˆa a= −

r

er r− r−

′ ≈

patientat r

ε ′θ

a′

z

y

x

e-

(a) Prove that the magnitude of the radiation electric field (for fixed ) is greatest at when .

r 90θ = °v c

(b) For as the electron slows to rest from a initial velocity prove that the intensity of x-rays at remains constant during the deceleration process, even though v in the early stages of deceleration.

constant acceleration relativistic90θ =

c∼°

2

(c) An observer whose teeth are at (r, θ ) sees the decelerating electron’s velocity as 0.5c at one particular instant in time. What value of θ, expressed in degrees, will yield the maximum electric field strength for this (retarded) velocity? (You will have to slog through a differentiation and a small amount of algebra to derive a useful quadratic equation, one of whose roots is the cosine of the desired angle. The fact that sin might be helpful.) 2 1 cosθ θ= −

217

Physics 212, Spring 2003. Unit 14 14.1 Oops! It doesn’t really work like that! (Quantum mechanics enters the picture)

Physics 212


Unit 14

Oops! It doesn’t really work like that! (Quantum mechanics enters the picture)

Spring, 2003

George Gollin


218


Unit 14: Oops! It doesn’t really work like that! (Quantum mechanics enters the picture) Recap of last week Introduction: The fields aren’t really fields, and you cannot know as much as you’d like You’ve probably heard of the Heisenberg Uncertainty Principle. Today we’ll investigate a plausibility argument for why it is true, as well as some of its consequences. We’re forced to head down this path because one of the reasonable assumptions we’ve made about electrodynamics is completely, totally, utterly wrong: the electromagnetic radiation streaming from an accelerating charge is not composed of fields which vary smoothly in space and time. Einstein’s Nobel Prize was not awarded for his work on relativity (which certainly did deserve one), but rather for his explanation of the photoelectric effect, one of the consequences of the “granularity” of electromagnetic radiation. Here’s what you’ll be working on: • a classical analysis of the rate at which electrons “should be” ejected from a conductor which

is exposed to an intense light source. • a classical analysis of the relationship between the maximum kinetic energy of electrons

ejected from a conductor and the intensity of the light source illuminating the conductor. • photons! (photon energy and momentum) • modeling the resolution of a simple magnifier • analyzing the “Heisenberg’s Microscope” thought experiment • estimating atomic sizes using the Uncertainty Principle We’ll also discuss the master equation of quantum mechanics, known as Schrödinger’s equation. Exercise 14.1: An estimate of the rate of photoelectron ejection from a metallic surface A charge above a classical conducting surface is attracted back towards the surface due to the fields created by the surface charge it attracts beneath it, as shown in the figure. The fields above the conductor are exactly the same as they would be if an image charge of equal magnitude, but opposite sign, were placed as shown, without the conductor being present.

219


these field linesdon’t really exist

+ + + + ++ ++++

conductorfictitious positive image charge

real surface charge

electron

(a) Since atomic scales are typically a tenth of a nanometer, it shouldn’t be too crazy to model an electron in the process of escaping from the conductor as sitting ~10 meters above the surface, fighting the attractive force of its image charge as it tries to wander off. Estimate the energy needed to fee an electron from its image charge, using this model. (Recall the expression for electrostatic potential you learned in physics 112.) Please express you answer in electron-volts (1 eV = 1.6×10 Joules).

-10

energy-19

(b) A bright light illuminates the metal surface with an intensity of 1 watt per square centimeter. (This corresponds to approximately 6×10 eV per second). Let’s make the entirely unreasonable assumption that electrons at the surface absorb all the incident light until they have accumulated sufficient energy to escape from the metal, and that there’s only one electron per atom that is capable of absorbing energy. How long after the light is switched on will electrons begin to escape the metal?

18

Discussion Our ridiculous model has the virtue that it underestimates by many orders of magnitude how long it should actually take for electrons to begin escaping from the metal. The reality is quite different! It was known in 1928 that it took less than 3 nanoseconds to begin observing electrons after illumination; in 1955 it was known to take less than 10 seconds. I’m sure the present upper limit is substantially smaller.

-10

220


Exercise 14.2: An estimate of the dependence of maximum photoelectron kinetic energy on illumination intensity Let’s continue to work with our flawed classical model. We illuminate our metallic surface with monochromatic light and crank up the intensity, measuring the kinetic energy of the ejected photoelectrons. Imagine that we find something like the following distribution for the kinetic energy spectrum of detected electrons:

.

conductor

light source

ejected electron

electron detector/spectrometer

kinetic energy (eV)

num

ber d

etec

ted maximum energy

Propose a sensible-sounding (classical) model that relates the maximum kinetic energy of detected photoelectrons to the intensity of the light source, then sketch what you expect to find on the graph below.

intensity of light source

max

imum

kin

etic

ene

rgy

(eV

)

221


Discussion: photons! Our ridiculous model gets this wrong too: the maximum kinetic energy is independent of the intensity of illumination. The number of ejected electrons does increase with intensity, but the photoelectron kinetic energy spectrum does not change with intensity! Einstein’s idea was that the energy of the electromagnetic field comes in discrete quanta, called photons. The energy carried by a single photon depends associated with the light shining on our metal surface: the shorter the wavelength, the higher the photon’s energy. The exact expression is

only on the classical wavelength

hcEγ λ=

where h = 6.626 × 10 Joule-second so that hc = 1.986 × 10 Joule-meter. It is more useful to express this in terms of electron volt-nanometers: hc = 1,240 eV-nm. For example, the energy carried by a photon of green light (~600 nm wavelength) is about 2 eV. If you’re viewing this in color (on the web, for example), the following spectrum will let you see the correspondence between color and wavelength.

-34 -25

viol

et

blue

gree

n

yello

w

oran

ge

red

The photoelectric effect involves a single photon being absorbed by an electron, which uses the acquired energy to escape from the surface of the metal. A brighter light source sends more photons per second towards the surface, but each photon still carries the same amount of energy. Photons carry momentum too. As is the case with classical electromagnetic waves, a photon with energy E carries momentum p with . (See the brief discussion of relativistic kinematics in the second week’s materials: since , massless must satisfy the energy-momentum relation E = pc: if it carries energy, it also carries momentum.)

p E c h λ= =2 2 2 2E p c m c= + 4 anything

Exercise 14.3: Resolution of a simple magnifier Imagine we build a simple optical system which images light from a small object onto a sensor. How well can it determine the position of a small object? Let’s begin with a very simple optical system, a pinhole camera. Note that determining the resolution isn’t a matter of “geometrical optics” since we’ll have to think about diffraction effects associated with the finite width of the pinhole. To further simplify the analysis, let’s use a horizontal slit instead of a pinhole so that the camera can only image light along the vertical direction. Here’s a diagram:

222


.

