toons

toonsMetaphysical Illustrations by Tomas Bunk

Physical Explanations by Arthur Eisenkraft and Larry D. Kirkpatrick

National Science Teachers AssociationArlington, Virginia

Claire Reinburg, DirectorJudy Cusick, Senior EditorJ. Andrew Cocke, Associate EditorBetty Smith, Associate Editor

Copyright © 2006 by the National Science Teachers Association.All rights reserved. Printed in the United States of America.

08 07 06 4 3 2 1

Library of Congress Cataloging-in-Publication DataBunk, Tomas, 1945- Quantoons : metaphysical illustrations by Tomas Bunk, physical explanations by Arthur Eisenkraft and Larry D.Kirkpatrick / by Tomas Bunk, Arthur Eisenkraft, and Larry D. Kirkpatrick. p. cm. ISBN 0-87355-265-2 1. Science—Problems, exercises, etc. 2. Physics—Problems, exercises, etc. 3. Mathematics—Problems, exercises,etc. 4. Science—Competitions. 5. Physics—Competitions. 6. Mathematics—Competitions. I. Eisenkraft, Arthur.II. Kirkpatrick, Larry D., 1941- III. Title. Q182.B855 2006 507'.1—dc22 2005024051 2002153474

The problems and artwork in this book were originally published in different form in Quantum magazine, Vol. 2, No. 1(September/October 1991), through Vol. 11, No. 6 (July/August 2001).

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Q U A N T O O N S v

Introduction ......................................................................viiA snail that moves like light .............................................. 2The leaky pendulum ........................................................... 6What goes up … ................................................................ 10The clamshell mirrors ...................................................... 14Shake, rattle, and roll ........................................................ 18Sources, sinks, and gaussian spheres ............................... 22The tip of the iceberg ........................................................ 26A topless roller coaster ..................................................... 30Row, row, row your boat .................................................. 34How about a date?............................................................. 38Animal magnetism ........................................................... 42Atwood’s marvelous machines ........................................ 46Thrills by design ............................................................... 50Electricity in the air .......................................................... 54Stop on red, go on green … ............................................... 58Fun with liquid nitrogen .................................................. 62Laser levitation ................................................................. 66Mirror full of water ........................................................... 70Rising star ......................................................................... 74Superconducting magnet .................................................. 78Cloud formulations........................................................... 84Weighing an astronaut ...................................................... 88The first photon ................................................................ 92Pins and spin ..................................................................... 96Split image ...................................................................... 100Gravitational redshift ..................................................... 104Focusing fields................................................................. 108

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v i Q U A N T O O N S

Sea sounds ....................................................................... 112Moving matter ................................................................ 116Boing, boing, boing … ..................................................... 120The bombs bursting in air .............................................. 124The nature of light .......................................................... 128Do you promise not to tell? ............................................ 132Mars or bust! ................................................................... 136Color creation ................................................................. 140A physics soufflé ............................................................. 144Cool vibrations ............................................................... 148Elephant ears ................................................................... 152Local fields forever .......................................................... 156Around and around she goes .......................................... 160Depth of knowledge ........................................................ 164Doppler beats .................................................................. 168Up, up and away.............................................................. 172Warp speed ...................................................................... 176Sportin’ life ...................................................................... 180Elevator physics .............................................................. 184The eyes have it .............................................................. 188Image charge.................................................................... 192Breaking up is hard to do ................................................ 196A question of complexity ............................................... 202Tunnel trouble ................................................................ 206Magnetic vee ................................................................... 210Rolling wheels................................................................. 214Batteries and bulbs .......................................................... 220Curved reality ................................................................. 226Relativistic conservation laws ....................................... 232A good theory .................................................................. 236The fundamental particles ............................................. 240

I N T R O D U C T I O N v i i

Introduction

QUANTOONS BRINGS PHYS-ics to you through masterful il-lustrations, quotes, text andchallenging problems. The

simple classic physics problem ofcrossing a raging river and determin-ing where you land on the othershore turns into a metaphor of tra-versing the river of life from birth todeath in a cartoon illustration filledwith humor and poignancy. TheHeraclitus quote “You can neverstep into the same river twice” addsanother aspect of appreciation to thephysics story of relative motion,movement in two dimensions, andcalculations of least time.

The colors in a rainbow, thesounds of rustling leaves, and thesplattering of waves crashing intorocks communicate to us throughour senses and our imaginations.Physics can add to our appreciationand understanding of these naturaloccurrences if the insights of phys-ics are made accessible.

