# Difference between revisions of "Hydrostatic Equilibrium"

### Central Topics

1. Force balance in stars: pressure vs. gravity
2. Energy transport by radiation, convection, and conduction
3. Energy generation by fusion

These combine to give the properties of the HR diagram, the main sequence, etc. Changes in mass and composition change each of the above. These three unifying principles used repeatedly in the course. We’ll look at each separately to see what we can learn, and then put them together to understand stars. Even without knowing anything about energy generation, we can understand a lot based on force balance and energy transport. \\

### Force Balance in Stars: Hydrostatic Equilibrium

\Throughout the course we will assume spherical stars, ignoring factors such as rotation that may cause a star to become oblate, for instance. \We will start by looking at a thin layer in a star, and over a small enough area that we can approximate the layer as being flat. The layer has thickness ${\displaystyle dr}$, density ${\displaystyle \rho }$, area ${\displaystyle A}$, and is at a distance ${\displaystyle r}$ from the center of the star and encloses a mass ${\displaystyle M_{r}}$. The mass of the shell is

${\displaystyle M_{\rm {shell}}=\rho Adr\,\!}$

The inward gravitational force on the region is then

${\displaystyle F_{g}=-{\frac {GM_{r}M_{\rm {shell}}}{r^{2}}}\,\!}$

The competing force (which provides force balance) is the pressure gradient within the star. That is, there is a slightly greater pressure on the bottom, ${\displaystyle P_{\rm {below}}}$ on the shell then there is on the top, ${\displaystyle P_{\rm {above}}}$. We then can multiply this pressure difference by the area of the thin shell to find the total force due to the pressure gradient. The net force is the sum of these two sources, and will give the acceleration of the shell.

${\displaystyle F_{\rm {net}}=M_{\rm {shell}}a\,\!}$
${\displaystyle F_{\rm {net}}=(P_{\rm {below}}-P_{\rm {above}})A-{\frac {GM_{r}M_{\rm {shell}}}{r^{2}}}=M_{\rm {shell}}a.\,\!}$

Since we are looking at a thin shell, we will say that

${\displaystyle P_{\rm {above}}=P_{\rm {below}}+dP.\,\!}$

Then, substituting above,

${\displaystyle -dpA-{\frac {GM_{r}}{r^{2}}}Adr\rho =Adr\rho a.\,\!}$

The area element cancels out, which is good because it was sort of arbitrarily chosen. Dividing through by the differential ${\displaystyle dr}$ gives

${\displaystyle \rho a=-{\frac {dP}{dr}}-{\frac {GM_{r}}{r^{2}}}\rho .\,\!}$

In the case where there is no acceleration, which is pretty common since stars generally appear to be neither contraction nor expanding, we have hydrostatic equilibrium (HE). In this case, the above reduces to

${\displaystyle {\frac {dP}{dr}}=-\rho {\frac {GM_{r}}{r^{2}}}=-\rho g.\,\!}$

To further justify this approximation, let’s imagine that ${\displaystyle a}$ is not equal to 0. Instead, it is roughly equal to the graviational acceleration. We can then estimate the amount of time it would take for the radius of a star to change appreciably. So

${\displaystyle a\sim -{\frac {GM_{r}}{r^{2}}}\,\!}$

We can look at the acceleration in terms of the size of the star, ${\displaystyle R}$, and the timescale of change, ${\displaystyle t}$. Then

${\displaystyle a\sim {\frac {R}{t^{2}}}.\,\!}$

For the purposes of this estimate, we can just take ${\displaystyle r=R}$ and ${\displaystyle M_{r}=M}$, which gives

${\displaystyle {\frac {R}{t^{2}}}\sim {\frac {GM}{R^{2}}},\,\!}$
${\displaystyle t\sim {\sqrt {\frac {R^{3}}{GM}}}\sim {\sqrt {\frac {1}{G\rho }}}.\,\!}$

This is often called the dynamical time, and applies equally well to planets or galaxies as it does to a star. It is is also called the free fall time, since it is the time it takes to move the size of the system at the free fall speed of the system. For the Sun, with an average density of ${\displaystyle 1}$ g cm${\displaystyle ^{-3}}$, this is about one hour, meaning any imbalances must quickly be eliminated. There are only brief periods in the life of a star where HE does not hold, such as the collapse to form a black hole. Even during convection, which violates the assumptions that went in to HE, this approximation holds very very well. \We should explicitly define ${\displaystyle M_{r}}$, which we use in HE.

