Plummer Sphere, Density from Potential

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SUMMARY

The discussion focuses on deriving the density function of a Plummer sphere from its gravitational potential, represented by the equation \(\Phi (r) = -\frac{GM}{\sqrt{r^{2}+a^{2}}}\). The correct density function is established as \(\rho = \frac{3a^{2}}{4\pi}\frac{M}{(r^{2} + a^{2})^{5/2}}\). The participant initially misapplied the Laplacian operator in their calculations, mistaking it for simple second derivatives. Clarification was provided regarding the proper use of the Laplacian in spherical coordinates.

PREREQUISITES
  • Understanding of gravitational potential and density functions
  • Familiarity with Poisson's equation in physics
  • Knowledge of calculus, specifically derivatives
  • Basic concepts of spherical coordinates
NEXT STEPS
  • Study the derivation of the Laplacian in spherical coordinates
  • Learn about gravitational potential in astrophysics
  • Explore the properties and applications of Plummer spheres
  • Investigate the relationship between density and gravitational potential
USEFUL FOR

Astronomy students, physicists, and anyone studying gravitational systems or astrophysical models will benefit from this discussion.

SHISHKABOB
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Homework Statement


It wants me to get the density function of a Plummer sphere from its gravitational potential.


Homework Equations


Plummer sphere potential:

\Phi (r) = -\frac{GM}{\sqrt{r^{2}+a^{2}}}

where phi is the potential as a function of radius from the mass, M. And a is a scale factor of the model. I think I'm just supposed to take M as a constant here.

I am supposed to end up with

\rho = \frac{3a^{2}}{4\pi}\frac{M}{(r^{2} + a^{2})^{5/2}}


The Attempt at a Solution



So according to Poisson's equation

\nabla ^{2} \Phi = 4\pi G \rho (x)

So to solve for ρ I just took the derivative of Phi twice with respect to r twice and then divided by 4πG

first derivative got me

2rGM(r^{2} + a^{2})^{-1/2}

and then the second derivative got me

\frac{GM}{2}\left[2(r^{2} + a^{2})^{-3/2} - (r^{2} + a^{2})^{-1/2}\right]


then after a bit of rearranging I have ended up with

\frac{M}{8\pi}(r^{2} + a^{2})^{-1/2}(2 - \frac{1}{r^{2} + a^{2}})

I'm not really sure if I'm on the right track...
 
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The Laplacian is not "taking the derivative with respect to r twice". Look up its expression in spherical coordinates (or derive it, if you are not supposed to know).
 
voko said:
The Laplacian is not "taking the derivative with respect to r twice". Look up its expression in spherical coordinates (or derive it, if you are not supposed to know).

ohhh yes you are right, thank you. That explains a lot :redface:
 

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