Special Relativity, light down a rocket

[CAF123]

Even though I already have the correct answer using the Lorentz transformations from the start, can you see where I made the error in the other method outlined above?

@BruceW: Did you catch my last message? (#48 bottom)

Thanks.

 

[BruceW]

I haven't looked back at the original problem for a while now. But now I do, uh... yeah, it looks like maybe you have gotten the velocity the wrong way around or something like that. To clarify, the rocket is moving at velocity v to the right, according to the S frame, yes? and the light beam is sent in a leftward direction from the nose of the rocket to it's end?

I think that maybe you made a slight mistake in the Lorentz transform which you used to derive your formula in post 7. Remember that the standard Lorentz formula
\Delta t' = \gamma (\Delta t - v\Delta x/c^2)
works on the assumption that $v$ is the velocity of frame S' with respect to frame S.

 

[CAF123]

Hi BruceW,
Just to clarify the formula in #7 is correct, (according to the answer key) so there is an error with what I did following Simon's method.

Yes, the inverse transform is $\Delta t = \gamma(\Delta t'+ v \Delta x '/c^2)$ where, as viewed in S', S moves with velocity -v. What do you mean for me to do?

Thanks.

 

[BruceW]

hmm. strange. your formula in post #7 agrees with the answer key? I got a different answer. hmm, I agree with
\Delta t = \gamma(\Delta t'+ v \Delta x '/c^2)
where $v$ is positive, since S' is moving right (which we're calling positive) with respect to S. And $\Delta x' = L_0$ and $\Delta t'=L_0/c$ Is that the same as you used? From here, I get a different answer than your formula in post #7. I think that maybe the problem is meant to be interpreted differently, which is why the answer key does not agree with this method.

edit: I am an idiot. $\Delta x' = -L_0$

 

[BruceW]

Wait, wait wait. I'm so sorry. Of course, $\Delta x' = -L_0$. OK, I'm with you now. Yeah, I agree with your answer in post #7. haha, it's been a long day.

 

[CAF123]

Ok, thanks for the check. But I would still be interested in finding the error that led me to a denominator of c+v via Simon's method. Do you have any ideas?

 

[BruceW]

Using purely Lorentz transformations, I derived the time for the signal to traverse the rocket in S as $t = \frac{L_o}{c} \sqrt{\frac{1-v/c}{1+v/c}}$. I noticed that this was less than $L_o/c$ for all v<c, and my interpretation was that in S, the rocket length is contracted so as viewed from S, it takes less time for the signal to reach the other end.

There are two things which contribute to the time being less according to S than it is according to S'. Firstly, the rocket length is contracted, as compared to the length according to S'. Secondly, in S, the rocket tail is moving toward the light beam. And in S', the rocket tail is not moving toward the light beam.

I get a similar answer by the method you outlined, but not quite the same. Here it is:
In frame S, $x_{\text{tail}} = -L + vt$ and $x_{\text{light}} = -ct$ so again equating gives $t=L/(c+v)\,\,\,\, (1)$. L is the length in S. To relate to S', use a Lorentz boost to that frame which in general is $\Delta x = \gamma(\Delta x' + v\Delta t')$ and in this case, we have $L = \gamma(-L_o + vL_o/c)$. Subbing this in for L into (1) gives $$t = \frac{L_o}{c+v} \sqrt{\frac{1-v/c}{1+v/c}},$$ a slight difference in the denominator there.

You need to take more care with your Lorentz transform. Remember to keep in mind the two spacetime events which you are taking the difference of. Here, you have chosen the light being emitted at the head of the spaceship and the light being received at the tail of the spaceship in the S' frame. But these two events, according to the S frame are not equal to the length of the spaceship according to the S frame. In fact, the spacetime path is light-like, so it cannot correspond to any kind of length as viewed by any frame. You need to decide which two spacetime events according to S frame would correspond to the length of the spaceship according to the S frame, and then do a Lorentz transform to get this Length in terms of things you know.

 

[CAF123]

You need to take more care with your Lorentz transform. Remember to keep in mind the two spacetime events which you are taking the difference of. Here, you have chosen the light being emitted at the head of the spaceship and the light being received at the tail of the spaceship in the S' frame. But these two events, according to the S frame are not equal to the length of the spaceship according to the S frame.

In my other method, I think I just did a Lorentz transformation to get the length of the rocket in the S frame. Why was it valid there?

In fact, the spacetime path is light-like, so it cannot correspond to any kind of length as viewed by any frame. You need to decide which two spacetime events according to S frame would correspond to the length of the spaceship according to the S frame, and then do a Lorentz transform to get this Length in terms of things you know.

I am not sure how I would go about finding such events; the only two events that seem to be of interest are the light emission and receiving at the ends of the rocket. I wrote down the coordinates of those spacetime events in post #7 as viewed in S.

 

[CAF123]

@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?

But I guess given that I then transform between frames in the next step (i.e substitute for L in terms of L_o), I should then be okay. But the result does not match up. 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. But I am not yet convinced given that I then make a transformation.

What do you think?
Thanks.

 

[BruceW]

In my other method, I think I just did a Lorentz transformation to get the length of the rocket in the S frame. Why was it valid there?

In your old method, you calculated
\Delta x = \gamma (\Delta x&#039; + v \Delta t&#039;) = \gamma(-L_o + v L_o/c)
which is the spatial length that the beam of light must travel to go from the front of the spaceship to the back of the spaceship, according to the S frame. But since the rocket is moving according to the S frame, this does not correspond to the length of the rocket according to the S frame.

I am not sure how I would go about finding such events; the only two events that seem to be of interest are the light emission and receiving at the ends of the rocket. I wrote down the coordinates of those spacetime events in post #7 as viewed in S.

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.

 

[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.