1  Hydrostatic Equilibrium

The fundamental physics of stars is determined by a handful of principles:

hydrostatic from Greek ὕδωρ, ‘water’, and στάσις, ‘standing’; just like water in a tank.

Let’s talk about hydrostatic equilibrium first.

An ordinary star like the Sun, throughout its whole body, is to a very good approximation an ideal gas, and is fully ionized except in its outermost layer. This means that the gas pressure satisfies the equation of state

\[ \textcolor{blue}{p} = n \textcolor{red}{k_B} \textcolor{orange}{T} \]

where \(\textcolor{blue}{p}\) is the pressure, \(n\) is the number density (particles per volume) of the gas, \(\textcolor{red}{k_B}\) is Boltzmann’s constant (\(1.38\times10^{-16}\) erg/K), and \(\textcolor{orange}{T}\) is the temperature in kelvin.

In a star like the Sun, the pressure is mostly provided by gas pressure. In hotter stars, photon or radiation pressure is dominant, but in the Sun this is \(\sim 10^{-3} p_\text{gas}\).

Even in the Sun, though, the gas is not quite a classical ideal gas: quantum mechanics is already relevant. There is an equation we will derive later in these notes for the pressure due to the degeneracy of a gas where the quantum wavefunctions of its constituent particles are nearly overlapping:

\[ p_\text{degeneracy} = \frac{\textcolor{purple}{\hbar}}{5\textcolor{green}{m_e}} (\frac{3}{8 \pi})^{2/3} n^{5/3} \] It turns out this is about a quarter of the gas pressure at the core of the Sun!

where \(\textcolor{purple}{\hbar}\) is the quantum of action \(h/2\pi\) (\(1.0546 \times 10^{-27} \text{erg}\cdot\text{s}\)), and \(\textcolor{green}{m_e}\) the mass of the electron.

1.1 The Equation of Hydrostatic Equilibrium

Consider a thin shell of radius \(r\) (and surface area \(A = 4\pi r^2\)), thickness \(dr\), and mass density \(\rho\), enclosing a mass \(M_r\).

i.e. \(M_r\) is the total mass integrated out up to a radius \(r\).

The mass of this shell is \(M_\text{shell} = \rho A dr\), and from Newton’s law of gravitation the magnitude of the gravitational force of the whole shell inwards is

Newton’s Shell Theorem states that the gravitational attraction of a symmetric shell of matter, and therefore by linearity of a ball of matter, can be treated as if the mass were concentrated at a point at the centre.

\[ \frac{- G M_r M_\text{shell}}{r^2} \]

So now we can calculate the net force on this shell (and therefore acceleration \(a\)), and require the forces to be in balance:

\[ F_\text{net} = M_\text{shell} a = P_\text{below} \cdot A - P_\text{above} \cdot A - \frac{G M_r M_\text{shell}}{r^2} \]

Letting \(P_\text{above} = P_\text{below} + dP\),

\[ M_\text{shell} a = a \rho A dr = - A dP - \frac{G M_r M_\text{shell}}{r^2} \]

and therefore rearranging, in equilibrium (\(a = 0\)) we have the equation of hydrostatic equilibrium:

\[ \frac{dP}{dr} = -\rho \frac{G M_r}{r^2} \tag{1.1}\]

Tattoo this equation on the back of your eyelids.

1.2 Plane-Parallel Approximation

This is actually a very familiar equation that we encounter not just in first-year physics, but in everyday life. Nearly all the mass of the Earth is in its body, and only a bit less than a millionth in its atmosphere. So we can consider a situation where

\[ r = R_{\textcolor{NavyBlue}{\oplus}} + \textcolor{green}{z} \]

where \(\textcolor{green}{z}\) is the \(\textcolor{green}{\text{height}}\) in the atmosphere \(\llless R_\oplus\). We often denote astronomical bodies by traditional symbols: \(\textcolor{NavyBlue}{\oplus}\) is the astronomical symbol for the \(\textcolor{NavyBlue}{\text{Earth}}\), and \(\textcolor{yellow}{\odot}\) the \(\textcolor{yellow}{\text{Sun}}\). There are many other traditional symbols that are now rarely used.

We can now write this in familiar terms with the acceleration due to gravity as

\[ \frac{dP}{dh} = -\rho \cdot \underbrace{\frac{G M_r}{r^2}}_{{\equiv g \text{, constant}}} \]

If we are dealing with an incompressible liquid like water, then \(\rho\) is a constant and we simply have

We can rearrange this to solve for the depth of water required to reach a gauge pressure \(\Delta P \equiv P-P_0\): \[ z = - \frac{\Delta P}{\rho g} \] For \(\rho_\text{water} = 1000 \text{kg\,m}^{-3}\), to get a 1 atmosphere (\(=10^6\) Pa) gauge pressure requires 10.34 m of water.

\[ P(z) = P_0 - \rho g z \]