Do

slit, width A

detector/sensor

object emits lightof wavelength λ image formed here

Dd/s

opaque screen If the object is a point source, how well localized is its image on the detector? Can we tell to arbitrary precision where the object was? Another way of asking this: is it possible to produce a single Euclidian point (or in this case Euclidian line) image at our detector/sensor? If we can’t (because of diffraction effects, for example), then the theoretical resolution of our optical system is limited by physical principles. It is tempting to think that the narrower the slit the better able we’ll be to localize the image of our point source on the detector surface. This is a geometrical-optics way of thinking! The problem is that we’ve neglected the role played by the opaque sheet in the propagation of the electromagnetic waves from the point source to the detector. The idea is simple: unless the opaque sheet does something to cancel the incident electromagnetic waves, they’ll go sailing straight through without being affected. The electric field in light waves from the point source accelerates electrons in the opaque screen, causing them to radiate their own fields which cancel the incident fields. It’s not hard to see how this works! The key is in the presence of the negative sign in front of the expression for the radiation electric field:

( )321 only0

1ˆ4 1r

Q

QaEc vr r cπε ε

⊥′= −′ − ⋅′ ′−

An electron in the opaque screen is accelerated by the incident electric field so that it experiences an acceleration . Neglecting fine points such as the distinction between

and , the radiation field is produced along the direction which is opposite to the incident field's direction.

incidenta QE∼a a⊥ incidentQa QE∼− −

Think of the opaque screen with the slit as a solid plane of antennae which radiate electromagnetic waves (perfectly canceling the incident electromagnetic wave downstream of the screen) combined with a narrow strip of identical antennae which generate electromagnetic

223


waves of the opposite polarity to cancel the perfect cancellation in the region where the actual screen has a slit. Sounds confusing, yes? Here’s a diagram. To make it easier to draw, I’ll only draw the polarization component of incident radiation parallel to the slit.

= +

What this all comes down to is that we’ll do fine if think of the electromagnetic radiation passing through the slit as if it had all been generated by a strip of antennae as wide as the slit which is cut into the opaque screen, all radiating coherently (in phase). No other sources of light need contribute to the signal observed by our detector. Another diagram:

224


enlarged view of slit, width A

detector/sensor

θ

A/2

extra distance for lower ray

Dd/s

opaque screen Since electromagnetic waves leaving all antennae are in phase, they can interfere constructively or destructively at the detector/sensor if the path lengths are different from antenna to antenna to a spot on the sensor. (a) Assuming the sensor is very far from the opaque screen (compared to the width of the slit) and that the wavelength of light is λ, at what angle θ will the two rays drawn in the diagram arrive at the sensor out of phase? (This angle corresponds to a half-wavelength difference in path length for the two rays, as indicated in the diagram.) You may assume that ,

, and that θ is small enough so that . /d sD A

/d sD λ sinθ θ≈ (b) In terms of λ, A, and D what is the half-width of the bright spot at the detector? d/s

225


Discussion Notice that all pairs of antennae interfere destructively at the angle you calculated. Discussion Resolution of our optical system: it would be difficult to tell two point sources apart if their bright spots overlapped at the detector. It's reasonable to identify the half-width of the bright spot from a point source as the system's resolution. This holds true even when we replace the slit with a lens of aperture A. Except for fine points concerning difference in how the interference zeroes behave for circular and slit apertures, it works the same way. Small wavelengths and large apertures give better position resolution. Exercise 14.4: Heisenberg’s microscope gedanken (thought) experiment Imagine we build a simple magnifier to locate a small object illuminated by a light source of wavelength λ. Our sensor is so quiet and so efficient that we can determine the position of the object when a single photon scatters from it, passes through the lens, and arrives at the sensor. Naturally, we are unable to tell the path the photon took as it traveled through the lens to the detector; all we know is where the photon ultimately arrived at the face of the detector.

light source

D

detector/sensor

A/2

photon’s path

Dd/s

θ x

z

226


(a) A photon, initially traveling along the z axis of our optical system, bounces off the object at an angle θ as shown. Calculate the x component of momentum transferred to the object when the photon scatters from it. (You may assume that the photon doesn’t lose any energy in the collision.) (b) Let's define the scale of momentum transfer to our object to be ∆p . The full range of possible x momentum transfers to the object, for all photon paths which pass through the lens, is . Calculate ∆p assuming that and that θ is small enough so that sin .

x

x

xp±∆D A θ θ≈

(c) Assuming that D = D for simplicity, calculate the position resolution ∆x for this optical system.

d/s

(d) Calculate the product ∆p ∆x for our locate-the-object optical system. x

227


Discussion The Heisenberg Uncertainty Principle! It’s true in general, not just for this particular (optical) system. The exact expression, which would require us to define more precisely what we mean by a position and momentum uncertainty is . Since appears so frequently in the equations of quantum mechanics, the convention is to use a new symbol (pronounced "h-bar"):

. Keep in mind that . The smaller the value of the smaller the domain in which

quantum mechanics will differ noticeably from the (fundamentally incorrect) classical description.

4xp x h π∆ ∆ ≥ 2h π

2 .h π≡1 Joule sec

34 161.0546 10 Joule seconds = 6.5822 10 eV seconds− −= × ⋅ × ⋅

( )ond = 1 kg m/sec m⋅ ⋅ ⋅

In three-dimensional systems we have three separate expressions: , ,

. 2xp x∆ ∆ ≥ 2yp y∆ ∆ ≥

2zp z∆ ∆ ≥ The uncertainty principle is built into the structure of our quantum mechanical universe. It holds true for ALL systems, not just those which employ photons in measurements. Exercise 14.5: Atomic stability, thanks to the uncertainty principle. An electron is known to be confined inside a hollow cubic of side length 2a so that a reasonable measure of its position uncertainty is . x y z a∆ = ∆ = ∆ = (a) In terms of a, the electron mass m , and , calculate the typical kinetic energy this electron will have. (You now know how to calculate typical magnitudes for the electron's momentum components. Use these to calculate the electron's kinetic energy.) You should find that

when a is expressed in nanometers.

e

( ) ( )2 23 8 .029 eV nmeK m a ≈ ⋅∼ 2a

228


(b) Here's a not-nearly-as-stupid-as-it-seems model for a hydrogen atom. Assume the electron in a hydrogen atom is confined by the Coulomb potential to rattle around inside a cube of side length 2a so that the electron's typical potential energy is . (q is the fundamental

charge; when a is expressed in nanometers.) As a result, our simple model predicts that the electron's total energy should be

( )204q aπε−

( ) ( )204 1.44 eV nmq aπε− ≈ − ⋅ a

.

The size of the volume in which the electron resides will be that which minimizes its total energy.

2

.029 eV nm 1.44 eV nm - E K Ua a

⋅ ⋅= + ≈

Sketch a graph of the electron's total energy as a function of a.

229


(c) Calculate the value of a which minimizes the electron's energy. (d) Calculate the electron's total energy at this preferred value for a.

)

Discussion Atoms don’t collapse: lower (more negative) potential energy is more than offset by the rising kinetic energy associated with better localizing the electrons! The full-blown quantum mechanical treatment of the hydrogen atom reveals that the probability to find the electron near the point ( is proportional to e . (The value of .053 nm is called the .) The energy needed to remove a bound electron from a hydrogen atom is 13.6 eV.

, ,r θ φ .053 nmr−

Bohr radius

Your results for the atom's size and the energy of the electron should have been surprisingly close to these values!

230


Discussion: what’s really going on It is not possible to describe the world correctly if we insist on using a theory which describes a system in terms of the positions and momenta of its constituent particles. The predictions of any such theory are guaranteed to be wrong whenever the theory crashes into the limits on precision set by the uncertainty principle. It’s a deeper problem than not just being able to the position of a particle without disturbing its momentum: a variety of effects which are absolutely impossible in a classical theory become allowed in quantum mechanics.

measure

We are forced to abandon a theory which makes statements like these:

• (which can be written ); F ma=dpVdt

−∇ =

• (which can be written ). K U E+ =2

2p U Em

+ =

In Newtonian mechanics we have the habit of solving directly for functions which tell us the values of the particle’s kinematic parameters. For example, in a uniform gravitational field,

( ) 20 0

12

x t x v t at= + + .

