the simple structure of our universe-physics

PHY2061 R. D. Field

Department of Physics quark_1.doc University of Florida

The Simple Structure of our Universe Elementary Particle: Indivisible piece of matter without internal structure and without detectable size or shape . . • Four Forces: • Gravity (Solar Systems, Galaxies, Curved Space-Time , Black Holes) • Electromagnetism (Atoms & Molecules, Chemical Reactions) • Weak (Neutron Decay, Beta Radioactivity) • Strong (Atomic Nuclei, Fission & Fusion) • Two Classes of Elementary

Particles: • Leptons: Do not interact

with the strong force (but may interact with weak, EM and gravity).

• Quarks: Do interact with the strong force (may also interact with weak, EM and gravity).

• Quarks and Leptons have very different properties:

1. Weak and EM forces much weaker that strong force. 2. Quarks have fractional electric charge. 3. Quarks are found only as constituents of composite particles

called hadrons (baryons have B not 0, mesons have B = 0). Leptons exist as free particles.

• Gauge Particles are the carriers (or mediators) of the forces:

• Electromagnetism – Photon γγγγ (massless) • Weak – Weak Vector Bosons W+, W-, Z (massive) • Gravity – Graviton • Strong – 8 Gluons (massless)

Mass and chage located inside sphere of radius zero!

Baryon Number

PHY2061 R. D. Field


Labeling the Particles – Quantum Numbers Elementary particles and hadrons are labeled by their quantum numbers. These labels characterize the properties of the particles.

Symbol Name Additive M Mass J Spin Angular Momentum C Charge Conjugation P Parity G G-Parity B Baryon Number Yes

Qem Electric Charge Q = Y/2 + Iz Q = Qweak + QU1 Yes QU1 U1 Charge Yes

Qweak Weak Charge Yes Qcolor Strong Charge

Y Hypercharge Y = B + S + Ch + Bo + To Yes S Strangness Yes

Ch Charmness Yes Bo Bottomness Yes To Topness Yes I Isospin Yes Iz 3rd component of Isospin Yes Le Electron Lepton Number Yes Lµµµµ Muon Lepton Number Yes Lττττ Tau Lepton Number Yes L Overall Lepton Number L = Le +Lµ +Lτ Yes

Not all particles carry every label. The particles are only labeled by the quantum numbers that are conserved for that particle. • Particles with integral spin J (J = 0, 1, 2, …) are called bosons. • Particles with half-integral spin J (J = ½, 3/2, …) are called fermions. • Particles with spin-parity JP = 0+ are refered to a scalars, 0- are

pseudo-scalars, 1- are vectors, 1+ are pseudo-vectors, 2+ are tensors, etc.

• Hadrons are labeled by IGJPC.

PHY2061 R. D. Field


Leptons & Anti-Leptons (J = ½ fermions, B = 0, Ch = 0, Bo = 0, To = 0)

Generation Qem = Qweak + QU1

Lepton Mass MeV

Qem Le Lµµµµ Lµµµµ QU1 Qweak

ννννe 1st ~ 0 0 1 0 0 -1/2 +1/2

e- 1st 0.5 -1 1 0 0 -1/2 -1/2

ννννµµµµ 2nd ~ 0 0 0 1 0 -1/2 +1/2

µµµµ- 2nd 106 -1 0 1 0 -1/2 -1/2

ννννττττ 3rd ~ 0 0 0 0 1 -1/2 +1/2

ττττ- 3rd 1777 -1 0 0 1 -1/2 -1/2

Generation Qem measured in units of the electron charge e

Anti-Lepton

Mass MeV

Qem Le Lµµµµ Lµµµµ QU1 Qweak

e+ 1st 0.5 +1 -1 0 0 +1/2 +1/2

ev 1st ~ 0 0 -1 0 0 +1/2 -1/2

µµµµ+ 2nd 106 +1 0 -1 0 +1/2 +1/2

µv 2nd ~ 0 0 0 -1 0 +1/2 -1/2

ττττ+ 3rd 1777 +1 0 0 -1 +1/2 +1/2

τv 3rd ~ 0 0 0 0 -1 +1/2 -1/2

SU(2) Weak Lepton Doublets:

=

=

= −−− τ

νµνν τµ

321 LLe

L e

SU(2) Weak Anti-Lepton Doublets:

=

=

=

+++

τµ ντ

νµ

ν 321 LLe

eL

PHY2061 R. D. Field


Quarks & Anti-Quarks (J = ½+ fermions, Le = 0, Lµµµµ = 0, Lττττ = 0)

Generation Qem = Qweak + QU1

Quarks Mass MeV

B Qem Y I Iz S Ch Bo To QU1 Qweak Qcolor

u, u, u 1st 5 1/3 2/3 1/3 1/2 1/2 0 0 0 0 +1/6 +1/2 R, B, G d, d, d 1st 10 1/3 -1/3 1/3 1/2 -1/2 0 0 0 0 +1/6 -1/2 R, B, G c, c, c 2nd 1,500 1/3 2/3 4/3 0 0 0 1 0 0 +1/6 +1/2 R, B, G s, s, s 2nd 200 1/3 -1/3 -2/3 0 0 -1 0 0 0 +1/6 -1/2 R, B, G t, t, t 3rd 175,000 1/3 2/3 4/3 0 0 0 0 0 1 +1/6 +1/2 R, B, G

b, b, b 3rd 4,700 1/3 -1/3 -2/3 0 0 0 0 -1 0 +1/6 -1/2 R, B, G

Anti-

Quarks Mass

MeV B Qem Y I Iz S Ch Bo To QU1 Qw Qcolor

dbar, dbar, dbar

1st 10 -1/3 1/3 -1/3 1/2 1/2 0 0 0 0 -1/6 +1/2 Rbar, Bbar, Gbar

ubar, ubar, ubar

1st 5 -1/3 -2/3 -1/3 1/2 -1/2 0 0 0 0 -1/6 -1/2 Rbar, Bbar, Gbar

sbar, sbar, sbar

2nd 200 -1/3 1/3 2/3 0 0 1 0 0 0 -1/6 +1/2 Rbar, Bbar, Gbar

cbar, cbar, cbar

2nd 150 -1/3 -2/3 -4/3 0 0 0 -1 0 0 -1/6 -1/2 Rbar, Bbar, Gbar

bbar, bbar, bbar

3rd 4,700 -1/3 1/3 2/3 0 0 0 0 1 0 -1/6 +1/2 Rbar, Bbar, Gbar

tbar, tbar, tbar

3rd 175,000 -1/3 -2/3 -4/3 0 0 0 0 0 -1 -1/6 -1/2 Rbar, Bbar, Gbar

SU(2) Weak Quark and Anti-Quark Doublets:

′

=

′=

′=

GBR

GBRGBR

GBR

GBRGBR

GBR

GBRGBR

bt

sc

du

Q,,

,,,,

3,,

,,,,2

,,

,,,,1 QQ

′=

′=

′=

GBR

GBRGBR

GBR

GBRGBR

GBR

GBRGBR

tb

cs

ud

Q,,

,,,,

3,,

,,,,2

,,

,,,,1 QQ

PHY2061 R. D. Field


Vector Bosons (J = 1-, B = 0, Ch = 0, Bo = 0, To = 0, Le = 0, Lµµµµ = 0, Lττττ = 0)

Qem = Qweak + QU1

Boson Name Mass GeV

Qem QU1 Qweak Qcolor

γγγγ Photon 0 0 0 0 none

W+ W-Boson 81 +1 0 +1 none

W- W-Boson 81 -1 0 -1 none

Z W-Boson 92 0 0 0 none G1 Gluon 0 0 0 0 RBbar G2 Gluon 0 0 0 0 RGbar G3 Gluon 0 0 0 0 BRbar G4 Gluon 0 0 0 0 BGbar G5 Gluon 0 0 0 0 GRbar G6 Gluon 0 0 0 0 GBbar G7 Gluon 0 0 0 0 RRbar

BBbar GGbar

G8 Gluon 0 0 0 0 RRbar BBbar GGbar

PHY2061 R. D. Field


Hadrons – PseudoScalar Meson Nonet (JP = 0- bosons, B = 0, Ch = 0, Bo = 0, To = 0)

Y = B + S +Ch +Bo + To Qem = Y/2 + Iz

Symbol Name Mass MeV

Qem Net Quarks

I Iz Y S Qcolor

ππππ+ pion 140 +1 udbar 1 +1 0 0 singlet

ππππ0 pion 135 0 uubar, ddbar

1 0 0 0 singlet

ππππ- pion 140 -1 dubar 1 -1 0 0 singlet

K+ kaon 494 +1 usbar ½ +1/2 +1 +1 singlet

K0 kaon 478 0 dsbar ½ -1/2 +1 +1 singlet

K0bar kaon 478 0 sdbar ½ +1/2 -1 -1 singlet

K- kaon 494 -1 subar ½ -1/2 -1 -1 singlet

ηηηη eta 549 0 uubar, ddbar,

ssbar

0 0 0 0 singlet

ηηηη’ eta-prime 958 0 uubar, ddbar,

ssbar

0 0 0 0 singlet

Iz

Y

SU(3)flavor Triplets

d u3

Iz

Y

SU(3)flavor Octet

ππππ+ππππ0ππππ-

K+

K0barK-

K0

ηηηη

8

Iz

Y

SU(3)flavor Singlet

ηηηη’

1

3 x 3 = 8 + 1

PHY2061 R. D. Field


Hadrons – ½+ Baryon Octet (JP = ½+ fermions, B = 1, Ch = 0, Bo = 0, To = 0)

Symbol Name Mass

MeV Qem/e Net

Quarks I Iz Y S Qcolor

ΣΣΣΣ+ Sigma 1189 +1 uus 1 +1 0 -1 singlet

ΣΣΣΣ0 Sigma 1193 0 uds 1 0 0 -1 singlet

ΣΣΣΣ- Sigma 1189 -1 dds 1 -1 0 -1 singlet

p Proton 938 +1 uud ½ +1/2 +1 0 singlet n Neutron 940 0 udd ½ -1/2 +1 0 singlet

ΞΞΞΞ0 Cascade 1315 0 ssu ½ +1/2 -1 -2 singlet

ΞΞΞΞ- Cascade 1321 -1 ssd ½ -1/2 -1 -2 singlet

ΛΛΛΛ Lambda 1116 0 uds 0 0 0 -1 singlet

Y = B + S +Ch +Bo + To Qem = Y/2 + Iz

Iz

Y

SU(3)flavor Octet

ΣΣΣΣ+ΣΣΣΣ0ΣΣΣΣ-

p

ΞΞΞΞ0ΞΞΞΞ-

n

ΛΛΛΛ

8

3 x 3 x 3 = 10 + 8 + 8 + 1

PHY2061 R. D. Field

Department of Physics chp27_1.doc University of Florida

Electrostatic Force and Electric Charge

Electrostatic Force (charges at rest): • Electrostatic force can be attractive • Electrostatic force can be repulsive • Electrostatic force acts through empty

space • Electrostatic force much stronger than

gravity • Electrostatic forces are inverse square law forces (proportional to

1/r2) • Electrostatic force is proportional to the product of the amount of

charge on each interacting object

Magnitude of the Electrostatic Force is given by Coulomb's Law: F = K q1q2/r2 (Coulomb's Law) where K depends on the system of units

K = 8.99x109 Nm2/C2 (in MKS system) K = 1/(4πεπεπεπε0) where εεεε0 = 8.85x10-12 C2/(Nm2) Electric Charge:

electron charge = -e e = 1.6x10-19 C proton charge = e C = Coulomb

Electric charge is a conserved quantity (net electric charge is never created or destroyed!).

q1 q2

r

PHY2061 R. D. Field


Units

MKS System (meters-kilograms-seconds): also Amperes, Volts, Ohms, Watts

Force: F = ma Newtons = kg m / s2 = 1 N Work: W = Fd Joule = Nm = kg m2 / s2 = 1 J Electric Charge: Q Coulomb = 1 C F = K q1q2/r2 K = 8.99x109 Nm2/C2 (in MKS system) CGS System (centimeter-grams-seconds): Force: F = ma 1 dyne = g cm / s2 Work: W = Fd 1 erg = dyne-cm = g cm2 / s2 Electric Charge: Q esu (electrostatic unit) F = q1q2/r2 K = 1 (in CGS system) Conversions (MKS - CGS): Force: 1 N = 105 dynes Work: 1 J = 107 ergs Electric Charge: 1 C = 2.99x109 esu Fine Structure Constant (dimensionless):

α = α = α = α = K 2ππππe2/hc (same in all systems of units)

h = Plank's Constant c = speed of light in vacuum

We will use the MKS system!

PHY2061 R. D. Field


Electrostatic Force versus Gravity Electrostatic Force : Fe = K q1q2/r2 (Coulomb's Law)

K = 8.99x109 Nm2/C2 (in MKS system) Gravitational Force : Fg = G m1m2/r2 (Newton's Law)

G = 6.67x10-11 Nm2/kg2 (in MKS system) Ratio of forces for two electrons :

e = 1.6x10-19 C m = 9.11x10-31 kg

e, m e, m

r

Fe / Fg = K e2 / G m2 = 4.16x1042 (Huge number !!!)

PHY2061 R. D. Field


Vector Forces

Q q

r

The Electrostatic Force is a vector: The force on q due to Q points along the radial direction and is

given by

!F

KqQr

r= 2 "

q1

QF1

q2

q3

F2

F3

Vector Superposition of Electric Forces:

If several point charges q1, q2, q3, … simultaneously exert electric forces on a charge Q then

F = F1 + F2 +F3 + …

Vector Form of Coulombs Law

PHY2061 R. D. Field


Vectors & Vector Addition

The Components of a vector:

x-axis

y-axis

θθθθ

A

Ax =A cos θθθθ

Ay =A sin θθθθ

Vector Addition:

x-axis

y-axis

AC B

To add vectors you add the components of the vectors as follows: !

!

! ! !

A A x A y A zB B x B y B z

C A B A B x A B y A B z

x y z

x y z

x x y y z z

= + += + +

= + = + + + + +

" " "

" " "

( ) " ( ) " ( )"

Vector Addition

PHY2061 R. D. Field


The Electric Dipole

+Q -Q

d

An electric "dipole" is two equal and opposite point charges separated by a distance d. It is an electrically neutral system. The "dipole moment" is defined to be the charge times the separation (dipole moment = Qd). Example Problem:

+Q

-Q

dqx

A dipole with charge Q and separation d is located on the y-axis with its midpoint at the origin. A charge q is on the x-axis a distance x from the midpoint of the dipole. What is the electric force on q due to the dipole and how does this force behave in the limit x >>d (dipole approximation)? Example Problem:

-Q +Qd

x

A dipole with charge Q and separation d is located on the x-axis with its midpoint at the origin. A charge q is on the x-axis a distance x from the midpoint of the dipole. What is the electric force on q due to the dipole and how does this force behave in the limit x >>d (dipole approximation)?

PHY2061 R. D. Field


The Electric Field

+Q q

E

The charge Q produces an electric field which in turn produces a force on the charge q. The force on q is expressed as two terms:

F = K qQ/r2 = q (KQ/r2) = q E The electric field at the point q due to Q is simply the force per unit positive charge at the point q:

E = F/q E = KQ/r2 The units of E are Newtons per Coulomb (units = N/C). The electric field is a physical object which can carry both momentum and energy. It is the mediator (or carrier) of the electric force. The electric field is massless. The Electric Field is a Vector Field:

!E

KQr

r= 2 "

Electric Field of a Point Charge

PHY2061 R. D. Field


Electric Field Lines

+Q -Q

Electric field line diverge from (i.e. start) on positive charge and end on negative charge. The direction of the line is the direction of the electric field. The number of lines penetrating a unit area that is perpendicular to the line represents the strength of the electric field.

+2Q+Q

PHY2061 R. D. Field


Electric Field due to a Distribution of Charge

dQ dE = K dQ/r2 r

r

The electric field from a continuous distribution of charge is the superposition (i.e. integral) of all the (infinite) contributions from each infinitesimal dQ as follows:

!E

Kr

r dQ= ∫ 2 " and Q dQ= ∫

Charge Distributions:

• Linear charge density λλλλ: λ(λ(λ(λ(x) ) ) ) = charge/unit length L

dQ = λλλλ dx

For a straight line dQ = λ(λ(λ(λ(x)))) dx and

Q dQ x dx= =∫ ∫ λ ( )

If λλλλ(x) = λλλλ is constant then dQ = λλλλ dx and Q = λλλλL, where L is the length.

PHY2061 R. D. Field


Charge Distributions Charge Distributions:

• Linear charge density λλλλ: λ(θ) λ(θ) λ(θ) λ(θ) = charge/unit arc length

RdQ = λλλλ ds = λλλλ R dθθθθ

For a circular arc dQ = λ(θ)λ(θ)λ(θ)λ(θ) ds = λ(θ)λ(θ)λ(θ)λ(θ) Rdθθθθ and

Q dQ ds Rd= = =∫ ∫∫ λ θ λ θ θ( ) ( ) If λλλλ(θθθθ) = λλλλ is constant then dQ = λλλλ ds and Q = λλλλs, where s is the arc length.

• Surface charge density σσσσ: σ(σ(σ(σ(x,y) ) ) ) = charge/unit area

dQ = σσσσ dA

For a surface dQ = σ(σ(σ(σ(x,y)))) dA and

Q dQ x y dA= =∫ ∫σ( , ) If σσσσ(x,y) = σσσσ is constant then dQ = σσσσ dA and Q = σσσσA, where A is the area.

• Volume charge density ρρρρ: ρ(ρ(ρ(ρ(x,y,z) ) ) ) = charge/unit volume

dQ = ρρρρ dV

For a surface dQ = ρ(ρ(ρ(ρ(x,y,z)))) dV and

Q dQ x y z dV= =∫ ∫ ρ( , , ) If ρρρρ(x,y,z) = ρρρρ is constant then dQ = ρρρρ dV and Q = ρρρρV, where V is the volume.

PHY2061 R. D. Field


Calculating the Electric Field

Example: P

x

L

A total amount of charge Q is uniformily distributed along a thin straight rod of length L. What is the electric field at a point P on the x-axis a distance x from the end of the rod?

Answer: !E

KQx x L

x=+( )

"

Example:

A total amount of charge Q is uniformily distributed along a thin straight rod of length L. What is the electric field at a point P on the y-axis a distance y from the midpoint of the rod?

Answer: !E

KQy y L

y=+2 22( / )

"

Example:

A infinitely long straight rod has a uniform charge density λλλλ. What is the electric field at a point P a perpendicular distance r from the rod?

Answer: !E

Kr

r=2 λ

"

P

y

L

P

r

λλλλ

PHY2061 R. D. Field


Some Useful Math Series Expansions:

....!6!4!2

1cos

....!7!5!3

sin

....!3!2!1

1

642

753

32

+−+−=

+−+−=

++++=

xxxx

xxxxx

xxxex

Approximations:

( )εεεεεεε

εε

ε

ε

ε

11

11tansin

111

<<<<

<<<<≈≈+≈±≈± epp

Indefinite Integrals:

∫

∫

+

−=+

+=

+

222/322

222/322

2

1)(

)(

axdx

axx

axxdx

axa

PHY2061 R. D. Field


Calculating the Electric Field

Example: A total amount of charge Q is uniformily distributed along a thin semicircle of radius R. What is the electric field at a point P at the center of the circle?

Answer: !E

KQR

x=2

2π"

Example: A total amount of charge Q is uniformily distributed along a thin ring of radius R. What is the electric field at a point P on the z-axis a distance z from the center of the ring?

Answer: ( )!E

KQzz R

z=+2 2 3 2/ "

Example: A total amount of charge Q is uniformily distributed on the surface of a disk of radius R. What is the electric field at a point P on the z-axis a distance z from the center of the disk?

Answer: !E

KQR

zz R

z= −+

212 2 2

"

RP

x-axis

RP z-axis

z

RP z-axis

z

PHY2061 R. D. Field


Calculating the Electric Field Example:

What is the electric field generated by a large (infinite) sheet carrying a uniform surface charge density of σσσσ coulombs per meter?

Answer: !E z=

σε2 0

"

Example:

What is the electric field at a point P between two large (infinite) sheets carrying an equal but opposite uniform surface charge density of σσσσ?

