Statistical Thermodynamics: Molecules to Machines

Statistical Thermodynamics: Molecules to MachinesVenkat Viswanathan May 5, 2015Module 4: Electrons, Phonons and PhotonsLearning Objectives: Analyze the statistical thermodynamics of Bose and Fermi particles Demonstrate consistency with our analysis for the ideal gas of Bose particles (ideal gas of diatomic molecules) Analyze the behavior of electrons in a metal Discuss the thermodynamic behavior of a crystalline solid. Find the role of lattice vibrations, resulting in the definition of quasi- particles called phonons. Compare two models for lattice vibrations Look at photons in the context of Bose-Einstein statistics Derive Plancks law of radiation and use density of states to derive thermodynamic properties of a photon gas.Key Concepts:Bose and Fermi statistics, bosons and fermions, electron gas, Fermi energy, Fermi momentum, Fermi temperature, electron pressure, elec- tron heat capacity, crystal lattice, lattice vibrations, phonons, Einstein model, Debye model, Black Body radiation, statistical mechanics of pho- tons, Bose occupation, Plancks law.Non-interacting particles obeying Bose and Fermi statisticsConsider a system of non-interacting, indistinguishable particles that can have energies ( = 1, 2, ...) associated with their quantum me- chanical states. The state of the system can be specified by the number of particles at each energy level, i.e. n is the number of particles at energy state . Total number of particles is . n = N . Total system energy is . n = E.In the canonical ensemble, we write the partition function:Q(T , V , N ) = n1,n2,...

..N n

exp

.. n

(1)where we include the delta-function constraint on the summation inorder to fix . n = N , and we define = 1/(kBT ).The indistinguishability of the particles is properly accounted for in this representation since any given set of n contributes a single term without over-counting indistinguishable states. In the grand canonical ensemble, we write the grand canonical partition function:(T , V , ) = Q(T , V , N ) exp(N )N =0=

.. N n

exp

.. n NN =0 n1,n2,.....= e .

n + .

= Y

ennn1,n2,...

nnThe Landau potential is written as:pV = kBT log = kBT log

.. ennn

(2)If the particles are Bosons, there are no restrictions on the number of particles in a given state n ( = 1, 2, ...), thus: .e +.n = 11 e +

(3)n =0If the particles are Fermions, any given state can only have eithern = 0 or n = 1 particles, thus: .e +.n = 1 + e +(4)n =0Therefore, we generally write:pV = kBT log .1 e +.(5)where "-" is for Bosons, and "+" is for Fermions.From this, we find the average number of particles:(N) =

. log .

e +=1e + =

(n)(6)where (n) is the average occupation number in the state.In the ideal gas limit, is very large and negative. Noting thate + 1, we have:(N) e + = eq and pV kBTeq = kBT(N)where we define the single-particle partition function q = .

e .The grand canonical partition function is: = epV /(kBT ) = exp .eq. = N =0

eN qN = Q(T , V , N )eNN !N =0thus Q = 1 qN for ideal gas (agrees with previous lecture).Fermi-Dirac statistics for conducting electrons in a metalElectrons in a metal can be modeled as a gas of non-interacting fermions: Electrons in a metal are at high densities (many atoms per volume with each contributing to the conducting electrons). Since no two electrons (fermions) can exist in the same state, a high density system fills many of the single particle energy levels The lowest unoccupied state will have a kinetic energy much larger than kBT , thus thermal excitations result in energetics with large kinetic energy and comparatively negligible potential energy of inter- action. The large kinetic energy associated with these electrons results in large conductivity of electrons in metalsUsing the results for the thermodynamic behavior of non-interacting fermions, find the behavior of an ideal gas of electrons in a metal. Elec- trons in a metal act as non-interacting particles with quantized energies given by:h2222En = 8mL2 .nx + ny + nz .translational modeswhich are the energy levels for a particle in a box for a particles with mass m = 9.10938 1031kg.The average number of electrons in the n state is:1(nn) = e(En) + 1 = F (En)(7)where we have defined the Fermi function F () = .e() + 1.1.The total number of electrons is given by:1

(N) = 2

= 2 (En)

