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# Chapter 2. Hydrostatic Equilibrium

**Approximation** - star \~ sphere (no special direction)

In this way, the stellar density $\rho$ is only a function of $r$ (and $t$ since the star evolves).

* **Mass shell**

  $$
  \text dm=4\pi \rho r^2\text dr\Rightarrow \frac{\partial m}{\partial r}=4\pi\rho r^2
  $$

  At given $t=t\_0$, $m(r,t\_0)$ is used for the radial coordinate

  $$
  \frac{\partial}{\partial r}=4\pi r^2\rho\frac{\partial}{\partial m}
  $$

## Gravity

The gravitational acceleration $g$ is given by

$$
g=-\frac{Gm}{r^2}
$$

where $G=6.67\times 10^{-8}$ in cgs units.

The gravitational field inside a star is given by a potential, which satisfies the Poisson equation

$$
\nabla^2\Phi=4\pi G\rho
$$

In a spherical system, we can rewrite it as

$$
\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial\Phi}{\partial r}\right)=4\pi G\rho
$$

And the total gravitational force onto a volume $V$ is

$$
\vec F\_G=\int\_V \rho\vec g\text dV
$$

## Pressure Gradient Force

The force onto a this mass shell per unit area $\text dS$ due to pressure is

$$
\vec f\_P=-P\vec n\text dS
$$

Thus the pressure gradient force onto a unit volume is given by

$$
\vec F\_P=-\oint P\vec n\text dS=-\int\nabla P\text dV
$$

## Hydrostatic Equilibrium

$$
0=\vec F\_G+\vec F\_p=\int\left(\rho \vec g-\nabla P\right)\text dV\Rightarrow -\frac1\rho \nabla P+\vec g=0
$$

Again, in a spherical system.

$$
\frac1\rho\frac{\partial P}{\partial r}=-\frac{Gm}{r^2}\iff\frac{\partial P}{\partial m}=-\frac{Gm}{4\pi r^4}
$$

This is known as the **hydrostatic equlibrium**, **one of the most important equations in astrophysics**.

* Estimate the central temperature of the Sun ($M*\odot=2\times 10^{33}$ g, $R*\odot=7\times10^{10}$ cm)

  First we estimate the central pressure $P\_c$, by assuming

  $$
  \frac{\partial P}{\partial m}\sim\frac{P\_0-P\_c}{M\_\odot}\sim -\frac{G\left(M\_\odot/2\right)}{4\pi \left(R\_\odot/2\right)^4}
  $$

  Here we adopt the median values of $m$ and $r$. $p\_0\sim 0$ is the surface pressure. Thus

  $$
  P\_c\sim\frac{2GM\_\odot^2}{\pi R\_\odot^4}
  $$

  Further assuming the ideal gas EoS

  $$
  P\_c=\frac{\rho\_ck\_BT\_c}{\mu m\_\text{p}}
  $$

  where $\mu\approx 0.5$ is the mean molecular weight (ionized Hydrogen), thus the central temperature is

  $$
  k\_BT\_c\sim\frac{2\mu m\_\text{p}GM\_\odot^2}{\pi \rho\_c R\_\odot^4}
  $$

  Since $\rho\_c>\bar\rho$, we have

  $$
  k\_BT\_c<\frac{8 GM\_\odot}{3 R\_\odot}\mu m\_\text{p}\sim 3\text{ keV}
  $$

  which means $T\_c<3\times10^7$ K.

  Currently the most updated value is $T\_c\simeq1.6\times10^7$ K, so our estimation is not bad.

So far, the acceleration of mass shells is neglected.

The EoM (only radial motion is considered) is give by

$$
\frac{\partial^2 r}{\partial t^2}=-\frac1\rho\frac{\partial P}{\partial r}-\frac{Gm}{r^2}
$$

* If there is no pressure

  $$
  \frac{\partial^2 r}{\partial t^2}=-\frac{Gm}{r^2}\equiv\frac{r}{t\_\text{ff}^2}\Rightarrow t\sim t\_\text{ff}\equiv\sqrt{\frac{r^3}{Gm}}\sim\frac1{\sqrt{G\rho}}
  $$

  $t\_\text{ff}$ is known as the **free-fall timescale**.
* If there is no gravity

  $$
  \frac{r}{t\_\text{sc}^2}\equiv\frac{1}{\rho}\frac{P}{r}\Rightarrow t\_\text{sc}\equiv\frac{r}{\sqrt{P/\rho}}\sim \frac{r}{c\_s}
  $$

  $t\_\text{sc}$ is known as the **sound-crossing timescale**, since $c\_s$ is the **sound speed**.

Therefore, hydrostatic equilibrium requires

$$
t\_\text{ff}\simeq t\_\text{sc}
$$

This is generally satisfied in stellar interior.