Quantoons is a compilation ofContest Problems that were pub-lished in Quantum magazine duringits 11-year run (1990–2001). Quan-tum was a collaborative effort of theUnited States and Russia. Publishedby the National Science TeachersAssociation (NSTA), this semi-monthly magazine was targeted at

students and teachers and anyoneelse interested in science and math-ematics. Borrowing from originalnotes and articles in the Russianpublication Kvant, Quantum addedarticles and problems from Ameri-can authors. The Contest Problemwas one such addition. In every is-sue, a physics problem was pre-sented and interested readers wereinvited to submit solutions. Thebest of these solutions were ac-knowledged in a subsequent issue,and often used as the basis for thepublished solution. The ContestProblems were also intended to be

descriptions of physics enhancedby the creative cartoons that ac-companied them.

In the first issue of Quantum,Bill Aldridge quotes the great Rus-sian scientist and poet MikhailLomonosov as he viewed theNorthern Lights: “Nature, whereare your laws? The dawn appearsfrom the dark northern climes!Does not the sun there set up itsthrone? Are not the ice-bound seasemitting fire?” For 11 years, Quan-tum attempted to answer some ofthese questions while illuminatingthe minds of so many who will oneday provide us with other glimpsesinto the wonder of the universe.

Bill Thurston, in that same firstissue, reflected on the beginning ofhis illustrious career as a mathema-tician. “As a child, I often hatedarithmetic and mathematics inschool. Pages of exercises were te-dious and dull … I stared out the win-dow and let my mind wander. Some-times I tried to puzzle somethingout … . Might the square root of 2eventually be periodic if you write itout in base 12 instead of base 10?How many ways are there to fold amap into sixteenths, in quarterseach way?” This spirit of inquirypervaded the many issues of Quan-tum and stimulated readers to in-

v i i i Q U A N T O O N S

vent their own questions and allowtheir minds to explore.

The history of Quantum is worthrecounting briefly. Arthur Eisen-kraft was first introduced to theRussian publication, Kvant, by hisfriend Sergey Krotov of the formerSoviet Union during their years asAcademic Directors for their respec-tive Physics Olympiad teams. Theproblems, articles, and humor inKvant seemed like something thatcould be imported and massaged forUnited States audiences. Lots of in-terested people stepped up to theplate. Bill Aldridge, then ExecutiveDirector of NSTA, led the charge tocreate a magazine of “the highestquality.” Bill came through on hiscommitment. He enlisted the helpof Sheldon Glashow, a Nobel Laure-ate in physics, William Thurston, aFields medalist in mathematics, andYuri Ossipyan, vice-president of theAcademy of Sciences of the USSR,to launch the magazine. EdwardLozansky served as an internationalconsultant and Tim Weber took onthe responsibility of managing edi-tor. NSTA, under Bill Aldridge’sleadership, with financial supportfrom the National Science Founda-tion (NSF), committed resources toinsuring that Quantum met theneeds of our intended audience. Healso brought the American Associa-tion of Physics Teachers (AAPT) andthe National Council of Teachers ofMathematics (NCTM) on board.Larry Kirkpatrick and Mark Saulbecame the field editors for physicsand mathematics, respectively.

Arthur Eisenkraft and LarryKirkpatrick collaborated from thebeginning on the physics ContestProblems. The physics was great, aswere the literary quotes accompany-ing each, but better illustrationswere needed. Tomas Bunk, a profes-sional cartoonist with credits in-cluding MAD magazine and Gar-bage Pail Kids, was approached. Hisfirst reaction was “But I don’t knowphysics.” Arthur responded, “Well,that might be an asset. The next ar-ticle is about light. How would youdraw light?” Tomas replied, “I guesslike a super-hero because it travels

so fast.” And so the collaborationbegan. Of course, as Tomas learnedenough physics to illustrate thatfirst picture, Arthur learned aboutart. The illustration had light at thecircus moving through hoops, rico-cheting off mirrors, and dartingthrough a water-filled aquarium.Arthur loved the sketch but thelight did not always obey the laws ofphysics. What to do? Tomas ex-plained that to shift the path of lightwould require a new illustrationbecause the balance and structure ofthe art would be off in this portrayal.Arthur decided it would be better tostate below the illustration, “Lightis misbehaving in this picture. Canyou find where?” The cartoon be-came another dimension to theContest Problem. As the ContestProblems evolved, so did Tomas’s il-lustrations. They began to take onpolitical commentary, historicalideas, and larger issues of philosophywhile always providing insights intothe physics with whimsy and hu-mor.

