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That the speed of light is constant for all inertial frames is the second postulate of special relativity but this does not means that nothing can travel faster than light.

  • so is it possible the point that nothing can travel faster than light was wrong?
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    $\begingroup$ That the speed of light is constant for all inertial frames is the second postulate of special relativity. $\endgroup$
    – Mark M
    Commented Aug 12, 2012 at 17:06

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There are various ways to formulate special relativity. The different approaches illustrate various different aspects of the theory so one of the tricks is to choose the formulation best suited to the question you're asking. My own favourite approach is based on the invarience of the proper time, and in fact this answers your question rather neatly.

If you think back to learning about Pythagorus' theorem, this tells you that the distance from the origin to the point in space (x, y, z) is:

$$d^2 = x^2 + y^2 + z^2$$

Special Relativity extends this idea and defines a quantity called proper time, $\tau$, defined by:

$$c^2\tau^2 = c^2t^2 - x^2 - y^2 - z^2$$

where $c$ is a constant that will turn out to be the speed of light.

The key thing about Special Relativity is that it states that the proper time is an invariant, that is all observers will calculate it has the same value. All the weird effects in SR like length contraction and time dilation come from the fact that $\tau$ is a constant.

So what about that constant $c$? Well the quantity $\tau^2$ can't be negative otherwise you can't take the square root - well, you can, but it would give you an imaginary number and this is unphysical. So suppose we let $\tau^2$ get as low as it can i.e. zero, then:

$$0 = c^2t^2 - x^2 - y^2 - z^2$$

and rearranging this gives:

$$c^2 = \frac {x^2 + y^2 + z^2}{t^2}$$

but $x^2 + y^2 + z^2$ is just the distance (squared) as calculated by Pythagorus so the right hand side is distance divided by time (squared) so it's a velocity, $v^2$, that is:

$$c^2 = v^2$$

or obviously

$$c = v$$

So that constant $c$ is actually a velocity, and what's more it's the fastest velocity that anything can travel because if $v > c$ the proper time becomes imaginary. That's why in special relativity there is a maximum velocity for anything to move. Although it's customary to call this the speed of light, in fact it's the speed that any massless particle will move at. It just so happens that light is massless.

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  • $\begingroup$ Just saw this. I think it is a bit of a tautology. Sure if proper time is zero then the velocity is the velocity of light, because it has been defined in the formula as such. But nothing in your argument identifies the left hand side c constant multiplying proper time square with the c of the right hand side? It must be postulated, imo. $\endgroup$
    – anna v
    Commented May 21, 2014 at 6:57
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    $\begingroup$ @annav: I think his point is that the postulate about the speed of light can be replaced by a postulate about the invariance of the space-time interval. $\endgroup$
    – MBN
    Commented May 21, 2014 at 7:14
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    $\begingroup$ @annav: as MBN says, I'm showing that the constancy of the speed of light can be derived from the postulate that the proper time is an invariant. After all, we have to have a postulate somewhere. The point is that the proper time invariance is a geometric property and this seems more elegant than just setting some speed to be a constant (or at least Einstein though so and who am I to disagree :-). $\endgroup$ Commented May 21, 2014 at 7:36
  • $\begingroup$ @John Rennie: I like your argument, but I don't find it entirely convincing as an argument why nothing can go faster than light. The proper time of a 4-vector or between two events is an invariant, and it does have a certain physical interpretation (as the minimal temporal separation between the events among all frames) but it remains abstract and I don't think it is obvious that it has to be a real number (rather than saying for example that beyond the speed of light the physical interpretation breaks down).... $\endgroup$
    – doetoe
    Commented May 21, 2014 at 9:19
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    $\begingroup$ @doetoe: it's a mathematical model intended as an approximate way of describing the universe, and of course experiment shows that it's a very good description. The requirement that the proper time be real, and can be zero, is part of the model. This is the best we physicists can do - we don't have the luxury of clear and well defined axioms. $\endgroup$ Commented May 21, 2014 at 9:29
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The fact that the speed of light is a maximum speed is a derived conclusion from the postulates of special relativity. it is not one of the axioms themselves.
You can Demonstrate this in a large variety of ways, the most convincing one is the fact that the energy required to create a particle and accelerate it to a speed $v$ is given by

$$E=\frac{mc^{2}}{\sqrt{1-\left(\frac{v}{c}\right)^{2}}}$$

which approaches infinity as $v\rightarrow c$.

