Is light the thing causing the universal speed limit to be $299\,792\,458\,\mathrm{m/s}$? So the universal speed limit would be different if light travelled faster or slower?
Or, is $299\,792\,458\,\mathrm{m/s}$ the universal speed limit anyway and light just goes that fast? Light is just something we commonly associate with it because it goes super fast.
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3$\begingroup$ Possible duplicate of: What is so special about speed of light in vacuum? $\endgroup$– John RennieCommented Jul 13, 2016 at 13:33
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1$\begingroup$ Related: physics.stackexchange.com/q/126694/2451 , physics.stackexchange.com/q/6406/2451 and links therein. $\endgroup$– Qmechanic ♦Commented Jul 13, 2016 at 13:41
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1$\begingroup$ I think it is "more right" to say the opposite, namely that the universal speed limit defines / determines the speed of light in vacuum. The later equals to the speed of causation merely because free photons are massless. Note that this could easily be otherwise, if conditions were changed, without any effect on the speed of causation... $\endgroup$– NewbieCommented Jul 13, 2016 at 18:57
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$\begingroup$ You might also wish to consider (separately or together) the question of Hubble expansion, which isn't limited in this fashion. $\endgroup$– Charles RockafellorCommented Jul 24, 2016 at 22:49
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6 Answers
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It's the second one: the reason the speed $299792458\ \mathrm{m/s} = c$ is special is because it's the universal speed limit. Light always travels at the speed $c$, whatever that limit may be.
The reason there is a "universal speed limit" at all has to do with the structure of spacetime. Even in a universe without light, that speed limit would still be there. Or to be more precise: if you took the theoretical description of our universe, and remove light in the most straightforward possible way, it wouldn't affect $c$.
There are many other things that depend on the speed $c$. A particularly important one is that it's the "speed of causality": one event happening at a particular time and place can't affect another event unless there's a way to get from the first event to the second without exceeding that speed. (This is sort of another way of saying it has to do with the structure of spacetime.)
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$\begingroup$ Could you also say how would you find the "universal limit speed" to be precisely 299792458 m/s in a universe without light? Or: how do you even discover SR in such a universe? $\endgroup$– c.p.Commented Jul 14, 2016 at 20:54
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$\begingroup$ @c.p. You can derive the value for c from Maxwell's Equations - it depends solely on the electrical permittivity and magnetic permeability of free space. $\endgroup$– user96070Commented Jul 15, 2016 at 5:47
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$\begingroup$ @Mego You don't have Maxwell's equations in a universe without light, which is the premise here. $\endgroup$– c.p.Commented Jul 15, 2016 at 7:03
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$\begingroup$ @PlasmaHH That's a smart mnemonic. I took to saying the archaic word "celerity" to myself many years ago for the same reason, but you idea is much more evocative of the right meaning. $\endgroup$ Commented Jul 19, 2016 at 23:48
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$\begingroup$ @c.p. The maximum speed could be discovered in particle accelerators (assuming these work in a universe without light). Adding more and more kinetic energy to tiny particles will result in their speed approaching some maximum. One early experiment is described here: en.wikipedia.org/wiki/… $\endgroup$– Mark HCommented Jul 27, 2016 at 2:17
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Even if nothing propagated at the speed $c$, it would still be a universal speed limit, and we could still measure it.
In fact, it's not impossible that light has a (very tiny) mass in reality. If it does, that wouldn't change anything about special relativity. It would make teaching it even more of a nightmare than it already is, because we'd have to deal with a century of textbooks and popularizations that made the mistake of calling $c$ "the speed of light", but other than that it wouldn't change anything.
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6$\begingroup$ How would you measure the speed limit if nothing propagated at that speed? $\endgroup$ Commented Jul 14, 2016 at 11:22
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6$\begingroup$ By interpolating from an asymptotic approach to the limit speed. This is quite easy to see with sub-atomic particles (electrons, protons) as you "accelerate" them. I remember reading somewhere that a particle accelerator should be called a particle ponderator, because it makes the particles heavier much more than it makes them faster! $\endgroup$– nigel222Commented Jul 14, 2016 at 15:10
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$\begingroup$ @LightnessRacesinOrbit Put is another way. We can measure time dilation for example, through increased lifetimes of unstable particles (such as muons) in particle accelerators. That phenomenon is governed by the universal signal speed limit, rather than the speed of light. So far, all results are consistent with the two being the same. Also, we might ultimately see a small, positive result for the Michelson-Moreley experiment, which give us a point on the relativistic velocity addition law. So we could work out the speed of light as a fraction of the universal signal speed limit ..... $\endgroup$ Commented Jul 19, 2016 at 23:45
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$\begingroup$ .... although in that case a positive MM result detection would be tantamount to our being able to measure the mass of the photon, which means that the photon's speed would depend on its energy, i.e. the speed of light could not be taken as a constant of nature and the experiment would give different results depending on the photon's energy. $\endgroup$ Commented Jul 19, 2016 at 23:46
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Above all, speed of light is the speed of propagation of fields through space. While light may be slowed down when crossing matter, fields (electromagnetic fields, gravity) are always propagated at c. One of the consequences is the "speed limit for causality" mentioned by DavidZ and the speed limit for transmission of information.
