Mechanics
Physics 151
Lecture 16
Special Relativity
(Chapter 7)
What We Did Last Time
„
Defined covariant form of physical quantities
„
„
„
Collectively 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
dp µ
= Kµ
„ Equation of motion Æ
dτ
„
EM potential Æ 4-vector (φ/c, A)
„
EM field Æ Faraday tensor Å I gave you a wrong one…
Faraday Tensor
⎛ ∂Aν ∂Aµ
K = e⎜
−
⎜ ∂x
⎝ µ ∂xν
µ
„
⎞
µν
u
e
F
uν
≡
⎟⎟ ν
⎠
Faraday Tensor
Fµν is derivatives of the EM potential Æ E and B fields
⎛ ∂A0 ∂A1 ⎞
∂φ ∂Ax
Ex = −
−
= c⎜
−
⎟
∂x ∂t
⎝ ∂x1 ∂x0 ⎠
F µν
∂Az ∂Ay
∂A3 ∂A2
Bx =
−
=−
+
∂y
∂z
∂x2 ∂x3
− Ex c − E y c − Ez c ⎤
⎡ 0
⎢E c
⎥
0
−
B
B
x
z
y
⎢
⎥
=
⎢ Ey c
− Bx ⎥
0
Bz
⎢
⎥
0 ⎥⎦
Bx
⎢⎣ Ez c − By
What
WhatIIgave
gaveyou
you
(and
(andthe
thetextbook)
textbook)
was
waswrong
wrongby
byaa
factor
factorcc
Multi-Particle System
„
Consider a system with particles s = 1, 2, …
„
Total momentum P µ = ∑ psµ
s
„
dpsµ
Equation of motion for each particle
= K sµ
dτ s
Different time for each particle!
dpsµ
dP µ
„ EoM for the total momentum is
= ∑γ s
= ∑ γ s K sµ
dt
dτ s
s
s
„ Not a very clean equation
Trouble ahead…
Momentum Conservation
dp
dP
= ∑γ s s = ∑γ s Ks
dt
dτ s
s
s
„
Imagine a 2-particle system with no external force
„
„
But there are internal forces between the particles
Law of action and reaction K1→2 = − K 2→1
dP1+ 2
= γ 1 K 2→1 + γ 2 K1→2
dt
„
This is zero only if
γ1 = γ 2
To conserve the total momentum in all frames,
the particles interacting with each other must have the
same velocity
Is this a weird restriction, or what?
Local Interaction
„
Suppose the particles interact only when in contact
„
„
Forces exchanged only when they collide
They share the rest frame momentarily Æ Same γ
Total momentum of a multi-particle system is conserved
if the interactions between the particles are local
„
Furthermore, Law of action/reaction K1→2 = − K 2→1
cannot work over a distance instantaneously
„
There must be delays Æ Conservation laws cannot hold
Relativity and non-local interactions don’t mix
Particle Collisions
„
Interactions between particles must be local
„
„
„
Force exchanged when they collide
Free 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 Æ Relativistic Kinematics
How much can we learn without opening the box?
Center-of-Momentum Frame
„
Local interactions conserve total 4-momentum
„
„
i.e., total energy and total 3-momentum are conserved
n
n
n
E
⎛
⎞
p µ = ∑ psµ = ⎜ , p ⎟
E = ∑ Es p = ∑ p s
⎝c ⎠
s=1
s =1
s =1
We know how to Lorentz transform it
as usual
p′µµ = ∑ ps′i µµ = ∑ Lνµνµ pνsνi = Lνµνµ pνν
si
„
si
Define the center-of-momentum frame in which p = 0
„
„
„
It’s the frame in which the total 3-momentum is zero
Or, the center of mass is at rest
Often called center-of-mass frame as well
CoM Energy and Boost
E′ ⎞
⎛E ⎞
⎛
µ
p = ∑ ps = ⎜ , p ⎟ CoM frame
p′ = ⎜ , 0 ⎟
⎝c ⎠
⎝ c ⎠
s =1
„ There are two particularly useful quantities
µ
„
µ
CoM energy E’
„
„
n
Lorentz invariance p pµ = p′ pµ′
„ E’ is the smallest possible E
µ
µ
E ′2 E 2
2
=
−
p
c2
c2
Boost β of CoM frame relative to the lab. frame
„
Lorentz transformation
γ E ′ γ βE ′ ⎞
⎛
µ
µ
ν
,
p = Lν p′ = ⎜
⎟
c
c
⎝
⎠
pc
β=
E
γ=
E
E′
Two-Particle Collision
„
Consider collision of particle 1 on particle 2 at rest
p1 = ( E1 c , p1 )
„
„
„
p2 = (m2 c, 0)
Fixed-target collision
Total 4-momentum is p = ( E1 c + m2 c, p1 )
Total CoM energy
E ′2 = p µ pµ c 2 = (m12 + m22 )c 4 + 2 E1m2 c 2
Boost of CoM frame is
p1
m1γ 1 v1
β=
=
E1 c + m2 c m1γ 1c + m2 c
E’ grows slowly with E1
Approaches v1/c for large E1
Creation Threshold
„
Suppose we are trying to create a new particle
„
„
„
„
In the best scenario, particles 1 and 2 merge to create a new
heavy particle 3
Total 4-momentum would be simply p = p1 + p2 = p3
Total CoM energy is
CoM energy must
2
2
µ
2 2
2
match the mass of
E ′ c = p3 p3 µ = m3 c
E ′ = m3c
How much energy E1 do we have to
give particle 1?