F (En)(8)nx =1 ny =1 nz =1 e+ 1nx =1 ny =1 nz =1where we include a factor of 2 since the electrons can exist in spin-up and a spin-down states.For sufficiently large V , the spectrum of translational wavemodes is effectively a continuum. Therefore, we can convert this summation to an integral over n, resulting in: (N) = 20

dnx

dny0

dnz F (En)0 = 2dkx

L

dky

L

dkz

LF [(k)]0 0= 2 L 1 dk

0 dk

dk F [(k)] = 2V

dkF [(k)] SHAPE \* MERGEFORMAT

3 23

xyz

(2)3where we have used a coordinate change from n to k = n, and theenergy is now written as (k) = k2k2 .Define the chemical potential at T = 0 to be (T = 0) = F (Fermi Energy). To proceed, consider the form of the Fermi function at T = 0,F () = 1 for F and F () = 0 for > F . Define the Fermimomentumaccording to= p2 = k2k2 . Thus, at= 0,isfound to be:2V43

8 V (2m)3/2

3/2

. 3 .2/3 h2(N) = (2)3 3 kF = 3

h3F F =82mA typical metal (Cu) has a mass density of 9g/cm3. Assuming each atom donates a single electron to the conducting electron gas, this den-sity has a Fermi temperature F = F 80, 000K. This verifies thatthe Fermi energy F is sufficiently large to make the ideal gas approx- imation valid at room temperature. At room temperature, only stateswith energy very near F will be affected by thermal energy kBT .The spread in the distribution is approximately 2kBT (Fig. 1).The pressure is found using the relationship for Fermi particlespV = 2kBT log .1 + e +.(9)

Figure 1: The Fermi function F () =.e() + 1.1Following a similar derivation as before, we write2VpV = (2)3

dk log{1 + e

[(k)]}4V (2m)2/3 =

d1/2 log

.1 + e().

(10)h30where we have used = k2k2 .In the limit T 0(or ), we have 1 + e() e(F ) for < F and 1 + e() 1 for > F .Therefore, the pressure is written as:pV =

4V (2m)3/2 F

d1/2(F ) =

16V (2m)2/3

5/2

(11)h30

15h3FThis pressure at T = 0 is approximately 106atm. This large pressure plays an important role in halting the collapse of a star (white dwarf) because this enormous pressure offsets the gravitational forces that oth- erwise drive the collapse.The average energy (E) is found using:e +(E) = 2

1 + e +

(12)Following a similar derivation as before, we write2V(E) = (2)3

dk(k)

1 SHAPE \* MERGEFORMAT

e[(k)] + 14V (2m)2/3 =h30

d3/2

e[] + 14V (2m)2/3 =

d3/2F ()h30where we have used = k2k2 .

2/3We apply integration by parts to this equation and use (N) = 8 V (2m)3/2to get:

3(N)

5/2 dF

3h3F(E) = 53/2

d

(13)dwhere dF

( ) [e()+1]2 .In the limit T 0, the function dF

becomes peaked near F ,and we can effectively expand the integrand near = F (Fig. 2).Expand the integrand about = F to get 5/2 5/2 + 5 3/2( F ) + 15 1/22

F2 F8 F ( F )

+ ....dis even about F in this limit, the odd-order terms will integrate to zero, leaving only the even terms. Thus, the final form of

Figure2:

Thederivativeof theFermifunction dF ()=the average energy is going to be:.

. T .2.

e()[e() + 1]2(E) = (N)F

A + BF

+ ...

(14)A precise calculation of this low-temperature expansion (outside scope of this module) gives:3.52 . T .2.(E) = 5 (N)F 1 + 12

+ ...

(15)Therefore, the heat capacity for a metal in the limit of small tem- perature is given by:352 T2TCV = 5 (N)F

2=122

2 (N)kB

(16)giving a linear temperature dependence in the small-T limit.The limiting behavior in the small-T limit suggests that a plot ofCV /T approaches a constant as T 0. This proves to be the behaviorof the heat capacity for metals in the limit of small T . As temperature increases, the fluctuations in the metal nuclei also contribute to the heat capacity (Fig. 3).Crystal line SolidThe atoms in a crystal are arranged in a regular array of points in space with a variety of possible crystalline lattices (Fig. 4). At zero temper- ature, the atomic coordinates are uniquely locked into spatial positions that minimize the energy. At finite temperature, the atoms fluctuate about the energy-minimum positions, leading to lattice vibrations that govern the thermodynamic behavior (Fig. 5).Thermodynamic contribution of lattice vibrationsConsider N atoms with positions {r} = r1, r2, ..., rN in a crystalline lattice. Define the potential energy V ({r}) that describes the energy for a given system configuration {r}. A minimum-energy configura-.

