Week 1

Question 1

When a process occurs in a fluid element that is in equilibrium at all times

Answer 1

Write dQ = Tds

Question 2

It is convenient to define quantities on a per baryon basis as baryon number is a conserved quantity. Then the first law of thermodynamics take on what form?

Answer 2

The general form dQ=d(E/n)+Pd(1/n), where dQ is the heat gained per baryon, P is the pressure, and 1/n is the volume per baryon.
This is addition to the simplifying assumption, dQ=Tds

Question 3

In general, the energy density of a system containing different species depends on the relative amounts of each species, as well as the volume (1/n) and the adiabat (value of s). Therefore we write the concentration of the ith species of particle by Yi = ni/n, where ni is the number density of particle i. Then what form does the energy take?

Answer 3

E = E(n,s,Yi), where E is the energy per baryon.
Therefore we should write in general d(E/n) = -Pd(1/n)+Tds+Sum(muidYi),
where P = -d(E/n)/d(1/n) = n^2*d(E/n)/dn,
T = d(E/n)/ds,
mui = d(E/n)/dYi = dE/dni.
The quantity mui is called the chemical potential of species i. It has the interpretation of the change in energy density of species i, while the volume, entropy, and all number densities are kept constant. Note that since E is defined to include the rest-mass energy, so is mui!

Question 4

In equilibrium reactions between the particles produce a state of detailed balance where each reaction is balanced by its inverse and the concentration of each species remains constant. Thus in equilibrium, the concentrations Yi are not independent of the other thermodynamic quantities. We can determine equilibrium relations in what way?

Answer 4

Consider first the special case where the system is infinitesimally close to equilibrium, but the sytem is kept thermally isolated (dQ=0) and at constant volume (dn=0) so that no work is done on the system. In this case, dQ=d(E/n)+Pd(1/n) gives d ( E / n ) = 0; that is, the energy of the system remains constant. Entropy is generated by the reactions, but because entropy is a maximum in equilibrium (from the second law of thermodynamics), ds = 0 to first order. Thus we get that Sum(muidYi)=0
Suppose, for example, one is considering the reaction
e-+p++ n+ve in equilibrium. Then dYe = dYP = - dYn = - dY, and so
mue + mup = mun + munu(e
Then the Sum(mui*Yi)=0 gives the concentrations n as a function s or E. Even wken the initial state is very far from equilibrium, if the system proceeds to equilibrium with dQ = 0 and dn = 0, then its energy still remains constant. Eventually it will be infinitesimally close to equilibrium and the above argument will apply. Thus the composition can again be determined if the chemical potentials are known at the fixed values of n and E.

Question 5

Now consider the general case where the system is not necessarily thermally isolated, and work may be done on the system. If the system were to achieve equilibrium by quasistatic reactions, then Tds = dQ. In general, however, the second law implies
dQ

Answer 5

d(E/n) + Pd(1/n)

Question 6

Quantities like energy, volume, entropy, and number of particles are extensive quantities: if one divides a given volume in two by means of a partition, the energy, entropy, and number of particles on each side are half of their values for the whole volume. Quantities like pressure and temperature which remain the same are called intensive quantities. The requirement that all the extensive quantities of the system scale with volume in the same way leads to the relation

Answer 6

g = Sum(mui*Yi), where g is the Gibbs free energy and is written as
dg = 1/ndP - sdT+Sum(mui*Yi)

Question 7

Let us be a little more precise about the number of independent thermody- namic quantities required to specify an equilibrium state. Consider, by way of illustration, an interacting mixture of baryons (including, e.g., neutrons and protons) and leptons (including electrons and muons and their associated neutri- nos). All reactions in a given volume will conserve baryon number density n, electron lepton number den~ity,n,~, muon lepton number density, nLp, and electric charge density nQ. Choose four basic chemical potentials corresponding to these four conserved quantities. For example, choose these to be p p (associated with n), pn(associated with n,,), p,, (associated with nLp),and p, (associated with no). Then in equilibrium all other chemical potentials will be linear combinations of these four. Now in equilibrium all thermodynamic quantities associated with species i are functions only of T and p,. (In the next section, we will see this explicitly for ideal gases.) Thus in general one needs to specify T and four of the p,’s for a complete description of the equilibrium state. Equivalently, one needs to specify any five independent thermodynamic quantities. Usually nQ = 0, so only four quantities are required.

Answer 7

In some situations the concept of a limited equilibrium applies. This means that certain reactions necessary to acheve complete equilibrium are too slow on the timescale of interest. This implies that there will be more than four conserved quantities, and one has to specify more n,’s to describe the system. For example, the dynamical timescale of a star is generally much shorter than the timescale to change its composition by nuclear reactions. To determine the pressure, internal energy, and so on in the star, one has to specify the concentration of H, He, and so on explicitly, and not simply the baryon density n . A similar situation prevails for chemical reactions in a terrestrial laboratory.

