- #1

user1139

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- 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)##?

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

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)##?

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