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As a change in magnetic flux results in induced EMF (electromotive force) likewise what is the result of a change in gravitational flux?

UPDATE: Gravitational flux according to me has only mathematical significance and not physical, But as gravitational field and magnetic field are mathematically similar. So is there any similarity in their change in flux?

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    $\begingroup$ Are you asking this for Newtonian gravity, or for general relativity? If the former, why should there be a "result", if the latter, define gravitaational flux. $\endgroup$
    – ACuriousMind
    Commented Apr 30, 2015 at 10:36
  • $\begingroup$ @ACuriousMind I have updated the question Please have a look $\endgroup$
    – Shubham
    Commented Apr 30, 2015 at 10:47
  • $\begingroup$ 1. You did not answer the question whether this is Newtonian gravity or general relativity. 2. The Newtonian gravitational field is much more similar to the electric field than it is to the magnetic field, but there is no full equivalence between EM and gravity because the of the interdependence of the electric and magnetic fields. There is no "second field" such as the magnetic field for the gravitational field, and the formalism gives no reason to believe there is, so it's unclear what you're asking. $\endgroup$
    – ACuriousMind
    Commented Apr 30, 2015 at 10:50
  • $\begingroup$ @ACuriousMind Thanks my doubt got cleared! Sorry for being unclear $\endgroup$
    – Shubham
    Commented Apr 30, 2015 at 10:55

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Actually you can formulate Gauss's law for the gravitational field as well:

$$\oint_S \vec g \cdot d\vec A=-4\pi G M, $$

where on the left you have the gravitational flux through a closed surface and $M$ is the mass inside the volume. $G$ is the gravitational constant. When you call this quantity on the left $\Phi_G$ and write the mass as an integral of the density $\rho$ over the volume and ask for the rate of change of the whole thing then you have

$$\frac{\partial}{\partial t}\Phi_G=-4\pi\int_{V(S)}\frac{\partial}{\partial t}\rho~dV $$

If you plug in the continuity equation on the right side and using the divergence theorem you get

$$\frac{\partial}{\partial t}\Phi_G=4\pi\oint_{S}(\rho \vec v)\cdot d\vec A, $$

where $(\rho \vec v)$ is the current of mass density.

So you have rate of change of the gravitational flux if mass goes in or out the considered volume, which is ... not ... really ... surprising. But there is no "gravitomotive force" or the like.

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In analogy with Gauss's law for the electric field $\nabla\cdot \vec{E}=\rho/\epsilon_0$, the flux of the gravitational field through a closed surface is proportional to the mass contained inside the surface.

There is an approximation to General Relativity called Gravitoelectromagnetism (see Wikipedia page of this name.

It's relationship with Newton's law of gravitation is almost exactly the same as the relationship between Maxwell's equations and the laws of electrostatics. That is, you begin with the inverse square law for the radial gravitational field from a point mass, and you assume that effects of gravity propagate at the finite speed $c$. You must do this so that your equations are Lorentz covariant(well almost: see footnote) - otherwise if you do this naively like Laplace did you come up with a failed theory that makes planetary orbits grossly unstable. Maxwell's equations can be derived from electrostatics using this approach.

In the Maxwell equation $\leftrightarrow$ gravitoelectromagetism mathematical analogy, the analogy of the Electric Field is the Gravitational field. The analogy with the Magnetic Field is the gravitomagnetic field.

So the time rate of change of the flux of the gravitomagnetic field through a loop is the line integral of the gravitational field around the loop. It seems, therefore, that the gravitomagnetic field, not the gravitational field itself, is the concept you are groping for.


In this approach, you can make Maxwell's equations truly Lorentz covariant because you postulate the source term $(c\,\rho,\,\vec{J})$, the four-current, to be a truly Lorentz covariant four-vector and experiment backs this postulate up overwhelmingly. You cannot do the same for the corresponding term in gravitoelectromagnetism and still be consistent with General Relativity and this is one of the ways wherein the two theories differ. For many weak field General Relativistic problems, though, GEM is amazingly accurate.

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