Advanced Thermodynamics: Lecture 19 Shivasubramanian Gopalakrishnan - - PowerPoint PPT Presentation

Advanced Thermodynamics: Lecture 19

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Bose–Einstein Statistics The thermodynamic probability Wk of a macrostate of an assembly depends on the particular statistics obeyed by the assembly. In Bose–Einstein (B–E) statistics, the particles are considered indistinguishable, and there is no restriction on the number of particles that occupy a particular energy state. The energy states themselves are however distinguishable. Considering a possible energy level j. The distribution of the particles in different states which are denoted by numbers in () can be represented as below. For convenience sake each particle is labeled with an alphabet though they are indistinguishable. [(1)ab][(2)c][(3)][(4)def ] . . . (1)

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

There are gj states, therefore there are gj ways in which the sequences can begin and in each of these sequences the remaining (gj + Nj − 1) numbers and letters can be arranged in any order. The number of different sequences in which N distinguishable

• bjects can be arranged is N!.

N(N − 1)(N − 2) . . . 1 = N! (2) For example 3 letters a, b, c can be arranged in following ways, abc, cab, bac, bac, cab, cab There are six possible sequences, which 3!.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

The factorial for a large number x is given by Stirling’s approximation, ln x! = x ln x − x Hence, for x = 70 ln 70! = 70 ln 70 − 70 = 245 log1070! = 106 70! = 10106

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

The number of different possible sequences of the (gj + Nj − 1) numbers and letters is therefore (gj + Nj − 1)! and the total number of possible sequences of gj numbers and Nj letters is gj[(gj + Nj − 1)!] (3) Even though each possible sequence represents a possible distribution of particles, many will represent the same distribution. For example sequences given by equations ( 4) and (1) are one and the same. [(3)][(1)ab][(4)def ][(2)c] . . . (4) There are gj groups in the sequence, one for each state, so the number of different sequences of groups is gj! and we must divide

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

equation (3) by gj! to avoid counting the same distribution more than once. Also, the particles are indistinguishable. Thus a different sequence

• f letters similar to

[(1)ca][(2)e][(3)][(4)bdf ] . . . (5) also represents the same distribution as equation (1). The Nj letters can be arranged in a sequence in Nj different ways, so equation (3) must be also be divided by Nj!. Hence the number of different distributions for the jth level is ωj = gj[(gj + Nj − 1)!] gj!Nj! (6) which can be written as

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

ωj = (gj + Nj − 1)! (gj − 1)!N! (7) since gj! = gj(gj − 1)! (8) Suppose an energy level j includes 3 states (gj = 3) and 2 particles (Nj = 2). The number of possible distributions are ωj = (3 + 2 − 1)! (3 − 1)!2! = 6

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

If a level is non–degenerate, that is, if there is only one state in the level and gj = 1, and there is only one possible way in which particles in the level can be arranged, hence ωj = 1. Then equation (3) becomes. ωj = Nj! 0!Nj! = 1 Therefore as a convention for the right answer, we must set 0! = 1. Also if a level j is unoccupied and Nj = 0 and ωj = 1 for that level, ωj = (gj − 1)! (gj − 1)!0! = 1

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

For each of the possible distributions in any level, we may have any

• ne of the possible distributions in each of the other levels, so the

total number of possible distributions, or the thermodynamic probability WB.E of a macrostate in the B–E statistics is the product over all levels of the values ωj for each level, given by WB.E = Wk =

• i

ωj =

• j

(gj + Nj − 1)! (gj − 1)!Nj! (9) For example, an assembly includes two level p and q, with gp = 3 ,Np = 2, gq = 2 and Nq = 1. The thermodynamics probability of the macrostate Np = 2, and Nq = 1 is WB.E = 4! 2!2! × 2! 1!1! = 6 × 2 = 12 (10)

