Environmental Engineering Reference
In-Depth Information
some kinetic energy which can be traded off for the additional mass. However it
would be a grave error to suppose that the photon needs a kinetic energy which
exactly compensates the mass difference between the initial and final state. This
is wrong because momentum conservation must hold. Therefore, since the initial
state has non-zero momentum so too must the final state. As a result, the final state
particles must always be produced in motion and therefore they will carry some
kinetic energy. How then shall we proceed?
We can neatly circumvent the problem of momentum conservation by thinking
not in the laboratory frame but in an inertial frame where the total momentum of
the system is zero, i.e. the incoming photon and proton have equal and opposite
momentum as illustrated in the lower pane in Figure 12.2. In this 'zero momentum
frame' it is possible to produce the final state pion and proton at rest and clearly
this is the configuration which corresponds to the smallest possible photon energy
since none of its energy is wasted on giving the final state particles some motion.
The challenge is to relate quantities in the zero momentum frame to those in the
laboratory frame. Jumping between frames can be a time consuming affair but not
if we restrict our attention to Lorentz invariant quantities. The relevant invariant
in this problem is the four-scalar associated with the total four-momentum of the
system, i.e.
M
2
c
2
P
p
)
2
,
=
(
P
γ
+
P
p
)
2
.
=
(
P
π
+
The quantity M is called the 'invariant mass' of the system and the second line
follows from the first by the conservation of four-momentum, i.e.
P
p
)
2
P
p
)
2
.
(
P
γ
+
=
(
P
π
+
(12.31)
The left hand side of this equation involves the photon energy, which we seek to
find, i.e. using
1
c
(E, E,
0
,
0
) and
=
P
γ
P
p
=
(m
p
c,
0
,
0
,
0
)
the left hand side is simply equal to
P
p
)
2
P
γ
+
P
p
+
(
P
γ
+
=
2
P
γ
·
P
p
m
p
c
2
=
+
2
Em
p
.
(12.32)
We are permitted to calculate the right hand side of Eq. (12.31) in any convenient
frame of reference since it is a four-scalar. At threshold it makes sense to compute
it in the zero momentum frame since in this frame we know that
P
π
=
(m
π
c,
0
,
0
,
0
) and
P
p
=
(m
p
c,
0
,
0
,
0
).
