Environmental Engineering Reference
In-Depth Information
out the photon momenta p
1
and p
2
. This is a two step process, firstly momentum
conservation in the y direction tells us that
7
p
1
y
+
p
2
y
=
0
⇒
p
1
=
p
2
.
We choose subsequently to define p
p
2
and our next task is to figure out p.
We can obtain this using the conservation of energy since this implies that
=
p
1
=
m
π
c
2
+
K
=
2
cp,
m
π
c
2
+
K
⇒
p
=
.
2
c
All that remains is for us to use conservation of momentum in the x direction:
(E
2
m
π
c
4
)
1
/
2
/c,
2
p
cos
α
=
−
which leads directly to (substituting for E)
((m
π
c
2
K)
2
m
π
c
4
)
1
/
2
+
−
cos
α
=
.
m
π
c
2
+
K
This is our final answer and we can substitute for the pion mass and its kinetic
energy to get a numerical answer, i.e.
1000
)
2
135
2
)
1
/
2
+
−
((
135
cos
α
=
=
0
.
993
+
135
1000
6
.
8
◦
.
⇒
α
=
As a final aside to this exercise, the decay π
γγ provides a very nice and direct
test of Special Relativity. Alvager et al (1962) showed that pions travelling at a
speed of
0
.
99975
c decayed to produce forward going photons of speed (
2
.
9977
±
→
10
8
ms
−
1
.
0
.
0004
)
×
Notice that the strategy for solving the last example is very similar to the one
we would use in classical mechanics. The only difference is that we should use the
relativistic forms for energy and momentum rather than the classical ones. It is well
worth stressing that we managed to get a numerical answer without ever needing to
multiply or divide by the speed of light. That happy circumstance arose because we
expressed all momenta and energies in MeV based units. In fact, particle physicists
often work in units where
c
=
1. It's not a bad idea to re-do the previous exercise
but putting
c
1 everywhere, although it is a pretty trivial exercise it should help
convince you that there is little point in forever writing factors of
c
all over the
place.
=
7
Strictly we can only conclude
p
1
=
p
2
if
α
=
0.