Figure 3: The heat capacity CV of T i3SiC2 exhibits the temperature de- pendence CV = 1T + 3T 3 + ... in the low-temperature limit [Ho et al., 1999]tion {r(0)} = r(0), r(0), ..., r(0) satisfies the condition

V .{r(0)} =12N

ri(0)ri V |0 = 0 for i = 1, 2, ..., N . The atomic positions {rregular crystalline lattice.

} define theExpand the potential energy about {r} = {r(0)}, such that:NV ({r}) V .{r(0)}. + .r V |0. .ri r( ). +1 N N .

i=1

ii. .(0). .

(0). 2i=1 j=1

ri rj V |0

: ri ri

rj rj

+ ...

Figure 4: Crystal lattice structuresThe linear term is zero (by definition), thus the energy is:1 N N V V0 + 2

sisj Kij(17)i=1 j=1 ,=x,y,zwhere si = e .ri r(0). and Kij =

V | .iri

rj0The Matrix Kij can be diagonalized into normal modes (eigen- vectors) with effective elastic constants (eigenvalues). Since there are3N 6 3N degrees of freedom, there are 3N normal modes.The potential energy is written as:1 3N

Figure 5: Schematic showing the atoms in a crystal in their locked, energy- minimum positions at T = 0 and the lattice vibrations that occur at finite temperature

2V = V0 + 2

l=1

Kll(18)where l is the magnitude of the lth normal mode.The total energy of the system is then written as:3NE =

p21 3N+ K 2 + V

(19)l=1

SHAPE \* MERGEFORMAT

2m l2 l l0l=1where pl and m l are the effective momentum and mass of the lth normal mode, respectively.The total energy E is decomposed into normal modes with an individual- mode energy in the form of a harmonic oscillator. These normal modes are called phonons.Phonons act as quasi-particles which means they are distinguishable and independent, i.e. they dont interact between each other.The Hamiltonian (sum of kinetic and potential energy) of the lthp2phonon is given by: Hl = 2 l + 1 Kl2. This harmonic oscillator energym l2lresults in the quantized energy:.1 .El =

jl + 2

kl(20). K lwhere jl = 0, 1, 2..., and the phonon frequency is l =The canonical partition function Q is given by:

m l .

.V0 .3N (j + 1 )kQ(T , V , N ) =

... e

l=1 l 2l.j1 =0 j2 =03N

j3N =0= eV0 Y e(jl + 1 )kll=1 jl =03N

1 kl= eV0 Y

e 2

(21)l=1

1 eklwhere we use the mathematical property .

n1 .1zThis gives the Helmholtz free energy:3N .1.F = kBT log Q = V0 + kBT l=1The average energy is given by:

log .1 ekl . +

2 kl log Q

3N . k ekl1.

l

(E) =

= V0 +

l=1

1ekl +

2 klThe next step is to analyze two competing models for the phonon frequencies l.Einstein modelIn the Einstein model, we assume there is a single characteristic fre- quency of the crystal, defined as the Einstein frequency E . The average energy for this model is given by:3N kE

3N kEekl(E) = V0 +

2+ 1

ekEFor this model, the heat capacity is given by:

kE

. .2E. (E) .

3NkB

. .2BkBT

3NkBE

e TCV =

=TV ,N

.kE .21 e kBT

=

E .21 e Twhere we define E = kE .In the limit T the heat capacity CV 3NkB (Dulong-PetitLaw). As we learned before, the equipartition theorem states that the energy receives kBT per thermally active degree of freedom for a har- monic oscillator (quantized energy is linear in the quantum index). Inthe limit T 0 the heat capacity CV 0. The heat capacity ap-proaches zero exponentially in the small-T limit (Fig. 6).Debye modelThe Debye model treats the solid as an elastic material. Vibrational modes in an elastic solid correspond to sound waves, thus the frequencies satisfy = ck, where c is the speed of sound in the solid and k = m/L, where m = 1, 2, ...

