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Suppose we could control gravitation inside a wire frame cube with a side length of $l$ at will (I assume some yet unknown technology). First we simulate empty space inside and when measuring how long a light beam takes to go through it is $t_0=l/c$.

Now we create a homogeneous gravitational field inside the cube. My hunch is that this would be a constant stress energy tensor field inside the cube. Again we measure the time the light beam takes through the cube and as far as I understand that would be $t_g = b\cdot l/c$ for $b>1$, meaning it takes longer than before, i.e. in coordinate time the light is slowed down.

Question: Will the $b$ inevitably be the same between any two pair of sides of the cube or could there be some directional dependence despite the constant stress energy tensor field?

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The coordinate speed of light is indeed directionally dependent in general. However, that has nothing to do with gravity or stress energy tensors. It is all about the coordinates chosen.

For example, in flat Minkowski spacetime you can use Anderson’s coordinate system described on p 105 here https://www.sciencedirect.com/science/article/abs/pii/S0370157397000513?via%3Dihub

For this coordinate system the metric is $ds^2 = - dt^2 - 2 \kappa \ dt dx + (1-\kappa^2) dx^2 + dy^2 + dz^2$ and the coordinate speed of light is $1/(1-\kappa)$ in the $+x$ direction and $1/(1+\kappa)$ in the $-x$ direction.

Conversely, in a spherically symmetric vacuum spacetime with spacetime curvature, you can use isotropic coordinates $$ds^2=-\left( \frac{1-m/2r}{1+m/2r} \right)^2 dt^2 - \left( 1+m/2r \right)^4 \left( dx^2 + dy^2 + dz^2 \right)$$ where the coordinate speed of light is isotropic.

So whether or not the coordinate speed of light is isotropic or not is not dependent on curvature or anything. It is simply an artifact of the chosen coordinate system.

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  • $\begingroup$ Coming back here I notice I asked the wrong question. I should not have talked about "coordinate speed", but just whether b is the same, inevitably, in all directions. But I should probably ask a new question. $\endgroup$
    – Harald
    Commented Apr 20, 2023 at 19:12

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