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 1: 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.

Note 2: Since it's been pointed out that an inertial frame can't be boosted into the frame of a photon and moreover there is no limiting behaviour here - no matter what frame you are boosted into, light will always appear to travel at c. Nevertheless, when an external observer watches a train chasing a photon by gradually increasing its speed, he will note that the train speed is approaching the speed of the photon (but will never match it) and when he looks at the clock inside the train he will not too that it is gradually slowing down (but never actually stops). Using this picture, we can attempt the thought experiment - if not one that can be carried out - of what a world looks like to a photon.

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    $\begingroup$ Are you sure that the phrase "an observer travelling with a photon" is meaningful? $\endgroup$ Commented Feb 17, 2013 at 2:33
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    $\begingroup$ 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) $\endgroup$
    – raindrop
    Commented Feb 17, 2013 at 2:40
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    $\begingroup$ 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. $\endgroup$ Commented Feb 19, 2013 at 20:02
  • $\begingroup$ Possible duplicate: physics.stackexchange.com/q/27794/2451 $\endgroup$
    – Qmechanic
    Commented Feb 20, 2013 at 15:54
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    $\begingroup$ Saying that no such observer exists is a rather boring way of not answering the question Not all answers should be answered. Some must be refuted. $\endgroup$
    – TaW
    Commented Feb 8, 2021 at 0:45

4 Answers 4


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?

  • $\begingroup$ A nice clear explanation. Thanks. Yes, it would be interesting if there is someway of making physical sense of this question. $\endgroup$ Commented Feb 18, 2013 at 18:35
  • $\begingroup$ 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? $\endgroup$ Commented May 23, 2016 at 16:48
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    $\begingroup$ 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. $\endgroup$ Commented Jun 29, 2016 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.

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    $\begingroup$ 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. $\endgroup$
    – David Z
    Commented Feb 17, 2013 at 3:13
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    $\begingroup$ Yes, that's what I'm saying. $\endgroup$
    – David Z
    Commented Feb 17, 2013 at 5:01
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    $\begingroup$ 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. $\endgroup$ Commented Feb 17, 2013 at 6:16
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    $\begingroup$ 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. ;-) $\endgroup$ Commented Feb 17, 2013 at 6:17
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    $\begingroup$ 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. $\endgroup$
    – Skúli
    Commented Feb 18, 2013 at 12:11

I really wish we could once and for all put an end to all the nonsense about "limiting behavior" that gets written whenever someone posts a question like this. (See for example "Note 2" in the present question.)

Every time you accelerate, your speed increases according to some observers and decreases according to others. Every time you accelerate you will see some clocks slow down and some clocks speed up. There is absolutely no meaningful frame-independent sense in which acceleration can get you "closer to the speed of light". So there is no sense whatsoever in which acceleration could make your experience "closer to that of a photon", even if that phrase were meaningful.

Check it out: Alice sits in her home in Wichita. Bob travels westward, toward California, at (according to Alice) 50 mph. Closer to California, also traveling westward, is Carla, traveling at 70 mph according to Alice (but of course stationary according to herself). Looking in her rearview mirror, Carla says "There's Bob, traveling eastward at 20mph".

Now Bob hits the accelerator. Alice says: Bob's speed just went up from 50 to 60. He's a little more like a photon! Carla says: Bob's speed just went down from 20 to 10. He's a little less like a photon! Bob, meanwhile, sees Alice's clock slow down and Carla's speed up. What limit is he approaching?

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    $\begingroup$ Funny analogy, but it holds only for a one-time, rather small, acceleration. If Bob did ten such jerks, he'd be "more like a photon" for both Alice and Carla. And the more he accelerates, the more observers will consider that to be an increase in speed. $\endgroup$
    – Ruslan
    Commented Feb 25, 2020 at 15:57
  • $\begingroup$ @Ruslan: Yes, the more he acccelerates, the more observers will consider that to be an increase in speed. Also, the more he accelerates, the more observers will consider that to be a decrease in speed. There is no meaningful sense in which one of these "mores" is bigger or smaller than the other. $\endgroup$
    – WillO
    Commented Feb 25, 2020 at 16:36
  • $\begingroup$ Among the earthly observers, eventually all of them will consider him going at almost the speed of light. $\endgroup$
    – Ruslan
    Commented Feb 25, 2020 at 16:37
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    $\begingroup$ @Ruslan: But "the earthly observers" is a purely arbitrary subset of "all observers". In any event, the main the point is this: Bob can approach the speed of light from Alice's viewpoint and therefore appear, in some sense "more like a photon" to Alice. At the same time, he'll appear "less like a photon" to someone else, and most importantly, he won't seem at all more or less like a photon to himself, contrary to what is frequently implied in these questions, including this one. $\endgroup$
    – WillO
    Commented Feb 25, 2020 at 16:40

As other answers have pointed out, there is no point of view or frame of reference that keeps up with a photon. Never the less, the idea that such a frame of reference exists as the limit of infinite boosts is a very natural one that comes up over and over. Here is why there are problems with that idea.

