# Is a photon “fixed in spacetime”?

From what I've read, according to relativity, a photon does not "experience" the passage of time. (Can we say there is no past/present/future for a photon?)

Would it be better to say a photon is "fixed in spacetime"? If so, how do we explain the apparent movement of a photon? Is everything else moving relative to it in spacetime?

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Because spacetime includes both multiple points in space and multiple moments in time, you have to think of a particle as a line (or a 1D curve) through spacetime, not a point. The line is called the world line. It's made up of all the $(x,t)$ points at which the particle exists: in other words, if you, as an external observer, measure the particle's position $x$ (using your own rulers) at a bunch of different times $t$ (using your own clock), and plot all those points on a graph, and connect them, you get a world line.

You could say that the world line itself is fixed in spacetime. But that's not a particularly useful statement to make, because there's no separate time outside of spacetime, so it's not like the world line can actually move. Besides, even if you did say the world line is fixed in spacetime, that applies equally well to massive particles and to photons.

You can pick any two points on a world line and figure out how much time elapses, by your clock, between those two points. It's just $t_2 - t_1$. You can also figure out how much time elapses, by some other observer's clock, between those two points: it's $\frac{t_2 - t_1}{\sqrt{1 - v^2/c^2}}$, where $v$ is the relative velocity between you and the other observer. And you can even try to apply this calculation treating the particle itself as the observer, which lets you figure out how much time passes for the particle between those two points. It works out to $\sqrt{(t_2 - t_1)^2 - (x_2 - x_1)^2/c^2}$. This works just fine for a massive particle. But for a photon, no matter which two points you pick, the answer you get is zero. This is why we say that photons don't experience time. I don't think it's accurate in general to say that photons are fixed in time, though. That might be a useful thing to say to make a particular point in a particular argument, but in most cases, it's probably more misleading than not.

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This makes a lot of sense. I'm still not sure how to understand "it just travels zero 'distance' per unit time" from Ron's answer though. Doesn't a photon travel at the speed of light which is a definite non-zero distance per unit time? Why does the answer come out to zero in the equation you posted when it seems like it should just be some very small non-zero number since the speed of light is so fast? Is it because we actually use the speed of light (c) in calculations? Is there anywhere I can learn more about the significance of it acting as a universal "speed limit"? – Ocsis2 May 5 '12 at 9:52
@Ocsis2: It's because David is using units where the slope of "zero length" is "c units over per unit up" which means the line for a photon is nearly flat if c is large. This is not convenient for relativity, but it's mathemtically equivalent to the usual description (where c=1 and the photon goes at 45 euclidean degrees in the diagram), it's just a rescaling of the horizontal axis. The formula for distance "delta x squared minus (c times) delta t squared" still gives zero when the line is sloped at the speed of light. – Ron Maimon May 5 '12 at 13:20
Okay, so regarding the "fixed in spacetime" statement, was he saying that the worldline of a photon is essentially flat (or as close to flat as is possible), meaning it's not really much of a "worldline" at all? It's just a photon, fixed in time and space. Or did he mean it's a photon fixed only in time (but not space)? Is there a degree of length contraction plus time dilation involved also in his statement? The "worldline" contracts at the speed of light into the photon itself? – Ocsis2 May 5 '12 at 13:59
Also, David, in the graph you posted, let's say the blue line corresponds to the speed of light. This would represent the boundary of spacetime, right? Which sort of corresponds to the whole light cone idea. So even though time might "stand still" for the photon, space and time are one continuum so space does not "stand still". In fact, time might not either but the way we measure time (since we have no other measuring stick) is dependent on the speed of light which is why it comes out to zero for a photon. – Ocsis2 May 5 '12 at 14:38
@Ocsis2 (4 up) Just like you can figure out the amount of time between two events going by a particle's clock (the proper time) using that formula $\sqrt{\Delta t^2 - \Delta x^2/c^2}$, you can also find the distance the particle sees the universe traveling past it by multiplying the proper time by the particle's speed. For a photon, since the proper time difference is always zero, the distance traveled is also zero. I suppose that's what Ron was talking about: the photon travels zero distance by its own measurement, in a nonzero time by an external observer's measurement. – David Z May 5 '12 at 14:47

A photon is not fixed in spacetime, it just travels zero "distance" per unit time. This is not a paradox because the photon doesn't do anything in its rest frame. If you chase it faster and faster, it redshifts to oblivion.

In relativity, the time experienced by an observer traveling from A to B along a straight line at a constant velocity is (defined as) the (timelike) relativistic distance between A and B, sometimes called the interval. This distance is zero when A and B are separated by an equal amount of distance and time (in units where the speed of light is 1), so that a light ray can go from A to B without turning a corner.

So the light ray goes along between different points, it can encounter different objects at different positions, even though there is no notion of time along the photon path. This just means that if you follow the photon ever more closely, you see the time between reaching the different objects along the path shrink to zero.

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That makes things clearer. So would you agree this person didn't put it correctly? – Ocsis2 May 5 '12 at 2:14
@Ocsis2: technically, DaveC426913 is correct, but his phrasing is rather misleading. – David Z May 5 '12 at 6:04
@Ocsis2: He is using words which are not precise enough to say that they are wrong or right, this is an example of what Pauli would call "not even wrong", but blathering. – Ron Maimon May 5 '12 at 13:04

$\newcommand{\dd}{\mathrm{d}}$ For a body in space-time the proper time is expressed is $c^2 \dd \tau^2 = \dd s^2\equiv g_{\mu\nu}\dd x^\mu \dd x^\nu$ along the world line of a body. For a photon, however, $\dd s^2=0$, hence the proper time doesn't change indeed.

However, you can introduce an analogue of time for photons, which is defined as $\dd\lambda=k^\mu\dd x_\mu$ (with $k_\mu$ being the 4-dimensional wave vector) and is called a canonical parameter and enumerates the points on the world line of the photon.

The conclusions: No, the photon is not fixed. It would be correct to say rather that the notion of proper time is applicable to non massless objects only, whereas for the photons the positions are specified by another measure called canonical parameter.

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Can you tell me more about this "canonical parameter" or link me to a Wikipedia article or something? – Ocsis2 May 5 '12 at 4:24
It's usually called an "affine parameter", and it is a limit of proper time which is nonzero as you approach the photon's worldline. It doesn't have to be labelled by "k", although it can be, but for photons without a definite frequency, you can't do that anyway. – Ron Maimon May 5 '12 at 13:21
(@Ocsis2) Thanks, @Ron, affine parameter is a better word for canonical parameter $\lambda$, I recommend MTW book for that. – Alexey Bobrick May 5 '12 at 23:17
@Ron: Concerning your limit statement, it is wrong: the norm of 4-velocity doesn't change under 4-acceleration, and hence timelike geodesics can not approach isotropic ones, as they have different norms. – Alexey Bobrick May 5 '12 at 23:19
@AlexeyBobrick: It's not wrong, but perhaps stated badly: you rescale the norm by a diverging factor as you approach the null limit. The "limit of proper time which is nonzero as you approach the photon's worldline" is the limit of the proper time rescaled by the difference between (say) the time component of each segment of the near-photon worldline and the true photon world line. This is not quite right if the photon is accelerating, the proper definition is that it is the parameter that obeys the geodesic equation (with arclength parametrization) in the limit that the velocity becomes null. – Ron Maimon May 6 '12 at 4:35