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I know my question sounds like a joke (and I suppose on some level it is) but I'm confounded by the following:

As the thought experiment goes, if I'm in a spaceship flying rapidly the people on earth are aging "faster" than me because relative to them my time is slower. As I approach light-speed time for me keeps slowing down.

So far so good.

But then we reach what *seems like a paradox (to me) in that if "I" (I get that physical things can't do this, but "I" can be a photon here) reach light speed then my frame = 0, but if my frame = 0 doesn't the rest of the universe around me immediately jump right to its "end of time" (whatever the heck that means).

So, am I not understanding the relationship between the way things are moving through time? Or does *nothing actually travel at the *actual (0 frame) speed of "light" including light?

[EDIT] I can tell that I'm not asking this in a way that triggers the explanation I'm seeking so I'll try it from a slightly different angle.

We conducted an experiment in which we flew atomic clocks around and they came back slightly behind the stationary clocks on earth.

This is attributed to both their speed and their distance from earth's mass.

So there's some equation by which we can calculate the rate of difference between the progress of time for me (on earth) and the progress of time for an atomic clock flying around.

So, if I understand correctly, if that atomic clock were to reach the speed of c its time would actually stop.

This seems paradoxical.

If I were a photon and I persisted at c for 1 hour of my time, how much time will have elapsed on earth? Intuitively it feels like the answer should be infinity (and further it shouldn't take an hour - it should be the instant I reach light speed and my time stops).

So, my desire is to forgo the analysis of light speed (as such) since the speed itself is not important. The only thing I'm curious about here is if time actually stops for particles traveling at light speed and if it does, then how does the surrounding universe (which is traveling forward in time) not leave those particles behind.

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  • $\begingroup$ Light travels at the speed of light. It does not have an associated frame, which means that if it were able to describe such things, it would not be able to assign numbers to your location in space or the time when you woke up this morning. That won't interfere with your ability to wake up. There are all kinds of things in this Universe that I am incapable of assigning numbers to, but they function just fine all the same. $\endgroup$
    – WillO
    Commented Feb 29, 2016 at 3:34
  • $\begingroup$ There is a fundamental difference between massless particles and massive particles that cannot be "crossed" in any way; massless particles always travel at $c$ and massive particles never travel at $c$, and since the effects of special relativity rely on a ratio of $v/c$ there is an asymptotically infinite gamma for massive particles and an undefined gamma for massless particles. In other words, massive particles and massless particles can never have the same reference frame (the same gamma with respect to another frame). $\endgroup$
    – Asher
    Commented Feb 29, 2016 at 3:39
  • $\begingroup$ @WillO, Ok, but if my thought experiment is correct then there's a problem of how it has time to "get" anywhere at all (relative to its surroundings which aren't moving at the speed of light). Consider what I'm describing. A photon "moves" at the speed of light (and is therefore stationary in time - it's not aging relative to us). So how is it that we perceive it at all? $\endgroup$
    – Yevgeny Simkin
    Commented Feb 29, 2016 at 3:40
  • $\begingroup$ @Asher, thanks... I apologize if I'm obtuse, but is it possible to explain this in some visualisation that doesn't require math for a layperson to understand? $\endgroup$
    – Yevgeny Simkin
    Commented Feb 29, 2016 at 3:42
  • $\begingroup$ @GeniaS: The photon follows a perfectly well-defined path through spacetime. You and I are perfectly capable of assigning a location and a time to every point on that path. The photon is not. But the path still exists, quite independent of who can and who can't label its points. $\endgroup$
    – WillO
    Commented Feb 29, 2016 at 3:46

4 Answers 4

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First of all, there are two things that you might want to consider:

-Speed of light is a constant and moves at a constant speed regardless of the relative speed of the source it's coming from.

-According to Special Relativity, there's a phenomenon called length contraction; which states that any moving object is shortened along the direction of motion.

So as an observer, it is theoretically impossible to reach the speed of light. Imagine this; two spaceships moving along the same direction, one of them is moving close to speed of light. When they measure the speed of light at the same time, they will get similar results. However, the ship that's moving closer to the speed of light will have covered more distance than the other, but because of length contraction, the light will reach at the same destination but the time frame will be shorter for the faster ship.

I hope this helps

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    $\begingroup$ That doesn't answer the question. $\endgroup$
    – pfnuesel
    Commented Feb 29, 2016 at 4:06
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    $\begingroup$ @Koray - it doesn't help because it has nothing to do with what I actually asked. I'm not curious about perception and the speed of light is just a patsy here - my question is about time. If time slows down as you approach the speed of light and stops when you get there, what happens to the rest of the universe relative to the light speed traveling particle? Doesn't it zip off into the "future" leaving the particle behind frozen in time? $\endgroup$
    – Yevgeny Simkin
    Commented Feb 29, 2016 at 7:41
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One lesson of special relativity is:

There is no frame of reference in which a photon is stationary.

We have this beautiful thing in special relativity. Every inertial reference frame is created equal. If you choose an inertial coordinate system, you can do your physics and calculate the evolution of the universe in that coordinate system, and everything will work out just fine. But this is a very special demand, because the coordinate system has to be mathematically well-defined.

