Boltzmann Distribution of Thermal Physics Lecture Notes

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We call this larger system a reservoir.  A common problem that we will encounter in thermal physics will be to find the probability that a system S, which is in thermal contact with a reservoir, is in a specific quantum state s of energy es.  Let U be the total energy of the combined system (reservoir and S).  When we specify that Sshould be in the quantum state s, the problem reduces to the question of determining what is the number of accessible states in the reservoir at the appropriate energy e?  This happens because we know that the probability of the system being in a state s is related to the multiplicity of the total system.  But the multiplicity of the total system is just the multiplicity of the reservoir times the multiplicity of S.  However, since we have already specified the state of S, the multiplicity of the total system is just the multiplicity of the reservoir.  If the system Shas an energy es, then the reservoir energy is U0 ‑ es.  Thus, the multiplicity of the reservoir is g(U0-es).  According to the fundamental postulate, the probability that the system is in any of the quantum states at a specific energy e1is then

P(e1) = g(U0-e1)

Notice that this is different from the relationship we encountered before between the probability and the multiplicity factor.  Before, we were asking what is the probability of finding the state in a specific quantum state, given an energy es.  There the probability was

P(specific state) = 1/g(es)

Now, we are asking what the probability of finding the system in any quantum state with the energy es(and satisfying any other conditions that we place on it), out of all of the states available to it.  Here the probability is

P(es) = g(es)

Returning to the system in contact with the reservoir, we can ask what is the ratio of the probability that the system is in one of the quantum states with energy elto the probability that the system is in one of the quantum states with energy e2. Then we get

                                                                                                       (5.1)

We can restate this in terms of the entropy.  Recalling the definition of entropy, we see that the ratio becomes

or

                                                                                                                   (5.2)

where Ds= s(U0-e1) – s(U0-e2).  If we expand s(U0-e1) and s(U0-e2) as Taylor series around the entropy of the reservoir, s(U0), we get

but , so this becomes

                                                                                        (5.3)

If we let the reservoir become infinitely large, all of the higher order terms vanish.   Substituting this into Ds, we see that

                                                                                                              (5.4)

Thus, the probability becomes

                                                                                                               (5.5)

A term of the form exp(‑e/t) is called a Boltzmann factor.

Using Boltzmann factors, we can build another function which is of great use to thermal physics. This is the partition function, and it is defined to be

                                                                                                               (5.6)

It is the sum over the Boltzmann factors associated with all of the allowed states.

Notice that the partition function acts as the normalization constant for the

Boltzmann factor to be used as a measurement of probability

                                                                                                                  (5.7)

This result is one of the most useful ones in statistical physics. As a result of this, we can determine the most likely result of any experimental measurement in thermal physics.

Example:

Given a system in contact with a reservoir, what is the average energy of the system?

                                                                                                   (5.8)

As a specific example, consider a single particle with two energy states.

so

as .

Heat Capacity

            We define the heat capacity of a system at constant volume as

                                                                                                                  (5.9)

Since sis dimensionless in fundamental units, we see that CV is also dimensionless in these units.  The specific heat is defined as the heat capacity per unit mass.  For the system discussed above, the heat capacity is

If we graph both /eand CV as functions of t/e, we get

The bump in the plot of CV verses t/eis called the Schottky anomaly.

Reversible Processes

For the rest of the discussion, we want to use a reversible process.  This is a process is which the system deviates at most by an infinitesimal amount from its equilibrium state.  Consider a cubical system in a quantum state associated with an energy es.  Compress the system from a volume V to a volume V ‑ DV.  Let the change occur slowly enough that the system remains in the quantum state throughout the process. This is called an isotropic process. The energy of the state after the compression is

Let the pressure associated with the change in volume be applied normal to all of the faces of the cube. Then the mechanical work done in changing the volume from V to V‑DVis equal to

Recall that the work is defined to be the scalar product of the force applied with the distance traveled, and that classically we define the pressure to be the force per unit area. Let A be the area of one face of the cube. Then the change in volume can be written as

DV= A(Dx+ Dy+ Dz)

This allows us to finally write

Thus, ps, which is the pressure of state es, is

                                                                                                                  (5.10)

Defining the general pressure to be the average over all of the states of the ensemble and letting the change in volume go to zero, yields

                                                                                                                 (5.11)

where U is the energy of the system, .  Notice that we kept the entropy constant in the derivative.  This is because the process was defined such that the number of states for the system remained unchanged.  Another useful expression for the pressure can be derived from the fact that the entropy is held constant.  Recall that the entropy is usually a function of both the number of particles and the energy.  The number of particles can be related to the volume of the system since each particle can be assumed to occupy some average volume of space.  Thus, s= s(U,V). For an infinitesimal change in s For an isotropic process this change is zero, so

or

                                                                                                              (5.12)

Where we have used the definitions  and .  Substituting this result back into dsyields the useful thermodynamic identity,

                                                          dU = tds- p dV                                              (5.13)

This is a generalized conservation of energy equation, where tdsis the heat put into the system and p dV is the work done by the system.

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