This doesn’t work for quantum systems! Our classical description assumes that we are able to say exactly where the particle is at a particular time. No good! We need new ways of thinking, described by a new equation. The idea is this: we need to give up an approach in which we talk directly about the values assumed by kinematic parameters as functions of position and time. Instead we need to work with the that we might find particular values for the various parameters if we happen to make a measurement. More exactly, we need to work with the associated with quantities like position, momentum, and energy.

probabilitiesprobability amplitudes

We’ll use our new equation to calculate the form of a particles : a function which contains information the particle. (It’s like an owner’s manual for a stereo: the manual contains information the stereo. The owner’s manual is NOT the stereo, just an information source the stereo.) The way we extract information about the particle from its wavefunction is this:

wave functionaboutabout

about( , , , )x y z tψ

• is the probability to find the particle in a small volume element dV centered

at position at time t. ( ) 2

, , ,x y z t dVψ( , , )x y z

231


• is the energy we will obtain if we measure the energy of many

identical systems

( )all

space

didtψψ ∗

∫ dV average

• is the x component of the momentum we will obtain if we

measure the energy of many identical systems

( )all

space

ixψψ ∗ ∂ − ∂ ∫ dV average

In general, can be complex: is the complex conjugate of .

We refer to as the “energy operator” and − as the “x-momentum operator.”

For three-dimensional systems, the momentum operator is .

( , , , )x y z tψ ψ ∗ ψ

it

∂∂

ix

∂∂

i ix y z

∂ ∂ ∂− + + = − ∇ ∂ ∂ ∂

We can replace momentum and energy in our equation: K U E+ =

2 2 2

22 2p dU E U im m x

∂+ = ⇒ − + =

∂

22

2dU i

m d− ∇ + =

dt in one dimension and in three.

t Clearly, we need to add something on which the differentiations can act: the correct choice is the wavefunction. This yields the “master equation” for quantum mechanics, called Schrödinger’s equation:

2 2

22d U i

m dx dtdψ ψψ− + =

22

2dU i

m dtψψ ψ− ∇ + = in one dimension and in three.

Solving this for will provide us with all the information about a nonrelativistic system that it is possible to obtain.

ψ

Summary and wrap-up discussion Here’s what you were working on: • a classical analysis of the rate at which electrons “should be” ejected from a conductor which

is exposed to an intense light source. • a classical analysis of the relationship between the maximum kinetic energy of electrons

ejected from a conductor and the intensity of the light source illuminating the conductor. • photons! (photon energy and momentum) • modeling the resolution of a simple magnifier

232


• analyzing the “Heisenberg’s Microscope” thought experiment • estimating atomic sizes using the Uncertainty Principle We also discussed the master equation of quantum mechanics, known as Schrödinger’s equation. That's it!

233


234

Physics 212, Spring 2003. .1 Notes on special relativity

Physics 212


Notes on special relativity

Spring, 2003


2003

235


Notes on Special Relativity

George Gollin University of Illinois

Contents The speed of light is finite ............................................................................................................1 The speed of light is constant, regardless of relative motion of source and observer ..................4 Nonsimultaneity of events viewed from different reference frames ............................................7 Things on which we can agree when comparing notes with other observers.............................12 Time dilation...............................................................................................................................16 Doppler shifts..............................................................................................................................18 Lorentz contraction .....................................................................................................................22 Quantitative description of nonsimultaneity...............................................................................24 Some simple paradoxes and their resolution ..............................................................................27 Coordinates of an event, viewed from different reference frames..............................................28 Time dilation in terms of events .................................................................................................29 Lorentz contraction in terms of events .......................................................................................30 Lorentz transformations ..............................................................................................................31 Addition of relativistic velocities................................................................................................33 Causality .....................................................................................................................................35

The speed of light is finite All the strange features of relativity are consequences of the fact that the speed of light is exactly 2.99792458 × 108 meters per second. It doesn't matter whether the source of light moves with respect to the measuring apparatus: any device measuring c (the speed of light) will obtain, within experimental error, this value. Nothing else works this way: sound travels at fixed speed with respect to the air, for example. The value 2.99792458 × 108 meters per second really is exact: it serves as a definition, along with some time standard, of the length of one meter. The speed of light is very close to 1 foot per nanosecond, and I'll use that as a convenient approximation periodically. Let's start by discussing how moving objects look to a single, stationary observer. Because the speed of light is finite, an object moves between the time light rays leave it and the time they arrive at an observer. For example, here's a diagram of what happens when a light bulb, traveling at 0.2 c, is seen by an observer:

236


v= 0.2c

true position

apparent position

10 feet

2 feet

. Things are more complicated when the light comes from an extended source instead of a point source. Imagine we put lights on the corners of a transparent rectangular block and observe it from far away like this:

1 foot Light from the back of the block takes about a nanosecond longer to reach the observer than light from the front. If the block is in motion, the observer will see its back at an earlier time from its front:

cube is moving up

. As a result, the block looks like a parallelogram and not a rectangle:

. Because the speed of the block is greater than the component of the light ray’s velocity along the cube’s direction of motion, light from the far, back corner doesn’t travel through the transparent

237


cube on its way to the observer. If the block had been opaque instead of transparent, light from the front side would not have been visible while light from the back side would have been. (Imagine a camera shutter.) There's nothing exotic about this: the same effect would happen if you tried to locate airplanes in formation at an air show by listening for their engine sounds. Objects moving towards or away from an observer will also appear distorted. For example, consider what happens when an observer looks at a square plate, as shown below:

2 feet

4.123'

4.000'

4.243'

2 feet

The center of the square is exactly 4 feet from the observer, while the centers of the sides and the corners are 4.123 feet and 4.243 feet away, respectively. Light rays leaving the square's center and corners simultaneously won't reach the observer simultaneously. If the square moves towards the observer, he'll see the corners of the square at an apparent position that corresponds to a time that's 0.243 nanoseconds earlier than the apparent position of the center of the square. A diagram, assuming the square moves with speed close to c (not really accurate-- the square looks curved in two dimensions):

apparent shape of thesquare's surface

0.243'

side view

. The outline of the square isn't "square" anymore, either:

238


. What does the square look like if it is moving away from the observer? What does a transparent rectangular solid look like when it moves towards the observer at a speed close to c? Keep in mind that the unusual appearance of the objects we’ve been discussing just arises from the finite speed of propagation of light. If several observers position themselves as shown below, they’ll all agree that each was passed by the closest part of the square at the same time (call it t=0).

In a sense, the objects just “look funny” but aren’t really distorted due to the finite speed of light.

The speed of light is constant, regardless of relative motion of source and observer

This is the tricky part in relativity. Let’s first consider how sound works, to show the contrast with the way light propagates. Sound travels at a fixed speed with respect to still air. The speed of sound is about 331 meters per second; let’s approximate this as 1000 feet per

239


second or one foot per millisecond. Imagine that an observer on the ground measures the propagation speed of the engine noise from a subsonic airplane:

1 2

100 feet Even though the airplane is moving, it will take about 100 milliseconds for any particular sound wave to travel between the two measuring stations labeled 1 and 2. The speed of sound is independent of the speed of the airplane. If we put the observer on a train, so that he’s moving towards the source of the sound with speed v, the time for a particular sound wave to travel from the first sensor to the second will be reduced.

