Degenerate Gases

or

But the summation is just the partition function of the internal states, Z_int.Thus the grand sum becomes

 

(14.1)

We can think of Z_int = Z_rot + Z_vib + Z_electronic.

but <N(e_n)> << 1 in the classical regime, so

 

(14.2)

What about the quantum regime, ?We can reach the quantum regime by increasing the concentration or lowering the temperature.The temperature for the quantum regime is found from n_q to be any temperature below

 

(14.3)

A gas in the quantum regime with a temperature t << t₀ is often called a degenerate gas.

Fermionic Degenerate Gases

where I have used the fact that m = e_F at t = 0.We would like to calculate e_F.To do this, recall that if we consider waves we found that the energy could be written in terms of the wave number k, , where k = (n_xi + n_yj + n_zk) with n_x, n_y, n_z = ±1, ±2, ±3, .…Thus, in k-space, each state is a point.We can apply this to our Fermi gas since we know from quantum mechanics that we can associate a wave with a particle.The density of states was calculated earlier and was found to be

D = 2V/(2p)³

where the factor of two comes from the fact that each fermion has two possible spin states associated with an energy state.Now let the temperature go to zero.At t = 0, all of the particles have fallen into the lowest energy available to them within the constraints of the Pauli exculsion priciple.This means that in k-space the sphere of radius k_F is totally occupied with particles.Then the total number of particles is given by

This can be inverted to find the radius k_F

 

KF = (3p2n)1/3

(14.4)

Substituting this into e_k yields

 

(14.5)

We can use e_k to determine the total energy in the ground state

 

(14.6)

Thus we see that the average kinetic energy per particle is 0.6 e_F.At constant N, as the energy increases the volume decreases, so that the Fermi energy gives a repulsive contribution to the binding of any material; that is, the Fermi energy tends to increase the volume.In most metals, and in white dwarf and neutron stars, it is the most important repulsive interaction.It is balanced in metals by the Coulomb attraction between electrons and ions and in stars by gravitational attraction.

where N(k) is the number of states with wave vector less than or equal to k.

So

Now recall that

so that

and thus

 

(14.7)

Notice that, since , and recalling that , we get that the average occupancy looks like

To derive the heat capacity of a degenerate fermionic gas, we need to consider the effect of raising the temperature.As the temperature goes from 0 to t, the energy in the system is raised from U₀ to U(t) .The difference in energy can be calculated from

 

(14.8)

Using the fact that

(14.8) can be rewritten as

or

 

(14.9)

The first integral on the right hand side gives the energy needed to take electrons from e_F to the orbitals of energy e > e_F and the second integral gives the energy needed to bring the electrons to e_F from orbitals below e_F.

It is a good approximation to evaluate D(e) = D(e_F), allowing it to be removed from the integral.Finally, we can also assume that the chemical potential varies slowly with t, allowing us to replace m with e_F in f(e).Substituting these two approximations into the integral and evaluating, we get

 

(14.10)

and so

Returning to (14.10), if we recall that , then we see that

whert_F = e_F.

Bosonic Degenerate Gases

When, the occupancy of the ground orbital becomes equal to the total number of particles in the system, so that

 

(14.11)

where we have used the series expansion .This can be rearranged to yield

 

m = t/N

(14.12)

as.Note that the chemical potential is negative.This is because the chemical potential in a boson system must always be lower in energy than the ground state orbital in order that the occupancy of every orbital be non-negative.

Example:

Consider an atom in a cube of side l. The energy states of the atom are given by

where n_x, n_y and n_z are positive integers.In units of , we see that the energy of the ground state is e(111) = 3, while the set of next lowest orbitals all have energy e(211) = 6.Thus the lowest excitation energy of the atom is

If we let the atom be a normal helium atom with a mass of m(⁴He) = 6.6 x l0^-24 g and the size of the cube be l = 1 cm, then De = 2.48 x l0^-30 erg.This corresponds to a temperature of 1.8 x l0^-14 K.This seems to be an insignificant temperature, but for N = 10²² atoms, at t = 1 mK, m = -1.4 x 10^-41 erg, where e(111) is taken as the zero energy orbital.Thus l is much closer to e(111) than to any other state, and e^[e(111) - m]/t is much closer to 1 than e^[e(211) - m]/t, so e(111) dominates the distribution function.The fractional occupancy of the first excited orbital is given by

Thus we see that the Bose-Einstein distribution favors the situation where the greatest part of the population is left in the ground orbital at sufficiently low temperatures.The particles left in the ground orbital, as long as their numbers are >> 1, are called the Bose-Einstein condensate.

