I must be certainly wrong, but there is something I don't understand when reading the paper on "On the Electrodynamics of Moving Bodies" from Albert Einstein. All the sources I found give the same equations as here.
I didn't check the original text, but I think a personal error is way more likely than a translation error.
In Chapter 1.3, when deriving the Lorentz transformation from the original assumptions of special relativity, the paper says:
Since tau is a linear function, it follows from these equations that $$tau=a\left(t-\frac{vx'}{(c^2-v^2)}\right)$$ where $a$ is a function $\phi(v)$ at present unknown, and where for brevity it is assumed that at the origin of $k$, $\tau=0$, when $t=0$ [...]
Then the value of $x'$ being $x'=x-vt$, the paper goes on:
If for x', we substitute its value, we obtain $$\tau = \phi(v)\beta\left(t-\frac{vx}{c^2}\right)$$ where $\beta=\frac{1}{\sqrt{1-v^2/c^2}}$
But if I do the calculus from the first equation of tau whatever I do I find:
$$\tau = \phi(v)\beta^2(t-(v/c^2)*x)$$
I don't understand. Why do I find $\beta^2$ instead of $\beta$? The calculus looks fairly simple so I don't see where I could have made a mistake. Can someone point me to the derivation of this calculus?
On my part I did like:
$$\tau=a(t-(vx'/(c^2-v^2)))$$
$$\tau=a(t-(v(x-vt)/(c^2-v^2)))$$
$$\tau=a(t(1+v^2/(c^2-v^2))-(vx/(c^2-v^2)))$$
$$\tau = a\beta^2(t-(v/c^2)x)$$