1 2

v

From practical experience, we know that snowballs don't work the same way sound does. A snowball thrown from a car hits a pedestrian a lot harder if it's thrown in the direction the car's traveling than if it's thrown in the opposite direction. Light has the (unusual??) property that any measurements made of its propagation speed always yield the same answer, regardless of relative motion between light source and observer. Each of the following observers will measure the same transit speed between the two sensors:

240


1 2

v=0.9c

laser beam from rocket 100 feet

(Here we see the source moving towards a fixed observer.)

1 2

v=0.9c

laser beam from rocket 100 feet

(Here we see the source moving away from a fixed observer.)

light ray from a distant star

v = 0.8c withrespect to star

21

(Here we see the observer as moving towards a fixed source.) This is not what we’re used to-- either in the way sound travels or in the way snowballs thrown from cars behave. You can read about some experimental tests of this in a variety of textbooks. Here's another, based on every-day things at Fermilab. We know that many different kinds of unstable elementary particles can be produced in violent collisions between high energy protons and target nuclei. The lightest of the particles made of quarks and antiquarks which can be produced this way are call pi mesons, or pions. There are three different kinds of pions: π+ and π- have identical masses of 139.57 MeV/c2 while the lighter π0 has mass 134.96 MeV/c2. (The proton mass is 938.28 MeV/c2.) It is known that nearly 100% of the time a π0 will decay into a pair of energetic photons (the quantum

241


mechanical version of electromagnetic waves-- light). Each gamma ray photon from a pion which decays at rest in the lab carries away 67.48 MeV of energy and travels travel with speed c (which isn't surprising). A source of energetic pions at Fermilab shoots π0's into my experiment, which measures the kinematic properties of their decays. These pions are traveling close to the speed of light-- useful π0's have values of v/c which range from 0.9996 to 0.9999993-- and their decay photons can have energies as high as 100 GeV (1 GeV = 1000 MeV). A diagram comparing a typical π0 decay at rest with what we observe at Fermilab follows:

pion with v/c = .999990

angle between the photons is about 1/2°

pion at rest

photons are180° apart

Even though the source of the photons is moving at high speed, we find that the speed with which these photons travel is exactly the same as the speed with which the photons from the stationary pion travel.

Nonsimultaneity of events viewed from different reference frames

The consequences of the constancy of the speed of light are dramatic. Here's the first of them. Consider the following hypothetical encounter between the rockets Nostromo and Exeter. Both are capable of traveling at speeds approaching c, and each carries a pair of periscopes. A spark jumps between the Nostromo’s and Exeter’s electrode as the two ships pass each other. Observers on each ship, equidistant from the electrode, see the light from the spark. Let’s view the creation of the spark from the perspective of passengers on the Nostromo. The spark is created at the electrodes when the Nostromo’s clocks read t = 0, as shown in the figure below.

242


N

Nostromo

EExeter

spark gap

?

periscopesperiscopes

?

As light from the sparks travels towards the Nostromo’s observer, the Exeter continues to move to the left. Here’s a “motion picture” of what happens, from the perspective of passengers on the Nostromo. Let’s assume that the Exeter is traveling with v = 0.5c.

243


Nostromo clocks read t=0 as the Exeter passes. Light from spark leaves spark gap. Nostromo crew thinks Exeter is moving to the left with v = 0.5c Nostromo clocks read t=100ns. Light from spark has traveled 100 feet, Exeter has traveled 50 feet. Light reaches aft Exeter periscope. Nostromo clocks read t=200ns. Light from spark has traveled 200 feet, Exeter has traveled 100 feet. Light has already reached both Nostromo periscopes and aft Exeter periscope.

NNostromo

EExeter

? ??

NNostromo

EExeter

? ??

NNostromo

EExeter

? ?

lightpulse

lightpulse

244


As you can see, the Nostromo’s observers see the spark simultaneously (light travels one foot per nanosecond, and the periscopes are equidistant from the spark gap). However, the Nostromo crew members think the Exeter’s aft observer sees the spark before the forward observer sees the spark. Let’s shift to the Exeter’s frame of reference now. As far as the Exeter’s crew is concerned, they are happily sitting around eating dehydrated food when the Nostromo glides past them, moving to the right with v = 0.5c. The Exeter’s periscopes are equidistant from the spark electrode, and will necessarily see the produced spark simultaneously, as far as the Exeter's crew is concerned. This is in disagreement with the conclusion of the crew of the Nostromo. Here is a "movie" of what things look like from the perspective of the Exeter’s crew. (There are certain inaccuracies in the drawing concerning the relative lengths of the ships that we'll neglect for the time being.)

245


Exeter clocks read t=0 as the Nostromo passes. Light from spark leaves spark gap. Exeter crew thinks Nostromo is moving to the right with v = 0.5c Light is about to reach aft Nostromo periscope. Light has already reached both Exeter periscopes and aft Nostromo periscope.

NNostromo

EExeter

? ?

? ?

NNostromo

EExeter

? ? ?

NNostromo

EExeter

? ? ?

lightpulse

lightpulse

Things are rather different in the Exeter's reference frame. So-- events which are simultaneous in one frame of reference are not necessarily simultaneous in another frame of reference!! Curious!

246


In a ship's rest frame, its clocks read the same value when the spark is produced (since they're synchronized) and also read the same value when the light from the spark arrives at the periscope. (The light travels the same distance from the spark gap to the forward periscope as it does to the aft periscope.) Imagine that each clock stops (breaks??) when the light reaches it. In a ship's rest frame, its own clocks will read the same value when they break. An observer in any frame of reference will agree with the ship's crew that the broken clocks read the same (stopped) time. A ship's crew will conclude that the aft clock on the other ship breaks before the forward clock on the other ship. However, once both its clocks are broken, they will be seen to read the same time. Consequently, as seen from the rest frame of one ship, the aft clock on the other ship reaches this time before the forward clock on the other ship. The crew of one ship, looking at the clocks on the other ship (which is seen to be moving) will conclude that the clocks on the other ship are not synchronized. In particular, the aft clock is set ahead (reads a later time) relative to the forward clock. Both crews will agree that the clocks at the spark gap (the center clocks in the pictures) read zero when the spark was created. (Imagine that these clocks break when current flows through the spark electrodes.) Here's how the clocks will look in the rest frame of the Exeter:

NNostromo

EExeter

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Things on which we can agree when comparing notes with other observers

The disagreement between the Exeter’s and Nostromo’s crews' observations about the (non)simultaneity of the observations of the spark comes about because of the fact that the observations take place at different points in space. Events which take place at the same point in space and at the same time will be seen by both crews as happening simultaneously. Here are some examples of pairs of events which work out this way: (Nostromo’s electrode passes Exeter’s electrode) .and. (Nostromo’s center clock reads t=0) (Nostromo’s rear clock breaks) .and. (light from spark arrives at Nostromo's rear periscope) (Exeter’s front clock breaks) .and. (light from spark arrives at Exeter's forward periscope) (I'm assuming that all clocks are tiny, and arbitrarily close to the electrodes or periscopes, etc. etc.) It’s easy to imagine how this works: imagine that one event (light arriving) causes some sort of explosion which breaks the clock immediately. All observers will agree on the value read by the (stopped) clock.

