Einstein's relativity tells us that light always travels at the speed of light relative to me, no matter how fast I'm going. Right? This really confuses me though. If light travels from A to B in one second while I'm standing at A, then where does light travel to when I'm going from A to B for one second as well, at half the speed of light? Surely the light is not further than B, is it? So what does it mean that light always travels with the speed of light relative to me?
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2$\begingroup$ It is really confusing. You're forgetting length contraction and time dilation though. It's only consistent when you take all of the effects into account. $\endgroup$– Brandon EnrightCommented Jun 1, 2013 at 22:07
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$\begingroup$ Actually, special relativity assumes (among other things) that the speed of light is independent of the speed of the source or observer. It then makes predictions based on these assumptions that in many cases seem absurd, but turn out to match reality quite well... $\endgroup$– DJohnMCommented Jun 2, 2013 at 2:52
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1 Answer
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When you are moving from A to B, the distance between A and B shrinks relative to you. This is known as length contraction.
The equation is as follows:
$$L' = L \sqrt{1 - v^2/c^2}$$
where L' is the length you will see at move. L is the length at the resting reference frame. v is your speed, and c the speed of light. In your case:
$$L' = L \sqrt{1 - (\frac{1}{2}c)^2/c^2} = L \frac{\sqrt{3}}{2}$$
So your distance between A and B would only be a fraction of the distance. And you will see the light travelling that distance in the speed of light.
But still, would the light beam not travel this in just a single second? It won't! Because in your reference frame the time is not the same as the time at the rest system where the distance between A and B is L. To calculate how long it will take the light to travel from A and B in your moving time frame, we need to calculate the time difference between those two reference frames. To calculate the time dilation you have the equation taken from here.
$$ \Delta t' = \frac{\Delta t}{\sqrt{1 - v^2 / c^2}}$$
In your case:
$$ \Delta t' = \frac{1}{\sqrt{1 - (\frac{c}{2})^2 / c^2}} = \frac{1}{\sqrt{3/4}}= \frac {2}{\sqrt{3}} sec $$
So as you can see, its going to take the light less time to cross that short distance in your moving reference frame.
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$\begingroup$ This answer is wrong, the main effect is the failure of simultaneity at a distance, this makes the speed of light the same in the backward and forward direction in the moving frame. $\endgroup$ Commented Aug 22, 2013 at 22:23
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$\begingroup$ @RonMaimon Can you point out a specific equation? My answer does imply non-simultaneity in some degree. Is there something wrong specifically? $\endgroup$– GuySoftCommented Aug 22, 2013 at 22:43
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$\begingroup$ Yes--- the time taken and the distance travelled are not just rescaled, you also have to add the distance travelled (or subtract) times the velocity, it's the failure of simultaneity that is the major effect, and the speed of light is constant no matter who measures it. $\endgroup$ Commented Aug 22, 2013 at 23:14
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$\begingroup$ To be precise, the light is travelling forward from A to B, but it is also going "up" in time in the moving frame, due to the "tilt" of the simultaneity lines in the moving frame. I can't explain without a picture, but you can see it in my answers to relativity questions--- you need to tilt the constant time line to explain why this happens. To see why, answer this: why is light moving backwards in the moving frame also moving at the speed of light as measured in the frame, same as light moving forward? $\endgroup$ Commented Aug 22, 2013 at 23:20
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$\begingroup$ @RonMaimon You can write an answer then :) $\endgroup$– GuySoftCommented Aug 24, 2013 at 18:27