Stellar Structure

First of all, what even is a star? A star is a luminous ball of plasma held together by its own gravity. These produce their light through nuclear fusion reactions in their cores. The nearest star to Earth is the Sun.

A star can shine steadily for thousands of millions of years. Thus, the equilibrium of a star must remain stable for such long periods. Mathematically, the conditions for the internal equilibrium of a spherically symmetric star can be expressed as four differential equations governing the distribution of mass, gas pressure, and energy production and transport in the star.

Mass Continuity

The first equation gives the mass contained within a given radius. Consider a spherical shell of thickness $dr$ at the distance $r$ from the centre. Its mass is

$$dm = 4 \pi r^2 \rho \, dr$$

which gives the mass continuity equation

$$\tag{4.1.1} \boxed{ \frac{dm}{dr} = 4 \pi r^2 \rho }$$

Energy Continuity

The second equilibrium condition expresses the conservation of energy, requiring that any energy produced in the star has to be carried to the surface and radiated away. Let $\varepsilon$ is the energy production coefficient, which is the amount of energy released in the star per unit time and mass. Let $l$ be the energy flux at radius $r$, then

$$\tag{4.1.2} \frac{dl}{dm} = \varepsilon$$

or

$$\tag{4.1.3} \boxed{\frac{dl}{dr} = 4\pi r^2 \rho \varepsilon}$$

If the mean free path of a photon in the star is $\lambda$, the probability of it getting absorbed in length $dr$ is $\frac{dr}{\lambda}$. Hence the change in flux is

$$\tag{4.1.4} dF = \frac{F}{\lambda} dr$$

The opacity $\kappa$ is defined as

$$\tag{4.1.5} \kappa = \frac{1}{\rho \lambda}$$

Hydrostatic Equilibrium

The force of gravity pulls the stellar material towards the centre. It is resisted by the pressure force due to the thermal motions of the gas molecules, as well as pressure from radiation. The third equilibrium condition is that these forces be in equilibrium.

Consider a cylindrical volume element at the distance $r$ from the centre of the star. If the mass enclosed inside radius $r$ is $m$, the force of gravity on the element is

$$dF_g = -\frac{Gm dm}{r^2} = -\frac{Gm \rho}{r^2}\,dA\,dr$$

If the pressure at the lower surface of the volume element is $P$ and at its upper surface $P + dP$, the net force of pressure acting on the element is

$$dF_p = -dP \, dA$$

Hence the equilibrium condition becomes $dF_g + dF_p = 0$, or

$$\tag{4.1.6} \boxed{ \frac{dP}{dr} = - \frac{Gm \rho}{r^2} = - g \rho }$$

or

$$\tag{4.1.7} \frac{dP}{dm} = - \frac{Gm}{4 \pi r^4}$$

Gas Pressure

Stars are huge spheres of fully ionized gas (or plasma). Hence to a good approximation, we can assume the ideal gas to apply to the contents of the star.

$$P = \frac{k}{\mu m_H} \rho T$$

where $\mu$ is the mean molecular weight. Hydrogen when ionized gives rise to two particles ($\ce{H^+}$ and $\ce{e}$), per nucleon (single proton in Hydrogen). Helium gives $3/4$ particles per nucleon when ionized (3 particles from one atom, each atom has 4 nucleons). For other elements having atomic number $Z$, give rise to ~$1/2$ particles per nucleon when ionized.

Let the relative mass fractions of Hydrogen, Helium and heavier elements (which are all called metals by astronomers) be $X$, $Y$ and $Z$ respectively, such that $X + Y+ Z = 1$. Then the expression for the mean molecular weight becomes

$$\tag{4.1.8} \mu = \frac{1}{2X + \frac{3}{4}Y + \frac{1}{2} Z }$$

For our Sun, $\mu_\odot \approx 0.6$.

Central Pressure

A lower limit on the central pressure can be calculated as

$$\int_0^S dP = \int_0^S - \frac{Gm \, dm}{4 \pi r^4} \gt \int_0^S - \frac{Gm \, dm}{4 \pi R^4} = -\frac{GM^2}{8 \pi R^4}$$

where $S$ represents the surface of the star, at $r= R$. Taking the pressure at surface to be zero, we get

$$\tag{4.1.9} P_c > \frac{GM^2}{8 \pi R^4}$$

Virial Theorem

Rearranging and integrating equation (4.1.7),

$$\frac{dP}{dm} = -\frac{Gm}{4\pi r^4} \implies \int_0^S 4 \pi r^3 dP = \int_0^S -\frac{Gm \, dm}{r} = \Omega$$

where $\Omega$ is the total gravitational potential energy of the star. Now since $4 \pi r^3$ = $3V$, where $V$ is the volume of the sphere upto radius $r$, using integration by parts we get

$$3 \int_0^S V dP = 3 \left( VP \vert_0^S - \int_0^S P dV \right) = -3 \int_0^S P dV$$

Since at $r=0$ we have $V=0$ and at $r=R$ we have $P =0$. Putting this back, we get the virial theorem

$$\tag{4.1.10} \boxed{ -3 \int_0^S P \, dV = \Omega }$$

Assuming an ideal gas star with an adiabatic index $\gamma$, the pressure $P$ is related to the energy density $u$ as

$$u = \frac{P}{\gamma - 1} \implies P = (\gamma - 1)u$$

Substituting this back into the virial theorem,

$$\tag{4.1.11} \quad \boxed{\, 3 (\gamma - 1) U + \Omega = 0 \,}$$

where $U = \int_0^S u dV$ is the total internal energy of the gas.

