The standard textbook approach in Newtonian gravitational dynamics is to derive the particle dynamics using the particle Lagrangian:

$$L = T-V = \frac 1 2 m \dot x_u \dot x_u -m\phi(x_u)\tag{1}$$

With the corresponding Euler-Lagrange equation taking the variation w.r.t. $x$:

$$m\ddot x_u=\frac{ \partial\phi}{ \partial x_u}\tag{2}$$

or $$\textbf{F} = m \ \textbf{a};\tag{3}$$ and, derive the Poisson field equation using the Lagrangian density:

$$ \mathfrak L = -\rho \phi - \frac {1}{8\pi G} (\partial_u\phi)(\partial_u\phi)\tag{4}$$

which is substituted in the Euler-Lagrange equation taking the variation w.r.t. $\phi$:

$$\frac{\partial\mathfrak L}{\partial\phi}-\frac{\partial}{\partial x_u}(\frac{\partial\mathfrak L}{\partial\phi_u})=0\tag{5}$$


$$-\rho +\frac {1}{4\pi G}\nabla^2\phi=0 \ \ \ or \ \ \ \nabla^2\phi=4\pi G\rho.\tag{6}$$

I have several questions about this.

First, is it essential to use two different Lagrangians? What is the physical basis and meaning of having two Lagrangians? Is it a rigorous approach?

Second, is it possible to use one Lagrangian density that encompasses both. Something like:

$$\mathfrak L = -\rho(\phi-\frac 1 2 m \dot x_u \dot x_u)-\frac {1}{8\pi G} (\partial_u\phi)(\partial_u\phi)\tag{7}$$

and taking variation w.r.t. to $x$ and $\phi$. This sort of works if, in the variation w.r.t. $x$, the

$$\frac{\partial}{\partial x_u}\tag{8}$$

only acts on the first term in parenthesis, giving $F = ma$, but this does not seem right because both $\rho$ and $\partial_u\phi$ are also functions of $x$ and would add additional terms.


3 Answers 3

  • It is important to mention that in this model the point particle masses $m_i$ serve as sources $$ \rho({\bf x},t)=\sum_i m_i \delta^3({\bf x}-{\bf x}_i(t)) $$ for the gravitational potential field $\phi$.

  • The full Lagrangian
    $$L=T_1-V_2-V_3$$ for both the point particle positions ${\bf x}_i(t)$ and the field $\phi({\bf x},t)$ consists of

    1. a kinetic term $T_1=\frac{1}{2}\sum_i m_i \dot{\bf x}_i^2(t)$.

    2. an interaction term $V_2=\sum_i m_i \phi({\bf x}_i(t),t)$.

    3. a potential energy term $V_3=\int_{\mathbb{R}^3}\! d^3{\bf x}~{\cal V}_3({\bf x},t)$, where ${\cal V}_3=\frac{1}{8\pi G}(\nabla\phi)^2$.

  • It is possible to integrate out the $\phi$ field to obtain a purely point-mechanical Lagrangian, cf. my related Phys.SE answer here.

  • $\begingroup$ I am confused by this last comment. I assume that you then get the EOM eq, by taking the x variation of the full Lagrangian. But, when you do that, won't the partial of del(phi) wrt to x contribute an extra, unwanted term? $\endgroup$
    – David
    Commented Jun 26, 2022 at 15:12
  • 2
    $\begingroup$ No. The space coordinates ${\bf x}$ and the point particle positions ${\bf x}_i(t)$ are different. $\endgroup$
    – Qmechanic
    Commented Jun 26, 2022 at 15:21
  • $\begingroup$ Thanks, that distinction along with the next comment are illuminating. $\endgroup$
    – David
    Commented Jun 26, 2022 at 20:00

First, is it essential to use two different Lagrangians? What is the physical basis and meaning of having two Lagrangians? Is it a rigorous approach?

No, it is not essential to use two different Lagrangians. In my opinion it is just a conceptional step before merging these two partial Lagrangians into a single common Lagrangian.

Second, is it possible to use one Lagrangian density that encompasses both.

Yes, you can derive the equations of motion for $N$ particles (with masses $m_i$ at positions $\mathbf{x}_i(t)$, for $i=1, ... N$) and for the gravitational field ($\Phi(\mathbf{x},t)$, for $\mathbf{x}\in\mathbb{R}^3$) from a common Lagrangian. However, we need to carefully distinguish between the discrete positions $\mathbf{x}_i$ of the particles and the continuous position $\mathbf{x}$ of the field.

The complete Lagrangian looks like this:

$$L[\mathbf{x}_i,\dot{\mathbf{x}}_i,\Phi,\nabla\Phi]= \sum_{i=1}^N \left(\frac{1}{2}m_i\dot{\mathbf{x}}_i(t)^2 -m_i \Phi(\mathbf{x}_i(t),t)\right) -\frac{1}{8\pi G}\int d^3 x\ (\nabla\Phi(\mathbf{x},t))^2 \tag{1}$$ It has 3 components:

  • a kinetic energy for each particle,
  • an interaction energy between each particle and the gravitational field,
  • the energy of the gravitational field.

We can define the density field made up by the particles as $$\rho(\mathbf{x},t)=\sum_{i=1}^N m_i\ \delta^3(\mathbf{x}-\mathbf{x}_i(t)) \tag{2}$$ which has Dirac-like peaks where the particles are.