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Although all microstates of an isolated, closed system are equally probable, the only possible microstates are those in which the total number of particles equals the number N in the system and in which the total energy of the particles equals the energy U of the system. For example, let us consider a system of 6 particles with all the permitted energy levels equally spaced, i.e ǫ0 = 0, ǫ1 = ǫ, ǫ2 = 2ǫ . . .. Let us assume that each energy level has 3 states, gj = 3 and the total energy U of the system is equals 6ǫ.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

k= 1 2 3 4 5 6 7 8 9 10 11 N=6 6 1 0.041 5 1 0.008Ω=1532 4 1 1 0.205 3 2 1 1 0.41 2 1 1 3 2 1 0.83 1 1 2 1 3 2 4 6 1.6 5 4 4 3 4 3 2 3 2 1 2.83 63 135 135 180 90 270 180 100 216 135 28 Nj εj/ε= U=6ε gj=3 Wk=

Eleven possible states of an assembly of 6 particles obeying B–E statistics.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Thus for macrostate k = 1, since gj = 3 is all levels. The thermodynamic probability is W1 = (3 + 1 − 1)! 2!1! × (3 + 5 − 1)! 2!5! = 3 × 21 = 63 (11) The single particle in level 6 could be in anyone of 3 states and the 5 particles in level 0 can be distributed in 21 different ways. Therefore there are 63 possible arrangements. The total number of possible microstates or the thermodynamic probability of the system is Ω =

• k

Wk = 1532 (12)

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

The average occupation number of level 2 is computed as ¯ N2 = 1 Ω

• k

N2kWk = 1272 1532 = 0.83 (13) The most probable macrostate is the one with the largest number

• f microstates (270), is the sixth state.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Fermi–Dirac Statistics Developed by Enrico Fermi and Paul Dirac. Applies to indistinguishable particles that obey Pauli’s exclusion principle, according to which there can be no more than one particle in each permitted energy state. Thus the number of particles Nj at an energy level j cannot exceed the degeneracy gj of that level. A possible arrangement for F–D statistics might be: [(1)a][(2)b][(3)][(4)c] . . . (14) The possible number of unique sequences are ωj = gj! (gj − Nj)!Nj! (15)

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

The thermodynamic probability in F–D statistics is WF.D =

• j

ωj =

• j

gj! (gj − Nj)!Nj! (16) Consider again a system of 6 particles with all the permitted energy levels equally spaced, i.e ǫ0 = 0, ǫ1 = ǫ, ǫ2 = 2ǫ . . .. And let us assume that each energy level has 3 states, gj = 3 and the total energy U of the system is equals 6ǫ.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

k= 1 2 3 4 5 N=6 4 1 0.123 3 1 1 0.494 Ω=73 2 1 3 2 1.15 1 2 1 3 2 1.73 3 3 2 3 2 2.51 9 27 9 1 27 Nj εj/ε= U=6ε gj=3 Wk=

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Five possible states of an assembly of 6 particles obeying F–D statistics. Thus in macrostate 1, W1 = 3! (3 − 1)!1! × 3! (3 − 2)!2! × 3! (3 − 3)!3! = 3 × 3 × 1 = 9 The total number of possible macrostates is Ω =

• k

Wk = 73

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Maxwell–Boltzmann Statistics In Maxwell–Boltzmann (M–B) statistics, the particles are considered distinguishable and like B–E statistics there is no restriction on the placement on the particles in various energy states. Consider a j level with a degeneracy of gj and Nj particles. The number of ways the particles can be arranged in gj states is, ωj = gNj

j

(17) The total number of distributions between all levels, with a specified set of particles in each level is,

• j

ωj =

• j

gNj

j

(18)

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Though, this will not be the thermodynamic probability Wk since the an interchange of particle between levels will also give rise to a different macrostate (by virtue of particles being distinguishable). The total number of ways in which N particles are distributed among levels with N1 particles in level 1, N2 particles in level 2 and so on, is given by. N! N1!N2! . . . = N!

• j Nj!

(19) The thermodynamic probability of M − −B statistics is the product of the above 2 equations and is given by WM.B = N!

• j Nj!
• j

gNj

j

= N!