Figure 6: Heat capacity of a crystal predicted by the Einstein model of lat- tice vibrationConvert the sum over normal modes to a sum over k using: (...) = (...) = dm

dm2

dm3

(...)lm1

m2 m3. L .3 kc=

dk1

kc

dk2

kc

dk3 (...)000= 4

. L .3 kc

dkk2 (...) = 4

. L .3 1 D

d2 (...)0

c3 0where kc is a cutoff wavemode (to be determined), and D = Ckc is called the Debye frequency.A complete conversion will include one longitudinal mode with cland two transverse modes with ct . This gives: (...) = 4 . L . . 1

2 . +

Dd2 (...)(22)33llt0To find D we must enforce that . 1 = 3N , thus we have: 1 = 4 . L . . 1

2 . +

Dd2 (...)l. L .3 . 1

33lt02 . 3

9N

1/3=+ D = 3N D =.. c3c33

L lt123ltFor convenience, use D as a parameter, thus:D (...) = 9N 1 D 0

d2(...)(23)lDefine the Debye temperature D = kD , which defines the temper- ature scale for vibrational fluctuations. To test this model, we find the heat capacity:CV = kB2

. (E) .

= 9NkB

T 3 D /Tdx

x4ex30

(1 ex)2

Figure 7: Heat capacity of a crystalIn the limit T the heat capacity approaches:

predicted by the Einstein model and Debye model of lattice vibrationT 3 D /TCV 9NkB 3

dxx2

= 3NkBD 0which is expected (Dulong-Petit Law).In the limit T 0 the heat capacity scales as:T 3 CV 9NkB 3dx

x4ex

T 3 NkB D 0(1 ex)23The heat capacity predicted by the Einstein and the Debye models are very similar. However, the low temperature of the heat capacityof non-conducting solids matches the Debye model, i.e. CV NT 3(Fig. 7).Black Body RadiationWe are all familiar with the idea that hot objects emit radiation, a light bulb, for example. In the hot wire filament, an electron, originally in an excited state drops to a lower energy state and the energy difference isgiven off as a photon, (s = h = s2 s1). We are also familiar with theabsorption of radiation by surfaces. For example, clothes in the summer absorb photons from the sun and heat up. Black clothes absorb more radiation than lighter ones. This means, of course that lighter colored clothes reflect a larger fraction of the light falling on them.A black body is defined as one which absorbs all the radiation in- cident upon it, i.e. a perfect absorber. It also emits the radiation subsequently. If radiation is falling on a black body, its temperature rises until it reaches equilibrium with the radiation. At equilibrium, it re-emits as much radiation as it absorbs so there is no net gain in energy and the temperature remains constant. In this case, the surface is in equilibrium with the radiation and the temperature of the surface must be the same as the temperature of the radiation.To develop the idea of radiation temperature we construct an en- closure having walls which are perfect absorbers (see Fig. 8). Inside the enclosure is radiation. Eventually this radiation reaches equilibrium with the enclosure walls, equal amounts are emitted and absorbed by the walls. Also, the amount of radiation travelling in each direction be- comes equal and is uniform. In this case the radiation may be regarded as a gas of photons in equilibrium having a uniform temperature. The enclosure is then called an isothermal enclosure.An enclosure of this type containing a small hole is itself a black body. Any radiation passing through the hole will be absorbed. The radiation emitted from the hole is characteristic of a black body at the temperature of the photon gas. The properties of the emitted radia- tion is then independent of the materials of the wall provided they are sufficiently absorbing that essentially all radiation entering the hole is absorbed. This universal radiation is called Black Body Radiation.An everyday example of a photon gas is the background radiation in the universe. This photon gas is at a temperature of about 5 K. Thus the earths surface, at a temperature of about 300 K, is not in equilibrium with this gas. The earth is a net emitter of radiation (excluding the sun) and this is why it is dark at night and why it is coldest on clear nights when there is no cloud cover to increase the reflection of the earths radiation back to the earth. A second example is a Bessemer converter used in steel manufacture containing molten steel. These vessels actually contain holes like the isothermal enclosure of Fig. 8. The radiation emitted from the hole is used in steel making to measure the temperature in the vessel, by means of an optical pyrometer.

