Statistical Mechanics Victor Naden Robinson vlnr500 3 rd Year MPhys 17/2/12 Lectured by Rex Godby Lecture 1: Probabilities Lecture 2: Microstates for system of N harmonic oscillators Lecture 3: More Thermodynamics, Boltzmann and Entropy Lecture 4: Entropy Lecture 5: Entropy and applications of statistical mechanics Lecture 6: 2-Level System Lecture 7: More harmonic oscillator and heat capacity Lecture 8: Modes Lecture 9: Debye continued Lecture 10: The ideal gas Lecture 11: More ideal gas Lecture 12: Maxwell-Boltzmann distribution Lecture 13: Summary and Gibbs approach Lecture 14: Identical particles Lecture 15: Bose-Einstein distribution Lecture 16: Continuation and condensation Lecture 17: Energy in EM modes Lecture 18: Black body radiation
Lecture 1: Probabilities {Note lots of graphs in this course} Probability of obtaining score S on: 1 die [Figure 1. Plotting P vs. S, dots on the P=1/6 line] 2 dice: Total Score Average Score Number of Configurations 2 1.0 1 3 1.5 2 4 2.0 3 5 2.5 4 6 3.0 5 7 3.5 6 8 4.0 5 9 4.5 4 10 5.0 3 11 5.5 2 12 6.0 1 Total: [Figure 2: Prob vs. avg score, 1 to 6 on x-axis, 0 to 1/36 to 1/6 on y-axis, triangle dot formation] 7 Dice: Total Average Score Number of ways Probability 7 1.0 1 8 8/7 7 9 9/7 21* *5x(1 dot) + 2x(2 dots): 6x(1 dot) + 1x(3dot): [Figure 3: Prob vs. score plotted as a Gaussian between scores 1 to 6, FWHM=245660] Scores obtained: 2.85 3.80 4.70 10 24 dice: [Figure 4: same as figure 3 but very thin (width ~10-12 )] With a large number of components we can describe properties with precision. We can make predictions about the collective behaviour of large systems without needing to predict the detailed motion of its components. Equally likely elementary outcomes 1.1 Microstates
Microstates are QM eigenstate solutions to S.E. but for a whole system rather than an elementary particle. For an isolated system the energy is fixed but this energy has degenerate states: W (large number) microstates with energy, E. [Figure 4: E on y axis then lots of dashed lines which I suppose is energy levels, circle enclosing some of the dashes labelled W microstates.] From Fermi s Golden Rule: Let p i be the probability of the system being in state I (among the W states): (1.1) (1.2) When a steady state has been reached, for all states. This can happen only if, i.e. all probabilities are equal and all equal. This is the Principle of Equal Equilibrium Probability (PEEP). Refresh: Possible E of single SHM, think about E of N SHM s. Lecture 2: Microstates for System of N harmonic oscillators Classically, Where is classical angular frequency (2.2) Energies where [Figure 2.1: of parabola with lines across it showing energy levels, at going up] A microstate of system of N oscillators is given by the states each oscillator is in. The energy of N oscillators is:
(2.3) What is the number of microstates W corresponding to a given value of Q (i.e. to a given total energy)? Consider: : Q dots, (N-1) partitions The no ways W is equal to no ways of arranging the Q dots and (N-1) partitions: 2.1 Thermal Equilibrium (2.4) Two systems of N harmonic oscillators The no microstates corresponding to distribution of energy is: From PEEP each of these W microstates is equally likely. (2.5) In the general case rather than W 1 we shall focus on the density of states, i.e. the number of microstates per unit energy: Note that where the band of energies within which E is fluctuating. (2.6) (2.7) Lecture 3: More thermodynamics System: E 1 W 1 energy E 1 W 1 G 1 G 1
The probability of this discussion of energy: For most likely partition, maximise, that is, maximise: (3.1) (3.2) Thus for most likely partition of energy (i.e. equilibrium): (3.3) (3.4) This quantity: is therefore equalised between two systems when they can exchange energy. [Figure 3.2: Ok I ll try and draw it!] lng E lng E Ok that went well. Define Statistical Mechanical temperature as: *For a large system This T is identical to the ideal gas temperature. (3.5)* 3.2 Boltzmann Distribution Visa vies systems of constant T. E R g R energy E 1 [Figure 3.3] T Microstate i Heat Reservoir
Because the reservoir is large, T is too good an approximation slowly varying with E R. By PEEP the probability of the situation shown is: (3.6) But: (3.7) Since (3.8) (3.9) (3.10) Z is the sum over all microstates j of the system of the Boltzmann Factor. Mean energy of a system at constant T The mean energy (3.11) Recall: (3.12) 3.3 Entropy (3.13) For an isolated system we define:
(3.14) For a non-isolated system exploring different microstates with probabilities p i (assumed to be known): {ends} Lecture 4: Entropy (4.1) For the general case where our system explores its microstates with given (generally equal) probabilities p i, consider N replicas of the system, where N is arbitrarily large. Then ~Np 1 systems will be in microstate 1, ~Np 2 in microstate 2, etc. Total Energy Np 1 E 1 + Np 2 E 2 + is effectively fixedm thus the system is thermally isolated in its behaviour, exploring states only with a total fixed energy. Thus the effective W is the number of distinct permutations of Np 1 identical objects, Np 2 identical objects, etc. (4.2) Using Stirling Approximation: (4.3) So the entropy of N replicas is: (4.4) For N=1, then: Entropy of an isolated system (4.5) (4.6) { (4.7)
