ttlt Ex. 66 Particles ol Zerc Rest Mass

C. PHOTONS

66. Particles of zero rost mass

On what arguments does the derivation of the im-portant relation E - p': rnr depend? Derive fromthis formula a relation between momentum andenerry that is valid in the case of zero rest mass(photons, gravitons, neutrinos). What does the re-sulting relation say about the slope of the world linefor such a particle and hence about its speed? Howdoes your result depend, as a limiting €$, on theequality of sinhd and coshd for large d? Is there a"rest frame" for particles of zero rest mass?

67. Einstein's derivation of theequivalence of energy and restmass-a worked oxamplo

Problem: From the fact that light exerts pressureand carries energy, show that this energy is equivalentto mass and hence-by extension-show the equiva-lence of all energy to mass. Commentary: Theequivalence of energy and mass is such an importantconsequenc€ that Einstein very early, after his rela-tivistic derivation of this result, sought and found analternative elementary physical line of reasoningt thatleads to the same conclusion. He envisaged a closedbox of mass M initially at rest (Fig. 106). A directedburst of electromagnetic enerry is emitted from thehft-hand wall. It travels down the length I, of the boxand is absorbed at the other end. The radiation carriesan enerry E. But it also carries momentum. This onesces from the following reasoning. The radiationercrts a pressure on the left-hand wall during theernbsion. In consequenc€ of this pressure the boxrcceivcs a push to the left, and a momentum -p. But

Flg. 105. Transfer of mass withoutnet transfer of particles or radia-tion.

the momentum of the system as a whole was zeroinitially. Therefore the radiation carries a momentump opposite to the momentum of the box. How can oneuse his knowledge of the transport of enerry andmomentum by the radiation to deduce the raassequivaknt of the radiation? Einstein got his answerfrom the argument that the center of mass of the sys-tem was not moving before the transport process andtherefore cannot be in motion during the transportproc€ss. But the box obviously carries mass to theldt. Therefore the radiation mwt cany mass to theright. So much for Einstein's reasoning in broad out-line. Now for the details.

From relativity Einstein knew that the momentump of a directed beam of radiation is equal to the energtE of that beam (both p and E measured in units ofmass; Section l0). However, to secure a derivatjonfree of all direct reference to relativity principles, hebased the conclusion p: E on the following ele-mentary argument. The pressure exerted on an idealemitter or absorber by a directed beam of radiation is

equal to the densitknows both fromnetic radiation arpressure exerted bvacuum. This mbeen carried out btween l90l and 19so simplified and icarried out in anresulting conclusic

/density of radianI in unis of ener\ unit of volu

and multiply bothwall and the lengvisaged as much stNote that / is equation pressure multthe multiplying far

/volume occupiedI by the pulse of\ radiant energy

- | ^r-of emil-

\ surface

Multiply correspoequations and find

( energy carried\\ by radiation /

: /force of rad\ on wal

tRobert Pollock,(1963). Pollock's rnlight makes use of reeasily measured mtexperiment in collatmen with whom thethe presentation ofauthors are gratefulmember of this groton several subsequer

/monpntum: / to wdll-an

\ of mon\ra

Fk. 106. Transfer of mass by radiation.rA. Fjnstein, Annalen der Physik, 20,627 (19o6).

Ex. 67 Einstein's Derivation of the Equivalence of Energy and Rest Mass 1/19

cf mass without:ticles or radia-

yhole was zeros a morl€ntum<. How qrn oncof energr andJuce the raasrgot his ans*rrnass of the sys-ort process andt the transportes mass to thcry mass to theE in broad out-

lhe momentumal to the enerpred in unis of:e a derivationl principles, hefollowing ele-

ed on an idealr of radiation is

equal to the density of energr in that beam. This oneknows both from Maxwell's theory of electromag-netic radiation and from direct observation of thepressure exerted by light on a mirror suspended in avacuum. This measurement had first successfullybeen carried out by E. F. Nichols and G. F. Hull be-tween l90l and 1903. By now the experiment has beenso simplified and increased in sensitivity that it can becarried out in an elementary laboratory.t Take theresulting conclusion,

/density of radiant energy\I in units of energy per I\ unit of volume /

fRobert Pollock, American Journal of Physics, 31,901(1963). Pollock's method of determining the pressure oflight makes use of resonance to amplify a small effect to aneasily measured magnitude. Dr. Pollock developed thisexperiment in collaboration with the same group of fresh-men with whom the authors had the privilege to work outthe presentation of relativity contained in this book. Theauthors are grateful in particular to Mark Wasserman, amember of this group, who made many helpful commentson several subsequent drafts.

and multiply both sides by the area A of the emittingwall and the length / of the pulse of radiation (en-visaged as much shorter than the length L ofthe box).Note that / is equal to the time of action of the radia-tion pressure multiplied by the speed of light. Thusthe multiplying factor is

/volume occupied\I by the pulse of I\ radiant energy /

- ( ur-of emiuing\ /time of action\ / .o \: \''*;;;*"'" )("t:":X

*/ (;tfi;,/Multiply corresponding sides of the foregoing twoequations and find the result

