by bgq
P: 136
 Quote by Doc Al Forget relativity for the moment. Imagine you are on a train moving at some speed. You walk from the rear of the car to the front of the car. As seen from the train tracks, would you not agree that you traveled a greater distance than just the length of the car? Same idea here.
This is OK according to Galilean relativity, where the speed is changed which ensures the change in the distance in the same invariant time. In SR there is a critical point is that the speed of light is not affected by either the observer or source or both. Anyway, even in Galilean relativity the distance is not changed between E and F, the speed does.
Mentor
P: 15,610
 Quote by bgq This is OK according to Galilean relativity, where the speed is changed which ensures the change in the distance in the same invariant time. In SR there is a critical point is that the speed of light is not affected by either the observer or source or both. Anyway, even in Galilean relativity the distance is not changed between E and F, the speed does.
You seem unable to grasp the fact that the distance traveled is not equal to the length of the tube. I suggest that you really think about the scenario that Doc Al posted and try to understand the difference between length and distance traveled. If you are unwilling to do that then it might help to write down the definitions of distance traveled and length.
 P: 292 From the stationary observer's perspective: If T1 is the time of light travel in the same direction as the motion af the train, then c*T1 = L/g +v*T1, which gives T1 = (L/g)/(c-v). Likewise, the time of light travel in the opposite direction is T2 = (L/g)/(c+v). Adding this gives, after a little algebra: T1+T2 = 2gT, where T is the one way time measured by the moving observer, just as it should be. Thus, the assumption that the stationary observer measures the velocity to c leads to the right answer for the time.
P: 136
 Quote by DaleSpam You seem unable to grasp the fact that the distance traveled is not equal to the length of the tube. I suggest that you really think about the scenario that Doc Al posted and try to understand the difference between length and distance traveled. If you are unwilling to do that then it might help to write down the definitions of distance traveled and length.
Hmm, I think I now understand the difference between length and the covered distance, but unfortunately this leads to the same result:

