Ideal Bose & Fermi Gases - 1 | Physics for IIT JAM, UGC - NET, CSIR NET PDF Download

Gases

Here we consider systems with the kinetic energy exceeding the potential energy of inter-particle interactions:  

Ideal Gases

We start from neglecting the potential energy of interaction completely. Note though that quantum effect does not allow one to consider particles completely independent. The absence of any interaction allows one to treat any molecule as subsystem and apply to it the Gibbs canonical distribution: the average number of molecules in a given state is

which is called Boltzmann distribution. One can also use grand canonical ensemble considering all molecules in the same state as a subsystem with a non-fixed number of particles. Using the distribution (35) with N = na and  one expresses the probability of occupation numbers via the chemical potential: 

Consider now a dilute gas, when all . Then the probability of no particles in the given state is close to unity,   and the probability of having one particle and the average number of particles are given by the Boltzmann distribution in the form

  (39)

 Boltzmann (classical) gas is such that one can also neglect quantum exchange interaction of particles (atoms or molecules) in the same state which requires the occupation numbers of any quantum state to be small, which in turn requires the number of states V p³/h³ to be much larger than the number of molecules N . Since the typical momentum is  we get the condition

 (40)

To get the feeling of the order of magnitudes, one can make an estimate with m = 1.6. 10^-24g (proton) and n = 10²¹cm^-3 which gives   Another way to interpret (40) is to say that the mean distance between molecules n^-1/3 must be much larger than the wavelength h/p. In this case, one can pass from the distribution over the quantum states to the distribution in the phase space:

(41)
In particular, the distribution over momenta is always quasi-classical for the Boltzmann gas. Indeed, the distance between energy levels is determined by the size of the box,

 which is much less than temperature according to (40). To put it simply, if the thermal quantum wavelength   is less than the distance between particles it is also less than the size of the box. We conclude that the Boltzmann gas has the Maxwell distribution over momenta. If such is the case even in the external field then 

 That gives, in particular, the particle density in space   where n₀ is the concentration without field. In the uniform gravity field we get the barometric formula   

Partition function of the Boltzmann gas can be obtained from the partition function of a single particle (like we did for two-level system and oscillator) with the only difference that particles are now real and indistinguishable so that we must divide the sum by the number of transmutations:

Using the Stirling formula ln   we write the free energy

   (42)
Since the motion of the particle as a whole is always quasi-classical for the Boltzmann gas, one can single out the kinetic energy: 

 If in addition there is no external field (so that  describes rotation and the internal degrees of freedom of the particle) then one can integrate over  and get for the ideal gas

To complete the computation we need to specify the internal structure of the particle. Note though that  depends only on temperature so that we can already get the equation of state 

Mono-atomic gas. At the temperatures much less than the distance to the first excited state all the atoms will be in the ground state (we put ε₀ = 0).
That means that the energies are much less than Rydberg   and the temperatures are less than   (otherwise atoms are ionized).
If there is neither orbital angular momentum nor spin (L = S = 0 — such are the atoms of noble gases) we get   as the ground state is non-degenerate and

Here  is called the chemical constant. Note that for F = AT + B T ln T the energy is linear  that is the specific heat, C_v = B , is independent of temperature. The formulas thus derived allow one to derive the conditions for the Boltzmann statistics to be applicable which requires . Evidently, it is enough to require exp where

Using such 

If there is a nonzero spin the level has a degeneracy 2S + 1 which adds

  to the chemical constant (45). If both L and S are nonzero then the total angular momentum J determines the fine structure of levels ε_J (generally comparable with the room temperature — typically,  Every such level has a degeneracy 2J + 1 so that the respective partition function

Without actually specifying ε_J we can determine this sum in two limits of large and small temperature. and z = (2S + 1)(2L + 1) which is the total number of components of the fine level structure. In this case 

In the opposite limit of temperature smaller than all the fine structure level differences, only the ground state with ε_J = 0 contributes and one gets

where J is the total angular momentum in the ground state.

Note that c_v = 3/2 in both limits that is the specific heat is constant at low and high temperatures (no contribution of electron degrees of freedom) having some maximum in between (due to contributions of the electrons).
We have already seen this in considering two-level system and the lesson is general: if one has a finite number of levels then they do not contribute to the specific heat both at low and high temperatures.

