Physics 301 16-Oct-2002 15-1Internal Degrees of FreedomThere are several corrections we might make to our treatment of the ideal gas. Ifwe go to high occupancies, our treatment using the Maxwell-Boltzmann distribution isinappropriate and we should start from the Fermi-Dirac or Bose-Einstein distributiondirectly.We have ignored the interactions between molecules. This is a good approximationfor low density gases, but not so good for higher densities (but these higher densities canstill be low enough that the MB distribution applies). We will discuss an approximatetreatment of interactions in a few weeks when we discuss phase transitions.Finally, we have ignored any internal structure of the molecules. We will remedy thisomission now. We imagine that each molecule contains several internal states with energiesint. Note that int is understood to be an index over the internal states. There may bestates with the same energy and states with differing energies. In our non-interactingmodel, the external energy is just the kinetic energy due to the translation motion of thecenter of mass, cm. Again, cm is to be understood as an index which ranges over all statesof motion of the cm. Although we are considering internal energies, we are not consideringionization or dissociation. When a molecule changes its internal state, we assume thenumber of particles does not change.Let’s consider the grand partition function for a single state of center of mass motion.That is, we’re going to consider the grand partition function for single particle states—withinternal degrees of freedom—in a box. The energy of the particle is cm+ int. Then thegrand partition function isZ =1+e(µ − cm− int,1)/τ+ e(µ − cm− int,2)/τ+ ···+ two particle terms + three particle terms + ··· ,=1+e(µ − cm)/τinte−int/τ+ two particle terms + three particle terms + ··· ,=1+e(µ − cm)/τZint+ two particle terms + three particle terms + ··· ,=1+e(µ − cm)/τZint+ e2(µ − cm)/τZ2int2!+ e3(µ − cm)/τZ3int3!+ ··· ,where Zintis the partition function for the internal states. The above expression is strictlycorrect only for bosons. For fermions, we would need to be sure that the multiple particleterms have all particles in different states which means that the internal partition functionsdo not factor as shown above.However, we really don’t need to worry about this because we’re going to go to theclassical limit where the occupancy is very small. This means we can truncate the sumCopyrightc 2002, Princeton University Physics Department, Edward J. Grot hPhysics 301 16-Oct-2002 15-2above after the second term,Z =1+e(µ − cm)/τZint.The mean occupancy of the center of mass state, whatever the internal state, isf(cm)=e(µ − cm)/τZint1+e(µ − cm)/τZint≈ e(µ − cm)/τZint,which is just the Maxwell-Boltzmann distribution with an extra factor of the internalpartition function, Zint.Now we should modify our previous expressions to allow for this extra factor of Zint.Recall that we chose the chemical potential to get the correct number of particles. Inthat calculation, exp(µ/τ) must be replaced by exp(µ/τ)Zint, and everything else will gothrough as before. Then our new expression for µ isµ = τ lognnQZint= τlognnQ− log Zint .The free energy becomesF = Fcm+ Fint= NτlognnQZint− 1 ,where Fcmis our previous expression for the free energy due to the center of mass motionof molecules with no internal degrees of freedom, andFint= −Nτ log Zint,is the free energy of the internal states alone. The expression for the pressure is unchangedsince in the normal situation, the partition function of the internal states does not dependon the volume. (Is this really true? How do we get liquids and solids? Under whatconditions might it be a good approximation?) The expression for the entropy becomesσ = σcm+ σint,where σcmis our previous expression for the entropy of an ideal gas, the Sackur-Tetrodeexpression, andσint= −∂Fint∂τ V,N=∂(Nτ log Zint)∂τ V,N= N log Zint+ Nτ∂(log Zint)∂τ V,N.The energy, U, and therefore the heat capacities, receive a contribution from the internalstates. The extra energy isUint= Fint+ τσint= Fint− τ∂Fint∂τ V,N= −τ2∂∂τFintτ .Copyrightc 2002, Princeton University Physics Department, Edward J. Grot hPhysics 301 16-Oct-2002 15-3To make further progress, we need to consider some specific examples of internalstructure that can give rise to Zint. Suppose the molecules are single atoms but these atomshave a spin quantum number S. Then there are 2S + 1 internal states that correspond tothe 2S + 1 projections of the spin along an arbitrary axis. In the absence of a magneticfield, all these states have the same energy which we take as int=0. ThenZint=2S +1andFint= −Nτ log(2S +1),σint= N log(2S +1),Uint=0,so the entropy is increased over that of a simple ideal gas, but the energy doesn’t change.The increase in entropy is easy to understand. What’s happening is that each atom has2S + 1 times as many states available as a simple atom with no internal structure. Theentropy, the logarithm of the number of states, increases by log(2S + 1) per atom.That was a fairly trivial example. Here’s another one: Suppose that each moleculehas one internal state with energy 1.ThenZint=exp(− 1/τ)andFint= −Nτ log Zint=+N1,σint=0,Uint= N1,and∆µ = −τ log Zint=+1.In this example, we didn’t change the entropy (each molecule has just one state), but weadded 1to the energy of each molecule. This change in energy shows up in the chemicalpotential as a per molecule change and it shows up in the free energy and energy as Ntimes the per molecule change. This example is basically a small test of the self-consistencyof the formalism!More realistic examples include the rotational and vibrational states of the molecules.Single atoms have neither rotational nor vibrational modes (they do have electronic exci-tations!). A linear molecule (any diatomic molecule and some symmetric molecules suchas CO2, but not H2O) has two rotational degrees of freedom. Non-linear molecules havethree rotational degrees of freedom. Diatomic molecules have one degree of vibrationalfreedom. More complicated molecules have more degrees of vibrational freedom. If themolecule has M atoms, 3M coordinates are required to specify the locations of all