Answer: !E z=

σε0

"

P

z

σσσσ

P

-σσσσ

σσσσ

PHY2061 R. D. Field


Flux of a Vector Field

Fluid Flow:

n

Flux = vA

n

Flux = 0

n

Flux = vA cosθθθθ

θθθθ

Consider the fluid with a vector

!v which describes the velocity of the fluid at every point in space and a square with area A = L2 and normal "n . The flux is the volume of fluid passing through the square area per unit time. Generalize to the Electric Field: Electric flux through the infinitesimal area dA is equal to

d E dAΦ = ⋅! !

where

dA An!

= " Total Electric Flux through a Closed Surface:

ΦE SE dA= ⋅∫! !

dA

E

θθθθ n

dΦΦΦΦ =E dA cosθθθθ

E

normal

Surface S

Electric Flux Surface Integral!

PHY2061 R. D. Field


Electric Flux and Gauss' Law

The electric flux through any closed surface is proportional to the net charge enclosed.

! !E dA

Qenclosed

S⋅ =∫ ε0

For the discrete case the total charge enclosed is the sum over all the enclosed charges:

Q qenclosed ii

N

==∑

1

For the continuous case the total charge enclosed is the integral of the charge density over the volume enclosed by the surface S:

Q dVenclosed = ∫ ρ

Simple Case: If the electric field is constant over the surface and if it always points in the same direction as the normal to the surface then

ΦE SE dA EA= ⋅ =∫! !

The units for the electric flux are Nm2/C.

E

normal

ClosedSurface S

Gauss’ Law

PHY2061 R. D. Field


Conductors in Static Equilibrium

Conductor: In a conductor some electrons are free to move (without restraint) within the volume of the material (Examples: copper, silver, aluminum, gold)

Conductor in Static Equilibrium: When the charge distribution on a conductor reaches static equilibrium (i.e. nothing moving), the net electric field withing the conducting

material is exactly zero (and the electric potential is constant). Excess Charge: For a conductor in static equilibrium all the (extra) electric charge reside on the surface. There is no net electric charge within the volume of the conductor (i.e. ρ ρ ρ ρ = 0). Electric Field at the Surface: The electric field at the surface of a conductor in static equilibrium is normal to the surface and has a magnitude, E = σσσσ/εεεε0, where σσσσ is the surface charge density (i.e. charge per unit area) and the net charge on the conductor is

Q d AS u r fa c e

= ∫ σ .

Conductor

Conductor instatic equilibrium

E = 0V = constant

Conductor instatic equilibrium

E = 0V = constant

ρρρρ = 0

Surface Charge Densityσσσσ

E

PHY2061 R. D. Field


Gauss' Law Examples

Problem: A solid insulating sphere of radius R has charge distributed uniformly throughout its volume. The total charge of the sphere is Q. What is the magnitude of the electric field inside and outside the sphere? Answer:

Problem: A solid conducting sphere of radius R has a net charge of Q. What is the magnitude of the electric field inside and outside the sphere? Where are the charges located? Answer: Charges are on the surface and

!

!E

K Qr

r

Eo u t

in

=

=2

0

"

Problem: A solid conducting sphere of radius b has a spherical hole in it of radius a and has a net charge of Q. If there is a point charge -q located at the center of the hole, what is the magnitude of the electric field inside and outside the conductor? Where are the charges on the conductor located? Answer: Charges are on the inside and outside surface with Qin=q and Qout=Q-q and

!

!E

K Q qr

r

E

EK q

rr

r b

a r b

r a

>

< <

<

=−

=

=−

( )"

"

2

2

0

Total Charge Qρρρρ = constant

Insulating Sphere

R

!

!

EK Qr

r

EK Q r

Rr

o u t

in

=

=

2

3

"

"

Net Charge Q

ConductingSphere

R

Net Charge Qon conductor

b

-q

a

PHY2061 R. D. Field

Department of Physics div_1.doc Univesity of Florida

Divergence of a Vector Function

Let zzyxFyzyxFxzyxFzyxF zyx ˆ),,(ˆ)..(ˆ),,(),,( ++=

! be a vector

function of position. The divergence of a vector function is the flux out of a volume, V, per unit volume, in the limit of infinitesimal V. It is the surface integral per unit volume as the volume enclosed by the surface goes to zero:

⋅=⋅∇= ∫→

SurfaceClosed

VAdF

VFFdiv

!!!!! 1lim)(0

Symbols: FgFdiv!!!

⋅∇==)(

In cartesian (or retangular) coordinates:

zz

yy

xx

∂∂+

∂∂+

∂∂=∇ ˆˆˆ

!

zF

yF

xFFg zyx

∂∂+

∂∂

+∂∂=⋅∇=

!!

Note:

3)( 2121

=⋅∇⋅∇+⋅∇=+⋅∇

rFFFF

!!

!!!!!!!

Vector F(x,y,z) Divergence Operator Scalar g(x,y,z)

Vector Operator

ClosedSurface S

Volume V enclosedby surface S

PHY2061 R. D. Field


Divergence Theorem

The divergence theorem states that the integral of the divergence of a vector function over a volume, V, is equal to the flux, ΦΦΦΦF, of the vector function through the closed surface, S, that encloses the volume V:

∫∫ ⋅=⋅∇SV

AdFdVF!!!!

)(

Proof (sketch):

∫∑ ∫∑ ∫∫ ⋅∇⇒

⋅=⋅=⋅=Φ

→∞→== VV

N

N

i Si

ii

N

i Si

SF dVFAdF

VVAdFAdF

iii

)(1

011

!!!!!!!!

The Laplacian Operator Suppose that the vector function, ),,( zyxF

!, is the gradient of the scalar

function f(x,y,z), fF ∇=!!

. Now suppose we construct a new scalar function g(x,y,z) that is the divergence of ),,( zyxF

!, Fg

!!⋅∇= . Then

g(x,y,z) is the divergence of the gradient of f(x,y,z) as follows:

ffg 2∇=∇⋅∇=!!


zz

yy

xx

∂∂+

∂∂+

∂∂=∇ ˆˆˆ

!

2∇=∇⋅∇!!

2

2

2

2

2

22

zyx ∂∂+

∂∂+

∂∂=∇

2

2

2

2

2

22

zf

yf

xffg

∂∂+

∂∂+

∂∂=∇=

ClosedSurface S


Laplacian Operator

Scalar f(x,y,z) Laplacian Operator Scalar g(x,y,z)

Divergence Theorem

PHY2061 R. D. Field


Divergence of the Electric Field

Consider a sphere with radius R, volume V = 4ππππR3/3, and surface area A = 4ππππR2 with a point charge Q at the center. The electric field from the charge is given by

rr

KQrE ˆ)( 2=!

,

and the electric flux through the surface of the sphere is

0

2222 )4()ˆˆ(

επ QR

RKQdA

RKQdAnr

RKQAdE

S SSE ===⋅=⋅=Φ ∫ ∫∫

!!.

The divergence of the electric field is given by

00

000

lim1lim1limερ

ε=

=

Φ=

⋅=⋅∇

→→→ ∫ VQ

VAdE

VE

VE

VS

V

!!!!.

Volume charge density: The electric charge density is the charge per unit volume,

dVdQ

VQzyx

V=

=

→0lim),,(ρ and dVdQ ρ= .

The volume charge density is a scalar function of position and has units of C/m3. Divergence of ),,( zyxE

!: The divergence of the electric field at every

point in space is equal to the charge density at that point (divided by εεεε0).

0ερ=⋅∇ E

!!

Sphere of radius Rand volume V

Point charge Qat the center

ClosedSurface

E

Differential form of Gauss’ Law

PHY2061 R. D. Field


Gauss’ Law (integral form)

The total electric flux, ΦΦΦΦE, through any closed surface S is equal to the total charge enclosed by the surface S (divided by εεεε0) as follows:

∫ =⋅=ΦS

enclosedE

QdAE0ε

Proof:

00

1)(ε

ρε

enclosed

VVSE

QdVdVEAdE ∫∫∫ ==⋅∇=⋅=Φ!!!!

,

where the total enclosed charge is given by

∫∫ ==VV

enclosed dVdQQ ρ .

Poisson’s Equation

The electrostatic field can be written as the gradient of the electric potential, V(x,y,z), as follows:

VE ∇−=!!

and 0ε

ρ=⋅∇ E!!

so that 0

)(ερ=∇−⋅∇ V

!! or

02

2

2

2

2

22

ερ−=

∂∂+

∂∂+

∂∂=∇

zV

yV

xVV .

Laplace’s Equation

Whenever 0=ρ , that is, in all parts of space containing no electric charge the electric potential must satisfy,

02

2

2

2

2

22 =

∂∂+

∂∂+

∂∂=∇

zV

yV

xVV .

Any ClosedSurface S

E

Gauss’ Law

Laplace’s Equation

Poisson’s Equation

PHY2061 R. D. Field

Department of Physics curl_1.doc University of Florida

Curl of a Vector Function

The curve C is the boundary of the surface S which it spans. Define the circulation ΓΓΓΓ as the line integral around the closed curve C as follows:

∫ ⋅=ΓC

rdF !!

.

Symbols: FGFcurl!!!!

×∇==)( Definition #1: The component of F

!!×∇ in the direction of the unit vector n is the limit

of the circulation ΓΓΓΓ per unit area, as the enclosed area goes to zero.

⋅=

Γ=⋅×∇ ∫→→

CAA

rdFAA

nF !!!! 1limlimˆ)(00

Definition #2: The curl of the vector function F

! is the limit of the ratio of the integral

of its cross product with the outward normal n , over a closed surface S, to the volume enclosed by the surface as the volume goes to zero.

×=×∇ ∫→

SV

dAFnV

F )ˆ(1lim0

!!!

nnormal

Boundry Curve C

Surface S

Surface S


Vector F(x,y,z) Curl Operator Vector G(x,y,z)

PHY2061 R. D. Field

Department of Physics curl_2.doc Univesity of Florida

Stokes’ Theorem

Stokes’ Theorem:

∫∫ ⋅=⋅×∇CS

rdFAdF !!!!!)(

Proof (sketch):

∫

∑

∑∫

⋅×∇⇒

⋅×∇=

Γ=⋅=Γ

→∞→

=

=

SAN

N

iii

N

i i

ii

C

AdF

nFA

AArdF

i

!!!

!!

!!

)(

ˆ)(

0

1

1


zz

yy

xx

∂∂+

∂∂+

∂∂=∇ ˆˆˆ

!

zyx FFFzyx

zyx

FG∂∂

∂∂

∂∂=×∇=

ˆˆˆ!!!

yF

xF

GxF

zFG

zF

yFG xy

zzx

yyz

x ∂∂−

∂∂

=∂∂−

∂∂=

∂∂

−∂∂=

Boundry Curve C

Surface S

Boundry Curve C

ΓΓΓΓ

Ai

ΓΓΓΓi

Vector Operator

Closed Curve C Surface bounded by closed curve

PHY2061 R. D. Field


The Curl of a Radial Function

Suppose ),,( zyxF!

is a radial (or central) function of r. Namely, rrfrF !!

)()( = , it points radially outward (or inward) along the radius vector zzyyxxr ˆˆˆ ++=!

and has a magnitude rf(r) that depends only on the distance

222 zyxr ++= from the origin. Theorem: The curl of a radial function is zero.

0=×∇ F!!

if rrfF !!)(=

Proof:

rfrfrfF !!!!!!!!×∇+×∇=×∇=×∇ )(

Term 1:

0

ˆˆˆ

=∂∂

∂∂

∂∂=×∇

zyxzyx

zyx

r!!

Term 2:

rr

drdfzzyyxx

rdrdf

zzry

yrx

xr

drdf

zzfy

yfx

xff

!

!

=++=

∂∂+

∂∂+

∂∂=

∂∂+

∂∂+

∂∂=∇

)ˆˆˆ(1

ˆˆˆ

ˆˆˆ

thus

01 =×=×∇ rrrdr

dfrf !!!!

Term 1

Term 2

r

F

PHY2061 R. D. Field

Department of Physics grad_1.doc Univesity of Florida

Gradient of a Scalar Function

Let f(x,y,z) be a scalar function of position and let zdzydyxdxrd ˆˆ ++= !"

be an infinitesimal displacement vector. Directional derivative:

−+++=

→ drzyxfdzzdyydxxf

rddf

dr

),,(),,(lim0

"

The directional derivative depends on the point (x,y,z) and the direction of rd" . Gradient: The gradient of a scalar function f(x,y,z) is a vector whose magnitude is the maximum directional derivative at the point being considered and whose direction is the direction of the maximum directional derivative at the point. Symbol:

ffgradF ∇==""

)( In cartesian (or rectangular) cordinates:

zz

yy

xx

∂∂+

∂∂+

∂∂=∇ ˆˆˆ

"

zfz

yfy

xfxfF

∂∂+

∂∂+

∂∂=∇= ˆˆˆ

""

zfF

yfF

xfF zyx ∂

∂=∂∂=

∂∂=

f(x,y)

x

y

Direction ofsteepest slope

P

Surface

Scalar Function f(x,y)

Scalar f(x,y,z) Gradient Operator

Vector F(x,y,z)

Vector Operator

PHY2061 R. D. Field

Department of Physics line_1.doc Univesity of Florida

P1

Closed Loop

Line Integral of a Vector Function

Let zzyxFyzyxFxzyxFzyxF zyx ˆ),,(ˆ)..(ˆ),,(),,( ++=!

be a vector function of position. In general, the line integral of ),,( zyxF

! depends of the start

point P1, and the end point P2 and the path chosen from P1 to P2 (curve C):

∫ ⋅=CurveC

rdFPathPPI !!),,( 21 ,

where I(P1,P2,Path) is the component of along the path integrated over the path,

∫ ∫∫ ++==⋅CurveC CurveC

zyxCurveC

dzFdyFdxFdrFrdF )(cosθ!!.

Remark: If 0=×∇ F

!!then I(P1,P2,Path) = I(P1,P2) and is only a function

of the end points P1 and P2 and does not depend on the path (curve C). In this case:

∫ =⋅=

LoopClosed

rdFPPI 0),( 11!!

Line Integral of the Gradient The line integral is the inverse of the gradient as follows:

∫ ⋅∇=−2

1

)()( 12

P

P

rdfPfPf !!

.

This line integral is independent of the path since the curl of the gradient is zero,

0=∇×∇ f!!

.

F Tangent tocurve

P2

P1

Fcosθθθθθθθθ

PHY2061 R. D. Field

Department of Physics line_2.doc Univesity of Florida

P1

Closed Loop

Question:

When can the vector function ),,( zyxF!

be written as the gradient of a scalar function f(x,y,z)? We know that

∫ ⋅∇=−2

1

)()( 12

P

P

rdfPfPf !!

.

This implies that

∫ =⋅∇=−

LoodClosed

rdfPfPf 0)()( 11!!

.

Thus if we demand that

fF ∇=!!

, then

∫ =⋅

LoopClosedAny

rdF 0!!

.

But Stokes’ Theorem tell us that

∫ ∫ ⋅×∇=⋅

LoopClosed Surface

AdFrdF!!!!!

)(.

Answer:

When can the vector function ),,( zyxF!

is derivable from a scalar function f(x,y,z) according to fF ∇=

!! provided 0=×∇ F

!! .

P2

P1

Definition of a conservative force!

Curl of F

PHY2061 R. D. Field


Curl of the Electrostatic Field

The electrostatic field rrKQE !!

3= is a radial (or central)

function of r. It points radially outward (or inward) along the radius vector zzyyxxr ˆˆˆ ++=!

and has a magnitude E that depends only on the distance

222 zyxr ++= from the origin. Thus

0=×∇ E!!

This means that the electrostatic field can be written as the gradient of a scalar function V as follows:

VE ∇−=!!

The electric potential V(x,y,z) is a scalar function of position and is equal to the potential energy per unit charge.

∫ ⋅−=−=∇P

P

rdEPVPVV0

)()( 0!!

.

P0 is some reference point and V(P0) is the potential at the reference point.

Electric Potential of a Point Charge:

)(ˆ)( 020

PVrdrr

KQPVP

P

+⋅−= ∫!

Taking P0 = infinity and V(P0) = 0 gives

rKQrV =)(

Electrostatic Field

Electric Potential

r

E

Q

P

P0

CurveC

Point Charge

PHY2061 R. D. Field

Department of Physics vector_0.doc Univesity of Florida

Vector Identities

0)(0=×∇⋅∇

=∇×∇Ff!!!

!!

ff 2∇=∇⋅∇!!

2121

2121

122121

2121

)()(

)()(

FFFFFFFF

ffffffffff

!!!!!!!

!!!!!!!

!!!

!!!

×∇+×∇=+×∇⋅∇+⋅∇=+⋅∇

∇+∇=∇∇+∇=+∇

FfFfFfFFFFFF

fFFfFf

!!!!!!

!!!!!!!!!

!!!!!!

×∇+×∇=×∇×∇⋅−×∇⋅=×⋅∇

∇⋅+⋅∇=⋅∇

)()(

)(

211221

Let zzyyxxr ˆˆˆ ++=! then

rr

drdfrf

rr

!!

!!

!!

=∇=×∇=⋅∇

)(03

Curl Grad = 0

Div Curl = 0

Div Grad = Laplacian

PHY2061 R. D. Field

Department of Physics vector_1.doc Univesity of Florida

Points enclose curveP2

P1

Curve C

Summary: GRAD

Relationship between the points P1 and P2 and the curve C enclosed by the points:

∫ ⋅∇=−C

rdfPfPf !!)()( 12

zz

yy

xx

∂∂+

∂∂+

∂∂=∇ ˆˆˆ

! z

fzyfy

xfxf

∂∂+

∂∂+

∂∂=∇ ˆˆˆ

!

Summary: STOKES Relationship between the closed curve C and the surface S enclosed by the curve:

∫∫ ⋅×∇=⋅SC

AdFrdF!!!!!

)(

zyx FFFzyx

zyx

F∂∂

∂∂

∂∂=×∇

ˆˆˆ!!

Summary: GAUSS Relationship between the closed surface S and the volume V enclosed by the surface:

∫∫ ⋅∇=⋅VS

dVFAdF )(!!!!

zF

yF

xFF zyx

∂∂+

∂∂

+∂∂=⋅∇

!!

Vector Operator

Boundry Curve C

Surface S

Curve encloses surface

ClosedSurface S

Volume V

Surface encloses volume

PHY2061 R. D. Field


Gravitational Potential Energy

Gravitational Force: F = G m1m2/r2 Gravitational Potential Energy GPE:

U = GPE = mgh (near surface of the Earth)

Kinetic Energy: KE = 12

2m v Total Mechanical Energy: E = KE +U Work Energy Theorem:

W = EB-EA = (KEB-KEA) + (UB-UA) (work done on the system)

Energy Conservation: EA=EB

(if no external work done on system) Example:

A ball is dropped from a height h. What is the speed of the ball when it hits the ground?

Solution: Ei = KEi +Ui = mgh Ef = KEf + Uf = mvf2/2

E E v ghi f f= ⇒ = 2

h

vi = 0

vf = ?

PHY2060Review

PHY2061 R. D. Field


Electric Potential Energy

Electrostatic Force: F = K q1q2/r2 Electric Potential Energy: EPE = U (Units = Joules)

Kinetic Energy: KE = 12

2m v (Units = Joules) Total Energy: E = KE + U (Units = Joules) Work Energy Theorem: (work done on the system)

W = EB - EA = (KEB - KEA) + (UB - UA) Energy Conservation: EA=EB (if no external work done on system) Electric Potential Difference ∆∆∆∆V = ∆∆∆∆U/q:

Work done (against the electric force) per unit charge in going from A to B (without changing the kinetic energy).

∆∆∆∆V = WAB/q = ∆∆∆∆U/q = UB/q - UA/q

(Units = Volts 1V = 1 J / 1 C)

Electric Potential V = U/q: U = qV Units for the Electric Field (Volts/meter):

N/C = Nm/(Cm) = J/(Cm) = V/m

Energy Unit (electron-volt): One electron-volt is the amount of kinetic energy gained by an electron when it drops through one Volt potential difference

1 eV = (1.6x10-19 C)(1 V) = 1.6x10-19 Joules

1 MeV = 106 eV 1 GeV=1,000 MeV 1 TeV=1,000 GeV

q

B

A

PHY2061 R. D. Field


Accelerating Charged Particles

Example Problem: A particle with mass M and charge q starts from rest a the point A. What is its speed at the point B if VA=35V and VB=10V (M = 1.8x10-5kg, q = 3x10-5C)? Solution: The total energy of the particle at A and B is

E KE U qV

E KE U Mv qVA A A A

B B B B B

= + = +

= + = +

012

2 .

