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LECTURE 10 Simple Applications of Statistical Mechanics We Have Seen That If We Can Calculate the Partition Function

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LECTURE 10 Simple Applications of Statistical Mechanics We Have Seen That If We Can Calculate the Partition Function LECTURE 10 Simple Applications of Statistical Mechanics We have seen that if we can calculate the partition function ¯Er Z = e¡ (1) r X then we can derive just about anything we want to know from the partition function such as the mean internal energy, the entropy, the pressure, the Helmholtz free energy, the speci¯c heat, etc. Let's give some examples of how this works. Paramagnetism You may remember from E&M that the magnetic behavior of materials usually arises from the angular momentum of the electrons. The electrons have 2 forms of angular momentum: spin and orbital. Classically their orbital angular momentum is associated with their orbital motion around the nucleus. Quantum mechanically the orbital angular momentum is associated with their spatial wavefunction. Even if an electron has zero orbital angular momentum, it still has acts like a tiny magnet and has a magnetic moment ¹~ B. Associated with this magnetic moment is the spin angular momentum h¹S~ of the electron.¼ Protons and neutrons also have angular momentum h¹J~ and magnetic moments ¹~. This gives rise to nuclear magnetic moments which are involved in NMR and MRI. The magnetic moment of an atom, ion or single elementary particle such as an electron or proton in free space is proportional to its angular momentum: ¹~ = g¹ J~ = γh¹J~ (2) B ¡ where the proportionality factor g is the \spectroscopic splitting" factor or just the g factor, ¹B the Bohr magneton, γ the gyromagnetic ratio, and h¹J~ the total angular momentum. Rather than considering the case of arbitrary J~ (see Reif 7.8, pages 257- 262), let us begin by considering the simple case of spin-1/2 where J = 1=2 and g = 2. Then the allowed quantum energy states are E = ¹~ H~ ¡ ¢ = g¹ J~ H~ ¡ B ¢ = g¹ J H (since H~ z^) ¡ B z k 1 = 2¹ ( )H ¡ B §2 = ¹ H (3) § B where H is the external magnetic ¯eld. Let us also assume that we have N such inde- pendent ions or atoms all in contact with a heat reservoir at temperature T . A physical system that approximates this model is a metal with magnetic impurities like Mn2+. If we consider only one atom or ion, the partition function ³ of that one atom or ion becomes ¯Er ¯¹B H ¯¹B H ³ = e¡ = e + e¡ (4) all statesX r and N N ¯¹B H ¯¹B H Z = ³ = e + e¡ (5) The mean energy can be calculated from³ ´ @ ln Z @ ln ³ N ¯¹B H ¯¹B H E = = N = ¹ H e e¡ (6) ¡ @¯ ¡ @¯ ¡ ³ B ¡ ³ ´ ¯¹B H ¯¹B H e e¡ E = N¹BH ¡ (7) ¯¹B H ¯¹B H ¡ e + e¡ ¹BH E = N¹BH tanh (8) ¡ µ kBT ¶ The total magnetic moment or magnetization can be calculated from M = N¹ = N (P ¹ + P ¹ ) " " # # ¯¹B H ¯¹B H e e¡ = N¹B ¡ (9) ¯¹B H ¯¹B H e + e¡ where we used ¹ = ¹ = ¹B, P = exp(¯¹BH)=³, and P = exp( ¯¹BH)=³. Thus " ¡ # " # ¡ ¹BH M = N¹B tanh (10) µ kBT ¶ Alternatively we can calculate the generalized force associated with the magnetic ¯eld. Recall that dW = Xdx (11) and 1 @ ln Z X = (12) ¯ @x From Reif 11.1 we ¯nd that magnetic work done by the sample is dW = M~ dH~ (13) ¢ where M~ is the magnetization. Suppose H~ is pointing along the z^ axis. We can identify H with the external parameter x, and M, the z{component of M~ , with the generalized force. Then the generalized force is 1 @ ln Z M = ¯ @H N @ ln ³ = ¯ @H ¯¹B H ¯¹B H e e¡ = N¹B ¡ ¯¹B H ¯¹B H e + e¡ ¹BH = N¹B tanh µ kBT ¶ E = (14) ¡H 2 We see that E = MH (15) ¡ We can examine M in the limit of high and low temperature. ¹BH ¹BH If T 0 or tanh 1 and M N¹B (16) ! µ kBT ! 1¶ µ kBT ¶ ! ! Physically at low temperatures all the magnetic moments are aligned with the magnetic ¯eld. tanh (x) 1 x M µ N B µB kT At high temperatures, ¹BH=kBT 0 and T . Since tanh(x) x as x 0, we obtain ! ! 1 ! ! 