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Unformatted text preview: Summary of last lecture Grand Canonical Ensemble and its relation to Thermodynamics Paramagnetism (to be continued) Chemistry 161: Statistical Thermodynamics
Lecture 11 Independent Particles. Maxwell Distribution, Classical Harmonic Oscillator
Key Concepts and Lessons: a) Paramagnetism b) CPF of Monoatomic Ideal Gas c) Gibbs paradox d) Maxwell Distribution of velocities, mean velocity e) Classical Harmonic Oscillator Reading: Reif, Ch6 Simple example of Canonical PF calculation: Paramagnetism Consider a system of N noninteracting particles with spin 1/2 arranged into a lattice in magnetic field H, Each spin can be in `'up'' orientation with spin magnetic moment aligned along the filed and in `'down'' orientation with its magnetic moment aligned antiparallel to the field. We first ask how does energy of such system depend on temperature? In order to do that we calculate the Canonical Partition Function:
Z=
1 N exp
= 1/ 2,
2 gH
i =1 = 1/ 2.... i =
1 N = 1/ 2, = 1/ 2.... i =1 exp [ gH i]= N gH
i =1
i 2 = 1/ 2 i = zN where z = exp 1 gH + exp 2 1 gH 2 The partition function sum contains 2N terms, the factorization into singlespin terms allows to sum over each spin's state independently. We therefore note a very important general result: CPF for a system of noninteracting Particles factorizes into CPF for each particle separately! Paramegnetism (Cont) Now we are in a position to calculate how energy depends on T. The Helmholtz free energy is:
1 gH + exp 2 1 gH 2 1 gH + exp 2 kT 1 gH 2 1 gH 2 1 gH 2 kT ln Z = N ln z = ln exp A = kT ln Z = kTN ln z = kTN ln exp exp E= 1 gH exp ln Z gH 2 = N 1 2 exp gH + exp 2 Let us play a bit with this result. The best way to understand something is to check limiting cases. In this case tow limits are interesting: of strong field gH 1 and that of weak field gH 1 In the former case only one exponential (with''+'' sign) is relevant and we get:
E NgH 2 Which makes perfect sense because in this case all spins are strongly aligned along the Field. Paramagnetism  Induced Magnetic Moment In the opposite case, of weak field we can expand exponentials in numerator and denominator and retain only terms up to first power. We get:
E= N ( gH )2
4 i.e. it is quadratic in applied field! Why? In order to figure that out let us find total magnetization, Again, according to canonical ensemble prescription:
N N exp M=
1 = 1/2, 2 gH
i =1 N = 1/2.... i = gH
i N i =1 1 = 1/2, 2 = 1/2.... i exp gH
i =1 N i =
i N
i = 1/2, i exp [ zN gH i
z N 1 =N exp
1 = 1/2, 2 exp
1 = 1/2, 2 gH
i =1 i =1 = 1/2.... i =1 = 1/2.... where 1 1 exp ( gH ) exp ( gH ) 1 2 = 2 = tanh ( gH ) 2 exp ( gH ) + exp ( gH )
1 Now we can again see what is happening here: At strong fields tanh=1 and = i.e all spins are aligned along the field. In the opposite case of weak field we get:2 = 1 1 tanh gH 2 2 1 gH gH = 4 4kT i.e. the field aligns spins weakly in a temperaturedependent fashion. Now, We see what happens at high T: the field interacts with magnetic moment E = gHM Induced by the field itself!: CPF in classical approximation There are many cases (high T) when QM description is not necessary. We already discussed the relationship of classical phase space to QM states. The relationship states that sum of all quantum states is equilvalent to integral over all phase space divided by volume of cells according to BornSommerfeld Quantization rules: quantumstates 1 = 3N h dpdq where h is Plancks constant and N is the number of particles (so that 3N is the number of degrees of freedom)
