- #1

LmdL

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## Homework Statement

Hello.

In my homework I was given to find a mistake in a derivation of hydrostatic equilibrium of a star.

In contrast to "regular" derivation, where a cylindrical segment is chosen, in HW a segment of sphere is used:

## Homework Equations

Derivation is given as following:

__a. Pressure on the outer surfase:__

[itex]P(r+dr)=P(r)+\frac{dP}{dr}dr[/itex]

__b. Area of the outer surface:__

[itex]A(r+dr)=d\Omega \left [ r+dr \right ]^2=r^2d\Omega+2rdrd\Omega+\left (dr \right )^2d\Omega[/itex]

__c. Therefore, force on the outer surface is:__

[itex]F(r+dr)=P(r+dr)A(r+dr)=\left [ P(r)+\frac{dP}{dr}dr \right ]\left [r^2d\Omega+2rdrd\Omega+\left (dr \right )^2d\Omega \right ]=[/itex]

[itex]=r^2P(r)d\Omega+2rP(r)drd\Omega+P(r)\left (dr \right )^2d\Omega+r^2\frac{dP}{dr}drd\Omega+2r\frac{dP}{dr}\left ( dr \right )^2d\Omega+\frac{dP}{dr}\left (dr \right )^3d\Omega=[/itex]

[itex]=r^2P(r)d\Omega+2rP(r)drd\Omega+r^2\frac{dP}{dr}drd\Omega+O\left ( \left ( dr \right )^2 \right )[/itex]

__d. The force on the inner surface is:__

[itex]F(r)=P(r)A(r)=P(r)A(r)=P(r)r^2d\Omega[/itex]

__e. Mass of the segment:__

[itex]dm=\rho r^2drd\Omega[/itex]

__f. Then, equation for equilibrium:__

[itex]gdm=F(r)-F(r+dr)[/itex]

[itex]g\rho r^2drd\Omega=-2rP(r)drd\Omega-r^2\frac{dP}{dr}drd\Omega[/itex]

[itex]g=-\frac{2}{\rho}\frac{P(r)}{r}-\frac{1}{\rho}\frac{dP}{dr}[/itex]

But we know that equation for hydrostatic equilibrium is:

[itex]g=-\frac{1}{\rho}\frac{dP}{dr}[/itex]

## The Attempt at a Solution

After going and checking the solution step by step I found only 2 places where error can occur:

1. The origin of the additional term is from the

__non-equal areas__of the segment in [itex]r[/itex] and [itex]r+dr[/itex]. This makes sense, since in "regular" derivation of the equilibrium equation a cylinder is used, and there both areas are equal. In addition, if we make [itex]r[/itex] to be very big (in limit - [itex]r\rightarrow \infty[/itex]) - then areas of outer and inner surfaces will be, indeed, identical, and as supposed the additional term ([itex]-\frac{2}{\rho}\frac{P(r)}{r}[/itex]) will be zero.

I told this to my teacher, but he said that this is not an origin of the problem.

2. So I checked the derivation again and found that in line "c", the term [itex]2r\frac{dP}{dr}\left ( dr \right )^2d\Omega[/itex] cannot be included inside [itex]O\left ( \left ( dr \right )^2 \right )[/itex]. But if I include this term in the derivation, I get:

[itex]g=-\frac{2}{r \rho} P(r)-\frac{1}{\rho}\frac{dP}{dr}-\frac{2}{r \rho}\frac{dP}{dr}dr[/itex]

which gets even worse than was before.

Can someone help me with that? What I'm missing here?

Thanks in advance.