Temperature and Orbital Occupancy

Then the total number of ⁴He atoms in the ground and excited orbitals is given by

 

(14.13)

where N₀(t) is the number of atoms in the ground orbital at temperature t.We call N₀ the number of atoms in the condensed phase and the number of atoms in the normal phase.Using the Bose-Einstein distribution function and setting e = 0, we see that

 

N0(t) = l/(1 - l)

(14.14)

Similarly, N_e is found to be

We can evaluate this integral by noting that whenever N₀ >> 1, i.e. at sufficiently low temperatures, the .Setting l = 1, we get 

 

(14.15)

We can write N₀ and N_e as fractional occupancies by dividing each by the total number of atoms.This leads to

 

(14.16)

and

 

Ne = 2.612 nq/n

(14.17)

Einstein Condensation Temperature

 

(14.18)

This allows (14.17) to be rewritten as

The number of atoms in the ground orbital is found from

 

(14.19)

A graph of the fractional occupancy of the two phases yields

Example:

where V_M is the molar volume in cm³/mole and M is the molecular weight.For liquid helium V_M = 27.6 cm³/mole and M = 4, so T_E = 3.1 K.In reality, a transition is observed in liquid ⁴He at 2.17 K.A graph of the heat capacity as a function of temperature shows this transition clearly

It is believed that at 2.17 K there is a condensation of liquid⁴He into the ground orbital of the system.This is different from the condensation in coordinate space that occurs in the condensation of a gas into a liquid.The interatomic forces that lead to liquefaction of ⁴He at 4.2 K does not destroy the effects of boson condensation, so in this respect the liquid behaves as a gas.The condensation is normally not permitted for fermions, but pairs of fermions may act as bosons, as in the superconductivity of electron pairs (Cooper pairs) in metals.In liquid ⁴He, we call the state above 2.17 K He I, and the state below 2.17 K He II.

The upper triple point is the transition of the solid, which does not exist below a pressure of 25 atm, into either He I or He II.The triple point is 1.74 K and about 30 atm.The second triple point is where the vapor and liquid phases meet.It is interesting to note that at normal pressure the solid phase does not ever occur, even at absolute zero.The two triple points are connected by a line which separates the regions of He I and He II.

Elementary Excitations

 
 

½M0V2 = ½M0V'2 + ek

(14.20)

where V', is the velocity after the excitation, and

 
 

M0V = M0V' +

(14.21)

In order to find the minimum value of V which can generate an excitation, lets solve(14.21) for V' and use it to eliminate the kinetic energy in (14.20).Solving (14.21) forV' we get

or

Substituting this into (14.20) yields

 

(14.22)

When, we can get the minimum of V as

This can be simplified if we assume that M₀ is large compared to the phonon momentum, so that

 

Vmin = ek/

(14.23)

A body moving slower than this will not be able to create excitations in the liquid, so the motion will be frictionless.We can see this result geometrically if we plot the energy of excitation as a function of momentum.If we construct a straight line which just meets the curve from below, the slope of that line is the minimum velocity.

If , as for the excitation of a free atom, the straight line has zero slope and the critical velocity is zero.The energy of a low energy phonon in liquid He II is , where v_s is the velocity of sound in liquid He II.The critical velocity is equal to sound if V_crit = v_s is valid for all wavevectors, which it isn't for He II.The observed critical flow velocities are considerably lower than the velocity of sound, presumably because the plot of e_kversesmay turn downward at very high .The actual spectrum of elementary excitations in liquid He II has been determined by observations on the inelastic scattering of slow neutrons.For the wavevectors covered by these neutron experiments, the critical velocity is V_C = 5 x 10³ cm/s.