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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?

Řídící
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Marco Prins
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3 Answers3

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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.

Jold
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  • 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 06 '14 at 08:14
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    @MarcoPrins because in that case the moving objects go in opposite directions. – Ruslan May 06 '14 at 08:35
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    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 06 '14 at 09:14
  • 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 06 '14 at 14:08
  • @brightmagus: if you're looking at this in terms of the Einstein postulates, yes. If you're looking at this in terms of a modern perspective, your postulate is Minkowski space, and boosts are hyperbolic rotations, and this forumla is easily derivable, and the constancy of the speed of light is a result, and not an assumption. – Zo the Relativist May 06 '14 at 15:18
  • There are some funny things about Minkowski, but this is not a place for this. – bright magus May 06 '14 at 15:39
  • @Jerry Schirmer: By the way, isn't $c$ in the $ct$ assumed constant? – bright magus May 06 '14 at 16:47
  • Hm... Interesting. So what happens if the reference frame we're measuring from is that of another photon? In other words, if we were seated at the front of a photon and saw another photon, would the second one still be observed as moving at c? That seems nonsensical, since we'd be moving at the same speed... – asteri May 06 '14 at 16:55
  • @brightmagus: there's a huge difference between saying "here's a constant parameter in your theory" and "I hold that it is a universal law that the relative velocity of light with respect to everything is a fixed constant". The $c$ in the $ct$ that you use for the time coordinate in Minkowski theory is a parameter. It is not necessarily the speed of anything. That it is is a derived result. – Zo the Relativist May 06 '14 at 17:14
  • @Jerry Schirmer: Would Minkowski (and Einstein) introduce this "parameter" if he didn't assume c=x/t and c=x'/t'? – bright magus May 06 '14 at 17:19
  • @Jeff Gohlke: You don't need to put yourself in the position of a photon. This problem is not much different than saying that if you are moving with velocity $v$ toward a ray of light which has velocity $c$, than you will still measure this light as moving $c$ relative to you. – bright magus May 06 '14 at 17:23
  • @brightmagus: Of course the theory has to reduce to special relativity. You're just replacing one set of postulates with a cleaner set of postulates. The key point is that you can get all of special relativity by assuming Minkowski spacetime. You don't need to assume Einstein's second postulate. You can derive it from this deeper geometrical structure. And the parameter is needed because you need some way to transform time into space in order to have spacetime at all. After deriving the structure, you are THEN free to say that this constant gives the speed of massless particles. – Zo the Relativist May 06 '14 at 17:28
  • @JerrySchirmer: But in order to assume Minkowski spacetime you need FIRST to assume c=x/t and c=x'/t'. As far as I know, you will not derive his equations for spacetime without this assumption. – bright magus May 06 '14 at 17:50
  • @brightmagus: you're wrong. Minkowski spacetime only relies on differential geometry. – Zo the Relativist May 06 '14 at 19:21
  • let us continue this discussion in chat – bright magus May 06 '14 at 19:28
  • Then why did he introduce the same $c$ on both sides of his equation and put it along $t$ and $t'$? He primed all other variables ($x, y, z, t$), but not $c$. This is clearly the assumption that in both frames of reference $c$ is the same constant. $c$ is unprimed from the very beginning. It is not that he proved along the way that $c=c'$. $c$ is always $c$ in his derivation. – bright magus May 07 '14 at 05:48
  • @brightmagus This seems to have spurred an interesting discussion, but to reply to your original comment: I never claimed this was a proof. Velocity addition was originally derived from the assumption that c=const, so it had better work that way. Marco Prins asked whether or not a photons velocity would still be c if one were trying to chase it at high speeds, and what the equation to calculate such a scenario is. I provided him with what he asked for. Of course, you can't "prove" that c=const must be true simply via theory - this is something that must be confirmed by experiment. – Jold May 08 '14 at 14:45
  • Sure thing, All I wanted to stress is that $c$ is not the result of maths. So my apologies for undeserved "reprimand" :) – bright magus May 08 '14 at 14:51
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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 $$u^{'} = \frac{u\pm v}{1 \pm \frac{uv}{c^2}}$$

PhotonBoom
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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 ... :-)

bright magus
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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 06 '14 at 09:20
  • 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 06 '14 at 09:28
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    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 06 '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 06 '14 at 14:30