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and thus in a general inertial frame
E 2 /c 2
p 2
m 2 c 2 ,
=
(12.30)
which is none other than Eq. (7.36). Also, since P is a four-vector we now
understand why it transforms just like the position four-vector under Lorentz trans-
formations, as we noted in Eq. (11.30).
12.3.1 Further examples in relativistic kinematics
The conservation of four-momentum is of great utility in studying relativistic
particle collisions, as is illustrated in the following examples.
Example 12.3.1 In Section 7.2.3 we explored the Compton scattering process γ
+
e
e . Re-derive Eq. (7.44) making use of the four-vector formalism.
γ
+
Solution 12.3.1 It is a good idea in problems like this to first write down the
four-momenta of the various particles before and after the collision, i.e.
1
c (E, E, 0 , 0 ),
P γ
=
P e
=
(m e c, 0 , 0 , 0 ),
1
c (E ,E cos θ,E sin θ, 0 ).
P γ
=
We would rather not write down the explicit representation for the four-momentum
of the scattered electron since we are aiming to express the scattered photon energy
purely in terms of the incoming photon energy and the photon scattering angle θ .
Four-momentum conservation informs us that
P γ +
P e .
P γ
+
P e
=
It is now clear how we should proceed if we would like to find a relationship between
E, E and θ : we should exploit the fact that P e ·
P e is Lorentz invariant and equal
to m e c 2 . Thus we write
P γ
P e
P γ
+
P e
=
and then 'square' each side (i.e. take the scalar product of each side with itself):
P γ ) 2
m e c 2 .
( P γ +
P e
=
The problem is essentially solved now and all that remains is for us to expand out
the left hand side. We could combine the three four-vectors into one big four-vector
and square that or we could stay in four-vector formalism for as long as possi-
ble. The latter has the advantage that we can make direct use of such results as
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