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It is known that $$x'=\gamma(x-Vt),$$

where $\gamma$ is the relativity multiplier. This equation means that $x'\to\infty$ if the frame is moving at the speed of light. Does this mean that if we treat light as a frame relative to some other frame with coordinates $x$, then the frame of light will always have coordinates at infinity? Something doesn't add up.

sequence
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2 Answers2

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What you've discovered is that the Lorentz transformations become singular if we try to apply them to a frame moving at the speed of light (or greater) relative to us. That's the fundamental reason we can never observe anything moving faster than light.

Light obviously moves at the speed of light, but light has no rest frame. That is, it is meaningless to talk about transformations into the rest frame of the light because no such frame exists. Thats why your question:

Does this mean that if we treat light as a frame relative to some other frame with coordinates xx, then the frame of light will always have coordinates at infinity?

has no answer.

Community
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John Rennie
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  • I didn't quite understand the argument why light has no rest frame. And even if we consider a frame with a speed close to the speed of light, why should its coordinates be moved so much further ahead of where they actually are? – sequence Dec 11 '16 at 20:25
  • @sequence: in SR the speed of light is always $c$. That's a fundamental principle. But the rest frame of light would be a frame in which the light was at rest i.e. its velocity would be zero. That would mean a frame exists where the speed of light isn't $c$ and this contradicts a basic principle of SR. So either special relativity is wrong or the rest frame of light cannot exist. – John Rennie Dec 13 '16 at 11:40
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There is no reference frame in which light doesn't move, because it always travels with the speed of light, so your question is meaningless.

Andrey Feldman
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