Suppose you start at rest in a certain frame of reference, and accelerate at $1$ g for $1$ sec. This give you a new speed. Do this again and again.

As you travel faster and faster an observer in your starting frame sees you traveling closer and closer to the speed of light, your clock running slower and slower, and your ruler getting shorter and shorter. The limit of these measurements is you traveling at the speed of light, your clock stopped, and your ruler contracted to $0$ length.

It is natural, but wrong, to suppose that at this point your frame of reference is the same as a photon. Therefore photons experience no time, and see the entire universe as contracted to a plane.

First, the limit state doesn't match what we see when we observe photons.

Photons travel at a finite speed. As they advance, they change phase. So the idea that they are in a frame where no time passes and all points along their path have been compressed into the same point is wrong.

Second, you don't get closer to traveling as fast as a photon.

This can be illustrated by a hyperbolic tesselation of the plane. The tesselation below uses 30, 45, 90 degree triangles. It became famous when Escher used it as the basis of his Circle Limit woodcuts. In this post, it represents a $2D$ velocity space.

H. S. M. Coxeter, Crystal symmetry and its generalizations, volume 51 of the Transactions of the Royal Society of Canada.

An observer is stationary in his reference frame. This velocity is the center point. The sides of the triangles represent boosts in various directions.

As you undergo boost after boost, the observer sees you travel faster and faster. Your velocity is a point farther and farther from the center of the circle. But each boost gives a smaller change to your velocity. You never reach the edge, which represents the speed of light.

After each boost, you can measure the speed of a photon. Each time, it is still passing you at the speed of light. You are no closer to its speed.

This video shows how velocity space appears to you as you try to accelerate to the speed of light. Or equivalently, try to reach the edge of velocity space. (It isn't quite the same tesselation, and the path isn't quite a straight line. But it gives the idea.)

All the triangles are the same, though the ones far away appear distorted. These triangles become normal as you approach them.

No matter how many boosts you undergo, you are still at rest in your own frame. From your point of view, you are in the center of the circle. You are a finite number of boosts from the observer and an infinite number from the circle. No part of the circle has become any closer or farther from you.

As you move from triangle to triangle, you stay at rest in your own frame, though you move far away from the observer. You see each boost as making the same change in velocity. Though the cummulative effect on the velocity of the observer gets smaller and smaller. The observer recedes at close to the speed of light, but never reaches it.

From your point of view, the observer is getting closer to a state where his ruler shrinks to $0$ and his clock stops. You might think the observer is closer to matching speed with a photon you send in his direction.

The observer thinks no such thing. He sees your photons arriving at the same speed as always, though they are increasingly red shifted.

Mathematically, you are advancing from 1 boost to 2 boosts to 3 boosts, etc. The limit of this sequence is an infinite number of boosts. This really means the sequence diverges and there is no limit. The definition of an infinite limit is that given any finite number, after enough steps you will pass that number. The limit is not a state where you are sitting on a point named infinity.

This means given any speed slower than light, after enough boosts you will be going faster than that speed. But there is never a state where you are going the speed of light.

If you try to construct a limit, it would go something like this: For any $\epsilon > 0$, there is a point $P$ in velocity space where you would measure the velocity as $v_p$ such that ($c - v_p) < \epsilon$. But an observer at $P$ would see a photon pass him at $c$. So the limit as "$P \rightarrow $ the edge" is a state where the observer at P sees a photon pass him at $c$. This really means at all large boosts, an observer at $P$ sees a photon pass him at $c$.

The separation between the interior of the circle and the edge is absolute. All points in the interior of the circle are an infinite number of boosts from the edge. No number of boosts brings an observer from the interior to the edge. Likewise, it never brings a photon from the edge into the interior.

Some references on the mathematics of the tesselation, and how Escher's woodcuts illustrate it:


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