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Imagine a universe without time, or more specifically without the Flow of Time. Everything will be a 2D projection and nothing more. No movement, no interaction, and in other words no Change.

But our universe is not like this. We have movement, we have interaction, and our world changes in many ways. But there is a catch. We have some 0 mass particles in our universe, which move at the speed of light e.g. photons and, according to the theory of relativity, these particles are not affected by time. So the universe is static from the point of view of a photon.

But it can't be like this. Our world is dynamic. And this kinda raises a contradiction. On the one hand, the universe is static and in another the universe is dynamic. Can anyone explain this?

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It's simple: there are only two elementary actions that can happen to a photon - it is generated at one point and then it is consumed at another. Its whole life consists of just these two events, two "instants" if you will - the "travel" is really just the delay between those two occurrences with regard to everyone else. Two instants, each of zero duration, comprise zero time: $0 \cdot 2 = 0$.

Because of this, it doesn't have the capability to define an "experience" that is anything more than those two events - and thus you cannot ascribe to it a "view" of the Universe. This is why, in a sense, you cannot define a reference frame for a photon that "makes sense", as you're running into. The "world view" is "I get emitted, I get absorbed". Boom. That's it.

Another thing that is often missed is that "reference frames" do not exist as an entity in their own apart from spacetime and its contents. What they are, instead, are just systems of coordinates, that is, ways of labelling events i.e. points in spacetime, and in that case are wholly arbitrary. The Lorentz frames in special relativity are arbitrary - what is not is the Lorentz symmetries that relate them, which are self-maps of the points on spacetime as points regardless of labels, and the fact that the dynamics, the physics, on that spacetime, respect those symmetries, meaning that if we want to describe a moving observer's view of the world, we can do so without altering the form of the equations we were already using - which is how you can talk hypothetically of a Lorentz symmetry violation without it tripping over itself in a contradiction. (Note that you can always create a "moving reference frame" in any sort of universe that has a notion of space and time, but it may be that you need to use different "physics" to talk about this: toy example is Conway's Game of Life. Its rules are not motionally symmetric in any way.)

And to link that back to the idea of "experience", any "way of experiencing" spacetime can be related to some coordinate system. In fact, the Lorentz frame is not any human's experience of spacetime: instead, what you want is a "light cone frame" where the "present" is taken as one's past light cone. The Lorentz symmetry still relates it to the moving case, but the coordinates are all different and the symmetry takes a different form. (Whose experience is it? Well, because its "present" involves knowledge of spacelike-separated events, a patch of it can only be seen by an observer well into the future of the happenings in question, that has reconstructed the view from suitable observations.)

Hence, the failure of the Lorentz transformation to meaningfully define a frame means that the situation in the case of a photon is fundamentally different from the situation in case of everyone else, and its experience is likewise different.

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  • $\begingroup$ +1, but a comment. Between being emitted and absorbed, the wave function evolves. It's location in space changes. Perhaps it passes through a double slit. This affects where it is likely to be absorbed. Perhaps this counts towards "experience." $\endgroup$
    – mmesser314
    Commented Jul 11, 2020 at 22:55
  • $\begingroup$ @mmesser314 : The wave function, I feel, is best understood as a subjective element describing the information held by an external agent regarding the photon. In this case, what it's saying is where, at a given time, is the "best projected guess" it can make regarding where the photon will land, i.e. "where & when will the absorption event occur?" In this regard a wave function should properly be assigned to a space-time point, then; but I'm not sure if that's been done before and/or could work. In other words, it's the quantum version of the "trip" I just referenced. The delayed "trip" $\endgroup$ Commented Jul 11, 2020 at 23:23
  • $\begingroup$ is also a feature of classical SR; hence it must carry over somehow to the quantum. $\endgroup$ Commented Jul 11, 2020 at 23:23
  • $\begingroup$ This is like its hinting to somekind of extrinsic geometrical explanation to spacetime that we might one day construct, and thus all of these conundrums like "the point of view of a photon" will finally be intuitively understood by us. $\endgroup$
    – Nuke
    Commented Aug 9, 2022 at 17:14
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There are various ways to answer this. See Would time freeze if you could travel at the speed of light?

I agree with David Z's answer, but I would put it differently. In short, there is no view of the world as seen by a photon.

There is an argument that as the speed of an object increases with respect to me, its clocks slow down with respect to mine. A faster object is closer to catching up to a photon. A photon must be like the limit of such objects. Its clock must be stopped.

This argument is misleading. Taking the limit does not lead to anything physical. There is no photon's view of the world. That is, there is no inertial frame of reference at which a photon is at rest. Saying that time does not pass in that non-existent frame is wrong.

You can get a hint of this by looking at photons in our frame. In the classical view, light has a wavelength and frequency. Light travels, and the phase changes as it does so. This is not consistent with photons seeing our world as compressed to a plane and having no time.

The quantum view is a little different, but the same argument can be made to apply.


This is not to say that the question deals with a completely unphysical situation. Consider an atom falling into a black hole. As it crosses the event horizon, it emits a photon straight up.

You can consider the world from the point of view of that atom. But the atom is not moving together with the photon. The atom is being sucked into the black hole. Naively, the photon hovers forever from the point of view of a distant observer. What that really means it the photon gets to the observer only after an infinite time.

The observer can assign coordinates to the event horizon. It stays at a constant location in his frame. But he cannot transform to find a momentarily comoving inertial frame of reference at the horizon. He gets a coordinate singularity if he tries.

A minor point is that a photon does not just exist in a plane. It is described by a wave function that would be partly inside the event horizon, and partly outside. Thus it would have a probability of being sucked in, and a probability of escaping. Taking a long time to escape corresponds to being in a narrow region just above the horizon. Taking an infinitely long time means being in an infinitely narrow region. The probability goes to $0$ as time to escape goes up.


There is a view of the world as a Block Universe. Time does not flow in this view. This is consistent with physics. See What is time, does it flow, and if so what defines its direction? But so is a view in which time does flow.

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  • $\begingroup$ It is entirely feasible to create a reference frame where a photon stands at rest. It's just that in that reference frame, the laws of physics will not look the same any more. $\endgroup$ Commented Jul 11, 2020 at 21:32
  • $\begingroup$ @The_Sympathizer - True. I should have been more careful. There is no inertial frame of reference. $\endgroup$
    – mmesser314
    Commented Jul 11, 2020 at 21:34
  • $\begingroup$ Maybe from the point of view of a photon, its speed through space is zero, its speed through time is C BUT, the distances between all points in spacetime are zero, and time for the outside world has reached infinity. Does that make any sense? Probably not, but to ease the curious mind, that's what the photon will "see". $\endgroup$
    – Nuke
    Commented Aug 9, 2022 at 17:19
  • $\begingroup$ @Nuke - One of the counter intuitive things about relativity is that there is no point of view of a photon. It is not correct to say that point of view compresses the world to a pancake and slows time to a stop. For more (with illustrations by Escher), see A photon travels in a vacuum from A to B to C. From the point of view of the photon, are A, B, and C at the same location in space and time?. $\endgroup$
    – mmesser314
    Commented Aug 10, 2022 at 0:25
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    $\begingroup$ I don't know why the universe can't be that way. But it isn't. I posted a link yesterday that talks about it. $\endgroup$
    – mmesser314
    Commented Aug 11, 2022 at 13:52

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