Density of star from hydrostatic equilibrium and pressure

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SUMMARY

The discussion focuses on deriving the relationship between density (\rho) and gravitational potential (\Phi) in a star using the equation of state p = \tfrac12 k\rho^2. Participants successfully derived the differential equation \frac{d^2[r\rho(r)]}{dr^2} = -\frac{4 \pi G}{k}r \rho(r) and confirmed that the radius (a) of the star is determined solely by the constant k, independent of mass (M). The mass is expressed as M = (4/\pi)a^4\rho(0). Key equations discussed include the fluid equilibrium equation and Poisson's equation.

PREREQUISITES
  • Understanding of hydrostatic equilibrium in astrophysics
  • Familiarity with Poisson's equation in gravitational contexts
  • Knowledge of spherical coordinates and their application in differential equations
  • Basic concepts of fluid dynamics and equations of state
NEXT STEPS
  • Study the derivation of Poisson's equation in spherical coordinates
  • Explore the implications of different equations of state in stellar structure
  • Learn about the stability of hydrostatic equilibrium in astrophysical contexts
  • Investigate the relationship between mass, radius, and density in stellar models
USEFUL FOR

Astronomy students, astrophysicists, and researchers focusing on stellar dynamics and structure will benefit from this discussion.

Dazed&Confused
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Homework Statement


Assume that the pressure p in a star with spherical symmetry is related to the density \rho by the (distinctly unrealistic) equation of state p= \tfrac12 k\rho^2, where k is a constant. Use the fluid equilibrium equation obtained in Problem 23 to find a relation between \rho and \Phi. Hence show that Poisson's equation yields
<br /> \frac{d^2[r\rho(r)]}{dr^2} = -\frac{4 \pi G}{k}r \rho(r)<br />
Solve this equation with the boundary conditions that \rho is finite at r=0 and vanishes at the surface of the star. Hence show that the radius a of the star is determinde solely by k and is independent of its mass M. Show also that M =(4/ \pi )a^4 \rho(0).

Homework Equations


\nabla p = \rho \textbf{g} and p + \rho \Phi = \textrm{constant}. And \nabla^2 \Phi = 4\pi G \rho.

The Attempt at a Solution


Assuming the differential equation, I am able to do the rest of the question. The relation they are looking for is \Phi = -k\rho and this can be easily shown by
\nabla p = k\rho&#039;(r) \rho(r) \hat{\textrm{r}} = \rho \textrm{g} and so
-\nabla \Phi = \textrm{g} = k\rho&#039;(r) \hat{\textrm{r}}

from which you can get \Phi. However by the second relevant equation this must also mean that -\tfrac12 k \rho^2 = \textrm{constant}, or I'm missing something. I've also not been able to obtain the differential equation.
 
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Not that it matters too much but it should be M = (4/\pi)a^3\rho(0). Any ideas anyone?

Edit 2: the equation p + \rho \Phi = \textrm{constant} is only for an incompressible fluid, so there is no contradiction.
 
Last edited:
Dazed&Confused said:
-\nabla \Phi = \textrm{g} = k\rho&#039;(r) \hat{\textrm{r}}

Note that this is just ##\nabla \Phi = -k \nabla \rho##.

To relate this to Poisson's equation, take the divergence of both sides.
 
Hi thanks for responding. I did realize this and found the relation they wanted. I took the Laplacian of this which did not work, so unsurprisingly taking the divergence of what you have said does not work either.
 
Yes I did use them. Could you try to see if you get the correct differential equation?
 
Yes, I am getting the correct differential equation. Can you show your work?
 
Certainly. I used Poisson's equation on the relation, so <br /> \nabla^2 \Phi = 4 \pi G \rho = -k \nabla^2 \rho = -k\frac{1}{r^2} \frac{d}{dr} \left ( r^2 \frac{d\rho}{dr} \right )

from which I can't get the correct equation.
 
OK, that looks good so far. Try re-expressing the right hand side.
 
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Thanks I was able to get it.
 

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