Suppose two observers, A and B, are moving away
from each other, and have synchronized their clocks as before.
Then suppose that after a period of time, a third observer, C,
starts moving away from B. (Examples of such a situation would be
if a spaceship flies past Earth and then launches a missile).
Using the previous results, it is easy to calculate times and
velocity of C relative to B or B relative to A, but what is the
velocity of C relative to A?
As in our previous derivations, suppose that A emits a beam of
lights for time T1, so the B receives light for a time
k1T1. Suppose also that B emits light for a
time T2 and C receives this light for a time k2T2.
Then all of the calculations from the previous lessons apply to
the A-B interaction and the B-C interaction.
Now assume that B does not capture all of the light which was
emitted by A. Some of the light continues on to hit observer C.
Then if A emits for a time T, C receives for time k3T
by the assumptions made before. But the speed of light is
constant, so the time experienced by C is the same regardless of
which of A and B emitted the light, so long as the time C
receives light is the same.
When A emits light for time T, then B receives
the light for time k1T. B then re-emits the light for
time k1T and so C receives light for a total time of k2(k1T).
Thus it is clear that
k3 = k2 k1
From this value of k3 and the formula of lesson 1, the
velocity of C relative to A can be calculated. (Notice that it is
not simply the sum of velocities as taught in non-relativistic
mechanics).
Lesson 4