MechanicsPhysics 151

Lecture 16Special Relativity

(Chapter 7)

What We Did Last Time

Defined covariant form of physical quantitiesCollectively called “tensors”

Scalars, 4-vectors, 1-forms, rank-2 tensors, …Found how to Lorentz transform them

Use Lorentz tensor & metric tensor

Covariant form of Newton’s equation with EM force

Equation of motion

EM potential 4-vector (φ/c, A)EM field Faraday tensor I gave you a wrong one…

dp Kd

µµ

τ=

Faraday Tensor

Fµν is derivatives of the EM potential E and B fields

A AK e e uFux x

ν µµ

ν νµ ν

µν⎛ ⎞∂ ∂= − ≡⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠

Faraday Tensor

00

00

x y z

x z y

y z x

z y x

E c E c E cE c B B

FE c B BE c B B

µν

− − −⎡ ⎤⎢ ⎥−⎢ ⎥=⎢ ⎥−⎢ ⎥−⎢ ⎥⎣ ⎦

0 1

1 0

xx

A A AE cx t x xφ ⎛ ⎞∂∂ ∂ ∂

= − − = −⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠

3 2

2 3

yzx

AA A ABy z x x

∂∂ ∂ ∂= − = − +∂ ∂ ∂ ∂

What I gave you (and the textbook)

was wrong by a factor c

What I gave you (and the textbook)

was wrong by a factor c

Multi-Particle System

Consider a system with particles s = 1, 2, …Total momentum

Equation of motion for each particle

EoM for the total momentum isNot a very clean equation

ss

P pµ µ=∑

ss

s

dp Kd

µµ

τ=

Different time for each particle!

ss s s

s ss

dpdP Kdt d

µµµγ γ

τ= =∑ ∑

Trouble ahead…

Momentum Conservation

Imagine a 2-particle system with no external forceBut there are internal forces between the particlesLaw of action and reaction

To conserve the total momentum in all frames,the particles interacting with each other must have the same velocity

ss s s

s ss

dpdP Kdt d

γ γτ

= =∑ ∑

1 2 2 1K K→ →= −

1 21 2 1 2 1 2

dP K Kdt

γ γ+→ →= + This is zero only if 1 2γ γ=

Is this a weird restriction, or what?

Local Interaction

Suppose the particles interact only when in contactForces exchanged only when they collideThey share the rest frame momentarily Same γ

Furthermore, Law of action/reaction cannot work over a distance instantaneously

There must be delays Conservation laws cannot hold

Total momentum of a multi-particle system is conserved if the interactions between the particles are local

1 2 2 1K K→ →= −

Relativity and non-local interactions don’t mix

Particle Collisions

Interactions between particles must be localForce exchanged when they collideFree motion between collisions

Consider the collision as a black box

We don’t know what happens in the box (not classical)Motion outside the box is easy Relativistic Kinematics

How much can we learn without opening the box?

?

Center-of-Momentum Frame

Local interactions conserve total 4-momentumi.e., total energy and total 3-momentum are conserved

We know how to Lorentz transform it

Define the center-of-momentum frame in which It’s the frame in which the total 3-momentum is zeroOr, the center of mass is at restOften called center-of-mass frame as well

1,

n

ss

Ep pc

µ µ

=

⎛ ⎞= = ⎜ ⎟⎝ ⎠

∑ p1

n

ss

E E=

=∑1

n

ss=

= ∑p p

i ii i

p p L p L pµ µ µ ν µ νν ν′ ′= = =∑ ∑ as usual

s ss s

p p L p L pµ µ µ ν µ νν ν′ ′= = =∑ ∑

0=p

CoM Energy and Boost

There are two particularly useful quantities

CoM energy E’Lorentz invariance

E’ is the smallest possible E

Boost β of CoM frame relative to the lab. frameLorentz transformation

p p p pµ µµ µ′ ′=

2 22

2 2

E Ec c′= −p

1

,n

ss

Ep pc

µ µ

=

⎛ ⎞= = ⎜ ⎟⎝ ⎠

∑ p CoM frame ,Epc

µ ′⎛ ⎞′ = ⎜ ⎟⎝ ⎠

0

,E Ep L pc c

µ µ νν

γ γ′ ′⎛ ⎞′= = ⎜ ⎟⎝ ⎠

β cE

=pβ E

Eγ =

′

Two-Particle Collision

Consider collision of particle 1 on particle 2 at rest

Total 4-momentum isTotal CoM energy

Boost of CoM frame is

1 1 1( , )p E c= p 2 2( , )p m c= 0

1 2 1( , )p E c m c= + p

2 2 2 2 4 21 2 1 2( ) 2E p p c m m c E m cµ

µ′ = = + +

Fixed-target collision

E’ grows slowly with E1

1 1 1 1

1 2 1 1 2

mE c m c m c m c

γβγ

= =+ +p v

Approaches v1/c for large E1

Creation Threshold

Suppose we are trying to create a new particleIn the best scenario, particles 1 and 2 merge to create a new heavy particle 3Total 4-momentum would be simplyTotal CoM energy is

How much energy E1 do we have togive particle 1?