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Here is another example of a “reference frame-invariant” occurrence. I’ve underlined the statement with which all observers will agree.

Left ruler is stationary, right ruler is moving. Rods at ends of right ruler pass through hoops at ends of left ruler when the rulers collide.

Left and right rulers are moving towards each other with equal speed. Rods at ends of right ruler pass through hoops at ends of left ruler when the rulers collide.

Note that this means that the two rulers will appear to have the same length, whether viewed from a frame in which one is at rest or in which both are moving (perpendicular to their lengths). Here’s another example: let’s say the Nostromo carries a pair of parallel mirrors. A crewmember sets a light beam bouncing between the mirrors.

Nostromo, as seen by a crewmember floating outside the rocket.

N

Nostromo

Light ray leaves bottom mirror

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N

Nostromo

Light ray bounces off top mirror

N

Nostromo

Light ray bounces off bottom mirror

The light beam bounces from mirror to mirror.

Nostromo, as seen from the Exeter.

N

Nostromo

Light ray leaves bottom mirror, moving up and to the right

N

Nostromo

Light ray bounces off top mirror, now moving down and to the right.

N

Nostromo

Light ray bounces off bottom mirror, moving up and to the right

The light beam bounces from mirror to mirror as the Nostromo moves to the right.

Time dilation Let’s examine the light-bouncing-off-a-mirror case in more detail. For convenience, let's assume that the mirrors are three feet apart. The roundtrip pathlength between the bottom and top mirrors is six feet; it takes a light beam about six nanoseconds to make the trip, as measured by a crewmember of the Nostromo. The path taken by the light is longer, when viewed by crewmembers of the Exeter who see the Nostromo as moving quickly to the right. Imagine that

250


the relative velocity between the Nostromo and Exeter is 0.8c. Here's how the light beam seems to move:

Nostromo speed is 0.8c

3 feet

An important point: an observer in either rocket who measures the speed with which the light beam in Nostromo’s clock travels will always find that it travels 2.997...× 108 meters/second. Let's calculate the round trip time (assuming c = 1 foot per nanosecond) for the moving mirrors. Let's say that the time for a half-tick is ∆t', so that the round trip takes 2∆t' (note the prime here), while the time for a half-tick of a stationary clock is ∆t (no prime): In time ∆t', the mirrors travel ' 'x v t∆ = ∆ . In this time, the light ray travels 2 2 2' ( ') ( ') ( ') (3)L x y v t= ∆ + ∆ = ∆ + 2 . Since c = 1ft/nsec in all frames, L' = c∆t' so... 2 2' ( ') (3)c t v t∆ = ∆ + . It's true that c∆t = 3 feet (no primes here: in the frame in which the clock is at rest), so we can rewrite 2' ( ') ( )c t v t c t∆ = ∆ + ∆ 2 . Solve for ∆t' in terms of ∆t:

2 2 2

2 2( ') ( ') ( ) '

1 /tc t v t c t or t

v c∆

∆ − ∆ = ∆ ∆ =−

.

Observers think the moving clock takes longer to tick by a factor of 2 21 /v c−1/ than do observers at rest with respect to the clock. The clock seems to be running slowly compared to a "normal" clock. When v=0.8c, 2 21 / 5 / 3v c− =1/ . The Nostromo's and Exeter's crews will not agree about the time required for Nostromo's "light clock" to tick once. You can imagine how this analysis works if the light clock had been on the

251


Exeter instead of the Nostromo: The Nostromo's crew would think the Exeter's light clock was ticking too slowly. There's complete symmetry between the Exeter and Nostromo. Each crew sees the other crew's rocket moving at high speed. In general, a light clock which requires time t for each tick (as viewed by observers stationary with respect to the clock) will require a time

2 2' 1 /t t v c= − for each tick when viewed from a frame in which the clock seems to be moving with

speed v. Here's a plot of t'/t as a function of v/c:

0 0.2 0.4 0.6 0.8 10

2

4

6

8

10

12

14Time dilation factor vs. v/c

v/c

2 21 1 v c−

So-- moving light clocks seem to run slow! What about other moving clocks, for example mechanical or biological clocks? They all have to seem to run slowly! Here's a plausibility argument for this. Let's say we stretch the distance between the mirrors in the Nostromo's light clock so that it takes one second, as determined by the Nostromo's crew, for the clock to tick. Imagine that one crewmember, whose pulse rate is 60 beats per minute, covers the bottom mirror with a black cloth, removing the cloth briefly each time his/her heart beats. The light ray bouncing back and forth will always arrive at the bottom mirror when it is uncovered, and the "clock" will continue to tick. A crewmember of the Exeter will see the Nostromo’s crewmember uncover the mirror in time with his/her heartbeat, even though the time between bounces will be longer than one second. The Exeter’s crewmember will conclude that the Nostromo’s crewmember’s pulse has slowed down by the same factor as

252


the rate of ticking of the light clock. All clocks, whether mechanical, biological, or electrical must slow down by the same factor. This “time dilation” effect is entirely consistent with observations of changes in average lifetimes of unstable particles which are moving at high speed. Here’s an example, based on what we see at Fermilab. K mesons are unstable particles which can be produced in collisions between protons and target nuclei at particle accelerators like Fermilab (which is in Batavia, about an hour’s drive west of Chicago.) A K meson weighs about half as much as a proton. There are two kinds of neutral K mesons and two kinds of charged K mesons; a “long-lived” neutral K, or KL, has an average lifetime of 51.83 nanoseconds. KL‘s can decay into various combinations of lighter particles such as pions, electrons, neutrinos, and muons. Fermilab shoots a beam of 1011 800 GeV protons per second into a beryllium rod located 128 meters “upstream” of my experiment. About 106 KL per second leave the target and travel towards the experiment. The K mesons are traveling close to the speed of light; typical values for their time dilation factors range from 50 to 300. If time dilation didn’t slow down the KL‘s internal clocks, the average number surviving after a time interval T would be 51.83T−e . Traveling at one foot per nanosecond, only 0.03% of the kaons (or about 300 per second) would survive to reach my experiment. In reality, more than 90% of the kaons manage to travel the 128 meters without decaying in flight.