Therefore in the classical approximation the canonical partition function is: 1 Z classical = 3N exp h E ( p1, p2 ....pN ;q1,q2 ...qN ) dp1dp2 ..dpN dq1dq2 ...dqN Where p and q are as usual coordinates and momenta of all N particles in the system And E is total energy of the system. CPF in classical approximation  cont Total energy of a classical system can be divided into two contributions: kinetic energy and potential energy: pi2 E ( p1, p2 ....pN ;q1,q2 ...qN ) = Ek ( p1, p2 ....pN ) +U ( q1,q2 ...qN ) = +U ( q1,q2 ...qN ) i=1 2m
N CPF also factorizes:
1 Z = 3N exp h where Z config = 1 h
3N 2 pi2 dp1dp2 ...dpN exp i =1 2m
N N U ( q1 , q2 ...qN ) dq1dq2 ...dqN = zkin Z config exp U ( q1 , q2 ...qN ) dq1dq2 ...dqN is a configurational part of CPF which depends only on coordinates (configurations) of all particles but not on their velocities zk = 1 h
3 2 exp p2 dp 2m 3 = 2 m h 3 = 2 mkT h 3 Is a contribution to the partition function from one particle The classical approximation (Cont) For Helmholtz free energy we get simply 3 2 mkT kT lnZconfig kTN ln 2 h and we can get entropy, energy and heat capacity from here A = kTN lnzk kT lnZconfig = 3 2 mkT 3 A + kN + Sconfig = kN ln 2 T V 2 h 3 E = A + TS = kTN + Econfig 2 3 CV = kN + CVconfig 2 S=
We therefore determined contributions from molecular motion to thermodynamic functions of ANY substance when it behaves classically CPF of Monoatomic Ideal Gas In the case of monoatomic ideal gas our analysis is very simple because molecules do not interact I.e. potential energy U=0. We can easily evaluate the configurational part of partition function in that case: = config 1 1 h
3N 2 exp U q1 ,q2 ,...qN dq1dq2 ,...dqN =
3N / 2 ( ) 1 h
3N 2 dq1 dq2 ,...dqN = 1 V N; 3N h2 Z= VN 3N h2 2 kTm h = V 2 kTm h2 3/ 2 N N = z particle A = kTN ln z particle N V kTN ln = kTN ln e N 3N 2 kTm kT ln 2 h2 Gibbs Paradox The above expression for CPF of ideal gas is almost correct. In order to see it consider a volume containing gas separated by a partition into two equal parts each having volume V and containing N particles. Now remove the partition and we get 2N particles in volume 2V. Apparently nothing changed and free energy must stay the same. Does it?
Apartition = 2 kTN lnV 3N 2 kTm kT ln 2 h2 3NkT ln = 2kTN lnV 3NkT ln 2 kTm h2 Anopartition = 2kTN ln 2V Apartition 2 kTm apparently 2 h Anopartition and the difference between the two is 2kTN ln 2 What happened? Apparently we forgot that particles are indistinguishable! In other words the correct expression of the CPF of ideal gas should contain N! as appropriate: Z ==
N z particle N! Gibbs Paradox (Cont) That resolves Gibbs paradox because now free energy of ideal gas is: N V 3N 2 kTm A= kTNlnz particle + kTNln = kTNln kT ln 2 e N 2 h
And clearly removing the partition does not change A Maxwell Distribution We can now focus on any particle (not necessarily in an ideal gas but in a classical system) and ask what is the probability distribution of its velocity. Consider, say, particle `'1'' out of N in our system:
P ( p1, p2 .......pN ) =
N dq1dq2 ...dqN e 1 2mkT N pi2
i=1 e 1 U ( q1 ,q2 ...qN ) kT Z = dq1dq2 ...dqN e 1 2mkT N pi2