So what are you doing when you say "the rest of the universe immediately jumps to the end of time"? You're doing a mathematical limiting procedure. As you travel in the direction of travel of a laser beam, you see the wavelength of the laser increase (redshifted), and you see things in front of you played in higher speed* (blueshifted). So in the limit as your velocity goes to $c$, yes, "everything in front of you plays out all at once". (In the form of infinite frequency light hitting your eyes!)

This sounds really dramatic, but why don't you hear physicists making statements like "the photon jumps to the end of time"?

It's because the physicist is forced to stop the limiting procedure at some point. The limit can't be achieved. It leaves you with a mathematically ill-defined "reference frame"/"coordinate system" which is not a frame of reference nor a coordinate system. You can't actually do any physics if you tried to work in this coordinate system. Since coordinate systems are just things humans impose on reality in their minds, this doesn't have any deep meaning.

*(By "see" I really mean "see". With your eyeballs. Sometimes people say "see" as in, "the mathematical positions in your coordinate system". But I mean "see" as in, the photons from whatever is in front of you are blue shifted and hit your retina at a higher frequency)

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  • $\begingroup$ I edited my question to make clearer what I'm trying to actually understand. I disagree that this question has been asked before - at least in the way in which I'm asking it. $\endgroup$
    – Yevgeny Simkin
    Commented Mar 1, 2016 at 0:36
  • $\begingroup$ @GeniaS. The way I see it my answer still stands. You want to know what a photon sees in its own frame of reference. Special relativity says that's an ill-defined question because there is no such frame of reference. You go wrong as soon as you say, "if I were a photon", because right then you presume a frame of reference exists. (And the big bold point of this post is that that frame doesn't exist!) $\endgroup$
    – user12029
    Commented Mar 1, 2016 at 1:00
  • $\begingroup$ I'm not sure what to do with that answer. But I can flip the observation and ask "how can I - who am not moving at the speed of light and therefore who is moving forward through time - observe a particle which *isn't moving forward through time? For that matter, what does it mean to not be moving forward through time??" $\endgroup$
    – Yevgeny Simkin
    Commented Mar 1, 2016 at 1:23
  • $\begingroup$ Ah! I found the answer to my question and it's here quora.com/Can-an-object-with-mass-travel-at-the-speed-of-light and (assuming that the first answere there is correct) then my assumption is correct! A conscious particle experiences the beginning of its journey and its end simultaneously with the entirety of the history of the universe passing by in a 0 time flash. $\endgroup$
    – Yevgeny Simkin
    Commented Mar 1, 2016 at 1:46
  • $\begingroup$ @GeniaS. I'm certain that, if pressed, Michael Enciso would tell you that he reached that conclusion through the limiting procedure I went through in my post. I covered all my bases! :) $\endgroup$
    – user12029
    Commented Mar 1, 2016 at 1:58
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The only way that light could travel at less than the speed of massless particles is if it has a tiny amount of rest mass. This would cause photons to be unstable meaning that they would decay over long periods of time. This would effect electromagnetism and mean that coulombs law would have to get modified for large distances meaning it would be possible to indirectly measure the rest mass of light if it has rest mass based on particular deviations in electromagnetic fields. So far there has been no detected deviation in electric or magnetic fields that would indicate that photons have rest mass which either means that photons have no rest mass or that they have a rest mass that is too small for us to detect the effects of this rest mass.

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  • $\begingroup$ thank, I edited my question to make it clearer what I'm actually asking. I'm not worried about the actual speed of light - I'm trying to get past the apparent paradox that I'm arriving at as I contemplate something ceasing to move forward through time. $\endgroup$
    – Yevgeny Simkin
    Commented Mar 1, 2016 at 0:37
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Note that light can easily travels at less than $c$... for c is the speed of light in the vacuum. In water, light travels at c/1.33 (the index of refraction).

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  • $\begingroup$ I seem to remember an explanation for why light slows down in mediums such as water being that it doesn't (i.e. it still travels at c, the only reason it appears to go slower is because it has to be absorbed and re-emitted by all the water particles as it travels). I can't remember if this interpretation is correct or not though $\endgroup$
    – QCD_IS_GOOD
    Commented Feb 29, 2016 at 7:19
  • $\begingroup$ @JoshuaLin the absorption/re-emission hypothesis is incorrect. See the responses to this question: physics.stackexchange.com/questions/11820/… $\endgroup$
    – JPattarini
    Commented Feb 29, 2016 at 7:41
  • $\begingroup$ @Joshua Lin: light in a transparent media (a) travels in the forward direction; (b) is coherent; (c) is scattered: the process is called forward coherent scattering. We know the light travels in the forward direction because it travels through the media; we know it is coherent because it can form an image cattied from the other side; we know it is scattering because it is consistent with the Huyghens model of wave propagation; that includes, as a special case, Snell's law. So the light travels at c everywhere inside the media, but the wavefront is slowed by the interference effects. $\endgroup$
    – Peter Diehr
    Commented Feb 29, 2016 at 7:52
  • $\begingroup$ ... as the interference between the source field and the reacting field produced by dielectrics (if I remind correctly). And this is also a perfect example of why it can be missleading sometime to think of light as photons (where is the "travelling photon" in the present case ? :-) ) $\endgroup$
    – Fabrice NEYRET
    Commented Feb 29, 2016 at 20:17

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