Physics2203,Fall2012ModernPhysics

.

 Monday,Aug.27th,2011:StartCh.2:Rela@vis@cMomentum,Energy,Conversionofmass

andEnergy.Doclassicallawsofmomentumandenergyconserva@onremainvalidinEinstein’srela@vity?

 Announcements:• Ipostedtheinclassexercise:mycalcula@onofγwaswrong!• WednesdayorFridaywewillhaveaQuiz.• Tutorialsession4:30pmTuesdayinNicholson102• IssacmayhitusonWednesday!

Physics2203ModernPhysics ClassAc5vity,Aug.24th,2012Twopowerlessrocketsareonacollisioncourse.Therocketsaremovingwithspeedsof0.80cand‐0.60candareini@ally2.52x1012mapart(t=0)asmeasuredbyLiz,anEarthobserver.Bothrocketsare50minlengthasmeasuredbyLiz.

a)  Whataretheirrespec@veproperLengths.b)  Whatisthelengthofeachrocketmeasuredbyanobserverintheotherrocket?c)AccordingtoLiz,howlongbeforetherocketscollide?d)Accordingtorocket#1howlongbeforetheycollide?

x ' = γ x − vt( )y ' = yz ' = z

t ' = γ t −vxc2

x = γ x '+ vt '( )y = y 'z = z '

t = γ t '+ vx 'c2

u 'x =ux − v

1− vuxc2

u 'y =uy

γ 1− vuxc2

u 'zuz

γ 1− vuxc2

ux =u 'x+ v

1+ vu 'xc2

uy =u 'y

γ 1+ vu 'xc2

uzu 'z

γ 1+ vu 'xc2

Considertheinelas@ccollisionoftwoequalmassobjectsshowninthefigure.

We will assume that p

= mu

p(before)=mv+m(-v)=0p(after)=0 Momentum is conserved

NowconsidertheS’systemmovingwithobject#1atavelocityofv.

Look at momentum p', before and after

p '(before) = mv1 '+ mv2 ' = 0 − 2mv

1− v2

c2

p '(after) = 2mV ' = −2mv : Momentum not conserved

Whatdidwedowrong?Howcanwefitthis?

u 'x =ux − v

1− vuxc2

we need to find v2 'and V' using the transform

v2 ' = v2 − v

1− v2vc2

=−v − v

1− v2

c2

=−2v

1− v2

c2

V ' = V − v

1− Vvc2

=0 − v

1− 0( )vc2

= −v

Whatdidwedowrong?Howcanwefitthis? u 'x =ux − v

1− vuxc2

Letsbackupandlookatthedefini@onof

momentum!

Classical definition

p= m

dr

dt: This did not work for relativistic transformations

p= m

γ dr

dt=

mu

1− u2

c2

Classicalmomentum

Rela@vis@cmomentum

Lets be very careful about a proper measurement of time, to

p= m dr

dt0

p= m dr

dt0: but dt0 =

dtγ

Considertheinelas@ccollisionoftwoequalmassobjectsshowninthefigure.

NowconsidertheS’systemmovingwithobject#1atavelocityofv.

Provethatusingthenewdefini5onofmomentumthatmomentumisconserved.

p=

mu

1− u2

c2

IfSisaniner@alframeandifasecondframeS’moveswithconstantvelocityrela@vetoS,thenS’isalsoaniner@alframe.

OR—thelawsofphysicsareinvariantaswechangefromonereferenceframetoanotheriner@alframe.

Inalliner@alframes,lighttravelsthroughthevacuumwiththesamespeed,c=299,792,458m/sinanydirec@on.

How does momentum transform?How can we conserve momentum?

p=

mu

1− u2

c2

What will the relativistic Energy look like?

E =mc2

1− u2

c2

IfSisaniner@alframeandifasecondframeS’moveswithconstantvelocityrela@vetoS,thenS’isalsoaniner@alframe.Wemustgetclassicallimit!

Momentum u<<<c?

px =mux

1− ux2

c2

= mux 1+ ux2

2c2

≈ mux

Relativistic Energy for u<<<c?

E =mc2

1− u2

c2

=mc2 1+ u2

2c2

= mc2 +

mu2

2 E = mc2

4‐meter‐tallsculptureofEinstein’s1905E=mc2

WalkofIdeas,Berlin,Germany

p=

mu

1− u2

c2

= γmu

Howshouldwethinkaboutthisequa@on??