(4.8) Entropy of a system at constant temperature (4.9) Recall Helmholtz free energy: (4.10) Thermodynamic entropy (4.11) Consider a system thermally isolated on which work can be done. The total internal energy is: When the volume of the system is changed: The RHS (and last term) must be associated with heat. (4.12) (4.13) Lecture 5: Entropy and applications of statistical mechanics If a system is insulated (thermally isolated): (5.1)
Now, (5.2) For a slow change in constraints, FMG shows that no additional transitions between microstates are induced, i.e. dp i = 0 for all For an insulated system: Since, the other term in (4.2) must be the heat: At constant temperature, the probabilities, p i, are given by the Boltzmann distribution: (5.3) (5.4) (5.5) Also (4.6): { } { } Thus statistical mechanical and thermodynamic entropy are identical. (5.6) 5.1 Applications of Statistical Mechanics Vacancies in crystals For N atoms, how many vacancies will there be at temperature T? Each vacancy costs an amount of energy due to not being bonded fully. Free energy F is to be minimised:
So when volume and temperature are held constant, hence we minimise not (5.7) What are and for a given? (5.8) Provided a system is sufficiently large, it s energy is effectively fixed as if it were thermally isolated so we may use: W is the number of arrangements of N atoms, and vacancies on a total of sites: { } (5.9) { } { } Since, (5.10) Note: At 300K, (5.11) At 3000K, Practical crystals at room temperature are not yet at thermal equilibrium but reflect vacancy concentrations from temperatures where the crystal first formed.
Lecture 6: 2-Level System E.g. each magnetic atom/ion in a dilute magnetic semi-conductor If then Because (6.1) B 0 μb μb Take (6.2) Low T: High T: P 1 p 1 (6.3) 0 KT μb KT μb T So, (6.4)
Diagram: E T μb Heat capacity C Schottky Anomaly (6.5) kt μb T Mean magnetic dipole moment μ μ T 0 So the magnetic polarisation with n magnetic ions per unit volume of crystal is Harmonic Oscillator ω ω ω [ ]
So This is a geometric series (6.6) (6.7) Lecture 7: More Harmonic Oscillator and Heat Capacity Harmonic oscillator continued, (7.1) (7.2) (7.3) This is the Planck oscillator formula (eqn. 7.4) (7.4) At low T ; At high T ; But
(7.5) An example of the classical equipartition theorem: A mean energy of the total energy is proportional to position 2 or velocity 2. per degree of freedom in E kt ω T Heat capacity of oscillator Graph: C k (7.6) ω k T Vibrational energy of a crystal First consider two atoms in one dimension λ m m 2 modes: ω λ/m
And ω Arbitrary motion of the atoms can be written as a linear combination of these normal modes. Next: 3 atoms in 1D Now 3 modes: ω ω ω N atoms in 3D have 3N modes each with a well-defined frequency. Lecture 8: Modes For a periodic crystal the modes can be classified by their wave vectors q. Displacement of atom at position is: (8.1) L L L Zero boundary conditions give a wave of zero when. Cubic lattice of points in q-space, each one representing an allowed value of q (8.2) C π L π L π L q y q x
For a realistic solid: ω boundary of s st Brillouin zone q We study two simplified models: Einstein model ω ω E N modes q Debye model ω slope c Einstein model q max q 3N simple harmonic oscillators (8.3) E C Nk (8.4) N ω q max q max
At low T, Einstein model good except at low T (see hand out). (8.5) Debye model The n. of q points per unit volume of q space is The volume of the sphere in q-space is The volume of the positive octant of this sphere is: (8.6) (8.7) Lecture 9: Debye model continued Picture of positive octant of sphere in q space with a strip on the surface For the Debye model We had, for a single oscillator {lecture 7}: Adding over the modes, with N. of q points per unit volume of q space is Contribution to heat capacity from modes in the shell of thickness is [1 st term from volume of octant shell, 2 nd term is n. of points per unit volume, 3 rd is modes per q] (9.1)
Looking at (9.2) Using: (9.3) Sub into (9.3) [Note the limits change too] (9.4) (9.5) [Graph showing this integrand vs. x and the tails and for low T and high T] Therefore for high T, integrand and so integral [ ] As expected for 3N oscillators with high T and heat capacity k. (9.6) For low T: let contour integration: since rest of tail (graph) contributes negligible amount to C. Law of
