The second law of thermodynamics - II.

Januay 21, 2013 The second law of themodynamics - II. Asaf Pe e 1 1. The Schottky defect At absolute zeo tempeatue, the atoms of a solid ae odeed completely egulaly on a cystal lattice. As the tempeatue inceases, the atoms can change thei positions: e.g., due to vibations. As the atoms migate, they leave vacant lattice sites, known as point defects. One kind of such defects ae Schottky defects, in which the atoms migate to the suface of the cystal (see Figue 1). We now calculate the numbe of Schottky defects fo a cystal in themal equilibium at some tempeatue T. Geneally, atoms inside the lattice have lowe enegy than atoms at the suface, due to the stonge binding enegy. Conside a lattice with N atoms and n Schottky defects, whee we assume n N, thus we can teat each defect sepaately. The enegy associated with the defects is E = nǫ. Fig. 1. Schottky defect. Left: a two dimensional pefect lattice. Right: the same lattice with Schottky defects, namely two atoms migated to the suface of the cystal leaving two vacancies. 1 Physics Dep., Univesity College Cok

2 In ode to ceate n Schottky defects, we must move n atoms out of total N, thus we have ( ) N N! Ω(n) = = n n!(n n)!, (1) and the associated entopy is S(n) = k B lnω(n) = k B ln We can now use Stiling s fomula, which is tue fo N 1, to wite N! n!(n n)! = k B(lnN! lnn! ln(n n)!). (2) lnn! N(lnN 1) (3) S(n) k B (N lnn nlnn (N n)ln(n n)). (4) Futhemoe, we know that the tempeatue is elated to the entopy via Diffeentiating Equation 4 gives and thus 1 T = S E = ds(n) dn dn de = 1 ǫ ds(n) dn ds(n) dn = k B( lnn+ln(n n)) = k B ln 1 T = k B ǫ ln ( N n n ( N n n (5) ), (6) ). (7) Obviously, it is much easie to measue T than n. We can thus use some algeba on Equation 7, to obtain N n = exp ǫ (8) k B +1. T In the limiting case ǫ k B T, exp(ǫ/k B T) 1 and n Ne ǫ/k BT (n N). At T = 0 K, n = 0 - all atoms ae in place. This is the state of lowest enegy, o gound state. At T > 0, n > 0. Typically, ǫ 1 ev (E.g., fo Ge, ǫ 1.7 ev). Thus, at oom tempeatue (290 K, k B T 1/40 ev), n N = e 1.7/0.025 3 10 30. (9) Howeve, at T = 1000 K, n/n 3 10 9 due to the exponential dependence.

3 2. Equilibium of a system in a heat bath So fa we have consideed the equilibium of isolated systems. Now let us look at a system which is in contact with a heat bath having tempeatue T. A heat bath is a body with a heat capacity vey lage compaed to the system in question. This implies that the system and the heat bath can come to themal equilibium without a significant change in the tempeatue of the heat bath. We can theefoe look at this combined system as two sub-systems (the oiginal system and the heat bath) which ae in contact. Howeve, since the heat bath is much lage than ou system, its tempeatue is not changed, because of its size. The tempeatue of the oiginal system, when in equilibium, is also T, but its enegy is not fixed. We assume that the combined system + heat bath ae fully isolated fom the est of the wold. Let us see how to calculate the enegy of the system. We assume that N and V ae constant (no numbe change o volume change), and so the macostate of the system in equilibium is specified by T, V and N. The system has many diffeent micostates, which we will denote 1,2,...,,... We will label thei enegies by E 1,E 2,...,E,... The enegy by itself is not enough to fully detemine the micostate. As we have seen in the paamagnet example, diffeent micostates can have the same enegy. Thus, we can only calculate the pobability P() of the system to be in micostate having enegy E. In ode to calculate P(), we ecall that the combined system + heat bath ae isolated fom the est of the wold. Thus, the total enegy of the combined system is constant - let us call it E 0. If the system has enegy E, then the enegy of the heat bath is E 0 E. Thus, the pobability P() must be popotional to the statistical weight of the heat bath to have enegy E 0 E, Ω HB (E 0 E ), P() Ω HB (E 0 E ). (10) Since the pobability of the system to be in a micostate is unity, the popotionality constant is obtained by summing ove all possible micostates of the heat bath, namely P() = Ω HB(E 0 E ) Ω HB(E 0 E ). (11) We can now use the definition of entopy, S = k B lnω, to wite Ω = e S/k B. We can thus wite the pobability by eshb(e0 E) k B P() =. (12) HB (E 0 E) es k B

