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I know Einstein was great and all. Why is it that exactly at the speed of light is where infinite energy is required to accelerate any object with mass? Is it simply because the math of relativity checks out and explains most of everything? Are there any physicists who disagree with Einstein's theory?

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marked as duplicate by John Rennie, DavePhD, Kyle Kanos, Jim, Alfred Centauri Jun 20 at 15:04

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The answer to most "why" questions in physics is: because that's what we observe in nature. –  pfnuesel Jun 20 at 8:15
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The point is not that speed of light itself is the limit ; it could be another particle it would not matter that much. The point is that there has to be a limit (otherwise modern physic would be pretty much messed up to the ground). –  Max Jun 20 at 12:44
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@Max Well, it has to be another massless particle. An it has to be sufficiently long ranged that it's speed can be measured and that leaves ... well, the graviton once (or if) we're ever equipped to detect the fool thing. –  dmckee Jun 20 at 14:19
    
Maybe you should add "speed of light in vacuum". See en.wikipedia.org/wiki/Cherenkov_radiation –  boucekv Jun 20 at 14:43

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For the non-physicists amongst us, Brian Greene gives a non-mathematical but intuitively satisfying explanation in 'The Fabric of the Cosmos': In a loose sense, when we are at rest, we are moving through time at the speed of light. As we start to move through space, what we are actually doing is diverting more and more of that motion to a spatial direction rather than a time direction, but the combined velocity remains constant. Thus we cannot exceed the speed of light, because at that point, we have diverted the entirety of our motion to the space direction. (I'm paraphrasing, but I believe I've caught the gist)

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Not bad +1. And you're quite right in a precise sense too: the velocity four vector does indeed have a constant "length" of $c$: the four velocity's components are $(c,\,v_x,\,v_y,\,v_z)\,\gamma$ where $\gamma = \sqrt{1-(v/c)^2}^{-1}$. The length of a four vector $x_0,\,x_1,\,x_2,\,x_3$ is $\sqrt{x_0^2-x_1^2-x_2^2-x_3^2}$, so you can explore Brian Greene's explanation a bit with these ideas. –  WetSavannaAnimal aka Rod Vance Jun 20 at 13:24

Let's assume, for argument's sake, that the Galilean transformation holds rather than the Lorentz transformation.

Then your questions would become

Why is infinite speed arbitrarily the limit? Why is it that exactly at infinite speed is where infinite energy is required to accelerate any object with mass?

I suspect that you wouldn't, in fact, think of asking such questions since they almost answer themselves.

Moreover, infinite speed would be an invariant speed - a speed that is measured to be the same in all reference frames - since (loosely speaking) $\infty + v = \infty$.

If we ask the question "what if there is a finite invariant speed", the mathematical answer is the Lorentz transformation where $c$ is the finite invariant speed.

Indeed, if we let $c \rightarrow \infty$ in the Lorentz transformation, we recover the Galilean transformation.

From this perspective, the result that accelerating to the invariant speed requires infinite energy and is thus impossible, doesn't seem so odd.

To summarize, without postulating that the speed of light is invariant, one can derive the form of the Lorentz transformations from just the principle of relativity. In this form, there is an undetermined finite invariant speed.

That light propagates at the invariant speed is then simply an empirical fact rather than a postulate.

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The problem is that the paper you cite assumes v in one ref. frame and −v in the other ref. frame, i.e. v'=-v. However, time dilatation and length (distance) contraction means that the two variables change in inverse proportions, i.e. if t' is larger than t then (absolute) x' is smaller than x. Now, the postulated constancy of the value of the velocity requires that v=x/t and v'=x′/t′=-v, and therefore x/t=-x′/t′. As t is always less than t', then x' must be in direct proportion to t' (in absolute numbers), and therefore larger than x. Which contradicts Lorentz transform equations. –  bright magus Jun 20 at 16:02

Why is it that exactly at the speed of light is where infinite energy is required to accelerate any object with mass? Is it simply because the math of relativity checks out and explains most of everything?

To say that the math checks out is the wrong way of putting what's going on because it separates the math and the explanation for the math. Einstein had explanations for why he picked the equations he picked. That is, he had an account of what was happening in reality to bring those outcomes about and without that explanation the math wouldn't have any relevance to physics. Einstein's explanation involved ideas about what kinds of measurements are allowed by the laws of physics, see:

http://www.fourmilab.ch/etexts/einstein/specrel/specrel.pdf

Are there any physicists who disagree with Einstein's theory?

General relativity is the best available explanation of gravity and the structure of spacetime. There is a problem with the theory in the sense that it is not a quantum theory of gravity. There is no reason to think that its replacement will involve going back to anything that resembles Newtonian mechanics.

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General Relativity (and a subset of it, Special Relativity) simply work, and have coherently explained the world at the macroscopic level. GR (and QM too) don't really offer a deterministic mechanism behind the observed effects (e.g., what exactly is gravity?), so they are unsatisfying in that sense (as well as not yet being unified). If you care to disagree with Dr Einstein's work, you'd better have a theory that explains everything just as well, and then goes beyond GR to explain other things (like QM). There's plenty more left to be done in physics! –  Phil Perry Jun 20 at 14:16

It's not actually. It's not like light has some special status in the Universe. It's just that there is a maximum speed, and light, among other things, tends to get very close to that limit. In practice light is slowed down by its environment, and there was even speculation at one point, that light mad a bit of mass and so neutrinos might move a tiny bit faster. This was retracted, but it goes to show that the thought is possible.

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So when do we get slow-glass? –  JAB Jun 20 at 13:53
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Light (any EM radiation) moving through a vacuum moves at c, not close to it. Through any non-vacuum, it slows down to c/IR (IR = index of refraction). I've never heard of any speculation that photons had actual rest mass (because it would be infinite mass at c.) The jury is still out on neutrinos, although it appears they may have a tiny rest mass. –  Phil Perry Jun 20 at 14:21
    
@PhilPerry I believe ThomasAhle may be referring to the retracted "faster than light" neutrinos in Italy, that turned out to be due to a faulty cable. –  gerrit Jun 20 at 14:56
    
@Phil There is no doubt that neutrinos (or at least all but the lightest masses states of...) have mass. The jury reported years ago. We even know a fair amount about the differences of the squares of the masses and there is a limit of the sum of the masses, but that leaves considerable room for uncertainty about the actual values. –  dmckee Jun 20 at 15:11
    
Yes, but I'm really just trying to say that relativity wouldn't fall on light not moving exactly at c, which seems to be the idea Derp is having. We just call the limit 'light speed' because it also happens to be so. –  Thomas Ahle Jun 20 at 15:13

Is it simply because the math of relativity checks out and explains most of everything?

Yes, the equations of special relativity have been tested against data innumerable times and have never been falsified.

Are there any physicists who disagree with Einstein's theory?

As far as I know, no.

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Well, I showed you before that this page hyperphysics.phy-astr.gsu.edu/hbase/relativ/muon.html presenting the so-called muon experiment actually invalidates SR. Time dilatation and length (distance) contraction means that the two variables change in inverse proportions, i.e. if Δt' is larger than Δt, than Δx' is smaller than Δx (because Δt′=Δtγ and Δx′=Δx/γ). This page claims, however (tables by the 4th and 5th figure) that Δt' is larger than Δt and Δx' is larger than Δx (afaik, what this page shows is accepted by mainstream). There couldn't be a better experimental falsification. –  bright magus Jun 20 at 16:04

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