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So I am inside a coach in a train that is fully sealed (with no windows and a locked door). I have a torch, photo & time sensor and a scale with me. So, I place the sensor at one of the walls and stand $L$ distance from it. At $t = 0$, I switch on the torch and measure the amount of time it takes for the light to reach the photosensor.

If the time it took light is, $t = \frac{L}{c}$ then I know that the train was stationary. If it takes more time than that, say $t = t_1$ then I know the train was moving and the light actually travelled a distance of $t_1 \times c$ and the extra distance it travelled is $(t_1 \times c) - L$ say $L'$ in time, $t' = t - \frac{L}{C}$ and I can know the speed of the train as $v_{train} = \frac{L'}{t'}$.

This would mean that I measured the speed of the train independent of any frame of reference. Now where did I go wrong?

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    $\begingroup$ The light will never take more time than $L / c$ to travel a distance you measure to be $L$. The speed of light is the same in all inertial reference frames $\endgroup$ – ACuriousMind Nov 14 '14 at 1:18
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    $\begingroup$ Remember that the length of the coach changes as your velocity increases. $\endgroup$ – Dave Coffman Nov 14 '14 at 1:19
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    $\begingroup$ Yes, velocity of light remain same in all inertial frame of reference. So it always takes same time $t={\frac{L}{c}}$. That's why it is not possible to say a inertial frame either moving or is at rest by doing experiment only on the frame. And as far as the length contraction of the coach is concerned, the length would be contracted only when the train move with a velocity respect to a observer. But here observer also moving with the train, so there will be no contraction of length of the train coach with respect to the observer. $\endgroup$ – Rajesh Sardar Nov 14 '14 at 1:50
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    $\begingroup$ @DaveCoffman Not in his frame (which is also the coach's frame). $\endgroup$ – dmckee Nov 14 '14 at 2:25
  • $\begingroup$ "Now where did I go wrong?" Your implicit assumption about the synchronization of the two clocks that are required for the measurement of the one way transit time - one clock at each end of the coach. More later... $\endgroup$ – Alfred Centauri Nov 14 '14 at 2:55
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Update and note: In the answer below, I do assume the OP and reader are aware of the Galilean relativity of motion but wonder why the invariance of the speed of light cannot be used to find an absolute rest frame.

If this isn't he case, then Rod Vance's excellent answer is more appropriate.

I switch on the torch and measure the amount of time it takes for the light to reach the photosensor.

With what apparatus?

Evidently, you have a clock, at one end of the coach, that records the time that the torch is activated. Call this time $t_1$.

Then, you have a clock at the other end of the coach that records the time that the light reaches the photo sensor. Call this time $t'_2$. The prime indicates here that this value is from a different clock.

So, to calculate the transit time, you take the difference in the readings of two, spatially separated clocks:

$$\Delta t = t'_2 - t_1$$

Your calculation assumes that both clocks are synchronized, according to some convention, such that the difference in the reading of the clocks is meaningful.

But how do you know the two spatially separated clocks are synchronized?

According to Einstein synchronization, one synchronizes spatially separated clocks with light signals which guarantees that one will measure the one-way speed of light to be $c$.

Put another way, before you perform you experiment, you must verify that the clocks are synchronized. What does this mean? For Einstein synchronization, we have:

According to Albert Einstein's prescription from 1905, a light signal is sent at time $\tau_1$ from clock 1 to clock 2 and immediately back, e.g. by means of a mirror. Its arrival time back at clock 1 is $\tau_2$. This synchronisation convention sets clock 2 so that the time $\tau_3$ of signal reflection is defined to be $\tau_3 = \tau_1 + > \tfrac{1}{2}(\tau_2 - \tau_1) = \tfrac{1}{2}(\tau_1 + \tau_2)$

When your clocks are synchronized in this way, the outcome of your experiment is guaranteed to be $\Delta t = \frac{L}{c}$, i.e., your result will be independent of the speed of the train relative to the tracks (or anything else).

Essentially, this is how the invariance of $c$ is made consistent with the relativity of motion. The Lorentz transformation assumes this synchronization convention in order to produce this result.

See the Wikipedia article "One-way speed of light" for more details.

The bottom line is that one cannot assume that elapsed time, as measured by two spatially separated clocks is independent of a synchronization convention.

Only elapsed times, as measured by one clock, e.g., a two-way speed of light measurement, are invariant (absolute).

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The fundamental postulate of special relativity, indeed of Galilean relativity, is that there is no experiment that determine the state of motion of any inertial frame relative to the outside world unless the measurement uses data gleaned from outside the frame.