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7$\begingroup$ This answer makes it sound like light propagates at speed v=c/n, but gravity propagates at speed v=c (regardless of index n). Do you have any evidence of that? $\endgroup$– amICommented Jul 13, 2016 at 22:42
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$\begingroup$ @amI your comment makes no sense to me. How do you conclude that v =c/n? All Moonraker is saying is that EM fields propagate at v = c! $\endgroup$– GuillCommented Jul 24, 2016 at 22:43
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$\begingroup$ Google "v=c/n" if you don't understand refraction. I was asking Moonraker why gravity wouldn't be subject to refraction. $\endgroup$– amICommented Jul 27, 2016 at 20:20
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The numerical value of $c$ does not have any fundamental significance. Rather it is the number we get based on the experimental fact (according to the number & unit system employed) . If some alien civilization ended with some different value of $c$ compared to us. Even that is not a problem. They will reach the conclusion that this is upper bound of the speed limit for any object, provided both civilizations governed by the same set of laws of physics. In that sense the speed of light is universal.
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There is quite a bit of ambiguity in the question(s), so let me start by substituting electro-magnetic (EM) wave for "light." Then, the "universal speed limit," is the speed at which EM waves propagate in "space." The reason I use space (not vacuum), is because it is the characteristics of space ($u_o, \epsilon_o$) that determine the speed of propagation of the EM waves. If these characteristics were different, the value of EM wave propagation would be different (larger, smaller) but it would still be the universal speed limit. As you can see, the correct option is #2, and since light happens to be an EM wave, it propagates at the universal speed limit.
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Articles published in Science and Nature say the speed of light is not constant:
http://science.sciencemag.org/content/347/6224/857 "Spatially structured photons that travel in free space slower than the speed of light" Science 20 Feb 2015: Vol. 347, Issue 6224, pp. 857-860
http://www.nature.com/nature/journal/v406/n6793/full/406277a0.html Nature 406, 277-279 (20 July 2000): "...a light pulse propagating through the atomic vapour cell appears at the exit side so much earlier than if it had propagated the same distance in a vacuum that the peak of the pulse appears to leave the cell before entering it."
answered Jul 13, 2016 at 14:58
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2$\begingroup$ This answer is slightly misleading. What we usually call "the speed of light (in vacuum)" is a parameter in the equations of electrodynamics. This is the speed at which monochromatic electromagnetic ways propagate in a vacuum. If you superimpose different frequencies (your first link) or add matter (your second link), all sorts of interesting things can happen. One then has to define what one means by "the speed of light", e.g. a group velocity of a wave packet. The "speed of light" thus defined for complicated wave phenomena may vary, but that does't mean the fundamental parameter $c$ varies. $\endgroup$– jorikiCommented Jul 13, 2016 at 16:27
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1$\begingroup$ @PentchoValev basically what he's getting at is the question is asking about the universal speed limit, whereas your answer is focused on the speed of light. The two conclusions to draw are either (incorrect) the universal speed limit varies, or (correct) the speed of light can't define
cbecausecis a constant and the speed of light varies. This ambiguity is what @joriki was attempting to clear up. $\endgroup$– Qwerty01Commented Jul 13, 2016 at 20:30 -
4$\begingroup$ The phrase "speed of light" appears only in the title, not in the question body, so focussing on it as normative for the question is an error. The phrase appearing in the title is "speed of light in vacuum". The first sentence in your first reference, "That the speed of light in free space is constant is a cornerstone of modern physics.", is contrary to your use of that reference to argue otherwise. Your second reference is not in a vacuum and is therefore nonresponsive to the Question. $\endgroup$ Commented Jul 13, 2016 at 21:04
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3$\begingroup$ This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review $\endgroup$ Commented Jul 14, 2016 at 6:01
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2$\begingroup$ @LightnessRacesinOrbit Yeah, I agree. It doesn't actually address the question, which makes it a terrible answer, but it seems to have posted with the intent of answering, and it would address the question if some logical steps are filled in, which I guess puts it just within the scope of what we consider an answer on SE. I wouldn't blame anyone for flagging it, though, since it's such a marginal case. $\endgroup$– David ZCommented Jul 14, 2016 at 14:28