Specific heat of diatomic molecules

We need to calculate the sum over the internal degrees of freedom in (43). We assume the temperature to be smaller than the energy of dissociation (which is typically of the order of electronic excited states). Since most molecules have S = L = 0 in the ground state we disregard electronic states in what follows. The internal excitations of the molecule are thus vibrations and rotations with the energy characterized by two quantum numbers, j and K :

        (46)

We estimate the parameters here assuming the typical scale to be Bohr radius  and the typical energy to be Rydberg ε₀ =  Note that m = 9 . 10^-28g is the electron mass here. Now the frequency of the atomic oscillations is given by the ratio of the Coulomb restoring force and the mass of the ion:

Rotational energy is determined by the moment of inertia   We may thus estimate the typical energies of vibrations and rotations as follows:

 (47)

Since  then that both energies are much smaller than the energy of dissociation  and the rotational energy is smaller than the vibrational one so that rotations start to contribute at lower temperatures:

The harmonic oscillator was considered in in Sect. 1.4.2. In the quasiclassical limit,    the partition function of N independent oscillators is  

 the free energy  and the mean energy from (24): E = N T . The specific heat C_V = N .

For a quantum case, the energy levels are given by 

where one sees the contribution of zero quantum oscillations and the breakdown of classical equipartition. The specific heat (per molecule) of vibrations is thus as follows:   we have  At large T we have classical equipartition (every oscillator has two degrees of freedom so it has T in energy and 1 in C_V ).
To calculate the contribution of rotations one ought to calculate the partition function

Again, when temperature is much smaller than the distance to the first level,  the specific heat must be exponentially small. Indeed, retaining only two first terms in the sum (49), we get  which gives in the same approximation

We thus see that at low temperatures diatomic gas behaves an mono-atomic.

At large temperatures,  the terms with large K give the main contribution to the sum (49). They can be treated quasi-classically replacing the sum by the integral:

That gives the constant specific heat c_rot = 1. The resulting specific heat of the diatomic molecule, is shown on the figure:

Note that for the specific heat (weakly) decreases because the distance between rotational levels increases so that the level density (which is actually c_v ) decreases.
For (non-linear) molecules with  atoms we have 3 translations, 3 rotations and 6 — 6 vibrational degrees of freedom (3n momenta and out of total 3n coordinates one subtracts 3 for the motion as a whole and 3 for rotations). That makes for the high-temperature specific heat c_v = c_tr + c_rot + c_vib = 3=2 + 3=2 + 3 - 3 = 3 . Indeed, every variable (i.e. every degree of freedom) that enters ε(p, q), which is quadratic in p, q, contributes 1/2 to c_v . Translation and rotation each contributes only momentum and thus gives 1/2 while each vibration contributes both momentum and coordinate (i.e. kinetic and potential energy) and gives 1.

Fermi and Bose gases

Like we did at the beginning  we consider all particles at the same quantum state as Gibbs subsystem and apply the grand canonical distribution with the potential

 (51)

Here the sum is over all possible occupation numbers n_a. For fermions, there are only two terms in the sum with n_a = 0; 1 so that

For bosons, one must sum the infinite geometric progression (which converges when depends on T ; V ; μ. The average number of particles in the state with the energy ε is thus

 (52)

Upper sign here and in the subsequent formulas corresponds to the Fermi statistics, lower to Bose. Note that at exp both distributions turn into Boltzmann distribution (39). The thermodynamic potential of the whole system is obtained by summing over the states

(53)
Fermi and Bose distributions are generally applied to elementary particles (electrons, nucleons or photons) or quasiparticles (phonons) since atomic and molecular gases are described by the Boltzmann distribution (with the recent exception of ultra-cold atoms in optical traps). For elementary particle, the energy is kinetic energy, ε = p²/2m, which is always quasi-classical (that is the thermal wavelength is always smaller than the size of the box but can now be comparable to the distance between particles). In this case we may pass from summation to the integration over the phase space with the only addition that particles are also distinguished by the direction of the spin s so there are g = 2s + 1 particles in the elementary sell of the phase space.
We thus replace (52) by

 (54)

Integrating over volume we get the quantum analog of the Maxwell distribution:

(55)
In the same way we rewrite (53):

Since also  we get the equation of state

(57)
We see that this relation is the same as for a classical gas, it actually is true for any non-interacting particles with ε = p²/2m in 3-dimensional space. Indeed, consider a cube with the side ℓ. Every particle hits a wall  times per unit time transferring the momentum  in every hit. The pressure is the total momentum transferred per unit time divided by the wall area ℓ² :

 (58)

In the limit of Boltzmann statistics we have E = 3N T /2 so that (57) reproduces P V = N T . Let us obtain the (small) quantum corrections to the pressure assuming exp

 Expanding integral in (56)

and substituting Boltzmann expression for μ we get

Non-surprisingly, the small factor here is the ratio of the thermal wavelength to the distance between particles. We see that quantum e®ects give some e®ective attraction between bosons and repulsion between fermions.

