2. States of stellar matter

MB statistics describes the distribution of speed or momentum of the particles in a gas constrained by three conditions: (i) all speeds in all three dimensions are equally probable, (ii) the total energy is limited and (iii) the total number of particles is limited. Number (i) ensures that the particles have a wide range of speeds and (ii) and (iii) ensure that there is enough variation in the speeds.

Consider a gas made of a limited number of particles. $\mathsf{P}(v_x)\ dv_x$ is the probability of finding a particle with speed $v_x$ in the interval $dv_x$, which can be shown to be

$$ \mathsf{P}(v_x)\ dv_x = \left(\frac{m}{2\pi kT}\right)^{1/2} e^{-mv_x^2/(2kT)} dv_x $$

where $m$ is the mass of the particle, $T$ the temperature of the gas and $k$ Boltzmann constant.

Momentum $p_x=mv_x$ and the probability $\mathsf{P}(p_x)dp_x=P(v_x)dv_x$. So

$$ \mathsf{P}(p_x)\ dp_x = \left(\frac{1}{2\pi mkT}\right)^{1/2} e^{-p_x^2/(2mkT)} dp_x $$

which is plotted in the figure (a) below. We can use figure (b) to transform this into three dimensions.

In three dimensions $\mathsf{P}(p)\ d^3p = \mathsf{P}(p_x)dp_x\ \mathsf{P}(p_y)dp_y\ \mathsf{P}(p_z)dp_z$ or

$$ \mathsf{P}(p)\ d^3p = \left(\frac{1}{2\pi mkT}\right)^{3/2} \exp\left(-\frac{p_x^2+p_y^2+p_z^2}{2mkT}\right) \ dp_x dp_y dp_z $$

which becomes

$$ \mathsf{P}(p)\ d^3p = \left(\frac{1}{2\pi mkT}\right)^{3/2} \exp\left(-\frac{p^2}{2mkT}\right) \ d^3p $$

which is the most used form of MBD. The quantity is dimensionless: $\mathsf{P}(p)$ has units (Ns)$^{-3}$ which is canceled by the unit (Ns)$^3$ of $d^3p$. That means the probability represents the number of particles expected per momentum cube.

MBD can also be written in terms of energy because $E=mv^2/2=p^2/(2m)$ which means

$$ \mathsf{P}(p)\ d^3p = \left(\frac{1}{2\pi mkT}\right)^{3/2} e^{-E/(kT)} \ d^3p $$

which means $\mathsf{P}(p) \propto e^{-E/(kT)}$ and the exponential becomes $e^{-1}$ when $E=kT$ . This form is also known as Maxwell-Boltzmann statistics or MBS.

Now as shown in figure ©, for a spherical shell in isotropic momentum space $d^3p=4\pi p^2\ dp$ and, hence, the probability of finding a particle in a thin shell of thickness $dp$ at a radius $p$

$$ \mathsf{P}(p)\ dp = \left(\frac{1}{2\pi mkT}\right)^{3/2} \exp\left(-\frac{p^2}{2mkT}\right) 4\pi p^2\ dp $$

which is more convenient because here the function is in 1D even though it is capable of describing a 3D gas. As in any probability, $\int \mathsf{P}(p)dp=1$ if the limit is from zero to infinity. This function is plotted in figure (d) above at two different temperatures $T_1$ and $T_2$.

We see that the function goes to zero at $p=0$ because of the $4\pi p^2$ term. At high momenta, $E > kT$, and the function again goes to zero because the exponential overcomes the $4\pi p^2$ term. So, unlike before, the maximum probability is not found at $p=0$. Also the $T^{-3/2}$ term ensures that the higher temperature curve has a lower amplitude.

Phase space is the 6D space $f(x,y,z,p_x,p_y,p_z)$ and MBS, or any other statistics, is best represented in this space.

If $N$ particles are distributed uniformly in volume $V$, then there position is described by $(x,y,z)$ in physical space (figure a), and their momentum is described by $(p_x,p_y,p_z)$ in momentum space (figure b). The number density of particles $n=N/V$ and hence, the phase space distribution function

$$ f_{MB} = n\mathsf{P}(p) = \frac{N}{V} \left(\frac{1}{2\pi mkT}\right)^{3/2} \exp\left(-\frac{p^2}{2mkT}\right) $$

where $f_{MB}$ has the units m$^{-3}$ (Ns)$^{-3}$, which results in (Js)$^{-3}$. So the phase space function gives probabilities of finding particles per joule-second cube. This $f$ is sometimes called phase-space density or just the distribution function.