2 ¹BH N¹BH 1 M »= N¹B = (17) µ kBT ¶ kB T We expect the magnetization to be proportional to the magnetic ¯eld H. This is what characterizes a paramagnet. How easy or hard it is to magnetize the system is reflected in the magnetic susceptibility Â: M = ÂH (18) From eq. (17) we can read o® the susceptibility: M N¹2 1  = = B T (high temperature limit) (19) H kB T ! 1 The fact that  varies as 1=T is known as the Curie law.  is called the Curie susceptibility and it goes as 1=T . 3 χ T We can also calculate the Helmholtz free energy using eqns. (5) and (4): F = k T ln Z = k T ln(³N ) = Nk T ln ³ (20) ¡ B ¡ B ¡ B where ¯¹B H ¯¹B H ¹BH ³ = e + e¡ = 2 cosh(¯¹BH) = 2 cosh (21) µ kBT ¶ So ¹BH F = NkBT ln 2 cosh (22) ¡ µ µ kBT ¶¶ We can also calculate the heat capacity. Recall from lecture 9 that @2F CV = T (23) ¡ @T 2 ¯ ¯V ¯ ¯ Taking two derivatives of (22) yields ¯ 2 ¹BH 2 ¹BH CV = NkB sech (24) µ kBT ¶ µ kBT ¶ The speci¯c heat per spin is 2 CV ¹BH 2 ¹BH cV = = kB sech (25) N µ kBT ¶ µ kBT ¶ This is called the Schottky speci¯c heat. It is the speci¯c heat characteristic of two state systems. A two state system has only 2 states available to it. For example a spin{1/2 object has spin{up and spin{down states. Another example is an object that can only access the lowest states in a double well potential. In considering such discrete states, we are thinking quantum mechanically. 4 Schottky Specific Heat 0.5 0.4 0.3 B /Nk V C 0.2 0.1 0.0 0.0 1.0 2.0 3.0 4.0 5.0 µB/kT Classical Harmonic Oscillator Consider a classical harmonic oscillator with a spring constant ·. The typical example is a mass on a spring. The energy is given by p2 1 E = + ·x2 (26) 2m 2 Assume the harmonic oscillator is in contact with a thermal reservoir at temperature T . The partition function is 1 ¯E Z = e¡ dxdp ho Z 1 ¯p2 ¯·x2 = 1 dp exp 1 dx exp (27) ho á 2m ! á 2 ! Z¡1 Z¡1 Recall from Reif Appendix A4 that 1=2 1 ®x2 ¼ e¡ dx = (28) ® Z¡1 µ ¶ Using this (27) becomes 1 2¼ 1=2 2¼m 1=2 1 2¼ m 1=2 Z = = (29) ho ï·! à ¯ ! ¯ ho µ · ¶ Or 2¼ m 1=2 ln Z = ln ¯ + ln (30) ¡ à ho µ · ¶ ! 5 and @ ln Z @ ln ¯ 1 E = = = = k T (31) ¡ @¯ @¯ ¯ B This makes sense; we expect the mean energy to be of order kBT . It turns out that E = kBT has equal contributions from the kinetic energy and from the potential energy. Each contributes kBT=2. One can show this explicitly. The mean kinetic energy is 2 p ¯p2=2m 1 2m e¡ dp ¡1 KE = ¯p2=2m (32) R 1 dpe¡ ¡1 Use Reif Appendix A4 to evaluate the nRumerator and denominator: 1=2 1 ®x2 ¼ e¡ dx = (33) ® Z¡1 µ ¶ and 1 ®x2 2 p¼ 3=2 e¡ x dx = ®¡ (34) 2 Z¡1 So 3=2 1 p¼ 2m 2m 2 ¯ KE = 1=2 ³2m ´ p¼ ¯ k T ³ ´ = B (35) 2 Similarly the mean potential energy is given by 1 2 ¯·x2=2 1 dx 2 ·x e¡ ¡1 P E = ¯·x2=2 R 1 dxe¡ ¡1 3=2 · pR¼ 2 2 2 ¯· = 1=2 ³ 2 ´ p¼ ¯· k T³ ´ = B (36) 2 The fact that the mean kinetic and potential energies of a harmonic oscillator equals kBT=2 is an example of the classical equipartition theorem. Classical Equipartition Theorem Let the energy of a system with f degrees of freedom be E = E(q1:::qf ; p1:::pf ) and assume 1. the total energy splits additively into the form E = ²i(pi) + E0(q1; :::; pf ) (37) where ²i involves only pi and E0 is independent of pi. 6 2. 2 ²i(pi) = bpi (38) where b is a constant. If we replaced pi by qi, the theorem is still true. We just want ²i to be a quadratic function of one component of q or p. Assume that the system is in equilibrium at temperature T . Then ¯E 1 ²ie¡ dq1:::dpf ¡1 ²i = ¯E R 1 e¡ dq1:::dpf ¡1 0 ¯(²i+E ) R1 ²ie¡ dq1:::dpf = ¡1 0 ¯(²i+E ) R 1 e¡ dq1:::dpf ¡1 0 ¯²i ¯E R1 ²ie¡ dpi 1 e¡ dq1:::dpf = ¡1 ¡1 0 (39) ¯²i ¯E R 1 e¡ dpi R1 e¡ dq1:::dpf ¡1 ¡1 R R where the last integrals in the numerator and denominator do not involve dpi. These integrals cancel leaving ¯²i 1 ²ie¡ dpi ²i = ¡1 ¯²i R 1 e¡ dpi ¡1 @ ¯²i R @¯ 1 e¡ dpi = ¡ ¡1 ¯²i 1R e¡ dpi ¡1 R@ 1 ¯²i = ln e¡ dp (40) ¡@¯ i µZ¡1 ¶ Now use the second assumption: 2 2 1 ¯²i 1 ¯bp 1=2 1 by e¡ dpi = e¡ i dpi = ¯¡ e¡ dy (41) Z¡1 Z¡1 Z¡1 where y ¯1=2p . Thus ´ i 2 1 ¯²i 1 1 by ln e¡ dp = ln ¯ + ln e¡ dy (42) i ¡2 µZ¡1 ¶ Z¡1 Notice that the integral does not involve ¯.
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