General relativity A to B. Calculate the rate of A’s clock (as compared to B’s clock) and the speed with B calculates A is approaching to him on the two time lines of A and B (see the figure below).
Consider the diagram below of time like world lines for the observers A and B.
Let first recall what we have learned about the A (Alice) observer deduces the relative rate of B‘s (Bob) clock (as compared to A’s) and the speed at which A calculates that B approaches her.
As concerns the relative rate of B’s clock as regarded by A, when A is in point s she seesthe B’s clock indicating the time T(q) =-6 seconds and from this she concludes that during 4 seconds of her time the B clock is ticking 6 times, hence it runs 50% (6-4)/4 =0.5 faster.
One can not take into account the fact that light take a time to get from point r on the A timeline to point q on the B timeline, and hence use the Geroch method to compare the times. This is happening simply because in the above geometric construction there is only an event described on the B timeline. To take the Geroch stand and way to compare the A and B times would need another event v on the B timeline that is simultaneous IN THE ACCEPTION OF B with the event r (and hence it is previous to q on the B timeline). To describe the position of this point p it would imply the presence of even another point x on the B timeline when light is sent from point x to point r to see the A clock. THE POSITION OF THIS POINT x WOULD NEED TO BE MENTIONED IN TIME (as the position of point r is mentioned as having T(r) =-9s). Since the above geometric construction only has one point q on the B timeline , only the old method of comparing time on the clocks can be applied!!!
Let however suppose one has a new geometric construction with points v and x on the B time lime. One must find the position of point x to determine the position of point v. For this one can write T(x)*T(q) = T(r)^2 or T(x)*(-6) =81, or T(x) = -13.5 seconds. Now one can determine the position of point v as being at the half way between points x and q. T(v) = 6 + (13.5-6)/2 = 9.75 seconds. Hence the points v and r are simultaneous as regarded by the observer B. From this B can say that in 9.75 seconds of his time the clock of A is ticking 9 times or in other words A’s clock is running 8% slower than B’s clock.
For the speed of B as calculated by A one must take into consideration the following facts. First the events u and q are simultaneous as regarded by A. Second the distance between A and B when A is at point u is delta_x(AB,u) = c*(T(s)-T(r))/2 = c*(t1+t2)/2 =c*(9-4)/2 =5*c/2. (here t1 = -4 seconds is the time between the events p and s, and t2 = +9 seconds is the time between the events r and p, considered exactly as in the textbook). Third the time indicated by the A’s clock in point u is (point u is at the half distance between events r and s) T(u) = T(s)+[T(r)-T(s)]/2 =4 – (9-4)/2 =-6.5 seconds. The time elapsed from event u to event p is therefore delta_t(u) = -T(u) =6.5 seconds. Hence the speed of B as regarded by A is simply V = delta_x(AB,u)/delta_t(u) = 5*c/2/6.5 =5*c/13.
Now let us solve our problems. As concerns the relative rate of A’s clock as regarded by B, when B is at point q he sees the A’s clock (in the point r) indicating T(r) = -9 seconds. Hence for him, A’s clock is ticking 9 times in the time interval of only 6 seconds (T(q) =-6 seconds). From this observation he can conclude that the A’s clock is running 50%, (9-6)/6 =3/6 =0.5 faster than his own clock.
To compute the speed of A as regarded by B one must raise two different questions. First question is how far apart are the observers A and B when B is located at point q. The answer at this questions is that this distance is equal to the distance between the events q and p on the B’s world line (since all the events r, u, s, p are on the same different time like world line, corresponding to observer A) and therefore delta_x(AB) = -T(q)*c =6*c. In other words the spatial distance between A and B as regarded by B is the space traveled by light in the time interval of 6 seconds as shown by B’s clock at point q. The second question is what is the time necessary for observer A to move on the distance delta_x(AB) as seen by observer B. This time is simply delta_t(AB) = T(r)+T(s) = t2-t1 =13 seconds. In other words it is the time necessary for the light to travel the distance between r and p plus the distance between p and s. From these two observations one can compute the speed of A as seen by B, V = delta_x(AB)/delta_t(AB) = 6*c/13.