It turns out all measurable or observable quantities in astronomy can be obtained by different integrations over the distribution function. For example, the particle number density

$$ n(x,y,z) = \int \int \int f\ d^3p $$

in units of particles m$^{-3}$. And the particle flux density

$$ \phi_p(x,y,z,t) = \int \int \int v f\ d^3p $$

in units of particles s$^{-1}$ m$^{-2}$ which is the number of particles streaming through a unit area per second.

Specific intensity, the most fundamental quantity in astronomy is also a similar integration. The definition of particle specific intensity

$$ J = \frac{dN}{dA\ d\Omega\ dE\ dt} $$

which is the number of particles passing through surface $dA$ with energy between $E$ and $E+dE$ within the solid angle $d\Omega$ per unit time $dt$.

Figure (a) above shows that the volume element in physical space $dV=v\ dt\ dA$ and the volume element in momentum space $dV_{mom}=p^2\ d\Omega_p\ dp$. So the number of particles in the 6D space

$$ dN = f p^2\ d\Omega_p\ dp\ v\ dt\ dA $$

which must be the same as the $dN$ given in the definition of $J$. So

$$ J dA\ d\Omega\ dE\ dt = f p^2\ d\Omega_p\ dp\ v\ dt\ dA $$

where $dA$ and $dt$ can be omitted from both sides and $d\Omega=d\Omega_p$ is also a good assumption. So

$$ J dE = f\ v\ p^2 dp. $$

Now, from special relativity, it can be shown that the total energy of a particle $E^2 = (pc)^2 + (mc^2)^2$. Differentiating this we get

$$ \frac{d}{dp}E^2 = c^2 2p + 0 \Rightarrow 2E\frac{dE}{dp}=c^2 2p \Rightarrow E\ dE = c^2 p\ dp. $$

And because $p=\gamma mv$ where $\gamma = E/(mc^2)$, we find $v=pc^2/E$. Replacing $dE$ and $v$ above we get

$$ J\frac{c^2 p\ dp}{E} = fp^2\ dp\ \frac{pc^2}{E} $$

which gives the very simple form:

\begin{equation}\label{J} J = p^2f \end{equation}

that is specific intensity is nothing but the distribution function multiplied by the momentum squared, they are essentially the same thing.

The unit of the particle-specific intensity $J(E)$ has to be m$^{-2}$ s$^{-1}$ J$^{-1}$ sr$^{-1}$ which is the particle flux density per unit energy interval. The unit of the energy-specific intensity $I(\nu)$ has to be J s$^{-1}$ m$^{-2}$ Hz$^{-1}$ sr$^{-1}$ which is the energy flux density per unit frequency ($\nu$) interval. By definition,

$$ I(\nu) = \frac{EJ(E)\ dE}{d\nu} $$

because this is the only way units would match. Now if $E=h\nu$ then $dE=h\ d\nu$ and, thus, $I=Jh^2\nu$. Now remembering that $J=p^2f$ and $p=h\nu/c$ we see that

\begin{equation}\label{I} I(\nu) = \frac{h^4\nu^3}{c^2} f \end{equation}

which is the equivalent of Eqn. \ref{J} for the energy-specific intensity.

Liouville’s theorem states that for an observer travelling with a stream of particle, the quantities $J$ and $I$ are conserved, they cannot change, their time derivative is zero. This, then, entails that the surface brightness of an astronomical object

\begin{equation}\label{B} B(\nu,\theta_s,\phi_s) = I(\nu,\theta_o,\phi_o) \end{equation}

where $(\theta_s,\phi_s)$ define a spherical coordinate at the source and $(\theta_o,\phi_o)$ define the same at the location of the observer.

Now we can derive the equation of state (EOS) of an ideal gas where the particles follow MB statistics. The EOS is the pressure $P(\rho,T)$ as a function of density $\rho$ and temperature $T$.

