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We could use any self-gravitating fluid for this question, but let's take a star as an example. Left alone, it's density is the source of it's potential and thus we have the simplest form of Poisson's equation:

$$ 4 \pi G \rho_0 = \nabla^2 \Phi_0 $$

Now let this system be perturbed by an outside potential, $U$ (which we assume is small compared to the star's own potential, $\Phi_0 \gg U$). An example physics case would be another star passing close by, and thus $U$ is approximately a point mass potential.

Let's take the first order perturbations to both the potential and the density: $$\rho = \rho_0 + \rho' + O(2)$$ and $$\Phi = \Phi_0 + \Phi' + U + O(2)$$

My question boils down to this, which of the following is the correct formulation of the first order Poisson equation? :

$$ 4 \pi G \rho' = \nabla^2 \Phi' $$ OR $$ 4 \pi G \rho' = \nabla^2 (\Phi' + U) $$

I'm inclined to say the first is correct, as it makes sense to me that the relationship only applies to quantities describing the self-gravitating fluid.

However, the perturbed density is certainly affected by the external potential. I imagine this effect could be written implicitly into the perturbed potential, but I'd be very interested to hear the thoughts of the community on this fine-detail point.

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  • $\begingroup$ What's the difference between $\Phi'$ and $U$? And what does $O(2)$ mean? $\endgroup$
    – probably_someone
    Nov 14, 2018 at 11:56
  • $\begingroup$ $\Phi'$ is the perturbation to the star's own potential (caused by the movement of fluid) and is the quantity normally ignored under the Cowling approximation. $U$ is the external perturbing potential. $O(2)$ means that any extra terms are 2nd powers or higher of small quantities (I think this is a slightly unusual but useful notation, let me know if you can think of a clearer alternative). $\endgroup$
    – zephyr
    Nov 14, 2018 at 12:27
  • $\begingroup$ If you're expanding in powers of small quantities, then you should be explicit about what those small quantities are. $\endgroup$
    – probably_someone
    Nov 15, 2018 at 11:55

1 Answer 1

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This is not a complete answer, but I thought I should add to this question the simple observation that if $U$ is a point mass potential then $\nabla^2 U = 0$ and thus the two forms in the question are the same.

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  • $\begingroup$ This isn't quite true. $\nabla^2 U$ is zero everywhere except the location of the point mass itself, where $\nabla^2 U$ is infinite; namely, for point mass position $\vec{x}_p$, we have that $\nabla^2 U = 4\pi G \delta(\vec{x}-\vec{x}_p)$ where $\delta$ is the Dirac delta function. This distinction sounds trivial, but is very important when solving the Poisson equation, because the delta function contributes significantly to an integral over any region enclosing the point mass. $\endgroup$
    – probably_someone
    Nov 15, 2018 at 11:46

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