Quantoons adds a feature thatwas not present in the originalQuantum series. Each illustrationnow has a brief commentary byTomas Bunk. This peek at the cre-ative mind of a visual artist not onlyprovides insight into how Tomasviews the world, but also howpeople who are not trained in phys-ics can appreciate the world of sci-ence and make it their own.

Some readers will move right tothe cartoons while others will beginwith the quotes. These readers maythen be curious enough to check outthe physics text that introduces thetopic. Still others will be intriguedby the complexity of some of thephysics problems, be tempted to in-vent a solution, and then may re-ward themselves with an investiga-tion of the illustration. Our hope isthat you will find your own, per-sonal way to enjoy Quantoons andbetter appreciate the world weshare.

—Arthur Eisenkraft andLarry D. Kirkpatrick

A S N A I L T H A T M O V E S L I K E L I G H T 1

toons

2 Q U A N T O O N S

scribed as “the angle of incidenceequals the angle of reflection.” Lighttraveling from a point in air to apoint in water is certainly morecomplicated. In this case, the lightbends (refracts) at the boundary be-tween the two surfaces. Theamount of bending is a property ofthe water and the color of the light.Light entering other transparentsubstances, like quartz or diamond,refract by different amounts.Willebrord Snell in 1621 was able togive a mathematical description of

the behavior of light, which is nowknown as Snell’s law:

n1 sin q1 = n2 sin q2,

where n1 and n2 are the indices of re-fraction. We can see that if the lightenters water (n = 1.33) from air (n =1.00) at an angle of 30°, the angle inwater would be 22°:

n1 sin q1 = n2 sin q2,1.00 sin 30° = 1.33 sin q2,

q2 = 22°.

Measuring the angle of refraction isone way to tell whether that’s a dia-mond or a piece of glass in that ring

you bought.What fascinates many people

about the study of physics is the al-ternative ways of explaining phe-nomena. The great mathematicianPierre de Fermat recognized (in1657) that the path of light is thepath that requires the least time.1 Ifyou try all possible paths from thelight source A to the object B afterthey hit the mirror, you’ll find thatthe shortest path, and so the quick-est, is the path through point D(fig. 1 on the next page), where theangle of incidence equals the angleof reflection.

1The “extremum path.”

MOST OF OUR READERSknow that light bouncing offa mirror travels along a paththat can be adequately de-

“The cause is hidden but the effect is known.”—Ovid, Metamorphoses

A snail that moves like light

P M S # 1 0 7

Light with its human face is having a super blast speeding in a flash through space and objects, breaking mirrorsand here and there some rules, shooting through a fishbowl—surprising the wet inhabitants—and through asolid glass prism without getting a headache. This spectacular high-speed performance is taking place in acosmic circus tent where the alligator and his dog assistant are in charge of the modest flashlight source. In theaudience we see the professional science community verifying the correct course and angles of light but notbeing very impressed by the seriousness of the performance. But that does not bother our lightweight spacecadet, who thinks that coming from a flashlight he is Flash Gordon himself.

—T.B.

A S N A I L T H A T M O V E S L I K E L I G H T 3

Light is bending the rules a bit here. (Can you see where?)

4 Q U A N T O O N S

You can demonstrate this foryourself by drawing lots of pathsand measuring them. You can alsoprove it with some simple geometryor by using some calculus.

Fermat’s theorem is also valid forrefraction: the path light takeswhen it passes from air to watermust be the path requiring the leasttime. In this case least time is notidentical to least distance, sincelight travels more slowly in waterthat in air. The speed of light in asubstance is equal to the speed oflight in a vacuum divided by thesubstance’s index of refraction n.

Proving that the path of the lightis the quickest one takes some inge-nuity. You can draw lots of paths oflight traveling from point A in air topoint B in water (fig. 2). You canthen measure the lengths of thelines in air and water. But Fermat’stheorem states that the path shouldtake the least time, not the leastdistance. We can multiply thelengths in water by 1.33, since the

light takes longer to travel in waterby a factor of 1.33. Then add thisdistance to the distance in air. Thepath that minimizes this sum is thepath the light takes. And—guesswhat? It’s the same path describedby Snell’s law! Those of you whohave some calculus background canprove it mathematically.

