go9ers14 said:
All that is necessary is movement. If neither person knows who moved, all that does is make it so that you don't know who is right.
In relativity there is no absolute notion of "movement", only movement relative to something else. In the Earth's frame, the train is moving, but in the train's frame, the Earth is moving, and both perspectives are equally valid.
go9ers14 said:
But what if they both start a stopwatch at the moment they pass each other. We will say the train is 10 units long and that light moves at one unit per second.
The distance that light travels in a second is known as a "light-second" (ls). So the train is 10 ls long...but is that in the train's frame, or in the Earth's frame? Keep in mind that each observer measures the length of objects in motion relative to themselves to shrink.
go9ers14 said:
If the person on the train is moving, he will see the bolt that he is moving towards strike five feet away in less than five seconds.
Did you mean "five units away" rather than "five feet away? One thing that's important to note is that the two observers disagree about the distance between the lightning strike and the train-observer at the moment the strike happens. If the Earth-observer measures the lightning strike to happen five light-seconds from the train-observer, the Earth-observer also measures the train-observer's ruler to be shrunk relative to his own, so that the lightning strike happens at a greater distance as measured by the train-observer. For example, if the train is moving at 0.6c (0.6 times the speed of light) relative to the Earth, an increment of 1-ls on the train-observer's ruler will appear to be shrunk to a length of 0.8-ls in the Earth-observer's frame*, so if the Earth-observer measures the lightning strike to happen at a distance of 5 ls according to his own ruler, that means the train-observer measured the lightning strike to happen at a distance of 5/0.8 = 6.25 ls using a ruler on the train (the train-observer also measures the length of the train to be 12.5 ls rather than 10 ls).
There is a further complication in that clocks which are synchronized in one frame are out-of-sync in another--this is known as "the relativity of simultaneity". Suppose the train-observer has one clock sitting next to him in the middle of the train, and another clock sitting on the end of the train where the lightning hits, and these two clocks are synchronized in his own frame. In the Earth observer's frame, these two clocks actually appear out-of-sync, with the clock at the center of the train ahead of the clock at the end by 3.75 seconds**. So if the lightning hits the spot near the far end when the train's clock there reads t=2 seconds, from the Earth's perspective the clock at the center of the train reads t=5.75 seconds at that moment. After 3.125 seconds in the Earth's frame, the light from the lightning has moved 3.125 light-seconds in the direction of the center, and the center itself has moved (3.125 ls)*(0.6c) = 1.875 ls in the direction of the strike...since the center of the train was originally 5 ls from the position of the strike in the Earth's frame, it is now 5 - 1.875 = 3.125 ls from the position of the strike, so this is how long it takes the light from the strike to meet up with the center of the train, in the Earth's frame. However, because of time dilation the clock at the center only appears to be ticking at 0.8 the normal rate in the Earth's frame***, so since it started off reading t=5.75 seconds at the moment of the strike in the Earth's frame, it reads t = 5.75 + (3.125)*(0.8) = 5.75 + 2.5 = 8.25 seconds at the moment the light from the strike reaches it.
From the perspective of the train's frame, the clocks at the end and center were synchronized, so if the lightning struck the end when the clock there read t=2 seconds, then the clock at the center of the train must also have read t=2 seconds "at the same moment" that the lightning struck (the relativity of simultaneity means that different frames disagree on whether events at different locations happened simultaneously, i.e. 'at the same moment'). Since the light reached the center of the train when the clock at the center read t = 8.25 seconds, that means the light must have taken 8.25 - 2 = 6.25 seconds to travel from the end of the train to the center, in the train's frame. And remember, in the train's frame the distance between the center and the end is 6.25 light-seconds! So there is no sense in which the train observer must consider himself to be "moving"--he measures light to travel at one light-second per second
relative to the train, just as the Earth observer measures light to travel at one light-second per second relative to the Earth.
*the length contraction formula says that if an object is moving at speed v in your frame, and its length is L in its own rest frame, its length in your frame will be L*\sqrt{1 - v^2/c^2}. So if the train is moving at v=0.6c in the Earth's frame, and there is a ruler on the train which is 1 light-second long in the train's frame, then its length will only be (1 ls)*squareroot(1 - 0.6^2) = (1 ls)*0.8 = 0.8 ls.
**If two clocks are a distance of x apart in their own frame, and synchronized in their own frame, and they're moving at speed v relative to you, then in your frame the back clock will be ahead of the front clock by a factor of vx/c^2. In this case the clock at the center and the clock at the end are a distance of 6.25 ls apart in their own frame, and moving at 0.6c in the Earth's frame, so in the Earth's frame the center clock is ahead by (0.6c)(6.25 ls)/c^2 = 3.75 seconds.
***The time dilation formula says that clocks in motion relative to you will be slowed down by a factor of \sqrt{1 - v^2/c^2} in your frame.
