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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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  • $\begingroup$ > "But in the Newtonian limit the total energy also reduces to M" How do you mean? That total energy of the system is $Mc^2$? This does not seem particularly connected to any Newtonian limit, it's just the relation between rest energy and mass in relativity. $\endgroup$ Commented Nov 15, 2023 at 12:10

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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.

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    $\begingroup$ Then it reduces the question to why Schwarzschild radius $R_s = 2M$ instead of $R_s= 2E$. $\endgroup$
    – Nayeem1
    Commented Nov 15, 2023 at 5:22
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    $\begingroup$ @Nayeem1 there is no “instead of”. It is both $R=2M$ and $R=2E$ in natural units. That is what I meant by “you could multiply energy by a multiple of $G/c^4$” (in conventional units). It is naturally a length scale, but you can change it into either a mass scale or an energy scale if you like with the right choice of constants $\endgroup$
    – Dale
    Commented Nov 15, 2023 at 11:21
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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.

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  • $\begingroup$ Sorry, I am not much familiar with ADM energy, looks like it has applications in quantum gravity. If we bound ourselves to pure general relativity, does it answer my question? How? $\endgroup$
    – Nayeem1
    Commented Nov 15, 2023 at 3:57
  • $\begingroup$ ADM is classical GR; not quantum. $\endgroup$
    – Qmechanic
    Commented Nov 15, 2023 at 4:00
  • $\begingroup$ Ok then, can you please explain how it infers that we should better use M instead of $E= \gamma M$? $\endgroup$
    – Nayeem1
    Commented Nov 15, 2023 at 4:07
  • $\begingroup$ @Nayeem1 the black hole is not moving so $\gamma=1$ $\endgroup$ Commented Nov 15, 2023 at 4:26
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    $\begingroup$ @Nayeem1 The parameter $M$ arises in the context of the Schwarzschild coordinate chart, in which the event horizon sits at $r=2M$. I suspect that you are being confused into thinking that coordinate charts correspond to the reference frames of some observers; this is not (generally) so in GR. In particular, it's not immediately clear what $\gamma$ would refer to here, since nothing in this chart is changing with (Schwarzschild) time. $\endgroup$
    – J. Murray
    Commented Nov 15, 2023 at 6:44
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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.

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  • $\begingroup$ No. That famous equation refers to relativistic mass in old notation. Now it is $\gamma m$, and the equation is $E = \gamma m$, taking c = 1. $\endgroup$
    – Nayeem1
    Commented Nov 15, 2023 at 3:53
  • $\begingroup$ So, for an infalling coordinate system (like Gullstrand–Painlevé) M is replaced by $\gamma M$? $\endgroup$
    – Nayeem1
    Commented Nov 15, 2023 at 6:19
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    $\begingroup$ @Nayeem1 Well I intended to apply it to a stationary black hole so these subtleties don't matter, but the way I would interpret $E=mc^2$ is in terms of the invariant mass, so the full equation for a moving particle is $E^2=m^2c^4 + p^2c^2$. The invariant mass of the black hole is the relevant quantity that determines the geometry; a boosted black hole spacetime is a coordinate transformation of the stationary case. $\endgroup$
    – Andrew
    Commented Nov 15, 2023 at 23:03
  • $\begingroup$ @safesphere I'm sorry but I have no idea what you're talking about. "In general relativity, a vacuum solution is a Lorentzian manifold whose Einstein tensor vanishes identically. According to the Einstein field equation, this means that the stress–energy tensor also vanishes identically, so that no matter or non-gravitational fields are present." - wikipedia $\endgroup$
    – Andrew
    Commented Nov 16, 2023 at 17:09
  • $\begingroup$ The statement in my answer is a tautology. Since the definition of a vaccum solution is that $T_{\mu\nu}=0$, it's equivalent to say that $T_{\mu\nu}=0$ since it is a vaccum solution or it is a vacuum solution since $T_{\mu\nu}=0$. I don't see that your objection raises any substantive physics issue, it just seems like semantics to me, I am sorry to say. But if you feel the need to downvote over that then of course feel free and I'll give that the weight I think it deserves. $\endgroup$
    – Andrew
    Commented Nov 16, 2023 at 17:10

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