Read Galileo's wonderful and very famous allegory of Salviati's Ship for a poetic and rock solid accurate description of what this postulate means. Then add to the list of things going on that light beams within the ship's cabin are unaffected by the ship's inertial motion.

Special relativity simply relaxes the assumption of absolute time in Galileo's relativity, which leads to the second postulate that The speed of light is the measured to be the same in all inertial reference frames, if the measurements are wholly local to that frame. Amongst other things, this means that your Galilean summing of velocities doesn't work for light.

But note that, even in the absence of special relativity's second postulate, your reasoning won't work. If you replace the light beam by a ping pong ball and time its flight, all you can do is measure its initial velocity relative to your frame if you are confined to measurements local to your frame, so there is no addition of this velocity with that of the train, even in Galilean relativity. So your seemingly Galilean velocity addition would be wrong even in Galilean relativity. The ping pong ball's flight time will be the same whatever the speed of the train. Your "no windows and a locked door" implies you sail in Salviati's Ship!

You can measure any acceleration of the train with your light beam, or ping pong ball, but the easiest way to do this is with an accelerometer, which can be a heavy ball on a spring, or a weight hanging from the ceiling on a flexible thread.

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By your assumptions, the train will always be stationary. Light will always take time $\frac{L}{c}$ to traverse the distance simply because $c$ is a constant.

Additionally, even if your train was moving at $c$, it would still not matter because you are still at zero velocity with respect to the coach. Relativistic measurements would come into picture if you were outside the coach, measuring it. Then there would be time dilations, contraction of the coach etc.

As it stands, there is no way you can determine whether your coach is moving, let alone measure its velocity.

[Also, it is for similar reasons that retarded time $t'=t-\frac{L}{c}$ (which, I gathered, was what you were referring to) makes no sense here.]

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An intuitive way to see why this method would not work (or any other method inside a truly sealed train) is that the speed of the train is always relative to something else. There is nothing stopping me from describing the situation as rails moving under a stationary train at a certain speed.

Now, you propose to measure the speed at which the rails move under the train by shining light on a wall inside your isolated train. If this would give you an actual result for the speed, why would that result not be applicable to a bird that flies by the train?

There is, from your perspective, absolutely no way to distinguish your trains movement relative to the rails from the movement relative to the flying bird.

Intuitively, any measurements inside your train will yield the same result as long as the train does not accelerate or decelerate. But that same result can not apply to rails and bird alike (assuming the bird does fly at some speed relative to the rails).

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    $\begingroup$ "There is nothing stopping me from describing the situation as rails moving under a stationary train at a certain speed." Specifically, in the case of British trains, that speed would be zero. *rimshot* $\endgroup$ – David Richerby Nov 14 '14 at 13:06
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    $\begingroup$ @DavidRicherby: that would explain why Einstein came to different results when contemplating trains than Newton. The latter was familiar with British trains, which, indeed, are unlikely to be subjected to relativistic phenomena. $\endgroup$ – oerkelens Nov 14 '14 at 13:18
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First off, you state

If the time it took light is, $t=\frac{L}{c}$ then I know that the train was stationary.

I must ask, stationary with respect to what? It may not seem so to you but answering this question with most likely highlight the error in your reasoning/understanding of the principles involved. Equivalently, a correct answer to this question will provide the correct resolution to the conundrum you present.

The basic postulate of special relativity is that the laws of physics are the same in all inertial reference frames. One of the consequences of this postulate is that the speed of light, $c$, is a constant in all inertial reference frames. Thus, whether you are "stationary" (again, with respect to what? I assume you mean the ground around you) or "moving" (and again, with respect to what?) as long as you're moving at a constant velocity you are in an inertial reference frame and the speed of light will be the same in either case. The coach of the train you are in will also be in the same reference frame as you since you and the coach are moving together (hopefully, at least, or you'll end up running into a wall at some point and possibly hurt yourself). The length of the coach is unchanging and, since you are always in the same reference frame as the coach, length contraction will not come into play. Thus the time $L$ and $c$ are both invariant across inertial reference frames and the time $t$ you measure will be as well.

Since the time $t$ you measure is independent of what inertial reference frame you're in, it will be impossible to measure speed using the method you've described.

In summary, since the speed of light is the same across all inertial reference frames there will be, from your point of view, no "extra distance" the light travels if the train were moving. An outside observer, on the other hand, would have a different story to tell about the distance light travels.

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Law of Special Relativity: The speed of light remains constant for all observers no matter their state of motions. $L$ and therefore $t$ would therefore remain constant irrelevant to your state of motion.

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