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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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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
  • $\begingroup$ 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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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 Mankad Sep 16 '18 at 18:38
  • $\begingroup$ 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
  • $\begingroup$ 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

protected by Qmechanic Feb 28 '13 at 7:12

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