Doppler shifts What would happen if we tried to make a “sound clock” which worked by bouncing the sound of a click between two reflectors mounted on a railroad flat car? Would we get the same sort of time dilation phenomena? (Of course we wouldn’t.) Let’s do the analysis. In the picture below, the reflectors are moving about half the speed of sound:

331m/sec

click!

stationary reflectors

331m/sec

click!

moving reflectors

253


Because the speed of sound is 331m/sec with respect to still air, it will take longer for the sound of the click to arrive at the upper reflector. Everyone will agree that the sound clock moving with respect to still air is ticking slowly. This time dilation effect will make Doppler shifts for relativistic motion work somewhat differently from Doppler shifts for nonrelativistic systems. Remember about Doppler shifts? Here’s how they work for sounds coming from nonrelativistic sources. Imagine a train emits periodic clicks, as in the following diagram:

1 2 3 4 5

1 2 3 4 5

331 m/sec

The wavefronts from the clicks travel away from their points of creation at 331 m/sec. A listener in front of the train will hear a shorter period between clicks than a listener behind the train. Let’s say the train is moving at half the speed of sound, emitting 331 clicks per second, and that the first click is emitted at x=0. Here’s a table describing what observers at x = +10 and x = -10 hear:

click (x,t) for emission of click

distance to x = +10 observer

arrival time at +10 observer

distance to x = -10

observer

arrival time at -10 observer

1 x=0, t=0 10m .0302 10m .0302 2 x=0.5, t=.003 9.5m .0317 10.5m .0347 3 x=1, t=.006 9m .0332 11m .0393 4 x=1.5, t=.009 8.5m .0347 11.5m .0438 5 x=2, t=.012 8m .0363 12m .0483

Because the source of the clicks is moving, the frequency with which the clicks arrive is shifted, due to the different distance traveled by each click. For a stationary observer listening to a

254


source which moves a distance D between clicks, the extra distance traveled by successive clicks is -Dcos(θ):

D

θ extra distance is

Dcos(θ)

observer If the source emits clicks separated by time t, it will move a distance D = vt between clicks. The difference in the transit time of successive clicks will be -Dcos(θ) / c where c is the speed of sound. As a result, the ratio of the apparent time interval between clicks to the “true” time interval will be

(t - Dcos(θ)/c) / t = (t - vtcos(θ)/c) / t = 1 - vcos(θ)/c.

If you prefer to deal in frequencies instead of periods, you can use the fact that frequency is the same thing as 1/period to conclude that

fapparent / fsource = c / (c - vcos(θ)). The same thing holds true for sinusoidal waves-- we’d just be describing the time between peaks, instead of time between clicks. There’s one other effect that has to be included when considering relativistic motion-- time dilation. Imagine the Nostromo transmits flashes of light, evenly spaced by a time interval t as measured on the Nostromo’s clocks. An observer on the Exeter will measure a different time interval between the flashes due to the same effect as described above for sound, but with an additional complication: the Nostromo’s clocks seem to be running slowly by a factor (1 - v2/c2)1/2. Naturally, c now represents the speed of light. The time interval between flashes is increased, or the frequency decreased, by this amount. As a result, the apparent frequency of flashes seen by the Exeter is

255


2 2apparent source 1 .

cos( )cf f v c

c v θ= × − ×

−

Some algebra reveals that this is equivalent to

apparent sourcec vf fc v

−= ×

+

for the case that the source is moving directly away from the observer and

apparent sourcec vf fc v

+= ×

−

for the case that the source is moving directly towards the observer. Notice that many of the algebraic expressions describing the effects associated with motion at high speeds misbehave when v = c. The fact that any observer measuring the speed of a light ray will determine it to be 2.997...× 108 m/sec makes it impossible for any massive object to travel faster than the speed of light. Here’s an illustration of why this is the case. Imagine that a nasty spaceship carrying a large bomb is constructed, as shown in the diagram below. The bomb will detonate if the attached photosensor is struck by light from the rocket’s internal laser. The laser shoots its beam upwards through a tube with black interior walls, from the perspective of the crew of the rocket.

Bombs 'r'Us

hollow tube withblack walls

Dr. Death Lasers, Inc.

Someone traveling on the rocket ship will get blown up a few nanoseconds after the laser is turned on, as measured by the ships clocks. If the rocket can travel faster than the light beam from the laser, when both are viewed from some other reference frame, the laser will never be able to detonate the bomb: the top of the tube will have moved horizontally farther than the laser beam, even if the beam travels horizontally, and the beam will be absorbed by the walls of the black tube. However, in the rest frame of the spaceship, the bomb certainly will go off. We can’t reconcile these two descriptions of events: either the ship blows up on its way to Alpha

256


Centauri, or it doesn’t. Later, we’ll see why it takes an infinite amount of energy to boost an object at rest to the speed of light.

Lorentz contraction Things we know: 1. A moving clock seems to run slow. 2. Two clocks synchronized in their rest frame appear to be out-of-synch when viewed from another frame. In particular, the front clock of a pair of moving clocks seems to be set back while the rear clock seems to be set ahead. What happens to (apparent) lengths of relativistic objects? Careful: you must state exactly what you mean to do when you say you'll measure something's length. One way to measure the length of a moving object:

N

Nostrom

Solaris

v

see which clocks are adjacent to the "moving ship's" nose and tail when t = 0 on all your clocks; call the distance between these two clocks the length of the moving object. Another way: use one clock and the fact that distance = speed × time; the moving ship is known to travel with speed v.

257


Solaris

N

finishSolaris

Nostromov

start

From the Nostromo's perspective:

N

Nostromo

Solaris

... ...v

First method (two clocks): Solaris' clocks are not synchronized so their measurement is not in agreement with the Nostromo's. Second method (one clock): Solaris' single clock runs slow by a factor of 2 21 v c− so its measurement is "inaccurate" due to this: it measures a length which is shorter than the Nostromo's "rest length" by this same factor. Both methods still have to agree, although they will produce a value which is smaller than the length determined by the Nostromo's crew.

258


Quantitative description of nonsimultaneity Let's calculate the amount of missynchronization in a pair of moving clocks (which are synchronized in their rest frame). The rest length of the ship is Lship.

1(in ship's rest frame)

vv


vv


vv

A B

A B

A B Clocks A and B are synchronized in their rest frame. Their spacing is chosen so that in their rest frame each reads zero when passed by the corresponding end of the space ship. Let's call their separation in their rest frame LAB. We will do this:

1. calculate the distance from clock A to the tail of the ship in picture 1, in the ship's rest frame

2. calculate how long it took in the ship's rest frame for clock A to travel this distance

(this is how much time has passed on the ship between pictures 1 and 2) 3. calculate the change in the reading on clock A between picture 1 and picture 2.

Since we know clock A reads zero in picture 2, this will let us calculate what it read in picture 1.

259


1. The clocks are positioned so that their rest-frame separation LAB matches the apparent length of the ship. In the clocks' rest frame, the ship is Lorentz contracted so

2 21 /AB shipL L v c= − . In the rest frame of the ship, the separation between the (moving) clocks is Lorentz

contracted. Call this L'AB: we have

( )2 2 2 2 2 22

' 1 / 1 / 1AB AB ship shipL L v c L v c L v c = − = − = −

/ .

As a result, the remaining distance to be traveled by clock A is

2

2' shipship AB

L vL L

c− = .

An illustration:

(in ship's restf )

vvA B

ABL'AB- L'Lship

Lship

1

2. In the ship's rest frame, the clock takes an amount of time 1 22

'ship AB shipship

L L Lt

v c→ −

= =v

to travel this distance. 3. The amount of time which passes on the face of clock A between pictures 1 and 2 is

reduced, because of time dilation:

2 2

1 2 1 2 2 22 2

1 /1 /

shipAB

AB ship

L v c v L vt t v cc c

→ →−

= − = = .

