Special Relativity, light down a rocket

[CAF123]

Hi BruceW,

Well, essentially, I'm just telling you to derive the standard formula for length contraction. But If you don't have so much time, you can just use the equation for length contraction.

I see, so Lorentz boost to a frame in which two events at either end of the rocket happen simultaneously in time in S. (E.g two events could be that a clock in the middle of the rocket in S fires two pulses of light towards each end and the set up is such that the light pulses reach either end in S at the same time. Then Δt = 0 and the length measured between these two events (reception of light at the tip and nose) corresponds to the length of the rocket in frame S, by construction.)

 

[Simon Bridge]

@Simon Bridge: Since my equations for $x_{\text{light}}$ and $x_{\text{nose}}$ were derived in frame S, the resulting equation should not imply a faster than light movement. Yet the result I find for t (=L/(c+v)) implies this. Do you agree?

It can do depending on the context.

In the S frame, the light travels a distance ct on one direction, and the ship (detector on the tail) travels a distance vt in the other direction, and they meet. No actual FTL movement here.
The result rests on the idea that ct+vt=L.

I spoke to my professor and he said the problem was the fact that my equation for t implied a faster than light movement as written above.

Well it doesn't - maybe he is checking your understanding?

It is the relative speed, in the S frame, between the light-pulse and the detector that is c+v. There's nothing wrong with that.

 

[BruceW]

Hi BruceW,

I see, so Lorentz boost to a frame in which two events at either end of the rocket happen simultaneously in time in S. (E.g two events could be that a clock in the middle of the rocket in S fires two pulses of light towards each end and the set up is such that the light pulses reach either end in S at the same time. Then Δt = 0 and the length measured between these two events (reception of light at the tip and nose) corresponds to the length of the rocket in frame S, by construction.)

yeah. That would be one way to do it. It takes a little bit more work to derive the Length contraction equation though. But anyway, you could derive it, or just look it up. And then you can use that in your 'new' method, and you will get the correct answer. (But it takes a little bit of rearranging to get it in the same form as the final answer of the old method).

 

[CAF123]

yeah. That would be one way to do it. It takes a little bit more work to derive the Length contraction equation though. But anyway, you could derive it, or just look it up. And then you can use that in your 'new' method, and you will get the correct answer. (But it takes a little bit of rearranging to get it in the same form as the final answer of the old method).

I got the answer, thanks. In the situation described previously, we want the reception of light at the tip and nose of the rocket to occur simultaneously in frame S. So then $\Delta t = 0$. So the spacetime coordinates are, in S:
Reception of light at tip: (L/2, 0) ; Reception of light at nose: (L/2, 0), so an observer in S reckons the length is L.

The Lorentz transformation, $\Delta x' = \gamma(\Delta x - v\Delta t)$ which relates the difference in space time spatial coordinates between the two frames, reduces to $L_o = \gamma(L)$ since $\Delta t = 0$ in S.

 

[BruceW]

yep. very good. and $\Delta t'$ is nonzero, but it doesn't matter, since we're saying the rocket is stationary in S', so we can identify $\Delta x'$ with $L_0$. I think it is good to state these things, rather than leave them as implied.