d = (L0/γ + vT1) + (L0/γ - vT1) = 2L0
Mentor
P: 40,276
 Quote by bgq hmm, i think i now understand the difference between length and the covered distance, but unfortunately this leads to the same result: D = (l0/γ + vt1) + (l0/γ - vt1) = 2l0/γ
t1 ≠ t2
Mentor
P: 15,610
 Quote by bgq Hmm, I think I now understand the difference between length and the covered distance, but unfortunately this leads to the same result: d = (L0/γ + vT1) + (L0/γ - vT1) = 2L0/γ
I cannot decipher that formula in this context. There should be two speeds, one for the speed of the object and one for the signal. Also, there should be two times, one for reaching the far end, and one for returning to the near end.
 P: 87 The situation is as following: A sees B moving at vrel, with a ruler/tube in front of him. At some point, an event happens. B shoots some photons towards the direction of the ruler/tube. Both draw a coordinate system where they consider this event to happen at x=0/t=0. As you correctly calculated, the time for the lightbeam to travel the ruler and back seen from B's point of view is 2Lo/c A will see the same ruler length contracted L = Lo*1/γ So LOGICALLY, as light travels at c for both observers. If the ruler was not moving, it would be trivial 2L/c. Unfortunately the ruler end is moving away of the lightbeam, while the back is moving towards it on the turn around trip. So we have L/(c-vrel) for the time it takes to move towards the end, while we have L/(c+v) for the time it takes to move back. L/(c-vrel) + L/(c+vrel) is the time it takes seen from A's point of view. Time-interval expansion(dilation) is measured LOCALLY in B's rest frame. Meaning that two events which happen at the SAME space but at different times and are measured to have a time interval of t seconds, are measured in A's rest frame to have a higher time-interval (dilation/expansion). They do NOT happen at the same space-distance in A's frame. In your example, the two local events seen from B's point of view are: E1) B shoots lightbeam at x=0/t=0 towards the back of the rocket. E2) Lightbeam returns to b after being reflected at x=0/t= 2Lo/c This is NOT equivalent to length contraction, because while similarly here we measure the distance between two events at the SAME time seen from B's rest frame. We however do NOT measure the distance between those two events in A's frame which of course happen at different times, just like in the case of time dilation the events A measures happen at a different space-distance. INSTEAD we measure the distance between two events which happen at the SAME time on the back and front of an object which is moving at vrel. I did not think this out, but i find it quite remarkable. It seems to be a human centric view. Whatever seems to make more sense to us.
 P: 137 Where are E and F with respect to A at the different points when the light reflects? How far does the light have to travel in frame A. Draw where the points would be with respect to A if you need to. I mean heck take a car moving at 60 mph. You have only a single point in the car lets call this point E. With respect to the car E never moves it's still in the same spot. But from the ground at the start there is spot E0, and spot E1, these after an hour are 60 miles away from each other.
 P: 87 To actually answer the question. The distance light will travel in A's frame is NOT 2L. You forgot that A sees the tube moving. This increases the distance on the trip towards the end, while shortening it on the round-trip. L/(c-v) = t1 is the time the lightbeam takes to the end of the tube, seen from A's point of view L/(c+v) = t2 the time it takes to the back. Light travels at c. So we have a total length of c*t1 + c*t2 = Ltotal So instead of just 2L/c you have, Ltotal/c should be T as seen from A's point of view. I did not calculate it, so there might be some mistake if my mind tricked me.
 P: 136 YES! I think I got it! Let t1 be the time from E to F, and t2 from F to E as measured by A. According to Lorentz transformation: t1 = γ(T0/2 + vL0/c2) t2 = γ(T0/2 - vL0/c2) d = (L0/γ + vt1) + (L0/γ - vt2) = 2L0/γ + v(t2 - t2) = 2L0/γ + v(2γvL0/c2) = 2L0/γ + 2γL0v2/c2 = 2L0(1/γ + γv2/c2) CA = d/t = 2L0(1/γ + γv2/c2)/γT0 = (2L0/T0)(1/γ2 + v2/c2) = (2L0/T0)(1 - v2/c2 + v2/c2) = (2L0/T0)(1) = 2L0/T0 = CB Thank you very much for your help, it was very difficult to me to solve it alone. Thank you very much. I am happy
P: 3,179
 Quote by bgq The light moves towards the extremity E with a speed c (not c+v), so the time will be the same (dilated of course). The issue here is that light does not behave in the same classical way. The motion of the tube has nothing to do with the propagation of the light (According to the postulates of SR).
Quite the contrary: c+v is used (or more precisely c-v), and that light behaves in a classical way is expressed by saying that the motion of the tube is independent of the light propagation (=second postulate).
I recently explained the use of c-v here:

But I see now that you already solved it by a different route - glad to see that you are happy.
P: 136
 Quote by harrylin Quite the contrary: c+v is used (or more precisely c-v), and that light behaves in a classical way is expressed by saying that the motion of the tube is independent of the light propagation (=second postulate). I recently explained the use of c-v here: - http://physicsforums.com/showthread.php?p=3992825 - http://physicsforums.com/showthread.php?p=4082811 But I see now that you already solved it by a different route - glad to see that you are happy.
Thanks
 P: 7 "bgq Posts: 59 YES! I think I got it! Let t1 be the time from E to F, and t2 from F to E as measured by A. According to Lorentz transformation: t1 = γ(T0/2 + vL0/c2) t2 = γ(T0/2 - vL0/c2) d = (L0/γ + vt1) + (L0/γ - vt2) = 2L0/γ + v(t2 - t2) = 2L0/γ + v(2γvL0/c2) = 2L0/γ + 2γL0v2/c2 = 2L0(1/γ + γv2/c2) so then solving, d = gamma * 2Lo, not divided by gamma as orginally posted, this confuses me, any explanations would be helpful, thanks
P: 3,179
 Quote by randyu [..] Let t1 be the time from E to F, and t2 from F to E as measured by A. According to Lorentz transformation: t1 = γ(T0/2 + vL0/c2) t2 = γ(T0/2 - vL0/c2) d = (L0/γ + vt1) + (L0/γ - vt2) = 2L0/γ + v(t2 - t2) = 2L0/γ + v(2γvL0/c2) = 2L0/γ + 2γL0v2/c2 = 2L0(1/γ + γv2/c2) so then solving, d = gamma * 2Lo, not divided by gamma as orginally posted, this confuses me, any explanations would be helpful, thanks
Welcome to physicsforums.

