# What is the physical meaning of a flux of gravitational field in classics?

I've stumbled upon an answer to a question about square power in Newton's law of gravity. After reading it I got a question whether the flux of gravitational field has actually any physical meaning.

Fluxes I know arise in a context of balance equations. The change of a certain physical quantity $a$ is comprised of change due to a flux $\boldsymbol j_a$ and due to a source $\sigma_a$:

$$\frac{\partial a}{\partial t} + \operatorname{div} \boldsymbol j_a = \sigma_a$$

But as for me the flux of gravitational field is actually nothing but a gravity field itself:

$$j_g = \mu(\boldsymbol g)$$

$\mu$ being the volume form. It doesn't bear the meaning of a flux propagating the gravity field $\boldsymbol g$ or anything else.

The question is specifically about gravity, and about classical gravity. Not about electromagnetic phenomena, general relativity or quantum gravity.

• The "flux" in this case just refers to the surface integral of the vector field. It has nothing really to do with flux in the sense of electromagnetic radiation. Have a look at this link. The math is clearly explained there. Dec 21, 2012 at 10:10
• Check the answer to this related question: physics.stackexchange.com/q/716807 Jul 4, 2022 at 14:13

The word "flux" is something of an accident of history. See for example it's use in Gauss' law or as a magnetic flux. Nothing is actually flowing e.g. for a static charge we would still refer to the flux through a surface surrounding the charge even though the system is time independant.

• so no physical meaning? Dec 21, 2012 at 10:34
• Yes, it has a physical meaning. It's the integrated field strength over the surface. Dec 21, 2012 at 10:38

If one considers the energy density of the gravitational field, given by:

$$\epsilon_G = \frac{1}{2} \frac{\boldsymbol{g}^2}{4\pi G}$$

It can be seen that its derivative equals that of the derivative energy density of matter:

$$\frac{\partial\epsilon_G}{\partial t}+ \frac{\partial\epsilon_m}{\partial t} = -\boldsymbol{g}\cdot\boldsymbol{j}_m +\boldsymbol{g}\cdot\boldsymbol{j}_m = 0$$

where $$\boldsymbol{j}_m = \rho_m\boldsymbol{v}$$ is the mass flux.

Thus, at each point in space, the energy lost by the field is recovered by matter instantaneously without the energy having to flow from one place to another (therefore, in Newtonian gravity there are no gravitational waves!).