Part 5: The Bose-Einstein Distribution

1 PHYS393 Statistical PhysicsPart 5: The Bose-Einstein DistributionDistinguishable and indistinguishable particlesIn the previous parts of this course, we derived the Boltzmanndistribution, which described how the number ofdistinguishableparticles in different energy states varied with the energy ofthose states, at different temperatures:nj=NZe jkT.(1)However, in systems consisting of collections of identicalfermions or identical bosons, the wave function of the systemhas to be either antisymmetric (for fermions) or symmetric (forbosons) under interchange of any two particles. With theallowed wave functions, it is no longer possible to identify aparticular particle with a particular energy state. Instead, allthe particles are shared between the occupied states. Theparticles are said to Physics1 Part 5: The Bose-Einstein DistributionIndistinguishable fermionsIn the case of indistinguishable fermions, the wave function forthe overall system must be antisymmetric under theinterchange of any two particles.

2 One consequence of this isthe Pauli exclusion principle: any given state can be occupiedby at most one particle (though we can t say which particle!)For example, for a system of two fermions, a possible wavefunction might be: (x1,x2) =1 2[ A(x1) B(x2) A(x2) B(x1)].(2)Here,x1andx2are the coordinates of the two particles, andAandBare the two occupied states. If we try to put the twoparticles into the same state, then the wave function Physics2 Part 5: The Bose-Einstein DistributionIndistinguishable fermionsFinding the Distribution with the maximum number ofmicrostates for a system of identical fermions leads to theFermi-Dirac Distribution :ni=gie i kT+ 1.(3)Here,niis the number of particles with energy i;giis thedegeneracy of the energy level ( the number of distinctstates with energy i);kis boltzmann s constant; andTis thethermodynamic temperature. The function (T) is known asthe chemical potential, and is determined from the condition: ini=N,(4)whereNis the total number of fermions in the Physics3 Part 5: The Bose-Einstein DistributionIndistinguishable fermionsSinceex 0 (for real values of the variablex), we find that forall values of iandT:nigi=1e i kT+ 1 1,(5)This tells us that the number of fermions with energy imustbe less than or equal to the number of statesgiwith thatenergy.

3 This is another statement of the Pauli Physics4 Part 5: The Bose-Einstein DistributionIndistinguishable bosonsThe wave function for a system of identical bosons must besymmetric with respect to the interchange of any two of thebosons. For example, with two bosons, a possible wavefunction might be: (x1,x2) =1 2[ A(x1) B(x2) + A(x2) B(x1)].(6)This time, if we put two particles into the same state, the wavefunction does not vanish: there is no limit on the number ofparticles we can put into any given Pauli exclusion principle does not apply to bosons: andthis implies that the Fermi-Dirac Distribution does not applyeither, since the Pauli exclusion principle follows from theFermi-Dirac Physics5 Part 5: The Bose-Einstein DistributionThe Bose-Einstein distributionIn this part of the lecture course, we shall derive theBose-Einstein Distribution for a system of indistinguishablebosons, and apply the Distribution to three cases: the Bose-Einstein gas; a collection of photons.

4 The vibrations of particles in a find a Distribution function to describe systems of bosons,we shall follow the same procedure that we did fordistinguishable particles, and for indistinguishable is, we write down the number of microstates for thepossible distributions, subject to constraints determined by themacrostate specifications. Then we look for the distributionwith the largest number of microstates: applying the principleof equal a priori probabilities, this Distribution is more likely tooccur than any Physics6 Part 5: The Bose-Einstein DistributionThe Bose-Einstein distributionConsider an energy level iwith degeneracygi, containingnibosons. The states may be represented bygi 1 lines, and thebosons bynicircles; distinguishable microstates correspond todifferent orderings of the lines and circles. For example, with 9particles in 8 states corresponding to a particular energy, aparticular microstate might be:The number ofdistinctorderings of lines and circles is:ti=(ni+gi 1)!

5 Ni!(gi 1)!.(7)Statistical Physics7 Part 5: The Bose-Einstein DistributionThe Bose-Einstein distributionA particular Distribution has a specified number of particlesniwithin each of the possible energy levels i. The total numberof microstates for a given Distribution is therefore:t({ni})= i(ni+gi 1)!ni!(gi 1)!.(8)Let us assume that each state has a high degeneracy, 1. Then we can make the approximation:t({ni}) i(ni+gi)!ni!gi!.(9)Statistical Physics8 Part 5: The Bose-Einstein DistributionThe Bose-Einstein distributionTo find the most probable Distribution , we follow the sameprocedure as for the boltzmann and the Fermi-Diracdistributions. That is, we maximise (9):t({ni})= i(ni+gi)!ni!gi!,subject to the constraint on the total number of particles: ini=N,(10)and the constraint on the total energy: i ini=U.(11)As usual, rather than maximisetdirectly, we maximise lnt.

