Suppose I want to know what the universe looks like from the perspective of a frame of reference moving at $c$ relative to my current frame. As discussed at length in various other questions on this forum, from the perspective of the current established theory of special relativity, the notion of such a frame is more or less nonsense.$^{[1]}$


Is it possible to modify the setting of special relativity, by adding dimensions to space-time, or any other hereto unobserved features, such that the restriction of the theory to our 4-D slice of space-time matches special relativity, and such that there is a meaningful way to describe the frame of a photon?

[1] In partiucular, the transformation we get by shoving $v=c$ into the usual formulae is not in the Lorentz group (it doesn't preserve volume of space-time and is not invertible).

  • $\begingroup$ If photon is moving at c relative to you, then you are moving at c relative to photon. Quite a normal-looking perspective, isn't it? Also, there can be a lot of objects found to move at nearly c relative to us, I.e. distant galaxies. Would the extra couple of km/s make a real difference? Unlikely. $\endgroup$ – bright magus Jul 6 '14 at 6:16
  • 4
    $\begingroup$ You can modify any important theory you want but if you modify a theory that is correct and essential, like relativity, you get a wrong and worthless theory. Why do you want to "modify" it? It's exactly like asking whether one may modify Darwin's theory so that animals never try to eat each other or compete with each other. Yes, you can "modify" it but the modification is wrong. The equations of relativity make clear conclusions about the value $v=c$ as well - and especially about it - and you can't cherry-pick or modify selective conclusions without destroying the whole structure. $\endgroup$ – Luboš Motl Jul 6 '14 at 6:33
  • $\begingroup$ @LubošMotl: SR is far from flawless. Let's take simple time dilatation and length/distance contraction. They mean that time and distance are in inverse proportions. And yet c=x/t and c=x'/t', which gives us x/t=x'/t' show they should change in direct proportions. Tables by the 4th and 5th pictures here: hyperphysics.phy-astr.gsu.edu/hbase/relativ/muon.html show SR does not influence x directly. On the other hand, it is possible to construct simple thought experiment showing time dilatation cannot be real. (And solving the twin paradox through acceleration is an obvious trick.) $\endgroup$ – bright magus Jul 6 '14 at 7:24
  • $\begingroup$ See my answer to this question about the non-existence of the $v=c$ frame. $\endgroup$ – Kyle Kanos Jul 6 '14 at 14:25
  • $\begingroup$ @LubošMotl: your statement "if you modify a theory that is correct and essential, like relativity, you get a wrong and worthless theory." is false. Suppose I have quantum mechanics but no notion of a density matrix. The theory is correct, but missing an important feature. When I add in the density matrix I do not spoil the rest of the theory, but I can now explain decoherence. This is a demonstration of modifying a correct theory to further explain natural phenomena. $\endgroup$ – DanielSank Jul 6 '14 at 15:15

No, and the reason is quite simple.

The proper time of a photon is zero (according to the principles of special relativity). That means that there is no time difference between the place of emission and absorption. By this, any hypothetical observation of a photon would be reduced to a time period of zero, and it would not be able to distinguish/ to measure time, and thus it cannot serve as a frame of reference.

By the way, the same principle applies also to distances, the proper distance of photons being zero. A rule yielding always the same measure is no rule.

A photon (from its "hypothetical point of view") is not participating in spacetime (except its places of emission and absorption), spacetime is reduced to one point, and thus it cannot be a frame within spacetime.

  • $\begingroup$ "That means that there is no time difference between the place of emission and absorption". That would mean an infinite speed. $\endgroup$ – bright magus Jul 6 '14 at 7:42
  • $\begingroup$ @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. $\endgroup$ – Moonraker Jul 6 '14 at 7:45
  • $\begingroup$ 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 ... $\endgroup$ – bright magus Jul 6 '14 at 7:49
  • 1
    $\begingroup$ No. The photon is not a reference frame. $\endgroup$ – Moonraker Jul 6 '14 at 8:29
  • 1
    $\begingroup$ 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. $\endgroup$ – DanielSank Jul 6 '14 at 15:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.