ASTR 498N                     Lecture 6                   Matters of State

 

(online at www.astro.umd.edu/~drabin/)

 

Well, isn’t this is a fine state.

 

                                                                     

A pressure equation of state (usually shortened to equation of state) relates the pressure to the temperature, mass density, and composition of a gas in thermodynamic equilibrium:  P = P(T,ρ,C).  Because T, ρ, and C vary within a star, and the star is radiating energy, a star cannot be in strict thermodynamic equilibrium (TE).  Usually, however, it makes sense to consider the gas to be at a particular location to be very nearly in TE; this approximation is called local thermodynamic equilibrium (LTE).

 

To justify the LTE approximation with one example, we consider the variation of temperature in the solar interior.  A characteristic microscopic length scale is the mean free path (mfp), , where σ is a cross section.  If we consider a point in the Sun at about half its radius, T 10^6.5 K and n_e   10³⁰ m³.  The mfp for photons is limited mainly by scattering off free electrons (Thomson scattering), for which σ πr₀² where r₀ = e²/m_ec² = 10^14.6 m is the so-called classical electron radius.  The resulting mfp is  10^1.3 m.  Comparing this with a characteristic temperature gradient, T_cen / R   10^2.6 K m³, we see that the temperature is expected to vary only minutely over the range of a typical photon.

 

 

Mean molecular weight

          (O&C §10.2)

 

The mean molecular weight μ in a gas is just the average mass of a free particle in units of m_H,  the mass of a hydrogen atom.  Thus, the number density of free particles is

 

                               

 

Let X_Z represent the fraction by mass of the element with atomic number Z, such that .  Let N_Z be the number of free particles contributed to the gas by each atom of element Z.  If all such atoms are neutral, N_Z = 1.  For complete ionization, N_Z = Z + 1.  The number density of atoms of element Z is

 

                               

 

where A_Z is the atomic weight. 

 

With these definitions, the total number density of free particles is

 

                               

 

Comparing this with n = ρ / μm_H , we have

 

                               

 

It is conventional in astrophysics to assign special labels to the following mass fractions:

 

                              X₁ ≡ X        X₂ ≡ Y                   1  X  Y ≡ Z´

 

(we use Z´ here instead of the usual Z to avoid confusion with the nuclear charge).  With these definitions,

 

                               

 

where  is an average over Z > 2, weighted by X_Z.   It is a convenient empirical fact that A_Z  2Z + 2 (check the periodic table).

 

In general, μ = μ(T, P, C) since temperature and pressure determine the degree of ionization and C determines what there is to ionize.  Consider two extreme cases:  a neutral gas and a completely ionized gas.  The completely ionized limit is a good approximation over much of a stellar interior.

 

Completely neutral

 

N_Z = 1 for all Z.  For the relative abundances of heavy elements in the Sun,  and

 

                               

 

To the extent that the relative heavy-element abundances vary in lockstep as Z´ (“metallicity”) changestrue to first approximationthis expression applies to any Z´. 

 

Completely ionized

 

Here N_H = 2, N_He = 3, N_Z = Z + 1.  With A_Z   2Z + 2,  = 0.5 and

 

                               

 

independent of T and p.  For typical Population I composition (X = 0.70, Y = 0.28, Z´ = 0.02),  μ  0.62.  For pure H, μ = 0.5.  For pure He, μ = 1.33.  For pure heavy elements with solar relative abundances, μ = 2.  Thus, 0.5 < μ < 2  for complete ionization.

 

 

Perfect gas

          (O&C §10.2)

 

A perfect gas is one in which there are no interactions between particles in the gas.  Although this criterion is never satisfied exactly, it is physically sound if the average interaction energy between particles is much smaller than their kinetic energies.  In a stellar interior, particles interact mainly via Coulomb potential energy, which is typically much smaller than kT.

 

There’s one immediate use for the μ-nastics abovethe perfect gas equation of state:

 

                                

         

P_gas depends on the composition of the gas through μ.  Note that the perfect gas equation of state holds for a classical gas even when the particles are relativistic.

 

P    Show that P_gas = 2 u_gas / 3 for non-relativistic particles and P_gas = u_gas / 3 for ultra-relativistic particles, where u_gas is the kinetic energy density.

 

         

Photon gas

          (O&C §10.2)

 

                               

 

where u_rad is the energy density.  P_rad  is a function of T only.

 

 

Equilibrium particle statistics

 

Recall that the Maxwellian speed distribution for a classical gas followed from the more fundamental velocity-space density, f (v) d³v.   In quantum or relativistic systems, it is more natural to work with the density of states in momentum space,  g (p) d³p, giving the number of states per unit spatial volume with momentum p in the range dp_x, dp_y, dp_z.

 

It is shown in most quantum mechanics texts that the number of distinct states of a free particle in a box of linear dimensions dxdydz and in the momentum range dp_xdp_ydp_z is 1/h³; that is, each state occupies a volume h³ in phase space.  This is a consequence of the uncertainty principle.  In the terminology above, g (p) = 1/h³.  However, protons, neutrons and electrons also have spin, an internal degree of freedom that doubles the number of possible states.  Similarly, photons can have two distinct states of polarization.  Thus, the correct expression is g (p) = 2/h³.

 

Because an equilibrium gas is isotropic, we may integrate over momentum shells to obtain the density function of p = |p|,

 

                               

 

Now, this expression enumerates all possible states.  Integrating g(p)dp over all p gives the (infinite) volume density of states, not the spatial number density of particles. To get the spatial volume density of particles, we have to multiply g(p)dp by the probability f (p) that a state of momentum p will be occupied by a particle.   That is,

                               

 

The occupation index f (p) is derived in statistical mechanics for the two fundamental forms of quantum statistics:

 

                Bose-Einstein (identical bosons)

 

                Fermi-Dirac (identical fermions)

 

where  is the relativistic expression for energy and ψ  is the chemical potential that appears in the thermodynamic relationship dE = TdS  PdV + ψ dN.  Note that ψ  has the dimensions of energy (per particle).  Note also that f (p) is not a density function:  , where ε_p is the energy corresponding to p.

 

The occupation index for every state will be small if it is small for even the lowest energy state, ; that is, if , in which case

 

                  Maxwell-Boltzmann (classical statistics)

 

Thus, for both fermions and bosons, the condition for a classical gas is

 

                              .

 

P    Use the normalization condition  to derive an expression for the chemical potential ψ  of a classical gas of non-relativistic particles and show that the condition for a classical gas demands that the average separation of gas particles is large compared with their typical de Broglie wavelength.