Cosider a particle moving to the right with momentum $\mathbf{p}$. It hits a wall head on and bounces off. The change in momentum

$$ \Delta\mathbf{p} = \mathbf{p}' - \mathbf{p} = - m\mathbf{v} - m\mathbf{v} = -2m\mathbf{v}. $$

So $+2m\mathbf{v}$ amount of momentum is transferred to the wall due to the collision. Force is the rate of change of momentum and pressure is the force per unit area of the wall. So the total pressure transferred to the wall due a total of $N$ particles

$$ P = N\frac{\Delta p}{\Delta t} \frac{1}{\Delta A} $$

where $N=V(n/6)$ and $n$ is the number of particles per unit volume $V=v\Delta t \Delta A$. The factor $6$ arises because there are six walls in the box containing the cylinder and only one-sixth of the particles hit a particular wall that we are considering. So

$$ P = \frac{nv}{6} 2mv = \frac{nmv^2}{3} = \frac{2}{3} n \frac{1}{2} mv^2 = \frac{2}{3} n E $$

where $E$ is the kinetic energy. So total pressure is proportional to the kinetic energy density $nE$. This is true if all the particles have the same mass and speed. However, if the particles have a distribution speeds, the general form of the pressure would be

$$ P = \frac{2}{3} n E_{av}. $$

Pressure and kinetic energy density even have the same units. Note that pressure is a tensor because it requires taking into account two vectors: the flow of particles and the normal force resulting from the bounce of the particles on a wall. But for an isotropic gas in equilibrium, the tensor becomes a scalar.

Now the average kinetic energy can be obtained by multiplying $p^2$ with its probability distribution MBD and integrating the product over all momenta, resulting in the expected value

$$ E_{av} = \left(\frac{p^2}{2m}\right)_{av} = \int_0^\infty \frac{p^2}{2m} P(p) dp = \frac{3}{2} kT $$

where the final step is given as a problem in APHB. The equation tells us that the average translational kinetic energy of the particles in a gas is a direct measure of its temperature.

We can find the color of the light emitted by a gas using this relation. A 12 MK monatomic hydrogen gas will have $E_{av}=kT=1.5$ keV and thus emit light at a frequency given by $h\nu\approx kT$, meaning at around 100 PHz, petahertz, falling in the x-ray range. For 6 kK, however, the radiation is in the visible range.

This equation is valid for monatomic gas that always has 3 degrees of freedom (DOF), motion in the 3 coordinates. Diatomic molecules have two additional DOFs, rotational kinetic energy and the vibrational kinetic energy. If energy per DOF is $E_{av}/3=kT/2$, then for diatomic gas

$$ E_{av} = \frac{5}{2} kT $$

which will not be important in case of stars. Molecules break down inside the hot interior of stars.

For stars, the Equation of State (EOS) for the gas inside a star is found by substituting $E_{av}$ in the equation of pressure:

$$ P = nkT = \frac{\rho}{m_{av}} kT $$

where the average mass $m_{av}=m_p/2$ for a hydrogen plasma. This is the preferred form of the ideal gas law. The more familiar form $PV=\mathsf{n}RT$ results from $n=\mathsf{n}N_0/V$ where $N_0$ is the Avogadro number.

The particles of the standard model of particle physics carry an intrinsic angular momentum expressed as the spin quantum number $S$. The momentum $\mathbf{L}$ can be either zero or $\gtrsim \hbar=h/(2\pi)$. The magnitude

$$ |\mathbf{L}|^2 = S(S+1)\hbar^2 $$

where $S$ can only be either zero or a multiple of $1/2$. The projection of an angular momentum on an axis is also quantized. Spin $1/2$ particles can have only two projections: $\pm 1/2\hbar$, called ‘spin up’ and ‘spin down’, respectively.

Particles with half-integer spins are called fermions and those with integer spins, bosons. Electrons, protons, neutrinos, or the nuclei with an odd number of nucleons are fermions. Photons and the $^4$He and $^{12}$C nuclei are bosons.

Pauli’s exclusion principle says that no more than one fermion of a given spin can occupy a single state in the phase space. A 1D version of of the phase space is shown above. In the area $\Delta A=\Delta p_x \Delta x = h$, only two electrons are allowed because electrons can have opposite spins. In the 6D phase space, the volume of a phase-space quantum state

$$ \Delta x \Delta y \Delta z \Delta p_x \Delta p_y \Delta p_z = h^3 $$

which has units of J$^3$ s$^3$. If two fermions occupy a state, they must have opposite spins. Note that Pauli’s principle dictates the electron configuration in atoms.

A gas of fermions with extremely low temperature or high density fills all the lowest momentum states, unlike MBS where very few particles occupy the lowest states. Such a gas is called degenerate and the distribution of fermions in that gas is described by Fermi-Dirac statistics (FDS).