Oneahrac@vewayistoconsiderγmasthemass.Thishastheadvantagethattheequa@onformomentumlooksliketheclassicalequa@on.

p=

m0u

1− u2

c2

= γm0u= muYouwillseethisdonemanyplaces.We

willdothisinsubsequentproblems.Butthisisnotreallyconsistent.

Lets write the kinetic energy using this form of the mass.

EK =mv2

2=p2

2m=γm0v( )2

2γm0

=γm0v

2

2 We will show thatEK = mc

2 γ −1( )

Ques5on:Forwhatvalueofu/cwillthemeasuredmassofanobjectγmexceedtherestmassbyagivenfrac@onf?

f =γm − mm

= γ −1 = 1

1− u2

c2

−1

Solving for u/c:

1− u2

c2 =1

f +1( )2 : u2

c2 = 1− 1f +1( )2

uc=

f f + 2)( )f +1

Wecangeneratethetableatthelej.

Howdowefindu/c??

f = 1: γm=2muc= 0.866

f = 3: γm=4muc= 0.968

Canu/c>1????

uc=

1+ 2f

1+ 1f

≈ 1+ 1f

1−

1f

=1−

2f

Todaywewillshowhowtocalculatev/c,buta1.3GeVelectronacceleratorproduceselectronswithv/c=0.99999992:γ=2545 f = γm −m

m= γ −1= 2545 −1= 2544!

f = γm −mm

= γ −1= 2545 −1= 2544!

Findγ forthe8GevSpring‐8inJapan.

c( ) : Classicallypx 0.4c( )px 0.8c)( )

= 0.5

Ques5on:Ahigh‐speedinterplanetaryprobewithamassm=50,000kghasbeensenttowardPlutoataspeedofu/c=0.8.(a)WhatisitmomentumasmeasuredbymissionControlonEarth?(b)If,preparatorytolandingonPluto,theprobe’sspeedisreducedto0.4c,byhowmuchdoesitsmomentumchange?(c)Howdoesthiscomparetheclassicalanswer?

a( ) : px =mux

1− u2

c2

=50,000kg( ) 0.8c( )

1− 0.8( )2= 2.0x1013kg •m / s

b( ) : px 0.4c( ) = mux

1− u2

c2

=50,000kg( ) 0.4c( )

1− 0.4( )2= 6.67x1012 kg •m / s

b( ) : px 0.4c( )px 0.8c)( )

= 0.33

p =mu

1− u2

c2

px =mux

1− u2

c2

WewillhavetworequirementsfortheformoftheRela@vis@cKine@cEnergy1)  ThetotalenergyEofanyisolatedsystemisconserved.2)  Ewillapproachtheclassicalvaluewhenu/capproacheszero.

Letusgobacktoourclassicaldefini@onofF,p,andE.

F=d p

dt=d γmu

( )dt

: This is the relativistic contribution

Work = EK = Fdx =d γmu

( )dt0

u

∫0

u

∫ dx = ud γmu

( )0

u

∫

You show that

d γmu

( ) = mdu

1− u2

c2

3/2 m 1− u2

c2

−3/2

udu0

u

∫ = mc21

1− u2

c2

−1

EK = mc2 γ −1( ) = γmc2 − mc2

EK = mc2 γ −1( ) = γmc2 − mc2

mc2 ⇒ Rest EnergyMass and Energy interchangeable

Make sure this works in classical limit

EK = γmc2 − mc2 = 1+ u2

2c2

mc2 − mc2

EK ≈mu2

2 OK

We can now define the total energy E

E = EK +mc2 = γmc2 =

mc2

1− u2

c2

E =mc2

1− u2

c2

E =mc2

1− u2

c2

p =mu

1− u2

c2Wecanderiveanotherveryusefulequa@on

pE

=u

c2β ≡

pcE

Squarebothequa@onsinyellow(pandE)

pc( )2 = m2c2u2

1− u2

c2

E2 =m2c4

1− u2

c2

E2 − pc( )2 = m2c4 1

1− u2

c2

−u2 / c2

1− u2

c2

E2 − pc( )2 = mc2( )2

E2 = pc( )2 + mc2( )2LookslikePythagoreanrela@onship.Ifpcismuchsmallerthanmc2,theenergyismostlyrestenergy

Ifpcismuchlargerthanmc2,theenergyismostlyKine@c.