Degenerate Fermi Gas

The main goal of the theory here is to describe the electrons in the metals (it is also applied to the Thomas-Fermi model of electrons in large atoms, to protons and neutrons in large nucleus, to electrons in white dwarf stars, to neutron stars and early Universe). Drude and Lorents at the beginning of 20th century applied Boltzmann distribution and obtained decent results for conductivity but disastrous discrepancy for the specific heat (which they expected to be 3/2 per electron). That was cleared out by Sommerfeld in 1928 with the help of Fermi-Dirac distribution. Since the energy of an electron in a metal is comparable to Rydberg and so is the chemical potential (see below) then for most temperatures we may assume  so that the Fermi distribution is close to the step function:

At T = 0 electrons fill all the momenta up to p_F that can be expressed via the concentration (g = 2 for s = 1=2):

(60)
which gives the Fermi energy

(61)
The chemical potential at T = 0 coincides with the Fermi energy (putting already one electron per unit cell one obtains   Condition is evidently opposite to (40). Note that the condition of ideality requires that the electrostatic energy Ze²/a is much less than where Ze is the charge of ion and  is the mean distance between electrons and ions. We see that the condition of ideality,  surprisingly improves with increasing concentration. Note nevertheless that in most metals the interaction is substantial, why one can still use Fermi distribution (only introducing an effective electron mass) is the subject of Landau theory of Fermi liquids to be described in the course of condensed matter physics (in a nutshell, it is because the main effect of interaction is reduced to some mean effective periodic field).

To obtain the specific heat,  one must find  E(T,V,N) i.e exclude μ from two relations, (55) and (56).

At  this can be done perturbatively using the formula

which gives

From the first equation we find μ(N, T) perturbatively

and substitute it into the second equation:

we see that

Photons

Consider electromagnetic radiation in an empty cavity kept at the temperature T. Since electromagnetic waves are linear (i.e. they do not interact) thermalization of radiation comes from interaction with walls (absorption and re-emission). One can derive the equation of state without all the formalism of the partition function. Indeed, consider the plane electromagnetic wave with the fields having amplitudes E and B. The average energy density is (E² +B²)/2 = E² while the momentum flux modulus is

The radiation field in the box can be considered as incoherent superposition of plane wave propagating in all directions. Since all waves contribute the energy density and only one-third of the waves contribute the radiation pressure on any wall then

PV =E/3

In a quantum consideration we treat electromagnetic waves as photons which are massless particles with the spin 1 which can have only two independent orientations  correspond to two independent polarizations of a classical electromagnetic wave). The energy is related to the momentum by = cp. Now, exactly as we did for particles [where the law = p²/2m gave PV = 2E/3 — see (58)] we can derive (65) considering that every incident photon brings momentum 2p cos θ to the wall, that the normal velocity is c cos θ and integrating   Photon pressure is relevant inside the stars, particularly inside the Sun.

Let us now apply the Bose distribution to the system of photons in a cavity. Since the number of photons is not fixed then minimumality of the free energy, F(T, V, N), requires zero chemical potential:0. The Bose distribution over the quantum states with fixed polarization, momentum is called Planck distribution

At  it gives the Rayleigh-Jeans distribution  which is classical equipartition. Assuming cavity large we consider the distribution over wave vectors continuous. Multiplying by 2 (the number of polarizations) we get the spectral distribution of energy

It has a maximum at  The total energy

    (68)

where the Stephan-Boltzmann constant is as follows:  The specific heat depends only on temperature, c_P does not exist (may be considered infinite). One can also derive the free energy (which coincides with  and entropy   that is the Nernst law is satisfied: S→0 when T → 0. Under adiabatic compression or expansion of radiation entropy constancy requires V T³ = const and PV⁴/3 = const.

Phonons

 The specific heat of a crystal lattice can be calculated considering the oscillations of the atoms as acoustic waves with three branches (two transversal and one longitudinal) where u_i  is the respective sound velocity. Debye took this expression for the spectrum and imposed a maximal frequency ω_max so that the total number of degrees of freedom is equal to 3 times the number of atoms:

Here we introduced some effective sound velocity u defined by   One usually introduces the Debye temperature

where a is the lattice constant.

