Notes on Relativity and Cosmology - Physics Department, UCSB

6.6. DERIVING THE EXPRESSIONS 163
in any frame of reference. In particular, we understand it in the lab frame where
the two particles (A and B) behave in a symmetric manner. (In physics speak,
we say that there is a ‘symmetry’ that relates the two particles.) Thus, ∆τ
is identical for both particles. Note that since ∆τ is independent of reference
frame, this statement holds in any frame – in particular, it holds in Alice’s frame.
Thus, the important point about equation (6.20) is that all of the quantities on
the right hand side can be taken to refer only to particle B!
In particular, the expression no longer depends on particle A, so the limit is
trivial. We have:
Since the motion of B is uniform after the collision, we can replace this ratio
with a derivative:
Thus, we have derived
dx⃗
B
|⃗p B | = m 0
dτ = m 1 dx⃗
B
0 √
1 − v2 /c 2 dt . (6.21)
the relativistic formula for momentum.
⃗v
|⃗p| = m 0 √
1 − v2 /c , (6.22)
2
Now, the form of equation (6.21) is rather suggestive. It shows that the momentum
forms the spatial components of a spacetime vector:
p = m 0
dx
dτ , (6.23)
where x represents all of the spacetime coordinates (t, x, y, z). One is tempted
to ask, “What about the time component m 0 dt/dτ of this vector?”
Recall that we have assumed that the momentum is conserved, and that this
must therefore hold in every inertial frame! If 3 components of a spacetime
vector are conserved in every inertial frame, then it follows that the fourth one
does as well. (Think about this from two different reference frames and you’ll
see why ...). So, this time component does represent some conserved quantity.
We can get an idea of what it is by expanding the associated formula in a Taylor
series for small velocity:
dt
m 0
dτ = m 1
0coshθ = m 0
1 − v 2 /c 2 = m 0(1+ 1 v 2
2 c 2 +...) = c−2 (m 0 c 2 + 1 2 m 0v 2 +.....)
(6.24)
In Newtonian physics, the first term is just the mass, which is conserved separately.
The second term is the kinetic energy. So, we identify this time component
of the spacetime momentum as the (c −2 times) the energy:
E = cp t =
m 0 c 2
√
1 − v2 /c 2 . (6.25)