# Simultaneity-Time Simultaneity-Time

by T L Hurst

### Velocity Effect continued...

At time t_{0}: A is a distance s_{0} from the transmitter, and is travelling away from it at velocity v. B is stationary with respect to the transmitter at a distance s_{2}. In B's rest frame, the time signal "t_{0}" is approaching him at the velocity c, such that he will receive the time signal "t_{0}" at time t_{2}.

At time t_{1}: A is a distance s_{1} from the transmitter when the time signal "t_{1}" is emitted. In A's rest frame, A is stationary and the time signal is approaching him at the velocity c, such that he will receive the "t_{1}" time signal at time t_{2}.

At time t_{2}: A and B are adjacent, but A receives the time signal "t_{1}", whilst B receives time signal "t_{0}".

So, we can say that the time taken for the t_{1} signal to reach A is t_{2} - t_{1}. In that time, A moves s_{2} - s_{1} at velocity v (with respect to the transmitter). Therefore:

**(t _{2} - t_{1}) = (s_{2} - s_{1})/v**

or

**(s _{2} - s_{1}) = v(t_{2} - t_{1})**

Note:

- The magnitude of the difference in the time signals received (in seconds) is equal to the distance "moved" by the observer (in light seconds) between the time that the time signal is emitted and when it is received.
- A velocity towards the transmitter causes the observer to receive an earlier time signal than that received by a "stationary" observer (who is co-located when the time signals are received).