Leaving light behind, we enterthe world of slow-moving mollusksto find our contest problem. A snailmust get from one corner of a room(dimensions 5 m × 10 m × 15 m) tothe diagonally opposite corner inthe least time. The snail can walkon any of the four walls but may notwalk on the floor or ceiling. What isthe path that the snail should take?In part B of the contest problem, forour more advanced readers, thesnail finds that the 15 meter wallthat must be traveled is sticky—

Figure 2

A

B

that is, the snail can only travel ata fraction of its normal speed. If thesnail on the sticky wall travels at1/3 of its normal speed, what is thepath that requires the least time forthe snail? Finally, in part C, for ourmost advanced readers, what hap-pens if the snail finds that thestickiness of the first wall is notconstant but increases linearlyalong one dimension of the wall? Spe-cifically, the speed at one end of thewall is the normal speed and thespeed at the far end of the wall is 1/3the normal speed. What will be thepath of least time? You may need touse graphical or computer techniquesto solve parts B and C. Our best read-ers are encouraged to see if they canfind general proofs for any room (di-mensions l × w × h) and a stickinessfactor of s. We are not sure ourselvesif such general proofs exist.

SolutionYou were asked to help a snail

find the quickest path from one cor-ner of a room to a diagonally oppo-site corner.

In the first case, in which allwalls were identical and the dimen-sions of the room were 5 × 10 × 15,there are at least three ways to solvethe problem. The first is to choosedifferent crossover points at theedge between the two walls and cal-culate the total distance that thesnail travels. This numericalmethod may appear to be tedious,but it will actually converge on thecorrect solution quickly. A secondmethod is to call the height of thecrossover point x, write the totaldistance traveled in terms of x, anddifferentiate. By setting the deriva-tive equal to zero, the minimumdistance will be revealed as the so-lution to the equation. The thirdmethod is the elegant solution. Inthis case, the wall is opened up. Theroom is now a large rectangle of di-mensions 25 × 5. The shortest dis-tance will be the diagonal connect-ing the two corners of the rectangle.If the snail starts at the lower cornerof the 15-meter wall, the crossoverpoint can be found by using similartriangles. The crossover point is

Figure 1

AB

D

A S N A I L T H A T M O V E S L I K E L I G H T 5

5,

15 103.

x x

x

-=

=

In the second case, one of thewalls was declared “sticky,” mean-ing that the snail could travel atonly 1/3 of its speed on this wall.Unlike the first case, the shortestdistance is no longer the shortesttime! Since the snail travels at dif-ferent speeds on the two walls, thequickest path will be the one wherethe snail travels a greater distanceon the faster wall. Once again, thestraightforward but tedious solutionwould be to assign the variable x tothe crossover point, write an equa-tion that describes all paths in termsof x, and the minimum time will berevealed.

The more elegant solution in thiscase is to realize that light alwaystakes the least time to travel, and thatthis snail traveling on a sticky wall islike light traveling in a slower me-dium. We then recognize that thesolution will be Snell’s law (or, ifyou’ll forgive us, “Snail’s law”). Evenwith this knowledge, we are facedwith a fourth-order equation, whichwe choose to solve by numericaltechniques:

n1 sin θ1 = n2 sin θ2.

Since the stickiness factor is 3, thenn1 = 3 and n2 = 1, and it follows that

( )2 2 2 2

53 .

15 5 10

x x

x x

-=+ - +

We’ll try different values of x andsee if the value of the left side of theequation is equal to the value of theright side.

x left side right side

1 0.1996 0.37142 0.3965 0.28731.5 0.2985 0.33041.7 0.3378 0.31341.6 0.3182 0.32191.63 0.3241 0.31941.62 0.3221 0.3202

This method can give us any accu-racy we desire. It would certainly be

easier to plug the equations into aspreadsheet program and have allvalues given “instantly.”

The third part of the problem,to solve for a wall whose sticki-ness varies along one dimension,was solved by Jason Jacobs ofHarvard University. We will leavethis problem as a tease.

1 8 Q U A N T O O N S

ONE PERSON DESCRIBEDhow the bedroom wall movedacross the room. Anotherwatched as a huge wave of con-

crete traveled along the highway.We all saw the massive destructionwhen one bridge roadway collapsedon top of another. The earthquakein the San Francisco area that coin-cided with the 1989 World Seriesgave us a glimpse of the power andenergy in our planet.

