In this wiki page we can read:

The Lagrangian density for Newtonian gravity is:

$$\mathcal{L}(\mathbf{x},t)= - \rho (\mathbf{x},t) \Phi (\mathbf{x},t) - {1 \over 8 \pi G} (\nabla \Phi (\mathbf{x},t))^2,$$

while in this other wiki page we can see:

The simplest classical field is a real scalar field — a real number at every point in space that changes in time. It is denoted as $ϕ(x, t)$, where $x$ is the position vector, and $t$ is the time. Suppose the Lagrangian of the field, $L$, is

$$L = \int d^3x\,\mathcal{L} = \int d^3x\,\left[\frac 12 \dot\phi^2 - \frac 12 (\nabla\phi)^2 - \frac 12 m^2\phi^2\right]. $$

I assume both expressions of Lagrangian density must agree(?). I can define $m=4 \pi G \rho$, but even after that there are two important differences:

  1. term in $\dot{\phi}$ not present in first equation.

  2. term in $\rho \phi$ of first equation seems to be $m^2 \phi^2$ in second one.

Must these two equations agree? if they must, how to pass from one to the other?

Related: Lagrangian potential for Newtonian gravity but curiously it talks about an expression not currently found in wiki page.

  • 1
    $\begingroup$ Those are two completely different lagrangians, as is stated in their names. Is there a reason you believe they refer to the same object? $\endgroup$
    – Triatticus
    Commented Sep 20, 2020 at 20:06
  • 2
    $\begingroup$ The two Lagrangians have nothing to do with each other, other than both involving (different kinds of) scalar fields. $\endgroup$
    – G. Smith
    Commented Sep 20, 2020 at 20:19

2 Answers 2


The first Lagrangian density $$\mathcal{L}(\mathbf{x},t)= -\rho(\mathbf{x},t)\Phi(\mathbf{x},t)-\frac{1}{8\pi G}(\nabla\Phi(\mathbf{x},t))^2$$ describes the gravity potential $\Phi(\mathbf{x},t)$ (a scalar field) in the presence of the density $\rho(\mathbf{x},t)$ (another scalar field).

The second Lagrangian density
(I have added the $(\mathbf{x},t)$ dependencies for more clarity)
$$\mathcal{L}(\mathbf{x},t)= \frac{1}{2}\dot\phi(\mathbf{x},t)^2-\frac{1}{2}(\nabla\phi(\mathbf{x},t))^2 - \frac 12 m^2\phi(\mathbf{x},t)^2$$ describes a single particle by a scalar field $\phi(\mathbf{x},t)$. Here $m$ is not a field, but just a constant number (the mass of the single particle).

These are two entirely different phenomena. Hence there is no reason to expect any similarity between the two.


These actions are logicaly different.

First Lagrangian describe nonrelativistic scalar field: $$ \mathcal{L}(\mathbf{x},t)= - \rho (\mathbf{x},t) \Phi (\mathbf{x},t) - {1 \over 8 \pi G} (\nabla \Phi (\mathbf{x},t))^2 $$

Second Lagrangian describe relativistic scalar field: $$ L = \int d^3x\,\mathcal{L} = \int d^3x\,\left[\frac 12 \dot\phi^2 - \frac 12 (\nabla\phi)^2 - \frac 12 m^2\phi^2\right] $$

First action can be deduced not from second action, but as non-relativistic Newtonian limit of Einstein-Hilbert action coupled to matter if one will use metric in leading order (see for example David Tong: Lectures on General Relativity, section 5.1):

$$ ds^2 = -(1+2\Phi)dt^2 + (1-2\Phi) \, d\mathbf{x} \cdot d\mathbf{x} $$


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