Suppose the observer $O'$ is moving to the right with speed $v$. If $x$ denotes the distance of an object from $O$, 1) then the distance between the object and $O'$ as measured by $O$ will be $x-vt$. But using the formula for length contraction derived above, namely $\Delta x'=\gamma\,\Delta x$, we see that \begin{equation} x' = \gamma\,(x-vt) \label{xlorentz} \end{equation}

By Postulate I, the framework used by each observer to describe the other must be the same. In particular, if we interchange the roles of $O$ and $O'$, nothing else should change — except for the fact that the relative velocity (of $O$ with respect to $O'$) is now $-v$ instead of $v$. By symmetry, we therefore have immediately that \begin{equation} x = \gamma\,(x'+vt') \end{equation} where we have been careful not to assume that $t'=t$. In fact, comparing these equations quickly yields \begin{equation} t' = \gamma \left(t-\vcsq x\right) \label{tlorentz} \end{equation} and a similar expression for $t$ in terms of $x'$ and $t'$.

Equations ($\ref{xlorentz}$) and ($\ref{tlorentz}$) together give the Lorentz transformation from the frame of the observer at rest ($O$) to the frame of the moving observer ($O'$). By symmetry, we can invert these expressions simply by replacing $v$ with $-v$.