E ′2 = (m12 + m22 )c 4 + 2 E1m2 c 2 = m32 c 4
„
For large m3, E1 grows with m32
the new particle
(m32 − m12 − m22 )c 2
E1 =
2m2
Fixed-Target vs. Collider
„
Consider hitting a proton with a proton to make a
Higgs particle, which is X times heavier than a proton
(m32 − m12 − m22 )c 2 X 2
E1 =
mpc2
≈
2m2
2
„
„
For X > 100, we’d need a >5000 GeV accelerator
Particle colliders are more energy-efficient
p1 = ( E1 , p1 )
„
„
p2 = ( E2 , p 2 )
Laboratory is CoM p = ( E1 + E2 , 0)
Just need E1 + E2 = m3c 2 = Xm p c 2
p1 = −p 2
50 GeV + 50 GeV
Elastic Scattering
„
Particle 1 hits particle 2 and get elastically scattered
y
p1
x
„
„
„
p3′
p3
ϑ
CoM
p4
p1′
Θ
p2′
p4′
Cross section is calculated in CoM frame
„ By 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
Elastic Scattering
„
First, what’s the boost?
„
„
Total momentum is p µ = p1µ + p2µ = ( E1 c + m2 c, p1 )
p
p1 c
β= 0 =
p
E1 + m2 c 2
Let’s get γ as well
p µ pµ = ( E1 c + m2 c) 2 − p1 = m12 c 2 + m22 c 2 + 2 E1m2
2
γ =
p0
µ
p pµ
=
E1 + m2 c 2
m12 c 4 + m22 c 4 + 2 E1m2 c 2
Elastic Scattering
„
Now we can boost p1 to CoM
p1µ = ( E1 c , p1 , 0, 0)
„
p1′µ = ( E1′ c , p1′, 0, 0)
E1′ = γ ( E1 − β p1c)
p1′ = γ ( p1 − β E1 c)
p3’ is given by rotating p1’ by Θ
p3′µ = ( E1′ c , p1′ cos Θ, p1′ sin Θ, 0)
„ Boost this back to get p3
E3 = γ ( E1′ + β p1′c cos Θ) p31 = γ ( p1′ cos Θ + β E1′ c)
„
p32 = p1′ sin Θ
Scattering angle in lab frame is
p
sin Θ
sin Θ
tan ϑ =
=
=
p
γ (cos Θ + β E1′ ( p1′c)) γ (cos Θ + β β1′)
2
3
1
3
Velocity of
1 in CoM
Elastic Scattering
„
What happens to the kinetic energy?
E3 = γ 2 ⎡⎣ (1 − β 2 cos Θ) E1 − β (1 − cos Θ) p1c ⎤⎦
Makes sense
At Θ = 0 E3 = γ 2 (1 − β 2 ) E1 = E1
„ With a little bit of work
Sign wrong in textbook
T3
2 ρ (1 + E1 2)
2
Kinetic
E
=
T
m
c
(1 − cos Θ) 1 1 1
= 1−
2
(1 + ρ ) + 2 ρ E1
T1
energies
ρ = m1 m2
„ Worst case is Θ = π
(T3 ) min
(1 − ρ ) 2
=
(1 + ρ ) 2 + 2 ρ E1
T1
„
Elastic Scattering
(T3 ) min
(1 − ρ ) 2
=
(1 + ρ ) 2 + 2 ρ E1
T1
„
Non-relativistic limit
(T3 ) min (1 − ρ ) 2
If m1 << m2, i.e., the target is heavy,
=
almost no energy is lost in the collision
(1 + ρ ) 2
T1
„
Ultra-relativistic limit
(T3 ) min (1 − ρ ) 2 (m2 − m1 ) 2 c 2
=
=
(T3)min is independent of T1
2 ρ E1
2m2T1
T1
„ As T1 increases, the energy loss becomes very large
Particle Decays
„
Some particles are unstable and decay after a while
p µ = ∑ piµ
i
„
Mass of the “mother” is known from the 4-momenta of the
“daughters” by calculating the total CoM energy
mc 2 = E ′ = p µ pµ
Also called the invariant mass of the system
„
This is how particle physicists find particles that do not live
long enough to be directly seen
„ Example: Discovery of J/ψ (November 1974)
Particle Decays
„
A day at the BABAR experiment at SLAC
„
„
„
„
„
„
„
Collide 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 particles
„ Masses known Æ Calculate 4-momenta
Rebuild the decay chain backwards and calculate invariant
masses of them all Æ Do they match the expected masses?
J/ψ mass
Combine e+e– or µ+µ– and to see if they make a J/ψ
B0 mass
Events / 2.5 MeV/c
2
Combine J/ψ and KS to make B0
180
160
BABAR
140
120
100
80
Found 440 signals
out of 30 million
Υ(4S) generated
60
40
20
0
5200
5210
5220
5230
5240 5250 5260 5270 5280 5290 5300
2
Beam-Energy Substituted Mass (MeV/c
)
Summary
„
Discussed multi-particle system
„
„
Relativistic kinematics for particle physics
„
„
„
„
„
Only local forces are amiable with relativity
CoM frame, CoM energy, invariant mass and boost
Particle creation and collider experiment
Elastic scattering
Particle decays
Next lecture: Lagrangian formalism