260


Since clock A reads zero in picture 2, it must have read 2ABL vc

− in picture 1. Recall that LAB is

the separation between the clocks in their rest frame, and that clock B read zero in picture 1. As a result, the amount of "missynchronization" between two clocks separated by a distance ∆x in their rest frame, when viewed from a frame in which the clicks are in motion, is v∆x/c2. The front clock reads an earlier time than the rear clock. Here's a summary of what we know about the effects of relative motion on distance and time:

v

v

L

From the perspective of the TOP ruler

x=0 x=+Lx=-L

t = 0 t = 0 t = 0

t = 0

t = 0 t = -vL/c2

t = +vL/c2

t = +vL/c2

t = -vL/c2

The top ruler is stationary with respect to us. The middle ruler is moving to our left, the lowest ruler is moving to our right. In each ruler's rest frame, the separation between clocks is L and the three clocks are synchronized. 1. Effect on distance between objects: The separation between adjacent clocks seems reduced. Clocks on the moving ruler will seem to be a distance apart given by

2 21L v× − c . 2. Effect on passage of time on a single clock: The rate of passage of time on moving clocks seems slowed. If a clock on the stationary ruler advances by an interval T, a clock on the moving ruler will advance by an amount of time

261


2 21T v× − c . 3. Effect on synchronization of clocks: Moving clocks which are synchronized in their rest frame will not appear synchronized when viewed from another frame. In general, a clock that is "in front" will seem to be set to an earlier time than a clock that is "in back". In the example with the rulers illustrated above, a clock at x = L (as measured in its rest frame) will read 2v c= −t L when the moving clock at x = 0 reads t = 0. Note that v < 0 corresponds to motion to the left.

Some simple paradoxes and their resolution I. Relativistic train/tunnel Train: rest length 1000 feet, travels with v/c = 0.8. Tunnel: rest length = 800 feet. Stationmasters try to slam entrance and exit doors to the tunnel simultaneously while the train is inside, trapping it in the tunnel.

Illinois CentralIllinois CentralIC

1000 foot rest length

800 foot rest length

doordoor

Illinois Central

a. Why does this seem like a paradox? b. What happens? Describe from both the train frame and the tunnel frame. II. Clock accelerators A "clock accelerator" consists of a vacuum pipe and a series of accelerating structures, as shown in the diagram. The vacuum pipe contains two rows of clocks; the upper row is moving to the

262


right at speed v = 0.8c while the lower row is stationary with respect to the accelerating structures. The accelerating structures (and stationary clocks) are 1000 feet apart. All clocks in a given rest frame are synchronized. The rightmost moving clock reads 0 as it passes the stationary clock, which also reads 0. Just as the stationary clocks read 0, all the accelerating structures fire, boosting the lower row of clocks into the same frame of reference as the upper row of moving clocks. Because the acceleration takes place in an extremely short period of time (let's say a picosecond), none of the clocks have moved an appreciable distance before the acceleration ends. In addition, the readings on the faces of all clocks are virtually unchanged compared with their pre-acceleration values. As a result, the apparent spacing between clocks in the (now moving) lower row will STILL be 1000 feet, and all these clocks will STILL read 0. How is this possible? What happened to all that Lorentz contraction stuff we've been discussing?

??

accelerating structures

Beam pipe

1000 feet

Coordinates of an event, viewed from different reference frames An event is something that happens at one point in space, and one instant in time. Examples: collision of clocks on board space ships Exeter forward clock reads 0 as the clock passes the Nostromo's aft clock Not an event: Exeter crew measures the length of the Nostromo (takes a finite time interval or finite distance interval ) Why is this event business useful? Time dilation, Lorentz contractions, etc. etc. can be described in terms of their effects on the separation between events when one shifts reference frames. These are special cases where either the space or time interval between events is zero in ONE frame of reference. For the general case where ∆x ≠ 0 and ∆t ≠ 0, describing measurements in terms of space-time intervals between events will make it easier for us to calculate correctly.

263


Time dilation in terms of events event 1: Nostromo nose and Solaris clock coincide and Solaris clock reads 0. event 2: Nostromo tail and Solaris clock coincide. Notice that these events happen in the same place, according to Solaris ...

Solaris

N

finishSolaris

Nostromov

start

Event 1 Event 2

...but occur in different places, according to the Nostromo:

N

Nostromo

Solaris

v

Event 1

264


N

Nostromo

Solaris

Event 2

If the time interval between two events in the frame in which the events occur at the same place is T, the time interval between the events in a different frame (in which the events occurred at different locations) is larger, namely 2 21 v−T . c

Lorentz contraction in terms of events

Event 1: Nostromo's tail passes by Solaris' forward clock. Event 2: Nostromo's nose passes by Solaris' aft clock. From the perspective of the Solaris, the events occur at the same time...

event 1

N

Nostromo

Solaris

vevent 2

...but from the perspective of the Nostromo, they don't.

265


If the distance interval between two events in the frame in which the events occur at the same time is L, the distance interval between the events in a different frame (in which the events occurred at different times) is larger, namely 2 21L v− c . Why bother? •Avoid sloppy thinking this way: describe all measurements in terms of intervals between events (back of rocket coasts past one part of my tape measure, front of rocket coasts past another part of my tape measure). •Also: gives us a natural language with which to describe things which happen neither at the same time nor at the same place in our reference frame, when viewed from other reference frames.

Lorentz transformations Let's work up a description of this: two events are separated by distance x, and time t in one frame; separated by a distance x' and a time t' in another frame. Imagine we are at rest with respect to a string of clocks as shown below. A second string glides past the first, moving with speed v. Two clocks in each string are attached to spark electrodes; a spark is created as the corresponding electrodes align. The left electrode in the (to us) stationary string is at x=0, the right electrode at x=L. In the rest frame of the moving string of clocks the left and right electrodes in this string of clocks are at x'=0 and x'=L'. In general, L and L' can be different. (For example, the electrodes are placed 11 clocks apart on the lower string, but 10 clocks apart on the upper string.) Event 1: the left clocks' spark electrodes touch and a spark is created.

? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?v

Event 1

x=0, t=0 x=L, t=0

x'=0, t'=0 x'=L', t'=?

Coordinates of event 1 from the perspective of the lower string: x=0, t=0. Coordinates of event 1 from the perspective of the upper string: x'=0, t'=0.

266


From the perspective of the lower clock string, the upper string's right clock is at 2 21x L v c′= − (note that I am using x, not x' here), and it reads 2L ct v′ ′= − (note that I am

using t', L', not t, L here).

Moving clocks look to be out-of-synch, and moving objects look shorter, right? Event 2: the right clocks' spark electrodes touch and a spark is created.

? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?v

Event 2

x=0, t=T x=L, t=T

x'=L', t'=??x'=0, t'=?

Coordinates of event 2 from the perspective of the lower string: x=L, t=T. Coordinates of event 2 from the perspective of the upper string: x'=L', t'=??. Let's sort out what the upper string's right clock is reading. We know it was reading -vL'/c2 when event 1 occurred. We know that a time interval T has passed on the lower string's clocks. We know the upper string's right clock is ticking more slowly than the lower string's clocks. As a result, the right clock must be reading 2 21t vL c T v c′ ′= − + − 2 . Do some work to replace L' with quantities known to the bottom clock string observers: Distance between bottom clocks is the same as the sum of the Lorentz-contracted rest-frame separation between top clocks and the distance the top clock string moves in time T. In

equations: 2 21L L v c vT′= − + or 2 21

L vT Lv c

− ′=−

.

Substitute this into the previous equation to get

22 2 2

2 2 2 21

1 1T vL cL vTt v c T v c

v c v c

−−′ = − + − = − −

after some algebra.