( eneryy carried\\ by radiation /: ( "'* s ;liation) ("Ti,l?$"" ) (i,fifl ,)

/monpntum transfer from radiationl: / to witt-and therefore magnitude \ / speeO \

\ of momentum carried by / \of light/\ radiation itself /

or, in mass units

/ energv of \ /momentum of\(114) E: l d i rectedbeam f : l d i rected beam f :a

\ of radiation / \ of radiation fThus the radiation carries momentum and energr- tothe right while the box carries momentum and rncssto the left. But the center of mass of the system, boxplus radiation, cannot move. So the radiation mustcarry to the right, not merely energy, but mass. Howmuch mass? To discover the answer is the object ofthese questions:

(a) What is the velocity of the box during the timeof transit of the radiation?

(b) After the radiation is absorbed in the other endof the box, the system is once again at rest. How farhas the box moved during the transit of the radiation?

(c) Now demand that the center of mass of the sys-tem be at the same location both before and after theflight of the radiation. From this argument, what isthe mass-equivalent of the energy that has beentransported from one end of the box to the other?

Solution: (a) During the transit of the radiationthe momentum of the box must be equal inmagnitude and opposite in direction to themomentum p of the radiation. The box moveswith a very low velocity d. Therefore the New-tonian formula Mp suffices to calculate its mo-mentum:

Mp: -p: -5

From this relation we deduce the velocity of thebox,

ts: -ElM(b) The transit time of the photon is very

nearly l: Z meters of light-travel time. In thistime the box moves a distance

Ax: Ft : _ELIM

(c) If the radiation carried with it no mass, andthe box were the sole object endowed with mass,then this displacement A-r would result in a netmotion of the center of mass of the system to theleft. But, Einstein reasoned, an isolated systemwith its center of rnass originally at rest can neverset itself into motion nor experience any shift inits center of mass. Therefore, he argued, theremust be some countervailing displacement of apart of the mass of the system. This transport ofmass to the right can be understood only as a newfeature of the radiation itself. Consequently, dur-

pressure exerted by directedlradiation on perfect emitter Ior absorber in units of force

fper unit of area

150 Ex. 67 Einstein's Derivation of the Equivalence of Energy and Rest Mass

ing the time that the box is moving to the left, theradiation must transport to the right some massm, as yet of unknown magritude, but such as toensure that the center of gravity of the system hasnot moved. The distance of transport is the fulllength Z of the box diminished by the distanceAx through which the box has moved to the leftin the meantime. But Ax is smaller than Z in theratio ElM. This ratio can be made as small asone pleases for any given transport of radiantenergy E by making the mass M of the box suf-ficiently great. Therefore it is legitimate to takethe distance moved by the radiation as equal tot itself. Thus, with arbitrarily high precision, thecondition that the center of mass shall not movebecomes

MAx I mL:0

Calculate the mass zr and find (using Ax frompart b)

m : - AXMIL : -(-ELIMXMIL)

or, finally m: E

We conclude that the process of emission, trans-port, and reabsorption of radiation of energ5r Eis equivalent to the transport of a mass m : Efrom one end of the box to the other end. Thesimplicity of this derivation and the importanceof the result makes this analysis one of the mostinteresting in all of physics.

Discussion: The mass equivalence of radiant energlimplies the mass equivalence of thermal energy and-by extension-of other forms of enerry, according tothe following reasoning. The energr which emergcsfrom the left wall of the box may reside there originatlyas heat enerS/. This thermal energy excites a typicalatom of the surface from its lowest energt state to Ihigher enerry state. The atom returns from this highcrstate to a lower state and in the course of this changrsends out the surplus energy in the form of radiatio.This radiant energy traverses the box, is absorbed, andultimately converted back into thermal energy. What-ever the details of the mechanisms by which light bemitted and absorbed, the net efect is the transfer ofheat energy from one end of the box to the other. Tosay that mass had to pass down the lenglh of the bqwhen radiation went from one wall to the other there-fore implies that mass moves when thermal energchanges location. The thermal energy in turn is dc-rived from chemical energy or the energy ofa nucleartransformation or from electrical enerry. Moreover,thermal energy deposited at the far end ofthe tube canbe converted back into one or another of these fornsof energy. Therefore these form of energy-and like'wise all other forms of energy-are equivalent in theirtransport to the transport of mass in the amount

m:E

How can one possibly uphold the idea that a pulseof radiation transports mass? One already knows thata photon has zero rest mass, by virtue of the relation

(rest mass)2 : (enerry)2 - (momentum)r : 0

tI

;I!

