Introduction to Relativistic Collisions

7/29/2019 Introduction to Relativistic Collisions

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IntroductiontoRelativisticCollisions

FrankW.K.FirkTheHenryKoernerCenterforEmeritusFaculty,YaleUniversity

NewHavenCT06520

Abstract:ThisintroductionisintendedforstudentstakingafirstcourseinNuclearandParticlePhysics.Electronpositronannihilationinflightisdiscussedasatopicalexampleofanexactrelativisticanalysis.

1.4-momentumandtheenergy-momentuminvariant

In Classical Mechanics, the concept of momentum is important

becauseofitsrleasaninvariantinanisolatedsystem.Wetherefore

introduce the concept of 4momentum in Relativistic Mechanics in

order to find possible Lorentz invariants involving this new quantity.

Thecontravariant4momentumisdefinedas:

P=mV (1)

wheremis themassof theparticle(it isaLorentzscalar themass

measured in the rest frame of the particle), and V[c, vN] is the 4

velocity. Here, is the relativistic factor and vN is theNewtonian 3

velocity,

Now,VV=c2,wherecistheinvariantspeedoflight,therefore

PP=(mc)2. (2)

The4momentumcanbewritten,

P=[mc,mvN](3)


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therefore,

PP=(mc)2(mvN)2.

Writing M=m,therelativisticmass,weobtain

PP=(Mc)2(MvN)2=(mc)2. (4)

Multiplyingthroughoutbyc2gives

M2c4M2vN2c2=m2c4. (5)

ThequantityMc2hasdimensionsofenergy;wethereforewrite

E=Mc2, (6)

thetotalenergyofafreelymovingparticle.

Thisleadstothefundamentalinvariantofdynamics

c2PP=E2(pc)2=Eo2 (7)

where

Eo=mc2istherestenergyoftheparticle,andpisitsrelativistic3-

momentum.

Thetotalenergycanbewritten:

E=Eo

=Eo

+T, (8)

where

T=Eo(1), (9)

therelativistickineticenergy.


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Themagnitudeofthe4momentumisaLorentzinvariant

|P|=mc. (10)

The4momentumtransformsasfollows: P=LP,whereListheLorentzoperator. (11)

2.TherelativisticDopplershift

Forrelativemotionalongthexaxis,theequationP=LPisequivalent

totheequations(here,=v/c)

E=Ecpx (12)

and,

cpx=E+cpx. (13)

UsingthePlanckEinsteinequationsE=handE=pxcforphotons,

theenergyequationbecomes

=

=(1)

=(1)/(12)1/2

={(1)/(1+)}1/2. (14)

ThisistherelativisticDopplershiftforthefrequency,measuredinan

inertialframe(primed)intermsofthefrequency measuredinanother

inertialframe(unprimed).


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3. Relativistic collisions and the conservation of 4-momentum

Considertheinteractionbetweentwoparticles,1and2,toform

twoparticles,3and4.(3and4arenotnecessarilythesameas1and2).Thecontravariant4momentaareP i:

BeforeAfter3P3P1P2124P41+23+4 Allexperimentsareconsistentwiththefactthatthe4momentum

of the system is conserved. We have, for the contravariant 4

momentumvectorsoftheinteractingparticles.

P1+P2=P3+P4 (15)____________initialfreestatefinalfreestate

andasimilarequationforthecovariant4momentumvectors,

P1+P2=P3+P4. (16)

IfweareinterestedinthechangeP1P3,thenwerequire

P1P3=P4P2 (17)


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Thisisthebasicequationforallinteractionsinwhichtworelativistic

entitiesintheinitialstateinteracttogivetworelativisticentitiesinthe

finalstate.Itappliesequallywelltointeractionsthatinvolvemassiveandmasslessentities.

3.1TheComptoneffect

Thegeneralmethoddiscussedintheprevioussectioncanbeused

toprovideanexactanalysisofComptonsfamousexperimentinwhich

thescatteringofaphotonbyastationary,freeelectronwasstudied.In

thisexample,wehave

E1=Eph(theincidentphotonenergy),E2=Ee0(therestenergyof

the stationary electron, the target), E3 = Eph (the energy of the

scatteredphoton),andE4=Ee(theenergyoftherecoillingelectron).

