Chasing a photon According to this article, the Theory of Special Relativity holds that if you were chasing a stream of light at half the speed of light, $c/2$, the light's speed relative to you would still be $c$.
Does this hold for any speed below $c$? For example, if you were travelling behind a photon at $0.9999999999\,c$, what would the photon's speed relative to you be?
Also, if you were travelling at $c/2$ and were chasing a particle at $0.9999999999\,c$, what would its speed relative to you be?
What is the equation that is used for this calculation?
 A: Yes, this holds for any speed below $c$. Even at $99.9999$% of speed of light you would still perceive photons to travel at c. This is a consequence of the relativistic addition of velocities:
The apparent velocity of an object relative to you is given by $$u^{'} = \frac{u\pm v}{1 \pm \frac{uv}{c^2}}$$
A: Marco Prins,
This has been proved by the famous Michelson-Morley experiment. Regardless of the velocity of the measurer, the speed of light is always $c$.
You need to remember that movement is relative. If you are moving with velocity $V$ relative to another object, than this object is moving with the same velocity relative to you; only the direction is opposite. (It's like when you are sitting in a train, and it suddenly starts to leave the station. For a while you might think it is the station that is leaving ...) If there is no third frame of reference (i.e. Earth), like in space, and the movement is inertial, than there is no way to tell which body is moving and which is stationary. In space there is no absolute reference frame, which you could call absolutely stationary.
Therefore you are always in movement relative to something, and you can easily find objects in space that move with very high velocity relative to you. (Or you are moving with a very high velocity relative to them, because how can you tell?) And yet the speed of light is still measured as exactly $c$ ...
Therefore you do not need any equation for this. You will always measure the speed of light as $c$.
Why? Good question ... :-)
A: Photons always travel at $c$ (not completely true, but a good simplification for this question's purposes). Common sense tells us that if person A running at velocity $v$ is chasing person B with velocity $u$, the velocity of person B with respect to person A ($w$) is:
$$w=u-v$$
But our common sense is misleading, and this equation is only an approximation that works well at low velocities. Special Relativity tells us that the correct equation is actually:
$$w=\frac{u-v}{1-uv/c^2}$$
So let's say that someone's running at velocity $v$ is chasing a photon traveling at $u=c$ with respect to the ground. The velocity of the photon with respect to the runner is:
$$w=\frac{c-v}{1-cv/c^2}=\frac{c(c-v)}{c-v}=c$$
So the photon is still traveling at $c$ with respect to the runner, regardless of how fast he's running.