The total energy of the star is thus

$$E = U + \Omega \:= -(3\gamma - 4) \, U \:= \frac{3\gamma - 4}{3 \gamma - 3} \, \Omega$$

For a monoatomic gas, $\gamma = 5/3$, giving $E = -U = \frac{1}{2} \Omega$.

As the total energy of a star increases, its gravitational potential energy increases (expands) and internal energy decreases (cools down). Hence, stars have negative heat capacity.

Energy transport

The total energy flux consists of energy flux due to radiation, convection and conduction. In stars, conduction is very inefficient, hence convection and radiation are the only processes at play.

Energy transport via radiation

The pressure due to radiation is

$$\tag{4.1.12} P_\text{rad} = \frac{1}{3}aT^4$$

where $a = 4\sigma /c$ is the radiation constant.

The energy flux due to radiation is

$$dF = d \left( \frac{1}{A} \frac{dE}{dt} \right) = d \left( \frac{1}{A} \frac{dp}{dt} c \right) = c \, dP_\text{rad}$$

Combining this with equation (4.1.4) and equation (4.1.12),

$$F \, \frac{dr}{\lambda} = \frac{4}{3} \, ac T^3 \, dT$$

Since $F = \frac{l_\text{rad}}{4 \pi r^2}$, we get

$$\tag{4.1.13} \boxed{ \frac{dT}{dr} = -\frac{3}{4ac} \: \frac{\kappa \rho}{T^3} \frac{l_\text{rad}}{4 \pi r^2} }$$

or

$$\tag{4.1.14} \frac{dT}{dm} = -\frac{3}{64 \pi^2 ac} \: \frac{\kappa l_\text{rad}}{r^4 T^3}$$

Energy transport via convection

To model convection, consider a small blob of gas inside the star. Initially, it is in hydrostatic equillibrium with the surroundings ($P_b = P_g$, $\rho_b = \rho_g$). After is rises up, it expands adiabatically. If in the new position, the blob has a density $\rho_b + d\rho_b > \rho_g + d\rho_g$, the blob will sink back down and the gas is stable to convection. If $\rho_b + d\rho_b < \rho_g + d\rho_g$, the blob is buoyant and will continue to rise; this marks the onset of convection. Hence the condition for stability against convection is

$$d\rho_b > d\rho_g$$

Given $\rho_b = \rho$ and $P_b = P$, and that in the new position it is in hydrostatic equillibrium ($dP_b = dP$), we get

$$\tag{4.1.15} - \left( 1 - \frac{1}{\gamma} \right) \frac{T}{P} \frac{dP}{dr} > -\frac{dT}{dr}$$

Assuming an ideal gas with adiabatic index $\gamma$. The LHS of the expression is the adiabatic temperature gradient, while the RHS is the actual temperature gradient. As long as equation (4.1.15) is satisfied, convection will not take place. Convection starts when these two are just equal, or

$$\tag{4.1.16} \boxed{ \frac{dT}{dr} = \left(1 - \frac{1}{\gamma} \right) \frac{T}{P} \frac{dP}{dr} }$$

Eddington Luminosity

The equation for temperature gradient due to radiation (4.1.13) can be written in terms of radiation pressure as

$$\frac{dP_\text{rad}}{dr} = - \frac{\kappa \rho}{c} \frac{L}{4 \pi R^2}$$

near the surface. Hydrostatic equillibrium demands that near the surface

$$\frac{dP}{dr} = - \frac{GM \rho}{R^2}$$

This puts an upper limit on a star’s luminosity, called the Eddington limit

$$\tag{4.1.17} L_\text{ED} = \frac{4\pi Gc}{\kappa} M$$

For pure ionized hydrogen, $\kappa = \sigma_T / m_p$.

Stars with luminosity higher than the Eddington limit will slowly lose their mass until they attain hydrostatic equillibrium.

