# Central Pressure for a star

1. Jul 25, 2008

### arunma

I have a question about the hydrostatic equation for a star's central pressure. I know that the central pressure for a star is,

$$\dfrac{dP}{dr} = \rho \dfrac{G M}{r^2}$$

My question is: why does this blow up at $$r = 0$$? Because of the singularity, I'm not sure how I can integrate the equation in order to obtain the total gravitational pressure on a star. Can anyone help? Thanks.

2. Jul 26, 2008

### Astronuc

Staff Emeritus
Well, firstly since M = M(r).

Think about the equation and what it is decribing, and the domain in which it is valid, i.e. pressure and density must increase as one descends toward the center of the star.

These might help.

http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/GravField.htm

http://www.yale.edu/phys180/lecture_notes/180_Lect_31/sld007.htm [Broken]

Last edited by a moderator: May 3, 2017
3. Jul 28, 2008

### arunma

Thanks, I think this takes care of my difficulty. I stared at that equation for quite awhile and completely forgot that M = M(r).

4. Aug 3, 2008

### Orion1

The gravitational force, due to spherical symmetry, points toward the center and therefore has a
negative sign.

The equation of state for hydrostatic equilibrium:
$$\boxed{\frac{dP}{dr} = - \rho(r) \frac{G m(r)}{r^2}}$$

Integrating for core pressure:
$$\boxed{P_c = - G \int_0^R \frac{\rho(r) m(r)}{r^2} dr}$$

Note that the density and mass functions must obey boundary conditions:
$$\rho(0) = \rho_c \; \; \; \rho(R) = 0$$
$$m(0) = M_t \; \; \; m(R) = 0$$

Reference:
http://www.jb.man.ac.uk/~smao/starHtml/stellarEquation.pdf" [Broken]

Last edited by a moderator: May 3, 2017
5. Aug 3, 2008

### MeJennifer

Does the above formula match the stituation under GR in the weak field limit?

An interesting document in this context might be: http://arxiv.org/PS_cache/arxiv/pdf/0705/0705.0825v3.pdf

Pressure under GR increases gravitational pull while in Newtonian physics this is not the case.

6. Aug 4, 2008

### Orion1

Does the Tolman-Oppenheimer-Volkoff (TOV) equation for gravitational hydrostatic equilibrium in General Relativity reduce to the classical Newtonian gravitational hydrostatic equilibrium equation under the General Relativity weak field limit?

Tolman-Oppenheimer-Volkoff (TOV) equation for gravitational hydrostatic equilibrium:
$$\frac{dP(r)}{dr}=-\frac{G(\rho(r)+P(r)/c^2)(m(r)+4\pi P(r) r^3/c^2)}{r^2(1-2Gm(r)/rc^2)}$$

Classical Newtonian equation for gravitational hydrostatic equilibrium:
$$\frac{dP(r)}{dr} = - \rho(r) \frac{G m(r)}{r^2}$$

The Schwarzschild solution analogue in classical Newtonian theory of gravitation corresponds to the gravitational field around a point particle. (ref. 1)

Static models for stellar structure must be based upon the Schwarzschild metric, which is the genesis solution of the TOV equation, in order to obey General Relativity. In models where the dimensionless quantities of each analogue are both much less than one, the model becomes non-relativistic, and deviations from General Relativity are small and reduces to Newton's law of gravitation: (ref. 2)

$$\frac{\Phi}{c^2}=\frac{GM_\mathrm{sun}}{r_\mathrm{orbit}c^2} \sim 10^{-8} \; \; \; \quad \left(\frac{v_\mathrm{Earth}}{c}\right)^2=\left(\frac{2\pi r_\mathrm{orbit}}{(1\ \mathrm{yr})c}\right)^2 \sim 10^{-8}$$

In situations where either dimensionless parameter is large, then the model becomes relativistic and General Relativity must be used to describe the system. General relativity reduces to Newtonian gravitation in the limit of small potential and low velocities, therefore Newton's law of gravitation is the low-gravitation non-relativistic weak field limit of General Relativity.

Reference:
http://www.iop.org/EJ/abstract/0264-9381/14/1A/010/"
Problems with Newton's theory - Wikipedia
http://en.wikipedia.org/wiki/Tolman-Oppenheimer-Volkoff_equation" [Broken]
http://en.wikipedia.org/wiki/Schwarzschild_metric" [Broken]

Last edited by a moderator: May 3, 2017