Help in derivation of Birkhoff's theorem

[user1139]

Homework Statement

The most general spherically symmetric line element is of the form $\mathrm{d}s^2=-A(r,t)\mathrm{d}t^2+2C(r,t)\mathrm{d}t\mathrm{d}r+B(r,t)\mathrm{d}r^2+r^2(\mathrm{d}\theta^2+\sin^2{\theta}\mathrm{d}\phi^2)$.
Show that with a transformation $t\rightarrow t+f(r,t)$ for some $f(r,t)$, the line element can be written as $\mathrm{d}s^2=-A(r,t)\mathrm{d}t^2+B(r,t)\mathrm{d}r^2+r^2(\mathrm{d}\theta^2+\sin^2{\theta}\mathrm{d}\phi^2)$.

Relevant Equations

Please see above.

Using the transformation for $t$, I obtained

$$\mathrm{d}t'=\left(1+\frac{\partial f}{\partial t}\right)\mathrm{d}t+\frac{\partial f}{\partial r}\mathrm{d}r$$.

Using this equation, I substituted it into the general line element to obtain

\begin{align*}
\mathrm{d}s^2=&-A(r,t)\left(1+\frac{\partial f}{\partial t}\right)^{-2}\mathrm{d}t'^{2}-A(r,t)\left(\frac{\partial f}{\partial r}\right)^2\left(1+\frac{\partial f}{\partial t}\right)^{-2}\mathrm{d}r^2+2A(r,t)\left(1+\frac{\partial f}{\partial t}\right)^{-2}\left(\frac{\partial f}{\partial r}\right)\mathrm{d}t'\mathrm{d}r\nonumber\\
&+2C(r,t)\left(1+\frac{\partial f}{\partial t}\right)^{-1}\mathrm{d}t'\mathrm{d}r-2C(r,t)\left(\frac{\partial f}{\partial r}\right)\left(1+\frac{\partial f}{\partial t}\right)^{-1}\mathrm{d}r^2+B(r,t)\mathrm{d}r^2+r^2(\mathrm{d}\theta^2+\sin^2{\theta}\mathrm{d}\phi^2).
\end{align*}

Is the equation that is immediately above correct? How do I proceed to show that the line element can be written as $\mathrm{d}s^2=-A(r,t)\mathrm{d}t^2+B(r,t)\mathrm{d}r^2+r^2(\mathrm{d}\theta^2+\sin^2{\theta}\mathrm{d}\phi^2)$?

 

[NateD]

The equation that you have obtained is correct. To show that the line element can be written in the form $\mathrm{d}s^2=-A(r,t)\mathrm{d}t^2+B(r,t)\mathrm{d}r^2+r^2(\mathrm{d}\theta^2+\sin^2{\theta}\mathrm{d}\phi^2)$, you need to make use of the fact that the transformation $t'=f(r,t)$ is a one-to-one transformation. That is, for any given value of $t$, there is only one corresponding value of $t'$, and vice versa.This implies that we can write $t=g(t')$, where $g$ is an inverse transformation of $f$. Now, if we substitute this expression for $t$ into the equation for the line element, we will obtain\begin{align*}\mathrm{d}s^2&=-A(r,t)\left(1+\frac{\partial f}{\partial t}\right)^{-2}\mathrm{d}t'^{2}-A(r,t)\left(\frac{\partial f}{\partial r}\right)^2\left(1+\frac{\partial f}{\partial t}\right)^{-2}\mathrm{d}r^2+2A(r,t)\left(1+\frac{\partial f}{\partial t}\right)^{-2}\left(\frac{\partial f}{\partial r}\right)\mathrm{d}t'\mathrm{d}r\\&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad+2C(r,t)\left(1+\frac{\partial f}{\partial t}\right)^{-1}\mathrm{d}t'\mathrm{d}r-2C(r,t)\left(\frac{\partial f}{\partial r}\right)\left(1+\frac{\partial f}{\partial t}\right)^{-1}\mathrm{d}r^2+B(