# Derivation of the Lorentz transformation

I have been reading Young's book called: University physics with modern physics and on page 1284 author states that we can derive Lorenz time transformation by eliminating two equations for Lorentz space transformation along the vector of relative speed $u$. First we eliminate them for $x$ and then for $x'$.

The equations to eliminate are:

\begin{equation} x=\gamma(x'+ut') \end{equation}

\begin{equation} x'=\gamma(x+ut) \end{equation}

After elimination we are supposed to get the following Lorentz transformations:

\begin{equation} t'=\gamma(t+ux/c^2) \end{equation}

\begin{equation} t =\gamma(t-ux/c^2) \end{equation}

There is also an article about this here. But neither in this article, neither in the Young's book the solution is shown. I have been trying to derive it myself, but failed and am now puting it as a topic on this forum.

$$x' = \gamma (x - vt)$$

$$x = \gamma (x' + vt')$$

Note the sign change of $v$ because going from $x'$ to $x$ is in the opposite direction to going from $x$ to $x'$. Rearrange equation 2 to give:

$$t' = \frac{1}{v} (\frac{x}{\gamma} - x')$$

Now use equation 1 to substitute for $x'$:

$$t' = \frac{1}{v} (\frac{x}{\gamma} - \gamma (x - vt))$$

and rearrange to give:

$$t' = \gamma \left(\frac{x}{v} \frac{1}{\gamma^2} - \frac{x}{v} + t\right)$$

and $1/\gamma^2$ is $1 - v^2/c^2$ so:

$$t' = \gamma \left(\frac{x}{v} - \frac{xv}{c^2} - \frac{x}{v} + t\right)$$

$$t' = \gamma \left(t - \frac{xv}{c^2}\right)$$

And likewise to get the expression for $t$ in terms of $t'$ and $x'$.