Question from the FAQ on Rest Frame of a Photon

[rede96]

Since it will take ten minutes for the image of B to reach A and another ten minutes for the laser pulse to travel from A to the intersection, A needs to pull the trigger twenty minutes before it appears that B would arrive at the intersection, in other words when B is 33.333 miles before the intersection.

Thanks ghwellsjr.

So as Harold pointed out here:

In order to understand it, stick with nearly the speed of light as I suggested, and reflect on that. After you understand what goes on for that case, then you can extrapolate to the unattainable limit for a clock with v=c, as Einstein did.

Harald

I though I would try and do just that.

Imagine that instead of A firing a laser, A just decides to set off in his super fast ship and fires himself at B.

So ignoring acceleration for now, if A was capable of traveling at 0.9999986111095911 c, then A would feel that he covered the 10 light minutes in just 1 second.

A must also see B travel the 33.3333 miles B was away from the intersection in just one second too, as A will still hit B, but will just be a little off target.

However B would still feel like it took him 10 minutes.

So if we imagine just for a moment what a photon might experience from A's laser in the original experiment, once the photon was created, it would be instantaneously at B's target where it would be absorbed.

B, which was 33.333 miles from the intersection would also have to be instantaneously at the intersection for the photon to hit the target.

Although A and B still feel the passing of time as normal, for the photon, as everything happens instantaneously, there is nothing for the photon to experience. Which I guess is why we can't look at this problem from the frame of a photon. There is nothing to look at!


I know that the above is not strictly in keeping with the principles of SR, but I thought it might help address DrDon's questions.