Given the Lagrangian density for a gravitation potential: $$\mathcal{L}(\mathbf{r},t) = -\rho(\mathbf{r},t)\,\phi(\mathbf{r})-\dfrac{1}{8\,\pi\,G}\,(\nabla\,\phi(\mathbf{r}))^2$$ how to derive the Newtons gravity law out of it?

It is demanded to use the Lagrangian for fields:

$$\sum_{\mu = 0}^3\,\dfrac{\mathrm{d}}{\mathrm{d\,x^{\mu}}}\,\dfrac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}-\dfrac{\partial\mathcal{L}}{\partial\phi} = 0$$

I were just introduced to this topic hence I am a little overwhelmed. Just plugging into the equation without giving a thought doesn't seem that intuitive, because I don't know how to deal with the expression $\fbox{$\dfrac{\partial(\nabla\,\phi(\mathbf{r})^2)}{\partial(\partial_\mu\phi)}$}$ for instance. this would require knowing $\phi(\mathbf{r})$ exactly but I'm fairy sure it should be solved without any concrete formula. According to Wiki after doing all the calculation it should collapse to: $$4\,\pi \,G\,\rho(\mathbf{r},t) = \Delta \phi(\mathbf{r},t) $$ but so far this is far in front of me. Any hints are absolutely welcome.

  • 2
    $\begingroup$ It's almost always better to express equations like this in index form, i.e. with Einstein summation the Lagrangian reads $L= -\rho\phi - \frac{1}{8\pi G} (\partial_\mu \phi)(\partial_\mu \phi)$ and then it's just a matter of applying the product rule $\endgroup$
    – Wihtedeka
    Commented Dec 4, 2021 at 14:16
  • $\begingroup$ Related: physics.stackexchange.com/q/571070/2451 $\endgroup$
    – Qmechanic
    Commented Dec 4, 2021 at 16:15


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