In the fury of the destruction, theEarth is whispering secrets about itscomposition. The Earth is not solidrock. The Earth is not of uniformdensity. Longitudinal and trans-verse waves, called P and S waves,

travel through the Earth as a resultof the quake. The differences in Pand S wave behaviors can give usclues about the structure of theEarth while also allowing us to lo-cate the epicenter of the quake.

Although the speeds of the P andS waves vary within the Earth, theP waves always travel faster thanthe S waves. This fact gives us theability to locate the epicenter of thequake. By knowing the relativespeeds of the P and S waves andmeasuring the delay in the arrival ofthe S waves, we can determine thedistance from the epicenter. Here’san analogy. If you run at 3 m/s anda friend walks at 1 m/s, you will al-

ways arrive at a given location be-fore your friend. If you arrive 10 sec-onds earlier, the distance traveledwas 15 meters. If you arrive 20 sec-onds earlier, the distance traveledwas 30 meters.

Let’s assume that the epicenter isnear the Earth’s surface and that theP and S waves have constant but un-equal velocities. If at one locationon the Earth the waves arrive witha time difference of 2 minutes, weknow that the epicenter of thequake must be situated a specifieddistance from this location. But inwhich direction? We don’t know.We therefore trace the circumfer-ence of a circle on the surface with

Shake, rattle, and roll“She stood in silence, listning to the voices of the ground . . .”

—William Blake, “The Book of Thel”

Here we are looking at the high drama of a tremendous earthquake, causing tsunamis to splash enormous wavesin all directions, and a gargantuan volcanic eruption, spitting out even the devil himself above a town surroundedby a calm agricultural landscape. In front we have numerous scientists with their top-notch equipment trying todetermine the Earth’s innermost secrets and where the epicenter of the quake might be. Some measure the Swaves and others the P waves, some take a look at the fiery inside and some listen to it. In the very front is theboss in charge, looking like some Old Testament prophet investigating with his assistants the seismographicchart, which looks like a Torah. Whether Mother Earth will reveal her deepest secrets is not clear yet, but thequest to learn where we come from and what we are made of will keep the curious among us on our toes and offthe streets for a long time.

—T.B.

S H A K E , R A T T L E , A N D R O L L 1 9

2 0 Q U A N T O O N S

a radius specified by this time delay.The epicenter can be located on anypart of this circumference. If wehave a second location with a (dif-ferent) time delay, this will provideus with a second circle. A third lo-cation and a third circle willuniquely determine the actual pointlocation of the epicenter.

The P waves are able to travelthrough solids and liquids, whilethe S waves travel only through sol-ids. The P waves arrive at locationson the opposite side of the Earth; theS waves do not. This informationleads us to conclude that a portionof the interior of the Earth is liquid.By carefully observing where the Pwaves travel and where the S wavesdo not, we can infer more about thesize of this liquid core of the Earth.

More curious is the observationthat there are positions on the Earthwhere neither the P nor the S wavesarrive. These shadow zones aresomehow protected from distur-bances at some locations. Whatcould cause such a shadow region?One explanation is that the P wavestravel at a different speed within theliquid core. A P wave traveling froma solid mantle into a liquid core willchange speeds and change direction(that is, they will refract). The resultof this refraction is the creation ofthe shadow region.

Professor Cyril Isenberg, academicleader of the British Physics Olym-piad Team, challenged studentsworldwide in the 1986 InternationalPhysics Olympiad with a problemabout the P and S waves of an earth-

quake. We present parts of that prob-lem as a challenge to our readers.

Let’s assume that the Earth iscomposed of a central liquid spheri-cal core of radius Rc that is sur-rounded by a solid, homogeneousmantle of radius R. The velocities ofthe S and P waves through themantle are vS and vP, respectively.An earthquake occurs at point E onthe surface of the Earth and pro-duces P and S seismic waves. A seis-mologist observes the waves at lo-cation X. The angular separationbetween E and X measured from thecenter of the Earth O is 2θ, as shownin figure 1.

A. Our beginning physics stu-dents should try to show that theseismic waves that travel throughthe mantle in a straight line arriveat X at a time t (the travel time af-ter the earthquake) given by t = 2Rsin θ/v for θ ≤ arccos (Rc/R), wherev = vP for the P waves and vS for theS waves.

