Timeline for Can special relativity be extended such that the frame of a photon makes sense?
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| Jul 6, 2014 at 15:49 | comment | added | Moonraker | @DanielSank: It is even not sure if there is any worldline, with proper time=0, proper distance=0, I would rather talk of a direct transmission of the momentum from the emitting particle to the absorbing particle, without any intermediate worldline (from the "hypothetical point of view" of the photon). Without movement, I really don't see any means for "co-moving". | |
| Jul 6, 2014 at 15:40 | comment | added | DanielSank | @Moonraker: I'm not looking for a research project, just curious :) Anyway, certainly there I can imagine a world line in a higher dimensional space which, when projected (or something like that) into a subspace, takes on the pathological properties of a world-line Lorentz boosted at c. If this sort of thing is not possible, an explanation of why would be a great answer to my question :) | |
| Jul 6, 2014 at 15:29 | comment | added | Moonraker | @DanielSank "frames co-moving with the photon" The problem is that it seems there is no movement at all from the "hypothetical point of view" of the photon (proper time=0, proper distance=0). The result is what we all know: It does not make any sense to talk about the frame of a photon. Thus even in an extended sense I think that there is no room for your research of a theory. | |
| Jul 6, 2014 at 15:22 | comment | added | DanielSank | This answer explains why there isn't a frame co-moving with a photon in the standard theory of SR. As explicitly indicated in my question, I already understand this. I would like to know if there can be an extended theory which recovers SR on our space-time but includes frames co-moving with the photon, perhaps in other dimensions or whatever. | |
| Jul 6, 2014 at 9:14 | comment | added | bright magus | OK, let's put my point of view clearer. You gave an answer, but in our conversation your entire reasoning boiled down to: The photon is not a reference frame. Fullstop. Not an explanation to my standards. (And I did see Lubos Motl comment - I even made my own in reply.) | |
| Jul 6, 2014 at 9:08 | comment | added | Moonraker | I don't agree with your point of view which is not quite clear. See the comment of Lubos Motl: "The equations of relativity make clear conclusions about the value v=c". I agree with this statement. | |
| Jul 6, 2014 at 8:37 | comment | added | bright magus | Classical circular reasoning. And great dodging of contradictions. | |
| Jul 6, 2014 at 8:29 | comment | added | Moonraker | No. The photon is not a reference frame. | |
| Jul 6, 2014 at 7:49 | comment | added | bright magus | How have you arrived at this conclusion (that everything is one and the same thing for photon - emitter being absorber, etc.)? Actually, the universe would be one big black hole for a photon in such case ... | |
| Jul 6, 2014 at 7:45 | comment | added | Moonraker | @bright magus: No, because distance is zero as well. Zero distance in zero time, that means that there is no kind of velocity at all. | |
| Jul 6, 2014 at 7:43 | history | edited | Moonraker | CC BY-SA 3.0 |
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| Jul 6, 2014 at 7:42 | comment | added | bright magus | "That means that there is no time difference between the place of emission and absorption". That would mean an infinite speed. | |
| Jul 6, 2014 at 7:09 | history | edited | Moonraker | CC BY-SA 3.0 |
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| Jul 6, 2014 at 7:03 | history | answered | Moonraker | CC BY-SA 3.0 |