# Spherical Symmetric Metrics

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.