B. After an earthquake an ob-server measures the time delay be-tween the arrival of the S wave andthe P wave as 2 minutes, 11 sec-onds. Deduce the angular separationof the earthquake from the observerusing these data:

R = 6,370 kmRc = 3,470 kmvP = 10.85 km/svS = 6.31 km/s

C. The observer in part B noticesthat at some time after the arrival ofthe P and S waves, there are two fur-ther recordings on the seismometerseparated by a time interval of 6minutes, 37 seconds. Explain thisresult and verify that it is indeed as-sociated with the angular separationdetermined in part B.

D. For those of you who wish toplunge deeper, draw the path of aseismic P wave that arrives at an ob-server, where θ ≤ arccos (Rc/R), aftertwo refractions at the mantle–coreinterface. Obtain a relation for Pwaves between θ and i, the angle ofincidence of the seismic P wave atthe mantle–core interface.

E. For our advanced problem

solvers, using the data above andthe additional fact that the speed ofthe P waves in the liquid core is 9.02km/s, draw a graph of θ versus i.Comment on the physical conse-quences of the form of this graph forobservers stationed at differentpoints on the Earth’s surface.

F. Sketch the variation of the traveltime taken by the P and S waves as afunction of θ for 0 ≤ θ ≤ 90 degrees.

SolutionA. In part A, you were asked to

calculate the time it would take forP or S waves emanating from theearthquake location E to reach anobservation point X. From figure 2we can see that

EX = 2R sin θ.

Therefore,

t Rv

= 2 sin ,θ

where v = vP for P waves and v = vSfor S waves. This is valid providedthat X is at an angular separationless than or equal to X′, defined bythe tangential ray to the liquid core.From figure 2, X′ has an angularseparation given by

1 c2 2cos .RR

- Ê ˆf = Á ˜Ë ¯

B. Given the delay time between Pand S waves, you were next asked todeduce the angular separation of Eand X. Using the result from part A,

t Rv

= 2 sin ,θ

Figure 2

O

R

R R

E X

X′

Rcφ

θ θ

Figure 1

E X

R

OR

c

2θ

S H A K E , R A T T L E , A N D R O L L 2 1

we can express the time delay as

1 1

2 sin .S P

t Rv v

Ê ˆD = q -Á ˜Ë ¯ (1)

Substituting the data given, we get

( ) 1 1131 2 6370 sin .

6.31 10.85Ê ˆ= - qÁ ˜Ë ¯

Therefore, the angular separation ofE and X is

2θ = 17.84°.

This result is less than

1 1c 34702cos 2cos 114 ,

6370RR

- -Ê ˆ Ê ˆ= = ∞Á ˜Á ˜ Ë ¯Ë ¯

and consequently the seismic waveis not refracted through the core.

C. If a second set of P and S waveshad a longer delay, readers first hadto hypothesize an explanation forthe second set of delayed waves andthen see if the result is consistentwith the time delay given in part B.

The observations are most likelydue to reflections from the mantle–core interface. Using the symbols infigure 3, we can express the time de-

lay ∆t′ as

( ) 1 1

1 12 .

S P

S P

t ED DXv v

EDv v

Ê ˆD = + -¢ Á ˜Ë ¯

Ê ˆ= -Á ˜Ë ¯

In triangle EYD,

(ED)2 = (R sin θ)2 + (R cos θ – Rc)2

= R2 + Rc2 – 2RRc cos θ,

since sin2 θ + cos2 θ = 1. Therefore,

2 2c c

1 12 2 cos .

S P

t

R R RRv v

D =¢

Ê ˆ+ - q -Á ˜Ë ¯

Using equation (1), we get

∆∆

′ =+ −

=

tt R R RR

R

2 2

396 7

c2

c

s = 6 min 37 s.

cos

sin

.

θθ

Thus, the subsequent interval pro-duced by reflection of seismic wavesat the mantle–core interface is consis-

Figure 3 Figure 4

O

R

E X

Rc

θ θ

Y

RD

O

E XRcθ

BC

A

α

ir

90°− r

tent with an angular separation of17.84°.

D. Since a P wave is able to travelthrough the core, you were asked todraw the path of the refracted Pwaves and derive the relation be-tween the angle of incidence andthe angular separation of E and X.

From figure 4, we get

θ = ∠AOC + ∠EOA= (90 – r) + (i – α). (2)

The law of refraction (Snell’s law)gives

sinsin c

ir

vv

P

P

= . (3)

From triangle EAO and the law ofsines, we get

R R

ic

sinα=

sin. (4)

Substituting equations (3) and (4)into (2) yields

1 c

1 c

90 sin sin

1 sin sin .