Collecting what we know, we can conclude the following about the space/time interval between the events, when viewed from either frame:

267


interval bottom string's frame top string's frame

space

2 1x x x L∆ = − = 2 1 2 21L vTx x x L

v c−′ ′ ′ ′∆ = − = =

−

time

2 1t t t T∆ = − =

2

2 1 2 21T vL ct t t

v c−′ ′ ′∆ = − =−

Slightly more compactly (and after some algebra to express ∆x, ∆t in terms of ∆x', ∆t'), we have the Lorentz Transformations:

2 21x v tx

v c∆ − ∆′∆ =

−

2

2 21t v x ct

v c∆ − ∆′∆ =

−

2 21x v tx

v c

′ ′∆ + ∆∆ =

−

2

2 21t v x ct

v c

′ ′∆ + ∆∆ =

−

It takes a bit of thought to keep the sign straight in front of the term proportional to velocity. Note that v refers to the velocity of the reference frame which measures "primed" quantities when viewed from the reference frame which measures "unprimed" quantities. Intervals along directions perpendicular to the relative velocity between the frames agree in both frames: ∆y' = ∆y, ∆z' = ∆z. A useful shorthand is to define 2 21 1 v cγ ≡ − and v cβ ≡ . Using this, we can write

( )x x c tγ β′∆ = ∆ − ∆ ( )t t xγ β′∆ = ∆ − ∆ c

( )x x c tγ β′ ′∆ = ∆ + ∆ ( )t t xγ β c′ ′∆ = ∆ + ∆ Given the space-time interval between two events as measured in one frame, we can calculate what observers in another frame will measure if we know the relative velocity of that frame with respect to the other frame.

Addition of relativistic velocities Let's use the Lorentz transformation formulas to derive a velocity addition formula. Here's the sort of problem we'll need this for: Imagine the Nostromo's shuttle craft leaves the Nostromo traveling at speed v, with respect to the Nostromo.

268


v

Nostromo

An observer on the Solaris sees the Nostromo coasting to the right with velocity u. What is the shuttle's apparent velocity according to this same observer?

Solaris

Nostromo

u

?

To use Lorentz transformations to sort this out, let's define two events like so: event 1: Shuttle leaves Nostromo and clock at Nostromo's nose reads t' = 0. event 2: Shuttle passes Nostromo crewmember on spacewalk who is at rest with respect to the Nostromo, a distance x' to the right of the Nostromo. The crewmember's watch reads t' = x'/v. The interval between the two events is ∆x' = x'-0 and ∆t' = x'/v - 0 From the perspective of the Solaris, the Nostromo rest frame is moving to the right with speed u, so we can calculate the separation between the two events using the Lorentz transformations

269


(with β = u/c here, not v/c): ( ) ( )x x c t x c x vγ β γ β′ ′ ′ ′∆ = ∆ + ∆ = + and

( ) ( )t t x c x v x cγ β γ β′ ′ ′ ′∆ = ∆ + ∆ = + . The shuttle's speed, according to the Solaris, is ∆x/∆t:

( )( )

( )( )

( )( )2 2

11 1

x c x v u v v uxt x v x c v u c vu c

γ βγ β

′ ′+ +∆= = =

′ ′∆ + + +

+.

In the limit that u → c and v → c, the apparent speed of the shuttle tends towards c, not 2c. Again-- we find that c is the maximum velocity for any object.

Causality Nonrelativistic version: Event 1 can only cause event 2 if event 1 happened before event 2. Problem with this version: nothing can go faster than c. Event 1: You pick up the phone at 9am on January 1, 3000AD and place a LONG distance call while you are coasting through space on the starship Alpha.

Alpha Omega

Event 1: x = 0, t = 9:00 am, Jan. 1, 3000AD. Alpha places call. Event 2: The starship Omega is 24 trillion miles (about four light years) away from the Alpha, but at rest with respect to the Alpha. Clocks on the two ships are synchronized. A phone rings on the Omega at 9:00am, January 1, 3003AD, waking up the Omega's captain.

Alpha Omega

Event 2: x = 24 trillion miles, t = 9:00 am, Jan. 1, 3003AD. Omega phone rings.

Is it possible that you woke up the Omega's captain? Of course not-- the signal from your telephone can't travel faster than c; it'll take slightly more than 4 years for your phone call to arrive at the Omega.

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Relativistic version of causality: event 1 can only cause event 2 if event 1 happened sufficiently early with respect to event 2 so that a light beam leaving x1 at time t1 could arrive at x2 no later than time t2: 2 1 2 1x x c t t− < − . We expect: events which are "causally connected" when described using x, t coordinates measured in one frame of reference will be causally connected when viewed from any other frame of reference. Is this really true? (We can Lorentz-transform to check that ∆x < c∆t → ∆x' < c∆t'.) How does the Alpha-Omega phone call look when viewed from a third ship, which sees both ships in motion? Let's say the ships seem to be moving to the right with speed v equal to βc. Use Lorentz transformations.

Solaris

Alpha Omega

Event 2

2

Solaris

Alpha Omega1

cosmonauts tethered to Solaris

Event 1

You might think that the Solaris could see the events as causally connected if the Alpha-Omega distance is Lorentz-contracted to less than three light years. Shouldn't a light beam be able to make it from the Alpha to the Omega before the Omega's phone rings, in this frame of reference? NO! It won't work. Non-simultaneity is at it again. Sure, The Omega's phone rings when its clocks read 9am, Jan. 1, 3003AD, but the Omega's clock didn't read 9am, Jan. 1, 3000AD when you placed the call from the Alpha. Use Lorentz transformations to work out the ratio of the space and time intervals between the events as seen from Solaris (primed coordinates are measured by the crew of the Solaris):

( )x x c tγ β′∆ = ∆ − ∆ and (t t xγ β′ ∆ − ∆ )c∆ = . After some algebra, one finds that

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( ) ( ) ( ) ( )2 2 2c t x c t x′ ′∆ − ∆ = ∆ − ∆ 2 : two events that are causally disconnected in one frame of reference will be causally disconnected in ALL frames of reference. The quantity ( ) ( )22 t xτ∆ ≡ ∆ − ∆ 2c is an invariant under Lorentz transformations: calculate it using values for the separation between two events as measured in one frame of reference and you'll obtain exactly the same value that someone else using measurements made in a different frame of reference will obtain. ∆τ2 is sometimes called the "invariant interval squared" or the "proper time squared" between two events. Note that ∆τ2 can be negative (as it is in the case we've been discussing): don't be mislead by the square. Events whose separations yield a positive ∆τ2 can be causally connected; events with negative ∆τ2 cannot be causally connected. A pair of events with ∆τ2 = 0 can only be causally linked if the first event used a light beam to make the second event take place. Some comments on ∆τ2: 1. If ∆τ2 > 0:

• Event 2 always takes place after event 1, regardless of your frame of reference (t2 > t1 always).

• It is possible to find a frame of reference in which event 1 and event 2 occur at the same

point in space.

• The interval between the events is said to be timelike (|∆t| > |∆x/c|).

2. If ∆τ2 < 0:

• In some frames, event 2 takes place after event 1, in some frames event 2 takes place before event 1. There is one particular frame of reference in which the two events occur simultaneously.

• It is impossible to find a frame of reference in which event 1 and event 2 occur at the same point in space.

• The interval between the events is said to be spacelike (|∆x/c| > |∆t|).

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honors topics in electrodynamics - high energy physics · physics 212, spring 2003. .1 introduction...

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