Things are different when your density depends on pressure. For an ideal gas, \(P = n k_B T\) and \(\rho = n \textcolor{OliveGreen}{\langle m \rangle}\) - so we have an equation of state that we can use to solve for the vertical structure of an \(\textcolor{orange}{\text{isothermal}}\) atmosphere:

where \(\textcolor{OliveGreen}{\langle m \rangle}\) is the \(\textcolor{OliveGreen}{\text{mean molecular mass}}\), and \(\textcolor{orange}{\text{isothermal}}\) means having the same temperature everywhere, from ἴσος, “same”, and θέρμη, “heat”.

\[ \begin{aligned} \frac{d}{dz} (n k_B T) &= - \langle m \rangle g n \\ \frac{dn}{dz} &= - \frac{\langle m \rangle g}{k_B T} \cdot N \\ n &= n_0 \exp{(- \frac{\langle m \rangle g}{k_B T} z)} \\ n &= n_0 \exp{- z/\textcolor{cyan}{h}} \end{aligned} \]

and we call \(\textcolor{cyan}{h}\) the \(\textcolor{cyan}{\text{scale height}}\) in the atmosphere. Let’s calculate this for some interesting situations!

First let’s try an isothermal Earth atmosphere:

mol_earth = 28.964 * m_p # average molecular weight for earth's atmosphere
T_earth = 300 # K; room temperature
g_earth = 9.8 # m/s

h = k_B * T_earth / mol_earth / g_earth

This scale height of 8.7 km is 1.4\(\,\times 10^{-3} R_\oplus\); the atmosphere really is very thin and can be treated as plane-parallel. This scale height would mean that the air is 96% as dense at the top of the \(\textcolor{Maroon}{\text{Q1 Tower}}\), and 36% as dense at the summit of \(\textcolor{gray}{\text{Mt Everest}}\).

\(\textcolor{Maroon}{\text{Q1 Tower}}\) on the Gold Coast is the tallest tower in Australia, at \(\textcolor{Maroon}{322.5}\) m; \(\textcolor{gray}{\text{Mt Everest}}\), Earth’s tallest mountain, stands \(\textcolor{gray}{8849}\) m tall.

Now let’s plug in some numbers for the Sun, using Python:

m = 0.5 * m_p # kg
# mean molecular weight for ionized hydrogen = 
# mean of electron & proton = m_p/2

g = G * M_sun / R_sun**2
h = k_B * Teff_sun / m / g

giving a gravity \(g =\) 28.0 Earth gravities, and a scale height of 5.0\(\,\times 10^{-4}\,R_\odot\). So we see that the solar atmosphere has a tiny scale height relative to the overall size of the Sun, and the plane-parallel approximation is even better than on Earth!

1.3 Scaling Relations

Most of the time, it is not possible to calculate properties of realistic stars in closed-form equations, and we will have to use computer models. But what understanding does this buy us? It is often just as important to have a sense of how these properties scale with one another in general terms, even if we might not be able to estimate actual numbers to better than an order of magnitude.

As a first example, let’s estimate the pressure at the centre of the Sun. Ideally, we would solve for the full stellar structure using some equation of state, and then integrate \(\frac{dP}{dr}\) from zero at the surface \(r=R_\odot\) all the way to the core. Instead, we’re going to do something very handwavy. Let’s rearrange the \(\frac{dP}{dr}\) in Equation 1.1 to instead be the mean gradient \(P/R_\odot\):

\[ \begin{aligned} \frac{dP}{dr} &= -\rho \frac{GM_r}{r^2} \\ \frac{P}{R_\odot} &\approx \langle \rho \rangle \frac{GM_\odot}{R_\odot^2}\\ P &\approx \langle \rho \rangle \frac{GM_\odot}{R_\odot}\\ &\approx \frac{M_\odot}{4/3 \pi R_\odot^3} \frac{GM_\odot}{R_\odot}\\ &\approx \frac{GM_\odot^2}{4/3 \pi R_\odot^4} &\propto \frac{GM_\odot^2}{R_\odot^4} \end{aligned} \]

Remarkably, it can be shown analytically (Milne 1936) that \(P > \frac{GM^2}{8\pi R^4}\) for an arbitrary star - so we are not even too far from the exact solution with these mathematical sins. It is often the case that very coarse approximations differ from exact solutions by a factor of order unity; we will therefore often drop these factors and just think about the scaling.

which for the Sun, gives \(1.1 \times 10^{15}\) Pa pressure, which is within an order of magnitude of the central pressure of \(2.3 \times 10^{16}\) Pa in the Standard Solar Model (Guenther et al. 1992).

We are going to do this kind of approximation a lot in these notes: to estimate how nuclear reactions scale with temperature, how stellar luminosities depend on mass, how stars form and how they die - and we will also tackle real computer models of these processes, so that we build both intuitive understanding and accurate models we can fit to data. Both are essential tools in astronomy.

“May God us keep From Single vision & Newtons sleep.” — William Blake, Letter to Thomas Butts, 22 November 1802.