Does the Newtonian gravitational field have momentum analogous to the Poynting vector?

Question

We can define the total energy of the electromagnetic field as:

$$\mathcal{E}_{EM}= \frac{1}{2} \int_V \left(\varepsilon_0\boldsymbol{E}^2+\frac{\boldsymbol{B}^2}{\mu_0}\right)dV$$

which satisfies the conservation law:

$$\frac{\partial\mathcal{E}_K}{\partial t} + \frac{\partial \mathcal{E}_{EM}}{\partial t} = 0$$

where $\mathcal{E}_K$ is the kinetic energy of the matter. In addition, we can define the electromagnetic momentum as the Poynting vector:

$$\boldsymbol{p} = \frac{1}{\mu_0}\boldsymbol{E}\times\boldsymbol{B}$$

For the newtonian gravitational field $\boldsymbol{g}$, it seems possible to define an gravitational energy as:

$$\mathcal{E}_{G}= \frac{1}{2} \int_V \frac{\boldsymbol{g}^2}{4\pi G}dV$$

satisfying the corresponding conservation law. However, it does not seem that from the field alone it is possible to construct a gravitational analogue to the Poynting vector Is there any intuitive idea why this would not be possible?

Answer 1

No, in Newtonian mechanics there is no gravitational momentum. Physically, the reason is that the gravitational field doesn't propagate: it responds instantaneously to changes in the matter distribution, so that there is no trackable movement of energy from one place to another.

Mathematically, this is related that there are no time derivatives of $\phi$ in the equations of motion or the Lagrangian, unlike in electromagnetism. This is because Newtonian gravity is really the very low speed limit of general relativity - it's like working in an approximation where charges move so slowly that the electric field is quasistatic and there's no magnetic field, and thus no Poynting vector.

Answer 2

The Lagrangian for Newtonian gravitation (ignoring the kinetic energy of the sources, which will not affect this argument) is $$ \mathcal{L} = - \frac{1}{8 \pi G} (\nabla \phi)^2 - \phi \rho $$ where $\phi$ is the gravitational field and $\rho$ is the density of the source matter. It is an easy exercise to confirm that this yields the Newtonian gravity equation $\nabla^2 \phi = 4 \pi G \rho$.

The above Lagrangian is invariant under spatial and temporal translations, so it will have a set of associated Noether currents that form a canonical stress-energy tensor: $$ T_\mu {}^\nu = \left( \frac{\delta \mathcal{L}}{\delta (\partial_\nu \phi)} \right) \partial_\mu \phi - \delta_\mu {}^\nu \mathcal{L}. $$ Specifically, the components $T_i {}^0$ correspond to the momentum density for any Lagrangian; and by definition, the integral of the Noether current over all of space, $$ P_i = \iiint T_i {}^0 \, d^3 \vec{r}, $$ is constant with respect to $t$. This is the general way in which momentum conservation manifests itself in field theories. In electrodynamics, the quantity $T_i {}^0$ is related to the Poynting vector, and the integral over $T_i {}^0$ over space does give a quantity that's conserved in time, which is the total momentum of the system (field momentum plus mechanical momentum.)

But in the Newtonian Gravity Lagrangian, $\partial_0 \phi = \dot{\phi}$ does not appear at all, and so we have $T_i {}^0 = [\delta {\mathcal{L}}/\delta(\partial_0 \phi)] \partial_i \phi = 0$ for the part of the Lagrangian that depends on the fields. In other words, the analogue of the Poynting vector in Newtonian gravity vanishes identically, and the statement of momentum conservation that is guaranteed by Noether's theorem only involves the mechanical momentum of the sources.

Answer 3

I will offer a tentative answer, although I think Michael Seifert's answer is actually excellent (although for an arbitrary system of masses it is impossible to construct a gravitational potential independent of time). On the one hand, if we derive the gravitational energy density equation, we have that:

$$\frac{\partial \epsilon_G}{\partial t} = \frac{\boldsymbol{g}}{4\pi G}\cdot \frac{\partial \boldsymbol{g}}{\partial t}, \tag{1}$$

On the other hand, the equation relating the field to the source $\nabla\cdot\boldsymbol{g} = 4\pi G \rho_m$ and the continuity equation imply that:

$$\frac{\partial (\nabla\cdot\boldsymbol{g})}{\partial t} = 4\pi G \frac{\partial \rho_m}{\partial t} = -4\pi G \nabla\cdot \boldsymbol{j}_m, \qquad \Rightarrow \frac{\partial \boldsymbol{g}}{\partial t} = -4\pi G \boldsymbol{j}_m, \tag{2}$$

By introducing this last result in $(1)$, we obtain:

$$\frac{\partial \epsilon_G}{\partial t} = -\boldsymbol{g}\cdot \boldsymbol{j}_m, \tag{3}$$

We can see that this part is cancelled out with kineteric energy of matter:

$$\frac{\partial \epsilon_m}{\partial t} = \rho_m\boldsymbol{g}\cdot\boldsymbol{v} = \boldsymbol{g}\cdot\boldsymbol{j}_m, \qquad \Rightarrow \frac{\partial \epsilon_G}{\partial t}+ \frac{\partial \epsilon_m}{\partial t} = 0, \tag{4}$$

This result contrasts with the electromagnetic case where we have that:

$$\frac{\partial \epsilon_{EM}}{\partial t}+ \frac{\partial \epsilon_m}{\partial t} = -(\boldsymbol{E}\cdot\boldsymbol{j}_q + \mu_0^{-1}\boldsymbol\nabla \cdot (\mathbf{E}\times\mathbf{B})) + (\boldsymbol{E}\cdot\boldsymbol{j}_q)$$

So, in general, we have:

$$\frac{\partial \epsilon_{EM}}{\partial t}+ \frac{\partial \epsilon_m}{\partial t} = -\frac{1}{\mu_0}\boldsymbol\nabla \cdot (\mathbf{E}\times\mathbf{B})) \neq 0, \tag{5}$$

Thus, the main difference seems to be that in newtonian gravity the flux of energy at each point of space is zero $(4)$, precisely because matter and the gravitational field are instantaneously readjusted, which is relativistically untenable.

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NOTE: In equation $(2)$ the more general solution is:

$$\frac{\partial \boldsymbol{g}}{\partial t} = -4\pi G \boldsymbol{j}_m + k\ \nabla\times(4\pi G\rho_m\boldsymbol{g})$$

although I originally considered $k = 0$, however, if one takes $k \neq 0$, then it opens the possibility of constructing a gravitational Poynting vector as:

$$ \boldsymbol{S}_G = \boldsymbol{p}_G = 4\pi k G \rho_m\boldsymbol{g}$$

That vector would only be nonzero if the gravitational field is NOT conservative, because it is not an irrotational field (this happens, for example, if $\boldsymbol{g}$ explicitly depends on time: $\boldsymbol{g} \neq -\nabla V_g$).