# Chasing a photon [duplicate]

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?

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## marked as duplicate by jinawee, Kyle Kanos, DavePhD, Brandon Enright, JimMay 6 '14 at 17:52

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Should 9.999999999c be 0.9999999999c? You can't travel faster than light. The equation used for the calculation is the relativistic addition formula. – John Rennie May 6 '14 at 7:15
"photon at $0.9999999999c$" Surely you mean to say that this is a particle, as such a particle cannot be a photon... – Steven Lu May 6 '14 at 17:23

## 3 Answers

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.

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Excellent, thank you! – Marco Prins May 6 '14 at 8:12
I notice this equation is sometimes given with u + v on top and 1 + uv/c^2 at the bottom, with addition instead of subtraction (example). Why is that? – Marco Prins May 6 '14 at 8:14
@MarcoPrins because in that case the moving objects go in opposite directions. – Ruslan May 6 '14 at 8:35
Proving that $w=c$ according to this equation seems circular. It was the assumption that $c=x/t$ and $c=x'/t'$ that lead to the derivation of this equation, and not the other way round. $c$ is constant not because we found it through this equation. The assumption of constant $c$ lead to the derivation of this equation, so obviously, it must show constant $c$. But this is not a proof. – bright magus May 6 '14 at 9:14
Note that the length of your yardstick and the rate of your clock change as you speed up (Lorentz Contraction), so that same photon is seen by both A (at "rest") and by B (at .5 c compared to A) as moving at c relative to both of them. – Phil Perry May 6 '14 at 14:08

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 ... :-)

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It is always good to have the experimental evidence. Specially when it is so simple to understand, even though it leads to such an counter-intuitive theory. – Davidmh May 6 '14 at 9:20
I would say this is the best way to do physics. Find something out through evidence and then build theory around it (that, in turn, allows to predict other things which proves it is true). – bright magus May 6 '14 at 9:28
Why? using maxwell equations you can deduce the wave function in which the speed depends only on μ and ε of the medium. – borjab May 6 '14 at 14:23
Providing the equations and constants and the rest of the theory we use are correct. But then knowledge moves on and after some time we discover we need to introduce dark matter or ghost fields or whatever, because observations do not much predictions. – bright magus May 6 '14 at 14:30

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 $$u^{'} = \frac{u\pm v}{1 \pm \frac{uv}{c^2}}$$

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What's $\sqrt{}$? doing here? – Stan Liou May 6 '14 at 7:42
That is a mistake, thank you for pointing it out. Corrected. – Constandinos Damalas May 6 '14 at 7:52