Free particles must satisfy this condition and are said to be on-shell, namely that there is
a well-defined relationship between their energy, momentum, and mass that can be written
as,
E 2 = |~p|2 c2 + m2 c4
(5.133)
Importantly, this relation is frame independent. Any observer measuring the energy and
momentum in her reference frame would conclude that this relation holds. We know that
with certainty because it was obtained from a manifestly Lorentz invariant quantity, pµ pµ ,
and thus it holds true in any frame. Finally, the Hamiltonian, which classically is just the
energy expressed in terms of momentum is given by,
p
H = c |~p|2 + m2 c2
(5.134)
5.6
Four-momentum conservation
In the previous section I derived for you the conservation of energy-momentum for a free
relativistic particle. It probably won’t surprise you to learn that Nature seem to also conserve the total energy and momentum of a collection of interacting particles. This follows
from Noether’s theorem and the fact that the interactions between particles appear to be independent of spacetime location. Relativistic energy-momentum conservation thus imposes
a constraint on any reactions. If we denote by pµi the four-momentum of each of the Ninc
incoming particles and by kjµ the four-momentum of each of the Nout outgoing particles then
the reaction is subject to the constraint,
Ninc
X
pµi
i=1
=
N
out
X
kjµ
(5.135)
j=1
with the on-shell conditions,
p2i ⌘ pµi piµ = m2i
kj2 ⌘ kjµ kjµ = m2j
for
for
i = 1, . . . , Ninc
i = 1, . . . , Nout
(5.136)
(5.137)
(note that i and j here are particle number indices, and I have introduced the shorthand
notation p2i and kj2 for the square of the four-momenta). It is important to keep in mind
that the energy-momentum conservation equation (5.135) is really a set of four independent
equations. Also, it holds as a four-vector equation and so it remains true in any reference
frame.
This is the most general statement I can make about the impact of energy-momentum
conservation on relativistic reactions. Just as in the case of non-relativistic reactions, this
94
constraint is rarely sufficient to fully fix the outgoing momenta associated with a given
reaction. This is a very positive thing since it leaves some momenta unconstrained. The
distribution of these unconstrained momenta gives us information about the nature of the
underlying reaction in the same way that the distribution of the scattering angle, ✓, was a
powerful probe of the nature of the forces in non-relativistic elastic collisions. I want to use
the remainder of this section to work out the consequences of four-momentum conservation
in a few simple, but important type of reactions that appear very frequently in particle
physics.
Almost without exception, there are only two type of initial states you will encounter: either
a particle decay into some number of particles,
1 ! Nout ;
(5.138)
or two particles collide and result in some number of final states,
2 ! Nout .
(5.139)
At the low densities relevant for particle physics, the probability of more than two particles
colliding at a point in three-dimensions is negligible and will be ignored. The constraints
imposed by the conservation of total four-momentum are most easily applied in the reference
frame where the total three-momentum of the initial state vanishes. In the case of particle
decay this frame is simply the rest frame of the decaying particle. In the case of two particle
collision this frame is the centre-of-mass frame, which we already encountered in the nonrelativistic case. Let me first work out the necessary Lorentz transformation that takes us to
these special frames, and then I will work out the constraints imposed by four-momentum
conservation in these frame.
Boosting to the rest frame
Let’s consider the decay of a particle of mass M into two other particles of mass m1 and
m2 . In the lab frame the decaying particle may be moving with some momentum in some
arbitrary direction, say P~ = P ẑ. But, I can always boost to the frame where that particle
decays at rest15 .
In the lab frame the decaying particle’s four-momentum is,
0 p
1
P 2 + M 2 c2
B
C
0
C
Pµ = B
@
A
0
P
15
(5.140)
The only obstacle for performing such a boost would occur when the particle moves at the speed of
light, but then it must have M = 0 and cannot decay into other particles, massless or not, without violating
energy-momentum conservation as you can easily prove.
95
We are interested in finding the appropriate Lorentz boost to take us into the rest-frame
where its four-momentum is
0
1
Mc
B 0 C
C
P 0µ = B
(5.141)
@ 0 A
0
Using the general transformation rule
0
1 0
Mc
B 0 C B 0
B
C B
@ 0 A=@ 0
0
for the four-momentum, Eq. (5.127) we have
10 p
1
0 0
P 2 + M 2 c2
B
C
1 0
0 C
0
CB
C
(5.142)
A
0 1
0 A@
0
0 0
P
This matrix equation yields two equations, which are not independent. Using the last row
we trivially solve
p
P
P 2 + M 2 c2
E
=p
)
=
=
(5.143)
2
2
2
Mc
M c2
P +M c
This is the required boost factor to move to the rest-frame of the particle where its fourmomentum is given by Eq. (5.141).
Boosting to the centre-of-mass frame
The next important reaction I would like to consider is the scattering of two particles into
two other particles, so called “two-to-two”. The kinematics of the reaction is particularly
simple in the frame where the total three-momentum vanishes, the centre-of-mass frame,
which we already encountered in the non-relativistic case in section 2.3. Just as in the case
of a decaying particle that we just analyzed, here too I can always find the appropriate
Lorentz transformation that takes me to the centre-of-mass frame. Instead of working out
the necessary transformation in the most general case of two-to-two collision16 , which is both
tedious and un-insightful, let me work it out in the important special case when one of the
colliding particles is at rest in the lab frame, as in so called “fixed target” experiments. In
this case, I will write the lab frame four-momenta of the colliding particles as,
0 p 2
1
0
1
m2 c
p1 + m21 c2
B
C
B 0 C
0
C
C
pµ1 = B
and
pµ2 = B
(5.144)
@
A
@ 0 A
0
0
p1
16
In general the particles in the lab frame may even be moving along di↵erent axes, which would require
boosting along two separate axes in order to reach the centre-of-mass frame.
