Relativistic Dynamics - Brown...

Gaitskell

PH0008Quantum Mechanics and Special Relativity

Lecture 11 (Special Relativity)

020322v3

Relativistic DynamicsCollision, Mass depends on velocity, energy-momentum

invariant, Compton Effect

Prof Rick Gaitskell

Department of PhysicsBrown University

Main source at Brown Course Publisher

background material may also be available at http://gaitskell.brown.edu

PH0008 Gaitskell Class Spring2002 Rick Gaitskell

Section: Special Relativity Week 4

• Homework (due for M 3/18)o Hand in now

• Reading (Prepare for 3/18)

o SpecRel (also by French)• Ch6 RelativisticKinematics

• Lecture 10 (M 3/18)o Relativistic Dynamics

• From Classical -> Relativistic

• What Conservation Laws can we rely on?• Expressions for m, E and p

• Collisions

• Lecture 11 (W 3/20)o Relativistic Dynamics

• Collisions

• Newton’s 2nd Law

• Compton Effect

• Lecture 12 (F 3/22)o Relativistic Dynamics

• Summary

• Review Problem Set

• Reading (Prepare for 4/1 after recess)

o SpecRel• Revise Ch2-6 (look at Ch 1 also)

o QuantMech• Ch1,2 & 3

• Homework #8 (M 4/1)o (see web “Assignments”)


Exam II Timing

• 4/8 (Monday) 8.30 am Start


Homework

• Please pick up your HW #1-4 from outside my office B&H 516o Definitely there


Question SectionQuestion Section


Question SpecRel L11-Q1

•Which relativistic expression correctly gives the totalmass of moving particle, velocity b(g) rest mass m0?

o(1)

o(2)

o(3)

o(4)

†

m = bm0

†

m = gbm0

†

m = gm0

†

m = g 2m0


Question SpecRel L12-Q2

•How big was the ice sheet that just fell off AntarcticIce Cap?

o(1) Block Island

o(2) Manhattan Island

o(3) Rhode Island

o(4) Who cares?Photo AFP / NY Times


Classical -> Relativistic Dynamics - “1 page”• Conservation of Energy & Mass are combined• Conservation of Momentum - still good

o but mass term is no longer invariant

†

m = g m0

E = g m0c2

p = gbm0c

E 2 = (pc)2 + (m0c2)2

KE = E - m0c2

= g -1( )m0c2

†

p = gbm0c

(cp)2 = g 2b 2(m0c2)2 g 2 =

11- b 2 fi b 2 =

g 2 -1g 2

= g 2 -1( )(m0c2)2

= g 2(m0c2)2 -(m0c

2)2

= E 2 -(m0c2)2

†

Consider KE for b <<1

KE = E - m0c2

= g -1( )m0c2

g =1

1- b 2( )12

ª 1+12

b 2 + ...Ê

Ë Á

ˆ

¯ ˜

ª 1+12

b 2 + ...-1Ê

Ë Á

ˆ

¯ ˜ m0c

2

ª12

b 2m0c2 + ...

ª12

m0v2 + ...

which is the classical form

†

b†

g

†

1

†

1

†

0

†

10

†

Consider p for b <<1p = gbm0c

g =1

1- b 2( )12

ª 1+12

b 2 + ...Ê

Ë Á

ˆ

¯ ˜

ª b +12

b 3 + ...Ê

Ë Á

ˆ

¯ ˜ m0c

ª bcm0 + ...ª m0v + ...



Collision treated inCollision treated inRelativistic DynamicsRelativistic Dynamics


Elastic Collision: (Extract m(v))

• Centre of Mass Frame S’

• Particle A Stationary Frame S

†

¢ b B

†

¢ b A†

¢ m B

†

¢ m A

†

bB

†

bA†

mB

†

mA

x

y

x’

y’b

†

(Assume ¢ S is moving at b in S.)

Usual velocity transforms apply

¢ b x =bx - b( )1- bxb( )

¢ b y =by g( )

1- bxb( )

bx =¢ b x + b( )

1+ ¢ b xb( )by =

¢ b y g( )1+ ¢ b xb( )

We will show that the masses of particles(used in momentum expression mv) are afunction of the particle velocities i.e. p=m(v)vMasses transform between frames in a waythat is similar to the transforms for lengthsand times


Elastic Collision: (Extract m(v)) (2)



†

¢ b B

†

¢ b A†

¢ m B

†

¢ m A

†

bB

†

bA†

mB

†

mA

x

y

x’

y’b

†



¢ b x =bx - b( )1- bxb( )

¢ b y =by g( )

1- bxb( )

bx =¢ b x + b( )

1+ ¢ b xb( )by =

¢ b y g( )1+ ¢ b xb( )

†

In ¢ S (Centre of Mass Frame) collision is symmetric¢

r b A = - ¢

r b B fi ¢ b xA = - ¢ b xB = -b( ) ¢ b yA = - ¢ b yB

¢ m A = ¢ m B

Conservation of Momentum in S(Elastic Collision)DpyA + DpyB = 0

DpyA = 2mAbyAcDpyB = 2mBbyBc

So mAbyA = -mBbyB

Use Velocity Transformation

mA

¢ b yA g

1+ b ¢ b xA

= -mB

¢ b yB g

1+ b ¢ b xB

mA

¢ b yA g

1- b 2 = mB

¢ b yA g

1+ b 2

mB

mA

=1+ b 2

1- b 2





†

¢ b B

†

¢ b A†

¢ m B

†

¢ m A

†

bB

†

bA†

mB

†

mA

x

y

x’

y’b

†



¢ b x =bx - b( )1- bxb( )

