The Timelessness of a photon https://www.youtube.com/watch?v=5ELA3ReWQJY

An observer's laws of physics are time based. "When you're traveling at the speed of light, time does not exist" - Neil deGrasse Tyson. So a photon's laws of physics can not include time because it does not exist.

How does the Photon exist in "space-TIME", if Time doesn't exist for the Photon? The Photon would not exist in "space time", it would exist only in "space". The physics of "space time", which is where we are, would not apply to "space without time", where the Photon is.

  1. The rules of physics are the same for all observers in uniform motion relative to each other.
  2. The Photon is never in uniform motion relative to any or all observer.
  3. A Photon in an observer's frame is not the same as a Photon in a Photon's frame.
  4. The Photon must have "a frame" (although it is not compatible with our own), because the Photon occupies coordinates in space as a particle.
  5. The Photon's frame would not define Time.
  6. The Photon does not exist in any observer's frame, because the Photon is never in uniform motion relative to any observer, but the Photon does exist.
  7. If the Photon does not exist in any observer's frame, then the Photon must have a frame of its own that it does its existing in.
  8. The laws of physics for all Observers includes a definition for time.
  9. The Photon's frame has no time coordinates, so time coordinates from any observer's frame can not be meaningfully associated with the Photon or the Photon's frame.

Time is relative to all "observers" but light is not an observer. Time can't be meaningfully applied to light because time doesn't exist to light. Time is not relative between an observer and light because they are not both "observers".

If one physicist says that a particle is instantaneous And another physicist says the same particle takes 8 minutes to travel from the Sun to the Earth That's a contradiction 8 Minutes > Instant

All physical laws should remain unchanged under a Lorentz transformation. An observer's physical laws include time laws. A Photon's physical laws do not include time laws. All physical laws can't remain unchanged when transforming an observer's physical laws into a photons physical laws. This would cause the application of time laws to the photon, but time laws and photons are incongruent.

Assume that "space time" is a prison. In the prison there are guards and inmates. Everyone in the prison follows rules, but the guards follow one list of rules and the inmates follow a different list of rules. Some of the rules for the inmates and the guards are the same, some of the rules for the inmates and the guards are not the same. Among the inmates, all inmates must follow all of the inmate rules, no inmate receives special treatment. Also among the guards, all guards must follow all of the guard rules, no guard receives special treatment. The two lists are written in different languages. All of the inmates speak different languages, but one inmate speaks the language that the inmate rules are written in, and all of the languages that the inmates speak. All of the guards only speak the language that the guard rules are written in.

It's possible for the one inmate who can read the inmate rules, to translate those rules into all of the languages that the inmates speak. And because all of the inmates follow exactly the same set of rules, the number of rules would be the same in each of the inmate languages. It's also possible for the inmate to translate the list of guard rules entirely into the language of the inmates, and the list of inmate rules into the language of the guards.

This doesn't matter though, because the inmates and the guards still follow different rules. Writing the rules of the guard in the language of the inmate does not allow the inmate to follow the rules of the guard, and writing the rules of the inmate in the language of the guard does not subject the guard to the rules of the inmate.

The laws of physics for any observer include laws that define the behavior of time. The entire system of physics for all observers then, is inherently based on the existence of time, and time is one of the defining features of the "existence" or "reality" that all observers occupy. It is the presence of time that allows an observer's laws of physics to establish a frame of reference. Because all observers follow exactly the same list of rules, like all inmates, all of the rules of "observer physics" can be translated (by the one inmate, "man") from one observer to the next, even though all observers speak a different language, or have differing personal frames of reference.

However, the laws of physics for "light" or "the speed of light", do not recognize the existence of time. The entire system of physics for "light", is inherently without time, and this lack of time is one of the defining features of the "existence" or "reality" of "light" or, "anything 'traveling' at the 'speed' of light". It is the lack of "time", and therefore the lack of a definition between "at rest" and "moving", which is THE REASON why a photon can't be assigned a frame of reference when trying to translate the photon's physics to the observer's physics.

