8Special relativity
IA Dynamics and Relativity
8.6 Particle physics
Many problems can be solved using the conservation of 4momentum,
P =
E/c
p
,
for a system of particles.
Definition
(Center of momentum frame)
.
The center of momentum (CM) frame,
or zero momentum frame, is an inertial frame in which the total 3momentum is
P
p = 0.
This exists unless the system consists of one or more massless particle moving
in a single direction.
Particle decay
A particle of mass m
1
decays into two particles of masses m
2
and m
2
.
We have
P
1
= P
2
+ P
3
.
i.e.
E
1
= E
2
+ E
3
p
1
= p
2
+ p
3
.
In the CM frame (i.e. the rest frame of the original particle),
E
1
= m
1
c
2
=
q
p
2

2
c
2
+ m
2
2
c
4
+
q
p
3

2
c
2
+ m
2
2
c
4
≥ m
2
c
2
+ m
3
c
2
.
So decay is possible only if
m
1
≥ m
2
+ m
3
.
(Recall that mass is not conserved in relativity!)
Example. A possible decay path of the Higgs’ particle can be written as
h → γγ
Higgs’ particle → 2 photons
This is possible by the above criterion, because m
h
≥ 0, while m
γ
= 0.
The full conservation equation is
P
h
=
m
h
c
0
= P
γ
1
+ P
γ
2
So
p
γ
1
= p
γ
2
E
γ
1
= E
γ
2
=
1
2
m
h
c
2
.
Particle scattering
When two particles collide and retain heir identities, the total 4momentum is
conserved:
P
1
+ P
2
= P
3
+ P
4
In the laboratory frame
S
, suppose that particle 1 travels with speed
u
and
collides with particle 2 (at rest).
1
2
21
φ
θ
In the CM frame S
0
,
p
0
1
+ p
0
2
= 0 = p
0
3
+ p
0
4
.
Both before and after the collision, the two particles have equal and opposite
3momentum.
1 2
p
1
p
2
1
2
p
3
p
4
The scattering angle
θ
0
is undetermined and can be thought of as being random.
However, we can derive some conclusions about the angles
θ
and
φ
in the
laboratory frame.
(staying in
S
0
for the moment) Suppose the particles have equal mass
m
.
They then have the same speed v in S
0
.
Choose axes such that
P
0
1
=
mγ
v
c
mγ
v
v
0
0
, P
0
2
=
mγ
v
c
−mγ
v
v
0
0
and after the collision,
P
0
3
=
mγ
v
c
mγ
v
v cos θ
0
mγ
v
v sin θ
0
0
, P
0
4
=
mγ
v
c
−mγ
v
v cos θ
0
−mγ
v
v sin θ
0
0
.
We then use the Lorentz transformation to return to the laboratory frame
S
.
The relative velocity of the frames is v. So the Lorentz transform is
Λ =
γ
v
γ
v
v/c 0 0
γ
v
v/c γ
v
0 0
0 0 1 0
0 0 0 1
and we find
P
1
=
mγ
u
c
mγ
u
u
0
0
, P
2
=
mc
0
0
0
where
u =
2v
1 + v
2
/c
2
,
(cf. velocity composition formula)
Considering the transformations of P
0
3
and P
0
4
, we obtain
tan θ =
sin θ
0
γ
v
(1 + cos θ
0
)
=
1
γ
v
tan
θ
0
2
,
and
tan φ =
sin θ
0
γ
v
(1 − cos θ
0
)
=
1
γ
v
cot
θ
0
2
.
Multiplying these expressions together, we obtain
tan θ tan φ =
1
γ
2
v
.
So even though we do not know what
θ
and
φ
might be, they must be related
by this equation.
In the Newtonian limit, where v c, we have γ
v
≈ 1. So
tan θ tan φ = 1,
i.e. the outgoing trajectories are perpendicular in S.
Particle creation
Collide two particles of mass
m
fast enough, and you create an extra particle of
mass M .
P
1
+ P
2
= P
3
+ P
4
+ P
5
,
where P
5
is the momentum of the new particle.
In the CM frame,
1 2
v v
P
1
+ P
2
=
2mγ
v
c
0
We have
P
3
+ P
4
+ P
5
=
(E
3
+ E
4
+ E
5
)/c
0
So
2mγ
v
c
2
= E
3
+ E
4
+ E
5
≥ 2mc
2
+ Mc
2
.
So in order to create this new particle, we must have
γ
v
≥ 1 +
M
2m
.
Alternatively, it occurs only if the initial kinetic energy in the CM frame satisfies
2(γ
v
− 1)mc
2
≥ M c
2
.
If we transform to a frame in which the initial speeds are
u
and 0 (i.e. stationary
target), then
u =
2v
1 + v
2
/c
2
Then
γ
u
= 2γ
2
v
− 1.
So we require
γ
u
≥ 2
1 +
M
2m
2
− 1 = 1 +
2M
m
+
M
2
2m
.
This means that the initial kinetic energy in this frame must be
m(γ
u
− 1)c
2
≥
2 +
M
2m
Mc
2
,
which could be much larger than
Mc
2
, especially if
M m
, which usually the
case. For example, the mass of the Higgs’ boson is 130 times the mass of the
proton. So it would be much advantageous to collide two beams of protons head
on, as opposed to hitting a fixed target.