• j

gNj

j

Nj! (20)

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Consider again a system of 6 particles with all the permitted energy levels equally spaced, i.e ǫ0 = 0, ǫ1 = ǫ, ǫ2 = 2ǫ . . .. The arrangement is similar to that of B–E statistics but has far more number of states since particles are now distinguishable. Thus for macro state k = 1, in which levels zero and six are occupied. W1 = 6!35 5! 31 1! = 18 × 35 (21) The total number of possible macrostates is given by Ω =

• k

Wk = 1386 × 35 = 3.37 × 105 (22)

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Sheet1 k= 1 2 3 4 5 6 7 8 9 10 11 N=6 6 1 0.013 5 1 0.065 4 1 1 0.195 3 2 1 1 0.455 2 1 1 3 2 1 0.91 1 1 2 1 3 2 4 6 1.64 5 4 4 3 4 3 2 3 2 1 2.73 18 90 90 180 45 360 180 60 270 90 3 Nj εj/ε= U=6ε =1386 x gj=3 Wk/35=

Eleven possible states of an assembly of 6 particles obeying M–B statistics.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Statistical Interpretation of Entropy Consider 2 equilibrium states of an open PVT systemic which the temperature and pressure are the same but in which energy,volume and number of particles are different, the principles of Thermodynamics lead to the result that the entropy difference between the states is given by T∆S = ∆U + P∆V (23) Entropy is an extensive property and the total entropy S of two independent systems is the sum of entropies S1 and S2 of the individual systems S = S1 + S2 (24)

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

But if we consider thermodynamic probabilities, the number Ω of possible microstates of a combination of two systems is the product of the number of independent possible states. Ω = Ω1Ω2 (25) It is logical that the entropy cannot be linearly proportional to thermodynamic probability. To find the relation between S and Ω, let us assume that S is some unknown function J of Ω, i.e., S = J(Ω. Since S = S1 + S2 and Ω = Ω1Ω2, J(Ω1) + J(Ω2) = J(Ω1Ω2) (26)

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Taking partial derivatives with first keeping Ω2 constant and then Ω1 constant, we have ∂J(Ω1) ∂Ω1 = dJ(Ω1) dΩ1 (27) The partial derivative of J(Ω2) with respect to Ω1 is zero, since Ω2 is constant. On the right side the partial derivative of J(Ω1Ω2), w.r.t to Ω1, equals the total derivative of J(Ω1Ω2) with respect to its argument (Ω1Ω2), multiplied by the partial derivative of its argument with respect to Ω1, which is simply the constant Ω2. Hence dJ(Ω1) dΩ1 = Ω2J′(Ω1Ω2) (28)

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Similarly, dj(Ω2) dΩ2 = Ω1J′(Ω1Ω2) (29) It follows that Ω1 dJ(Ω1) dΩ1 = Ω2 dJ(Ω2) dΩ2 (30) and since Ω1 and Ω2 are independent, the equation can be satisfied only if LHS and RHS are equal to a constant, assumed to be kB. Then for any system we have ΩdJ(Ω) dΩ = kB (31) dJ(Ω) = kB dΩ Ω (32)

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

J(Ω) = kBlnΩ (33) and hence S = kBlnΩ (34) The only function of Ω which satisfies the condition that the entropies are additive while thermodynamic probabilities are multiplicative is the logarithm. It turns out that the numerical value chosen for kB, so that statistical and classical thermodynamics agree, is the Boltzmann constant.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Additional insight into the connection between statistical and classical thermodynamics can be gained by considering two neighboring states of a closed system, in which the values of the internal energy U, the energy level ǫj, and the average occupation numbers ¯ Nj are slightly different. The energy difference between the two states is given by, dU =

• j

ǫjd ¯ Nj +

• j

¯ Njdǫj (35) that is, the difference in energy results in part from the differences d ¯ Nj , in the average occupation numbers, and in part from the differences dǫj in the energy levels. if the values of the energy levels are functions of some extensive parameter X, such as the volume V , then

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

dǫj = dǫj dX dX (36) and

• j

¯ Njdǫj =  

j

¯ Nj dǫj dX   dX (37) Let us define a quantity Y as Y ≡ −

• j

¯ Nj dǫj dX (38) Then

• j

¯ Njdǫj = −YdX (39)

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

if, for example the parameter X is the volume V , the quantity Y will represent YdX = PdV (40) The energy difference is then dU =

• j

ǫjd ¯ Nj − YdX (41) if for two states the parameter X is same, i.e., dX = 0, then dUX =

• j

ǫjd ¯ Nj (42) The result when X is constant according to laws of thermodynamics is,

