This is probably a naive question and I'm missing something really simple. The Schwarzschild solution has been constructed in consideration of the following requirements:

  1. The field equations $ \frac{\partial \Gamma_{\mu\nu}^\alpha}{\partial x^\alpha}+\Gamma_{\mu\beta}^\alpha \Gamma_{\nu\alpha}^\beta=0. $
  2. The coordinate condition $Det(g_{\mu \nu}) = -1.$
  3. All the metrical components are independent of time $x^4$.
  4. The equations $g_{\rho 4}=g_{4\rho}=0$ hold exactly for $\rho=1,2,3$.
  5. The solution is spatially symmetric with respect to the origin of the coordinate system in the sense that one finds again the same solution when $x_1, x_2, x_3$ are subjected to an orthogonal transformation (rotation).
  6. The $g_{\mu\nu}$ vanish at infinity, with the exception of the following four limits different from zero: $$g_{44}=1,~g_{11}=g_{22}=g_{33}=-1.$$

The problem is to determine a line element with coefficients such that the 1) ... 6) are satisfied, with the exception of the point $x_1=x_2=x_3=0$, the location of a point mass, where the requirements are undefined.

Now, it appears to me that the line element $$ds^2=(dx^4)^2-(dx^1)^2-(dx^2)^2-(dx^3)^2$$ satisfies the requirements 1) ... 6). Then, what condition or requirement prevents the line element from being like above, everywhere in the vicinity of a point mass?

  • $\begingroup$ The Einstein Field Equations formulated for a point mass, maybe? $\endgroup$ Dec 22, 2022 at 8:19
  • $\begingroup$ If you mean, 'EFE at the location of the point mass', then one may argue that Schwarzschild doesn't take that into consideration anyhow. Yet he deduces his line element. My confoundment is - this means there should be an additional requirement beyond 1) ... 6). $\endgroup$
    – John Doe
    Dec 22, 2022 at 8:29
  • $\begingroup$ Related: How do we know the Schwarzschild solution contains an object of mass $M$? $\endgroup$
    – Qmechanic
    Dec 22, 2022 at 9:23

2 Answers 2


It doesn't match the Newtonian limit. The Minkowski metric is, in a certain sense, a particular case of Schwarzschild (namely $M = 0$), but it does not recover the Newtonian result.

The usual derivation of the Schwarzschild solution would include, after your six points, finding the most general solution possible, which has an undetermined constant. This constant is typically fixed by imposing that it should match the Newtonian prediction very far away from the mass.

Notice that none of your points makes reference to the actual mass of the point mass, and hence there surely is something missing.


The famous Schwarzschild solution describes only that part of spacetime where stress energy tensor is zero. However, solution of Einstein field equations (EFE), the metric $g_{\mu\nu}$, should describe the whole spacetime, like in example https://physics.stackexchange.com/a/679431, see equations 7,8,9 and 10.

Missing in your six points are the two (EFE is second order ODE) physical boundary conditions, expressed as pressure and energy density on the surface where vacuum begins. Setting boundary pressure to zero, $p(R)=0$, ensures the asymptotic flatness of the solution, and $\varepsilon(R)$ defines the constant in Schwarzschild metric. In my understanding, Einstein’s general relativity has no need for Newton's theory to fix its constants. (see https://physics.stackexchange.com/a/707944).


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