# Relative to the speed of light

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?

• 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. – Brandon Enright Jun 1 '13 at 22:07
• 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... – DJohnM Jun 2 '13 at 2:52

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.

$$\Delta t' = \frac{\Delta t}{\sqrt{1 - v^2 / c^2}}$$

$$\Delta t' = \frac{1}{\sqrt{1 - (\frac{c}{2})^2 / c^2}} = \frac{1}{\sqrt{3/4}}= \frac {2}{\sqrt{3}} sec$$