4 Equation 12 is completely geneal, and is tue fo any system composed of two sub-systems. Now, we make use of the assumption that one of the sub-systems is a heat bath, namely its enegy is much lage compaed to that of the system: E E 0. This assumption enables us to expand S HB (E 0 E ) into a Taylo seies: S HB (E 0 E ) = S HB (E 0 ) E S HB (E 0 ) E 0 Using the definition of the tempeatue: and 2 S HB (E 0 ) = ( ) SHB (E 0 ) E0 2 E 0 E 0 hence we can wite Equation 13 as + E2 2 2 S HB (E 0 ) E 2 0 +... (13) S HB (E 0 ) E 0 = 1 T, (14) = ( ) 1 = 0, (15) E 0 T S HB (E 0 E ) = S HB (E 0 ) E T, (16) o S HB (E 0 E ) = S HB(E 0 ) βe, (17) k B k B whee β 1 (18) k B T is known as the tempeatue paamete, and natually occus in many statistical mechanics equations. Using this infomation we can now wite the pobability P() in Equation 12 in the fom eshb(e0 E) k B P() = = e S HB (E 0 ) βe k B = e βe HB (E 0 E) HB es k B (E 0 ) βe es k B = 1 Z e βe. (19) e βe whee Z e βe (20) is called the patition function. It plays a cental ole in studying the popeties of systems at a fixed tempeatue. Equation 19 is known as Boltzmann distibution. It gives the pobability that a system, when placed in a heat bath at tempeatue T be in the micostate with enegy E. We see that this pobability depends on the enegy E of the state. The only popety of the heat bath on which it depends is its tempeatue, T.

5 2.1. The patition function and the mean enegy The patition function in Equation 20 is a sum ove all mocostates of the system. As we noted in the paamagnet example, diffeent micostates may have the same enegy. We can theefoe e-wite the patition function in Equation 20 as a sum ove all enegies: Z e βe = E g(e )e βe. (21) In Equation 21, the summation is ove all diffeent enegies, E, and g(e ) is the numbe of micostates all having the same enegy E. The numbe g(e ) is known as the degeneacy of the enegy E. Using the same line of easoning, we can wite the pobability P(E ) of the system to be in a state with enegy E, P(E ) = g(e )P() = 1 Z g(e )e βe. (22) Fom Boltzmann distibution we can obtain the mean enegy Ē of the system in contact with a heat bath: E = P()E = 1 E e βe (23) Z Using the definition of Z fom Equation 20, we find that Z/ β = ( E )e βe, which is just the summation in Equation 23. We can thus e-wite Equation 23 as E = 1 Z Z β = lnz β. (24) Thus, we obtained the aveage enegy of the system in tems of the patition function. Note that we used patial deivatives as the enegy levels E ae held constant: these enegies do not depend on the tempeatue, but puely on the macoscopic stuctue of the system. (Recall that in the paamagnet example, the enegy levels depend on the extenal magnetic field). 2.2. Fluctuations aound the mean enegy Equation 24 gives the mean enegy of the system. Howeve, its actual enegy will fluctuate, as it is in contact with the heat bath. How lage ae these fluctuations?

6 by The magnitude of the fluctuations is measued by the standad deviation, E, defined In ode to calculate E 2, we calculate ( 2 lnz β 2 ( E) 2 (E E) 2 = E 2 E 2 (25) = β = β = 1 Z Z 2 β lnz β ) ( 1 Z E ) e βe E e βe + 1 Z = E 2 + E2 P() E2 e βe Since the aveage of a function f(e ) is f(e ) f(e )P(), the last tem in Equation 26 is just E 2. We thus obtain (26) ( E) 2 = 2 lnz β 2 = E β = dt E dβ T = k BT 2 C, (27) whee C E/ T is the heat capacity of the system at constant extenal paametes. The magnitude of the fluctuations is theefoe given by E E = (k BT 2 C) 1/2 E (28) The cucial point is that both C and E ae extensive paametes, namely they ae popotional to the numbe of atoms in the system, N; while k B T 2 is independent of N. Theefoe, the dependence of E/E on the size of the system is E E 1 N. (29) Thus, fo macoscopic systems with N 10 23, E/E 10 11!. This means that the fluctuations ae extemely small; the enegy of a macoscopic body in a heat bath is completely detemined, fo any pactical pupose. This is why we can use statistical physics to detemine quantitatively the popeties of macoscopic systems: the elative fluctuations ae always of the ode N 1/2, which is tiny. The fact that E/E is so small implies that the pobability distibution (Equation 22) has an extemely shap maximum at enegy E. This is tue fo N 1; in the limit of small N, this no longe hold.