We can now write the energy of lattice vibrations using the Planck distribution (since the number of phonons is indefinite, μ = 0)

At  for the specific heat we have the same cubic law as for photons:

    (73)

For liquids, there is only one (longitudinal) branch of phonons so C = which works well for He IV at low temperatures.

At  we have classical specific heat (Dulong-Petit law) C = 3N. Debye temperatures of different solids are between 100 and 1000 degrees Kelvin. We can also write the free energy of the phonons

and find that, again, al low temperatures  Nernst theorem. An interesting quantity is the coefficient of thermal expansion To get it one must pass to the variables P, T, μ introducing the Gibbs potential G(P, T) = E − T S + PV and replacing  At high temperatures,  It is the Debye temperature here which depends on P, so that the part depending on T and P in both potentials is linearly proportional to 

That makes the mixed derivative

independent of temperature. One can also express it via so-called mean geometric frequency defined as follows: When the pressure increases, the atoms are getting closer, restoring force increases and
so does the frequency of oscillations so that

Note that we’ve got a constant contribution  in (72) which is due to quantum zero oscillations. While it does not contribute the specific heat, it manifests itself in X-ray scattering, M ̈ossbauer effect etc. Incidentally, this is not the whole energy of a body at zero temperature, this is only the energy of excitations due to atoms shifting from their equilibrium positions. There is also a negative energy of attraction when the atoms are precisely in their equilibrium position. The total (so-called binding) energy is negative for crystal to exists at T = 0.
One may ask why we didn’t account for zero oscillations when considered photons in (67,68). Since the frequency of photons is not restricted from above, the respective contribution seems to be infinite. How to make sense out of such infinities is considered in quantum electrodynamics; note that the zero oscillations of the electromagnetic field are real and manifest them-selves, for example, in the Lamb shift of the levels of a hydrogen atom. In thermodynamics, zero oscillations of photons are of no importance.

Bose gas of particles and Bose-Einstein condensation

We consider now an ideal Bose gas of massive particles with the fixed number of particles. This is applied to atoms at very low temperatures. As usual, equaling the total number of particles to the sum of Bose distribution over all states gives the equation that determines the chemical potential as a function of temperature and the specific volume. It is more convenient here to work with the function z= exp(μ/T) which is called fugacity:

We introduced the thermal wavelength  and the function

One may wonder why we split the contribution of zero-energy level as it is not supposed to contribute at the thermodynamic limit V → ∞. Yet this is not true at sufficiently low temperatures. Indeed, let us rewrite it denoting n₀ = z/(1-z) the number of particles at p =0

     (76)

The function  behaves as shown at the figure, it monotonically grows while z changes from zero  . Remind that the chemical potential of bosons is non-positive (otherwise one would have infinite occupation numbers).  and the derivative is infinite. When the temperature and the specific volume v= V/N  are such that   (notice that the thermal wavelength is now larger than the inter-particle distance) then there is a finite fraction of particles that occupies the zero-energy level. The solution of (76) looks as shown in the figure. When V → ∞ we have a sharp transition at  we have

Therefore, at the thermodynamic limit we put  and as it follows from (76). All thermodynamic relations have now different expressions above and below T_c (upper and lower cases respectively):

At low   it decreases faster than   for electrons (since the number of over-condensate particles now changes with T as for phonons and photons and μ = 0 too) yet slower than (that we had for ε_p= p²/2m, are denser at lower energies. On the other hand, since the distance between levels increases with energy so that at high temperatures c_v decreases with T as for rotators

At T < T_c the pressure is independent of the volume which promts the analogy with a phase transition of the first order. Indeed, this reminds the properties of the saturated vapor (particles with nonzero energy) in contact with the liquid (particles with zero energy): changing volume at fixed tem- perature we change the fraction of the particles in the liquid but not the pressure. This is why the phenomenon is called the Bose-Einstein condensation. Increasing temperature we cause evaporation (particle leaving condensate in our case) which increases cv; after all liquid evaporates (at T = T_c) c_v starts to decrease. It is sometimes said that it is a “condensation in the momentum space” but if we put the system in a gravity field then there will be a spatial separation of two phases just like in a gas-liquid condensation (liquid at the bottom).

We can also obtain the entropy [above Tc by usual formulas that follow from (56) and below Tc just integrating specific heat  

   (79)

The entropy is zero at T = 0 which means that the condensed phase has no entropy. At finite T all the entropy is due to gas phase. Below T_c we can write S/N = (T/T_c)^3/2 s= (v/v_c) s where s is the entropy per gas particle:  The latent heat of condensation per particle is T s that it is indeed phase transition of the first order.