Question 8

In lunetic theory, the number density in phase space for each species of particle, dN/d3x d3p,provides a complete description of the system. Equivalently,one can specify the dimensionless distribution function in phase space, f(x, p, t ) , defined by

Answer 8

dN/d3xd3p = (g/h^3)*f, where h is Plancks constant, g the statistical weight (2S+1), for photons g=2 and for neutrinos g=1. The functionfgives the average occupation number of a cell in phase space. d3xd3p is a Lorentz invariant (i.e., scalar under Lorentz transformations) and hence f is also.

Question 9

Convenient form of velocity

Answer 9

v = pc2/E

Question 10

What are two examples of f (the distribution function)

Answer 10

The fermi and bose distributions, f(E) = 1/(exp((E-mu)/kT)+/-1)

Question 11

For completely degenerate Fermions we use what function?

Answer 11

f(E)=1 if EEFermi

Question 12

An isolated whte dwarf or neutron star ultimately cools to zero temperature, and it is the pressure associated with matter at T = 0 that supports these stars against gravitational collapse. The simplest cold, degenerate equation of state is that due to a single species of ideal (noninteracting) fermions. For definiteness, let us take the gas to be one of electrons at zero temperature. The gas can be treated as ideal if we ignore all electrostatic interactions. Then how do we define the Fermi Momentum?

Answer 12

EF =(pF2c2+me2c4)1/2

Question 13

What is the relativity parameter?

Answer 13

x=pF/(me*c)

Question 14

Even in applications where degenerate electrons contribute most of the pres- sure, the density is usually dominated by

Answer 14

the rest-mass of the ions. This density is Sum(ni*mi)

Question 15

What is the mean Baryon rest mass

Answer 15

mB=1/n*Sum(nimi) = Sum(nimi)/Sum(niAi). This allows us to define the density of the gas in terms of the abundance of baryons times the mean Baryon rest mass. or also as the abundance of electrons times the mean baryon rest mass divided by the mean number of electrons.

Question 16

What is the mean molecular weight per electron?

Answer 16

mue = mB/m_u/Ye, and this allows one to define the density as muem_une. ne is the abundance of electrons.

Question 17

What is the total density of relativistic electrons?

Answer 17

rho = rho0 + Ee/c^2

Question 18

What is Ye? What is an expression for Ye in terms of Integer Atomic Number and Integer Atomic Weight?

Answer 18

The mean number of electrons per baryon.

Ye = Z/A

Question 19

What is the mean molecular weight?

Answer 19

mu, and it is defined by
rho0 = (ne+Sum(ni))mum_u
Therefore, we have that 1/mu = (Ye+Sum(Yi))m_u/mB. Except for very accurate work, m_u = mB. The mean molecular weight, mu, is particularly useful in the nondegenerate limit, when the pressure is given by the perfect gas law,
P=(ne + Sum(ni))kT = rho0/mu/m_u*(kT)

Question 20

What are examples of Electrostatic Corrections to the Equation of State?

Answer 20

The ideal Fermi gas equation of state of the previous section was employed by Chandrasekhar (1931a, b) in his pioneering analysis of equilibrium white dwarfs. There are two important corrections to this equation of state in practice. One of them is inverse Beta-decay. Corrections due to electrostatic interactions among the ions and electrons are the other. The principal electrostatic correction arises because the positive charges are not uniformly distributed in the gas but are concentrated in individual nuclei of charge 2. This decreases the energy and pressure of the ambient electrons: the repelling electrons are, on the average, further apart than the mean distance between nuclei and electrons, so repulsion is weaker than attraction. In a nondegenerate gas, Coulomb effects become more important as the density increases: the ratio of Coulomb Energy to Thermal Energy is
E/kT = Ze^2//kT = Ze^2*n^(1/3)/kT, which increases with ne. Here = ne^1/3 is the characteristic electron ion separation. In contrast, for a degenerate ga, we have EC/EF = (Ze^2/)/(pF^2/2me). EC«EF in most astrophysical degenerate cases.

Question 21

What is the Bohr Radius?

Answer 21

a0 = h^2/me^2

Question 22

We can derive an approximate expression for the correction to the ideal degenerate equation of state using the fact that n, is to first approximation uniform.
As T->0, the ions are located in a lattice that maximizes the inter-ion separation. Consider a “spherical” cell of the lattice of volume 4pir03/3 = l/nN, where n N = number density of nuclei. In this Wigner-Seitz approximation‘ the gas is imagined to be divided into neutral spheres of radius r, about each nucleus, which contain the 2 electrons closest to the nucleus. The total electrostatic energy of any one sphere is the sum of the potential energies due to electron-electron (e-e) interactions and electron-ion ( e - i ) interac- tions. To assemble a uniform sphere of 2 electrons requires energy?