E =mc2

1− u2

c2

p =mu

1− u2

c2Wecanderiveanotherveryusefulequa@on

pE

=u

c2β ≡

pcE

Squarebothequa@onsinyellow(pandE)

pc( )2 = m2c2u2

1− u2

c2

E2 =m2c4

1− u2

c2

E2 − pc( )2 =m2c4 1− u

2

c2

1− u2

c2

E2 − pc( )2 = mc2( )2

E2 = pc( )2 + mc2( )2

E = γmc2 =mc2

1− u2

c2

EK = mc2 γ −1( ) = γmc2 − mc2

Ques@on:Whatistherestenergyofanelectron.Wecanthinkofmc2astheproperenergy,measuredatrest.

mc2 = 9.11x10−31kg( ) 3x108m / s( )2= 8.19x10−14 J

Terrible Units

1.0eV = e 1.0V( ) = 1.602x10−19C( ) 1.0V( ) =1.602x10−19 J

mc2 (electron) =8.19x10−15 J( )eV1.602x10−19 J

= 5.11x105eV

mc2 (electron) = 0.511MeVEe = γmc

2 =0.511MeV

1− u2

c2

E = γmc2 =mc2

1− u2

c2

EK = mc2 γ −1( ) = γmc2 − mc2

Ques@on:Whatistherestenergyofaproton.Wecanthinkofmc2astheproperenergy,measuredatrest.

mc2 = 1.67x10−27 kg( ) 3x108m / s( )2=15.03x10−11J

Terrible Units

1.0eV = e 1.0V( ) = 1.602x10−19C( ) 1.0V( ) =1.602x10−19 J

mc2 (proton) =15.03x10−11J( )eV

1.602x10−19 J= 938 MeV

mc2 (proton) = 938 MeVEe = γmc

2 =938 MeV

1− u2

c2

p/mv

p =mu

1− u2

c2pmu =

1

1− u2

c2

E2 = pc( )2+ mc2( )2

= K + mc2( )( )2

K 2 + 2Kmc2 + mc2( )2= p2c2 + mc2( )2

p2

2K= m +

K2c2 EK = K( )

EK =1

1− u2

c2

−1

0.511MeV( )K = p2c2 + mc2( )2 − mc2

E(electron) = 0.511MeV

1− u2

c2

EK = γ −1( ) 0.511MeV( )Ques@on:Whatisthevelocity,momentum,andtotalenergyofanelectroninthestorageringatCAMD?Themachineenergyis1.3GeV.

1.3 Gev is the Kinetic Energy, so1.3Gev = γ −1( ) 0.511MeV( )γ −1= 2545 : rest mass irrelevant

γ = 2545 = 11− β2

1− β2 =1.54x10−7

v = 0.99999992c

MomentumE =1.3Gev+ 0.511MeV

pc = E2 − mc2( )2 =1.3005GeVAlmost like a massless particle

pc = E2 − mc2( )2=1.3005GeV

pc ≈ E

Massmc = γm0 = 2545m0

Ques5on:Rememberthatweusedmuonsproducedbycosmicraystoillustratebothlengthcontrac@onand@medila@on.Theyenterouratmospherewithv=0.998c.Therestenergyofthemuonis~2075mesthatofanelectron.(a)  WhatwillanobserveronEarthmeasureasthetotalenergy?(b) Whatwillanobservermeasureasthemassofthemuon?

The rest Energy mc2 is:207x 0.511MeV( ) = 105.7MeV

β = 0.998cγ = 15.8

(a) E = γmc2 = 15.8 105.7MeV( ) = 1670MeV

(b) measured m = γm =Ec2 = 1670 MeV

c2

Beverycarefulwiththis!

Conserva5onofmass‐energy

Classicallymomentumisconservedinthetotalinelas@ccollisionshownabove,buttheKine@cenergyisnotconserved.Beforeitis2mvbutajeritis0.

