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#19
Oct2112, 01:50 PM

P: 144




#20
Oct2112, 02:27 PM

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#21
Oct2112, 03:15 PM

P: 345

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)/(cv). 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. 


#22
Oct2112, 04:13 PM

P: 144

d = (L_{0}/γ + vT_{1}) + (L_{0}/γ  vT_{1}) = 2L_{0}/γ 


#24
Oct2112, 04:31 PM

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#25
Oct2212, 10:12 AM

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/(cvrel) 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/(cvrel) + L/(c+vrel) is the time it takes seen from A's point of view. Timeinterval 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 timeinterval (dilation/expansion). They do NOT happen at the same spacedistance 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 spacedistance. 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. 


#26
Oct2212, 10:36 AM

P: 140

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. 


#27
Oct2212, 10:48 AM

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 roundtrip. L/(cv) = 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. 


#28
Oct2212, 03:05 PM

P: 144

YES!
I think I got it! Let t_{1} be the time from E to F, and t_{2} from F to E as measured by A. According to Lorentz transformation: t_{1} = γ(T_{0}/2 + vL_{0}/c^{2}) t_{2} = γ(T_{0}/2  vL_{0}/c^{2}) d = (L_{0}/γ + vt_{1}) + (L_{0}/γ  vt_{2}) = 2L_{0}/γ + v(t_{2}  t_{2}) = 2L_{0}/γ + v(2γvL_{0}/c^{2}) = 2L_{0}/γ + 2γL_{0}v^{2}/c^{2} = 2L_{0}(1/γ + γv^{2}/c^{2}) C_{A} = d/t = 2L_{0}(1/γ + γv^{2}/c^{2})/γT_{0} = (2L_{0}/T_{0})(1/γ^{2} + v^{2}/c^{2}) = (2L_{0}/T_{0})(1  v^{2}/c^{2} + v^{2}/c^{2}) = (2L_{0}/T_{0})(1) = 2L_{0}/T_{0} = C_{B} Thank you very much for your help, it was very difficult to me to solve it alone. Thank you very much. I am happy 


#29
Oct2212, 03:48 PM

P: 3,187

I recently explained the use of cv 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. 


#30
Oct2312, 04:08 PM

P: 144




#31
Nov812, 07:53 PM

P: 20

"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 


#32
Nov912, 01:14 PM

P: 3,187

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 + γv^{2}/c^{2}) = 1/γ(1 + γ^{2}v^{2}/c^{2}) ≠ γ [EDIT: I was wrong, see next] 


#33
Nov912, 01:40 PM

P: 20

hi harrlylin,
Y=1/sqt(1v2/c2) > v2/c2=(Y21)/Y2 then, d=2Lo * 1/Y+Y(Y21)/Y2= 2Lo*Y is this not right, thanks 


#34
Nov912, 02:49 PM

P: 3,187

And yes you are right. Regretfully I did not follow that conversation, 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 MichelsonMorley 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 


#35
Nov912, 04:15 PM

P: 20

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. 


#36
Nov1012, 02:55 AM

P: 3,187

Cheers 


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