The $M$ is Schwarzschild metric comes from Newtonian limit. But in the Newtonian limit the total energy also reduces to $M$. Is there any particular reason for using mass instead of total energy? Doesn't it feel unnatural that total energy is used in stress-energy tensor, but not in Schwarzschild metric? And yes, I understand that Schwarzschild is a Ricci-Flat solution.
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Doesn't it feel unnatural that total energy is used in stress-energy tensor, but not in Schwarzschild metric?
Neither energy nor mass are particularly natural for the Schwarzschild metric. The natural quantity is length, specifically the Schwarzschild radius, $R$. In terms of $R$ the metric is $$ds^2=-\left(1-\frac{R}{r}\right)c^2 dt^2 + \left(1-\frac{R}{r}\right)^{-1}dr^2 + r^2 (d\theta^2+\sin^2\theta \ d\phi^2)$$ It makes sense that length, $R$, is the most natural parameterization, since the metric describes the geometry of spacetime.
It is always possible to take a simple length parameter and obtain a mass parameter by multiplying the mass parameter by some multiple of $G/c^2$ to get the length. Similarly you could multiply energy by a multiple of $G/c^4$. The net result would be a length that is dressed up to look like a mass or an energy, but under all the dressing it is still a length.
Dale
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One may show that the total (ADM) energy of a Schwarzschild black hole (at rest) is $E=Mc^2$.
For more details, see e.g. my related Phys.SE answer here.
Qmechanic
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The short answer is the famous equation $$ E=Mc^2 $$ There's no meaningful distinction between energy and mass in this context.
In slightly more words, the Schwarzschild metric is stationary, so the only source of energy is the mass of the black hole. In fact $T_{\mu\nu}=0$ everywhere in the Schwarzschild spacetime since it is a vacuum solution, so the "mass" of the black hole is measured by looking at its effect on the spacetime far away from the black hole.
Andrew
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