7 3. Extending the definition of entopy Peviously, we defined entopy fo isolated systems. Now, let us genealize this definition, so that it would be applicable also fo systems in contact with a heat bath. Conside a geneal macoscopic system. Let us label its micostates by 1,2,...,,... Let P() be the pobability that the system is in the micostate. At this point, we know nothing about P(), except fo the fact that the system must be at a state, hence P() = 1. (30) What is the entopy of the system? In ode to answe this question, let us conside an ensemble of vey lage numbe, say ν of identical eplicas of ou system. Thus, all these systems ae in themal equilibium with the heat bath. We can now ask: how many of these systems ae in the micostate? - the answe is ν = νp(). (31) The statistical weight Ω ν of the ensemble in which ν 1 of the systems ae in state 1, ν 2 ae in state 2,... is the numbe of ways in which this paticula distibution can be ealized (same logic as in the paamagnetic solid example), which is Ω ν = ν! ν 1!ν 2!...ν!... We can now use Boltzmann s definition of entopy, to wite S ν = k B lnω ( ν = k B ln ν! ν 1!ν 2!...ν!... = k B (lnν! lnν!) = k B (νlnν ν ν lnν + ν ) = k B (νlnν ν lnν ), ) (32) (33) whee we have used Stiling s fomula. Using now ν = νp(), we get S ν = k B (νlnν νp()ln(νp())) = k B (νlnν νlnν P() ν P()lnP()) = k B ν P()lnP() (34) whee in the last line we used the nomalization, P() = 1.

8 This entopy is the entopy of the total ensemble of systems; since entopy is an additive quantity, the entopy of a single system is given by S = 1 ν S ν = k B P()lnP(). (35) We can now calculate the entopy of a system in contact with a heat bath with tempeatue T: using the pobability P() fom Equation 19, we find ( ) ( ) e S = k B βe e ln βe Z Z ( ) (36) = k B (βe +lnz) e βe Z We can use now the definition of Z fom Equation 20 and E fom Equation 23 to wite S = k B βe +k B = E +k T BlnZ e βe Z lnz (37) Note that in the new definition of the entopy in Equation 37, the entopy is a function of T, V and N, while in the definition fo isolated systems, S = k B lnω(e,v,n) is a function of E, V and N. Howeve, as we saw, fo macoscopic systems at tempeatue T, the enegy fluctuations ae tiny. Theefoe the enegy is well defined by its aveage, E. 4. Helmholtz fee enegy When we discussed isolated systems, the basic statistical quantity was the statistical weight Ω(E,V,N), and the basic themodynamic quantity was the entopy, S(E,V,N). When consideing a system in a heat bath the basic statistical weight is the patition function, Z(T, V, N). Thus, we can find the coesponding themodynamics quantity, which is the Helmholtz fee enegy, defined by Using Equation 37, we can wite F(T,V,N) k B T lnz(t,v,n) (38) S = E T F T F = E TS (39) In fact, in macoscopic systems we can safely eplace E with E, as the fluctuations ae so tiny.

9 Thus, fo a system in equilibium in contact with heat bath and having a constant volume, the Helmholtz fee enegy plays a simila ole to that of entopy in an isolated system. In paticula, just as the entopy S is maximal in equilibium in an isolated system, so the Helmholtz fee enegy F obtains its minimum in equilibium state of a system having constant volume in a heat bath (namely, T, V and N ae constants). 4.1. Example: Helmholtz fee enegy and the Schottky defect Let us etun to the example of Schottky defect intoduced ealie. Recall that we had a solid cystal with N atoms and n defects. The enegy pe defect is ǫ, and so the total enegy is E = nǫ. The entopy was calculated in Equation 4 to be and thus the Helmholtz fee enegy is S(n) k B (N lnn nlnn (N n)ln(n n)), (40) F = E TS = nǫ k B T (N lnn nlnn (N n)ln(n n)). (41) We know that F obtains its minimum value in Equilibium state, so we need to find df/dn and equate to 0, The solution is Fo ǫ k B T, df = dn ǫ TdS = ǫ k dn B T ( lnn+ln(n n)) = ǫ k B T ln ( ) N n (42) n = 0. n = N ( e ǫ/k BT +1 ) 1. (43) n Ne ǫ/k BT. (44) These ae the same esults found when maximizing the entopy (see Equation 8).