(9.7) [Graph comparing C vs. T for Debye and Einstein] [Graph of omega vs. q showing straight line relationship and something about not being frozen] (9.8) Sometimes we prefer to use as the parameter in the Debye model not c but the Debye frequency Lecture 10: The ideal gas Gas molecules move around independently: Ignore long-range Van der Waals attraction Ignore scattering (except to allow molecules to explore all accessible microstates) Initially we study monatomic gases. Partition function of ideal monatomic gas is required. For one atom in a box: For two non-identical atoms: (10.1) For two identical atoms: (10.2) (10.3) { For three atoms: (10.4) For N atoms: { (10.5) { (10.6)
In an ideal gas, most of the states are empty so as long as the temperature is not extremely low, an atom has so many particle states to choose from that a state with more than one atom in it is extremely unlikely. To a good approximation we can apply the correction factor to all cases. Thus: (10.7) ( ) ( ) ( ) Thus Z for N atoms is: (10.8) Schrödinger eqn: (10.9) Choosing boundaries for which ; the solution is: (10.10) Where (10.11) The energy Eigen function is: Volume of the octant shell of radius thickness in q space is: (10.12) Number of states per unit q space is: (10.13) (10.14) (10.15) Lecture 11: More ideal gas! Continues from L10
Continuing on, define ( ) (11.1) ( ) And since ( ) (11.2) So Free energy is given by [ ] [ ] Note that F is extensive: doubling both V and N simply doubles F. (11.3) Pressure of ideal gas Recall (11.4) [ ]
(11.5) This establishes the equivalence of stat mech. and (ideal gas) thermo-dynamical temperatures. Energy of ideal gas [ ] (11.6) I.e. energy of per gas molecule 1 atom, as predicted by classical equipartition theorem: per K.E. term Note that our non-degenerate approximation that led to assumes that T is not too low, hence we are already sufficiently hot enough for equipartition to apply. Heat capacity of an ideal gas: Maxwell-Boltzmann distribution (11.7) The probability distribution function of c: the speed of the gas molecule If speed c corresponds to then corresponds to (11.8) Lecture 12: starts next page
Stat Mech cont. Lecture 12: Opens with another three dimensional picture (12.1) From the Boltzmann distribution, probability of an atom (only one for now) being in a particular state q is: (12.2) (12.3) Now (12.4) (12.5) (12.6) Total probability that speed is in range is:
Cancelling and subbing in c (12.7) Leading to the Maxwell-Boltzmann distribution: (12.8) NB: is the Boltzmann factor for a state; is proportional to number of states ; the prefactor could be deduced from normalisation of the probability distribution: [Two figures of f(c) vs c for various Temps] (12.9) Molecular gases: e.g. Vibration: single SHO with frequency (12.10)
(12.11) Typically therefore we can use a low temp approximation: (12.12) Rotation: [Picture of degrees of freedom for diatomic molecule] Eigenvalues at normal temperatures, therefore use high T approximation From classical equipartition theorem (12.13) So for a gas of N such molecules, the total heat capacity is: Where the 1 st term is from translation, 2 nd is from rotation and 3 rd is vibration (12.14) Entropy of ideal gas (monatomic) Get S from the free energy eqn using
So (12.15) There was a problem question which helps practice this part well. As but our non-degenerate approximation breaks down at low temperatures Lecture 13: Key Theory so far Bit of recap with some new stuff too, aids course completeness. If you can t remember what these are then go back and learn them: Principle of equal equilibrium (PEEP) at equilibrium Boltzmann distribution: With partition function Recall For an isolated system the entropy is Other it is (Gibb s formula)
At constant, Harmonic oscillator (Planck oscillator formula) Now lecture moves on to new topic based on everything so far Systems with variable numbers of particles Chemical potential [Figure similar to before ill copy later] Most likely division of energy will maximise: (13.1) Consider (13.2) (13.3) Define
(13.4) Thus both temperature and the chemical potential are equalised in equilibrium. has dimensions of energy. Since For a large system, then (13.5) Comes from [Now adds or amends original diagram to concern as well] So now (13.6) New version of eqn (13.7) (13.8)
(13.9) (13.10) (13.11) Gibbs distribution For a system able to exchange energy and particles with a reservoir [Figure of a system connected to a heat reservoir] For the reservoir (13.12) This gives / (13.13) From PEEP:
/ Since But thus / (13.14) Absorbing constants into proportionality (Gibbs distribution) Since : / / / (13.15) Where is the Grand Partition Function Lecture 14: Identical particles We ignore, for simplicity any interaction between particles (e.g. coulomb repulsion between electrons). Specify how many particles in each 1-particle quantum state. For Bosons (spin 0, 1, 2,...) (e.g.) photon, hydrogen atom) there is no restriction on the occupation. For Fermions (spin ½, 3/2,...) (e.g. electrons, protons, neutrons) the Pauli Exclusion Principle prohibits two or more fermions in the same 1-particle state. [Figure of particles in states for bosons] [Similar figure isolating all other particles (bar one and its state) as a reservoir (for fermions)] N.B.1-particle state means spatial wave function together with spin, e.g. 1s state with spin up Key concept: Treat the 1-particle state as our system with the other states playing the role of the particle reservoir (figure). The whole system is to be held at temperature T. Apply Gibbs distribution: (1) For fermions
