Trying to understand relativity

[KFC]

Hi there,
I am quite new this relativity, I read some book about this field, and I want to check if I am understanding this topic by considering the following example.

Assume I am staying in a train which is moving ahead with velocity u (u=3/5 c) relative to somebody staying on the platform. I call my frame S' and the frame attached on the platform S. Now I am sitting in the middle of the train and switch the light on. I measure the time when the light hits the front and back end of the train, we call this time t' (same for light hit the front and back end). Of course, the length of the train I measure is L' .

Now someone standing in the platform also measure the length of the train and times for light travel to the front and back end of the train in his frame. He will see the length of the train will be contracted due to

$$L = L' \sqrt{1-u^2/c^2}$$

And for measuring the time when light hit the front and rear end, in his frame, the distance for the light moving ahead is half of L, and the distance for the light traveling backward (to the rear end) is also half of L. But for the moving-ahead light, the speed is (c-u) and for the backward light, the speed is c (c+u>c, impossible, just use c). So we have

$$t_f = \frac{L/2}{c-u}, \qquad t_b = \frac{L/2}{c}$$

I don't quite sure this is correct or not. But if I start from the Lorentz transformation, I found

$$t_f = t'\sqrt{1-\dfrac{u^2}{c^2}} + \dfrac{|u|x}{c^2}$$

$$t_b = t'\sqrt{1-\dfrac{u^2}{c^2}} - \dfrac{|u|x}{c^2}$$

where t' is the time measure in S' frame, x is half of L.

Obviously, these two methods give different answer. But |t_f-t_b| is the same from these two method. I wonder which one is correct? or both are incorrect?

 

[JesseM]

And for measuring the time when light hit the front and rear end, in his frame, the distance for the light moving ahead is half of L, and the distance for the light traveling backward (to the rear end) is also half of L. But for the moving-ahead light, the speed is (c-u) and for the backward light, the speed is c (c+u>c, impossible, just use c).

In his frame the light must move at c in both directions, since light must move at c in all directions in all inertial frames (this is one of two postulates of special relativity, the other being that all laws of physics work the same in all inertial frames). However, in his frame the back of the train is moving towards the point where the light was set off at speed u, while the front of the train is moving away from the point where the light was set off at speed u. So, the "closing speed" between the back end and the light (the speed with which the distance between them is decreasing in his frame, which is different from either of their individual speeds in his frame) must be u+c, while the closing speed between the front end and the light must be u-c. Sing the light was first set off at a distance of L/2 from each end in his frame, the time between when the light was sent out and when it reaches the back end must be L/[2*(u+c)] while the time between when the light was sent out and when it reaches the front end must be L/[2*(u-c)] in his frame.

Assume the light was sent out at position x=0 and time t=0 in the outside observer's frame. In that case, the light must reach the back end at time t=L/[2*(u+c)] and position x=-L*c/[2*(u+c)] (we know that must be the position since the light moving towards the back was moving at c in the -x direction in this frame, so after time t the light must be at position x=-ct.) Likewise, the light must reach the front at time t=L/[2*(u-c)] and position x=L*c/[2*(u-c)]. Now you can use the Lorentz transform to find the x' and t' coordinates of these two events in the train's own frame--you should find that they are simultaneous, unless I made a mistake.

 

[jtbell]

And for measuring the time when light hit the front and rear end, in his frame, the distance for the light moving ahead is half of L, and the distance for the light traveling backward (to the rear end) is also half of L.

No, because the train is not stationary in this frame. The rear end of the train is moving towards the point where the light was emitted, and the front end is moving away from the point where the light was emitted. The two ends travel some distance while the light is "in flight." Therefore, the light that is going forward has to travel further than the light that is going backward, in this frame.

But for the moving-ahead light, the speed is (c-u) and for the backward light, the speed is c (c+u>c, impossible, just use c).

No, the speed of the light is the same in both directions, regardless of the fact that the source is moving in this frame. This is one of Einstein's fundamental postulates of relativity.