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The gravitational field of a point mass can be shown to be solenoidal (i.e. its divergence is zero) at all points except where the point mass actually is. This implies that there exists a vector field $\vec{F}$ whose curl is the point mass's gravitational field. I know in magnetostatics, due to Gauss's law for magnetism (i.e. $\vec{\nabla}\cdot\vec{B}=0)$, there exists a (not necessarily unique) vector field $\vec{A}$ called the magnetic vector potential such that $\vec{\nabla}\times{\vec{A}}=\vec{B}$. My question is: is there a similar name for that vector field $\vec{F}$ such that $\vec{\nabla}\times{\vec{F}}=\vec{g}$, where $\vec{g}$ is the gravitational field? And how come I don't think I ever really see this vector field get used? I mean, since the gravitational field of a point mass is also irrotational everywhere (except where the point mass is), it is quite common to be using a scalar field $\Phi$ called the gravitational potential such that $-\vec{\nabla}\Phi=\vec{g}$.

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  • $\begingroup$ In fluid mechanics the two dimensional version of ${\bf A}$ is called a stream function. It still works in three dimensions, but I've not seen it used much in fluids, but there is also the Hertz vector in E&M theory. $\endgroup$ – mike stone Mar 18 at 21:57
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The gravitational field is not a solenoidal field. See the definition. The difference between the magnetic field and the gravitational field is that the magnetic field is source-free everywhere, while the gravitational field (just like the electric field) ist only source-free almost everywhere. While this might seem a minor difference, it is actually of topological relevance: the magnetic field lines are closed (they do not end at a source), while the gravitational and electric field lines begin and end at sources.

Of course you can represent the gravitational field (just like the electric field) locally (i.e. in a compact, simply connected region that does not contain sources) by a curl even though it is not a solenoidal field. But for what purpose? That might be the reason why you don't see it: because there is no use case for it.

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  • $\begingroup$ Are physicists sure though that there is no use for some kind of "local gravitational vector potential"? At this moment I don't see any good use for it either, but is it possible that it has some hidden potential (pardon the pun) that hasn't been thought of or exploited yet? $\endgroup$ – SurfaceIntegral Mar 18 at 22:51
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    $\begingroup$ I can't talk about physicists in general. But from my point of view, everything that looks interesting to anyone deserves to be investigated by him. In the best case you might find something new, but always keep in mind that there are thousands of people around the world (and in history) who may have looked at it the same way. $\endgroup$ – oliver Mar 18 at 22:54
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In order to describe any vector field, you need to know its divergence and curl in order to have it uniquely determined. When we are dealing with gravitational field $$ \vec{g}=-Gm\frac{\vec{r}-\vec{r}'}{|\vec{r}-\vec{r}'|^3} $$ One can find that its divergence is $$ \nabla·\vec{g}=-4\pi G\rho_m $$ This divergence is $4\pi G\rho_m$ at mass sources and $0$ everywhere else. You can check this by calculating divergence in spherical coordinates and by integrating it over a sphere of constant radius.

Its curl, though, is exactly equal to zero (check it!) and that is precisely the reason why we can write the field as the gradient of a potential function $$ \nabla\times\vec{g}=0 \to \vec{g}=-\nabla\phi $$ This kind of fields are called conservative or irrotational. Thus, we have the field completely defined by its divergence and curl.

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  • $\begingroup$ I think you mean $-4\pi{G}\rho_{m}$. And yes, I understood all the other things you talked about, but you didn't really answer my question of why we don't have some kind of gravitational vector potential. $\endgroup$ – SurfaceIntegral Mar 18 at 22:48
  • $\begingroup$ Actually, I see now thanks to another user's answer that the gravitational field is, strictly speaking, not solenoidal (at the location of the point mass for instance) and therefore the theorem that it can be expressed as the curl of another vector field breaks down. $\endgroup$ – SurfaceIntegral Mar 18 at 22:59

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