# Meaning of zeros in the metric tensor

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?

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 $$\eta(\cdot, \cdot)$$

• And what is the physical meaning that they are orthogonal? That one does not affect the other? How would you deduce it in a non-static time metric like Vaidya metric and having $g_{0i}=0$ (as with Vaidya's)? Jul 1, 2023 at 0:35
• They have zero physical meaning. The mean absolutely nothing. Since a basis is an absolutely arbitrary choice, even though some of them are more natural, they cannot have a physical meaning. Only thing which do not depend on our human choices are physically relevant. If I can set anything I want zero and expect it to have physical significance what I'm expecting is that a choice I made on a paper affects the meaning of nature. It's clear that this is not the case Jul 1, 2023 at 0:49
• I'm sorry, i don't understand what you want to deduce from Vaidya's metric, could you be a bit more clear? :) Jul 1, 2023 at 0:50
• @Antoniou In this related MSE post we see a metric that is the identity matrix in polar coordinates but has non zero off diagonal terms and very ugly Christoffel symbols in Cartesian coordinates. To rephrase LolloBoldo's comment: The components of a metric depend on the reference frame and have zero geometrical meaning. In pseudo-Riemannian geometry (nothing else is GR) replace geometrical by physical. That's it. Jul 1, 2023 at 11:27
• Yes, that is another way of saying this. The metric tensor components tell you how the basis vector you choose overlap, nothing more. The physical meaning of it is the scalar in produces when contracted. The golden rule in physics is: index$\rightarrow$not physical Jul 1, 2023 at 13:43

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.