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First some assumptions.

1) Photons travel at the speed of light. 2) From the photon's reference spacetime is contracted to 0 length in the direction of photon travel. 3) From the photon's reference it is emitted and absorbed at the same time and in the same place.

Now imagine a setup with 1 emitter and 2 detectors (E, D1, D2). E emits photons toward the center of D1. D2 is directly behind D1 in the direction in which E emits (D1 is in the line-of-site of E, D2 is hidden from E behind D1.

E -> D1 D2

From our frame of reference photons are emitted and always absorbed by D1.

From the photon's reference, aren't the emitter and detectors sitting in the same place in its perceived 2D space? Wouldn't it then be equally valid for the photon to be absorbed by D2, or both D1 and D2? Maybe neither? Does contracted spacetime maintain some sort of z-order?

It's been my understanding that when things happen doesn't remain constant between reference frames, but the effects, what actually happens does.

Is there a paradox here or am I looking at it wrong? Is it ever valid to ask what happens in an external frame of reference?

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    $\begingroup$ 1/x may approach infinity as x approaches zero, but 1/0 is undefined, not zero. $\endgroup$ – Jimmy360 Jun 18 '15 at 5:15
  • $\begingroup$ So what does that mean? That asking about the experience from a photons perspective is simply undefined/nonsensical? $\endgroup$ – Scott Jun 18 '15 at 5:29
  • $\begingroup$ Perhaps this shows my ignorance but I don't see how this question relates to the rest frame of photons in a vacuum. $\endgroup$ – Scott Jun 18 '15 at 5:34
  • $\begingroup$ Photons are not tiny hard balls. They don't "travel" the way you imagine based on a classical corpuscular model. There is no "reference frame of photons", there are only planes of equal distance, which correspond to the phase of the photon. To say that photons can't distinguish space and time is patently false, both are simply the same and they are being measured by the number of wavelengths between two points A and B. $\endgroup$ – CuriousOne Jun 18 '15 at 5:52
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    $\begingroup$ Scott, there is no frame of reference for a photon, as the linked question explains, thus your question doesn't really make sense. $\endgroup$ – ACuriousMind Jun 18 '15 at 6:08
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Let's start with your assumptions:

1) Photons travel at the speed of light.

Right. No problem with that.

2) From the photon's reference spacetime is contracted to 0 length in the direction of photon travel.

Wrong. The photon doesn't have any kind of reference frame. To appreciate why, imagine you're travelling at the speed of light. We know you can't actually do this, but just imagine it. Some would say that in your reference frame everything happens at once. But I know your course, and I can nudge an asteroid into your path: BLAM! It's game over for you. You didn't see the asteroid because you don't see anything when you're moving at the speed of light. Light would have to move faster than light for that, and it doesn't. So you don't see everything happening at once, you don't see anything. Or think anything either. You're a popsicle.

3) From the photon's reference it is emitted and absorbed at the same time and in the same place.

Wrong. Like I said, the photon doesn't have any reference frame. Its wavelength isn't length-contracted to zero either. It gets emitted, it moves through space at the speed of light with its E=hc/λ wave nature, then it gets absorbed. And these three events didn't happen at the same time. Note that I measure my time using my optical clock, which has light moving inside it. The elapsed time relates to how much the light has moved.

Now imagine a setup with 1 emitter and 2 detectors (E, D1, D2)... ...From the photon's reference, aren't the emitter and detectors sitting in the same place in its perceived 2D space?

No. It doesn't perceive anything. Just like you didn't perceive that asteroid.

Wouldn't it then be equally valid for the photon to be absorbed by D2, or both D1 and D2?

No.

Does contracted spacetime maintain some sort of z-order?

Space doesn't change one iota just because you changed your speed through space. Nor do the things in space. Instead you change, along with your measurements of space and time and things.

Is there a paradox here or am I looking at it wrong?

You're looking at it wrong I'm afraid.