${\displaystyle M_{r}=\int _{0}^{r}4\pi r^{2}\rho dr,{\rm {\;or}}\,\!}$
${\displaystyle {\frac {dM_{r}}{dr}}=4\pi r^{2}\rho .\,\!}$

There are still more variables here than equations. We need to know what the source of pressure is to solve for the structure of a star, as well as how energy is transported. We can, however, use just these equations to understand the outermost layers of the Sun, or even the atmosphere of the Earth. We use the approximation of a plane parallel atmosphere, which holds very well since the thickness ${\displaystyle z}$ of the atmosphere is much smaller than the size of the star (or planet) ${\displaystyle R}$. Then we can assume that the surface gravity is set by the total mass and radius of the star, without worry about the mass or size of the atmosphere. Then

${\displaystyle g={\frac {GM}{R^{2}}}={\rm {constant\;at\;surface}}.\,\!}$

Then HE is

${\displaystyle {\frac {dP}{dz}}=-\rho g.\,\!}$

The pressure is due to that of an ideal gas, so

${\displaystyle P=nk_{B}T,\,\!}$

where ${\displaystyle n}$ is the number density, ${\displaystyle k_{B}}$ is Boltzmann’s constant, and ${\displaystyle T}$ is the temperature. We need to know then the average mass per particle, ${\displaystyle m}$, so that ${\displaystyle \rho =nm}$. This gives

${\displaystyle {\frac {d(nk_{B}T)}{dz}}=-mgn.\,\!}$

This is still not something we can solve, so we have to make the further approximation that the atmosphere is isothermal, i.e. ${\displaystyle T}$ is constant as a function of ${\displaystyle r}$. Then

${\displaystyle {\frac {1}{n}}{\frac {dn}{dz}}=-{\frac {mg}{k_{B}T}}.\,\!}$

The solution of this is

${\displaystyle \left[\ln(n)\right]_{\rm {surface}}^{z}=-{\frac {mgz}{k_{B}T}},\,\!}$
${\displaystyle n(z)=n_{\rm {surf}}e^{-z/h},{\rm {\;where}}\,\!}$
${\displaystyle h={\frac {k_{B}T}{mg}}.\,\!}$

We call ${\displaystyle h}$ the scale height of the atmosphere, and it is given by the ratio of the thermal energy at the surface to the gravitational potential energy at the surface. We can think of it as the distance over which the density changes by an appreciable amount (1 / ${\displaystyle e}$). Thus the thermal energy is trying to puff up the atmoshpere, while the gravitational potential is trying to keep it close to the surface. For the Sun, the scale height is

${\displaystyle h=2\times 10^{7}{\rm {cm}}=3\times 10^{-4}R_{\odot }.\,\!}$

This is much smaller than the radius of the Sun. There is an interesting statistical mechanics interpretation of the scale height. Particles with an energy ${\displaystyle E}$ are distributed in energy levels according to the Boltzmann factor, with

${\displaystyle n\propto e^{-E/k_{B}T}.\,\!}$

Since the gravitational potential energy for our particles at height ${\displaystyle z}$ is ${\displaystyle mgz}$, this Boltzmann argument gives the same radial dependence of number density of the atmosphere as our argument above. \\

### Mean Molecular Weight

\One of the major ways fusion affects the structure of stars is by changing the average mass per particle. If it starts out as ionized hydrogen, there are basically two particles per ${\displaystyle m_{p}}$, while for neutral hydrogen there is about one particle per ${\displaystyle m_{p}}$, and for ionized helium it is 3 particles per ${\displaystyle 4m_{p}}$, roughly. Each of these particles is at temperature ${\displaystyle T}$, and thus contributes to the total pressure in the star. \\To account for this, we will define a quantity ${\displaystyle \mu }$ called the mean molecular weight to formally define our average mass per particle and encapsulate the summing of pressures from each element. Each element has a mass fraction ${\displaystyle X_{i}}$ and an atomic number ${\displaystyle A_{i}}$ that represents the total number of protons and neutrons. If the total mass density is ${\displaystyle \rho }$, then the number density of a given species ${\displaystyle n_{i}}$ is

${\displaystyle n_{i}={\frac {\rho X_{i}}{A_{i}m_{p}}}.\,\!}$

The pressure from all ions is

${\displaystyle P=\sum _{i}n_{i}k_{B}T.\,\!}$

Substituting in the expression for ionic number density,

${\displaystyle P={\frac {\rho k_{B}T}{m_{p}}}\sum _{i}{\frac {X_{i}}{A_{i}}}.\,\!}$

We can then define the ion mean molecular weight to be

${\displaystyle {\frac {1}{\mu _{i}}}=\sum _{i}{\frac {X_{i}}{A_{i}}}.\,\!}$

The ions also contribute electrons, which we also need to account for. The number density for electrons ${\displaystyle n_{e}}$ depends upon the charge ${\displaystyle Z_{i}}$ of the atom (assuming complete ionization), with

${\displaystyle \sum _{i}=Z_{i}n_{i}.\,\!}$

Then

${\displaystyle P_{e}=n_{e}k_{B}T=k_{b}T\sum _{i}n_{i}Z_{i}={\frac {\rho k_{B}T}{m_{p}}}\sum _{i}{\frac {Z_{i}X_{i}}{A_{i}}}.\,\!}$

This allows us to define the electronic mean molecular weight as

${\displaystyle {\frac {1}{\mu _{e}}}={\frac {Z_{i}X_{i}}{A_{i}}}.\,\!}$

The final mean molecular weight is then

${\displaystyle {\frac {1}{\mu }}={\frac {1}{\mu _{i}}}+{\frac {1}{\mu _{e}}}.\,\!}$