Meaning of zeros in the metric tensor

Question

I'm trying to find the $g_{0i}$ components of the metric I mentioned here, but it has turned extremely difficult. My current strategy is to equate Ricci tensor components gotten from the Christoffel symbols to the cosmological constant (the result of finding $R=4\Lambda$ in $$R_{\mu\nu}-\dfrac{1}{2}Rg_{\mu\nu}+\Lambda g_{\mu\nu}=0$$ without considering $\Lambda$ as negligible).

Here it's the metric for easier consultancy:

$g_{\mu\nu}=\begin{pmatrix} -c^2\left(1-\dfrac{2GMa}{c^2r}\right) & A & B & C \\ A & a^2\left(1-\dfrac{2GMa}{c^2r}\right)^{-1} & 0 & 0 \\ B & 0 & a^2r^2 & 0 \\ C & 0 & 0 & a^2r^2\sin^2{\theta} \end{pmatrix}$

Where $A$, $B$ and $C$ are unknown functions that are time and radius dependent (see the other post for more information).

I was thinking that all three functions should be zero, but I cannot find any way to justify it. In Wikipedia they prove that for Schwarzschild's metric $g_{0i}=0$ considering the state of the body at $t$ equal to $-t$ (reversing time doesn't change the state of the body), and the only way that happens is having zeros there. But that does not work necessarily in the metric I'm referring to.

So the question is: what is the meaning of zeros in the metric tensor? What does it take to have a zero in one component? Can the $A\left(t,r\right)$, $B\left(t,r\right)$ and $C\left(t,r\right)$ functions be zero? How could we get there or why is it not necessarily the case?

Answer 1

Given a basis $e_{\mu}$ for your manifold, the metric tensor component $g_{\mu \nu}(x) $ is just $\langle e_{\mu}(x), e_{\nu}(x)\rangle$ with $\langle \cdot , \cdot\rangle$ the inner product defined on the (tangent) vector space at $x$ where the basis is defined.

In your case $g_{0i}(x)$ is the inner product of the timelike basis vector with each of the 3 spatial ones at point $x$.

To evaluate if it is sensible to take them as $0$ you need to understand if the 3 spatial vectors are always $\langle \cdot , \cdot\rangle$-orthogonal for each $x$ in your manifold.

Note: By $x$ i mean the 4-position $(x_0, \vec{x})$, not the Cartesian coordinate $x$, and in GR $\langle \cdot , \cdot\rangle$ corresponds to the Minkowsky metric tensor $\eta(\cdot, \cdot)$

Answer 2

If the cross terms are 0 then the axes are orthogonal to each other. In case of Vaidya-like metrics you can do that by setting your local set of observers to fixed spatial coordinates.