Hi everyone.
Right now, I'm studying time dilation. I think I understand the thought experiment which involved a laser reflecting from a mirror in a train car, (Page 15 of Modern Physics, Serway, Moses, Moyer) and the derivation of
$$\Delta t = \gamma \Delta t_p$$
using a right triangle.

My problem is, I made up a different (most probably flawed) thought experiment, through which I cannot arrive at the same formula. Here it is: There is a stationary observer in O, who has a light source which emits in +x direction, and I call his reference frame S. There is another observer in O, with a constant velocity u in +x direction. Call his reference frame S'. Now, after $$\Delta t$$ time for a clock in S, the second observer has moved $$u \Delta t$$, the light $$c \Delta t$$. So, the light went $$(c - u)\Delta t$$ according to the observer in S'. If $$\Delta t'$$ is the time passed in S', $$\frac{(c - u)\Delta t} {\Delta t'} = c$$ because the observer in S' should see the speed of light as c too. But this is not equivalent to $$\Delta t = \gamma \Delta t_p$$ mathematically. What am I missing here?

Thank you.

(I'm not experienced with LaTeX, so I will edit the post until the tex code is correct.)
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 Quote by kishtik Hi everyone. Right now, I'm studying time dilation. I think I understand the thought experiment which involved a laser reflecting from a mirror in a train car, (Page 15 of Modern Physics, Serway, Moses, Moyer) and the derivation of $$\Delta t = \gamma \Delta t_p$$ using a right triangle. My problem is, I made up a different (most probably flawed) thought experiment, through which I cannot arrive at the same formula. Here it is: There is a stationary observer in O, who has a light source which emits in +x direction, and I call his reference frame S. There is another observer in O, with a constant velocity u in +x direction. Call his reference frame S'. Now, after $$\Delta t$$ time for a clock in S, the second observer has moved $$u \Delta t$$, the light $$c \Delta t$$. So, the light went $$(c - u)\Delta t$$ according to the observer in S'. If $$\Delta t'$$ is the time passed in S', $$\frac{(c - u)\Delta t} {\Delta t'} = c$$ because the observer in S' should see the speed of light as c too. But this is not equivalent to $$\Delta t = \gamma \Delta t_p$$ mathematically. What am I missing here? Thank you. (I'm not experienced with LaTeX, so I will edit the post until the tex code is correct.)
The emitted light signal generates in S the event 1((0,0) when the light signal is emitted and the event 2(x=ct,t=x/c) after a given time of propagation. Detected from S’ the same events are 1’(0,0) and 2’(x’=ct’,t’=x’/c). In accordance with the Lorentz-Einstein transformations we have

I never recommend to use u-V or u+V.

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 Quote by kishtik Right now, I'm studying time dilation. I think I understand the thought experiment which involved a laser reflecting from a mirror in a train car, (Page 15 of Modern Physics, Serway, Moses, Moyer) and the derivation of $$\Delta t = \gamma \Delta t_p$$ using a right triangle.
I don't have that text, but I assume this is the standard derivation of time dilation using a light clock. Realize that the path of the light in the train's frame is perpendicular to the direction of travel. There's a good reason for that--it is much simpler to analyze since only a single moving "clock" is involved.

 My problem is, I made up a different (most probably flawed) thought experiment, through which I cannot arrive at the same formula. Here it is: There is a stationary observer in O, who has a light source which emits in +x direction, and I call his reference frame S. There is another observer in O, with a constant velocity u in +x direction. Call his reference frame S'. Now, after $$\Delta t$$ time for a clock in S, the second observer has moved $$u \Delta t$$, the light $$c \Delta t$$.
Realize that these measurements are all according to frame S. I'm going to restate these observations a bit more carefully: At the very moment when "moving" observer O' passes O, O emits a beam of light in the +x direction. After a time $$\Delta t$$ (measured on the O clock), S frame observers will agree that the light moved a distance $$c \Delta t$$ while O' moved a distance $$u \Delta t$$.
 So, the light went $$(c - u)\Delta t$$ according to the observer in S'.
No. That's the difference in distance traveled according to S frame observers.

Here's what O' and the S' frame observers will measure: When the O clock has measured $$\Delta t$$, S' frame observers will see that a time of $$\gamma \Delta t$$ has passed. During that time the light has traveled a distance of $$c \gamma \Delta t$$. Of course, S' frame observers see the light as traveling at speed c.

This thought experiment of yours doesn't allow you to derive time dilation in a simple way, since it involves multiple observers at different positions in the moving frame. (In the standard derivation, only a single observer at one position is needed to operate the light clock in the moving frame.) When multiple positions and observers are involved, each with their own clocks, you have to worry about all the relativistic effects at once: time dilation, length contraction, and (most important) clock synchronization.

Thank you for replies.
 Quote by Doc Al I don't have that text, but I assume this is the standard derivation of time dilation using a light clock.
Well, I've seen the same thing on another book, so I guess it is something standard.
 Quote by Doc Al No. That's the difference in distance traveled according to S frame observers. Here's what O' and the S' frame observers will measure: When the O clock has measured $$\Delta t$$, S' frame observers will see that a time of $$\gamma \Delta t$$ has passed. During that time the light has traveled a distance of $$c \gamma \Delta t$$. Of course, S' frame observers see the light as traveling at speed c.
So, in the standard derivation, the stationary observer knows how much distance the light travels as viewed from the train car's reference frame, without using $$\gamma$$, since the path of the light is perpendicular to the direction of the relative motion, it doesn't "contract", and because the two events happen at the same point. And because he doesn't have to use $$\gamma$$ in both "differences of displacement", he can actually derive it. Am I correct?
But I couldn't get why there is "multiple observers at different positions in the moving frame" in the thought experiment of mine.

Edit: Sorry for the sentence structures...

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 Quote by kishtik So, in the standard derivation, the stationary observer knows how much distance the light travels as viewed from the train car's reference frame, without using $$\gamma$$, since the path of the light is perpendicular to the direction of the relative motion, it doesn't "contract", and because the two events happen at the same point. And because he doesn't have to use $$\gamma$$ in both "differences of displacement", he can actually derive it. Am I correct?
I think you've got it.

 But I couldn't get why there is "multiple observers at different positions in the moving frame" in the thought experiment of mine.
In your thought experiment, there are two events that happen at different places in the S frame:
(1) The light is emitted at (t=0, x=0)
(2) The light reaches a point (t = $$\Delta t$$, x = $$c \Delta t$$)

Since those events happen at different places, different clocks (thus multiple observers with their own clocks) are used to measure the times of departure (1) and arrival (2). And according to the S' frame, those separated clocks are not synchronized, so S' will not agree that event (2) took place at the same time that the O clock reads $$\Delta t$$.

Make some sense?
 Yes, I think I got what you meant after considering the thought experiment about loss of simultaneity.