P

P

vi

v

Ri

R

-

-

Ê ˆq = - Á ˜Ë ¯

Ê ˆ+ - Á ˜Ë ¯ (5)

E. Our most talented readers werethen asked to draw a graph of the re-lationship expressed in part D and tocomment on the physical conse-quences.

Substituting i = 0° into equation(5) gives θ = 90°; i = 90° gives θ =90.8°. Substituting numerical valuesfor i = 0° to i = 90°, one finds a mini-mum value at 55° and the corre-sponding minimum value of θ: θmin= 75.8° (fig. 5). As θ has a minimumvalue of 75.8°, observers at positionsfor which 2θ < 151.6° will not ob-serve the earthquake as seismicwaves. For 2θ < 114°, however, thedirect, nonrefracted waves willreach the observer.

F. Finally, readers were asked tosketch a comparison of the traveltimes for P and S waves for all angles.In this sketch (fig. 6) we can get a bet-ter sense of the “shadow” regionwhere no earthquake waves will beobserved.

0 55° 90°

90°

75.8°

i

θ

Figure 5

refr

acte

d

wav

es

S

P

P

t

0 90°

180° 90°

θ57° 75.8°

Figure 6

1 5 2 Q U A N T O O N S

W

“Sir Isaac Newton was very much smaller than a hippopotamus,but we do not on that account value him less.”

—Bertrand Russell (1872–1970)

Elephant ears

HY DO ELEPHANTS HAVEsuch big ears? And why dothey have such thick legs? Inother words, why do el-

ephants have different shapes thanhorses? These questions and morecan be answered using the laws ofscaling that we learn in physics.

Elephant bones are made fromthe same basic material as humanbones. Therefore, the bones must bethicker to support the extra mass ofthe elephant. But how muchthicker? Let’s compare an elephantto a horse. A typical horse has amass around 600 kg and a typicalelephant has a mass around 4200 kg,or some 7 times larger. Because allmammals have a density near thatof water, the elephant must have

7 times the volume of the horse. Ifwe assume that the two have thesame shape (they both have fourlegs!), the linear dimensions of theelephant must be 3 7 = 1.9 timesthe corresponding dimensions of thehorse.

Each elephant leg must support7 times as much weight as a horseleg. Because the compressionstrength of a beam depends on itscross-sectional area, an elephant legbone must have 7 times the cross-sectional area of a horse leg bone. Inother words, the elephant leg bonemust have 2.6 times the diameter ofa horse leg bone. Notice that the el-ephant and the horse cannot havethe same shape; the legs must beproportionately larger than the other

dimensions. The comparison wouldbe even more dramatic if we com-pared the elephant to a mouse!

This explains why elephants havesuch thick legs, but what about theears? Let’s assume for the momentthat an elephant eats 7 times asmuch as a horse because it has 7times as much mass. As this food isused by the body, it generates heat.Therefore, an elephant must dissi-pate 7 times as much heat as a horse.We know that the thermal loss isproportional to the difference in thetemperature across the skin and tothe area of the skin. The surface areaof any solid depends on the square ofits linear dimensions, so the el-ephant only has 1.92 = 3.6 times thesurface area. This means that the

Tonight at the circus an elephant is being measured to find out if it makes sense to be so big or to be differentfrom a horse or piglet, for example. A handful of specially appointed clowns are investigating with scientificunseriousness his ears, legs, memory, blood pressure, weight, height, inside mechanism that makes him tick, andthe important capacity of his trunk to hold water. The results will be collected, may be analyzed, but will defi-nitely and thoroughly be ignored. Even Einstein and Galileo stepped down from the heights of their scientificstatus to join the fun of circus life, performing daring stunts on the tightrope while a mouse tries to get hold ofa horse on the flying trapeze. Not an easy task compared to the little piglet dancing on top of the world.

—T.B.

E L E P H A N T E A R S 1 5 3

1 5 4 Q U A N T O O N S

elephant must have a much higherbody temperature or some otherway of getting rid of the thermalenergy. This is one of the roles of thebig ears. They increase the surfacearea and, by moving in the air, keepthe air temperature near the skinfrom climbing very much. The el-ephant also eats less per unit massthan a horse.

It is interesting to note that al-though elephants communicate byultrasound, it is not necessary forthem to have big ears for this purpose.

