Lecture 21. Boltzmann Statistics (Ch. 6) We have followed the following logic: 1. Statistical treatment of isolated systems: multiplicity ⇒ entropy ⇒ the 2nd Law. 2. Thermodynamic treatment of systems in contact with the heat reservoir ⇒ the minimum free energy principle. However, the link between G and the process of counting of accessible microstates was not straightforward. Now we want to learn how to “statistically” treat a system in contact with a heat bath. The fundamental assumption states that a closed (isolated) system visits every one of its microstates with equal frequency: all allowed states of the system are equally probable. This statement applies to the combined system (the system of interest + the reservoir). We wish to translate this statement into a statement that applies to the system of interest only. Thus, the question: how often does the system visit each of its microstates being in the thermal equilibrium with the reservoir? The only information we have about the reservoir is that it is at the temperature T. Reservoir R U - ε Combined system 0 U0 = const System S ε a combined (isolated) system = a heat reservoir and a system in thermal contact The Fundamental Assumption for an Isolated System Isolated ⇒ the energy is conserved. The ensemble of all σ i equi-energetic states ⇒ a mirocanonical ensemble The ergodic hypothesis: an isolated system in thermal equilibrium, evolving in time, will pass through all the accessible σ 2 microstates states at the same recurrence rate, i.e. all accessible σ 1 microstates are equally probable. The average over long times will equal the average over the ensemble of all equi- energetic microstates: if we take a snapshot of a system with N microstates, we will find the system in any of these microstates with the same probability. Probability of a particular microstate of a microcanonical ensemble = 1 / (# of all accessible microstates) The probability of a certain macrostate is determined by how many microstates correspond to this macrostate – the multiplicity of a given macrostate macrostate Ω Probability of a particular macrostate = (Ω of a particular macrostate) / (# of all accessible microstates) Note that the assumption that a system is isolated is important. If a system is coupled to a heat reservoir and is able to exchange energy, in order to replace the system’s trajectory by an ensemble, we must determine the relative occurrence of states with different energies. Systems in Contact Reservoir R System S ε with the Reservoir U0 - ε The system – any small macroscopic or R microscopic object. If the interactions between the system and the reservoir are S weak, we can assume that the spectrum of ε2 energy levels of a weakly-interacting system is ε1 the same as that for an isolated system. Example: a two-level system in thermal contact with a heat bath. We ask the following question: under conditions of equilibrium between the system and reservoir, what is the probability P(εk) of finding the system S in a particular quantum state k of energy εk? We assume weak interaction between R and S so that their energies are additive. The energy conservation in the isolated system “system+reservoir”: U0 = UR+ US = const According to the fundamental assumption of thermodynamics, all the states of the combined (isolated) system “R+S”are equally probable. By specifying the microstate of the system k, we have reduced ΩS to 1 and SS to 0. Thus, the probability of occurrence of a situation where the system is in state k is proportional to the number of states accessible to the reservoir R . The total multiplicity: Ω()ε k ,U 0 −ε k = ΩS (ε k )×ΩR (U 0 −ε k )(=1×ΩR U 0 −ε k )= ΩR (U 0 −ε k ) Systems in Contact with the Reservoir (cont.) The ratio of the probability that the system is in quantum sate 1 at energy tε1 to the probability that the system is in quantum state 2 at energy ε2 is: PS ()ε1 ΩR ()U0ε − 1 expS[ R ( U0−ε 1)/ B ] k SUSU⎡ R ()−0ε 1 −R ( 0ε − 2)⎤ ⎛ ΔSR ⎞ = = = exp⎢ ⎥ = exp⎜ ⎟ PS ()ε 2 ΩR ()U0ε − 2 expS[]R () U0−ε 2 / B k ⎣ kB ⎦ ⎝ kB ⎠ SR S(U0) Let’s now use the fact that S is much smaller than R (US=εk << UR). S(U0- ε1) l consider the so, we’lcase lAof fixed vlume and numbero of (particlattles limitat ter ion will be removed later, when we’ll S(U - ε ) 0 2 w the system to exchange paallorticles with the heat bath 1 2 ε 0 ε UR 1 (Pr. 6.9 addresses the case - - U 0 0 dS=RdU()R + PdVR μ − R dN nd U U when the 2 is not negligible) TR 0 ⎡ ∂SR ⎤ ⎡ dSR ⎤ SUSUR ()≈R (0 ) +⎢ UUU()0 ⎥ (− 0 ) SUSU)R −0 ε (i ≈ R ()0 ε −i ⎢ ()U 0 ⎥ ⎣∂U R ⎦VN, ⎣dU R ⎦VN, ⎛ ∂S ⎞ ⎛ ∂S ⎞ ⎡⎛ ∂S ⎞ 1 ⎤ ε− ε ΔS =⎜ R ()U ⎟ ε − ⎜ R ()U ⎟ ε = ⎢⎜ R ()U ⎟ = ⎥= −1 2 R ⎜ ∂U 0 ⎟ 2 ⎜ ∂U 0 ⎟ 1 ⎜ ∂U 0 ⎟ TT ⎝ R VN⎠ , ⎝ R VN⎠ , ⎣⎢⎝ R VN⎠ , R ⎦⎥ R Boltzmann Factor PS ()ε1 ⎛ ΔSR ⎞ ⎛ ε1− ε 2⎞ ⎛ ⎞ = exp⎜ ⎟=exp⎜ − ⎟ ε1 ⎜ ⎟ ⎜ ⎟ exp⎜− ⎟ PS ()ε 2 ⎝ kB ⎠ ⎝ kB T ⎠ P ()ε k T S 1 = ⎝ B ⎠ T is the characteristic of the heat reservoir PS ()ε 2 ⎛ ε 2 ⎞ exp⎜− ⎟ ⎝ kB T⎠ exp(- εk/kBT) is called the Boltzmann factor This result shows that we do not have to know anything about the reservoir except that it maintains a constant temperature T ! The corresponding probability distribution is known as the reservoir canonical distribution. An ensemble of identical systems all of which are in contact with the same heat reservoir and T distributed over states in accordance with the Boltzmann distribution is called a canonical ensemble. The fundamental assumption for an isolated system has been transformed into the following statement for the system of interest which is in thermal equilibrium with the thermal reservoir: the system visits each microstate with a frequency proportional to the Boltzmann factor. Apparently, this is what the system actually does, but from the macroscopic point of view of thermodynamics, it appears as if the system is trying to minimize its free energy. Or conversely, this is what the system has to do in order to minimize its free energy. One of the most useful equations in TD+SP... Firstly, notice that only the energy difference Δε = ⎛ ⎞ ⎛ ⎞ P()ε1 ΔSR ε1− ε 2 ε - ε comes into the result so that provided that = exp⎜ ⎟=exp⎜ − ⎟ i j P()ε 2 ⎝ kB ⎠ ⎝ kB T ⎠ both energies are measured nfrom the same origi it does not matter what that origin is. what matters in determy, lcondSe tionaining the ratio of the occup P(ε i ) 9 numbers is the ratio of the energy difference Δε to kBT. Suppose =e ≈8000 P()ε that εi = kBT and εj = 10kBT . Then (εi - εj ) / kBT = -9, and j The lowest energy level ε0 e to a system (e.g., a llabavai molecule) is referred to as the “ground state”. If we measure all n(ε ) ⎛ ε ⎞ =exp⎜ − ⎟ energies relative to ε0 and n0 s esin the number of moleculithis ⎜ ⎟ n0 ⎝ kB T⎠ state, than the number molecules with energy ε > ε0 Problem 6.13. At very high temperature (as in the very early universe), the proton and the neutron can be thought of as two different states of the same particle, called the “nucleon”. Since the neutron’s mass is higher than the proton’s by Δm = 2.3·10-30 kg, its energy is higher by Δmc2. What s the waratio of the number of protons to the number of neutrons at T=1011 K? ⎛ m2 c⎞ exp⎜− n ⎟ ⎜ ⎟ 2 −30 8 2 P() n k T ⎛ Δ2mc . 3⎞ 10⎛ ⋅ kg × 3 ⋅ 10⎞ m/s = ⎝ B ⎠=exp⎜ − ⎟=exp⎜ − ()⎟0= . 86 P() p ⎛ m2 c⎞ ⎜ k1 T .⎟ 38⎜ 10⋅ −23 J/K × 11 10⎟ K ⎜ p ⎟ ⎝ B ⎠ ⎝ ⎠ exp⎜− ⎟ ⎝ kB T⎠ More problems Problem 6.14. Use Boltzmann factors to derive the exponential formula for the density of an isothermal atmosphere. The system is a single air molecule, with two states: 1 at the sea level (z = 0), 2 – at a height z. The energies of these two states differ only by the potential energy mgz (the temperature T does not vary with z): ⎛ ε 2 ⎞ ⎛ ε1 + mgz ⎞ exp⎜− ⎟ exp⎜− ⎟ P()ε k T k T ⎛ mgz ⎞ ⎛ mgz ⎞ 2 ⎝ B ⎠ ⎝ B ⎠ ⎜ ⎟ ⎜ ⎟ = = =exp⎜ − ⎟ ρ()z= ρ0 exp⎜ − ⎟ P()ε1 ⎛ ε1 ⎞ ⎛ ε1 ⎞ ⎝ kB T⎠ ⎝ kB T⎠ exp⎜− ⎟ exp⎜− ⎟ ⎝ kB T⎠ ⎝ kB T⎠ At home: A system of particles are placed in a uniform field at T=280K. The particle concentrations measured at two points along the field line 3 cm apart differ by a factor of 2. Find the force F acting on the particles. Answer: F = 0.9·10-19 N A mixture of two gases with the molecular masses m1 and m2 (m2 >m1) is placed in a very tall container. The container is in a uniform gravitational field, the acceleration of free fall, g, is given. Near the bottom of the container, the concentrations of these two types of molecules are n1 and n2 respectively (n2 >n1) . Find the height from the bottom where these two concentrations become equal. k Tln( n / ) n h = B 2 1 Answer: m()2− m 1 g The Partition Function For the absolute values of probability (rather 1 ⎛ ε ⎞ 1 ⎜ i ⎟ than the ratio of probabilities), we need an P()ε i =exp⎜ − ⎟=exp() −βε i explicit formula for the normalizing factor 1/Z: Z⎝ kB T⎠ Z 1 β ≡ - we will often use this notation kB T The quantity Z, the partition function, can be found from the normalization condition - the total probability to find the system in all allowed quantum states is 1: 1 ZTVN,, =exp −βε The Zustandsumme 1=∑P()ε i = ∑exp()− βε i or ( ) ∑ ( i ) i i Z i in German The partition function Z is called “function” because it depends on T, the spectrum (thus, V), etc.