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?

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    $\begingroup$ In electrodynamics, the Poynting vector forms part of the canonical stress-energy tensor, specifically the component $T^{0i}$. I suspect (though it's been a while since I've done a similar calculation, so caveat lector) that the analogous quantities vanish identically in Newtonian gravity because $\partial \mathcal{L}/\partial \dot{\phi} = 0$, i.e., the Lagrangian does not depend on $\dot{\phi}$. $\endgroup$ Jul 4, 2022 at 14:26
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    $\begingroup$ Yes, in Gravitoelectromagnetism, but with the 4D representation of space-time (GR): en.wikipedia.org/wiki/Gravitoelectromagnetism $\endgroup$
    – The Tiler
    Jul 4, 2022 at 15:51

3 Answers 3


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.

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    $\begingroup$ Is there a Poynting equivalent in relativity? $\endgroup$ Jul 5, 2022 at 3:33
  • $\begingroup$ Concerning "trackable movement of energy": if I'm not mistaken, the $T_0 {}^i$ components of the canonical stress-energy tensor tell you about the fluxes of energy. And if you calculate them, they're non-zero: $T_0 {}^i = - \frac{1}{4\pi G} \dot{\phi} (\partial^i \phi)$. If so, saying that "there's no trackable movement of energy" is a bit misleading. $\endgroup$ Jul 5, 2022 at 11:36
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    $\begingroup$ @BlueRaja-DannyPflughoeft Sort of. The closest analog is the Landau-Lifshitz pseudotensor. However, as its name implies, it's not a tensor - it's designed to vanish in an inertial frame. The issue is that defining the energy and momentum of the gravitational field is a tricky thing, because gravity defines the very spacetime which you want to use to define energy and momentum. You either get zero, or weird non-tensorial constructions like the LL pseudotensor. $\endgroup$
    – Javier
    Jul 5, 2022 at 13:51
  • $\begingroup$ @MichaelSeifert That's a good point. I'm also not completely sure on what to do with a non symmetric energy-momentum tensor - and also not completely sure on how to interpret an energy flux for a field that responds instantly to changes. $\endgroup$
    – Javier
    Jul 5, 2022 at 13:54
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    $\begingroup$ Even with instantaneous propagation, I think that the statement of energy conservation would still be that $$\int_\mathcal{V} \left( \frac{1}{4 \pi G} (\nabla \phi)^2 + \rho_m \right) d^3r + \oint_{\partial \mathcal{V}} \left( -\frac{1}{4 \pi G} \dot{\phi} \nabla \phi \right) \cdot d\vec{a} = 0,$$where $\rho_m$ is the mechanical energy density. In other words, if the energy density changes in some volume, there's a corresponding change in the surface term (instantaneously) that can be interpreted as a flux of energy in or out of the volume. $\endgroup$ Jul 5, 2022 at 14:55

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.

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    $\begingroup$ To the OP: please don't accept this until a couple of days have gone by, as I am not 100% sure of the argumentation here (particularly the interpretation of the components of the canonical stress-energy tensor, and which of $T_i {}^0$ or $T_0 {}^i$ is the momentum density and which is the energy flux in situations like this one where the canoncial stress-energy is not symmetric.) To everyone else: please poke holes in this argument if you see any. $\endgroup$ Jul 4, 2022 at 15:20

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.

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: $\boldsymbol{g} \neq -\nabla V_g$).

  • $\begingroup$ I believe there is a flaw in your logic, in line 2, the curl of some arbitrary vector function is present . Since the divergence of the curl is zero. However if I remember correctly when playing around with this myself integral form, this term is dependant on v, I could be wrong though. $\endgroup$ Jul 4, 2022 at 19:21
  • $\begingroup$ @jensenpaull I am aware that I arbitrarily took that possible vector equal to zero, but that in fact the most general solution is $\dot{\boldsymbol{g}} = -4\pi G \boldsymbol{j}_m + k \boldsymbol{\nabla}\times \boldsymbol{b}$, with $k\neq 0$, then $k\boldsymbol{b}$ would be a kind of "graviatory Poynting vector". $\endgroup$
    – Davius
    Jul 4, 2022 at 19:44
  • $\begingroup$ @jensepaull Infact, the election $\boldsymbol{b} = \rho\boldsymbol{g}$ and $k \propto G$ is dimensionally consistent. $\endgroup$
    – Davius
    Jul 5, 2022 at 22:47
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    $\begingroup$ Less of a poynting vector since its not in the form div [blank], the closed surface integral of this is always zero. $\endgroup$ Jul 6, 2022 at 5:58

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