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

dUX = Tds (43) and

• j

ǫjd ¯ Nj = TdS (44) Thus the equation dU =

• j

ǫjd ¯ Nj +

• j

¯ Njdǫj (45) is the statistical form of the combined first and second law of thermodynamics for a closed system. dU = TdS − YdX (46)

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

If the system is taken from one state to another by a reversible process, then TdS = d′Qr and YdX = d′Wr (47) Hence for such a process, dU = d′Qr − d′Wr (48) and

• j

ǫjd ¯ Nj = d′Qr,

• j

¯ Njdǫj = −d′Wr (49) It is mostly assumed that the sum

j ǫjd ¯

Nj is always equal to reversible heat flow d′Q into the system and the sum

j ¯

Njdǫj is

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

always equal to the negative work −d′W . This is valid only for reversible processes.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Bose–Einstein Distribution function In general systems the number of particles is quite large and one needs an expression for the average occupation numbers at different energy levels. Such and expression is called the distribution function. Let us assume 2 systems which have the same permitted energy levels and the number of particles in the second is less than that of the first by a number n, such that n << N. Also the energy of the second system is less compared to the first by nǫr where ǫr is an arbitrary system. The unprimed symbols refer to the first system and the primed symbols refer to the second system. N′ = N − n, U′ = U − nǫr (50)

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Essentially, the occupation numbers at all levels in the two systems are the same, except at the level r. Where for the primed system the occupation is less than that of the unprimed system by a number n. N′

j = Nj(j = r),

N′

r = Nr − n

(51) The thermodynamic probability Wk of a macrostate k in unprimed system is Wk =

• j

(gj + Njk − 1)! (gj − 1)!Njk! (52) and in the primed system is W′

rk =

• j

(gj + N′

jk − 1)!

(gj − 1)!N′

jk!

(53)

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

The subscript rk denotes that one particle at level r has been

• removed. If a macrostate in the unprimed system has the level r

unoccupied i.e., Nrk = 0, then according to our assumptions in the primed system N′

rk = −1. Therefore the rth term in W′ rk will look

like (gr − 2)! (gr − 1)!(−1)! = 1 (gr − 1)(−1)! (54) Since W′

rk is zero, we must assume the convention (−1)! = ∞

The ration of thermodynamic probabilities is W′

rk

Wk =

• j

(gj + N′

jk − 1)!

(gj + Njk − 1)! Njk! N′

jk!

(55)

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

At all levels except level r, N′

jk = Njk. At the level r,

N′

rk = Nrk − 1, Therefore

Nrk! = Nrk(Nrk − 1)! = NrkN′

rk!

(56) and (gr + Nrk − 1)! = (gr + N′

rk! = (gr + N′ rk)(gr + N′ rk − 1)!

(57) Hence W′

rk

Wk = Nrk gr + N′

rk

(58) and summing over all macro states after rearranging terms

• k

NrkWk = gr

• k

W′

rk +

• k

N′

rkW′ rk

(59)

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

• k

NrkWk = gr

• k

W′

rk +

• k

N′

rkW′ rk

(60) The term on the left equals ¯ NrΩ. On the right, the first term equals grΩ′

r and the last term equals ¯

N′

rΩ′

• r. Therefore

¯ NrΩ = (gr + ¯ N′

r)Ω′ r

(61) and ¯ Nr gr + ¯ Nr = Ω′

r

Ω (62) for large systems, it is reasonable to assume that ¯ Nr = ¯ N′

r, so that

¯ Nr gr + ¯ Nr = Ω′

r

Ωr (63)

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Taking logarithms on both sides ln ¯ Nr gr + ¯ Nr = lnΩ′

r

Ω (64) but lnΩ′

r

Ω = lnΩ′

r − lnΩ

(65) and since, S = kBlnΩ ln ¯ Nr gr + ¯ Nr = S′ − S kB = ∆S kB (66) The entropy difference ∆S between 2 states of a open constant volume system is related to energy difference ∆U and the ∆N, difference in number of particles by,

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

T∆S = ∆U − µ∆N (67) Assume that µ is chemical potential per particles, For the two states we have ∆U = −ǫr ∆N = −1 (68) and hence ∆S = µ − ǫr T (69) since level r was arbitrarily chosen, we could say the same for any level j. ln ¯ Nj gj + ¯ Nj = µ − ǫj kBT (70)