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The total incoming momentum is P µ = pµ1 + pµ2 . Since P µ is itself a four-momentum it
transforms in the same way as Eq. (5.127) under Lorentz boosts. We are interested in a boost
such that the total three-momentum in the new frame vanishes, or P 01 = P 02 = P 03 = 0.
Using the usual Lorentz boost along the ẑ-direction Eq. (5.84), I obtain the condition,
✓q
◆
2
2 2
0 = p1
p1 + m1 c + m2 c
(5.145)
Solving for
I get
is immediate and writing E1 = c
=
p1 c
E 1 + m 2 c2
)
p
p21 + m21 c2 for the energy of the first particle
=p
E 1 + m2 c2
(E1 + m2 c2 )2
p21 c2
(5.146)
Notice that when p ⌧ m1 c, m2 c we have E1 ⇡ m1 c2 and this reduces nicely to the nonrelativistic velocity boost required to move to the centre-of-mass frame, Eq. (2.27) with
~v2 = 0. The total energy available in the centre-of-mass frame is a very important quantity
and is given by,
✓q
◆
Ecm
2
2 2
=
p1 + m1 c + m2 c
p
(5.147)
c
p1
=
where in the last step I used Eq. (5.145). Using Eq. (5.146) for the boost factor I finally
obtain,
q
q
2 2
2
2
Ecm = (E1 + m2 c )
p1 c = c 2E1 m2 + (m21 + m22 )c2
(5.148)
Despite its unassuming character, this relation is of extreme importance. At sufficiently high
incoming energy it is given approximately by,
Ecm ⇡
p
2E1 m2 c2
(at fixed target experiments).
(5.149)
This means that the energy available for the reaction in the centre-of-mass frame only grows
as the square-root of the incoming energy, which is very slow growth. As we shall soon
see it is the energy in this frame that determines the character of the collision and the
smallest length scales we can probe via scattering is limited to ⇠ ~c/Ecm . Thus, probing
higher energies and smaller length scales is very inefficient with fixed-target experiments.
Nevertheless, fixed target experiments are of great importance in particle physics since they
allow to probe low-energy processes with much higher luminosities.
A better way to probe higher energies and shorter length scales exists with colliders. Colliders
are facilities where both incoming particles are accelerated to high energies, say E1 and
E2 . As p
you can easily convince yourself, the centre-of-mass energy in this case scales like
Ecm ⇡ 4E1 E2 and thus rises linearly as the energy of the two beams is simultaneously
increased.
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Two-body decays and two-body scattering
Above we established that both in the case of decay and in the case of scattering we can
always boost to a frame where the total incoming energy is,
0
1
Ecm /c
B 0 C
C
P 0µ = B
(5.150)
@ 0 A
0
where Ecm = M c2 in the case of decay, or in the case of scattering it is related to the energy of
the incoming particles in the lab frame as in Eq. (5.148). Let’s now work out the kinematics
of the reaction in this frame. In the rest-frame of the decaying particle four-momentum
conservation requires,
Ecm
0
0
0
E1 + E2
(5.151)
kx1 + kx2
(5.152)
ky1 + ky2
(5.153)
kz1 + kz2
(5.154)
q
q
The on-shell conditions demand that E1 = c |~k1 |2 + m21 c2 and E2 = c |~k2 |2 + m22 c2 . Momentum conservation in the rest frame implies that ~k1 = ~k2 . Then the energy conservation
condition yields an equation for |~k1 | alone,
q
q
Ecm
2 2
2
~
=
|k1 | + m1 c + |~k1 |2 + m22 c2
(5.155)
c
E4
|~k1 |2 = cm
)
=
=
=
=
2
2Ecm
(m21 + m22 ) c4 + (m21
2 c2
4Ecm
2
m22 ) c8
with |~k1 |2 = |~k2 |2 and the corresponding energies for particles 1 and 2 given by,
E12
E22
=
=
2
4
2
Ecm
+ 2Ecm
(m21
m22 ) c4 + (m21
2 c2
4Ecm
m22 ) c8
4
2
Ecm
+ 2Ecm
(m22
m21 ) c4 + (m21
2 c2
4Ecm
m22 ) c8
(5.156)
2
(5.157)
These expressions are not particularly illuminating (although it would be instructive for
you to examine a few limiting cases, for example the case when Ecm
m1 , m2 or when
m1 = m2 ). But, they do carry a few important and simple messages:
1. In the decay of an unstable particle into two other particles the energies of the decay
products are fixed in the rest frame! This is how we know that the beta decay of
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the neutron must involve a neutrino in addition to the electron and the proton —
otherwise, the energy of the electron and the proton should have been fixed in the rest
frame of the neutron, contrary to observation.
2. Similarly, the energies of the reaction products in two-to-two scattering are fixed by
the energy of the reaction in the centre-of-mass! Since the energy of a particle must
at least include its rest mass, it means that we cannot produce particles heavier than
the total energy in the centre-of-mass frame. This is why we must build high-energy
colliders if we want to look for heavier particles.
3. On the other hand, the orientation of the decay products or reaction products is not
fixed by four-momentum conservation. Their orientation is denoted by two angles
(✓, ), and just as in the non-relativistic case these angles are determined by the forces
involved in the reactions are known. Thus, the distribution of these angles may teach
us something about the reaction.
99