Setting EA = EB (energy conservation) yields (Note: the particle gains an amount of kinetic energy equal to its charge, q, time the change in the electric potential.)

Solving for the particle speed gives

(Note: positive particles fall from high potential to low potential VA >VB, while negative particles travel from low potential to high potential, VB >VA.)

Plugging in the numbers gives

vC V

kgm sB =

××

=−

−

2 3 10 251 8 10

9 15

5

( )( ).

. / .

VB = 10VVA = 35V

A B

q

+ -E

12

2Mv q V VB A B= −( )

vq V V

MBA B=

−2 ( )

PHY2061 R. D. Field


Potential Energy & Electric Potential

Mechanics (last semester!): Work done by force F in going from A to B:

W F drA BbyF

A

B

→ = ⋅∫! !

Potential Energy Difference ∆∆∆∆U:

W U U U F drA BagainstF

B AA

B

→ = = − = − ⋅∫∆! !

! !F U

Ux

xUy

yUz

z= −∇ = − − −∂∂

∂∂

∂∂

" " "

Electrostatics (this semester): Electrostatic Force:

! !F qE=

Electric Potential Energy Difference ∆∆∆∆U: (work done against E in moving q from A to B)

∆U U U qE drB AA

B

= − = − ⋅∫! !

Electric Potential Difference ∆∆∆∆V=∆∆∆∆U/q: (work done against E per unit charge in going from A to B)

∆V V V E drB AA

B

= − = − ⋅∫! !

! !E V

Vx

xVy

yVz

z= −∇ = − − −∂∂

∂∂

∂∂

" " "

PHY2061 R. D. Field


The Electric Potential of a Point Charge

+Q E

r

V(r) V(r) = KQ/r

Potential from a point charge:

V(r) = ∆∆∆∆V = V(r) - V(infinity) = KQ/r U = qV = work done against the electric force in bringing the charge q from infinity to the point r.

+Q q

E

Potential from a system of N point charges:

VKqr

i

ii

N

==∑

1

PHY2061 R. D. Field


Electric Potential due to a Distribution of Charge

dQ dV = K dQ/r

r

The electric potential from a continuous distribution of charge is the superposition (i.e. integral) of all the (infinite) contributions from each infinitesimal dQ as follows:

VKr

dQ= ∫ and Q dQ= ∫

Example: A total amount of charge Q is uniformily distributed along a thin circle of radius R. What is the electric potential at a point P at the center of the circle?

Answer: VKQR

=

Example: A total amount of charge Q is uniformily distributed along a thin semicircle of radius R. What is the electric potential at a point P at the center of the circle?

Answer: VKQR

=

RP

x-axis

RP

x-axis

PHY2061 R. D. Field


Calculating the Electric Potential Example: A total amount of charge Q is uniformily distributed along a thin ring of radius R. What is the electric potential at a point P on the z-axis a distance z from the center of the ring?

Answer: V zKQ

z R( ) =

+2 2

Example: A total amount of charge Q is uniformily distributed on the surface of a disk of radius R. What is the electric potential at a point P on the z-axis a distance z from the center of the disk?

Answer: ( )V zKQR

z R z( ) = + −2

22 2

RP z-axis

z

RP z-axis

z

PHY2061 R. D. Field


Electric Potential Energy

For a system of point charges: The potential energy U is the work required to assemble the final charge configuration starting from an inital condition of infinite separation. Two Particles:

U Kq q

rq

Kqr

qKq

r= =

+

1 21

22

112

12

so we see that

U q Vi ii

==

∑12 1

2

where Vi is the electric potential at i due to the other charges.

Three Particles:

U Kq qr

Kq qr

Kq qr

= + +1 2

12

1 3

13

2 3

23

which is equivalent to

U q Vi ii

==

∑12 1

3

where Vi is the electric potential at i due to the other charges. N Particles:

U q Vi ii

N

==

∑12 1

q1 q2

r

q1q2

r12

q3

r13 r23

PHY2061 R. D. Field


Stored Electric Potential Energy For a conductor with charge Q: The potential energy U is the work required to assemble the final charge configuration starting from an inital condition of infinite separation.

For a conductor the total charge Q resides on the surface

Q d q d A= = ∫∫ σ Also, V is constant on and inside the conductor and

dU dQV V dA= =12

12

σ

and hence

U V d Q V d A V QSurface Surface

= = =∫ ∫12

12

12

σ

Stored Energy: U Q Vc o n d u c to r =12

where Q is the charge on the conductor and V is the electric potential of the conductor. For a System of N Conductors:

U Q Vi ii

N

==

∑12 1

where Qi is the charge on the i-th conductor and Vi is the electric potential of the i-th conductor.

dQ=σσσσdA

V = constant

E=0

PHY2061 R. D. Field


Capacitors & Capacitance Capacitor: Any arrangement of conductors that is used to store electric charge (will also store electric potential energy). Capacitance: C=Q/V or C=Q/∆∆∆∆V Units: 1 farad = 1 F = 1 C/1 V 1 µµµµF=10-6 F 1 nF=10-9 F 1 pF=10-12 F

Stored Energy:

U Q VQ

CC Vc o n d u c to r = = =

12 2

12

22

where Q is the charge on the conductor and V is the electric potential of the conductor and C is the capacitance of the conductor. Example (Isolated Conducting Sphere): For an isolated conducting sphere with radius R, V=KQ/R and hence C=R/K and U=KQ2/(2R). Example (Parallel Plate Capacitor):

For two parallel conducting plates of area A and separation d we know that E = σσσσ/εεεε0 = Q/(Aεεεε0) and ∆∆∆∆V = Ed = Qd/(Aεεεε0) so that C = Aεεεε0/d. The stored energy is U = Q2/(2C) = Q2d/(2Aεεεε0).

E

d

+σσσσ

−σ−σ−σ−σ

Q Area A

-Q Area A

E=Q/(Aεεεε0)

PHY2061 R. D. Field


Capacitors in Series & Parallel Parallel: In this case ∆∆∆∆V1=∆∆∆∆V2=∆∆∆∆V and Q=Q1+Q2. Hence, Q = Q1 + Q2 = C1∆V1 + C2∆V2 = (C1+C2)∆V so C = Q/∆∆∆∆V = C1 + C2, where I used Q1 = C1∆V1 and Q2 = C2∆V2.

Capacitors in parallel add. Series: In this case ∆∆∆∆V=∆∆∆∆V1+∆∆∆∆V2 and Q=Q1=Q2. Hence, ∆V = ∆V1 + ∆V2 = Q1/C1+Q2/C2 = (1/C1+1/C2)Q so 1/C = ∆∆∆∆V/Q = 1/C1 + 1/C2, where I used Q1 = C1∆V1 and Q2 = C2∆V2.

Capacitors in series add inverses.

∆∆∆∆VC1 C2

∆∆∆∆V

C1

C2

PHY2061 R. D. Field


Energy Density of the Electric Field Energy Density u: Electric field lines contain energy! The amount of energy per unit volume is

u = e0E2/2, where E is the magnitude of the electric field. The energy density has units of Joules/m3. Total Stored Energy U: The total energy strored in the electric field lines in an infinitessimal volume dV is dU = u dV and

U udVVolum e

= ∫

If u is constant throughout the volume, V, then U = u V.

Example: Parallel Plate Capacitor Think of the work done in bringing in the charges from infinity and placing them on the capacitor as the work necessary to produce the electric field lines and that the energy is strored in the electric field! From before we know that C = Aεεεε0/d so that the stored energy in the capacitor is

U = Q2/(2C) = Q2d/(2Aεεεε0). The energy stored in the electric field is U = uV = e0E2V/2 with E = σσσσ/e0 = Q/(e0A) and V = Ad, thus

U=Q2d/(2Aεεεε0), which is the same as the energy stored in the capacitor!

Volume

E

+Q

-Q

d

Area A

E-field

PHY2061 R. D. Field


Electric Energy Examples Example: How much electric energy is stored by a solid conducting sphere of radius R and total charge Q?

Answer: UKQ

R=

2

2

Example: How much electric energy is stored by a two thin spherical conducting shells one of radius R1 and charge Q and the other of radius R2 and charge -Q (spherical capacitor)?

Answer: UKQ

R R= −

2

1 221 1

Example: How much electric energy is stored by a solid insulating sphere of radius R and total charge Q uniformly distributed throughout its volume?

Answer: UKQ

RKQ

R= +

=1

15 2

35

2 2

R

Charge Q

E

R2

Q

E

R1

-Q

R

Charge Q

EE

PHY2061 R. D. Field


Charge Transport and Current Density Consider n particles per unit volume all moving with velocity v and each carrying a charge q. The number of particles, ∆∆∆∆N, passing through the (directed) area A in a

time ∆∆∆∆t is ∆ ∆N nv A t= ⋅!!

and the amount of charge, ∆∆∆∆Q, passing through the (directed) area A in a time ∆∆∆∆t is

∆ ∆Q nqv A t= ⋅!!

. The current, I(A), is the amount of charge per unit time passing through the (directed) area A:

I AQt

n q v A J A( )! ! ! ! !

= = ⋅ = ⋅∆∆ ,

where the “current density” is given by ! !J nqv drift= .

The current I is measured in Ampere's where 1 Amp is equal to one Coulomb per second (1A = 1C/s). For an infinitesimal area (directed) area dA:

dI J dA= ⋅! !

and !J n

d Id A

⋅ =" .

The “current density” is the amount of current per unit area and has units of A/m2. The current passing through the surface S is given by

I J dAS

= ⋅∫! #

.

The current, I, is the “flux” associated with the vector J.

q

A

v

PHY2061 R. D. Field


Electrical Conductivity and Ohms Law

Free Charged Particle: For a free charged particle in an electric field,

! ! !F ma qE= = and thus

! !a

qm

E= .

The acceleration is proportional to the electric field strength E and the velocity of the particle increases with time! Charged Particle in a Conductor:

However, for a charged particle in a conductor the average velocity is proportional to the electric field strength E and since

! !J nqvave= we have

! !J E= σ ,

where σσσσ is the conductivity of the material and is a property of the conductor. The resistivity ρρρρ = 1/σσσσ. Ohm's Law: ! !J E= σ

I JA E A= = σ

∆ V E LIA

LLA

I R I= = =

=

σ σ

∆∆∆∆V = IR (Ohm's Law) R = L/(σσσσA) = ρρρρL/A (Resistance) Units for R are Ohms 1ΩΩΩΩ = 1V/1A

q m E

q E

Conductor

Current Density JElectric Field E

Conductor σσσσ

Length L

Current I

A

V1 V2Potential Change ∆∆∆∆V

PHY2061 R. D. Field


Resistors in Series & Parallel Parallel: In this case ∆∆∆∆V1=∆∆∆∆V2=∆∆∆∆V and I=I1+I2. Hence, I = I1 + I2 = ∆V1/R1 + ∆V2/R2 = (1/R1+1/R2)∆V so 1/R = I/∆∆∆∆V = 1/R1 + 1/R2, where I used I1 = ∆V1/R1 and I2 = ∆V2/R2. Also, ∆V = I1R1 = I2R2 = IR so I1 = R2I/(R1+R2) and I2 = R1I/(R1+R2).

Resistors in parallel add inverses.

Series: In this case ∆∆∆∆V=∆∆∆∆V1+∆∆∆∆V2 and I=I1=I2. Hence, ∆V = ∆V1 + ∆V2 = I1R1+I2R2 = (R1+R2)I so R = ∆∆∆∆V/I = R1 + R2, where I used ∆V1 = I1R1 and ∆V2 = I2R2.

Resistors in series add.

∆∆∆∆VR1 R2

I

∆∆∆∆V1

I1 I2

∆∆∆∆V2

∆∆∆∆V

R1

R2

I

∆∆∆∆V1

∆∆∆∆V2

PHY2061 R. D. Field


Direct Current (DC) Circuits

Electromotive Force: The electromotive force EMF of a source of electric potential energy is defined as the amount of electric energy per Coulomb of positive charge as the charge passes through the source from low potential to high potental.

EMF = εεεε = U/q (The units for EMF is Volts)

Single Loop Circuits:

ε - IR = 0 and I = εεεε /R (Kirchhoff's Rule)

Power Delivered by EMF (P = εεεεI):

dW dq PdWdt

dqdt

I= = = =ε ε ε

Power Dissipated in Resistor (P = I2R):

dU V dq PdUdt

Vdqdt

V IR R R= = = =∆ ∆ ∆

∆∆∆∆V

+EMF

-

I

EMF + ∆∆∆∆V = 0

V

I

R+

εεεε-

I

PHY2061 R. D. Field


DC Circuit Rules

Loop Rule: The algebraic sum of the changes in potential encountered in a complete traversal of any loop of a circuit must be zero.

∆Viloop

=∑ 0 .

Junction Rule: The sum of the currents entering any junction must be equal the sum of the currents leaving that junction.

I Iiin

iout

=∑ ∑

Resistor: If you move across a resistor in the direction of the current flow then the potential change is ∆∆∆∆VR = - IR.

Capacitor: If you move across a capacitor from minus to plus then the potential change is

∆∆∆∆VC = Q/C, and the current leaving the capacitor is I = -dQ/dt.

Inductor (Chapter 38): If you move across an inductor in the direction of the current flow then the potential change is

∆∆∆∆VL = - L dI/dt.

∆∆∆∆V

+EMF

-

I

EMF + ∆∆∆∆V = 0

∆∆∆∆V=-IR

I

∆∆∆∆V =Q /C

I

Q +

-

∆∆∆∆VL=-LdI/dt

I

L

PHY2061 R. D. Field


Charging a Capacitor After the switch is closed the current is entering the capacitor so that I = dQ/dt, where Q is the charge on the capacitor and summing all the potential changes in going around the loop gives

ε − − =IRQC

0 ,

where I(t) and Q(t) are a function of time. If the switch is closed at t=0 then Q(0)=0 and

ε − − =RdQdt

QC

0 ,

which can be written in the form

( )dQdt

Q C= − −1τ

ε , where I have define ττττ=RC.

Dividing by (Q-εC) and multipling by dt and integrating gives

( )dQ

Q Cdt

Qt

−= −∫ ∫ε τ0

0

1 , which implies ln

Q CC

t−−

= −

εε τ .

Solving for Q(t) gives

( )Q t C e t( ) /= − −ε τ1 . The curent is given by I(t)=dQ/dt which yields

I tC

eR

et t( ) / /= =− −ετ

ετ τ. The quantity ττττ=RC is call the time

constant and has dimensions of time.

Switch

+εεεε

-

R

C

0.00

0.25

0.50

0.75

1.00

1.25

1.50

0 1 2 3 4

Time

Q(t)

Charging a Capacitor

PHY2061 R. D. Field


Discharging a Capacitor After the switch is closed the current is leaving the capacitor so that I = -dQ/dt, where Q is the charge on the capacitor and summing all the potential changes in going around the loop gives

QC

IR− = 0 ,

where I(t) and Q(t) are a function of time. If the switch is closed at t=0 then Q(0)=Q0 and

QC

RdQdt

+ = 0 ,

which can be written in the form dQdt

Q= −1τ , where I have defined ττττ=RC.

Dividing by Q and multiplying by dt and integrating gives

dQQ

dtQ

Qt

0

1

0∫ ∫= −

τ , which implies lnQQ

t0

= −

τ .

Solving for Q(t) gives

Q t Q e t( ) /= −0

τ . The current is given by I(t)=-dQ/dt which yields

I tQRC

e t( ) /= −0 τ.

The quantity ττττ=RC is call the "time constant" and has dimensions of time.

Switch

+

-R

C

Discharging a Capacitor

0.00

0.25

0.50

0.75

1.00

0 1 2 3 4

Time

Q(t)

PHY2061 R. D. Field

Department of Physics relativity_1.doc University of Florida

Postulates of Special Relativity

Consider two frames of reference the O-frame and the O'-frame moving at a constant velocity V, with respect to each other at let the origins coincide at t= t' = 0. Follow the path of a light ray that was emitted at the origin of the two frames at the instant they crossed. First Postulate of Special Relativity (“Relativity Principle”): • The basic laws of physics are identical in all systems of reference

(frames) which move with uniform (unaccelerated) velocity with respect to one another. The laws of physics are invariant under a change of inertial frame. The laws of physics have the same form in all inertial frames. It is impossible to detect uniform motion.

Second Postulate of Special Relativity (“Constant Speed of Light”): • The speed of light in a vacuum has the same value, c, in all inertial

frames. The speed of light in a vacuum is always independent of the velocity of the source of the light or the velocity of the observer.

The entire theory of special relativity is derived from these two postulates.

Light Path in O-frame Light Path in O'-frame

0)( 2222

222

=−−−++=

=

zyxctzyxd

ctd

0)( 2222

222

=′−′−′−′′+′+′=′

′=′

zyxtczyxd

tcd

Must find the transformation that results in

22222222 )()( zyxtczyxct ′−′−′−′=−−−

2.99792458x108 m/s

y

x

z

y'

z'

x'

V

Light Ray O: (t,x,y,z)

O': (t'.x',y',z')

O O'

PHY2060Review

Experimental observation!

Invariant!

inertial

PHY2061 R. D. Field


Lorentz Transformation

Consider two frames of reference the O-frame (label events according to t,x,y,z) and the O'-frame (label events according to t',x',y',z') moving at a constant velocity V, with respect to each other at let the origins coincide at t= t' = 0. The Lorentz transformations tell us how the O and O' frame are related.

zzyy

tcxxxtcct

′=′=

′+′=′+′=)()(

βγβγ

zzyy

ctxxxcttc

=′=′

−=′−=′

)()(

βγβγ

zzyy

tcxxtcct

′=′=

′+′=′=

β

where 21/1/ βγβ −== cV . 4-vector Notation:

=

''''

100001000000

zyxct

zyxct

γβγβγγ

rLr ′= ~~

−

−

=

′′′′

zyxct

zyxtc

100001000000

γβγβγγ

rLr ~~ 1−=′

y

x

z

y'

z'

x'

V

Event O: (t,x,y,z)

O': (t'.x',y',z')

O O'

Lorentz Transformations: Special Relativity

Galilean Transformations: Classical Physics

PHY2060Review

PHY2061 R. D. Field


4-Vector Notation

4-vector “dot product”: Define the 4-vector dot product as follows:

23

22

21

20

20

~~ xxxxrrxrr −−−=⋅−≡⋅ !!

where

=

3

2

1

0

~

xxxx

r and

=

3

2

1

xxx

r!

Space-Time 4-vectors:

=

zyxct

r~

′′′′

=′

zyxtc

r~

=

100001000000

γβγβγγ

L

−

−

=−

100001000000

1 γβγβγγ

L

21/1/ βγβ −== cV Lorentz Transformations:

rLr ′= ~~ rLr ~~ 1−=′

Lorentz Invariant: A “Lorentz invariant” is any quantity that is the same in all inertial frames of reference (i.e. same in O and O' frame). The square of a Lorentz 4-vector is a Lorentz invariant (i.e. Lorentz scalar).

22 )~(~~~~~ rrrrrr ′=′⋅′=⋅=

Any four quantities that transform from O' to O according to Lorentz forms a Lorentz 4-vector

Same in all inertial frames of reference

4-vector 3-vector

PHY2060Review

3-vector dot product

PHY2061 R. D. Field


Space-Time Intervals

Consider two events A=(tA,xA,yA,zA) and B=(tB,xB,yB,zB) and define ∆∆∆∆t=tB-tA, ∆∆∆∆x=xB-xA, ∆∆∆∆y=yB-yA, ∆∆∆∆z=zB-zA. These space-time intervals also transform according to the Lorentz transformations.

A

BFrame O

x

ct

c∆∆∆∆t=c(tB-tA)

∆∆∆∆x=xB-xA

Light Cone

45o

A

BFrame O'

x'

ct'

c∆∆∆∆t'=c(t'B-t'A)

∆∆∆∆x'=x'B-x'A45o

zzyy

tcxxxtctc

′∆=∆′∆=∆

′∆+′∆=∆′∆+′∆=∆)()(

βγβγ

zzyy

tcxxxtctc

∆=′∆∆=′∆

∆−∆=′∆∆−∆=′∆

)()(

βγβγ

The following are Lorentz 4-vectors:

∆∆∆∆

=∆

zyxtc

r~

′∆′∆′∆′∆

=′∆

zyxtc

r~ and

=

dzdydxcdt

rd~

′′′′

=′

zdydxdtcd

rd~

Space-time Separation (∆∆∆∆S)2:

222 )~()~()( rrS ′∆=∆=∆ 2222 )()()()( xtcxtc ′∆−′∆=∆−∆

The quantity (∆∆∆∆S)2 is a Lorentz invariant (same in all inertial frames). If (∆∆∆∆S)2 > 0 the two events A and B are said to be “time-like” and there exists an inertial frame where the two events occur at the same spacial point (i.e. ∆∆∆∆x'=0). If (∆∆∆∆S)2 < 0 the two events A and B are said to be “space-like” and there exists an inertial frame where the two events occur simultaneously (i.e. ∆∆∆∆t'=0). If (∆∆∆∆S)2 = 0 the two events A and B are said to be “light-like” they can only be connected by light (travelling at speed c).