Using this density we can rewrite the Lagrangian (1) as:

$$L[\mathbf{x}_i,\dot{\mathbf{x}}_i,\Phi,\nabla\Phi]= \sum_{i=1}^N \frac{1}{2}m_i\dot{\mathbf{x}}_i(t)^2 +\int d^3 x\underbrace{\left( -\rho(\mathbf{x},t)\Phi(\mathbf{x},t)-\frac{1}{8\pi G}(\nabla\Phi(\mathbf{x},t))^2 \right)}_{\mathcal{L[\Phi,\nabla\Phi]}} \tag{3}$$

From Lagrangian (1) or (3) we get the equations of motion.

From Lagrangian (1) the Euler-Lagrange equations with respect to $\mathbf{x}_i(t)$ give Newton's law: $$\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{\mathbf{x}}_i}\right) =\frac{\partial L}{\partial\mathbf{x}_i}$$ $$m_i\ddot{\mathbf{x}}_i=-m_i\nabla\Phi(\mathbf{x}_i) \tag{4}$$

And from Lagrangian (3) the Euler-Lagrange equation with respect to $\Phi(\mathbf{x},t)$ gives Poisson's equation: $$\nabla\left(\frac{\partial{\mathcal{L}}}{\partial(\nabla\Phi)}\right) =\frac{\partial\mathcal{L}}{\partial\Phi}$$ $$-\frac{1}{4\pi G}\nabla^2\Phi=-\rho \tag{5}$$

  • $\begingroup$ Wow! This was just what I was looking for. Thank you. $\endgroup$
    – David
    Commented Jun 26, 2022 at 20:02

The first Lagrangian is the Lagrangian for the motion of a point particle in a Newtonian gravitational potential (or field), whereas the second is the Lagrangian for the gravitational potential (or field), the latter which eventually leads to the Newtonian field equation. So, actually there is no real relationship. The only connection one might think of is that it is about Newtonian gravity. But the equation of motion (EOM) of a particle is fundamentally different from the field equation (FE). In the first (EOM) the motion of a single particle is searched for in a given known Newtonian gravitational potential (or field) --- a problem of mechanics, whereas in the second, an unknown Newtonian gravitational potential is searched for in a given mass density distribution --- a problem of field theory.

Also, the combined Lagrangian makes not much sense: in particular in the first phrase of the post $\phi(x)$ is considered as potential, whereas in the "combined" Lagrangian, it has the function of potential energy.

Lagrangians with different variables to be varied, actually exist. But I wonder if a Lagrangian where, one varies in the first part of the coordinates and their time derivatives and in the second part of the field components, and their spacial derivatives, makes any sense?

By the way, the correct EOM (even in Newtonian gravity) should be

$$\ddot{x}_u = -\frac{\partial \phi}{\partial x}$$

So, according to Newton, the coincidental equality of inertial and gravitational mass, can have the mass factored out completely.

This can indeed be found by varying the coordinates and their time derivates in the first Lagrangian:

$$L = \frac{1}{2}m \dot{x}_u \dot{x}_u - m \phi(x_u)$$

where $\phi$ being the potential that is not equal to the potential energy $\phi \neq V$.

  • 2
    $\begingroup$ Thank you for that detailed answer. Just a follow up. If one writes the particle Lagrangian in terms of the density instead of the mass: L=−ρ(ϕ−1/2 v˙^2) then when you take the x variation, I assume you must ignore the x dependence of ρ to get the correct EOM (a=-partial(phi)/partial(u). Why can this x dependence by ignored, or, is it incorrect to write Lagrangian in terms of density? $\endgroup$
    – David
    Commented Jun 25, 2022 at 14:29
  • 2
    $\begingroup$ @David, upon varying the particle Lagrangian, the EOM of a point particle is searched for. In that case it would not make much sense to use a mass density. Because the mass is the mass of the point particle. May be it is possible, but it would be a delta-function $\rho(x) = \delta(x-x_u)$. To try this would lead to a couple of computational complications and I guess for no benefit. Consider the particle as a test particle in the gravity field. It samples locally the field. Actually there is no genuine mass distribution in play because the potential is already given, i.e. known. $\endgroup$ Commented Jun 25, 2022 at 15:21
  • 1
    $\begingroup$ Again, thanks so much. I was interested in this question because Mordehai Milgrom, in his derivation of Q-Mond [link] (academic.oup.com/mnras/article/403/2/886/1184577) includes the 1/2 v^2 terms in the general Lagrangian density and derives F=ma using that. But, from your comments, I now realize that actually need two Lagrangians and this is just a short cut that, presumably, the reader should be aware of. $\endgroup$
    – David
    Commented Jun 25, 2022 at 18:45
  • $\begingroup$ Well, I would have preferred to know from the beginning that you are actually interested in the MOND modification of Newtonian gravity, because this changes the perspective on gravity. However, I argued from the perspective of Newtonian gravity. Unfortunately I am not a MOND expert, and I dont' have the time to read the paper. $\endgroup$ Commented Jun 25, 2022 at 21:24
  • $\begingroup$ Sorry for confusion. I thought it would complicate things to ask the question in relation to the Mond paper. But, I do not think anything you said does not apply to the Q-Mond case. I am convinced that the 1/2 v^2 term does not belong in the general Langrangian density relation (as it is in eq. 3 in Milgrom's paper) that is used to derived the potential, but rather should simply be in the different EOM Lagrangian as you discussed. $\endgroup$
    – David
    Commented Jun 25, 2022 at 22:07

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