# How does a photon experience space and time?

To an an external observer it appears that time has stopped for photon. But this relation is reflexive, so for an observer travelling with the photon it appears the universe has stopped everywhere.

Is this right?

Space also gets distorted parallel to the direction of motion, but not perpendicular to it.

Does this mean that for an observer travelling with a photon sees spacetime as a flat plane?

NOTE: I'm using language vividly not literally when I say a photon experiences space and time. Not that I'm against idealist or panpyschist interpretations of matter or energy come to that.

-
Are you sure that the phrase "an observer travelling with a photon" is meaningful? – Mark Mitchison Feb 17 '13 at 2:33
An observer can't travel at speed of light. Photons always travel at the speed of light relative to an observer. (I don't know any exceptions) – raindrop Feb 17 '13 at 2:40
Saying that no such observer exists is a rather boring way of not answering the question. One obvious point is that a timelike observer cannot attain the speed of light. Yet another obvious point, is that spacelike 'observers' are perfectly well defined. One could well hypothesise on what a light-like observer would be, by using (arbitrary) canonical parameter of light world line instead of time. – Alexey Bobrick Feb 19 '13 at 20:02
Possible duplicate: physics.stackexchange.com/q/27794/2451 – Qmechanic Feb 20 '13 at 15:54
Nice question. +1 – N.S.JOHN May 23 at 16:39

There is a more precise sense in which the question is ill-posed (at least mathematically); namely, it is a fundamental assertion of relativity (special and general) that the time 'measured' (counted, experienced, observed...) by an observer between two events occurring on her worldline is the length of her worldline-segment joining the two events (that's how we connect the physical notion of (personal) time with the mathematics of the theory). The way she determines motion depends on this notion of time. Equivalently, proper time is measured by the arc-length parameter of the observer. Now, since null curves have zero length (hence no arc-length parameter) the concept of proper time is not defined for null observers. Hence neither is (proper) relative motion (i.e from the photon's perspective').

Also, the relation you describe between timelike and null (instantaneous) observers isn't reflexive at all (whereas it is for the timelike ones, via the Lorentz boosts'): no isometry of Minkowski space can take a timelike vector to a null one.

Although the question doesn't make sense, in this strict sense, mathematically, perhaps there are other physical or mathematical tricks for interpreting it?

-
A nice clear explanation. Thanks. Yes, it would be interesting if there is someway of making physical sense of this question. – Mozibur Ullah Feb 18 '13 at 18:35
There tend to be various "trace" operators to handle behaviors on measure zero subspaces that really need a non-zero measure, such as with Sobolev spaces. Different concepts, ultimately, but maybe similar things exist on Lorentzian manifolds? – zibadawa timmy May 23 at 16:48
I think of the emission and absorbtion of a photon as one and the same event. That seems to be the clearest way of thinking of it. – Robert Frost Jun 29 at 16:08

There is no such thing as an observer traveling with a photon. Photons don't have experiences. So there's really no valid answer to this question.

-
Nope. An observer in the sense of special relativity must be (at least instantaneously) traveling along the time axis of a reference frame, which is not true of a photon. There are no reference frames whose time axes correspond to light-speed motion. – David Z Feb 17 '13 at 3:13
Yes, that's what I'm saying. – David Z Feb 17 '13 at 5:01
Dear Mozibur, there's no inertial system where a photon is at rest simply because you would need to boost regular inertial frames that do exist by an infinite amount, to get to $v=c$. But the number "infinity" doesn't exist or, at least, isn't an element of real numbers. If you try to calculate with it, you get singular values of everything: infinite Lorentz contraction, infinite time dilation etc. In some contexts, we may extend the real numbers by the number infinity but it makes no sense to allow it here because everything becomes ill-defined. – Luboš Motl Feb 17 '13 at 6:16
I would also add a "neuroscience" comment. Collections of photons propagating in some direction, because they move by $v=c$ exactly, can't have brains that would send signals back and forth. If an electric signal were sent back, against the direction of motion of the photon, it could never get back because to do so, these signals would have to travel faster than the original photons - faster than light - and that's not allowed. So nothing moving at the speed of light can actively think, at least not a nonzero number of operations per second. ;-) – Luboš Motl Feb 17 '13 at 6:17
It sounds a little bit like cheating to state that asking how a photon observes the world is an invalid question. After all, a photon interacts with the world. I think that if our current physics can't give any insight into this it simply means our physical understanding is limited. But I may well be wrong. – Skúli Feb 18 '13 at 12:11

## protected by Qmechanic♦Feb 28 '13 at 7:12

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).