For large m3, E1 grows with m32

1 2 3p p p p= + =

2 2 2 23 3 3E c p p m cµ

µ′ = = 23E m c′ =

CoM energy must match the mass of the new particle

2 2 2 4 2 2 41 2 1 2 3( ) 2E m m c E m c m c′ = + + =

2 2 2 23 1 2

12

( )2

m m m cEm

− −=

Fixed-Target vs. Collider

Consider hitting a proton with a proton to make a Higgs particle, which is X times heavier than a proton

For X > 100, we’d need a >5000 GeV accelerator

Particle colliders are more energy-efficient

Laboratory is CoMJust need

2 2 2 2 223 1 2

12

( )2 2 p

m m m c XE m cm

− −= ≈

1 1 1( , )p E= p 2 2 2( , )p E= p 1 2= −p p

1 2( ,0)p E E= +2 2

1 2 3 pE E m c Xm c+ = = 50 GeV + 50 GeV

Elastic Scattering

Particle 1 hits particle 2 and get elastically scattered

Cross section is calculated in CoM frameBy treating it as a central-force problem

Experiment is done in the laboratory frame

We need to learn how to translate between the CoM and the laboratory frames

CoMϑ Θ1p

3p

4p

1p′3p′

2p′

4p′x

y

Elastic Scattering

First, what’s the boost?Total momentum is

Let’s get γ as well

1 2 1 2 1( , )p p p E c m cµ µ µ= + = + p

22 2 2 2 21 2 1 1 2 1 2( ) 2p p E c m c m c m c E mµ

µ = + − = + +p

0 21 2

cp E m c

= =+1p pβ

0 21 2

2 4 2 4 21 2 1 22

p E m cp p m c m c E m cµ

µ

γ += =

+ +

Elastic Scattering

Now we can boost p1 to CoM

p3’ is given by rotating p1’ by Θ

Boost this back to get p3

Scattering angle in lab frame is

1 1 1( , ,0,0)p E c pµ = 1 1 1( , ,0,0)p E c pµ′ ′ ′=

1 1 1( )E E p cγ β′ = − 1 1 1( )p p E cγ β′ = −

3 1 1 1( , cos , sin ,0)p E c p pµ′ ′ ′ ′= Θ Θ

3 1 1( cos )E E p cγ β′ ′= + Θ 23 1 sinp p′= Θ1

3 1 1( cos )p p E cγ β′ ′= Θ +

2313 1 1 1

sin sintan(cos ( )) (cos )

pp E p c

ϑγ β γ β β

Θ Θ= = =

′ ′ ′Θ + Θ+

Velocity of 1 in CoM

Elastic Scattering

What happens to the kinetic energy?

At Θ = 0With a little bit of work

Worst case is Θ = π

2 23 1 1(1 cos ) (1 cos )E E p cγ β β⎡ ⎤= − Θ − − Θ⎣ ⎦

2 23 1 1(1 )E E Eγ β= − = Makes sense

3 12

1 1

2 (1 2)1 (1 cos )(1 ) 2

TT

ρρ ρ+

= − − Θ+ +

EE

1 2m mρ =

21 1 1T m c=EKinetic

energies

23 min

21 1

( ) (1 )(1 ) 2

TT

ρρ ρ−

=+ + E

Sign wrong in textbook

Elastic Scattering

Non-relativistic limit

Ultra-relativistic limit

As T1 increases, the energy loss becomes very large

2 2 23 min 2 1

1 1 2 1

( ) (1 ) ( )2 2

T m m cT m T

ρρ− −

= =E

23 min

21

( ) (1 )(1 )

TT

ρρ

−=

+If m1 << m2, i.e., the target is heavy,almost no energy is lost in the collision

23 min

21 1

( ) (1 )(1 ) 2

TT

ρρ ρ−

=+ + E

(T3)min is independent of T1

Particle Decays

Some particles are unstable and decay after a while

Mass of the “mother” is known from the 4-momenta of the “daughters” by calculating the total CoM energy

This is how particle physicists find particles that do not live long enough to be directly seen

Example: Discovery of J/ψ (November 1974)

ii

p pµ µ=∑

2mc E p pµµ′= = Also called the invariant mass of the system

Particle Decays

A day at the BABAR experiment at SLACCollide e+ and e– to generate a few 100,000 Υ(4S) particles… each of which decays into two B0 mesons… some of which decays into a J/ψ and a KS

… J/ψ decays into e+ and e–, or µ+ and µ–

… KS decays into π+ and π–

Measure 3-momenta of the stable particlesMasses known Calculate 4-momenta

Rebuild the decay chain backwards and calculate invariant masses of them all Do they match the expected masses?

J/ψ mass

Combine e+e– or µ+µ– and to see if they make a J/ψ

)2Beam-Energy Substituted Mass (MeV/c5200 5210 5220 5230 5240 5250 5260 5270 5280 5290 5300

2E

vent

s / 2

.5 M

eV/c

0

20

40

60

80

100

120

140

160

180

BABAR

B0 mass

Combine J/ψ and KS to make B0

Found 440 signals out of 30 million Υ(4S) generated

Summary

Discussed multi-particle systemOnly local forces are amiable with relativity

Relativistic kinematics for particle physicsCoM frame, CoM energy, invariant mass and boostParticle creation and collider experimentElastic scatteringParticle decays

Next lecture: Lagrangian formalism