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My understanding of relativity isn't very sophisticated, but it seems to me that relative to a photon moving at the speed of light, we are moving at the speed of light. Is this the case?

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Related: . – dmckee Aug 12 '13 at 16:21
up vote 10 down vote accepted

When we say that object A is moving at speed v relative to an object B it means that there is a reference from where B is at rest and A is moving at speed v in that reference frame.

If A is a photon then it moves at the speed of light c in all reference frames so if B is is us then in our reference frame it is moving at c, so it makes sense to say that the photon moves at speed c relative to us, but is it OK to say it the other way round?

If A and B are both objects that have mass so that they move at less than the speed of light, and if A is moving at speed v relative to an object B then in the frame where A is at rest B will be moving at speed v relative to A in the opposite direction. So for speeds less than the speed of light, the speed of A relative to B equals the speed pf B relative to A.

It is tempting to extrapolate this to the case where A is a photon and conclude that therefore B (us) is also moving at speed c relative to the photon. However this would mean that we were moving at speed c in a reference frame where the photon is at rest. This is not possible because we cannot move at speed c in any reference frame and a photon cannot be at rest in any reference frame.

So the answer is "no", it is not correct to say that we move at speed c relative to the photon.

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Yes... Kind of... There is no preferred reference frame (this is Einstein's postulate of relativity), so if one observer measures another to move at the speed of light (or any other velocity), this other observer must measure the same velocity relative to the first.

However, if you move at the speed of light things get weird. For example, how can we move at the speed of light if we have mass? That would require infinite energy, which is impossible. On the other hand, when moving at the speed of light, no time is passing. So this photon wouldn't even be able to measure our speed. Also, what would the photon's velocity relative to itself be? None. But then it can't have any energy, and since a photon doesn't have mass either, there is no measurable quantity of itself that the photon can observe. So to the photon, the photon doesn't even exist. Also, since the entire universe is moving at the speed of light relative to the photon, the entire universe must experience infinite length contraction, which means the universe doesn't even have a size. Wait... does that mean the universe doesn't exist?!

The bottom line is, for the photon, no time passes, which means it couldn't possibly make any observations, which means discussing any of the observations a photon would make is pointless, which means the universe probably does exist.


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That is a perfectly sensible sounding question that can be answered (1) in a simple way, (2) in a way where it looks like physics is totally broken, and (3) in a smarter way that gives the same answer as (1) but is also much crazier.

The (1) answer is yes it does.

The (2) answer is that light doesn't really experience time the same way according to special relativity... in fancy language, the proper time along a null geodesic is 0. So if you try to calculate dx/dt in the light cone's rest frame, you'll get 0/0. So physics is broken.

The (3) answer is to consider taking an observer moving at a speed near the speed of light and then take a limit as his speed approaches the speed of light. At every step, if we see him moving at $v$, he will see us moving at $-v$. So in the limit, we indeed are moving at the speed of light relative to the light beam.

But, the really mindblowing thing (that also explains (2)) is to consider another observer, moving relative to us at some speed $u$. At what speed will the light beam see that observer?

If you think about it, the only answer that makes sense with relativity is that the light beam ALSO sees that observer moving at speed $c$ relative to it! This is crazy! An observer moving at the large speed $v$ will see us and the $u$ observer moving at different speeds, but as $v$ increases the difference between how fast he sees us going and how fast he sees the $u$ observer going tends to zero.

That's why $dx/dt$ in the light beams coordinates gives $0/0$--the only two velocities that the light beam detects are 0 and c!

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The ONLY velocity that the light beam detects is $c$. In what situation would it possibly measure a different speed? – JSQuareD Aug 12 '13 at 16:29
I had in mind an observer holding a cat and then accelerating them both towards $c$ at the same rate, so in the limit there is no relative velocity between the observer and the cat. But maybe this isn't a very good example. I see your point. – Andrew Aug 12 '13 at 16:31
Hmm.. I guess. The problem is, a cat can't actually accelerate towards $c$ because a cat has mass. If two massless particles move alongside one an other, they will still measure each other's velocity to be $c$. The problem is, there is no possible way any observer could accelerate to $c$, which means taking the limit is fundamentally different from actually considering the end situation. Formally, the formulas describing the behavior aren't continuous in $c$. This is the problem I have with your answer (3) – JSQuareD Aug 12 '13 at 17:02
OK, I concede. Your point that there are no observers moving at the speed of light is a good one, and conceptually it is definitely important to say that massless particles have no rest frame. I'm really answering a different question, what happens if you are moving much faster than everything else in your environment. Thanks for pointing this out. – Andrew Aug 12 '13 at 21:06

This is not true. I will prove that we are traveling at the speed of light.

Step 1: $$E=Fd=mad=mv/t\cdot d=mv\cdot v=mv^2$$ but $$E=mc^2$$ so $$c^2=v^2$$ but if object is at rest then $C$ cannot be zero(it is a constant) and that simply implies that object at rest are traveling at the speed of light!

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wow really amazinging it breaks down all theories of relativity – anonymous Dec 15 '14 at 14:59
Except $E=mc^2$ is a particular limit, it's really $E^2=(pc)^2+(mc^2)^2$. Note also that $a=dv/dt$ (or if you're not familiar with calculus, $a=\Delta v/\Delta t$), so the conversion $mad=mv/t\cdot d$ is wrong. – Kyle Kanos Dec 15 '14 at 15:21

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