N E 0 0 1 / (14.1) Gives Fermi-Dirac distribution: (14.2) Note Where is the average number of particles in a state with energy. (2) For bosons / / / (14.3) (14.4) (14.5) ( )
[I think that step is done using another geometric progression (the square term etc.)] (14.6) This is the Bose-Einstein distribution (14.7) [Two graphs describing each dist. For a range of temperatures] NB. For the purpose of graph only the T-dependence of is ignored Lecture 15: Bose Einstein Distribution (15.1) [Figure plotting with asymptote] Asymptote at. There fore must be less than the lowest energy of the 1-particle states. is the number of particles with energy The total number of particles, N, in the system is known. This means that all the energies,, must yield a total of N: summed over (15.2) This applies for both fermions and bosons depends on (albeit weakly).
The free-electron metal is given by solving the Schrödinger equation: [Cubic box of lengths L] (15.3) The density of these single-particle quantum states is: (15.4) Number of states: 1 st term is the positive octant of spherical shell in q-space, 2 nd term is number of q-points per unit volume of q-space) (15.5) / / / (15.6) (15.7)
BUT: for each solution to the S.E. eqn there are 2 modes (spin) (15.8) [Two graphs showing a step function and a rising graph] (15.9) This fixes at any temperature. At, f is a step function: / / (15.10) (15.11) This value of at is called the Fermi Energy, [Figure displaying Fermi energy on a Fermi-Dirac graph] Lecture 16: Continues really Roughly half of the electrons within below the Fermi energy,, have been raised in energy by ~. The number of such electrons is: (16.1)
In fact careful integration gives 1 st term is the electrons, 2 nd is vibrational [Graph showing these two terms plotted] Bose-Einstein condensation [Figure of moving for BE distribution; for details, see Mandl.] Below what critical temperature quantum state? are a significant fraction of the bosons in the very lowest For particles in a box: / / (16.2) Where the factor of 2 {sure a half?} undoes spin degeneracy For all states but the lowest so the total of particles in all the states but the lowest is: / Approximate done as the step function has width of (16.3) For condensation, this is, the total of particles (16.4)
(16.5) (16.6) Careful integration (some method, Canto?) gives: [Figure with two graphs showing proportion of occupied states with temperature, with being the fraction of bosons in very lowest state] Blackbody radiation [Cube again with lengths L] Electric field At boundary so (16.7) Mode has Each mode is a simple harmonic oscillator we can apply Planck oscillator formula: Lecture 17: Energy in EM modes Useful in Electrons in solids to know these parts well actually S.M. ties into most modules
Using all this (17.1) Integral of 1 st term gives infinity, but this is the zero point energy and is independent of T with no observable consequences. (17.2) (17.3) Spectral density Exclude (17.4) Since (17.5)
Low : High Lecture 18: Blackbody Radiation Continued Energy density: (18.1) Spectral density: (18.2) Absorption and emission [Figure of temperature being absorbed and emitted in a square, also a figure of a semi circle with various parts labelled in polar co-ordinates] Blackbody: Zero reflection, absorbs all incident radiation (and subsequently re-emits) Consider EM radiation arriving at angle range between and to the normal. The fraction of the total modes considered is: (18.3) [Another figure with some light hitting a surface and reflecting possibly, the angles and lengths are labelled. In retrospect the previous diagram was also describing light hitting a surface, see audio lecture for description until diagrams drawn here] Energy arriving on area A (figure) per unit time is:
(18.4) Total for all modes (therefore all angles) / (18.5) The energy flux is then And is called the Stefan-Boltzmann constant This flux is also the energy flux EMITTED by a blackbody at temperature T. Classical statistical mechanics Phase Space: For N particles in D dimensions there are DN position variables and DN momentum coordinates, giving 2DN dimensional phase space. For a 1D particle in an infinite well: [Figure with lengths 0 to L and then P vs x plot related to it] QM shows that the volume between successive quantised states is. Boltzmann and Gibbs and co. assumed one can integrate over phase space to achieve what in QM is given by a sum over microstates. This was later justified by the above result. Fin