Unformatted text preview:

Quantum Statistical MechanicsInitial questions: What holds up Jupiter’s core? Does the composition matter? Howconductive are white dwarf and neutron star interiors?We’ve talked about the interactions of photons, and (briefly) how they differ fromthe interactions of electrons in a fundamental way related to their quantum statisticalproperties. It’s the difference between bosons and fermions, specifically whether particles“like” to occupy the same quantum state (bosons) or don’t (fermions). In this lecture we’llexplore that a bit more. To start, we need to define the concept of a chemical potential.Chemical potential:µi=Ã∂E∂Ni!S,V. (1)Here Niis the number density per gram of the ith species, so that Ni= ni/ρ.Thermodynamic and chemical equilibrium require that if there are reactions that mightchange the Ni, thenPiµidNi= 0. Photon number is not conserved, so µγ= 0. Notethat here we’re interested in reactions that take place fairly rapidly, so things like nuclearreactions (which usually take years to billions of years) aren’t included. Technically, thesereactions mean that the system is not in equilibrium, but this is another example of how wesimplify by dropping small terms. However, we should remember that it is possible to getto a lower-energy state via these reactions, and that at some point if the reactions are fastcompared to chemical reactions then they must be included.The dNiare related to each other by particular reactions, so that means we can alsowritePiµiνi= 0, where νiare the stoichiometric coefficients. From this, can see in anotherway that µγ= 0: if a reaction of the form A → B + γ is allowed, so is A → B + γ + γ. Ifµγwere nonzero, equilibrium would not be possible.Before we delve into quantum statistical mechanics, let’s recall classical statisticalmechanics. To do this, we’ll follow chapter 40 in Volume 1 of the Feynman Lectures onPhysics.Suppose that we have a gas, and we imagine two parallel planes in the gas separated bydistance dx. Let there be some force F on the atoms in the gas. Suppose that the gas has anumber density of n atoms per volume. We will consider an ideal gas, which has a pressureof P = nkT at temperature T . In order for the system to be in equilibrium, the force dueto the pressure must be balanced by the effect of the force F . However, Ask class: willthere be any net force if the pressure is constant? No, it is a change in pressure (gradientof pressure) that can exert a net force. Therefore, the force F must balance the gradient ofthe pressure P .Let’s assume that the system has a constant temperature T , as it will in thermalequilibrium. Pressure (and change in pressure dP ) has units of force per area. Therefore,we haveF n dx = dP = kT dn . (2)Note that −F dx is the work involved in moving a molecule from x to x + dx. Now supposethat the force is derived from a potential V . Ask class: how is the force related to thepotential? We know that F = −∇V = −dV/dx in our case. Therefore, we havekT dn = −n dVdn/n = −dV /kTn = const × exp(−V/kT ) .(3)Therefore, if the force derives from a potential, the number of molecules in a particularlocation depends on the negative exponential of the potential energy, divided by kT . This iscalled Boltzmann’s law. It is remarkably general. Any force that is derived from a potentialleads to Boltzmann’s law, in classical physics.But what if the force isn’t derived from a potential? In that case, thermal equilibriumisn’t possible at all! Sound bizarre? Consider the following. Suppose you have a bunch ofballs in a container, and the balls have friction with each other. To be truly dissipative,the energy released in friction must leave the container entirely. Then you can see that nothermal equilibrium is possible, because energy continues to be lost from the system. Inthe general case, if a force isn’t derived from a potential then there are closed loops in thesystem that lose or gain net energy, so there is no equilibrium.We now have the distribution in space, but what about in velocity? It turns outthat there is a remarkable (and not accidental) parallel with the distribution in space.Following Feynman again, let’s consider a column of gas in thermal equilibrium in auniform gravitational field. Let’s say that there are no collisions between molecules. Letus consider arbitrarily some height h = 0 (a reference height; there could be molecules atnegative h). How many molecules make it from h = 0 to h = h0> 0? It’s not all of them,since some molecules don’t have the required energy and will fall down before getting toh = h0. In fact, it is exactly the molecules with enough kinetic energy that populate thehigher level. We saw from before that the relative number of molecules at each heightis n(h = h0)/n(h = 0) = exp(−mgh0/kT ). Therefore, the relative fraction of moleculeswith enough energy to make it from h = 0 to h = h0must also be exp(−mgh0/kT ). Byconservation of energy, the kinetic energy of those molecules must be at least12mv20, wherev0is the vertical speed needed to just make it to h0from 0. Therefore,n(v > v0)n(v > 0)= exp(−kinetic energy/kT ) . (4)Adding collisions doesn’t change this; we could, for example, imagine moving only a tinyheight dh such that no collisions occurred in the meantime.Combined, these two results imply something remarkable. The number of moleculesin a particular state (which now includes a position, velocity, and any internal states) inthermal equilibrium is proportional to the exponential of the negative of the total energydivided by kT . Therefore, schematically, in classical physicsN(state) ∝ exp(−Etot/kT ) . (5)But what about in quantum mechanics? A lot of the interesting thermodynamicquantities can be calculated from the distribution function in 6-dimensional phase space(three coordinate, three momentum). This is also represented by the occupation numberdivided by h3, where h3is a unit of phase-space volume. For a given species, the distributionfunction for a particular momentum isn(p) =1h3Xstates1eEtot(state)/kT± 1. (6)Here Etot= −µ + E(state) + E(p), where E(state) is the energy of the state of that speciesrelative to some level (often the ground state energy) and E(p) is the kinetic energy formomentum p. Note that the energy level relative to which one determines Estateis a freeparameter, but −µ + E(state) isn’t, which can lead to varying definitions for µ (beware!).For


View Full Document

UMD ASTR 601 - Quantum Statistical Mechanics

Download Quantum Statistical Mechanics
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Quantum Statistical Mechanics and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Quantum Statistical Mechanics 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?