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

and gj + ¯ Nj ¯ Nj = gj ¯ Nj + 1 = exp ǫj − µ kBT

• (71)

Which can written as ¯ Nj gj = 1 exp ǫj−µ

kBT

• − 1

(72) This equation gives the Bose–Einstein distribution function. It gives us an expression for the average occupation number per state at any level j, in terms of energy ǫj, chemical potential µ, temperature T and universal constant kB.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Fermi–Dirac Distribution Function Let us consider a similar exercise for F–D statistics by assuming two systems one unprimed and other primed. The thermodynamic probabilities of the macrostates are Wk =

• j

gj! (gj − Njk)!Njk! (73) and W′

rk =

• j

gj! (gj − N′

jk)!N′ jk!

(74) Then W′

rk

Wrk =

• j

(gj − Njk)!Njk! (gj − N′

jk)!N′ jk!

(75)

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Which after cancellation reduces to W′

rk

Wrk = Nrk gr − N′

rk

(76) summing over all values of k, we have

• k

NrkWk = gr

• k

W′

rk −

• k

N′

rkW′ rk

(77) and ¯ Nr gr − ¯ N′

r

= Ω′

r

Ω (78)

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Letting ¯ Nr = ¯ N′

r, we get

¯ Nj gj = 1 exp ǫj−µ

kBT

• + 1

(79) Which is the Fermi–Dirac distribution function

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

The Classical Distribution Function In many systems with indistinguishable particles, the average number of particles ¯ Nj is usually much smaller than the degeneracy

• gj. By virtue of the denominator being quite large compared to the

numerator, one can neglect the 1. Hence the expression for distribution function for both B–E and F–D reduces to ¯ Nj gj = 1 exp ǫj−µ

kBT

• (80)

which is the classical distribution function. In general one can write ¯ Nj gj = 1 exp ǫj−µ

kBT

• + a

(81) where a = −1 for B–E, a = 1 for F–D and a = 0 for classical.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

The Maxwell–Boltzmann Distribution Function Doing a similar exercise for M–B statistics. The thermodynamic probabilities for corresponding macro states for primed and unprimed states are. Wk = N!

• j

gj Nj Nj! (82) W′

k = N′!

• j

gj

N′

j

N′

j!

(83) Taking their ratios W′

rk

Wk = N′! N!

• j

gj

N′

jk

gj Njk Njk! N′

jk!

(84)

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

which simplifies to W′

rk

Wk = Nrk Ngr (85)

• r

NrkWk = Ngr W′

rk

(86) summing over all macrostates we will get ¯ Nr Ngr = Ω′

r

Ω (87) and using a similar procedure as previous two analysis, we get ¯ Nj/N gj = exp µ − ǫj kBT

• (88)

which is the Maxwell-Boltzmann distribution function.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

The partition function The distribution function in M–B statistics can be written as ¯ Nj = N

• exp

µ kBT

• gjexp −ǫj

kBT (89) Since

j ¯

Nj = N, and the chemical potential µ does not depend

• n energy level.
• j

¯ Nj = N − N

• exp

µ kBT

j

gjexp −ǫj kBT (90) The sum term in the last part of the equation is called the partition function or sum over states and is denoted by Z Z ≡

• j

gjexp −ǫj kBT (91)

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

The partition function depends only on the temperature T and on parameters which determine energy levels. For M–B statistics, exp µ kBT = 1 Z (92) and hence the M–B distribution function can be written as ¯ Nj gj = N Z exp −ǫj kBT (93) This equation tells us that the average occupation number decreases exponentially with the energy level. The rate of decrease is inversely proportional to the temperature.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

The classical distribution function can be written as ¯ Nj =

• exp

µ kBT

• gjexp −ǫj

kBT (94) and summing over all levels we have,

• j

¯ Nj = N =

• exp

µ kBT

j

gjexp −ǫj kBT (95) Thus if the partition function Z is defined in the same way as in M–B statistics, we have exp µ kBT = N Z (96)

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

And the classical distribution function can be written as ¯ Nj gj = N Z exp −ǫj kBT (97) which has the same form of the M–B distribution. The significance

• f the partition function Z is that in M–B and classical statistics,

all the thermodynamic properties of a system can be expressed in terms of Z and its partial derivatives.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661