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Since photons move at $c$, do they experience time or distance? If they don't, doesn't this explain action at a distance? From the point of view of the photons, there is no time, so the action at a distance must happen instantly. It only seems weird from our point of view.

Qmechanic
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devhl
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  • Photons don't move, at all. They are units of energy exchange between quantum fields. What "moves" are the fields and they move at the speed of light for the massless case. That these things don't happen instantly is a tested fact. That radiation doesn't form an observer system has little to do with that, though. – CuriousOne Jul 25 '16 at 02:00
  • @CuriousOne I agree these things don't happen instantly but why does it make more sense to take a single photon oscillating 600 trillion times per second and changing it into 600 trillion waves passing by for second? Isn't that much more complicated and harder to explain? – Bill Alsept Jul 25 '16 at 02:23
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    @BillAlsept: A photon doesn't oscillate. the field does. A photon is just a small amount of energy and angular momentum. When electromagnetic fields interact, they can only do it by means of one of these small units. A rapidly oscillating field exchanges more energetic photons than a slowly oscillating field does. There is nothing complicated about this, people are just not reading the instructions that come with quantum field theory. – CuriousOne Jul 25 '16 at 02:31
  • You're just passing one solution off for another one that's far more complicated. A photon coming from a distant star makes far more sense than trillions and trillions and trillions of waves coming from that star just to make One photon when it gets here. There needs to be something traveling from that start to here. As you would say can you write your name on that field ? I've never seen one. I still don't see why you need to introduce the wave to explain it. – Bill Alsept Jul 25 '16 at 03:05
  • Based on the duplicate listing, it seems there are two opinions. – devhl Jul 27 '16 at 00:16
  • Based on the duplicate listing, it seems there are two opinions: time is zero for protons; it is meaningless to ask the question since no reference frame can be assigned. If there is no time, then the bigger picture I'm trying to get at is that we can explain action at a distance and the delayed choice experiment with a shockingly simple answer. If the only problem is the frame of reference, can it be so simple as there isn't one due to time and space being points? I'm not sure I understand what a reference frame is, but I can look it up. – devhl Jul 27 '16 at 00:22

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There is no "point of view of the photons", you can't attach a frame of reference to them. Best if you imagine the photons as waves, and as these waves are propagating with $c$.

The classical time dilatation, length contraction formulas (you know, everywhere the $\frac{1}{\sqrt(1-\frac{v^2}{c^2})}$ in them) are defined only for macro-sized objects going below $c$.

peterh
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  • As an object moves closer to c, time slows down; so why is it that at c the time dilation suddenly stops? It seems like the formula was only written for matter with mass. Perhaps matter without mass has different formulas. – devhl Jul 25 '16 at 01:37
  • @devhl Yes as you near $c$, the time dilation nears to $0$. But the same formulas would also show that the mass of the photon is $\frac{0}{0}$. Btw, photons are microscopic things and their formulas aren't so easy and beautiful. Quantumelectrodynamics plays with them. In it, the photons are described as the "waves" of a quantum field, and these waves are propagating with c. – peterh Jul 25 '16 at 01:49
  • It is not so bad, the problem is that there are no such easy formulas in the QFT as in the SR. But understanding them on a basic level is not so hard if you know derivation of many-dimensional functions, and complex numbers, and matrices. – peterh Jul 25 '16 at 01:58
  • Since photons have no mass, isn't it a moot point that the mass dilation is undefined? I still feel like there should be time dilation involved to the most extreme. – devhl Jul 25 '16 at 02:06
  • @devhl I think you feel it well, but every theory is only an approximation and has its limits. On my opinion, your thoughts reached the limits of the Special Relativity and hit the barrier to the QFT (quantum field theory). :-) – peterh Jul 25 '16 at 02:16
  • @devhl Here I've wrote an answer to somebody saying that the length of the photon is zero because of the length contraction. – peterh Jul 25 '16 at 02:19
  • If they do not experience time or distance, then the action at a distance is explained, as well as the delayed choice slit experiment. It seems too elegant to not be true. In fact, I wondered if Einstein himself thought of this as an answer. – devhl Jul 25 '16 at 02:24
  • @devhl Huhh, it is too phylosophical to me. I only played with the math of the SR and the QM on an enthusiast level. This is what I suggest also to you, despite the common concept, only to understand what the formulas say is not so hard as it seems. The main problem as enthusiasts with glowing eyes are talking with scientists, that the enthusiasts are using (and thinking in) phylosophical terms, while the scientists see what it means on the language of the formulas. – peterh Jul 25 '16 at 02:30
  • @devhl It is also very useful to read the old posts of big names here. It is also very useful if you buy some QM, SR, GR books, most Universities have some specific book store dedicated for them. Learning them makes very enjoyable that you don't need to pass exams from them. :-) Unfortunately, wikipedia is not so good, because it continously uses terms whose meaning it forgets to explain. – peterh Jul 25 '16 at 02:39