Therestenergyofthephotoniszero:

EphEph=pphcEe0>

EeThegeneralequationisnow

02(EphEphEphEphcos)=Ee022Ee0(Eph+Ee0Eph)+Ee02(23)

or


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2EphEph(1cos)=2Ee0(EphEph)

sothat

EphEph=EphEph(1cos)/Ee0

. (24)

Comptonmeasuredtheenergylossofthephotononscatteringand its

cosdependence.

4.Relativisticinelasticcollisions

Weshallconsideraninelasticcollisionbetweenaparticle1anda

particle2(initiallyatrest)toformacompositeparticle3.Insucha

collision,the4momentumisconserved(asit isin anelasticcollision)

however,thekineticenergyisnotconserved.Partofthekineticenergy

of particle 1 is transformed into excitation energy of the composite

particle3.Thisexcitationenergycantakemanyformsheatenergy,

rotational energy, and the excitation of quantum states at the

microscopiclevel.Theinelasticcollisionisasshown:

BeforeAfter123

p1p2=0p3 Restenergy:E10E20E30 Totalenergy:E1E2=E20E3 3momentum:p1p2=0p3 Kineticenergy:T1T2=0T3


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In this problem, we shall use the energymomentum invariants

associatedwitheachparticle,directly:

i)E12(p1c)2=E102 (25)

ii)E22=E202 (26)

iii)E32(p3c)2=E302. (27)

Thetotalenergyisconserved,therefore

E1+E2=E3=E1+E20. (28)

Introducingthekineticenergiesoftheparticles,wehave

(T1+E10)+E20=T3+E30. (29)

The3momentumisconserved,therefore

p1+0=p3. (30)

Using

E302=E32(p3c)2, (31)

weobtain

E302=(E1+E20)2(p3c)2

=E12+2E1E20+E202(p1c)2

=E102+2E1E20+E202

=E102+E202+2(T1+E10)E20 (32)

or


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E302=(E10+E20)2+2T1E20(E30>E10+E20). (33)

UsingT1=1E10E10,where1=(112)1/2and1=v1/c,wehave

E30

2

=E10

2

+E20

2

+21E10

E20

. (34)

Iftwoidenticalparticlesmakeacompletelyinelasticcollisionthen

E302=2(1+1)E102. (35)

5.TheMandelstamvariables

In discussions of relativistic interactions it is often useful to

introduceadditionalLorentzinvariantsthatareknownasMandelstam

variables.Theyare,forthespecialcaseoftwoparticlesintheinitialand

finalstates(1+23+4):

s = (P1 + P2)[P1 + P2], the total 4momentum

invariant

=((E1+E2)/c,(p1+p2))[(E1+E2)/c,(p1+p2)]

=(E1+E2)2/c2(p1+p2)2 (36)

Lorentzinvariant,

t = (P1 P3)[P1 P3], the 4momentum transfer

(13)invariant

=(E1E3)2/c2(p1p3)2 (37)

Lorentzinvariant,


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and

u = (P1 P4)[P1 P4], the 4momentum transfer

(14)invariant

=(E1E4)2/c2(p1p4)2 (38)

Lorentzinvariant.

Now,

sc2=E12+2E1E2+E22(p12+2p1p2+p22)c2

=E102+E202+2E1E22p1p2c2

=E102+E202+2(E1,p1c)[E2,p2c]. (39) _____________ Lorentzinvariant

The Mandelstam variable sc2 has the same value in all inertial

frames. We therefore evaluate it in the LAB frame, defined by the

vectors

[E1L,p1Lc]and[E2L=E20,p2Lc=0], (40)

sothat

2(E1LE2Lp1Lp2Lc2)=2E1LE20, (41)

and

sc2=E102+E202+2E1LE20. (42)


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Wecanevaluatesc2inthecenterofmass(CM)frame,definedby

thecondition

p1CM

+p2CM

=0(thetotal3momentumiszero): sc2=(E1CM+E2CM)2. (43)

ThisisthesquareofthetotalCMenergyofthesystem.

