Birkhoff's theorem proof of step

Question

In E. Poisson's book "A relativist's toolkit" he says we can set $\psi$ to $0$ without off of generality and i can't work out why.

Picture attached.

Answer 1

The equations imply $\partial_r\psi=0$ so that $\psi$ is a function only of $t$. Then, the metric takes the form $$ ds^2 =- e^{2\psi(t) } f dt^2 + \cdots $$ We can now redefine the coordinate $t$ so that $$ dt' = e^{\psi(t)} dt $$ Then, $$ ds^2 =-f dt'^2 + \cdots $$ We can now remove the prime superscript. We then retrieve the original metric with $\psi = 0$.

In other words, the author means that $\psi$ can be set to zero by choosing an appropriate time coordinate.