# How would an observer perceive movement on a train that's travelling near the speed of light?

Person A is on Earth and a train (or whatever you want to imagine) travels past him at near the speed of light. How would person A perceive movement on the ship? If time is slowed on the ship from the perspective of Person A, then if Person B were washing the windows or something (moving his hands up and down), would they look like they were going really fast, slow, or normal? If time is slowed from the perspective of Person A, then Person B's hands would be moving a certain distance (let's say the window is a meter) in a really short amount of time, no? And that would mean it looks like it is going fast. But then since Person A is perceiving the time as moving slower, it would look slower, no? Would they cancel and movement on the ship looks normal?

I tried to see if the Wikipedia page could help, but it didn't seem to say anything about a stationary observer and a fast-moving vessel.

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Here is a Wikipedia page on the train thought experiment. – Qmechanic Nov 11 '12 at 22:13

Let's call the time interval of washing the windows in the train's reference frame $\Delta t$ and the time interval in the stations reference frame $\Delta t'$. As you alude to, the observers at the train station will measure this time interval to be longer than those on the train, specifically $\Delta t' = \Delta t \gamma$ where $\gamma > 1$ when the relative motion is not $0$.

This essentially means that the time interval that would pass from the person on the train started washing his windows till he ended would seem longer from the perspective of the observer on the station. The closer to the speed of light he is moving at, the slower he would seem to be moving.

As a more illustrative example, consider the following: Lightning strikes the end of a moving train car. The train car has a length of L. To an observer located at the other end of the train car, the flash will reach him at the time,

$$t = \frac{L}{c}$$

where $c$ is the speed of light, which is always the same for all observers in relativity. For an observer on the ground though, the train will have moved slightly between the two events, meaning that the light flash will have traveled a slightly different distance. This distance could be longer or shorter depending on the direction of the trains movement. Let's call this change in distance $\Delta L$. We then get the time as measured in the ground reference frame

$$t' = \frac{L+\Delta L}{c}$$

which illustrates that, if the speed of light is the same for all observers, the time it takes light to travel the length of the train is different for the two observers. This is analogous to your example of washing windows, except instead of the train moving slow compared to $c$ and the movement inside the train (lightning flash) being close to $c$, the situation is reversed (train speed close to $c$, speed of washing hands not close to $c$).

The important thing to remember here is that special relativity only works on inertial reference frames (i.e. no acceleration), so the two observers could never meet to compare clocks, except for possibly one single occasion when they pass each other (which is generally considered the point where $t = t' = 0$). This means there is no "true time" in special relativity. They are both correct in their analysis, even though they would disagree on the time interval if they could meet which, again, they can't unless one of them accelerates. (In the case of acceleration, general relativity would be required to analyze the scenario, which I am not familiar with... although I know time is still funky. There is no absolute time).

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