i=1 e
1 2mkT 1 U ( q1 ,q2 ...qN ) kT
N =
pi2 dq1dq2 ...dqN dp1dp2 ...dpN e = e
1 2mkT pi2
i=1 N i=1 e 1 U ( q1 ,q2 ...qN ) kT dp1dp2 ...dpN e 1 2mkT =
pi2 N i=1 1 2 mkT 3/2 e pi2 2mkT i=1 and we get famous Maxwell distribution for velocities of each individual particle (hence tilde sign below):
m 2 2 2 m (vx +vy +vz ) 2kT e dvx dvy dvz P vx ,vy ,vz dvx dvy dvz = 2 kT which gives us the probability that velocity of the particle lies in the range ( )
x 3/2 (v ,v + dv ;v ,v + dv ;v ,v + dv ) (note change of normalization as we switched from p to v)
x x x y y x z z Maxwell Distribution, Cont Distribution of speeds. Now consider a system of N molecules and ask what fraction of them have absolute value of speed between v and v+dv. This is given by the relation:
f (v)dv = 4 P vx , vy , vz v dv = 4
2 ( ) 2 mkT h2 3 ve 2 mv 2 2 kT dv we used a relation 4 v2 dv for an element of volume in 3dimensional vspace f (v) This distribution is not symmetric! Maxwell Distribution mean velocity We can evaluate mean velocity of a particle as:
m 2 kT
3/2 mv 2 2 kT v = vf (v)dv = 4
0 e
0 v dv = 4
3 m 2 kT 3/2 1 m 2 2kT 2 ; v= 8kT m On the other hand means SQUARED speed (related to kinetic energy): 1 2 mv = Ek = 4 2 m 2 kT 3/2 v e
0 4 mv 2 2 kT dv = 4 m 2 kT
3/2 2 m kT e
0 mv 2 2 kT 3 dv = kT 2 Here we encounter the manifestation of an equipartition theorem: Each degree of freedom (I.e. in this case motion along x, y and z axis) contributes 1/2kT to mean kinetic energy (we will soon discover a similar result for potential energy) Maxwell Distribution  Most probable speed Finally we determine the value of most probable speed from the relation:
d 2 v f (v) = 0 corresponding to the maximum of the distribution for v : dv
2 mvm . p . ( ) 2vm. p.e vm. p. = 2 kT +v 2 m. p. mvm. p. kT 2 mvm . p . e 2 kT =0 2kT m Due to the asymmetry of f(v) we get different results fo various averages v2 v vm. p. Classical Harmonic Oscillator Consider a system consisting of N Harmonic oscillators (H.O) (we will see soon that it is a good model of crystals under certain conditions) Energy of a classical H.O is:
2 x pz2 1 2 1 2 1 2 x p y z EHO = Ek +U = + + + kx + ky + kz = EHO + EHO + EHO 2 2 2m 2m 2m 2
And frequency of oscillations is given by a wellknown relation
=2 k m py2 Partition function of such system of N noninteracting 3D HO is given as before
Z= z 3N where N! e
p2 2 mkT 1 z= h e 1 kx 2 2 kT dpdx is partition function of a onedimensional H.O we get z = 1 kT 2 mkT 2 h k Classical Harmonic Oscillator TD functions Now we can determine Helmholtz free energy oof the system of H.O and its energy and heat capacity:
A = kT ln Z = 3NkT ln z + kTN ln A 3 3 = NkT + NkT; T 2 2 E 3 3 CV = = Nk + Nk = 3Nk T 2 2 E=A T
Heat capacity of the system of classical H.O is twice as much as monoatomic ideal gas. Why? Because in H.O we have contribution from potential energy which appears to be equal To that of kinetic energy  Equipartition at work! So each degree of freedom for H.O Gives 1/2kT to energy due to potential energy and 1/2 kT due to kinetic energy 3 N 3 = kT lnT kT lnT + Const(T ) 2 e 2 Next Lecture: Quantum oscillator, Einstein model and its critique, Boltzmann, FermiDirac and BoseEinstein Statistics. Reading: Reif Ch.7. McQuarrie. Ch4 ...
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 Spring '10
 Shaklovich

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