PHY2060Review

differentials

PHY2061 R. D. Field

Department of Physics relativity_4a.doc University of Florida

Analogy with Rotations

Consider two frames of reference the O-frame (label points according to x,y) and the O'-frame (label points according to x',y'). Let the origins of the two frames coincide and rotate the O'-frame about the z-axis by an angle θθθθ. The two frames are related by the following transformation (i.e. by a rotation).

θθθθ

cossinsincos

yxyyxx′+′=′−′=

θθθθ

cossinsincos

yxyyxx

+−=′+=′

Vector Notation:

′′

−=

yx

yx

θθθθ

cossinsincos

rRr ′= rr

−=

θθθθ

cossinsincos

R

=

yx

rr

′′

=′yx

rr

Rotational Invariant: 222222 rrryxyxrrr ′=′⋅′=′+′=+=⋅= rrrr

Lorentz Transformation: Let coshθθθθ = γγγγ and sinhθθθθ = βγβγβγβγ then

′′

=

xtc

xct

θθθθ

coshsinhsinhcosh

rLr ′= ~~

=

θθθθ

coshsinhsinhcosh

L

=

xct

r~

′′

=′xtc

r~

Lorentz Invariant: 222222 ~~~)()(~~~ rrrxtcxctrrr ′=′⋅′=′−′=−=⋅=

y

x

y'

x'

θθθθ

Point P: (x,y)

P': (x',y') O

O'

r = r'

sin2θθθθ + cos2θθθθ = 1

PHY2060Review

cosh2θθθθ - sinh2θθθθ = 1

Hyperbolic cosine Hyperbolic sine

Length of vector invariant under

rotations

“Length” of vector invariant under

“rotations”

PHY2061 R. D. Field


Moving Clocks & Simultaneity

Consider two clocks in the O-frame located a distance ∆∆∆∆x = L apart (simultaneous in the O-frame) and one clock at the origin of the O'-frame. Let event A be the comparison of the O and O' clock at the origin and let the two clocks agree. Let event B be the comparison of the O' clock with the second O clock at ∆∆∆∆x = L.

A

BFrame O

x

ct

c∆∆∆∆t=cγτγτγτγτ

∆∆∆∆x=L

A

B

Frame O'

x'

ct'

c∆∆∆∆t'=cττττ

∆∆∆∆x'=0

proper time =time at restwith clock

)()(

tcxxxtctc′∆+′∆=∆′∆+′∆=∆

βγβγ

0=′∆=′∆

xctc τ

where 21/1/ βγβ −== cV . Thus, τγβcx =∆ and γτ=∆t Simultaneity: Suppose that two events A and B a distance ∆∆∆∆x = L appart occur simultaneously in the O-frame (i.e. ∆∆∆∆t = 0).

A B

Frame O

x

ct

c∆∆∆∆t=0

∆∆∆∆x=L

Events simultaneous

A

B

Frame O'

x'

ct'

c∆∆∆∆t'=-βγβγβγβγL

∆∆∆∆x'=γγγγL

Events NOT simultaneous

LtcxxLxtctc

γβγγββγ

=∆−∆=′∆−=∆−∆=′∆

)()(

PHY2060Review

y

x∆∆∆∆x = L

y'

B

x'

V

Simltaneous in O-frame

O O'

A

Moving clocks are slower!

Not simultaneous in O'-frame, event B occurs first!

PHY2061 R. D. Field


Moving Lengths

Parellel Moving Lengths: The “proper length”, L0, of a rod is defined to be its length at rest. Suppose the rod is at rest in the O'-frame so that ∆∆∆∆x' = L0. The length of the moving rod as observed in the O-frame, L, is defined be marking both ends of the rod simultaneously in S (i.e. ∆∆∆∆t=0). Thus,

)( tcxx ∆−∆=′∆ βγ implies that

LxLx γγ =∆==′∆ 0 and γ/0LL =

where 21/1/ βγβ −== cV . Perpenduclar Moving Lengths: Perpendicular moving distances are invariant under uniform velocities since

0LyyL =′∆=∆=

y

x

z

y’

z’

x’

V Rod at rest inthe O' frame

O'O

Mark endssimultaneously in

the O frame

L0

PHY2060Review

y

x

z

y’

z’

x’

VRod at rest inthe O' frame

O'OL0

Parallel moving lengths are shorter!

Perpenduclar moving lengths are invariant!

Length Contraction!

PHY2061 R. D. Field


Velocity Transformations

Consider a particle moving with velocity v in the O-frame and v' in the O' frame.

yy

xx

vvVvv

′=+′=

In the O-Frame In the O'-Frame

dtdzvdtdyvdtdxv

z

y

x

=

=

=

tdzdvtdydvtdxdv

z

y

x

′′

=′′′

=′′′

=′

Lorentz Transformations: 21/1/ βγβ −== cV

)(

)'(

xdtcdcdtzddzyddy

cdtxddx

′+′=′=′=

+′=

βγ

βγ

Velocities:

′+

′=

′′

+

′′=

′+′′

==

′+

+′=

′′

+

+

′′

=′+′

′+′==

x

yy

x

xx

vcV

v

tdxd

c

tdydcxdtd

yddtdyv

vcV

Vv

tdxd

c

ctdxd

cxdtdtcdxd

dtdxv

2

2

11

/)/(

1

)(

1)/()(

γβγβγ

β

β

βγβγ

y

x

z

y'

z'

x'

V

Particle O: (vx,vy,vz)

O': (vx',vy',vz')O O'

PHY2060Review

Galilean Transformation Classical Physics

Relativity!

PHY2061 R. D. Field


Relativistic Energy and Momentum

Relativistic Energy: The total relativistic energy is the sum of the kinetic energy (energy of motion) plus the rest mass energy (RME = m0c2).

20cmKERMEKEE +=+=

Also, the relativistic energy is equal to the relativistic mass, m, times c squared.

20

2 cmmcE γ== with 0mm γ=

where 21/1/ βγβ −== cv . Relativistic Kinetic Energy:

201

20

20

20

2

21)1(

)1(

vmcmKE

cmcmmcRMEEKE

→−=

−=−=−=

<<βγ

γ

Relativistic Momentum: The relativistic momentum p is the relativistic mass, m, time the velocity.

vmvmp !!!0γ== 0mm γ=

Energy Momentum Connection:

220

22 )()( cmcpE += with 222zyx pppp ++=

Speed ββββ of a particle: The speed of an object with rest mass m0 is given by

220

2 )()( cmcpcp

Ecp

cv

+===β .

PHY2060Review

Relativistic Mass

Mass of the object at rest

Classical KE

Relativistic energy and momentum are conserved!

PHY2061 R. D. Field

Department of Physics relativity_8b.doc University of Florida

Relativistic Kinetic Energy (derivation)

Relativistic Force: The force is equal to the rate of chance of the (relativistic) momentum as follows:

vdtdm

dtvdm

dtvmd

dtpdF !

!!!!+=== )(

where 0mm γ= is the relativistic mass. Relativistic Kinetic Energy: The kinetic energy of a particle is (as classical) the total work done in moving particle from rest to the speed v as follows:

RMEEcmmcdmcdmvdmvc

dmvmvdvmvvddxdtmvdFdxKE

m

m

m

m

−=−==+−=

+====

∫∫

∫∫∫∫2

022222

2

00

))((

)()()(

where I used )/1( 22

022 mmcv −= and dmvcmvdv )( 22 −= .

Energy Momentum Connection:

)/1/( 2220

2 cvmm −= and 20

222 )/1( mcvm =− which implies that

20

2222 / mcvmm =− and 420

22242 cmcvmcm =− thus

220

22 )()( cmcpE += . Speed ββββ of a particle: Since mvp = and 2/ cEm = we get 2/ cEvp = and thus

220

2 )()( cmcpcp

Ecp

cv

+===β .

PHY2060Review

Classically this term is

zero and F = ma

Time t = 0

F

Particle at rest:v = 0, m = m0

Later time t

v

Particle moving atspeed v, m = γγγγm0

PHY2061 R. D. Field


Relativistic Energy & Momentum

Consider two frames of reference the O-frame (label energy and momentum according to E,px,py,pz) and the O'-frame (label energy and momentum according to E',px',py',pz') moving at a constant velocity V, with respect to each other at let the origins coincide at t= t' = 0. The Lorentz transformations tell us how the frames are related.

zz

yy

xx

x

pccppccp

EpccppcEE

′=′=

′+′=′+′=

)()(

βγβγ

zz

yy

xx

x

cppccppc

EcppccpEE

=′=′

−=′−=′

)()(

βγβγ

where 21/1/ βγβ −== cV . The following are Lorentz 4-vectors:

=

z

y

x

cpcpcpE

p~

′′′′

=′

z

y

x

pcpcpcE

p~ and

=

z

y

x

cdpcdpcdpdE

pd~

′′′′

=′

z

y

x

pcdpcdpcdEd

pd~

Invariant Mass:

220

22 )()~(~~~~~ cmpppppp =′=′⋅′=⋅=

y

x

z

y'

z'

x'

V

Object O: (E,px,py,pz)

O': (E',px',py',pz')O O'

PHY2060Review

differentials

Same in all frames!

PHY2061 R. D. Field


Force Transformations

Consider a particle moving with velocity v' in the x' direction in the the O' frame. Let ββββ = V/c and ββββ' = v'/c. Hence, ββββ' = px'c/E' and

tdpdc

tdpd

pdEd

tdEd

cEpc

pdEd

cmpcE

xx

x

x

x

x

′′′=

′′

′′

=′′

′=′′

=′′

+′=′

β

β2

220

22 )(

In the O-Frame In the O'-Frame

dtdp

F

dtdpFdtdxv

yy

xx

=

=

=

tdpd

F

tdpdFtdxdv

yy

xx

′′

=′

′′

=′′′

=′

Lorentz Transformations: 21/1/ βγβ −== cV

)(

)'(

xdtcdcdtyddy

cdtxddx

′+′=′=

+′=

βγ

βγ

)(

)(

x

yy

xx

pcdEddEpddp

Edpcdcdp

′+′=′=

′+′=

βγ

βγ

Forces:

( )( )

yyyyy

y

xxx

x

xxx

Ftdpd

tdxd

c

tdpdcxdtd

pddt

dpF

Ftdpd

tdpd

tdxd

c

tdEd

ctdpd

cxdtdcEdpd

dtdpF

′′+

=′′

′+=

′′

+

′′=

′+′′

==

′=′′

=′′

′+′+=

′′

+

′′

+′′

=′+′′+′

==

)1(1

)1(1

1

/)/(

11

1)/()/(

ββγββγβγβγ

ββββ

β

β

βγβγ

⊥⊥ ′′+

=′= FFFF)1(

1|||| ββγ

y

x

z

y'

z'

x'

V

Particle O: (E,px,py,pz)

O': (E',px',py',pz')O O'

v'

PHY2060Review

PHY2061 R. D. Field


Invariance of Electric Charge

In O-frame charge is defined using Gauss’ Law as follows:

∫ ⋅=)(

0tS

AdEQ!!

ε

In O'-frame charge is also defined using Gauss’ Law as follows:

∫′′

′⋅′=′)(

0tS

AdEQ!!

ε

Experimental result: Q = Q'

Electric charge is a Lorentz invariant quantity (it is a Lorentz scalar)!

∫∫′′

′⋅′=⋅)()( tStS

AdEAdE!!!!

Measure total flux through surface S at

time t.

Measure total flux through surface S' at

time t'.

y

x

z

y'

z'

x'

VSurface S

O O'

Q

Surface S'

PHY2061 R. D. Field


Moving Charge Density

Linear Charge Density: Let λλλλ0 be the linear charge density in the frame at rest with respect to the charges (O'-frame). Then,

λλλλ0 = charge/length = Q'/L0. Let λλλλ be the charge density observed in the O-frame. Then, λλλλ = Q/L, but Q = Q' and L = L0/γγγγ. Thus,

00

γλγλ =′

==LQ

LQ

and 0γλλ =

where 21/1/ βγβ −== cV . Surface Charge Density: Let σσσσ0 be the surface charge density in the frame at rest with respect to the charges (O'-frame). Then,

σσσσ0 = charge/area = Q'/(L0)2. Let σσσσ be the charge density observed in the O-frame. Then, λλλλ = Q/(LxLy), but Q = Q' and Lx = L0/γγγγ and Ly = L0. Thus,

020

γσγσ =′

==LQ

LLQ

yx and 0γσσ =

where 21/1/ βγβ −== cV .

y

x

z

y'

z'

x'

VCharges at rest

in O' frame

O O'+ + + + + + +

L0

y

x

z

y'

z'

x'

V

Charges at restin O' frame

O O'

L0

+ + + + + ++ + + + + ++ + + + + ++ + + + + +

L0

PHY2061 R. D. Field


Electric Field Measured in Different Frames

Transverse Components: Consider two (large) parallel sheets of opposite charge density at rest in the x'-z' plane O'-frame. In the O'-frame, Ey' = σσσσ0/εεεε0. In the O-frame, Ey = σσσσ/εεεε0, but we know that

0γσσ =

with 21/1/ βγβ −== cV . Thus,

yy EE ′= γ and similarly zz EE ′= γ . ⊥⊥ ′= EE γ Parallel Component: Consider two (large) parallel sheets of opposite charge density at rest in the y'-z' plane O'-frame. In the O'-frame, Ex' = σσσσ0/εεεε0. In the O-frame, Ex = σσσσ/εεεε0, but in this case

0σσ = .

Thus, xx EE ′= .

|||| EE ′=

y

x

z

y'

z'

x'

V


O O'

+ + + + + + + + + + + ++ + + + + +

- - - - - - - - - - - - - - - -- - - - - - - - -

-σσσσ0000

+σσσσ0000E'

y

x

z

y'

z'

x'

V


O O'

- - - - - - - - - - - - - - - - - - --σσσσ+σσσσ

++ + + + + + + + + + +

+ + E'

Transverse components of the electric field transform acccording to

The parallel component of the electric field is

invariant!

PHY2061 R. D. Field


Electric Field of a Moving Charge Particle

Consider a point charge Q at rest at the origin in the O'-frame and consider the electric field in the O-frame at the moment the origins of the two frames coincide (t = t' = 0). We know that,

rrKQE ′

′=′ ˆ

)( 2

!.

In the O'-frame:

2/3222

2/3222

)(sin

)(

)(cos

)(

yxyKQ

rKQE

yxxKQ

rKQE

y

x

′+′′

=′′

=′

′+′′

=′′

=′

θ

θ

22

/sin/cos

yxrryrx

′+′=′′′=′′′=′

θθ

In the O-frame:

22

/sin/cos

yxrryrx

+===

θθ

21/1/ βγβ −== cV

2/3222

2

2/32222/322

2/3222

2

2/32222/322

)sin1(sin)1(

)()(

)sin1(cos)1(

)()(

θβθβ

γγγγ

θβθβ

γγ

−−=

+=

′+′′

=′=

−−=

+=

′+′′

=′=

rKQ

yxyKQ

yxyKQEE

rKQ

yxxKQ

yxxKQEE

yy

xx

The magnitude of E is given by 222yx EEE += and hence,

2/3222

2

)sin1()1(θβ

β−

−=r

KQE

E'y'

Qx'

V

Charge Q at rest in O' frame

O' P'=(x',y')

θθθθ'r'

Ey

Qx

Charge Q moving at speed Vin O frame

OP=(x,y)

θθθθ

r

E-field of a moving charge!

PHY2061 R. D. Field


Electric Field of a Moving Charge Particle

There is the same amount of electric flux lines but lines are shifted. In both cases

∫ ⋅=)(

0tS

AdEQ!!

ε

The E-field of a moving charge is a remarkable electric field. It is a field that no stationary charge distribution, in whatever form, can produce. For this electric field has the property that

0≠×∇ E!!

and the line integral of E is not zero around every closed loop. The field of a moving charge is not an electrostatic field which has

0=×∇ E!!

.

E

Q

Stationary Charge:Spherically Symmetric E-field

E

θθθθ

Moving Charge:E-field NOT Spherically Summetric

Q

V

E

θθθθQ

V

Closed Loop

Electrostatic Field

PHY2061 R. D. Field


Force on a Charged Particle in an E-Field Consider the interaction of a moving charge with stationary charges (Case I, O-frame, lab frame) or a stationary charge with moving charges (Case II, O'-frame, particle frame). Case I: Parallel Force

||||

||||

||||

||||

EQFQEFFFEE

′=′=

′=′=

Case II: Perpendicular Force

⊥⊥

⊥⊥

⊥⊥

⊥⊥

′=′=

′==′

EQFQEFFF

EEγ

γ/

Hence,

EQFrr

= in any frame of reference. The laws of physics are invariant under change of frame of reference!

E

QV

++++++++

--------

Lab FrameO-frame

E'

Q

V ++++++++

--------

Particle FrameO'-frame

V

EQ V

Lab FrameO-frame

+ + + + + + + + + +

- - - - - - - - - - - - - -

E' Q

V

Particle Frame O'-frame

+ + + + + + + + + +

- - - - - - - - - - - - - -

V

Same in all frames!

PHY2061 R. D. Field


Interaction of a Moving Charge with a Current Carrying Wire

Consider the force on of a moving charge particle with speed, V, due to a (neutral) wire carring a current, I. Even though the wire is electrically neutral (no electric force), the charge particle experiences a force, Fnew. This new force is the magnetic force and can be calculated from our knowledge of the electric force and special relativity. Case I: V and I in the same direction

0

2

41

2

πε=

=

Kr

VIcKQFnew

Case II: V and I in the opposite directions

0

2

41

2

πε=

=

Kr

VIcKQFnew

The Magnetic Force:

22

cKk

rkIBBVQFF wireBnew ==×==

!!!!

Fnew

QV

Current INeutral Wire

r

Fnew

QV

Current INeutral Wire

r

Attractive Toward Wire

Repulsive Away from Wire

PHY2061 R. D. Field


Interaction of a Moving Charge with other Moving Charges – Case I

Lab Frame (O-frame): Charge particle Q moving to the right with velocity V. Neutral wire with current I = λλλλ+V moving to the right with λλλλ =λλλλ+-λλλλ- =0, where λλλλ is the net charge density of the wire. Particle experiences a “magnetic” force F.

Particle Frame (O'-frame): Charge particle Q at rest. Wire is now negatively charged with λλλλ' =λλλλ'+-λλλλ'- <0, and the particle experiences an electric field, E' = 2Kλλλλ'/r' and an (attractive) electric force F’ = 2KQλλλλ'/r'.

Calculate the force in the lab frame from knowledge of the electric force in the particle frame!

Lab Frame: 000 =−=== −+−−++ λλλλλγλλ

++−+ === 000 γβλλλγλ cVI Particle Frame:

+−+−+−−++ −=−=′−′=′=′=′ 022

0000 λγβγλλλλλγλλλλ

rKQrKQF ′=′′=′ +⊥ /2/2 0

22 λγβλ Lab Frame:

rcVIKQ

crIKQrKQFF 20

2 22/2/ ===′= +⊥⊥

βγλβγ

2//2 cKkrkIBQVBF ===⊥

y

x

z

V

Neutral Wire

O

+ + + + + + + + + + + + + + + + + + + + + +- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - V

QMoving Charge

F

y'

x'

z'

O'

+ + + + + + + + + + + + + + + + + + + + + +- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -V

Q

Charge at rest

Negatively Charged Wire

F'

Magnetic Force!

“Rest” Densities Neutral Wire

Negatively Charged Wire!

Attractive Force

Magnetic Field!

PHY2061 R. D. Field


Interaction of a Moving Charge with other Moving Charges – Case II

Lab Frame (O-frame): Charge particle Q moving to the right with velocity V. Neutral wire with current I = λλλλ+V moving to the left with λλλλ =λλλλ+-λλλλ- =0, where λλλλ is the net charge density of the wire. Particle experiences a “magnetic” force F.

