# Schwarzschild metric in 3-dimensional space-time

No. When you solve for the Schwartzshild metric in 4-d spacetime, you actually omit 2 of the 3 spherical coordinates, namely the polar and azimuthal angles.
By omitting coordiante I've meant your suggestion
Another option is to place $x_{3} = 0$
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In my reply I was referring to that approach.

In my reply I was referring to that approach.
Yes, but I gave another option as well.

Yes, but I gave another option as well.
You mean, to make $x_3$ invariant? But both setting it to zero and making it invariant would still keep us in 3d space (where we no longer want to be). So, I rather was calling 'omitting' either of those approaches.
Actually, I've been trying to do the same at first. But I was told that such derivation would be incorrect.

Ok, I said what I had to say. Play with (1+2)-spacetime. However, it is a mathematical exercise with no connection to Physical Reality.

Just as a sidenote. If the Ricci tensor:

$$R_{\mu \nu} = 0$$

is zero, it still does not necessarily mean space-time is flat. There is something called Riemann curvature tensor $R^{\mu}{}_{\nu \rho \pi}$, which is a 4-fold tensor. This one has to be equal to zero in order for space-time to be flat.

As you vary the dimensions, the number of independent components of the two change. In 2-d (total number of dimensions), they have the same number and are uniquely expressible between each other. There, it would mean that space-time is flat. However, in higher number of dimensions, I think there are more components to the Riemann tensor, then there are to the Ricci tensor and no such unique relationship exists.

Ok, I said what I had to say. Play with (1+2)-spacetime. However, it is a mathematical exercise with no connection to Physical Reality.

Just as a sidenote. If the Ricci tensor: $$R_{\mu \nu} = 0$$ is zero, it still does not necessarily mean space-time is flat.
Well, I've already made a disclaimer about its real-world value. Nevertheless, it might prove helpful as an analogy for a more complex 3d world.
As for the Ricci tensor, you are right: in an ordinary (4-d spacetime) Schwarzschild solution $R_{\mu \nu} = 0$ too, but the metric is not flat.

Stingray
$$ds^2 = - \rho^{4 \alpha} d t^2 + \rho^{4 \alpha (2 \alpha-1)} ( d \rho^2 + d z^2 ) + \beta^{-2} \rho^{2-4\alpha} d \phi^2 .$$
$\alpha$ can be interpreted as a mass/length and $\beta$ is an angular defect parameter.