# Newtonian Gravity on a Riemannian $3$-Manifold

To solve the Poisson equation for the Newton Potential, say $\phi$, one can use the divergence theorem, such that $$\int_U \nabla^2 \phi \sqrt{g}~ dV= \int_{\partial U} <\nabla \phi,n> \sqrt{g_\Sigma}~ d\Sigma=1,$$ where $\sqrt{g_\Sigma}$ represents the induced metric on the border.

Considering that $\phi$ is classic it will only depend on the relative position between the source and the point where we want to compute the potential, so, in order to use the divergence theorem, we need to find the set of points that are at the same distance of a given point (where we consider the source of the field to be) and that will be the border, $\partial U$, of the considered volume that contains the source.

Well, if space is the three dimensional Euclidean space then the volume that contains the source is straightforwardly the 3-ball whose center is at the source position and the border $\partial U$ the corresponding 2-sphere.

But if space is a different 3 Riemannian Manifold things are a "little" bit harder and finding the set of points that are at the same (geodesic) distance from the source is in general quite difficult or even impossible... But when searching for the geodesic curves on a manifold one can, in some circunstances, instead of searching for the curve that minimizes the distance

$$d~=~\int_a^b [g(\dot{c}(s),\dot{c}(s))]^{1/2} ds,$$

search for the curve that minimizes the energy functional

$$E~=~\int_a^b [g(\dot{c}(s),\dot{c}(s))] ds,$$

since in those circunstances the curves that minimize the energy are the same as the curves that minimize the distance.

Well, finding the set of points that are at the same energy is much easier...I would then ask: what is the significance of the energy defined above and would it make sense, physically, to say that equipotetial surfaces are not the set of points that at the same distance from the source but the set of points that have the same energy.

PS: Note that in the Euclidean Space, the set of points that are at the same distance is the same as the set of points that are at the same energy.

-
Could you please edit your question so that it is more readable? Breaking it up into more sentences with a little more detail would greatly enhance the clarity. Thanks! – jeffdk Mar 5 '13 at 16:47
@jeffdk Just edited the question. Hope it's better now. Thank you for the suggestion. – PML Mar 5 '13 at 17:39

You can use the energy functional to compute geodesics in any circumstances (even in GR!). The geodesic computed using the energy functional "differs" from a geodesic computed from the distance functional only in that it's automatically parametrised so that $g(\dot{c}(s),\dot{c}(s))$ is constant along the geodesic (this is automatic by conservation of energy if you consider the "energy" functional as the action for a particle). This last fact is very useful if you actually attempt to perform such a calculation by hand.
EDIT: I forgot to specify how to compute the distance function... once you've solved the geodesic equation for curves joining some distinguished point $p^*$ to an arbitrary point $x$, the distance functional effectively becomes a distance function (of $x$) on the manifold so, in principle, you've solved the problem.
I skimmed the question in maths SE and the person answering that has got it right. Anyway, if you consider $L(\dot{c}(s))=g(\dot{c}(s),\dot{c}(s))$ as a Lagrangian defining the dynamics for a mechanical system, then the Hamiltonian $H= \frac{\partial L}{\partial \dot{c}} \dot{c} - L$ is automatically conserved in any system without explicit time dependence, and in this case evaluates to $L$. – alexarvanitakis Mar 5 '13 at 23:38
Well, I feel dumb... For example here when searching for geodesics they just minimize the energy functional imposing that the <$\dot{c},\dot{c}$> is constant. Well, then if the curves that minimize the energy are also extremum of the distance everything is much easier and I can move on. Thank you very much, truly, for your help! – PML Mar 5 '13 at 23:58
This is important: they do not a priori impose that $\langle \dot{c} , \dot{c} \rangle$ is constant; this is a consequence of the equations of motion. – alexarvanitakis Mar 6 '13 at 0:04