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A thought experiment I was pondering - Say there's a photon source and two observers, one at rest and one in motion relative to the photon source.

It would seem that at the point the photon is emitted, it 'knows' where it will be observed, because it will travel at c for the particular observer.

I was thinking that this might be because the wavefunction for the photon hasn't collapsed until it is actually observed. Does that mean that the wavefunction of a photon contains the possibilities of all paths including all observers at the endpoints of each path, and is that any different to saying the wavefunction includes all paths?

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  • $\begingroup$ Why the minus 1? $\endgroup$
    – joshua.thomas.bird
    Commented Apr 29, 2018 at 22:37
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    $\begingroup$ It will travel at c for all observers, not just one particular observer. Even if that means different observers have to perceive a different distance from the point of emission to the point of absorbtion. $\endgroup$
    – The Photon
    Commented Apr 29, 2018 at 22:53
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    $\begingroup$ What's more difficult is for any observer to know when a particular photon was created. $\endgroup$
    – The Photon
    Commented Apr 29, 2018 at 23:21
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    $\begingroup$ On that oscilliscope example - wouldn't the observer be the oscilliscope? I.e the photon that hit the oscilliscope in the context of the two observers would be a virtual photon (math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/…), and the observers would be observing the photon emitted from the oscilliscope? $\endgroup$
    – joshua.thomas.bird
    Commented Apr 29, 2018 at 23:30
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    $\begingroup$ Related: Can a photon get emitted without a receiver?. $\endgroup$
    – Stéphane Rollandin
    Commented Apr 30, 2018 at 16:03

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  1. you are right, the photon will be observed to have traveled with speed c by any observer

  2. yes the wavefunction is not collapsed until observed

  3. the wavefunction is the probability distribution of the photon

  4. there is no new information created by observing the photon, because like you write, the wavefunction already has the probability of the photon being at all locations

  5. when you observe the photon, you are just mapping the probability distrbution

  6. yes as you write, in this case you could say that the wavefunction maps the paths probability distribution

  7. you are right, it does not matter if an observer is in motion or at rest, they will all see the photon travel at speed c

  8. you are thinking this the wrong way, because you think the photon moves faster then the observers. But it is the photon that moves at speed c, that is constant, and all the observers are just relative to that.

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  • $\begingroup$ All of these points make sense, but can you go into a bit more detail on 8? I'm not sure I understand the premise that the observers move relative to the photon $\endgroup$
    – joshua.thomas.bird
    Commented Apr 30, 2018 at 14:44
  • $\begingroup$ Sure. It took me a long time until I understood, but when you do, your view of the Universe will change. So at the Big bank, there was a sea of photons, that was created first as energy. Just photons, flying around. Those were all traveling at the speed c. There was no other speed, since there were no other particles, just particles with no rest mass. Now these particles were not moving in the time dimension, their speed was 0. Their spatial speed was c. Now this will sound very hard to imagine and most physicists would say this is not commonly accepted, but imagine the photon box. $\endgroup$
    – Árpád Szendrei
    Commented Apr 30, 2018 at 16:00
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    $\begingroup$ A photon box is a photon in confinement. If you want to see time flow, you need to slow down in space. Your spatial speed needs to be slower then c to start moving in the time dimension. You just have to say OK, this is how the universe works, the magnitude of the four speed vector has to be c. Now if you want to see time flow, so like we do, you need to slow down in the spatial dimensions, and since your four speed vector's magnitude stays c, your speed vector in the time dimension will compensate, and you will start moving in the time dimension. $\endgroup$
    – Árpád Szendrei
    Commented Apr 30, 2018 at 16:03
  • $\begingroup$ So after the sea of photons, there was pair creation, and baryon asymmetry, and pointlike particles were created, with rest mass. Now everything that you see as matter is built up of these, quarks, and electrons. These particles have rest mass, they move in the time dimension, so everything they build up will move in the time dimension too, so matter as you see it. Now of course the higgs field gives these rest mass, I understand, but what makes up quarks and electrons? Again, here a physicist would say there is no commonly accepted theory. $\endgroup$
    – Árpád Szendrei
    Commented Apr 30, 2018 at 16:06
  • $\begingroup$ But look at the photon box. If you imagine the photon box, why does it have rest mass? Because the photon inside is in confinement. Like photons absorbed by the electron. Or maybe quarks too are built up like that, energy in confinement. So now if the only speed at which energy can move (because permettivity and permeability makes it) is c, then imagine everything that has rest mass moving relative to the photon. $\endgroup$
    – Árpád Szendrei
    Commented Apr 30, 2018 at 16:11

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