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Integral of pressure over Newtonian star

by Jonathan Scott
Tags: integral, newtonian, pressure, star
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Jonathan Scott
#1
Feb21-07, 04:09 PM
PF Gold
P: 1,156
1. The problem statement, all variables and given/known data

Last part of MTW Gravitation exercise 23.7:

Calculate in Newtonian theory the energy one would gain from gravity if one were to construct a star by adding one spherical shell of matter on top of another, working from the inside outward. Use Laplace's equation d(r^2 (d phi/dr))/dr = 4 pi r^2 rho and the equation of hydrostatic equilibrium dp/dr = -rho d phi/dr to put the answer in the following equivalent forms:

(energy gained from gravity) is equivalent to -(gravitational potential energy)
= [various things which all come to Gm^2/2R]
= 3 times integral from 0 to R of 4 pi r^2 p dr.

Variables assumed:
r = radius variable
R = total radius of star
m = total mass of star
p = pressure (function of r)
rho = density
phi = Newtonian gravitational potential (function of r)

2. Relevant equations:

Laplace's equation in radial coordinate:
d(r^2 (d phi/dr))/dr = 4 pi r^2 rho

Equation of hydrostatic equilibrium:
dp/dr = -rho d phi/dr

3. The attempt at a solution

It's not clear from the context whether rho is assumed to be able to vary with radius, but if it works in that case it should also work when rho is constant, so I tried that first.

Assuming rho constant then by integrating the equation of hydrostatic equilibrium we have:

p = -rho phi(r) + k where k is constant of integration

I assume that the pressure must be zero at the surface of the star (and increases with decreasing radius), so

k = rho phi(R)

giving

p = -rho (phi(R) - phi(r))

(Convention in MTW is to use units where G=1 but I'll include it explicitly anyway)

phi(r) = -G/r * mass within radius r
= -G/r * mr^3/R^3
= -Gm/R * r^2/R^2

phi(R) = -Gm/R

This gives

p = -rho (-Gm/R (1 - r^2/R^2))

The quantity requested on the last line of the question is:

3 * integral from 0 to R of 4 pi r^2 p dr

Filling in the above expression for p, this is

3 * integral from 0 to R of 4 pi r^2 rho (Gm/R) (1-(r^2/R^2)) dr

= 3 * 4 pi Gm/R rho integral from 0 to R r^2 (1-(r^2/R^2)) dr
(moving the constants outside the integral)

= 12 pi rho Gm/R (r^3/3 - r^5/5R^2) between r = 0 and R
(integrating)

= 12 pi rho Gm/R R^3 (1/3 - 1/5)

= 12 pi rho Gm/R R^3 (2/15)

Using m = 4/3 * pi R^3 rho, this becomes

12 * 3/4 * (4/3 pi rho R^3) Gm/R (2/15)

= 9 * 2/15 * Gm^2/R

= 6/5 Gm^2/R

However, all of the other cases gave 1/2 Gm^2/R. Where did I go wrong, please? (I've tried various other ways of looking at it, and all of them give the same result, so at least I'm consistent). This isn't for homework - I'm just trying to understand the relationship to the pressure term in the Komar mass.
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Jonathan Scott
#2
Feb21-07, 06:29 PM
PF Gold
P: 1,156
Hmmm. I've just realized (while trying to get to sleep) that I've used a bad expression for phi(r), the Newtonian potential. I've used the expression Gm/r based only on the amount of mass within radius r, but it should be the integral of the local acceleration, so that it continues to add up inside the object.
Jonathan Scott
#3
Feb21-07, 08:07 PM
PF Gold
P: 1,156
I also got the wrong answer for the other integrals whenever it was I did them (must be just over 20 years ago but my notes are not dated - looks like that was just a guess at the answer rather than the result of a calculation - at the time I was studying thin shells rather than solid spheres so that may have something to do with it). Now I've got the right potential and redone the other integrals, they seem to be coming out the same.


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