Particle Frame (O'-frame): Charge particle Q at rest. Wire is now positively charged with λλλλ' =λλλλ'+-λλλλ'- >0, and the particle experiences an electric field, E' = 2Kλλλλ'/r' and an (repulsive) electric force F’ = 2KQλλλλ'/r'.

Calculate the force in the lab frame from knowledge of the electric force in the particle frame!

Lab Frame: 000 =−=== −+−−++ λλλλλγλλ

++−+ === 000 γβλλλγλ cVI Particle Frame:

++−+−−++ =−′=′−′=′=′′=′ 022

02

00 )( λγβλγγλλλγλλλγλ )1()1/(2)/1/()( 2222 βγγβ +=′+−=−−=′ VcVvVvv xxx

rKQrKQF ′=′′=′ +⊥ /2/2 0

22 λγβλ Lab Frame:

rcVIKQ

crIKQrKQFF 20

2 22/2/ ===′= +⊥⊥

βγλβγ

2//2 cKkrkIBQVBF ===⊥

y

x

z

V

Neutral Wire

O

+ + + + + + + + + + + + + + + + + + + + + +- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -V

QMoving ChargeF y'

x'

z'

O'

+ + + + + + + + + + + + + + + + + + + + + +- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

V

QCharge at rest

Positively Charged Wire

F'

v'x

Magnetic Force!

“Rest” Densities Neutral Wire

Positively Charged Wire!

Repulsive Force

Magnetic Field!

PHY2061 R. D. Field


Transformation Properties of E and B

Consider two frames of reference the O-frame and the O'-frame moving at a constant velocity V, with respect to each other at let the origins coincide at t= t' = 0. Note that B has units of Tesla = N/(C m/s) and E has units of N/C so that E and cB have the same units. Lorentz Transformations:

)()( yzzzyyxx cBEEcBEEEE βγβγ +=′−=′=′

)()( yzzzyyxx EcBBcEcBBccBBc βγβγ −=′+=′=′ Lorentz Invariants:

BErr

⋅ and 222 BcE − are Lorentz invariants

Special Case: Suppose that 0=Br

everywhere in the O-frame. Then, in the O'-frame

zzyyxx EEEEEE γγ =′=′=′

yzzyx EBcEBcBc βγγβ −=′=′=′ 0

yzzyx EBcEBcBc ′−=′′=′=′ ββ0 and thus

Evc

B ′×′=′rrr

21

where xVv ′−=′ ˆr.

y

x

z

y'

z'

x'

V

O : (Ex,Ey,Ez,Bx,By,Bz)O': (E'x,E'y,E'z,B'x,B'y,B'z)

O O'Point P

True if B = 0 in O-frame

Same in all inertial frames

PHY2061 R. D. Field


Magnetic Field of a Moving Charged Particle

Particle Frame:

0ˆ2 == Br

rKQE

!!

Lab Frame:

EVc

B

rr

KQE

!!!

!

×=

−−=

2

2/3222

2

1

ˆ)sin1(

)1(θβ

β

B is perpenducular to both E and V and has a magnitude

2/32222

2

)sin1(sin)1(θβ

θβ−−=

rcVKQB

E

Q

Particle Frame:Particle at rest

B = 0

E

θθθθ

Lab Frame:Particle moving at speed V

Q V

EQ

VB

E

Q

V out of paper

B

PHY2061 R. D. Field


Force Between Two Moving Charged Particles

Consider the force on q moving with velocity v due to Q moving with velocity V. The moving charge Q produces both an electric and magnetic field given by

EVc

B

rr

KQE

!!!

!

×=

−−=

2

2/3222

2

1

ˆ)sin1(

)1(θβ

β

where ββββ = V/c, which produce a force on q given by

BvqEqF!!!!

×+= and thus

)ˆ()sin1(

)1(ˆ)sin1()1(

2/32222

2

2/3222

2

rVvrc

KqQrr

KqQF ××−

−+−

−=!!!

θββ

θββ

Classical Result: If is assume that v << c and V << c then we arrive at the classical formula (approximation)

)ˆ(ˆ 222 rVvrc

KqQrr

KqQFFF BE ××+=+=!!!!!

where

BvqrVvr

kqQF

Eqrr

KqQF

B

E

!!!!!

!!

×=××=

==

)ˆ(

ˆ

2

2

with k = K/c2 and

)ˆ(

ˆ

2

2

rVrkQB

rr

KQE

×=

=!!

!

r

r

q

Q

V

v

θθθθ

Classical Approximation

PHY2061 R. D. Field

Department of Physics chp34_1.doc Univesity of Florida

The Electromagnetic Force

The Force Between Two-Charged Particles (at rest): The force between two charged particles at rest is the electrostatic force and is given by

!F

KQqr

rE = 2 " (electrostatic force) ,

where K = 8.99x109 Nm2/C2.

The Force Between Two Moving Charged Particles: The force between two moving charged particles is the electromagnetic force and is given by

! ! !F

KQqr

rKQqc r

v V rEM = + × ×2 2 2" "

(electromagnetic force) where K = 8.99x109 Nm2/C2 and c = 3x108 m/s (speed of light in a vacuum). The first term is the

electric force and the second (new) term is the called the magnetic force so that

! ! !F F FEM E B= + , with

! !

! ! ! ! ! ! !

FKQqr

r qKQr

r qE

FKQqc r

v V r qvKQc r

V r qv B

E

B

= =

=

= × × = × ×

= ×

2 2

2 2 2 2

" "

" "

The electric and magnetic fields due to the particle Q are

!

! !

EKQr

r

BKQc r

V r

=

= ×

2

2 2

"

"

The electromagnetic force on q is given by ! ! ! !F qE qv BEM = + × (Lorenz Force).

r

r

q

QV =0

v=0

r

r

q

Q

V

v

E

B

Electric and Magnetic Fields of aCharged Particle Q moving with

Speed V (out of the paper)

Q

Classical Approximation

PHY2061 R. D. Field


The Magnetic Force The Force on Charged Particle in a Magnetic Field:

The magnetic force an a charged particle q in a magnetic field B is given by ! ! !

F qv BB = × . The magnitude of the magnetic force is FB = qvB sinθθθθ and B = FB/(qv sinθθθθ) is the definition of the magnetic field. (The

units for B are Tesla, T, where 1 T = 1 N/(C m/s)). The magnetic force an infinitesimal charged particle dq in a magnetic field B is given by

dF dqv BB

! ! != × .

The Force on Wire Carrying a Current in a Magnetic Field:

A current in a wire corresponds to moving charged particles with I = dq/dt. The magnetic force on the charge dq is

dF dqv BB

! ! != × ,

and the speed v=dl/dt. Hence,

dqv dqdldt

Idl!!

!= = ,

and the magnetic force on a infinitesimal length dl of the wire becomes dF Idl BB

! " != × . The total

magnetic force on the wire is ! ! " !F dF Idl BB B= = ×∫ ∫ ,

which for a straight wire of length L in a uniform magnetic field becomes ! ! !F IL BB = × .

q

B

v

θθθθ

B-out

I

dF

dq

dl

PHY2061 R. D. Field


Vector Multiplication: Dot & Cross Two Vectors: Define two vectors according to !

!A A x A y A zB B x B y B z

x y z

x y z

= + += + +

" " "

" " " .

The magnitudes of the vectors is given by !

!A A A A A

B B B B Bx y z

x y z

= = + +

= = + +

2 2 2

2 2 2

Dot Product (Scalar Product): The dot product, S, is a scalar and is given by

S A B A B A B A B A Bx x y y z z= ⋅ = = + +! ! ! !

cosθ

Cross Product (Vector Product): The cross product,

!C , is a vector and is given by ! ! !

C A B A B A B x A B A B y A B A B zy z z y x z z x x y y x= × = − − − + −( ) " ( ) " ( )"The magnitude of the cross product is given by ! ! ! ! !

C A B A B= × = sinθ The direction of the cross product can be determined from the "right hand rule". Determinant Method: The cross product can be constructed by evaluating the following determinant:

! ! !C A B

x y zA A AB B B

x y z

x y z

= × =" " "

B

A

θθθθ

PHY2061 R. D. Field


Motion of a Charged Particle in a Magnetic Field Consider a charged particle q with velocity !v v x v yx y= +" " , and kinetic energy

E mv mv vkin = = ⋅12

12

2 ! !,

in a uniform magnetic field !B Bz= − " .

The magnetic force on the particle is given by ! ! !

F qv BB = × . The magnetic force does not change the speed (kinetic energy) of the charged particle. The magnetic force does no work on the charged particle since the force is always perpendicular to the path of the particle. There is no change in the particle's kinetic energy and no change in its speed.

Proof: We know that ! ! ! ! !

! !F qv B m

dvdt

mdvdt

qv BB = × = = × . Hence

( )dEdt

mdvdt

md v v

dtmv

dvdt

qv v Bkin = =⋅

= ⋅ = ⋅ × =12

12

02 ! !

!!

! ! !,

and thus Ekin (and v) are constant in time. The magnetic force can change the direction a charged particle but not

its speed. The particle undergoes circular motion with angular velocity ωωωω = qB/m.

vdFm

dtqvBm

dtddt

qBm

θ

ωθ

= =

= =

x-axis

v

R

y-axis

B-in

q

v(t+dt)

v(t)

dθθθθFdt/m

PHY2061 R. D. Field


Circular Motion: Magnetic vs Gravitational Planetary Motion: For circular planetary motion the force on the orbiting planet is equal the mass times the centripetal acceleration, a = v2/r, as follows:

FG = GmM/r2 = mv2/r Solving for the radius and speed gives, r = GM/v2 and v = (GM/r)1/2. The period of the rotation (time it takes to go around once) is given by

T=2ππππr/v=2ππππGM/v3 or TGM

r=2 3 2π / . The angular velocity, ω ω ω ω = dθθθθ/dt,

and linear velocity v = ds/dt are related by v = rωωωω, since s = rθθθθ. Thus, 2/3/ rGM=ω . The angular velocity an period are related by T = 2ππππ/ωωωω

and the linear frequency f and ωωωω are related by ω ω ω ω = 2ππππf with T = 1/f. Planets further from the sum travel slower and thus have a longer period T.

Magnetism: For magnetic circular motion the force on the charged particle is equal its mass times the centripetal acceleration, a = v2/r, as follows:

FB = qvB = mv2/r. Solving for the radius and speed gives,

r = mv/(qB) = p/(qB) , and v = qBr/m. The period of the rotation is given by T = 2ππππr/v =

2ππππm/(qB) and is independent of the radius! The frequency (called the cyclotron frequency) is given by f = 1/T= qB/(2ππππm) is the same for all particles with the same charge and mass (ωωωω = qB/m).

M m

v

r

v

r

B-in

q

PHY2061 R. D. Field


The Magnetic Field Produced by a Current

The Law of Biot-Savart: The magnetic field at the point P due to a charge dQ moving with speed V within a wire carrying a current I is given by

dBKdQc r

V r! !

= ×2 2 "

where K = 8.99x109 Nm2/C2 and c = 3x108 m/s (speed of light in a vacuum).

However, we know that I = dQ/dt and !

!

Vd ld t= so that dQ V Idl

! !=

and,

dBkIr

dl r! !

= ×2 " (Law of Biot-Savart),

where k = K/c2 = 10-7 Tm/A. For historical reasons we define µµµµ0 as follows:

kKc

= =µπ0

24 , (µµµµ0 = 4π π π π x 10-7 Tm/A).

Example (Infinite Straight Wire):

An infinitely long straight wire carries a steady current I. What is the magnetic field at a distance r from the wire?

Answer: B rkIr

( ) =2

r

r

P

dQ d lI

W ire

B

Magnetic Field of an Infinite WireCarrying Current I (out of the paper)

I-out

I

r

PHY2061 R. D. Field


Calculating the Magnetic Field (1)

Example (Straight Wire Segment): An infinitely long straight wire carries a steady current I. What is the magnetic field at a distance y from the wire due to the segment 0 <x < L?

Answer: B rkIy

Ly L

( ) =+2 2

Example (Semi-Circle): A thin wire carrying a current I is bent into a semi-circle of radius R. What is the magnitude of magnetic field at the center of the semi-circle?

Answer: BkIR

=π

Example (Circle): A thin wire carrying a current I is forms a circle of radius R. What is the magnitude of magnetic field at the center of the semi-circle?

Answer: BkI

R=

2π

I

y

L

RIP

R

I

P

PHY2061 R. D. Field


Calculating the Magnetic Field (2) Example (Current Loop): A thin ring of radius R carries a current I. What is the magnetic field at a point P on the z-axis a distance z from the center of the ring? Answer:

( )B zkI R

z Rz ( ) /=+

2 2

2 2 3 2π

Example (Magnetic Dipole): A thin ring of radius R carries a current I. What is the magnetic field at a point P on the z-axis a distance z >> R from the center of the ring?

Answer: B zkz

I R IAzB

B( ) = = =2

32µ

µ π

The quantity µµµµB is called the magnetic dipole moment, µµµµB = NIA,

where N is the number of loops, I is the current and A is the area.

RP z-axis

z

I

R P z-axis

zI

PHY2061 R. D. Field


Ampere's Law

Gauss' Law for Magnetism: The net magnetic flux emanating from a closed surface S is proportional to the amount of magnetic charge enclosed by the surface as follows:

ΦBS

enclosedMagneticB dA Q= ⋅ ∝∫

! !

.

However, there are no magnetic charges (no magnetic monopoles) so the net magnetic flux emanating from a closed surface S is always zero,

ΦBS

B dA= ⋅ =∫! !

0 which implies 0=⋅∇ B!!

Ampere's Law:

The line integral of the magnetic field around a closed loop (circle) of radius r around a current carrying wire is given by

B dl rB r kI ILoop

⋅ = = =∫!

2 4 0π π µ( ) .

This result is true for any closed loop that encloses the current I.

The line integral of the magnetic field around any closed path C is equal to µµµµ0 times the current intercepted by the area spanning the path:

B dl IenclosedC

⋅ =∫!

µ0 Ampere's Law

The current enclosed by the closed curve C is given by the integral over the surface S (bounded by the curve C) of the current density J as follows:

I J dAenclosedS

= ⋅∫! !

B

Magnetic Field of an Infinite WireCarrying Current I (out of the paper)is B(r) = 2kI/r.

I-out

r

Curve C

PHY2061 R. D. Field


Ampere's Law Examples

Example (Infinite Straight Wire with radius R): An infinitely long straight wire has a circular cross section of radius R and carries a uniform current density J along the wire. The total current carried by the wire is I. What is the magnitude of the magnetic field inside and outside the wire? Answer:

B rkIr

B rkrIR

out

in

( )

( )

=

=

2

22

.

Example (Infinite Solenoid): An infinitely long thin straight wire carrying current I is tightly wound into helical coil of wire (solenoid) of radius R and infinite length and with n turns of wire per unit length. What is the magnitude and direction of the magnetic field inside and outside the solenoid (assume zero pitch)? Answer:

B rB r nI

out

in

( )( )

==

0

0µ .

Example (Toroid): A solenoid bent into the shape of a doughnut is called a toriod. What is the magnitude and direction of the magnetic field inside and outside a toriod of inner radius R1 and outer radius R2 and N turns of wire carrying a current I (assume zero pitch)? Answer:

B r

B rkNIr

out

in

( )

( )

=

=

02

IR

BR

Infinite SolenoidI

R1

R2

Toriod

PHY2061 R. D. Field


Electromagnetic Induction Conducting Rod Moving through a Uniform Magnetic Field:

The magnetic force on the charge q in the rod is ! ! !

F qv BB = × . The induced EMF, εεεε, is equal to the amount of work done by the magnetic field in moving a unit charge across the rod,

ε = ⋅ = × ⋅ =∫ ∫1q

F dl v B dl vLBB

! ! ! ! !

.

In Steady State: In steady state a charge q in the rod experiences no net force since, ! !

F FE B+ = 0 , and thus, ! ! !

E v B= − × . The induced EMF (change in electric potential across the rod) is calculated from the electric field in the usual way,

ε = ⋅ = − × ⋅ =∫ ∫! ! ! ! !E dl v B dl vLB ,

which is the same as the work done per unit charge by the magnetic field.

B-out

vFB

q

Rod

L

B-out

vFB

q

Rod

L

FE

+ + +

- - -

PHY2061 R. D. Field


Induced Electric Fields

Lab Frame: Rod Frame:

xVVzBB rodext ˆˆ ==!!

0ˆ =′′=′ rodext VzBB!!

γ

yqVBBVqFF B ˆ−=×==!!!!

ycBEext ˆγβ−=′!

yVBqEqFF E ′−=′=′=′ ˆγ!!!

In Steady State: In Steady State:

0=+ Bind FEq!!

0=′+′ Eind FEq!!

yVBEind ˆ=!

yVBEind ′=′ ˆγ!

B'-out

E'

Rod

q

Rod Frame (O'-frame) x'

y'B-out

v

Rod

q

Lab Frame (O-frame) x

y

V

FB

q

Rod

F=qEind

+ + + + +

- - - - -

F'E

q

Rod

F' =qE'ind

+ + + + +

- - - - - -

Observer O: Inside the rod there has developed and induced electric field that exerts a force which just balances the magnetic force.

Observer O': Inside the rod there is no net electric field although there is a uniform magnetic field, no force arises from it because no charges are moving.

PHY2061 R. D. Field


Electromagnetic Induction Conducting Loop Moving through a Uniform Magnetic Field:

The magnetic force on the charge q in the loop on side 1 is, ! ! !

F qv BB 1 1= × , and for a charge q on side 2 to it is, ! ! !

F qv BB 2 2= × . However, because the magnetic field is uniform,

! !B B1 2= ,

and the induced EMF's on side 1 and side 2 are equal, εεεε1 = εεεε2, and

the net EMF around the loop (counterclockwise) is zero,

ε ε ε= ⋅ = − =∫1

01 2qF d lB

L o o p

! !.

Conducting Loop Moving through a Non-Uniform Magnetic Field:

If we move a conducting loop through a non-uniform magnetic field then induced EMF's on side 1 and side 2 are not equal, εεεε1 = vLB1, εεεε2 = vLB2, and the net EMF around the loop (counterclockwise) is,

ε ε ε= ⋅ = − = −∫1

1 2 1 2qF d l v L B BB

L o o p

! !( ) .

This induced EMF will cause a current to flow around the loop in a counterclockwise direction (if B1 > B2)!

B-out

vFB

q

Loop

1

FE

+ + +

- - -

2

B1

B2

FB1FB2

1 2

vL

PHY2061 R. D. Field


Faraday's Law of Induction Magnetic Flux: The magnetic flux through the surface S is defined by,

Φ BS

B d A= ⋅∫! !

.

In the simple case where B is constant and normal to the surface then ΦΦΦΦB = BA.

The units for magnetic flux are webbers (1 Wb = 1 Tm2). Rate of Change of the Magnetic Flux through Moving Loop:

The change in magnetic flux, dΦΦΦΦB, in a time dt through the moving loop is,

dΦΦΦΦB = B2dA-B1dA, with dA = vdtL so that d

d tvL B BBΦ = − − = −( )1 2 ε

where εεεε is the induced EMF. Hence,

ε = − dd t

BΦ (Faraday's Law of Induction).

Substituting in the definition of the induced EMF and the magnetic flux yields,

ε∂∂

= ⋅ = − = − ⋅

= − ⋅∫ ∫ ∫

! ! ! !!

!E d l

dd t

dd t

B d ABt

d AC lo se dL o o p

B

S u r fa c e S u r fa c e

Φ

We see that a changing magnetic field (with time) can produce an electric field!

B1

B2

vdt vdt

v

L

PHY2061 R. D. Field


Lenz's Law Example (Loop of Wire in a Changing Magnetic Field):

A wire loop with a radius, r, of 1 meter is placed in a uniform magnetic field. Suppose that the electromagnetic is suddenly switched off and the strength of the magnetic field decreases at a rate of 20 Tesla per second. What is the induced EMF in the loop (in Volts)? If the resistance of the loop, R, is 5 Ohms, what is

the induced current in the loop (in Amps)? What is the direction of the induced current? What is the magnitude and direction of the magnetic field produced by the induced current (the induced magnetic field) at the center of the circle?

