The argument with [math]\sqrt{1-\frac{v^2}{c^2}}[/math] in the denominator isn't bad, but it's purely mathematical and it doesn't really explain why this is in the denominator in the first place. I'll try to give an approach that is a bit more physical, but it will require a pretty long story, and it will also require a lot of math
For the sake of simplicity, let's limit ourselves to a 2-dimensional spacetime - with 1 spatial dimension and 1 time dimension.
So, let's say we have 2 observers, one moving with velocity [math]v[/math] with respect to the other. Each of these observers can assign his own coordinates to events - let's say one of them has coordinates [math](t,x)[/math] and the other one has [math](t',x')[/math]. Let's also say [math](0,0)[/math] is the same event in both coordinate systems - for example, it's the moment and place where they met. We want to find out how to transform one set of coordinates to the other one.
We make one simple assumption - that spacetime is homogeneous and isotropic. This just means that no point is better than other points, and that no direction is better than other directions. As can be shown, this implies that the transformation between coordinates must be linear, or, if you prefer, it can be expressed with a matrix:
[math]\left[ \begin{array}{c}
t' \\
x'
\end{array} \right] = \left[ \begin{array}{cc}
A(v) & B(v) \\
C(v) & D(v)
\end{array} \right] \left[ \begin{array}{c}
t \\
x
\end{array} \right][/math]
Let's call the matrix [math]L(v)[/math].
Now, what can we say about A, B, C and D?
For those, who don't want to follow the maths - just skip to
large bold text
First, let's notice that if we change coordinates in such a way:
[math]x \to -x[/math][math]x' \to -x'[/math][math]v \to -v[/math]then it's just reversing the x axes of both observers, so:
[math]\left[ \begin{array}{c}
t' \\
-x'
\end{array} \right] = \left[ \begin{array}{cc}
A(-v) & B(-v) \\
C(-v) & D(-v)
\end{array} \right] \left[ \begin{array}{c}
t \\
-x
\end{array} \right][/math]
This leads us to a conclusion, that:
[math]A(-v) = A(v)[/math][math]B(-v) = -B(v)[/math][math]C(-v) = -C(v)[/math][math]D(-v) = D(v)[/math]
Now let's note that [math]L^{-1}(v) = L(-v)[/math] (converting coordinates to an observer with velocity [math]v[/math] is an opposite transformation to converting coordinates to an observer with velocity [math]-v[/math] - this is the principle of relativity). So, we get:
[math]\left[ \begin{array}{cc}
A(-v) & B(-v) \\
C(-v) & D(-v)
\end{array} \right] = \frac{1}{A(v)D(v)-B(v)C(v)} \left[ \begin{array}{cc}
D(v) & -B(v) \\
-C(v) & A(v)
\end{array} \right][/math]
So:
[math]\frac{D(v)}{A(v)D(v)-B(v)C(v)} = A(-v) = A(v)[/math]and:
[math]\frac{A(v)}{A(v)D(v)-B(v)C(v)} = D(-v) = D(v)[/math]
This implies that:
[math]A(v) = \pm D(v)[/math][math]AD - BC = \pm 1[/math]
The transformations with minus change the orientation of spacetime. They are just as good as the transformations with plus, but again for the sake of simplicity let's limit ourselves to just those with plus. We get:
[math]L(v) = \left[ \begin{array}{cc}
A(v) & B(v) \\
C(v) & A(v)
\end{array} \right][/math][math]A^2 - BC = 1[/math]
Now, again some physics. Since the second observer moves with velocity [math]v[/math] with respect to the first, he satisfies:
[math]x = vt[/math]Points on this line in his coordinates will correspond to [math]x'=0[/math] (because he to himself is always at x'=0). We get:
[math]\left[ \begin{array}{c}
t' \\
0
\end{array} \right] = \left[ \begin{array}{cc}
A(v) & B(v) \\
C(v) & A(v)
\end{array} \right] \left[ \begin{array}{c}
t \\
vt
\end{array} \right][/math]So:
[math]0 = Ct + Avt[/math]Hence:
[math]C = -Av[/math]
So now we have:
[math]L(v) = \left[ \begin{array}{cc}
A(v) & B(v) \\
-vA(v) & A(v)
\end{array} \right][/math][math]A^2 + ABv = 1[/math][math]B = \frac{1-A^2}{Av}[/math]
[math]L(v) = \left[ \begin{array}{cc}
A(v) & \frac{1-A(v)^2}{vA(v)} \\
-vA(v) & A(v)
\end{array} \right][/math]
Now comes the hardest part.
Let's call our (t,x) observer O, and our (t',x') observer O'. Now let's say we have an observer O'' with coordinates (t'',x'') moving with velocity u relative to O'. What is the velocity V of O'' relative to O?
Well... If O'' moves with u relative to O', then it satisfies [math]x' = ut'[/math]. Relative to O, it will satisfy [math]x = Vt[/math]. So:
[math]\left[ \begin{array}{c}
t \\
Vt
\end{array} \right] = L(-v) \left[ \begin{array}{c}
t' \\
ut'
\end{array} \right][/math](We transform coordinates from O' to O.)