Is it ever valid to ask what happens in an external frame of reference?

Yes, that's no problem. The problem comes when you think there is a frame of reference when actually there isn't. And you are not the only person who has this problem.

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  • $\begingroup$ I guess part of the confusion is the definition of "reference frame". Is sub-light speed part of the definition (or implied by the definition) of reference frame? $\endgroup$ – Scott Jun 18 '15 at 7:31
  • $\begingroup$ I understand that a photon can't "perceive" the world around it. Your answer seems to be that it's nonsensical to try to image what happens when traveling at the speed of light. $\endgroup$ – Scott Jun 18 '15 at 7:35
  • $\begingroup$ @Scott it's nonsensicial in terms of it does it make sense within SR or GR theory, yes. Nonsensical, as in the way I myself have asked questions in the past, i.e. stupid or badly researched, no. As John Rennie refers to him, his Albertness asked himself questions in the same mould as yours. $\endgroup$ – user81619 Jun 18 '15 at 10:35
  • $\begingroup$ It's simpler than you think Scott. A reference frame is an abstract thing associated with your motion and your measurements. Only when you're moving at the speed of light, you can't measure anything. It isn't nonsensical to try to imagine what happens, it's nonsensical to say you see everything when you actually see nothing. As willyW said the events are well separated and ordered. That's because the zero spacetime interval is little more than a light-clock reading. And don't blame yourself, there are celebrity cosmologists who get this stuff wrong, especially when it comes to black holes. $\endgroup$ – John Duffield Jun 18 '15 at 12:37
  • $\begingroup$ One last thing - and feel free to tell me to go away and do my own research. $\endgroup$ – Scott Jun 18 '15 at 18:40
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As John Duffield pointed out in his answer, there is no reference frame in which the photon is at rest. This is the simple reason why most parts of your statements do not make sense.

Still, there is an interesting perspective to the issue you raise: Normally one considers the events or states "photon emitted", "photon is traveling", "photon absorbed" as well separated, and, indeed, from our perspective (from the perspective of a real frame of reference), there can be a large amount of space and time between the emission and the absorption of a photon. Only the spacetime interval which separates these events is 0. So guess, some photon is emitted from a star in the Andromeda galaxy and is on the way to our planet. Isn't there plenty of time or occasion to distract this photon from its original way? Not really. Guess we want to decide how and if to detect a photon while it is on its way. That means that first we need to get the information that the photon is on its way (has been emitted and not absorbed yet) and then we need to decide if and how to detect it. Such an experiment cannot be executed since it would be necessary that the information that the photon is on its way gets to us before the photon arrives at our location. And this would require the information to travel faster than light.

So, indeed, the absorption and emission events seem to be tightly connected. There is not much which fits in between even if these events are separated by a large distance in space and time from our perspective. On the other side, there is not really a frame of reference where the order of these events is reversed. The order remains well maintained.

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The photon does not have an rest frame because that would imply it's velocity is 0, when by postulate it must be moving at C in all frames. Rather pointless to wax on about not experiencing any time or what not.

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Your question has nothing to do with the (hypothetical) photon's reference frame.

Even if the spacetime interval of lightlike movements is reduced to zero, the order of events is persisting - In the same way as when you cover one sheet of paper by another, you cannot reach the hidden sheet even if their respective distance approaches zero.

Edit: By the way, there is no zero distance: the spacetime interval is reduced to zero, but not the space interval. Remember that length contraction is only perceived by the moving frame, in this case by the (hypothetical) reference frame of the photon. But from the point of view of any other frame the distance is not zero.

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  • $\begingroup$ Doesn't the entire concept of "depth", "hidden" and covering one thing with another require a non-zero length in the given dimension? $\endgroup$ – Scott Jun 18 '15 at 7:38
  • $\begingroup$ No, see my edit. $\endgroup$ – Moonraker Jun 18 '15 at 8:21

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