Answers: If I choose my orientation to be counterclockwise then ΦΦΦΦB = BA and

εεεε = -dΦΦΦΦB/dt = -A dB/dt = -(πr2)(-20T/s) = 62.8 V. The induced current is I = εεεε/R = (62.8 V)/(5 Ω) = 12.6 A. Since εεεε is positive the current is flowing in the direction of my chosen orientation (counterclockwise). The induced magnetic field at the center of the circle is given by Bind = 2ππππkI/r = (2π x 10-7 Tm/A)(12.6 A)/(1 m) = 7.9 µµµµT and points out of the paper. Lenz's Law: It is a physical fact not a law or not a consequence of sign conventions that an electromagnetic system tends to resist change. Traditionally this is referred to as Lenz's Law:

Induced EMF's are always in such a direction as to oppose the change that generated them.

B-out changing with time

r

Loop

PHY2061 R. D. Field


Induction Examples

Example (simple generator): A conducting rod of length L is pulled along horizontal, frictionless, conducting rails at a constant speed v. A uniform magnetic field (out of the paper) fills the region in which the rod moves. The rails and the rod have negligible resistance but are connected by a resistor R. What is the induced EMF in the loop? What is the

induced current in the loop? At what rate is thermal energy being generated in the resistor? What force must be applied to the rod by an external agent to keep it in uniform motion? At what rate does this external agent do work on the system? Example (terminal velocity): A long rectangular loop of wire of width L, mass M, and resistance R, falls vertically due to gravity out of a uniform magnetic field. Instead of falling with an acceleration, g, the loop falls a constant velocity (called the terminal velocity). What is the terminal velocity of the loop? Example (non-uniform magnetic field):

A rectangular loop of wire with length a, width b, and resistance R is moved with velocity v away from an infinitely long wire carrying a current I. What is the induced current in the loop when it is a distance c from the wire?

B-out

v

Rod

R

B-out

Mg

L

a

b

v

c I

PHY2061 R. D. Field


Mutual & Self Inductance

Mutual Inductance (M): Consider two fixed coils with a varying current I1 in coil 1 producing a magnetic field B1. The induced EMF in coil 2 due to B1 is proportional to the magnetic flux

through coil 2, Φ 2 1 22

2 2= ⋅ =∫! !B dA N

coil

φ ,

where N2 is the number of loops in coil 2 and φφφφ2 is the flux through a single loop in coil 2. However, we know that B1 is proportional to I1 which means that ΦΦΦΦ2 is proportional to I1. The mutual inductance M is defined to be the constant of proportionality between ΦΦΦΦ2 and I1 and depends on the geometry of the situation,

MI

NI

N M I= = = =Φ

Φ2

1

2 2

12 2 2 1

φφ . The induced EMF in coil 2 due

to the varying current in coil 1 is given by, The units for inductance is a Henry

(1 H = Tm2/A = Vs/A).

Self Inductance (L): When the current I1 in coil 1 is varying there is a changing magnetic flux due to B1 in coil 1 itself! The self inductance L is defined to be the constant of proportionality between ΦΦΦΦ1 and I1 and depends on the geometry of the situation,

LI

NI

N LI= = = =Φ

Φ1

1

1 1

11 1 1 1

φφ ,

where N1 is the number of loops in coil 1 and φφφφ1 is the flux through a single loop in coil 1. The induced EMF in coil 2 due to the varying current in coil 1 is given by,

B1I1

Coil 1 Coil 2

I2

ε22 1= − = −

ddt

MdIdt

Φ

B1I1

Coil 1

ε11 1= − = −

ddt

LdIdt

Φ

PHY2061 R. D. Field


Energy Stored in a Magnetic Field When an external source of EMF is connected to an inductor and current begins to flow, the induced EMF (called back EMF) will oppose the increasing current and the external EMF must do work in order to overcome this opposition. This work is stored in the magnetic field and can be recovered by removing the external EMF. Energy Stored in an Inductor L: The rate at which work is done by the back EMF (power) is

P I LIdIdtback = = −ε ,

since εεεε = -LdI/dt. The power supplied by the external EMF (rate at which work is done against the back EMF) is

PdWdt

L IdId t

= = ,

and the energy stored in the magnetic field of the inductor is

U Pdt LIdIdt

dt LIdI LIt I

= = = =∫∫ ∫0 0

212 .

Energy Density of the Magnetic Field u: Magnetic field line contain energy! The amount of energy per unit volume is

u BB =1

2 0

2

µ

where B is the magnitude of the magnetic field. The magnetic energy density has units of Joules/m3. The total amount of energy in an infinitesimal volume dV is dU = uBdV and

U u dVBV o lum e

= ∫ .

If B is constant through the volume, V, then U = uB V.

BI

Coil

B

PHY2061 R. D. Field


RL Circuits "Building-Up" Phase: Connecting the switch to position A corresponds to the "building up" phase of an RL circuit. Summing all the potential changes in going around the loop gives

ε − − =IR LdId t

0 ,

where I(t) is a function of time. If the switch is closed (position A) at t=0 and I(0)=0 (assuming the current is zero at t=0) then

d Id t

IR

= − −

1τ

ε , where I have define ττττ=L/R.

Dividing by (I-ε/R) and multiplying by dt and integrating gives

( )dI

I Rdt

It

−= −∫ ∫ε τ/0

0

1 , which implies ln

//

I RR

t−−

= −

εε τ .

Solving for I(t) gives

( )I tR

e t( ) /= − −ε τ1 .

The potential change across the inductor is given by ∆∆∆∆VL(t)=-LdI/dt which yields ∆V t eL

t( ) /= − −ε τ .

The quantity ττττ=L/R is call the time constant and has dimensions of time.

"Collapsing" Phase: Connecting the switch to position B corresponds to the "collapsing" phase of an RL circuit. Summing all the potential changes in going around the

loop gives − − =I R Ld Id t

0 , where I(t) is a function of time. If the

switch is closed (position B) at t=0 then I(0)=I0 and

d Id t

I= −1τ and I t I e t( ) /= −

0τ .

Switch

+

εεεε

-

R

L

A

B

0.00

0.25

0.50

0.75

1.00

1.25

1.50

0 1 2 3 4

Time

I(t)

"Building-Up" Phase of an RL Circuit

PHY2061 R. D. Field


Simple Harmonic Motion

Hooke's Law Spring: For a Hooke's Law spring the restoring force is linearly proportional to the distance from equilibrium, Fx = -kx, where k is the spring constant. Since, Fx = max we have

− =kx md xdt

2

2 or d xdt

km

x2

2 0+ = , where x = x(t).

General Form of SHM Differential Equation: The general for of the simple harmonic motion (SHM) differential equation is

d x tdt

Cx t2

2 0( )

( )+ = ,

where C is a positive constant (for the Hooke's Law spring C=k/m). The most general solution of this 2nd order differential equation can be written in the following four ways:

x t Ae Bex t A t B t

x t A tx t A t

i t i t( )( ) cos( ) sin( )

( ) sin( )( ) cos( )

= += +

= += +

−ω ω

ω ωω φω φ

where A, B, and φφφφ are arbitrary constants and ω = C . In the chart, A is the amplitude of the oscillations and T is the period. The linear frequency f = 1/T is measured in cycles per second (1 Hz = 1/sec). The angular frequency ωωωω = 2ππππf and is measured in radians/second. For the Hooke's Law Spring C = k/m and thus ω = =C k m/ .

x(t) = Acos(ωωωωt+φφφφ)

-1.0

-0.5

0.0

0.5

1.0

0 1 2 3 4 5 6 7 8

ωωωωt+φφφφ (radians)

TA

PHY2060Review

PHY2061 R. D. Field


SHM Differential Equation The general for of the simple harmonic motion (SHM) differential equation is

d x tdt

Cx t2

2 0( )

( )+ = ,

where C is a constant. One way to solve this equation is to turn it into an algebraic equation by looking for a solution of the form

x t Aeat( ) = . Substituting this into the differential equation yields,

a Ae CAeat at2 0+ = or a C2 = − . Case I (C > 0, oscillatory solution): For positive C, a i C i= ± = ± ω , where ω = C . In this case the most general solution of this 2nd order differential equation can be written in the following four ways:

x t Ae Bex t A t B t

x t A tx t A t

i t i t( )( ) cos( ) sin( )

( ) sin( )( ) cos( )

= += +

= += +

−ω ω

ω ωω φω φ

where A, B, and φφφφ are arbitrary constants (two arbitrary constants for a 2nd order differential equation). Remember that e ii± = ±θ θ θcos sin

where i = −1 . Case II (C < 0, exponential solution): For negative C, a C= ± − = ±γ , where γ = − C . In this case, the most general solution of this 2nd order differential equation can be written as follows:

x t A e B et t( ) = + −γ γ,

where A and B arbitrary constants.

PHY2060Review

PHY2061 R. D. Field

Department of Physics chp38_5b.doc University of Florida

Re(z)

A

z = Aeiωωωωt Im(z)

φ φ φ φ = ω ω ω ωt t

t

Im(z) = Asinωωωωt

Re(z) = A

cos ωωω ωt

The Complex Plane

φ

φ

i

i

ezzizzezzizz−=−=

=+=)Im()Re()Im()Re(

)(21cos)Re( ∗+=== zzzzx φ

)(21sin)Im( ∗−=== zzi

zzy φ

)/arctan( xy=φ ∗=+= zzyxz 22

φφφ sincos ie i ±=± 1=± φie 22 1

πieii

±=±−=

Using Complex Numbers to Represent SHM We can use complex numbers to represent simple harmonic motion. If we let

tiAez ω= then

tAz ωsin)Re( =tAz ωcos)Im( =

zA =

Re(z)

|z| y

x

z = x+iy Im(z)

φφφφ

SHM with amplitude A and

“angular” frequency ωωωω

Phase

Magnitude

PHY2060Review

PHY2061 R. D. Field


Capacitors and Inductors

Capacitors Store Electric Potential Energy:

UQ

CE =2

2

Q C VC= ∆ ∆V Q CC = /

u EE =12 0

2ε (E-field energy density)

Inductors Store Magnetic Potential Energy:

U LIB =12

2

Φ B LI= L IB= Φ /

ε L LdIdt

= −

u BB =1

2 0

2

µ (B-field energy density)

Q

C E

BI

L

PHY2061 R. D. Field


An LC Circuit

At t = 0 the switch is closed and a capacitor with initial charge Q0 is connected in series across an inductor (assume there is no resistance). The initial conditions are Q(0) = Q0 and I(0) = 0. Moving around the circuit in the direction of the current flow yields

QC

LdIdt

− = 0 .

Since I is flowing out of the capacitor, I dQ dt= − / , so that d Qdt LC

Q2

21

0+ = .

This differential equation for Q(t) is the SHM differential equation we studied earlier with ω = 1 / LC and solution

Q t A t B t( ) cos sin= +ω ω . The current is thus,

I tdQdt

A t B t( ) sin cos= − = −ω ω ω ω .

Applying the initial conditions yields Q t Q tI t Q t

( ) cos( ) sin

==

0

0

ωω ω

Thus, Q(t) and I(t) oscillate with SHM with angular frequency ω = 1 / LC . The stored energy oscillates between electric and magnetic according to

U tQ t

CQ

Ct

U t LI t LQ t

E

B

( )( )

cos

( ) ( ) sin

= =

= =

202

2

202 2 2

2 212

12

ω

ω ω

Energy is conserved since Utot(t) = UE(t) + UB(t) = Q02/2C is constant.

Q

C L+ + + + + +

- - - - - -

Switch

PHY2061 R. D. Field


LC Oscillations

Q t Q tI t Q t

( ) cos( ) sin

==

0

0

ωω ω

U tQ

Ct

U tQ

Ct

E

B

( ) cos

( ) sin

=

=

02

2

02

2

2

2

ω

ω

Q

C L+ + + + + +

- - - - - -

t = 0

E

I

C L

t = T/4

I

B

-1.0

-0.5

0.0

0.5

1.0

0 1 2 3 4 5 6 7 8

ωωωωt (radians)

Q(t)

I(t)

0.0

0.5

1.0

0 1 2 3 4 5 6 7 8

ωωωωt (radians)

UE(t) UB(t)

Utot = UE + UB

PHY2061 R. D. Field


Mechanical Analogy

At t = 0: At t = 0:

E kxv=

=

12

002

U

CQ

I=

=

12

002

At Later t: At Later t:

vdxdt

x t x tkm

E mv kx

==

=

= +

( ) cos0

2 212

12

ω

ω

IdQdt

Q t Q t

LC

E LIC

Q

= −=

=

= +

( ) cos0

2 2

1

12

12

ω

ω

Correspondence: x t Q tv t I t

m Lk C

( ) ( )( ) ( )

/

↔↔↔

↔ 1

k

t = 0

x0

mx-axis

Q

C L+ + + + + +

- - - - - -

t = 0

E

I

Constant

PHY2061 R. D. Field


Maxwell's Equations (integral form) I. (Gauss' Law):

Φ Eenclosed

Surface Volume

E dA Q dV= ⋅ = =∫ ∫! !

ε ερ

0 0

1

Volume Enclosed by Surface

II. (Gauss' Law for Magnetism):

Φ BSurface

B dA= ⋅ =∫! !

0 No Magnetic Charges!

III. (Faraday's Law of Induction):

ε∂∂

= ⋅ = − = − ⋅∫ ∫! !

!!

E dld

dtBt

dAB

Curve Surface

Φ

Surface Bounded by Curve

IV. (Ampere's Law):

! ! ! !B dl I J dAenclosed

Curve Surface

⋅ = = ⋅∫ ∫µ µ0 0


E

Charge Q

E

ChangingMagnetic

Field

B

CurrentDensity J

Two Sources of Electric Fields

One Source of Magnetic Fields

PHY2061 R. D. Field


ClosedSurface S

Volume ChargeDensity ρρρρ

J

Curl of the Electric Field Consider a fixed coil with varying current I. If C is a closed curve, stationary in the O-frame, and if S is a

surface spanning C, and ),,,( tzyxBr

is the magnetic field measured in the O-frame at time t, then

∫ ⋅=C

rdE rrε and ∫ ⋅=Φ

SB AdB

rr

. Faraday’s

Law of induction tells us that

dtd BΦ−=ε and Stoke’s Theorem says ∫∫ ⋅×∇=⋅

SC

AdErdErrrrr

)( .

Thus,

∫∫ ∫∫ ⋅∂∂−=⋅−=⋅×∇=⋅

SS SC

AdtBAdB

dtdAdErdE

rr

rrrrrrr)(

which implies that

tBE

∂∂−=×∇r

rr

Charge Conservation: Consider a volume charge density ρρρρ enclosed within a fixed closed surface S and a current density J flowing through the surface. Since electric charge is conserved we know that

dtdQI −= where ∫ ⋅=

S

AdJIrr

and ∫=V

dVQ ρ .

Thus,

∫∫∫∫ ∂∂−=−=⋅∇=⋅

VVVS

dVt

dVdtddVJAdJ ρρ)(

rrrr

which implies that

tJ

∂∂−=⋅∇ ρrr

B(x,y,z,t)I

CoilCurve C

Divergence Theorem

Differential form of Faraday’s Law!

Charge Conservation!

PHY2061 R. D. Field


Maxwell's Equations (differential form)

I. (Gauss' Law): Integral Differential

∫ =⋅Surface

enclosedQAdE0ε

!!

0ε

ρ=⋅∇ E!!

II. (Gauss' Law for Magnetism): Integral Differential

∫ =⋅Surface

AdB 0!!

0=⋅∇ B!!

III. (Faraday's Law of Induction): Integral Differential

∫Φ−=⋅

Curve

B

dtdldE

!!

tBE

∂∂−=×∇!

!!

IV. (Ampere's Law): Integral Differential

∫ =⋅Curve

enclosedIldB 0µ!!

JB!!!

0µ=×∇

0. (Charge Conservation): Integral Differential

dtdQI −= t

J∂∂−=⋅∇ ρ!!

Something Missing! Equation IV is not consistant with Equation 0 since BJ

!!×∇=

0

1µ

implies 01

0

≡×∇⋅∇=⋅∇ BJ!!!!!

µ(since Div Curl = 0)

and charge conservation says that t

J∂∂−=⋅∇ ρ!!

. Hence Equation IV cannot

be correct as it stands!

Electric charges are a source (and sink) of E-field!

No magnetic monopoles!

Changing magnetic fields are a source

of E-field!

Current is a source of B-field!

Changing charge is the source of

current!

PHY2061 R. D. Field


Finding the Missing Term (differential method) Must make Ampere’s Law consistant with charge conservation. Look for the missing piece as follows:

XJB!!!!

+=×∇ 0µ . Solution:

tJXBJ

∂∂−=⋅∇−×∇= ρ

µ!!!!!!

0

1

tXBJ

∂∂−=⋅∇−×∇⋅∇=⋅∇ ρ

µµ!!!!!!!

00

11

Thus,

tX

∂∂=⋅∇ ρµ0

!! but E

!!⋅∇= 0ερ

so that

∂∂⋅∇=⋅∇

∂∂=⋅∇

tEE

tX

!!!!!!

0000 )( εµεµ

and hence

tEX

∂∂=!

!00εµ and t

EJB∂∂+=×∇!

!!!000 εµµ

Corrected Ampere’s Law (integral form):

∫∫∫∫ ⋅∂∂+⋅=⋅×∇=⋅

SSSC

AdtEAdJAdBldB

!!

!!!!!!!000)( εµµ

Thus,

tIldB E

enclosedC ∂

Φ∂+=⋅∫ 000 εµµ!!

where

∫∫ ⋅=Φ⋅=S

ES

AdEAdJI!!!!

Missing term

Missing term!

Corrected Ampere’s Law!

Corrected Ampere’s Law!

Electric Flux!

0

PHY2061 R. D. Field


Finding the Missing Term (integral method) We are looking for a new term in Ampere's Law of the form,

! !B dl I

ddtC

E⋅ = +∫1

0µ δΦ

,

where δδδδ is an unknown constant and I J d A E d A

SE

S

= ⋅ = ⋅∫ ∫! ! ! !

Φ ,

where S is any surface bounded by the curve C1. Case I (use surface S1): If we use the surface S1 which is bounded by the curve C1 then

! ! !!

!B dl I

ddt

JEt

dA IC

E

S

⋅ = + = +

⋅ =∫ ∫

10 0

10µ δ µ δ

∂∂

µΦ

,

since E = 0 through the surface S1. Case II (use surface S2): If we use the surface S2 which is bounded by the curve C1 then

! ! !!

!B dl I

ddt

JEt

dAI

C

E

S

⋅ = + = +

⋅ =∫ ∫

10 0

1 0

µ δ µ δ∂∂

δε

Φ ,

since J = 0 through the surface S2 and

EQ

AEt A

dQdt

IA

= = = =σε ε

∂∂ ε ε0 0 0 0

1.

Ampere's Law (complete): ( )! ! !

!!

B d l Id

d tJ

Et

d A I IC u rve

E

S u rfa ced⋅ = + = +

⋅ = +∫ ∫µ µ ε µ ε

∂∂

µ0 0 0 0 0 0Φ

,

I J d A JEtd d

Sd= ⋅ =∫

! ! !!

ε∂∂0 .

S1+

εεεε

-

R

C

Q

S2

II

C1

E

Must be equal, hence δδδδ=µµµµ0εεεε0.

"Displacement Current" "Displacement Current" Density

0

0

PHY2061 R. D. Field


Complete Maxwell's Equations (integral form)

I. (Gauss' Law):

Φ Eenclosed

Surface Volume

E dA Q dV= ⋅ = =∫ ∫! !

ε ερ

0 0

1

II. (Gauss' Law for Magnetism):

Φ BSurface

B dA= ⋅ =∫! !

0 No Magnetic Charges!

III. (Faraday's Law of Induction):

ε∂∂

= ⋅ = − = − ⋅∫ ∫! !

!!

E dld

dtBt

dAB

Curve Surface

Φ


IV. (Ampere's Law):

! ! !!

!B dl I

ddt

JEt

dAencCurve

E

Surface

⋅ = + = +

⋅∫ ∫µ µ ε µ ε

∂∂0 0 0 0 0

Φ


E

Charge Q

E

ChangingMagnetic

Field

B

CurrentDensity J

B

ChangingElectric

Field

Two Sources of Magnetic Fields

Volume Enclosed by Surface

Two Sources of Electric Fields

PHY2061 R. D. Field


Complete Maxwell's Equations (differential form)

I. (Gauss' Law): Integral Differential

∫ =⋅Surface

enclosedQAdE0ε

!!

0ε

ρ=⋅∇ E!!

II. (Gauss' Law for Magnetism): Integral Differential

∫ =⋅Surface

AdB 0!!

0=⋅∇ B!!