Figure 8: Isothermal Enclosure.Statistical ThermodynamicsTo derive Plancks radiation law directly from our statistical mechanics, we note that number of photons in the gas is not fixed. The photons are absorbed and re-emitted by the enclosure walls. Since the photons are non-interacting it is by this absorption and re-emission that equilib- rium is maintained in the gas. Since, also the free energy F (T , V , N ) is constant in equilibrium (at constant T and V ) while N varies it follows that F /N = 0, that is = F ..V ,N

= 0(24)The photon gas is then a Bose gas with = 0 so that the canonical partition function is given directly asrQ = Y(1 exp(ss))1(25)s=1And the expected Bose occupation isns = (exp(ss) 1)1(26) Using s = h and the equation of density of states, we then obtain:u() =or

1s()n ()g()(27)V8u() =

h2(28)c3 (exp(h) 1)Which is Plancks Radiation Law. We may also recover Wien and Rayleigh-Jeans laws as limits of Plancks law,(a) Long wavelength, hc > 1. Heres = hc exp( hc/kT)(31)And Eq. 30 becomesu() =

1V s()g() c

8hc 5

exp(hc/kT)(32)Which is the Wiens law valid at short wavelength. Employing our statistical mechanics we readily obtained Plancks radiation law. We may also derive the Stefan-Boltzmann law for u =du() =

8

h2(33)0c3 0

(exp(h) 1)Introducing x = h, this reduces toWhere

8k4 u = (hc)3

x34dx exp(x) 1 T8k4 4

= aT 4

(34)a = (hc)3 15(35)In this way we obtain, using statistical mechanics, a law derived previously using thermodynamics including all the numerical factors. This gives Stefans constant in14E = 4 caTAs

= T 4

(36)2ck4 4 = (hc)3 15 = 5.67 10

5ergcm2 Sec K

4(37)A measurement of could then be used, for example, to determine Plancks constant. Planck, in fact, determined h as the constant needed in his radiation law to fit the observed spectral distribution law. Thisgave him the value h = 6.55 1027 erg.sec which compares with thepresent value of h = 6.625 1027 erg.secThermodynamic PropertiesWe may calculate all the thermodynamic properties of black body radia- tion using statistical mechanics through the partition function Q, whereF = kT log(Q)(38)And Q was derived as shown in Eq. 25. This is the basic method of statistical thermodynamics. The aim is to reproduce all the ther- modynamic properties with all factors and constants evaluated. This givesF = kT logr

. r.Y(1 exp(ss))1s=1F = kT log(1 exp(ss))1s=1 d1F = 2kT

h3 log(1 exp(ss))

(39)Where the ss are the single photon states and the factor of 2 arises from the two polarizations available to each photon. This can be in- tegrated in a variety of ways. Perhaps the most direct is to integrate over phase space (d = dV 4p2 dp) and write s = pc. Introducing the dimensionless variable x = s = pc, the Helmholtz free energy is:1 .F = 3

8k4 (hc)3 0

d(x)3

.log(1 exp(x))

V T 4

(40)The dimensionless integral here can be transformed into that appear- ing in the constant of a of Eq. 35, by an integration by parts, i.e., I = 0

d(x)3 log(1 exp(x)).= (x3) log(1 exp(x))..0

+x3d[log(1 exp(x))](41)0The first term vanishes because:33(a) limx x3

log(1 exp(x)) c limx x3

exp(x) = 0(b) limx0 xAnd

log(1 exp(x)) c limx0 x

log(x) = 0

x3 I =dx exp(x) 1 = 15

(42)Comparing Eq. 40 and Eq. 35, we get:14F = 3 aV T

(43)From F we may determine all other thermodynamic properties by differentiation. For example, the entropy is. F . .43S = TThe internal energy is:

. =aV T.V

(44)U = F + TS = avT 4(45)The pressure is:

. F . .14p = V

. =aT.T

(46)Finally, the Gibbs free energy is: s14G = F + pV = 3 aV T

14+ 3 aV T

= 0(47)This is zero as required G = N and the chemical potential = 0. We may use these expressions to further verify thermodynamic consis-.tency, for example that CV = T S ..V

.U .T ..VIn summary, we have obtained the spectral distribution from theBose occupation in much the same way as we obtained the Maxwell- Boltzmann distribution for a classical gas. The only other required ingredient was the density of states. We have also obtained all the thermodynamic properties using the partition function Q.ReferencesJ. C. Ho, H. H. Hamdeh, M. W. Barsoum, and T. El-Raghy. Low temperature heat capacity of Ti3SiC2. Journal of Applied Physics, 85 (11):7970, June 1999.

1

N !

3

L

2m

m

pFF2F

2m

FT(N)

kB

2m

1

2m

F

0

d

=e

d

Since dF ()

d

F

F

F

0

l

2

n=0

z =

k Ee k

T

T

.

kB

1

3

c

c

l

3

c

c

c3

+

c

kB

D

D

N .

.

kT

s =

kT

0

.

0

.

3

.

3

T .

=

.