We get:
[math]t = \left( A(v) + \frac{A(v)^2-1}{vA(v)}u \right) t'[/math][math]Vt = A(v) \left( v+u \right) t'[/math]
So:
[math]V = \frac{v+u}{1 + \frac{A(v)^2-1}{A(v)^2} \frac{u}{v}}[/math]
Still following me? Now let's reverse the situation
O moves with -V relative to O'', and with -v relative to O'. So, it satisfies [math]x' = -vt'[/math] and [math]x'' = -Vt''[/math]. We get:
[math]\left[ \begin{array}{c}
t'' \\
-Vt''
\end{array} \right] = L(u) \left[ \begin{array}{c}
t' \\
-vt'
\end{array} \right][/math](We transform from O' to O''.)
Hence:
[math]t'' = \left( A(u) + \frac{A(u)^2-1}{uA(u)}v\right) t'[/math][math]-Vt'' = -A(u)(v+u)t'[/math]
So we get:
[math]V = \frac{v+u}{1 + \frac{A(u)^2-1}{A(u)^2}\frac{v}{u}}[/math]
Okay... We have two expressions for V, which are supposed to be equal... So:
[math]\frac{A(u)^2-1}{A(u)^2}\frac{v}{u} = \frac{A(v)^2-1}{A(v)^2}\frac{u}{v}[/math]
Divide both sides by uv:
[math]\frac{A(u)^2-1}{u^2 A(u)^2} = \frac{A(v)^2-1}{v^2 A(v)^2}[/math]
Left-hand side depends only on [math]u[/math], right-hand side only on [math]v[/math]. This means they are both equal to some constant [math]\alpha[/math]:
[math]\frac{A(v)^2-1}{v^2 A(v)^2} = \alpha[/math]This gives:
[math]A(v) = \frac{1}{\sqrt{1-\alpha v^2}}[/math]
So, we can write our
complete transformation:
[math]L(v) = \left[ \begin{array}{cc}
\frac{1}{\sqrt{1-\alpha v^2}} & \frac{-\alpha v}{\sqrt{1-\alpha v^2}} \\
\frac{-v}{\sqrt{1-\alpha v^2}} & \frac{1}{\sqrt{1-\alpha v^2}}
\end{array} \right][/math]
Now, the point is - what exactly is [math]\alpha[/math]? There are two ways to go about it:
1. We know from theory of electromagnetism, that [math]c[/math] should be independent of the observer. We can plug that in and we'll get [math]\alpha = \frac{1}{c^2}[/math].
2. From this we have some law of adding velocities. We can use that to perform an experiment - measure the speed of light in water, and then the speed of light in moving water. Then we can see how the velocities of light and water add up and calculate [math]\alpha[/math]. We will again get the right result.
Actually there was a physicist, Fizeau, who performed this experiment and got results which would be consistent with special relativity - except that special relativity didn't exist yet
Ok, but why did I do all this? I want to show something nice. Let's introduce [math]\eta[/math] such that:
[math]v = c \tanh \eta[/math]
"tanh" is hyperbolic tangent, [math]\eta[/math] is called "rapidity". It's nice, because then:
[math]\left[ \begin{array}{c}
ct' \\
x'
\end{array} \right] = \left[ \begin{array}{cc}
\cosh \eta & -\sinh \eta \\
-\sinh \eta & \cosh \eta
\end{array} \right] \left[ \begin{array}{c}
ct \\
x
\end{array} \right][/math]
Probably it reminds some of you of the 2D rotation matrix:
[math]\left[ \begin{array}{c}
x' \\
y'
\end{array} \right] = \left[ \begin{array}{cc}
\cos \phi & -\sin \phi \\
\sin \phi & \cos \phi
\end{array} \right] \left[ \begin{array}{c}
x \\
y
\end{array} \right][/math]
And quite correctly, because it behaves in a similar way. Just as rotations don't change [math](\Delta x)^2+(\Delta y)^2[/math], these "hyperbolic rotations" don't change [math](c \Delta t)^2 - (\Delta x)^2[/math]. This value can be then interpreted as a kind of "length" in the spacetime.
Ok, but what is the connection to [math]c[/math] as the speed limit? Well, imagine a body going at a velocity [math]v = \frac{\Delta x}{\Delta t} < c[/math]. Let's calculate the "length" of the path it went along:
[math]d^2 = (c \Delta t)^2 - (\Delta x)^2 = (c^2-v^2)(\Delta t)^2 > 0[/math]But this means that for every observer this will be greater than 0, because it can't change when we change the observer. So it means that always we will get [math]v<c[/math]. [math]c[/math] appears to be the speed limit.
There is one "but". We assumed, that the body moved slower than c relative to one observer and this implied that it moves slower than c relative to every observer. If we assumed that it was going faster than c for one observer (because why not, it wouldn't give a contradiction), it would appear to be going faster than c relative to every observer. So actually, the conclusion is:
nothing can cross the barrier of c - but it doesn't exclude objects that are always faster than c.
I wonder if somebody will actually read this whole post