III. (Faraday's Law of Induction): Integral Differential

∫Φ−=⋅

Curve

B

dtdldE

!!

tBE

∂∂−=×∇!

!!

IV. (Ampere's Law): Integral Differential

tIldB E

Curveenclosed ∂

Φ∂+=⋅∫ 000 εµµ!!

tEJB

∂∂+=×∇!

!!!000 εµµ

0. (Charge Conservation):

Integral Differential

dtdQI −= t

J∂∂−=⋅∇ ρ!!

Electric charges are a source (and sink) of E-field!

No magnetic monopoles!

Changing magnetic fields are a source

of E-field!

Current is a source of B-field!

Changing charge is the source of

current!

Changing electric field is a source of B-field!

PHY2061 R. D. Field


Electric & Magnetic Fields that Change with Time

Changing Magnetic Field Produces an Electric Field:

A uniform magnetic field is confined to a circular region of radius, r, and is increasing with time. What is the direction and magnitude of the induced electric field at the radius r? Answer: If I choose my orientation to be counterclockwise then ΦΦΦΦB = B(t)A with

A = ππππr2. Faraday's Law of Induction tells us that ! !E dl rE r

dd t

rdBdt

B

C ircle

⋅ = = − = −∫ 2 2π π( )Φ

,

and hence E(r) = -(r/2) dB/dt. Since dB/dt > 0 (increasing with time), E is negative which means that it points opposite my chosen orientation. Changing Electric Field Produces a Magnetic Field:

A uniform electric field is confined to a circular region of radius, r, and is increasing with time. What is the direction and magnitude of the induced magnetic field at the radius r? Answer: If I choose my orientation to be counterclockwise then ΦΦΦΦE = E(t)A with

A = ππππr2. Ampere's Law (with J = 0) tells us that ! !B dl rB r

ddt

rc

dEdt

E

Circle

⋅ = = =∫ 2 0 0

2

2π ε µπ

( )Φ

,

and hence B(r) = (r/2c2) dE/dt. Since dE/dt > 0 (increasing with time), B is positive which means that it points in the direction of my chosen orientation.

B-out increasing with time

r

E

E-out increasing with time

r B

PHY2061 R. D. Field


Traveling Waves

A “wave” is a traveling disturbance that transports energy but not matter.

Constructing Traveling Waves: To construct a wave with shape y = f(x) at time t = 0 traveling to the right with speed v simply make the replacement x x vt→ − .

Traveling Harmonic Waves: Harmonic waves have the form y = A sin(kx) or y = Acos(kx) at time t = 0, where k is the "wave number" (k = 2ππππ/λλλλ where λλλλ is the "wave length") and A is the "amplitude". To construct an harmonic wave traveling to the right with speed v, replace x by x-vt as follows: y = Asin(k(x-vt) = Asin(kx-ωωωωt) where ω ω ω ω = kv (v = ωωωω/k). The period of the oscillation, T = 2ππππ/ωωωω = 1/f, where f is the linear frequency (measured in Hertz where 1Hz = 1/sec) and ωωωω is the angular frequency (ωωωω = 2ππππf). The speed of propagation is given by v = ωωωω/k = λλλλf .

y = y(x,t) = Asin(kx-ωωωωt) right moving harmonic wave y = y(x,t) = Asin(kx+ωωωωt) left moving harmonic wave

y = f(x-vt)

x = vt

v

y = f(x) at time t=0

x = 0

y=Asin(kx)

-1.0

-0.5

0.0

0.5

1.0

kx (radians)

λλλλ

A

PHY2061 R. D. Field


The Wave Equation

∂∂

∂∂

2

2 2

2

21

0y x t

x vy x t

t( , ) ( , )

− =

Whenever analysis of a system results in an equation of the form given above then we know that the system supports traveling waves propagating at speed v. General Proof: If y = y(x,t) = f(x-vt) then

∂∂

∂∂

∂∂

∂∂

yx

fy

xf

yt

vfy

tv f

= ′ = ′′

= − ′ = ′′

2

2

2

22

and

∂∂

∂∂

2

2 2

2

2

10

y x tx v

y x tt

f f( , ) ( , )

− = ′′ − ′′ = .

Proof for Harmonic Wave: If y = y(x,t) = Asin(kx-ωωωωt) then

∂∂

ω∂∂

ω ω2

22

2

22y

xk A kx t

yt

A kx t= − − = − −sin( ) sin( )

and

∂∂

∂∂

ωω

2

2 2

2

22

2

2

10

y x tx v

y x tt

kv

A kx t( , ) ( , )

sin( )− = − +

− = ,

since ωωωω = kv.

PHY2061 R. D. Field


Light Propagating in Empty Space

Since there are no charges and no current in empty space, ρρρρ = 0 and J = 0, and Maxwell’s Equartions take the form:

(1) tBE

∂∂−=×∇!

!!

(2) tEB

∂∂=×∇!

!!00εµ

with 0=⋅∇ E!!

and 0=⋅∇ B!!

.

Look for a solution of the form: ytxEtxE y ˆ),(),( =!

Equation (1) implies that

zx

E

txEzyx

zyx

EtB y

y

ˆ

0),(0

ˆˆˆ

∂∂

−=∂∂

∂∂

∂∂−=×∇−=

∂∂ !!!

Thus let, ztxBtxB z ˆ),(),( =

!

and equation (2) gives

yx

B

Bzyx

zyx

BtE z

z

ˆ

00

ˆˆˆ

00 ∂∂−=

∂∂

∂∂

∂∂=×∇=

∂∂ !!!

εµ

Coupled Differential Equations for E & B:

xB

tE

xE

tB

zy

yz

∂∂−=

∂∂

∂∂

−=∂

∂

00εµ

x-axis

y-axis

z-axis

E

B

PHY2061 R. D. Field


Electromagnetic Plane Waves (1) We have the following two differential equations for Ey(x,t) and Bz(x,t):

∂∂

∂∂

Bt

Ex

z y= − (1) and

∂∂ µ ε

∂∂

Et

Bx

y z= −10 0

(2)

Taking the time derivative of (2) and using (1) gives ∂∂ µ ε

∂∂

∂∂ µ ε

∂∂

∂∂ µ ε

∂∂

2

20 0 0 0 0 0

2

21 1 1E

t tBx x

Bt

Ex

y z z y= −

= −

=

which implies ∂∂

µ ε∂∂

2

2 0 0

2

2 0Ex

Et

y y− = .

Thus Ey(x,t) satisfies the wave equation with speed v = 1 0 0/ ε µ and has a solution in the form of traveling waves as follows:

Ey(x,t) = E0sin(kx-ωωωωt), where E0 is the amplitude of the electric field oscillations and where the wave has a unique speed

v ck

f m s= = = = = ×ω

λε µ1

2 99792 100 0

8. / (speed of light).

From (1) we see that ∂∂

∂∂

ωBt

Ex

E k kx tz y= − = − −0 cos( ) ,

which has a solution given by

B x t Ek

kx tEc

kx tz ( , ) sin ( ) sin ( )= − = −00

ωω ω ,

so that Bz(x,t) = B0sin(kx-ωωωωt),

where B0 = E0/c is the amplitude of the magnetic field oscillations.

x-axis

y-axis

z-axis

E

B

Wave equation for Ey!

PHY2061 R. D. Field


Electromagnetic Plane Waves (2) The plane harmonic wave solution for light with frequency f and wavelength λλλλ and speed c = fλλλλ is given by

!

!E x t E kx t yB x t B kx t z

( , ) sin( ) "( , ) sin( ) "

= −= −

0

0

ωω

where k = 2ππππ/λλλλ, ωωωω = 2ππππf, and E0 = cB0.

Properties of the Electromagnetic Plane Wave: • Wave travels at speed c (c =1 0 0/ µ ε ). • E and B are perpendicular (

! !E B⋅ =0).

• The wave travels in the direction of ! !E B× .

• At any point and time E = cB.

The Electromagnetic Spectrum:

10-15

1 fm10-12

1 pm10-9

1 nm10-6

1 µµµµm10-3

1 mm1

1 m103

1 km106 109

Wavelength (in meters)

103

1 kHz

Infrared

Long radio wavesMicrowaves TV AMFMX-rays

UltravioletGamma rays

Frequency (in Hertz)

Visible

11 Hz

106

1 MHz10910121015101810211024

x-axis

y-axis

z-axis

E

B

Direction of Propagation

Relative Sensitivity of the Human Eye

0%

20%

40%

60%

80%

100%

400 450 500 550 600 650 700

Wavelength (in nm)

BlueViolet Green Yellow Orange Red Visible spectrum 400nm – 700nm

PHY2061 R. D. Field


Energy Transport - Poynting Vector Electric and Magnetic Energy Density: For an electromagnetic plane wave

Ey(x,t) = E0sin(kx-ωωωωt), Bz(x,t) = B0sin(kx-ωωωωt),

where B0 = E0/c. The electric energy density is given by

u E E kx tE = = −12

120

20 0

2 2ε ε ωsin ( ) and the magnetic energy density is

u Bc

E E uB E= = = =1

21

2120

2

02

20

2

µ µε ,

where I used E = cB. Thus, for light the electric and magnetic field energy densities are equal and the total energy density is

u u u E B E kx ttot E B= + = = = −εµ

ε ω02

0

20 0

2 21sin ( ) .

Poynting Vector (! ! !S E B= ×

1

0µ ):

The direction of the Poynting Vector is the direction of energy flow and the magnitude

S EBE

c AdUdt

= = =1 1

0

2

0µ µ

is the energy per unit time per unit area (units of Watts/m2). Proof:

dU u V E Acdttot tot= = ε02 so

SA

dUdt

cEE

cE

ckx t= = = = −

10

22

0

02

0

2εµ µ

ωsin ( ) .

Intensity of the Radiation (Watts/m2): The intensity, I, is the average of S as follows:

I SA

dUdt

Ec

kx tE

c= = = − =

12

02

0

2 02

0µω

µsin ( ) .

x-axis

y-axis

z-axis

E

B

x-axis

y-axis

z-axis

E

B

Energy Flow

A

cdt

PHY2061 R. D. Field


Momentum Transport - Radiation Pressure

Relativistic Energy and Momentum:

E2 = (cp)2 + (m0c2)2

energy momentum rest mass For light m0 =0 and

E = cp (for light)

For light the average momentum per unit time per unit area is equal to the intensity of the light, I, divided by speed of light, c, as follows:

1 1 1 1A

dpdt c A

dUdt c

I= = .

Total Absorption:

Fdpdt c

dUdt c

IA= = =1 1

PFA c

I= =1

(radiation pressure)

Total Reflection:

Fdpdt c

dUdt c

IA= = =2 2

.

PFA c

I= =2

(radiation pressure)

Light

Total Absorption

Light

Total Reflection

PHY2061 R. D. Field


The Radiation Power of the Sun

Problem: The radiation power of the sun is 3.9x1026 W and the distance from the Earth to the sun is 1.5x1011 m. (a) What is the intensity of the electromagnetic radiation from the sun at the surface of the Earth (outside the atmosphere)? (answer: 1.4 kW/m2) (b) What is the maximum value of the electric field in the light coming from the sun? (answer: 1,020 V/m) (c) What is the maximum energy density of the electric field in the light coming from the sun? (answer: 4.6x10-6 J/m3) (d) What is the maximum value of the magnetic field in the light coming from the sun? (answer: 3.4 µµµµT) (e) What is the maximum energy density of the magnetic field in the light coming from the sun? (answer: 4.6x10-6 J/m3) (f) Assuming complete absorption what is the radiation pressure on the Earth from the light coming from the sun? (answer: 4.7x10-6 N/m2) (g) Assuming complete absorption what is the radiation force on the Earth from the light coming from the sun? The radius of the Earth is about 6.4x106 m. (answer: 6x108 N) (h) What is the gravitational force on the Earth due to the sun. The mass of the Earth and the sun are 5.98x1024 kg and 1.99x1030 kg, respectively, and G = 6.67x10-11 Nm2/kg2. (answer: 3.5x1022 N)

P = 3.9 x 1026 W

d = 1.5 x 1011 mSun

Earth

PHY2061 R. D. Field


Relativistic Doppler Shift (source moving away)

Consider a flashlight at rest in the O'-frame shinning its light in the –x' direction. Since the O'-frame is at rest with the source of light f' = f0, λλλλ' = λλλλ0, and T' = T0, where f0, l0, and T0 are the “proper frequency”, “proper wavelength”, and “proper period” of the light. Let event A be the occurrence of wavefront 2 reaching the O-frame and let event B be the occurrence of wavefront 3 reaching the O-frame, etc..

A

Frame O'

x'

ct'

cT0

proper periodat rest with light

source

B

1

3

2

∆∆∆∆x'=-V∆∆∆∆t'

45o

Light Path

O-frame Path

c∆∆∆∆t'

A

Frame O

x

ct

B

1

3

2

45o

Light Path

O'-frame Path

c∆∆∆∆t=cT

00

0

0

0

/11

1

)(

Tffcf

VTtVxAxBx

cTtc

tVcTttctc AB

==′=′′−

−=′∆−=−=′∆

−=′∆

′∆+=−=′∆

λβ

β

Tfcf

cTtcxx

cTcTxtctc

/1

0)(1

)(

)1()1(1

)(

0

020

==

=−−

=

′∆+′∆=∆

+=−−

=

′∆+′∆=∆

λ

βββ

γβγ

βγββ

γβγ

where 21/1/ βγβ −== cV .

Thus, 0)1( TT βγ += and 0)1( λβγλ += )1(0

βγ += ff

y

x

y'

x'

VO O'

4321

Light source at restin O'-frame

Relativistic Doppler shift (source moving away)

PHY2061 R. D. Field


Relativistic Doppler Shift (general case)

Consider an observer at rest in the O-frame and a light source moving with velocity cV /

!!=β .

Then the observed period, wavelength, and frequency are given by

)ˆ1(

)ˆ1()ˆ1(

0

0

0

rff

rTrT

⋅+=

⋅+=⋅+=

βγ

λβγλβγ

!

!

!

)ˆ1(

)ˆ1()ˆ1(

0

0

0

rff

rTrT

⋅+=

⋅+=⋅+=

β

λβλβ

!

!

!

with 21/1/ βγβ −== cV"!

and where T0, λλλλ0, and f0 are the period, wavelength, and frequency of the light in the frame at rest with the source. Case I “away” ( 1ˆ =⋅ rβ

!):

)1()1( 0

0 βγλβγλ

+=+= ffawayaway

Case II “toward” ( 1ˆ −=⋅ rβ

!):

)1()1( 0

0 βγλβγλ

−=−= fftowardtoward

Case III “transverse” ( 0ˆ =⋅ rβ

!):

γγλλ 0

0fftransversetransverse ==

y

x

V

O-frame

r

Relativistic Doppler Shift!

Classical Doppler Shift!

Transverse Doppler Shift!

PHY2061 R. D. Field


Geometric Optics

Fermat's Principle: In traveling from one point to another, light follows the path that requires minimal time compared to the times from the other possible paths. Theory of Reflection: Let tAB be the time for light to go from the point A to the point B reflecting off the point P. Thus,

tc

Lc

LAB = +1 1

1 2 ,

where

L x aL d x b

12 2

22 2

= += − +( ) .

To find the path of minimal time we set the derivative of tAB equal to zero as follows:

dtdx c

dLdx c

dLdx

AB = + =1 1

01 2,

which implies dLdx

dLdx

1 2= − ,

but dLdx

xL

dLdx

d xL

i

r

1

1

2

2

= =

=− −

= −

sin

( )sin

θ

θ

so that the condition for minimal time becomes

sin sinθ θ θ θi r i r= = .

θθθθi θθθθr

P

A

a

B

b

x d-x

d

L1L2

θθθθi θθθθr

P

A

a

B

b

x d-x

d

L1L2

PHY2061 R. D. Field


Law of Refraction

Index of Refraction: Light travels at speed c in a vacuum. It travels at a speed v < c in a medium. The index for refraction, n, is the ratio of the speed of light in a vacuum to its speed in the medium,

n = c/v, where n is greater than or equal to one.

Theory of Refraction: Let tAB be the time for light to go from the point A to the point B refracting at the point P. Thus,

tv

Lv

LAB = +1 1

11

22 ,

where

L x aL d x b

12 2

22 2

= += − +( ) .

To find the path of minimal time we set the derivative of tAB equal to zero as follows: dtdx v

dLdx v

dLdx

AB = + =1 1

01

1

2

2, which implies

1 1

1

1

2

2

vdLdx v

dLdx

= − , but

dLdx

xL

dLdx

d xL

1

11

2

22

= =

=− −

= −

sin

( )sin

θ

θ

so that the condition for minimal time becomes

1 1

11

22 1 1 2 2v v

n nsin sin sin sinθ θ θ θ= =.

θθθθ1

θθθθ2

P

A

a

B

b

x d-x

d

L1

L2

n1

n2

Snell's Law

PHY2061 R. D. Field


Total Internal Reflection

Total internal refection occurs when light travels from medium n1 to medium n2 (n1 > n2) if θθθθ1 is greater than or equal to the critical angle, θθθθc, where

sinθcnn

= 2

1 .

Problem: A point source of light is located 10 meters below the surface of a large lake (n=1.3). What is the area (in m2) of the largest circle on the pool's surface through which light coming directly from the source can emerge? (answer: 455)

θθθθc

θθθθ2=90o

n1

n2

θθθθc

R

n1=1.3

n2=1

PHY2061 R. D. Field


Refraction Examples

Problem: A scuba diver 20 meters beneath the smooth surface of a clear lake looks upward and judges the sun to be 40o from directly overhead. At the same time, a fisherman is in a boat directly above the diver. (a) At what angle from the vertical would the fisherman measure the sun? (answer: 59o) (b) If the fisherman looks downward, at what depth below the surface would he judge the diver to be? (answer: 15 meters)

θθθθ1

θθθθ2

20 m

n1=4/3

n2=1

PHY2061 R. D. Field


Spherical Mirrors

Vertex and Center of Curvature: The vertex, V, is the point where the principal axis crosses the mirror and the center of curvature is the center of the spherical mirror with radius of curvature R.

Real and Virtual Sides: The "R" or real side of a spherical mirror is the side of the mirror that the light exits and the other side is the "V" or virtual side. If the center of curvature lies on the R-side then the radius of curvature, R, is taken to be positive and if the center of curvature lies on the V-side then the radius of curvature, R, is taken to be negative.

Focal Point: A light ray parallel to the principal axis will pass through the focal point, F, where F lies a distance f (focal length) from the vertex of the mirror. For spherical mirrors a good approximation is f = R/2.

Concave and Convex Mirrors: A concave mirror is one where the center of curvature lies on the R-side so the R > 0 and f > 0 and a convex mirror is one where the center of curvature lies on the V-side so that R < 0 and f < 0.

concave f > 0 convex f < 0

Flat Mirror: A flat mirror is the limiting case where the radius R (and thus the local length f) become infinite.

Principal Axis

C F

Light Ray Enters

R = Radius of CurvatureC = Center of CurvatureF = Focal PointV = Vertex

Concave Mirror

R-side V-sideV

Light Ray Exits

R

Principal Axis

CF

Light Ray Enters

Convex Mirror

R-side V-side

R = Radius of CurvatureC = Center of CurvatureF = Focal PointV = Vertex

Light Ray Exits

R

V

PHY2061 R. D. Field


Mirror Equation

Object and Image Position: For spherical mirrors,

1 1 1p i f

+ = ,

where p is the distance from the vertex to the object, i is the distance from the vertex to the image, and f is the focal length. Focal Length: For spherical mirrors the focal length, f, is one-half of the radius of curvature, R, as follows:

f = R/2. Magnification: The magnification is

mip

= − , (magnification equation)

where the magnitude of the magnification is the ratio of the height of the image, hi, to the height of the object, hp, as follows:

mhh

i

p= .

