How can we find the Lorentz Factor using these equations?

Question

---

$$ct'=\gamma(ct-vt)$$ $$t'=\gamma\Bigg(t-\dfrac{\gamma^2-1}{\gamma^2v}x\Bigg)$$

I'm trying to understand how did the textbook derivate the Lorentz Factor just using these two equations.

Answer 1

We consider the transformations of Lorentz assuming that the lengths transverse to the direction of motion are the same.

\begin{cases} x'=\gamma ( x-\beta c t)\\ y'=y\\ z'=y\\ ct'=\gamma (c t-\beta x) \end{cases}

Being that $\beta=v/c$ (speed parameter) we have:

$1/\gamma^2=1-\beta^{2}$, equivalent to $$\beta^{2}=1-\frac{(\gamma^2-1)}{\gamma^2}=\frac{(\gamma-1)(\gamma+1)}{\gamma^2}$$ or $$\frac{\gamma-1}{\beta^{2}}=\frac{\gamma^2}{\gamma+1} \iff \beta^2=\frac{\gamma^2-1}{\gamma^2} \tag 1$$

After using the identity $(1)$ you have:

$$t'=\gamma t-\gamma\frac\beta c x= \gamma t-\gamma \frac{\beta}c\frac v v x=\gamma t-\frac{\gamma\beta^2}{v}x=\gamma t-\frac{\gamma \dfrac{\gamma^2-1}{\gamma^2}}{v}x=\gamma\Bigg(t-\dfrac{\gamma^2-1}{\gamma^2v}x\Bigg)$$