Notes on Relativity and Cosmology - Physics Department, UCSB

6.6. DERIVING THE EXPRESSIONS 161
where particle A has the same speed before as it does after, as does particle B.
Also, the angle θ is the same both before and after. Such a symmetric situation
must be possible unless there is an inherent breaking of symmetry in spacetime.
Now, the velocity of particle A in this frame is to be slow enough that its
momentum is given by the Newtonian formula p A = m 0 v A . For convenience, I
have indicated coordinate directions x and y on the diagram in Alice’s reference
frame. It’s velocity in the x direction is zero, so its momentum in this direction
must also be zero. Thus, particle A only has momentum in the y direction. As
a result, the change in the momentum of particle A is 2m 0 dy/dt, where dy/dt
denotes the velocity in the y direction after the collision.
If momentum is to be conserved, the total vector momentum must be the
same before as after. That is to say, if (in Alice’s frame of reference) we add
the arrows corresponding to the momentum before, and the momentum after,
we must get the same result:
p before
A
p before
A
θ
θ
Total P
p before
B
p B before
For the next part of the argument notice that if, after the collision, we observe
particle B for awhile, it will eventually hit the south wall. Let us call this event
Y, where B hits the south wall after the collision. Recall that the collision takes
place in the middle of the box. Event Y and the collision will be separated by
some period of time ∆t B (as measured by Alice) and some displacement vector
∆⃗x B = (∆x B , ∆y B ) in space as measured by Alice. If the box has some length
2L in the north-south direction, then since the collision took place in the middle,
∆y B = L.
Also, if we trace particle B back in time before the collision, then there was
some event before the collision when it was also at the south wall. Let us call
this event X, when B was at the south wall before the collision. By symmetry,
this event will be separated from the collision by the same ∆t B and by −∆⃗x B .
What I want you to notice, is that the displacement ∆⃗x B points in the same
direction as the momentum of particle B, since that is the direction in which B
moves. Thus, we can draw another nice right triangle:
L
θ
∆ x B
Note that this triangle has the same angle θ as the one drawn above. As a
result, we have
L
|∆x⃗
B | = sin θ = p A
|⃗p B | . (6.17)
Note that, since p A has no x component, I have not bothered to represent it as