It only APPEARS that photons exist by the same laws of physics as an observer because some of the laws cross over from one reality to the next. It appears that the physics of photons and the physics of observers are the same because the existence of time in our reality prevents you from observing the photon in it's "natural habitat" to see how we different we really are.


closed as off-topic by ACuriousMind, Brandon Enright, Ali, Kyle Kanos, John Rennie Jul 20 '14 at 4:54

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  • 3
    $\begingroup$ You seem to be trying to reason out the meaning of relativity in words without understanding the underlying math. That is a mistake. Words are blunt tool with imprecise meaning. The math matters, because that is where all the answers are. The answers to all the question you've been asking in such awkward ways are right there on the surface of the math. Insisting that you can scramble among the words and get valid answers that contradict the well tested math is only going to annoy people. $\endgroup$ – dmckee Jul 19 '14 at 14:02
  • $\begingroup$ Haha. The question is, "Does your math, which requires time, apply to something that doesn't exist within the frame of time?" Or, "Can you apply time to something that is timeless?" $\endgroup$ – Derek Roberts Jul 19 '14 at 22:44
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    $\begingroup$ At this point you are not even wrong. Your wanting to impose your interpretation on the words that physicists use is pointless, because when we use those words they take the sense that we mean. Relativistic physics works. That's the final test from which there is no appeal. $\endgroup$ – dmckee Jul 19 '14 at 22:45
  • $\begingroup$ Relativistic physics works for all observers in uniform motion relative to each other but light is not included in that. Relativistic physics excludes light because it can't be framed so it can't be an observer. $\endgroup$ – Derek Roberts Jul 19 '14 at 22:46
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    $\begingroup$ 4. is completely false. $\endgroup$ – Kyle Kanos Jul 20 '14 at 1:39

No, it doesn't mean that.

One must distinguish two things: "laws of physics that apply to an object" and "laws of physics formulated from an object's viewpoint".

These are two different things. Laws of physics apply to all objects. And the behavior of the objects may be described relatively to many coordinate systems or "frames of reference". The special theory of relativity postulates that the form of the laws of physics is the same in all inertial systems.

The light – or anything else that moves exactly by the speed of light – doesn't have its own inertial system or frame of reference. The transformation needed to convert the data from a legitimate reference frame to the light's reference frame would be singular because the factor $$\gamma =\frac{1}{\sqrt{1-v^2/c^2}} $$ is singular (infinite) for $v=c$. So light isn't a valid "observer" from whose "viewpoint" we could describe all processes just like we do so from the viewpoint of the legitimate reference frames.

But light is still an object that behaves according to the laws of physics, and these laws are pretty much the same. In particular, light and other particles moving at the speed of light obey laws that may be obtained as the $m\to 0$ limit of the laws obeyed by massive (mass $m$) particles. For any nonzero $m$, we may actually use the massive particles' frames. It's just the strict value $m=0$, the limit, where the inertial system becomes singular.

One should point out that professional physicists sometimes do use the "light-cone coordinates" that are very similar (and could be interpreted) as the description of the world from the viewpoint of a light ray – and this coordinate system is actually very helpful and many things simplify in it. But these coordinates $x^\pm=(x^0\pm x^{3})/\sqrt{2}, x^1,x^2$ aren't organized in the same way as the usual coordinates used by a massive observer in a normal inertial system so we can't expect the exact form of the laws of physics in these coordinates to be "the same" as in the massive observers' inertial systems.

  • $\begingroup$ "So light isn't a valid "observer" from whose "viewpoint" we could describe all processes just like we do so from the viewpoint of the legitimate reference frames." Doesn't this reinforce my question? In the laws of physics, light can not be treated as anything else in this way, that it doesn't have a valid frame of reference in the current laws of physics? Or in other words, physics works for anything with a frame of reference but you will need some other system for understanding and explaining anything that doesn't have a frame of reference. $\endgroup$ – Derek Roberts Jul 19 '14 at 6:16
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    $\begingroup$ Dear Derek, I don't sufficiently understand what it means to "reinforce" your question. Instead, I tried to answer your question. Light doesn't have its own inertial system because it can't be at rest. It always moves at the speed $c$ relatively to any inertial system. This is not a contradiction with special relativity. Instead, it is one of the two fundamental postulates of special relativity, the constancy-of-speed-of-light postulate! As I have already told you, your (non-existent) "light's inertial system" isn't needed for anything in science. Light is understood without it. $\endgroup$ – Luboš Motl Jul 19 '14 at 7:10
  • $\begingroup$ It's a simple logic argument: If - The laws of physics are the same for all "observers" And - Light is "not a valid observer" Then - Why do you assume that the laws of physics are the same for light (non observer) as they are for anything else (observers)? $\endgroup$ – Derek Roberts Jul 19 '14 at 7:40
  • $\begingroup$ Dear Derek, the laws of physics have the same form for all observers - in all inertial systems. Light isn't an observer - it doesn't have any inertial system associated with it. There is no contradiction whatever in between the two sentences, is there? Light obeys the laws of physics we know but we don't formulate these laws from the light's inertial system because the latter doesn't exist. I have already answered this question of yours, now it's time for you to stop writing confused comments and read the answer. $\endgroup$ – Luboš Motl Jul 19 '14 at 7:54
  • $\begingroup$ There IS a contradiction in those two statements. The laws of physics are the same for all "observers" Light is NOT an "observer". That is the contradiction. $\endgroup$ – Derek Roberts Jul 19 '14 at 8:16

You vastly overestimate the meaning of frame.