You referred to post #28.

How did you get from
d= 2L0(1/γ + γv2/c2)
to
d = γ * 2Lo ?

Perhaps you did not look at it carefully.
(1/γ + γv2/c2) = 1/γ(1 + γ2v2/c2) ≠ γ
[EDIT: I was wrong, see next]
 P: 7 hi harrlylin, Y=1/sqt(1-v2/c2) > v2/c2=(Y2-1)/Y2 then, d=2Lo * 1/Y+Y(Y2-1)/Y2= 2Lo*Y is this not right, thanks
P: 3,179
 Quote by randyu hi harrlylin, Y=1/sqt(1-v2/c2) > v2/c2=(Y2-1)/Y2 then, d=2Lo * 1/Y+Y(Y2-1)/Y2= 2Lo*Y is this not right, thanks
Oops - I must admit that I did not check bqq's derivation.
And yes you are right. Regretfully I did not follow that conversation, and I won't look into that now. Maybe someone else will, or I will later.
What I can say already, is that I see nowhere claimed that d= 2L0/Y

Oh OK I see it: the answer is in post #22. Did you read that?

d is the path length that the light ray travels, and as d>2Lo I suppose that it is the path length in the rest system, with respect to which the apparatus is moving. Do you follow that?

It's like the Michelson-Morley experiment. And it is very well explained in post #17.
So, to elaborate: You walk from the rear of the car to the front of the car and back. As seen from the train tracks, the distance is greater.
- http://en.wikipedia.org/wiki/Michels...ley_experiment
P: 7
 Quote by bgq Hi, there is something I can't understand: Consider a stationary observer at A. Now consider an observer B in a train that moves with constant velocity v with respect to A. In the train, B tries to measure the speed of light using an empty tube of length L0 (proper length as measured by B). He sends a light signal at extremity E, the signal reaches extremity F (where a mirror exists) and return back to the extremity E. B measure this duration T0 (proper period as measured by B). Now B measure the speed of light as: CB = 2L0/T0 Now according to A the length of the tube is contracted and the time is dilated, so he measure the speed of light as: CA = 2L/T = (2L0/γ)/(γT0) = (L0/T0)/(γ^2) = CB/(γ^2) which is different from the value measure by B (divided by Gamma squared)
thanks harrylin, what I was thinking is from post #1, d=2Lo/Y and t=YTo which gave c as "divided by gamma squared".
This was wrong from the beginning, the length in the stationary frame is longer than the moving length Lo as you state, confusing because each is moving relative to the other. And also the time is YTo in the "stationary" frame, longer.
I guess this all makes sense, I just get confused about time dilation/expansion using t in different ways it seems to me, sometime an interval, sometimes ticks. Oh well, will come together sooner or later.
Thanks.
P: 3,179
 Quote by randyu thanks harrylin, what I was thinking is from post #1, d=2Lo/Y and t=YTo which gave c as "divided by gamma squared". This was wrong from the beginning, the length in the stationary frame is longer than the moving length Lo as you state, confusing because each is moving relative to the other. And also the time is YTo in the "stationary" frame, longer. I guess this all makes sense, I just get confused about time dilation/expansion using t in different ways it seems to me, sometime an interval, sometimes ticks. Oh well, will come together sooner or later. Thanks.
That's why it is necessary to make sketches.
Cheers

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