5.1ThetotalCMenergyandtheproductionofnewparticles

Thequantitycsistheenergyavailablefortheproductionofnew

particles,orforexcitingtheinternalstructureofparticles.Wecannow

obtaintherelationbetweenthetotalCMenergyandtheLABenergyof

theincidentparticle(1)andthetarget(2),asfollows:

sc2=E102+E202+2E1LE20=(E1CM+E2CM)2=W2,say.(6.44)

Here,wehaveevaluatedthelefthandsideintheLABframe,andthe

righthandsideintheCMframe!

Atveryhighenergies,cs>>E10andE20,therestenergiesofthe

particlesintheinitialstate,inwhichcase,

W2=sc22E2LE20. (45)

The total CM energy,W, available for theproductionof new particles

thereforedependsonthesquarerootoftheincidentlaboratoryenergy.

Thisresultledtothedevelopmentofcolliding,orintersecting,beamsof

particles (such as protons and antiprotons) in order to produce


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sufficient energy to generate particleswith restmasses typically 100

timestherestmassoftheproton(~109eV).

6.Positron-electronannihilation-in-flight

Adiscussionoftheannihilationinflightofarelativisticpositron

and a stationary electron provides a topical example of the use of

relativisticconservationlaws. Thisprocess, inwhichtwophotonsare

spontaneously generated, has been used as a source of nearly

monoenergetic highenergy photons for the study of nuclear photo

disintegration since 1960. The general result for a 1 + 2 3 + 4

interaction,giveninsection3,providesthebasisforanexactcalculation

ofthisprocess;wehave

E1=Epos(theincidentpositronenergy),E2=Ee0(therestenergyof

the stationary electron), E3 = Eph1 (the energy of the forwardgoing

photon),andE4=Eph2(theenergyofthebackwardgoingphoton).The

restenergiesofthepositronandtheelectronareequal.Thegeneral

equationnowreads

Ee02

2{EposEph1cpposEph1(cos)}+0=0Ee02

2Ee0

(EposEph1)(46)

therefore

Eph1{Epos+Ee0[Epos2Ee02]1/2cos}=(Epos+Ee0)Ee0,

giving


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Eph1=Ee0/(1kcos) (47)

where

k=[(EposEe0

)/(Epos+Ee0

)]1/2

. Themaximum energyof thephoton,Eph1maxoccurswhen=0,

correspondingtomotionintheforwarddirection;itsenergyis

Eph1max=Eoe/(1k). (48)

If,forexample,theincidenttotalpositronenergyis30MeV,and

Ee0=0.511MeVthen

Eph1max=0.511/[1(29.489/30.511)1/2]MeV

=30.25MeV.

Theforwardgoingphotonhasenergyequaltothekineticenergyofthe

incidentpositron(T1=300.511MeV)plusaboutthreequartersofthe

totalrestenergyofthepositronelectronpair(2Ee0=1.02MeV).Using

the conservation of the total energy of the system, we see that the

energyofthebackwardgoingphotonisabout0.25MeV.

Themethodofpositronelectron annihilationinflight provides oneof

theveryfewwaysofgeneratingnearlymonoenergeticphotonsathigh

energies.ThemethodwasfirstproposedbyTzara(1957).


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References

Thefollowingreferencesplacethesubjectofrelativisticcollisionsinthe

general context of Special Relativity, and provide many specificexamplesoftheapplicationofthetheorytoexperiments.

1.G.StephensonandC.W.Kilmister,SpecialRelativityforPhysicists,

DoverPublications,Inc,NewYork(1987).

2.WolfgangRindler,IntroductiontoSpecialRelativity,2ndedition,

ClarendonPress,Oxford(1991).

3.C.Tzara,Compt.Rend.245,56(1957).

4.C.SchuhlandC.Tzara,Nucl.Instr.Meth.10,217(1961).

5.A.M.Baldin,V.I.GoldanskiiandI.L.Rozenthal,KinematicsofNuclear

Reactions,Oxford,(1961).

6.C.F.Powell,P.H.FowlerandP.H.Perkins,TheStudyofElementary

ParticlesbythePhotographicMethod,PergamonPress,Oxford

(1959).

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Introduction to Relativistic Collisions