A frame is a (local) choice of coordinates on the spacetime manifold $\mathcal{M}$. All physical laws can be directly formulated on the manifold itself, without referring to frames at all. That is at the heart of relativity, and that is what Lorentz invariance means. Let's go through your numbered points one by one:

$1.$ Partially true. The laws of physics do not require observers at all. Maxwell's equations, to take one example, are

$$ \mathrm{d}F = 0 \; \wedge \; \mathrm{d}\star F = J$$

and that is a coordinate free description of them. No mention of frames, no mention of observers, nothing. The specific way of calculating these things may be different in different coordinate system, but they do not depend on a single one of them. The laws of physics are invariant under arbitrary coordinate transformations, no matter whether the chosen frame is "inertial", "light cone" or whatever.

$2.$ True, if you ignore the part about "relative to observers". Uniform motion is defined by there being a reference frame in which the object considered is stationary, i.e. there must exist a frame whose coordinate is a multiple of the arc length of the worldline. As the arc length of the world line of a photon is $0$, such a frame cannot exist, thus the notion of uniform motion can never apply to a photon.

$3.$ False, a photon is a photon is a photon. It's description is different, but the photon itself lives on the spacetime manifold, which is indifferent to coordinate descriptions.

$4.$ False. Again, when we talk about the frame of an object, we mean the choice of coordinates in which the object is stationary. As such a frame does not exist for the photon, it has no frame.

$5.$ Meaningless, as the photon has no frame.

$6.$ False, as things exist if they have a world line. Again, this is independent of frames, since the worldline is just a map $[a,b] \rightarrow \mathcal{M}$, no frames necessary.

$7.$ Meaningless, as existence is not coupled to frames.

$8.$ False, the laws of physics when formulated on the spacetime manifold make no reference to particular choices of coordinates, especially not to those which have the arc length of a world line as a coordinate.

$9.$ Meaningless, as the photon as no frame.

One might quibble which of the things above I called false and which I called meaningless, but basically the answer to your whole confusion boils down to this: Frames (i.e coordinate choices on $\mathcal{M}$ which possess the arc length of a world line as a coordinate direction) are not the fundamental object in relativity, the insight that spacetime is a manifold endowed with the Minkowski metric is. All physics may be purely formulated on the manifold itself, without ever explicitly choosing a coordinate system. Of course, when you want to know what you will measure in any given situation, you have to choose the coordinates you use for the measurement - most often, these will be the coordinates constituting your own or any other inertial frame, but they need not be.

  • $\begingroup$ When you apply the concept of time to the photon, you say that it travels, because that is how particles behave when they co-exist with time, when time is one of their dimensions. At one moment in the time, the particle is at one location in space, and at another moment in time, the particle is at another location in space. Now imagine that there is no time though, because for you, time does not exist. You would not be able to use time to distinguish where you were, from where you are, from where you will be. $\endgroup$ – Derek Roberts Jul 19 '14 at 23:32
  • $\begingroup$ How does the Photon exist in "space-TIME", if Time doesn't exist for the Photon? The Photon would not exist in "space time", it would exist only in "space". The physics of "space time", which is where we are, would not apply to "space without time", where the Photon is. $\endgroup$ – Derek Roberts Jul 19 '14 at 23:35
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    $\begingroup$ @Derek Roberts: "spacetime" is just a 4D manifold $\mathcal{M}$, some of whose 3D submanifolds you may call "space" and some of whose 1D submanifolds you may call "time" (though these choices are not unambiguous, and not every 3D submanifold is "space" and not every 1D submanifold is "time"). It is not simply "space" + "time", that's the entire point of relativity. You are constantly trying to argue with words whose meaning is simply not what you think it is. $\endgroup$ – ACuriousMind Jul 19 '14 at 23:40
  • $\begingroup$ You're still in the box. All of what you just described is the reality where time exists, but that is not the reality where light exists, because light and time don't coexist. $\endgroup$ – Derek Roberts Jul 19 '14 at 23:41
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    $\begingroup$ I cannot decide if you are making fun of me, high, or simply that obstinate. Either way, this conversation is over. $\endgroup$ – ACuriousMind Jul 19 '14 at 23:46

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