Not all scaling deals with lengths.We can use any factor as a scalingparameter. For instance, the Bohrradius for the hydrogen atom isgiven by

2

0 2 0.0529 nm,amke

= =

where � is Planck’s constant di-vided by 2π, m is the mass of theelectron, k is Coulomb’s constant,and e is the electronic charge. Whatwould be the new radius if the elec-tron were replaced by a muon witha mass 207 times as large? (We as-sume that the mass of the proton islarge compared to the mass of themuon.) We do not need to solve forthe new radius from scratch; all weneed to know is that the radiusscales inversely with mass. There-fore, the radius of the muonic hydro-gen atom is

00 0.256 pm.

207em a

a amm

m

Ê ˆ= = =Á ˜

Ë ¯

This was one of a series of fiveproblems on scaling that made up oneof the three theory questions at theInternational Physics Olympiad heldin Sudbury, Canada, in July 1997. Thetheory problems were developed un-der the direction of Chris Waltham,who is a faculty member at the Uni-versity of British Columbia. Three ofthe other scaling problems make upthe problem we offer below.

A. The mean temperature on theEarth is T = 287 K. What would thenew mean temperature T′ be if themean distance between the Earth

and the Sun were reduced by 1%?B. On a given day, the air is dry

and has a density ρ = 1.2500 kg/m3.The next day the humidity has in-creased and the air contains 2%water vapor by mass. The pressureand temperature are the same as theday before. What is the new air den-sity ρ′? Assume ideal gas behavior.The mean molecular weight of dryair is 28.8 g/mol, and the molecularweight of water is 18 g/mol.

C. A type of helicopter can hoverif the mechanical power output ofits engine is P. If another helicopteris produced that is an exact half-scale replica (in all linear dimen-sions) of the first, what mechanicalpower P′ is required for it to hover?

SolutionA correct solution to the first

question was submitted by TalCarmon from Technion, Israel.

A. The first question asked whatwould happen to the mean tempera-ture T of Earth if the mean distanceR between Earth and the Sun de-creased by 1%. To do this we matchthe input radiation to the output ra-diation because Earth is in thermalequilibrium. If the power output ofthe Sun is P, the radiation reachingEarth per unit area is P/4pR2. If wedenote Earth’s radius by RE and itsreflectance by r, the input power Pinto Earth is

( ) 2in E21 .

4

PP r R

R= - p

p

Stefan’s Law gives the output power

2 4out E4 ,P R T= p es

where e is the Earth’s emissivity ands is Stefan’s constant. Although theemissivity is a function of tempera-ture, the change in temperature isexpected to be small, and we canneglect this dependence. Therefore,

1,T

Rµ

and a reduction of 1% in R gives a0.5% rise in T. For a mean tempera-ture of 287 K, we get a rise of 1.4 K.

B. The second question askedabout the change in the density ofdry air with an increase in humiditywhen the temperature and pressureremain the same. Let’s use the sub-scripts d and m for dry and moist air,respectively. Then the number ofmolecules Nd in the dry air is

dd ,

28.8M

N µ

where Md is the mass of dry air in aunit volume and the mean molecu-lar mass of dry air is 28.8 g/mol. Formoist air, we must account for theproportions of dry air and water va-por. For 2% humidity, we have

m mm 0.02 0.98 ,

18 28.8M M

N µ +

where the mean molecular mass ofwater is 18 g/mol.

We know that identical volumesof ideal gases with the same tem-perature and pressure have the samenumber of molecules. Therefore,

Md = 1.012Mm.

Because the densities of equal vol-umes are proportional to the respec-tive masses,

m m

d d0.988,

MM

r= =

r

and using rd = 1.25 kg/m3, we getour answer:

rm = 1.235 kg/m3.

C. The last question asked how thepower required for a helicopter tohover depends on the size of the he-licopter. The mechanical power P ofthe helicopter is equal to the thrust Ttimes the downward velocity compo-nent v of the air below the blades. Thethrust is given by the change in mo-mentum of the air per unit time

,dm

T vdt

=

with

,dm

Avdt

= r

E L E P H A N T E A R S 1 5 5

where r is the density of the air andA is the cross-sectional area coveredby the blades. Thus,

T = rAv2.

When the helicopter is hovering, thethrust must be equal to thehelicopter’s weight. Therefore,

2 .T W

vA A

= =r r

If the size of the helicopter is char-acterized by a linear dimension L,then W µ L3, A µ L2, and v µ L0.5.Thus,

P = Tv = Wv µ L3.5.

For a half-scale helicopter, the requiredpower is 0.53.5P = 0.0884P.