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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