Conservation of total enery mass-energy

Ei∑ =mic

2

1− ui / c( )2i∑ conserved

Lets not assume that mass is conserved, only that Ebefore = Eafter

mic2

1− ui / c( )2+

mic2

1− ui / c( )2= Mc2

Lets not assume that mass is conserved only that Ebefore = Eafter

M =2m

1− ui / c( )2: M>2m

Define ΔM = M − 2m

ΔM = 2m 1− 1

1− ui / c( )2

ΔM =2mc2

c2 1−γ( )

ΔM =2Ki

c2

ThelostKine@cEnergyisconvertedintomass!!

E = γmc2 =mc2

1− u2

c2

=E0

1− u2

c2

= EK + E0Asimpleexperiment:twoblocksofwoodwithequalmassmandkine@cenergyK,aremovingtowardeachotherwithvelocityv.Aspringplacedbetweenthemiscompressedandlocksinplaceastheycollide.Letslookattheconserva@onofmass‐energy

Mass − Energy before: E=2mc2 + 2KMass − Energy after: E=Mc2

Since Energy is conserved we haveE=2mc2 + 2K=Mc2

M is greater than 2m because K went into mass

ΔM = M − 2m =2Kc2

fr =M − 2m

2m=

Kmc2

for real systems ~10−16

HW:Whatisthemassofacompressedspring?

ExamplesofEnergytoMassExchangethatfollow• Ioniza@on• ChemicalBindingEnergy• Fusion• Fission

BindingEnergyoftheHydrogenAtom:Thebindingenergiesofelectronstothenucleiofatomsaremuchsmallerthannuclearbindingenergies.Thebindingenergyforanelectrontoaproton(Bohrmodel)is13.6eV.Howmuchmassislostwhenanelectronandprotonfromahydrogenatom?

Δmc2 = E(binding) = 13.6eVmHc

2 ≈ mpc2 = 938.3MeV

ΔmmH

≈ 1.4x10−8

Thisisagainabeforeandajerproblem:Beforeyouhaveanisolatedelectronandproton.ATeryouhaveaHatomwhosebindingenergyis13.6eV.

(a)  Howmuchlighterisamoleculeofwaterthantwohydrogenatomsandanoxygenatom?Thebindingenergyofwateris~3eV.

(b)  Findthefrac@onallossofmasspergramofwaterformed.(c)  Findthetotalenergyreleased(mainlyasheatandlight)when1gramofwaterisformed?

(a) ΔM= mH + mH + mO( ) − MH2O=E(binding)

c2

ΔM=3.0eV( ) 1.6x10−19 J / eV( )

3.0x108m / s( )2 = 5.3x10−36 kg Reallysmall!

(b) ΔMMH2O

=E(binding)MH2O

c2

MH2O=18u : u is 1/12 of the mass of Carbon (6 protron, 6 neutrons): Wednesday!

MH2O=18 1.66x10−27 kg( )

ΔMMH2O

= 5.3x10−36 kg18 1.66x10−27 kg( )2 =1.8x10−10 S@llsmall!

(c) E=Δmc2 = 1.8x10−10( ) 10−3kg( ) 3x108m / s( )2= 16kJ Big!

Fusion:Energyisgainedbytakingtwolightatomsandcombiningthemintoanatomwithaheaviernuclei‐laterthissemester.Forexample:

12H + 1

2H = 24He + Energy

Doyouunderstandthisnota@on?ForHe,thereare4nucleons,andtwoareprotons.

Ec2= mass 1

2H + 12H( ) − mass 2

4He( )E = 3751.226 − 3727.379( )MeVE = 23.9MeV

FissionReac5ons:Thedecayofaheavyradioac@venucleusatrestintoseverallighterpar@clesemihedwithlargekine@cenergiesisagreatexampleofmass‐energyconversion.AnucleusofmassMundergoesfissionintopar@cleswithmassesM1,M2,andM3,withspeedsofu1,u2,andu3.

The conservation or Relativistic Energy Requires that

Mc2 =M1c

2

1− u12

c2

+M2c

2

1− u22

c2

+M3c

2

1− u32

c2

This is a very important equation to remember

The Equation above is true if M > M1 + M2 + M3( )Disintegration Energy Q defined

Q= M − M1 + M2 + M3( ) c2

Example in our text (pg 61)232Th→ 228Ra + 4HeThe offspring have 4 Mev Kinetic Energy

AppendixD:1u=1.66x10‐27kg:Wednesday

ΔM = 0.004u( )1.7x10−27 kg( )

uΔM = 7x10−30 kgΔMc2 = 6.3x10−13J = 4MeV