Why does v = 0 in the Lorentz Transformation equation?

[xwolfhunter]

So I've been reading Einstein's theory of relativity, and at one point when discussing the Lorentz equations' proof that light remains constant, he just states it without mathematically doing it. Probably because it wasn't the super scientific version (?) but I wanted to see how he did it, so I did it, but the only way it works out to exactly what it needs to be is if $v=0$. So here's the math:
$$x^1=\frac{(c-v)t}{\sqrt{1- \frac{v^2}{c^2}}}$$
$$t^1=\frac{(1-\frac{v}{c})t}{\sqrt{1-\frac{v^2}{c^2}}}$$
This is after you state that $x=cv$ and then substitute it into the Lorentz equations. So then you solve for $t$, and get $t=\frac{\Big(\sqrt{1-\frac{v^2}{c^2}}\Big)t^1}{(1-\frac{v}{c})}$. Substituting $t$ into the other equation, and simplifying it, I end up with:
$$x^1=\frac{t^1(c-v)}{1-\frac{v}{c}}$$
The only way it simplifies to $x^1=ct^1$ is if, as previously stated, $v=0$. And I'm sure it does, but I just cannot figure out why. First I thought $v=c$, since that was what was being measured, but then I discarded it, since mathematically it made $x^1=0$. If somebody could briefly explain why $v=0$, I would greatly appreciate it!

Thanks!

 

[Bill_K]

So here's the math:
$$x^1=\frac{(c-v)t}{\sqrt{1- \frac{v^2}{c^2}}}$$
$$t^1=\frac{(1-\frac{v}{c})t}{\sqrt{1-\frac{v^2}{c^2}}}$$

Multiply the second equation by c, and the right-hand side will be exactly the same as the right-hand side of the first equation, showing immediately that ct¹ = x¹.

 

[xwolfhunter]

Aha, that's beautiful! Thanks!

 

[Dale]

$$x^1=\frac{t^1(c-v)}{1-\frac{v}{c}}$$
The only way it simplifies to $x^1=ct^1$ is if, as previously stated, $v=0$.

Or$$x^1=\frac{t^1(c-v)}{1-\frac{v}{c}}$$$$x^1=\frac{c\;t^1(c-v)}{c\;(1-\frac{v}{c})}$$$$x^1=\frac{c\;t^1(c-v)}{(c-v)}$$$$x^1=c\;t^1$$