Sign Conventions:

Variable Assigned a Positive Value Assigned a Negative Value p (object distance) almost always positive special compound lens case i (image distance) if image is on R-side (real image) if image is on V-side (virtual image) R (radius of curvature) if C is on R-side (concave) if C is on V-side (convex) f (focal length) if C is on R-side (concave) if C is on V-side (convex) m (magnification) if the image is not inverted if the image is inverted

C

C = Center of Curvature F = Focal Point p = Object Position i = Image Position

R-side V-side

p

i

PHY2061 R. D. Field


Mirror Examples (1)

Mirror Equations:

fR

p i fm

ip

= + = = −2

1 1 1

Example:

R = 2, p = 3

f = 1, i = 3/2, m = -1/2 Example:

R = 2, p = 3/2

f = 1, i = 3, m = -2

CR-side V-side

pi

F

Concave Mirror

Reduced Inverted Real Image

R-side V-side

pi

F

Concave Mirror

Magnified Inverted Real Image

C

PHY2061 R. D. Field


Mirror Examples (2)

Mirror Equations:

fR

p i fm

ip

= + = = −2

1 1 1

Example:

R = 2, p = 1/2

f = 1, i = -1, m = 2 Example:

R = -2, p = 3

f = -1, i = -3/4, m = 1/4

CR-side V-side

p iF

Concave Mirror

Magnified Non-inverted Virtual Image

R-side

V-side

p i F

Convex Mirror

Reduced Non-inverted Virtual Image

C

PHY2061 R. D. Field


Thin Lenses Formula Lens Makers Equation: The lens makers formula is

11

1 1

1 2fn

R R= − −

( ) ,

where f is the focal length, n is the index of refraction, R1 is the radius of curvature of side 1 (side that light enters the lens), and R2 is the radius of curvature of side 2 (side that light exits the lens). Lens Equation:

1 1 1p i f

+ =

Magnification:

mip

= −

Sign Conventions:

Variable Assigned a Positive Value Assigned a Negative Value p (object distance) almost always positive special compound lens case i (image distance) if image is on R-side (real image) if image is on V-side (virtual image) R1 (radius of curvature) if C1 is on R-side if C1 is on V-side R2 (radius of curvature) if C2 is on R-side if C2 is on V-side f (focal length) if f > 0 then converging lens if f < 0 then diverging lens m (magnification) if the image is not inverted if the image is inverted

Example (converging lens): R R R R fR

n1 2 2 10= = − =

−>

( )

Example (diverging lens): R R R R fR

n1 2 2 10= − = =

−−

<( )

C2 R-sideV-side F C1F

Side 1 Side 2

Converging Lens

n

PHY2061 R. D. Field


Thin Lenses (converging) Example:

f = 1, p = 2

i = 2, m = -1

Example:

f = 1, p = 1/2

i = -1, m = 2

iR-sideV-side Fp F

1 2

Converging Lens

n

Inverted Real Image

iR-sideV-side F p F

1 2

Converging Lens

n

Magnified Non-inverted Virtual Image

PHY2061 R. D. Field


Thin Lenses (diverging) Example:

f = -1, p = 2

i = -2/3, m = 1/3 Example:

f = -1, p = 1/2

i = -1/3, m = 2/3

i R-sideV-side Fp

F

1 2

Diverging Lens

n


i R-sideV-side F p F

1 2

Diverging Lens

n


PHY2061 R. D. Field


Two-Lens Systems

Consider two lenses separated by a distance L. Step 1: Let p1 be the distance of the object from lens 1. Find the location of the image, i1, from lens 1 (ignoring lens 2) using

111

111fip

=+

and m1 = -i1/p1. Step 2: Take the image formed in step 1 as the object for lens 2. Take

p2 positive if image 1 is to the left of lens 2 (regardless of whether image 1 is real or virtual) and take p2 negative if image 1 is to the right of lens 2 (opposite the side of the entering light). Find the location of the

image, i2, from lens 2 (ignoring lens 1) using

222

111fip

=+ and m2 = -i2/p2. Image 2 is the overall image

with overall magnification M = m1m2.

L p1

Lens 1 Lens 2

Object

Light

|i1|

p1

Lens 1

Image 1

Object

Step 1

L

p2

Lens 2 Image 1 = Object 2

Image 2 = Overall Image

|i1|

Step 2

PHY2061 R. D. Field


L = 1 p1 = 3/2

Lens 1 f1 = 1

Lens 2 f2 = 1

Object

Light

Reduced Inverted Real Image

Two-Lens Systems Examples Example 1: Consider the case f1 = 1, f2 = 1, L = 1, and p1 = ½. We

see that i1 = -1, m1 = 2. Thus, p2 = 2 and i2 = 2, m2 = -1, and M = m1m2 = -2. Hence, the resulting image is an enlarged real inverted image located 2 units to the right of lens 2.

Example 2: Consider the case f1 = 1, f2 = 1, L = 1, and p1 = 3/2.

We see that i1 = 3, m1 = -2. Thus, p2 = -2 and i2 = 2/3, m2 = 1/3, and M = m1m2 = -2/3. Hence, the resulting image is a reduced real inverted image located 2/3 units to the right of lens 2.

L = 1 p1 = 1/2

Lens 1 f1 = 1

Lens 2 f2 = 1

Object

Light

Enlarged Inverted Real

Image

PHY2061 R. D. Field


λλλλ1

λλλλ2

n1 = c/v1

n2 = c/v2

v2 < v1

v2

v1 λλλλ1

λλλλ2

n1 = c/v1

n2 = c/v2

v2 < v1v2

v1

θθθθ1

θθθθ2

L

Light as a Wave - Interference Huygen’s Principle: All points on a wavefront serve as point sources of spherical wavelets. After a time t, the new position of the wavefront will be the surface tangent to the secondary wavelets.

The Law of Refraction:

Zero incident angle: Incident angle θθθθ1:

1

1

2

2

22

1

1

vv

tvv

t

λλλ

λ

=

∆=

=∆

2

2

1

1

22

11

sinsin

sin

sin

λθ

λθ

λθ

λθ

=

=

=

L

L

2211 λλ nn = 2211 sinsin θθ nn =

ct

ctWavefront at t = 0 Spherical

wavelets

New wavefrontat time t

Snell’s Law

PHY2061 R. D. Field


Interference

Wave Superposition: Consider the addition (superposition) of two waves with the same amplitude and wavelength:

21

2

1

)sin())(sin()sin(

yyykxArxkAy

kxAy

sum +=∆+=∆+=

=φ

The "phase shift", ∆φ∆φ∆φ∆φ, between the two waves is related to the quantity ∆∆∆∆r by

rk∆=∆φ or λπφ r∆=∆

2

where k=2ππππ/λλλλ is the wave number and λλλλ is the wavelength. Maximal Constructive Interference: The condition for maximal constructive interference is

!,2,1,02 ±±==∆=∆ mmrm λπφ (max constructive) Maximal Destructive Interference: The condition for maximal destructive interference is

!,2,1,0212 ±±=

+=∆+=∆ mmrm λππφ (max destructive)

Wave Superposition

-2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0

0 1 2 3 4 5 6 7 8

kx (radians)

ysum = y1 +y2 y2=Asin(k(x+∆∆∆∆r)))))

y2=Asin(kx)

∆∆∆∆r

“Lateral” shift

PHY2061 R. D. Field


Interference Examples

Wave Superposition (∆∆∆∆r = λλλλ; max constructive):

Wave Superposition (∆∆∆∆r = λ/2λ/2λ/2λ/2; max destructive):

Wave Superposition (∆∆∆∆r = λλλλ/4):

Wave Superposition

-2.0-1.5-1.0-0.50.00.51.01.52.0

0 1 2 3 4 5 6 7 8

kx (radians)ysum = y1 + y2

Wave Superposition

-1.0

-0.5

0.0

0.5

1.0

0 1 2 3 4 5 6 7 8


W ave Superposition

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

0 1 2 3 4 5 6 7 8


PHY2061 R. D. Field


Re(ΨΨΨΨ)

A

ΨΨΨΨ = Aei(kx-ωωωωt) Im(ΨΨΨΨ)

φ φ φ φ = kx- ω ω ω ωt t

Im(ΨΨΨΨ) = Asin(kx-ωωωωt)

A

A

Crest

Trough

A

ΨΨΨΨ(0,t) = Ae-iωωωωt

Distance r φ φ φ φ = kr- ω ω ω ωt

x = 0 φ φ φ φ =- ω ω ω ωt

A

x = r

ΨΨΨΨ(r,t) = Aei(kr-ωωωωt)

Representing Waves as Complex Numbers

We can use complex numbers to represent traveling waves. If we let )( tkxiAe ω−=Ψ then )sin()Re( tkxA ω−=Ψ

is a traveling plane wave with wave number k = 2ππππ/λλλλ, “angular” frequency ωωωω = 2ππππf, and amplitude A. The intensity, I, is proportional to A2.

∗ΨΨ=Ψ=A 22 Ψ=∝ AI

Phase-Shift Due to a Path Length Difference

Consider two traveling wave that are in phase at their source, but wave 1 travels a distance r1 and wave 2 travels a distance r2 to the point P. The phase difference between the two waves at the point P is given by

rrrk ∆=−=−=∆λπφφφ 2)( 1212

The condition for maximal constructive interference is !,2,1,02 ±±==∆=∆ mmrm λπφ (max constructive)

The condition for maximal destructive interference is

!,2,1,0212 ±±=

+=∆+=∆ mmrm λππφ (max destructive)

Phase

Amplitude

Intensity

Wave Function

PHY2061 R. D. Field


Double Slit Interference The simplest way to produce a phase shift a difference in the path length between the two wave sources, S1 and S2 is with a double slit. The point P is located on a screen that is a distance L away from the slits and the slits are separated by a distance d.

If L >> d then to a good approximation the path length difference is,

θsin12 drrr =−=∆ .

Maximal Constructive Interference: The condition for maximal constructive interference is

!,2,1,0sin ±±== md

m λθ

(Bright Fringes - max constructive) Maximal Destructive Interference: The condition for maximal destructive interference is

!,2,1,021sin ±±=

+= m

dm λθ

(Dark Fringes - max destructive)

S1 r2

P

S2

r1

d

L

y

Double Slit

S1

r2 S2

r1

d θθθθ

Double Slit

θθθθ

d sinθθθθ

S1 θθθθ

P

S2 d

L

y

Double Slit

y = L tanθθθθ

r

Order of the Bright Fringe

PHY2061 R. D. Field


Double Slit Intensity Pattern We form the superposition of the two waves at the point P on the screen as follows.

21

)(2

)(1

2

1

Ψ+Ψ=Ψ=Ψ=Ψ

−

−

tot

tkri

tkri

AeAe

ω

ω

and thus

( ) ( ))cos(2

1

212/)(

2/2/2/)()(

)()(21

1

11

21

rkeAeeeeAeeAe

AeAe

riktkri

rikrikriktkririktkri

tkritkritot

∆=

+=+=

+=Ψ+Ψ=Ψ

∆−

∆−∆∆−∆−

−−

ω

ωω

ωω

where k = 2p/l. The path difference (for L >> d) is given by θsin12 drrr =−=∆

The distance from the center of the two slits and the point P on the screen is given by r = r1 + ∆∆∆∆r/2 and hence

)/sincos(2),( )( λθπθ ω dAer tkritot

−=Ψ . The intensity is proportional to the amplitude squared and hence

)/sin(cos4)( 20 λθπθ dII =

where I0 is the intensity of a single wave (wave 1 or wave 2).

S1

r2 S2

r1

d θθθθ

Double Slit

θθθθ

∆∆∆∆r =d sinθθθθ

S1 θθθθ

S2 d

L

0

Double Slit

Intensity

r

1/2

1/2

1

-1/2

d sin θθθ θ/ λλλ λ

2

5/2

-1

-3/2

-2

P

Central Bright Spot

PHY2061 R. D. Field


Thin Film Interference Thin film interference occurs when a thin layer of material (thickness T) with index of refraction n2 (the "film" layer) is sandwiched between two other mediums n1 and n3. The overall lateral shift between the reflected waves 1 and 2 is given by, ∆ ∆ ∆overall T= + +2 1 2 , where it is assumed that the incident light ray is nearly perpendicular to the surface and the lateral shifts ∆∆∆∆1 and ∆∆∆∆2 are given the table.

Maximal Constructive Interference: The condition for maximal constructive interference is

∆ ∆ ∆overall filmT m m= + + = = ± ±2 0 1 21 2 λ , , ,! (max constructive) where λλλλfilm = λλλλ0/n2, with λλλλ0 the vacuum wavelength.

Maximal Destructive Interference: The condition for maximal destructive interference is

∆ ∆ ∆overall filmT m m= + + = +

= ± ±2

12

0 1 21 2 λ , , ,!(max destructive)

Incident Light 12

n1

n2

n3

"film"T∆∆∆∆1

∆∆∆∆2

Shift Condition Value ∆∆∆∆1 n1 > n2 0 ∆∆∆∆1 n1 < n2 λλλλfilm/2 ∆∆∆∆2 n2 > n3 0 ∆∆∆∆2 n2 < n3 λλλλfilm/2

PHY2061 R. D. Field


Interference Problems Double Slit Example: Red light (λλλλ = 664 nm) is used with slits separated by d = 1.2x10-4 m. The screen is located a distance from the slits given by L = 2.75 m. Find the distance y on the screen between the central bright fringe and the third-order bright fringe.

Answer: y = 0.0456 m

Thin Film Example: A thin film of gasoline floats on a puddle of water. Sunlight falls almost perpendicularly on the film and reflects into your eyes. Although the sunlight is white, since it contains all colors, the film has a yellow hue, because destructive interference has occurred eliminating the color of blue (λλλλ0 = 469 nm) from the reflected light. If ngas = 1.4 and nwater = 1.33, determine the minimum thickness of the film.

Answer: Tmin = 168 nm

PHY2061 R. D. Field


Narrow Slit Diffraction

Consider a plane wave with wavelength λ incident on a narrow slit of width W. Case 1 (λ << W) corresponds to little or no diffraction with a bright spot on the screen of width not much larger than W. Case 2 (W << λ) corresponds to a lot of diffraction resulting in the spreading out of the light across the screen in the form of a “diffraction pattern” with a central bright

spot and a series of dark and light fringes similar to the two-slit interference pattern. Actually diffraction is a form of interference in which each of the infinite number of points along the wavefront acts as a point source of secondary spherical wavelets (Huygen’s Principle). Each of these waves has a different path length to the point P on

the screen and hence they interfere with each other and produce an “interference” (diffraction) pattern. At y = 0 (central axis) all the paths are roughly equal resulting in constructive interference and a central bright spot. If is mathematically difficult to calculate the position of the bright fringes, but the dark fringes are located at

!,2,1sin ±±== mW

m λθ where Ly=θtan .

The bright fringes are roughly (but not exactly) at the midpoint of the dark fringes.

Case 1: Without Diffraction λλλλ << W

Slit width W Wavefronts Wavelength = λλλλ

Bright spot on Screen

Case 2: With Diffraction W << λλλλ

Slit width W Wavefronts Wavelength = λλλλ

Diffraction Pattern on

Screen

Dark Fringe

W

Each point along wavefront acts as a source!

Screen

y

P

L

r

θθθθ

y = L tanθθθθ

Position of dark fringes

PHY2061 R. D. Field


Single-Slit Diffraction Intensity Pattern (1)

Consider N point sources a distance ∆∆∆∆y = W/(N-1) apart across the width of the slit such that W = (N-1)∆∆∆∆y (we will let N become large). In the limit L >> W,

θθθθ

θ

sinsin)1(sin)1(sin

sin

11

11

12

WryNrrynryrr

yrr

N

nn

+=∆−+=∆−+=∆+=

∆+=

−

Thus, the phase difference between the source n and source n-1 is given by

θλπ

φφφ

sin2)( 11

y

rrk nnnn

∆=

−=−=∆ −−

and the phase difference between source 1 and source N is

φθλπφφ ∆−==−=−=∆Φ )1(sin2)( 11 NWrrk NN

The overall wave function at the point P on the screen is the sun of all the individual wavelets as follows:

( )( ) N

ikrtirrikrrikikrti

ikrikrikrtiN

nnN

SeAeeeeAe

eeeAe

N

N

11121

21

)()(

1

...1

...

ωω

ω

−−−−

−

=

=+++=

+++=Ψ=Ψ ∑

where 12 ...1 −++++= N

N aaaS with θφ sinyiki eea ∆∆ == . Note that aSN – SN = aN –1 and hence

( )( )

( )( )

( )( )φ

φφφ φ

φ

φ

φφφ

φφφ

φ

φ

∆∆=

∆∆=

−−=

−−=

−−=

∆−∆

∆

∆−∆∆

∆−∆∆

∆

∆

21

21

2/)1(

212/

212/

2/2/2/

2/2/2/

sinsin

sinsin

11

11

Nee

Neeeeeee

ee

aaS

Nii

iN

iii

iNiNiN

i

iNN

N

W =

(N-1

) ∆ ∆∆∆y

W sinθθθθ

∆∆∆∆y 1

2

n

n-1

N-1

N

θθθθ

∆∆∆∆y sinθθθθ

r1

r2

rn

rn-1

center axis

Maximum Phase Shift

Overall Wave Function at P

PHY2061 R. D. Field


Single-Slit Diffraction Intensity Pattern (2) The overall wave function at the point P on the screen is given by

)sin()sin(

21

21

2/)1(

1

11

φφφωω

∆∆==Ψ=Ψ ∆−−−

=∑ NeeAeSeAe Niikrt

Nikrti

N

nnN .

The distance where r from the center of the slit to the point P is given by, φθ ∆−+=+= )1(sin 2

112

11 NrWrr k .

Now as N becomes large φφ ∆ →∆−=∆Φ >> NN N 1)1(

and

∆Φ∆Φ≈

∆Φ∆Φ≈

∆∆=Ψ

−

−−

21

21

)(

21

21

)(

21

21

)(

)sin()/sin(

)sin()sin()sin(

tkri

tkritkriN

NAe

NAeNAe

ω

ωω

φφ

where I used

NN N 22sin 1

∆Φ →

∆Φ

>> .

Thus for large N

∆Φ∆Φ≈Ψ −

21

21

)( )sin(),( tkriN NAer ωθ

where θλπ sin2 W=∆Φ .

The intensity is proportional to 2

NΨ and hence

221

212

max )()(sin)(

∆Φ∆Φ= II θ

where Imax = I(θθθθ=0). Note that I(θθθθ) vanishes when

,...2,1sin22 ±±===∆Φ mWm θλππ

,...2,1sin ±±== mmW λθ

W

Central Bright Spot

0

P

L

W sin θθθ θ/ λλλ λ

1

2

-2

-1

θθθθ

r

Single Slit

Intensity

Single-Slit Intensity Pattern

Dark Fringes

Note:

...1sin 261 +−= x

xx

1sin =x

x at x = 0

PHY2061 R. D. Field


Diffraction Summary Single Slit-Diffraction: Angular position of the dark fringes:

!,2,1sin ±±== mW

m λθ

(Dark Fringes - max destructive) Round Hole-Diffraction: Angular position of the first dark ring:

Dλθ 22.1sin =

(Dark Ring - max destructive) Diffraction Grating: Angular position of the bright fringes:

!,2,1,0sin ±±== md

m λθ

(Bright Fringes - max constructive)

θθθθ

P

W

L

y

Single Slit

y = L tanθθθθ

θθθθ

P

d

L

y

DiffractionGrating

y = L tanθθθθ

Diameter of the Hole

Width of the Slit

Slit Separation

PHY2061 R. D. Field


Resolving Power

The fact that round holes produce a diffraction patter in which the light spreads out on the screen is important when we wish to “resolve” (distinguish) two distant objects whose angular separation is small. Rayleigh’s Criterion for resolvability states that the minimum angle θmin such that two objects can be distinguished as two separate objects occurs when the center of the central bright spot of object 1 is located at the first diffraction minimum of object 2

(or visa-versa). For angles less that θmin the two objects will appear as one. For round hole diffraction the first diffraction minimum of object 2 occurs at

Dλθ 22.1sin = .

Thus (assuming small angles, θ << 1), we get

Dλθ 22.1min ≈

This is only an approximation (the actual resolution is usually worse than this), but it allows us to make calculations. Example Problem: A hang glider is flying at an altitude of 120 m. Green light (λλλλ0 = 555 nm) enters the pilot’s eye through a pupil that has a diameter D = 2.5 mm. If the average index of refraction of the material in the eye is n = 1.36, determine how far apart two point objects on the ground must be if the pilot is to have any hope of distinguishing between them.

Answer: 2.4 cm

Object 1

Diffraction Patterns on

Screen

Round hole diameter D

Object 2

Intensity

θθθθmin

Rayleigh’s Criterion for resolution

PHY2061 R. D. Field


Diffraction Problems Single Slit Example: Light passes through a slit and shines on a flat screen that is located L = 0.4 m away. The width of the slit is W = 4x10-6 m. The distance between the middle of the central bright spot and the first dark fringe is y. Determine the width 2y of the central bright spot when the wavelength of light is λλλλ = 690 nm.

Answer: 2y = 0.14 m

the simple structure of our universe-physics

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