In his paper Über das Gravitationsfeld eines Massenpunktes nach der Einsteinsche Theorie Schwarzschild used this equation (Paper eq. (4)):

$$\frac{\partial}{\partial x_\alpha}\Gamma^\alpha{}_{\mu\nu}+\Gamma^\alpha{}_{\mu\beta}\,\Gamma^\beta{}_{\nu\alpha}=0$$ for Einstein Field Equation $G_{\mu\nu}=0$ where $\Gamma$ is the Christoffel symbol. How is this form of the equation obtained?

  • 3
    $\begingroup$ The vacuum equations can be shown to be just $R_{\mu\nu}=0$. I believe he is also demanding the metrict to have determinant equal to $-1$. In that case using the formula for the Ricci tensor you may find here en.m.wikipedia.org/wiki/List_of_formulas_in_Riemannian_geometry the equation follows. I believe it is this. $\endgroup$
    – Gold
    Commented Jul 1, 2019 at 17:04
  • $\begingroup$ I may be wrong on my history, but I think your question may amount to "Why didn't Schwarzschild use notation that hadn't been standardized at the time that he wrote his paper?" $\endgroup$
    – Brick
    Commented Jul 1, 2019 at 17:13
  • $\begingroup$ In this paper Schwarzschild develop his famous metric that solve the Einstein Field equation , so I don’t think there is any simplifications $\endgroup$
    – Eli
    Commented Jul 1, 2019 at 17:16
  • $\begingroup$ @Brick I think that the Schwarzschild Field equations are equivalent to Einstein field equations ? this is my question . $\endgroup$
    – Eli
    Commented Jul 1, 2019 at 17:22
  • 2
    $\begingroup$ @Eli Did you understand the first comment? It is the explanation. $\endgroup$
    – G. Smith
    Commented Jul 1, 2019 at 17:28

1 Answer 1


As @user1620696 in the first comment to your question pointed out, the Einstein field equations in vacuum can be shown to be just $$ R_{\mu\nu} = 0, \tag{a}$$

and in the List of formulas of Riemannian geometry - Ricci and scalar curvatures you find (among others) this formula for the Ricci curvature tensor: $$ R_{ik} = \frac{\partial \Gamma^l{}_{ik}}{\partial x^l} - \Gamma^m{}_{il} \Gamma^l{}_{km} - \nabla_k \left( \frac{\partial}{\partial x^i}\left(\log\sqrt{|g|}\right)\right). \tag{b}$$

Because Schwarzschild also demands the "determinant equation" (equation (5) in his paper) $$ |g| = -1, $$ the last summand in equation (b) vanishes, and you are left with

$$ R_{ik} = \frac{\partial \Gamma^l{}_{ik}}{\partial x^l} - \Gamma^m{}_{il} \Gamma^l{}_{km} . \tag{c}$$

Putting equations (a) and (c) together, you get $$ \frac{\partial \Gamma^l{}_{ik}}{\partial x^l} - \Gamma^m{}_{il} \Gamma^l{}_{km} = 0 $$

This is apart from naming of the indexes the same as Schwarzschild's equation (4).

  • $\begingroup$ Thank you for your answer $\endgroup$
    – Eli
    Commented Jul 1, 2019 at 19:24

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