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

 

[sardar]

Dear reader,

Thank you for sharing your thoughts and calculations on the Lorentz Transformation equation. It is always important to question and understand the mathematical reasoning behind scientific theories.

In the Lorentz Transformation equation, v represents the relative velocity between two frames of reference. This means that v is the speed at which one frame of reference is moving relative to the other. Therefore, if v=0, it means that the two frames of reference are not moving relative to each other. In other words, they are at rest with respect to each other.

In the context of Einstein's theory of relativity, this is significant because it shows that the speed of light is constant in all inertial frames of reference, regardless of their relative motion. This is a fundamental principle of the theory and has been experimentally verified.

I hope this helps to clarify why v=0 in the Lorentz Transformation equation. Keep questioning and exploring the world around you!

Best regards,
A scientist