Boundary Conditions

We have four differential equations for stellar structure

$$\begin{aligned} \frac{dm}{dr} = 4 \pi r^2 \rho \qquad \qquad& \frac{dP}{dr} = - \frac{Gm \rho}{r^2} \\ \frac{dl}{dr} = 4\pi r^2 \rho \varepsilon \qquad \qquad& \frac{dT}{dr} = \begin{cases} -\frac{3}{4ac} \: \frac{\kappa \rho}{T^3} \frac{l_\text{rad}}{4 \pi r^2} &\small{\text{(for radiation)}} \\ \frac{dT}{dr} = \left(1 - \frac{1}{\gamma} \right) \frac{T}{P} \frac{dP}{dr} &\small{\text{(for convection)}} \end{cases} \end{aligned}$$

But seven variables. Hence we need three more equations. Generally these are

• Equation of state: $P = P(\rho, T, [X])$
• $\kappa = \kappa(\rho, T, [X])$
• $\varepsilon = \varepsilon(\rho, T, [X])$

Here $[X]$ represents the chemical composition of the star. In addition to this, the boundary conditions usually taken are:

• $m(r=0) \,= 0,\quad\:\: l(r=0) \:= 0$
• $m(r=R) = M, \:\:\: l(r=R) \,= L$
• $T(r=R) \,= 0,\quad P(r=R) = 0, \quad \rho(r=R) = 0$

Energy generation in stars

Stars are fueled by nuclear fusion reactions which take place in their cores. The binding energy of an atom $\ce{^A_ZX}$ is

$$Q = Z m_p + (A-Z) m_n - m_X$$

The binding energy per nucleon is defined as $Q/A$, which is maximum for iron (Z = 56).

The main reactions which take place in stars are the following.

Proton-Proton (pp) Chain

$$\begin{align} \text{ppI}: \qquad & \ce{^1H + ^1H &->& \quad ^2H + e^+ + \nu_e} \\ & \ce{^2H + ^1H &->& \quad ^3He + \gamma} \\ & \ce{^3He + ^3He &->& \quad ^4He + 2 ^1H} \\ \end{align}$$$$\begin{align} \text{ppII}: \qquad \tag{3} & \ce{^3He + ^4He &->& \quad ^7Be + \gamma} \\ & \ce{^7Be + e^- &->& \quad ^7Li + \nu_e} \\ & \ce{^7Li + ^1H &->& \quad ^4He + ^4He} \\ \end{align}$$$$\begin{align} \text{ppIII}: \qquad \tag{3} & \ce{^3He + ^4He &->& \quad ^7Be + \gamma} \\ \tag{4} & \ce{^7Be + ^1 H &->& \quad ^8B + \gamma} \\ \tag{5} & \ce{^8B &->& \quad ^8Be + e^+ + \nu_e} \\ \tag{6} & \ce{^8Be &->& \quad ^4He + ^4He} \\ \end{align}$$

Reaction (1) has a very small probability of occuring, once every ~ $10^{10}$ years on average. The ppI chain is responsible for ~$91$% of sun’s energy production.

CNO Cycle

$$\begin{align} \tag{1} \ce{^{12}C + ^1H &-> ^{13}N + \gamma} \\ \tag{2} \ce{^{13}N &-> ^{13}C + e^+ + \nu_e} \\ \tag{3} \ce{^{13}C + ^1H &-> ^{14}N + \gamma} \\ \tag{4} \ce{^{14}N + ^1H &-> ^{15}O + \gamma} \\ \tag{5} \ce{^{15}O &-> ^{15}N + \gamma + \nu_e} \\ \tag{6} \ce{^{15}N + ^1H &-> ^{12}C + ^4He} \\ \tag{7} \end{align}$$

Triple Alpha Reaction

$$\begin{align} \tag{1} \ce{^{4}He + ^4He &<-> ^{8}Be} \\ \tag{2} \ce{^{4}He + ^8Be &-> ^{12}C + \gamma} \\ \end{align}$$

This is commonly written as $\ce{3 ^4He -> ^{12}C + \gamma}$

Carbon Burning

$$\begin{aligned} \ce{^{12}C + ^{12}C &-> ^{24}Mg + \gamma} \\ \ce{&-> ^{23}Na + ^1H} \\ \ce{&-> ^{20}Ne + ^4He} \\ \ce{&-> ^{23}Mg + n^0} \\ \ce{&-> ^{16}O + 2 ^4He} \\ \end{aligned}$$

Carbon Burning occurs when Helium is exhausted.

Oxygen Burning

$$\begin{aligned} \ce{^{16}O + ^{16}O &-> ^{32}S + \gamma} \\ \ce{&-> ^{31}P + ^1H} \\ \ce{&-> ^{28}Si + ^4He} \\ \ce{&-> ^{31}S + n^0} \\ \ce{&-> ^{24}Mg + 2 ^4He} \\ \end{aligned}$$

Silicon Burning

$$\begin{aligned} \ce{^{28}Si + ^{28}Si &-> ^{56}Ni + \gamma} \\ \ce{^{56}Ni &-> ^{56}Fe + 2e^+ + 2\nu_e} \\ \end{aligned}$$