Because all the low-momentum states are filled in FDS, some particles find themselves at a momentum higher than usual. These particles create a huge extra pressure, called the degenracy pressure, which, for example, is responsible for keeping a white dwarf in equilibrium against inward gravity.

To explain degeneracy, consider a gas of 37 electrons occupying the states shown in the 1D phase-space diagram of Fig. (a) following MBS. They are not degenerate. Most of them are occupying the lowest states as dictated by 1D MBD. The momentum at which kientic energy is equal to $E_{av}=kT/2$ is indicated using a dashed line.

If this gas loses energy and cools down, the electrons will lose energy and settle down at the bottom of the diagram as in Fig. (C). All the low states are completely filled. Many of the electrons thus have a higher energy than they would normally have at this low temperature. This unusually higher energy creates the degeneracy pressure. The momentum at the upper end of the FDD is called the Fermi momentum $p_F$. The FD distribution $\mathsf{P}(p_x)$ looks almost rectangular, there are almost no electrons with $p>p_F$.

We can make a gas degenerate by squeezing it (decreasing $\Delta X$) instead of cooling, i. e. by moving from Fig. (a) to (b). Moving from (b) to (d) we see that the electrons remain almost in the same states. That means in a completely degenerate gas, the pressure becomes independent of temperature.

How can a degenerate gas cool down if no lower momenta are available for the electrons to reach? The answer lies in the protons and neutrons. The equipartition of energy dictates that electrons and protons have equal average energy. The energy $p^2/(2m)$ is the same for all kinds of particles, but the protons have 43 times more momentum than electrons. Even after the electrons reach degeneracy, the protons keep following MBS and losing energy.

In the realistic 3D case, the fermions will try to gather near the origin of the momentum space in a spherical region as shown below because the momentum is lowest there. The region is delimited by the Fermi momentum $\mathbf{p}_F$ which is a radial vector. The physical dimenions $x,y,z$ are not shown because the 6D phase space is difficult to visualize.

Now to find the Fermi momentum and distribution function, first note that the total volume of the region in 6D phase space $V_{px}=V_pV_x$ where $V_p=4\pi p_F^3/3$ is the volume in momentum space and $V_x=4\pi R^3/3$ is the volume in physical space. The number of allowed quantum states in the volume is $V_{px}/h^3$. If each of these states has two electrons, then the total number of electrons within the volume

$$ N_e = \frac{2}{h^3} \frac{4}{3} \pi p_F^3 V_x $$

and, hence, the number density of electrons

$$ n_e = \frac{N_e}{V_x} = \frac{2}{h^3} \frac{4}{3} \pi p_F^3 $$

which means the Fermi momentum

$$ p_F = h\left(\frac{3}{8\pi}n_e\right)^{1/3} $$

depends only on the electron density if a volume is completely degenerate.

However, matter is usually not completely degenerate and we need the FDS for understanding the distribution of momenta. According to FDS, the probability of finding a particle

$$ f_{FD} = \frac{2}{h^3} \frac{1}{e^{\alpha+E/(kT)}+1} $$

where $\alpha(n,T)$ is a complicated function of $n$ and $T$ and, hence, constant for a particular density and temperature. In the non-degenerate limit $f_{FD}$ becomes $f_{MB}$; note the similarity with MBS.

The quotient $2/h^3$ is the number of particles in complete degeneracy and the second term modifies this number as a function of energy when the degeneracy is partial. Near complete degeneracy, $\alpha=-E_F/(kT)$ which leads to the Fermi function

$$ F(E) = \frac{1}{e^{(E-E_F)/(kT)}+1} $$

and, finally, to the shorter form the distribution

$$ f_{FD} = \frac{2}{h^3} F(E) $$

which is plotted in Fig. (b) above. Note that at zero temperature the factor $(E-E_F)/(kT)$ changes from $-\infty$ suddenly to $+\infty$ at $E=E_F$ and, hence, the Fermi function changes from unity to zero.

In Fig. (C), the Fermi function is plotted for two more temperatures $T_1$ and $T_2$. As temperature increases, more particles become available at higher energies. For $T>0$, the function becomes $1/2$ at $E=E_F$ because the exponential becomes unity ($e^0=1$).

In non-relativistic conditions, the relation between Fermi energy and Fermi momentum is simply $E_F=p_F^2/(2m)$, but in relativistic conditions

$$ E_F = U_F - mc^2 = \sqrt{p_F^2c^2+m^2c^4} - mc^2 $$

where $U_F$ is the total energy associated with Fermi level and $mc^2$ the rest mass of energy of a particle. This reduces to the classical value if the momentum $p_F$ is low. Fermi energy is the highest kinetic energy we find for a particle in a completely degenerate gas.

