Reverse Derivation

Travel of Light Back from B to A

Here we derive the time interval t'_{_A} - t_{_B} of the reverse travel of the photon from B back to A.

The laws of motion of the photon F and of A are as follows:

F(t) = F(t_{_B}) - c(t - t_{_B})
A(t) = A(t_{_A}) + v(t - t_{_A})

Thus, for time t = t'_{_A} we have

F(t'_{_A}) = F(t_{_B}) - c(t'_{_A} - t_{_B})
A(t'_{_A}) = A(t_{_A}) + v(t'_{_A} - t_{_A})

At time t'_{_A} the positions of the photon and A coincide, therefore the distance between F and A is zero and we may write:

0 = F(t'A) - A(t'_{_A}) = F(t_{_B}) - c(t'_{_A} - t_{_B}) - A(t_{_A}) - v(t'_{_A} - t_{_A}) =
= F(t_{_B}) - A(t_{_A}) - c(t'_{_A} - t_{_B}) - v(t'_{_A} - t_{_B} + t_{_B} - t_{_A}) = F(t_{_B}) - A(t_{_A}) - v(t_{_B} - t_{_A}) - (c + v)(t'_{_A} - t_{_B}).

However, from the law of motion of A we have, that

A(t_{_B}) = A(t_{_A}) + v(t_{_B} - t_{_A}),

which we substitute in the above expression to obtain

0 = F(t_{_B}) - A(t_{_B}) - (c + v)(t'_{_A} - t_{_B}).

t'_{_A} - t_{_B} = (F(t_{_B}) - A(t_{_B}))/(c + v)

When the photon reaches B the position of the photon and B coincide and, therefore, F(t_{_B}) = B(t_{_B}), which leads to

t'_{_A} - t_{_B} = (B(t_{_B}) - A(t_{_B}))/(c + v) = r_{_AB}/(c + v) (see why B(t_{_B}) - A(t_{_B}) = r_{_AB} ...).