Answer 22

E = integral(qdq/r), where q = -Zer^3/r0^3 and there are two contributions to the energy - the electron electron energy and the electron ion energy. Since the cells are neutral, interactions between electrons and nuclei of different cells can be ignored.

Question 23

Although the above Coulomb correction is relatively small, we shall find it is nevertheless important for hgh-density white dwarfs and low-density neutron stars. When is it not useful?

Answer 23

At low densities it is no longer a good approximation to treat the electron gas as uniform. An accurate equation of state at laboratory densities is very complicated to derive, because electron shell effects mask the simpler statistical effects. Above a few times laboratory densities, however, a statistical approach to the equation of state works well and is sufficient to treat the low-mass range of white dwarfs (and even the gross structure of large planets). The simplest statistical treatment of atomic structure is the Thomas-Fermi method. One assumes that within each Wigner-Seitz cell, the electrons move in a slowly varying spherically symmetric potential V(r ) . Because the potential is approximately constant at any point, we can use free-particle Fermi-Dirac statistics. This means we are taking the interaction energy between the electrons to be much less than the kinetic or potential energies of an individual electron. Thus at any r . all states up to E = EF are occupied. The energy E , is independent of r, otherwie electrons would migrate to a region of smaller EF.

Question 24

At high densities, the most important correction to the equation of state is due to inverse Beta-decay” :
e-+p-> n+v. The proton and neutron are generally bound in nuclei. In this section, however, we will get an overview of the effects of inverse P-decay by examining the case of a gas of free electrons, protons, and neutrons. We assume that the neutrinos generated in reaction escape from the system. In more complicated models we will treat the more interesting case of bound nuclei. The reaction can proceed whenever the electron acquires enough energy to balance the mass difference between proton and neutron, ( m N - m P ) c 2= 1.29 MeV. This process is effective for transforming protons into neutrons if Beta-decay, that is,
n-+p+e+F
does not occur. It is blocked if the density is high enough that all electron energy levels in the Fermi sea are occupied up to the one that the emitted electron would fill. Thus there is a critical density for the onset of reactions converting protons into neutrons.

Answer 24

The result ne:np:nn= 1:1:8 in the limit of very large densities is a trivial consequence of charge neutrality,Beta-equilibrium, and extreme relativistic degeneracy. One can compute the maximum momentum of an electron emitted in the reaction, showing that pF is greater than this value for all densities above a critical value, thus verifying that the equilibrium is stable. The results of this section are not, strictly speaking, applicable to the case of gravitational collapse with escaping neutrinos, even when the collapse occurs quasistatically and at zero temperature. Thermodynamic equilibrium is not achieved in an open system, and the n-p-e composition must be determined by solving appropriate rate equations for the various reactions.
The above equations do precisely describe an equilibrium system of fixed charge (zero), baryon number, -and lepton number. The lepton number in this system has been chosen to have its minimum possible value; that is, we have taken a limit where nv->0(muv->0) while preserving detailed balance.

Question 25

Cold Equatfon of State Relow Neutron Drip: Summary

Answer 25

I. The equation of state for zero-temperature matter is well understmd in the density regime below “neutron drip,” where pdrip= 4 x 10^11 g cm-3 . The pressure is dominated by degeneraie electrons that become Fully
relativistic above - 10^7g CM-3’. Positive charge is concentrated in separated nuclei, which form a regular Coulomb lattice embedded in the electron gas.
2.) If the matter has settled into its ground state, it may be assumed to be in nuclear equilihrium; that is, its energy cannot he lowered by changing its composition via strong, weak, or electromagnetic interactions. The equilihrium nuclide can be determined as a function of density. BeIow 10^7 g cm-3 the ground state corresponds to 56:26:Fe nuclei. At higher densities the equilibrium nuclide becomes increasingly neutron-rich. The nuclei are stabi-
lized against Beta-decay by the filled Ferrni sea of electrons.
3.) The equation of state helow neutron drip determines the structure of planets and stable white dwarfs. In white dwarfs, matter has probabIy not achieved complete equilibrium. The Chandrasekhar equation of state with Coulomb Carrections (Salpeter. I961) therefore applies. The composition dcpends on the evolutionary history of the star and determines the mean molecular weight per electron, mu_e (cl. Chapter 3). In neutron stars. matter in this density regime is in complete equilibrium and an equilibrium equation of state (e.g., BPS) is applicable.