¢ b y =by g( )

1- bxb( )

bx =¢ b x + b( )

1+ ¢ b xb( )by =

¢ b y g( )1+ ¢ b xb( )

†

So we havemB

mA

=1+ b 2

1- b 2

Re - express in terms of bxB

bxB =¢ b xB + b( )

1+ ¢ b xBb( )Given ¢ b xB = b

bxB =2b( )

1+ b 2( )So

mB

mA

=1

1- bxB2( )

12

= g xB

†

If we wish to ensure Conservation of Momentumin S frame, we must conclude that apparentmass in a given frame is transformed

m = g m0where m0 is the rest mass (in the rest frame of the particle)




• Particle A Stationary Frame S†

¢ b B

†

¢ b A†

¢ m B

†

¢ m A

†

bB

†

bA Æ 0†

mB

†

mA = m0

x

y

x’

y’b

†



¢ b x =bx - b( )1- bxb( )

¢ b y =by g( )

1- bxb( )

bx =¢ b x + b( )

1+ ¢ b xb( )by =

¢ b y g( )1+ ¢ b xb( )

†

So we havemB

mA

=1+ b 2

1- b 2

Re - express in terms of bxB

bxB =¢ b xB + b( )

1+ ¢ b xBb( )Given ¢ b xB = b

bxB =2b( )

1+ b 2( )So

mB

mA

=1

1- bxB2( )

12

= g xB

†

So if we wish to ensure Conservation of Momentumin S frame, we must conclude that apparentmass in any given frame is transformed

m = g m0where m0 is the rest mass (in the rest frame of the particle)

†

The final "trick" is to imagine the collisionin the limiting case that the y componentsof velocities go to zero...

In this case byB , ¢ b yB Æ 0 bxB Æ bB

byA , ¢ b yA Æ 0 bxA Æ 0 (rest frame of A)The eqn

mB

mA

=1

1- bxB2( )

12

= g xB

can be re - expressed as the more general relationmB

m0

=1

1- bB2( )

12

= gB


Newton’s 2nd Law & Kinetic Energy

†

Remember !m = g m0

E = mc2 = g m0c2

p = mv = g m0( ) bc( ) = gbm0c

†

Consider KE for b <<1

KE = E - m0c2

= g -1( )m0c2

g =1

1- b 2( )12

ª 1+12

b 2 + ...Ê

Ë Á

ˆ

¯ ˜

ª 1+12

b 2 + ...-1Ê

Ë Á

ˆ

¯ ˜ m0c

2

ª12

b 2m0c2 + ...

ª12

m0v2 + ...


†

r F = dr p

dt=

d gm0r u ( )

dt

K =r F .d

r l

v= 0

v= u

ÚConsider special case along x

= Fdxv= 0

v= u

Ú

=d mu( )

dtdx

0

u

Ú

= ud mu( )0

u

Ú since dxdt

= u

= u mdu + udm( )0

u

ÚCan show (see next slide)

mudu + u2dm = c 2dm

= c 2dmm0

m

Ú= mc2 - m0c

2

= g -1( )m0c2 since m = g m0

K = E - m0c2


Newton’s 2nd Law & Kinetic Energy (2)

†

Discussion of the following steps

(1) ud mu( )0

u

Ú

= u mdu + udm( )0

u

Ú

(2) mudu + u2dm = c 2dm

= c 2dmm0

m

Ú †

(1) Differentiation, by parts gives, d mu( ) = mdu + udm

The full (technically correct) expression is,

d mu( ) =∂(mu)

∂u m constdu +

∂(mu)∂m u const

dm

= mdu + udm

†

(2) This step is a little trickier mudu + u2dm = c 2dm

We know that

m = gm0 =m0

1- u2 c 2( )12

Som2c 2 - m2u2 = m0

2c 2

Form the derivatived(m2c 2) - d(m2u2) - d(m0

2c 2) = 0d(m2c 2)

dmdm -

∂(m2u2)∂m u const

dm +∂(m2u2)

∂u m const

duÊ

Ë Á Á

ˆ

¯ ˜ ˜ - 0 = 0

The last term is d(m02c 2) is zero as argument is a constant

2mc 2dm - 2mu2dm + m22udu( ) = 0fi 2m(c 2 - u2)dm - 2m2udu = 0Divide by 2m and rearrangefi u2dm + mudu = c 2dm


Energy and Momentum

†

Givenm = g m0

E = g m0c2

p = gbm0c

Then(cp)2 = g 2b 2(m0c

2)2

But g 2 =1

1- b 2 fi b 2 =g 2 -1

g 2

So(cp)2 = g 2 -1( )(m0c

2)2

= g 2(m0c2)2 -(m0c

2)2

= E 2 -(m0c2)2

orE 2 = (pc)2 + (m0c

2)2

E 2 = p2 + m02 with c =1

†

Lorentz InvariantE 2 - (pc)2 = (m0c

2)2

(m0c2)2 = ¢ E 2 - ( ¢ p c)2

= ¢ E 2 - ( ¢ p c)2

= ¢ ¢ E 2 - ( ¢ ¢ p c)2...(Consider also rest frame E and p)

†

Massless particleE 2 = (pc)2

E = pcNote also that

pE

=gbm0cg m0c

2 =bc

=vc 2

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