In a degenerate gas, the protons and the nuclei are responsible for the inward gravitational pull and the electrons are almost totally responsible for the outward pressure that balances gravity. This is because pressure

$$ P \approx nE_{av} $$

and the kinetic energy of degenerate electrons are much higher than their normal average. In non-degenerate matter, the average kinetic energy of electrons and protons are equal dictated by the equipartition of energy.

As in the case of MBS, we find the $E_{av}$ first to find the EOS. For electrons in nonrelativistic degeneracy, the average kinetic energy

$$ E_{av} = \frac{1}{N_e} \int_0^{p_F} \frac{p^2}{2m_e} \frac{2V_x}{h^3} 4\pi p^2 dp $$

where $dp$ is the thickness of a shell in momentum space and $V_x$ is the volume in physical space. Here $2V_x/h^3$ is the total number of particles in the physical space. So

$$ E_{av} = \frac{4\pi V_x}{N_e m_e h^3} \int_0^{p_F} p^4 dp = \frac{4\pi V_x}{N_e m_e h^3} \frac{p_F^5}{5} = \frac{4\pi p_F^5}{5m_e h^3} \frac{1}{Ne/V_x}. $$

Substitute the previously found $N_e/V_x$ here to find

$$ E_{av} = \frac{4\pi p_F^5}{5m_e h^3} \frac{3h^3}{8\pi p_F^3} = \frac{3}{5} \frac{p_F^2}{2m_e}. $$

We can also replace $p_F$ with what we found before and, so, write the pressure as

$$ P_e = \frac{2}{3} n_e E_{av} = \frac{2}{3}n_e \frac{3}{10 m_e} h^2 \left(\frac{3n_e}{8\pi}\right)^{2/3} = \frac{1}{20} \left(\frac{3}{\pi}\right)^{2/3} \frac{h^2}{m_e} n_e^{5/3}. $$

The number density $n_e=\rho/(\mu_e m_p)$ where

$$ \mu_e = \frac{\sum_i m_i/m_p}{N_e} $$

is the electron molecular weight, roughly the number of free electrons per nucleon. So

$$ P_e = \frac{1}{20} \left(\frac{3}{\pi}\right)^{2/3} \frac{h^2}{m_e} \left(\frac{\rho}{\mu_e m_p}\right)^{5/3} $$

which means the pressure is proportional to $\rho^{5/3}$ and does not depend on temperature.

For particles with relativistic velocity ($v\approx c$), $E=pc$ similar to photons and the pressure

$$ P_e = \frac{1}{3}nE_{av}. $$

We can find the average kinetic energy as

$$ E_{av} = (pc)_{av} = \frac{1}{N_e} \int_0^{p_F} pc \frac{2}{h^3} V_x 4\pi p^2 dp = \frac{3}{4}cp_F. $$

Similar to what we have done for the nonrelativistic case, replace $p_F$ with its full form and find that

$$ P_e = \frac{1}{8} \left(\frac{3}{\pi}\right)^{1/3} ch \left(\frac{\rho}{\mu_e m_p}\right)^{4/3} $$

meaning the pressure is proportional to $\rho^{4/3}$ and it does not depend on the temperature.

When the electrons inside a white dwarf reach relativistic speeds, the pressure varies softly with density and cannot be high enough to sustain the dead star against gravity leading to a gravitational collapse. This can occur when the mass of a white dwarf approaches the Chandrasekhar limit of $\sim 1.4$ M$_\odot$.

We have discussed three different EOS for non-degenerate gas following MBS and the relativistic and nonrelativistic degenerate gas following FDS. Photons follow a different statistics, the Bose-Einstein Statistics (BES) and we will see in a later chapter that in that case the EOS is $P=aT^4/3$, meaning the pressure depends only on temperature.

The four EOS are depicted above in a logarithmic temperature-density ($T-\rho$) plot. At low density and high temperature, radiation pressure dominates and at high density and low temperatures, degeneracy pressure dominates. The ideal gas pressure is dominant in between.

Neutron stars are mainly supported by repulsive nuclear forces, but neutron degeneracy pressure also plays a role. When all support mechanisms fail, gravity overtakes and turns a star into a black hole.