I

(earlier derivatiexercise; also Sis true of the inradiation madeand momentunrest mass of tthere not a funthe rest mass ofbreath that rad:m: E from on

The source otween two quiltime componerand (2) rest nWhen the systetion going to tlthe componentsthe recoiling bponents of theemission (Fig.4-vectors (magNo one dealingthe length of orsum of the lengLorentz geomelnot to be consimass of the ra<recoiling box (ltors are additiv

/ energy \\of system/

Thus we see tlM-E.Notorthe emission o:mass is reduc€diagram). Thufrom the wallhas zero rest rr

/rest mass\\of systemr/-

is as natural in5#3+4inE

What aboutthe system ontion of mass armakes some dtest object beredistribution IIn other wordttest object is ti

Fig. lfi. Radiation transfers rest mass from placcto place even though the rest mass of radiation iszero !

Ex. 68 Photon Integrity 151

adiant energyenergy and-according to

hich emergesrere originally:ites a typicalrs/ state to arm this highcrrf this changpof radiation.

rbsorbed, and:nergr. What-rhich light bhe transfer ofthe other. To[h of the bore other thcre'ermal etrcrg-n turn is dcy of a nuclcary. Moreorcr.f the tube canrf these fornrgy-and lile-alent in thirI amount

r that a pubcly knom ttntrf the relatio

umF:0

-sr fr@ ?hof radinb r

(earlier derivation in this problem; see also precedingexercise; also Section 12 of the text). Moreover, whatis true of the individual photon is true of the pulse ofradiation made up of many such photons: The energyand momentum are equal in magnitude, so that therest mass of the radiation necessarily vanishes. Isthere not a fundamental inconsistency in saying thatthe rest mass of the pulse is zero and saying in the samebreath that radiation of energy .E transports the massm : E from one place to another?

The source of our difficulty is some confusion be-tween two quite different concepts: (l) energy, thetime component of the momentum-energy 4-vector,and (2) rest mass, the magnitude of this 4-vector.When the system divides itself into two parts (radia-tion going to the right and box recoiling to the left)the componenls of the 4-vectors of the radiation and ofthe recoiling box add up to identity with the com-ponents of the original 4-vector of the system beforeemission (Fig. 107). However, the magnitudes of the4-vectors (magrritude : rest mass) are not additive.No one dealing with Euclidean geometry would expectthe length of one side of a triangle to be equal to thesum of the lengths of the other two sides. Similarly inLorentz geometry. The rest mass of the system (M) isnot to be considered as equal to the sum of the restmass of the radiation (zero) and the rest mass of therecoiling box (less than M). But components of 4.vec-tors are additive; for example,

( en rgy \ _ /ene.gvof\ -,_ / energv of \\of system/ - \radiation/ ' \recoiling box/

Thus we see that the energr of the recoiling box isM - E. Not only is the energy of the box reduced bythe emission of radiation from the wall; also its restmass is reduced (see shortened length of 4-vector indiagram). Thus the radiation takes away rest massfrom the wall of the box even though this radiationhas zero rest mass. The result

/rest mass\ , / rest mass of \ , / restmassof \\of system/ - \radiation [zerol )

r \ recoiling box/

is as natural in spacetime geometry as is the inequality513 + 4 in Euclidean geometry.

What about the gravitational attraction exerted bythe system on a test object? Of course the redistribu-tion of mass as the radiation moves from left to rightmakes some difference in the attraction. But let thetest object be at a distance / so great that any suchredistribution has a negligible effect on the attraction.In other words, all that counts for the pull on a unittest object is the total mass M as it appears in New-

ton's formula for gravitational force/ . \I rorce per \: 674112\ unlt mass /

Even so, will not the distant detector momentarily ex-perience a less-than-normal pull while the radiation isin transit down the box? Is not the rest mass of theradiation zero, and is not the rest mass of the recoilingbox reduced below the original mass M of the system?So is not the total attracting mass less than normalduring the process of transport? No! The rest mass ofthe system-one has to say again-is not equal to thesum of the rest masses of its several parts. It is insteadequal to the magnitude of the total momentum-energy 4-vector of the system. And at no time doeseither the total momentum (in our case zero!) or thetotal energy of the system change-it is an isolatedsystem. Therefore neither is there any change in themagnitude M of the total momentum-energy 4-vector(Fig. 107). So, finally, there is never any change in thegravitational attraction.

There is one minor swindle in the way this problemhas been presented: the box cannot in fact move as arigid body. If it could, then information about theemission of the radiation from one end could be ob-tained from the motion of the other end long beforethe arrival of the radiation itself-this informationwould be transmitted at a greater speed than that oflight! Instead, the recoil from the emission of theradiation travels along the sides ofthe box as a vibra-tional wave, that is, with the speed of sound, so thatthis wave arrives at the other end long after the radia-tion does. In the meantime the absorption of theradiation at the second end causes a second vibra-tional wave which travels back along the sides of thebox. The addition of the vibration of the box to theproblem requires a more complicated analysis butdoes not change in any essential way the results foundabove.

68.' Photon integrity

Show that an isolated photon cannot split into twophotons going in directions other than the originaldirection. (Hint: Apply the laws of conservation ofmomentum and energy and the fact that the third sideof a triangle is shorter than the sum of the other twosides. What triangle?)

69.' The pressure of l ight

(a) Calculate the total force exerted by a one-wattflashlight beam.