# Lagrangian of Newtonian gravity

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.

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

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.

• Thanks for your answer. This time wiki has made things really difficult to understand. Some cites from second expression chapter "m is a real parameter (the "mass" of the field)", "Suppose the Lagrangian of the field, L, is ", "the equations of motion for the field". No where the word "particle". – pasaba por aqui Sep 21 at 18:29

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):

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