In the case where all books try to illustrate a spherical metric, the procedure goes this way:

First they impose isotropy in terms of polar coordinates so that one can write:

$$ds^2=-A(r)dt^2 + B(r)dr^2+2F(r)drdt + D(r)r^2(d\theta^2+sin^2\theta d\phi^2)$$

They then eliminate the off-diagonal term in the metric by choosing $\psi$ to satisfy the differential equation $$\frac{d\phi(r)}{dr}=-\frac{C(r)}{A(r)}$$

The last statement is what I do not understand, how is it that this differential equation lead to the elimination of the term $F(r)dtdr$?


You just need to plug your expression for $dT^2$ back into the original metric, with the substitution $\psi' = -C/A$. This gets you \begin{align*} ds^2 &= -A(r) \left[ dT^2 - {\psi'}^2 dr^2 + 2 \frac{C}{A} dr dt \right] + B(r) dr^2 + 2 C(r) \, dr \, dt + D(r) r^2 \, d\Omega^2 \\ &= -A(r) dT^2 + \left[ B(r) + A(r) \left( \psi'(r) \right)^2 \right]dr^2 + D(r) r^2 \, d \Omega^2, \end{align*} and the cross term is eliminated as desired. One would presumably then redefine $B(r)$ to be the quantity in square brackets, and proceed from there.

  • $\begingroup$ Why is it okay for us to choose the differential equation that suits us? Why there are no restrictions on our choice? $\endgroup$ – Beyond-formulas Jun 9 '15 at 21:13
  • $\begingroup$ My total guess, open to correction, we have not defined a particular function in the metric yet, so until we do that, no problem. Does that make sense? $\endgroup$ – user81619 Jun 9 '15 at 21:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.