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    $\begingroup$ This does not preclude the existence of "tachyons", i.e., particles born moving faster than light. $\endgroup$
    – Siyuan Ren
    Commented Aug 13, 2012 at 4:39
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    $\begingroup$ @KarsusRen: such things would have imaginary energies. And, like I said, there are other ways of getting at this result. A better argument against tachyons is that you can always boost to a reference frame where a tachyon is not travelling through time at all, or travelling into the past. $\endgroup$ Commented Aug 13, 2012 at 12:40
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    $\begingroup$ No, such things will not have imaginary energies but instead their rest mass (and rest energy) are imaginary. Because they can never be at rest, the imaginary nature of rest mass is not a problem. $\endgroup$
    – Siyuan Ren
    Commented Aug 14, 2012 at 6:06
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    $\begingroup$ @KarsusRen: like I said, there is a reference frame where they will sit a constant time, and ''evolve'' in space. There's no sense to be made of a dynamic particle that behaves like that. $\endgroup$ Commented Aug 14, 2012 at 12:36
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It says, "The speed of light in vacuum is constant in all inertial frames of reference (i.e. for all inertial observers)". To explain in a simple manner, "Light cannot be measured relative to any objects and is always constant in all inertial frames. In other words, Light is the maximum velocity allowed by nature. If something approaches near $c$ - Time, Length and even Mass changes.

No... Of course there's no possibility for the second principle of Special relativity to be wrong. And for this reason, it was accepted and has been in existence for nearly a century. This restriction increased the interest for Physicists to concentrate on Tachyons

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So is it possible the point that nothing can travel faster than light was wrong?

No. The "nothing can travel faster than light" restriction logically follows from the two postulates of special relativity.

I'll try to briefly show you how to get to the conclusion.

  1. First you have to convince yourself that the two postulates imply the phenomenon called the relativity of simultaneity. That is the first thing discussed in every textbook on special relativity, so I'm not getting into it.

  2. Now we use a following claim from p.1: "If one would be able to get from event A to event B only if he could move with faster-than-light speed (spacelike events). Then we can change the time order of the events A and B just by changing our reference frame." We can make A and B simultaneous, make A precede B or make B precede A -- all that just by moving to different reference frame.

  3. Now we can start a proof by contradiction. Suppose that we have some way to transmit faster-than-light signals. It then immediately follows follows from p.2 that we can transmit instantaneous signals (by making emission and reception events simultaneous) and even signals that are received before they are transmitted (by swapping the order of emission and reception events).

  4. Imagine that we have two guys $\alpha$ and $\beta$, equipped with such a spectacular communication channel. Then $\alpha$ could send a signal to $\beta$ "back in time", and then $\beta$ will return the signal to $\alpha$ instantaneously. Which means that $\alpha$ will receive his own signal from the future. Such ability instantly leads one to lots of self-contradictory situations. Hence our assumption was false.

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  • $\begingroup$ But 4. requires changing the reference frame, doesn't it? Why does this not invalidate the argument? Also it is the outside observer that perceives simultaneity etc., right? what do a and b observe? $\endgroup$
    – kutschkem
    Commented May 21, 2014 at 9:10
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From a purely theoretical point of view, the Special Relativity (SR) is based on a space-time metric $$\eta=\begin{bmatrix}+&0&0&0\\0&−&0&0\\0&0&-&0\\0&0&0&-\end{bmatrix}$$ The most general transformation to preserve metric $\eta$ is global Poincaré group which is the limit of the de sitter group with sphere radius $R\rightarrow \infty$. There is an other type of de Sitter transformation with $R \rightarrow$ finite which also leads to a special relativity theory. Basically one plays with cphoton and c.

But, keep in mind that if it is possible that SR be a finite large $R$ de Sitter transformation, it has not been experimentally confirmed, and as far as we know we can use Einstein special relativity.

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