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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? $\endgroup$ – Mark Mitchison Feb 17 '13 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 Feb 17 '13 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. $\endgroup$ – Alexey Bobrick Feb 19 '13 at 20:02
  • $\begingroup$ Possible duplicate: physics.stackexchange.com/q/27794/2451 $\endgroup$ – Qmechanic Feb 20 '13 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 Feb 8 at 0:45
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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?

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  • $\begingroup$ A nice clear explanation. Thanks. Yes, it would be interesting if there is someway of making physical sense of this question. $\endgroup$ – Mozibur Ullah Feb 18 '13 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? $\endgroup$ – zibadawa timmy May 23 '16 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. $\endgroup$ – samerivertwice Jun 29 '16 at 16:08
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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 Feb 17 '13 at 3:13
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    $\begingroup$ Yes, that's what I'm saying. $\endgroup$ – David Z Feb 17 '13 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. $\endgroup$ – Luboš Motl Feb 17 '13 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. ;-) $\endgroup$ – Luboš Motl Feb 17 '13 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 Feb 18 '13 at 12:11
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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 Feb 25 '20 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 Feb 25 '20 at 16:36
  • $\begingroup$ Among the earthly observers, eventually all of them will consider him going at almost the speed of light. $\endgroup$ – Ruslan Feb 25 '20 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 Feb 25 '20 at 16:40
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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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TL;DR

There is no legitimate observer comoving with a photon in special relativity (and thus, in physics). Thus, any question about what such an observer would feel, see, or observe are really meaningless questions.


To an external observer it appears that time has stopped for a photon.

No--not really. I believe that this statement is meant to imply that a clock moving alongside a photon stops ticking (as observed by an inertial observer with respect to whom the clock is moving at the speed of light--and thus, with respect to all the inertial observers as anything moving at the speed of light moves at the speed of light with respect to all the inertial observers). But this is a meaningless statement because there can be no clock that moves alongside a photon. And I would say that here is where the story can end for good.

But this relation is reflexive, so for an observer traveling with the photon, it appears the universe has stopped everywhere.

So, as I explained, the claim that to an external observer the time appears to have stopped for a photon is not right--it is not even a properly posed claim that can actually be tested to be either right or wrong. The premise itself cannot be realized.

But on top of that, the argument about reflexivity is also flawed here. Simply because there is no legitimate notion of an observer comoving with a photon so as to give rise to any possibility of reflexivity. (Also, I think the correct word to be used here would be "symmetry" rather than "reflexivity" as reflexivity refers to the relation of an element with itself but the OP evidently seems to refer to the supposed symmetric nature of the relation between two distinct elements.)

Space also gets distorted parallel to the direction of motion, but not perpendicular to it. Does this mean that for an observer traveling with a photon sees spacetime as a flat plane?

Again, since there is no legitimate observer comoving with a photon, the question is moot--rather, the question is just a grammatically correct structure with no actual meaning in physics.


So, the bottom line is simply what I wrote in the TL;DR that there is no legitimate observer comoving with a photon in special relativity (and thus, in physics). Thus, any question about what such an observer would feel, see, or observe are really meaningless questions.

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    $\begingroup$ Apparently Einstein asked what would happen if we were comoving along with an electromagnetic wave. He then claimed that we would see a standing wave and remarked such a standing wave isn't a solution of Maxwells equations. Nevertheless, the question isn't completely meaningless. As it helped him to understand the structure of electromagnetism. $\endgroup$ – Mozibur Ullah Sep 16 '18 at 5:49
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    $\begingroup$ @MoziburUllah Yes, Einstein indeed asked the question as to what would happen if he traveled alongside an EMW. But that piece of information is not really physics. That is just history. The physics is what came after, namely, special relativity. And that answered Einstein's question fully by clarifying that the question is meaningless as no observer can ever travel alongside an EMW. Was the question useful? Maybe yes, to Einstein. I am not saying that there is something wrong in asking that question--I am just saying that the question is meaningless and that is just physics! $\endgroup$ – Dvij D.C. Sep 16 '18 at 18:38
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    $\begingroup$ I can't say I wholly agree with you here - it is actually physics, because it helped Einstein to understand the role of symmetry and this helped him formulate SR. It's wrong to say that this question of Einstein was 'meaningless' as it wasn't SR that answered this question - because this suggests Einstien merely had misunderstood relativity, when in fact it was Einstein that developed and formulated SR. $\endgroup$ – Mozibur Ullah Sep 16 '18 at 20:41
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    $\begingroup$ In the context of SR, the question may be meaningless, but if we are always restricted by what our current theories say then we can never develop new physics. For example, look at the attempts of modified gravity to explain certain anomalies in the rotation curves of galaxies. $\endgroup$ – Mozibur Ullah Sep 16 '18 at 20:43
  • $\begingroup$ Anyway, thanks for taking the trouble to answer my question. $\endgroup$ – Mozibur Ullah Sep 16 '18 at 20:44

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