I know that this experiment to measure unidirectional speed of light will fail: attempting to measure it by first synchronizing the clocks of both the emitter of light and the receiver and then moving the receiver away from the emitter, no matter how slowly, will lead to time slowing down for the receiver exactly so that, once we run the experiment, the time that the receiver has noted it has received the light will be the same as the time the emitter will claim to have emitted the light (therefore, it will appear, if we don't take special relativity into account, that light has infinite speed).
However, let's propose a very similar experiment: There is an equilateral triangle, in which there is a clock with a light receiver in one corner, a light emitter in another corner, and a light receiver and emitter in the third corner. The emitter sends a signal to both the clock+receiver and the emitter+receiver at the same time. As soon as the emitter+receiver receives the signal, it sends light towards the clock+receiver. Now, the clock+receiver, as soon as it receives the signal from the emitter (at the same time as the receiver+emitter receives it), starts measuring time. And it stops measuring time as soon as it receives the light from the emitter+receiver.
What will that time measured by the clock+receiver in the second experiment be? I know it cannot be the distance from the emitter divided by the unidirectional speed of light, since that would give us the unidirectional speed of light then, and I know special relativity does not allow us to measure that. So, what would that time be then? Will it again be zero? If so, why?
Best Answer
The time measured will be $\Delta t = L/c$ where $L$ is the length of one side of the equilateral triangle.
It is not that you cannot measure it, but that you cannot do so without assuming a synchronization convention. Here, you are using the initial flash of light as a synchronization signal. By assuming that the flashes arrive at the two destinations simultaneously you are already assuming that the one way speed of light is isotropic.
Suppose instead that the one way speed of light is assumed to be anisotropic. In that case then the light pulses would arrive at their destinations at different times. The clock would start counting at a different time than the emitter emitted. Accounting for the anisotropic speed of light on the final leg of the triangle and the offset due to the unsynchronized start, then the outcome of the measurement would still be $\Delta t = L/c$ even though the one way speed of light on that leg was not c.
Both assumptions are compatible with the observation. So the observation $\Delta t = L/c$ does not allow us to reject either assumption.