6 Ifwe assume that bothgiandniare large enough for Stirling sapproximation to hold for lngi! and lnni!, we find that lntisgiven by:lnt i[(ni+gi)ln(ni+gi) gilngi nilnni].(12)Statistical Physics9 Part 5: The Bose-Einstein DistributionThe Bose-Einstein distributionThe change in lntresulting from changesdniin each of thepopulationsniis then:dlnt i[ln(ni+gi)dni lnnidni].(13)From the constraints (10) and (11), we find: idni= 0, i idni= 0.(14)Combining (13) and (14) with Lagrange multipliers and ,we have:dlnt i[ln(ni+gini)+ + i]dni.(15)For appropriate values of and , equation (15) is true for alldni, hence:ln(ni+gini)+ + i= 0.(16)Statistical Physics10 Part 5: The Bose-Einstein DistributionThe Bose-Einstein distributionWe then find that the most probable Distribution can bewritten:ni=gie i 1.(17)Equation (17) is the Bose-Einstein disbribution. It gives thepopulation of an energy level that has energy iand degeneracygi.

7 The constants and are determined from the constraints(10) and (11) on the total number of particles and the totalenergy. can, as usual, be related to the thermodynamictemperature, so that the Bose-Einstein Distribution takes theform:ni=giBe ikT 1.(18)Statistical Physics11 Part 5: The Bose-Einstein DistributionThree distributionsWe have now completed our collection of three , a collection of distinguishable particles follows theBoltzmann Distribution :nj=NZe jkT.(19)Note thatjindexes particlestates, not energy , for a collection of indistinguishable fermions, we foundthe Fermi-Dirac Distribution :ni=gie i kT+ 1.(20)In this case,iindexes energylevels, not individual , we found the Bose-Einstein Distribution for a collectionof indistinguishable bosons (18):ni=giBe ikT Physics12 Part 5: The Bose-Einstein DistributionFermions and bosonsThe formulas for the Fermi-Dirac Distribution (20):ni=gie i kT+ 1and the Bose-Einstein Distribution (18):ni=giBe ikT very similar.

8 However, the difference in sign in thedenominator is critical. Sinceex>0 for allx, for theFermi-Dirac Distribution , we always haveni< gi; in otherwords, fermions always obey the Pauli exclusion for bosons, it is possible to haveni gi, if i kT, andB 1. In other words, it is possible for many identical bosonsto exist in the same Physics13 Part 5: The Bose-Einstein DistributionBose-Einstein Distribution : low-density limitWhen we derived the Fermi-Dirac Distribution , we showed thatfor systems with low density, the Distribution could be closelyapproximated by the boltzmann Distribution . Using the samereasoning, we can show that this is also the case for theBose-Einstein write the Bose-Einstein Distribution (18) in the form:nigi=1Be ikT 1.(21)Ifni gi, thenni/gi 1, which implies that:Be ikT 1 1,(22)so we can approximate:Be ikT 1 Be ikT.(23)We then have:ni giBe ikT.

9 (24)Statistical Physics14 Part 5: The Bose-Einstein DistributionThe Bose-Einstein gasAs an example of the Bose-Einstein Distribution , let us considera boson gas. This consists of a large number of identicalbosons in a box with rigid walls and fixed volume. The bosonsare free to move within the box, but cannot move beyond itswalls. Bosons exchange energy in occasional collisions witheach other. For simplicity, we assume that the bosons havespin zero, so there is no degeneracy associated with differentspin us first consider the behaviour of the constantBthatappears in the expression for the Bose-Einstein quantity corresponds to the partition function in theBoltzmann Distribution , or the chemical potential in theFermi-Dirac determined by the constraint: ini=N,(25)whereNis the total number of particles. Let us find howBdepends on Physics15 Part 5: The Bose-Einstein DistributionThe Bose-Einstein gasFrom our previous analysis of the Maxwell- boltzmann gas andthe Fermi gas, the density of states for spin-0 particles movingfreely in a box of volumeVis:g( ) =V4 2(2m~2)32.

10 (26)Let us assume that we can replace a summation over discreteenergy levels by an integration over a continuum of energylevels. Then the constraint (25) becomes: ini 0n( )d =N.(27)Using the Bose-Einstein Distribution (18) and the density ofstates (26), this becomes:V4 2(2m~2)32 0 Be kT 1d =N.(28)Statistical Physics16 Part 5: The Bose-Einstein DistributionThe Bose-Einstein gasTo simplify things a little, we define a new variable,y, suchthat:y= kT.(29)In terms ofy, the equation forBbecomes:V4 2(2mkT~2)32 0 y dyBey 1=N.(30)We define the functionF(B):F(B) =2 0 y dyBey 1.(31)In terms ofF(B), the equation forBbecomes:V(mkT2 ~2)32F(B) =N.(32)Statistical Physics17 Part 5: The Bose-Einstein DistributionThe Bose-Einstein gasTo understand the functionF(B), let us draw a graph of it:Statistical Physics18 Part 5: The Bose-Einstein DistributionThe Bose-Einstein gasFor largeB, we see thatF(B) 1/B.