# Determinant and inverse of metric tensor in Eddington-Finkelstein coordinates

I need to make sure I'm not going crazy here. We define the metric of Eddington-Finkelstein (EF) coordinates $$(v,r, \theta,\phi)$$ as $$g=-\left(1-\frac{2m}{r}\right)dv^2+2dvdr+r^2d\Omega^2$$ where $$d\Omega^2$$ is the usual metric on the sphere. My book claims $$\text{det} g=-r^4\sin^2\theta,$$ but I think this is off by a factor of 4. This is true in the usual Schwarzchild coordinates, but I don't think this is true in EF coordinates. One can see this by writing out $$g$$ as a block matrix with off-diagonals $$0$$. Namely, $$g=\begin{bmatrix} -\left(1-\frac{2m}{r}\right)&2&0&0\\ 2&0&0&0\\ 0&0&r^2&0\\ 0&0&0&r^2\sin^2\theta \end{bmatrix}$$ Calling the top left block $$A$$ and the bottom right block $$B$$, we have $$\det g= \det A \cdot\det B=-4\cdot(r^4\sin^2\theta).$$ Normally I'd let this go, but then the author goes on to use fact nontrivially but stating the inverse metric is $$g^{uv}\partial_u\partial_v=2\partial_v\partial_r+\left(1-\frac{2m}{r}\right)\partial_r^2+r^{-2}\partial_\theta^2+r^{-2}\sin^{-2}\theta\partial_\phi^2.$$ It seems to me that the correct form should be $$\frac{1}{2}\partial_v\partial_r+\frac{1}{4}\left(1-\frac{2m}{r}\right)\partial_r^2+r^{-2}\partial_\theta^2+r^{-2}\sin^{-2}\theta \partial_\phi^2.$$ So, my question is are my compuations of the determinant and inverse metric correct, or is the author's correct?

• Books do make mistakes, but not as often as students do. Many GR students make this mistake the first time they encounter a non-diagonal metric; I certainly did. Commented Sep 18, 2022 at 20:39
• @Ghoster I fell victim to one of classic blunders
– user344261
Commented Sep 18, 2022 at 20:50

You got the off-diagonal matrix elements of $$g$$ wrong.
The line element $$ds^2=-\left(1-\frac{2m}{r}\right)dv^2 + 2dv\ dr+r^2(d\theta^2+\sin^2\theta \ d\phi^2)$$ can be written as $$ds^2= \begin{bmatrix} dv & dr & d\theta & d\phi \end{bmatrix} \begin{bmatrix} -\left(1-\frac{2m}{r}\right) & \color{red}{1} & 0 & 0 \\ \color{red}{1} & 0 & 0 & 0 \\ 0 & 0 & r^2 & 0\\ 0 & 0& 0 & r^2\sin^2\theta \end{bmatrix} \begin{bmatrix} dv \\ dr \\ d\theta \\ d\phi \end{bmatrix}$$ Notice that the off-diagonal matrix elements need to be $$g_{rv}=\color{red}{1}$$ and $$g_{vr}=\color{red}{1}$$.
Together they contribute $$g_{rv}dr\ dv + g_{vr}dv\ dr = 2\ dv\ dr$$ to the line element $$ds^2$$.