The Boltzmann Factor

Consider a macroscopic physical system, made up of a very large number of microscopic parts; the domain of statistical mechanics. In particular, we note that most macroscopic states correspond to many possible microstates, and that a macrostate becomes more likely the more distinct microstates correspond to it. This insight is encapsulated in the Second Law of Thermodynamics, which can be stated as: An isolated physical system will tend toward the allowable macrostate with the largest number of possible microstates. This form is reflected in the Boltzmann definition of entropy:

, where *S* is the entropy of a macrostate, *N* is the number of microstates, and *k* is Boltzmann’s constant. In particular, this helps us to express entropy as a function of energy, *S*(*E*).

Now, suppose you have two systems, and place them into contact; if the systems have energies *E*_{1} and *E*_{2}, these energies may change, but their sum, the total energy *E*=*E*_{1}+*E*_{2}, is constant. While the two individual systems are not isolated systems, the larger system of their combination is, and so the second law can be applied. If system 1 and system 2 are in macrostates with *N*_{1} and *N*_{2} microstates, respectively, then the combination system has *N*=*N*_{1}*N*_{2} microstates, so the total entropy is ; this should be maximized. Now, taking these as a function of energy, and noting that as *E*=*E*_{1}+*E*_{2} is conserved, we have

,

so that we have the entropy of the total system as a function of the energy of the first sub-system. Now, the maximization of entropy tells us that . This, in turn, tells us that

or thus

.

Thus, we expect that is equal to some property which must be the same for two systems in equilibrium. Considering the older thermodynamic definition of entropy given by , we then see that , the reciprocal of temperature, and the above confirms that systems in thermodynamic equilibrium have the same temperature (see here for a previous post using this relation). In fact, this can be used as a definition of temperature (see here).

So now, let us consider a system (system 1), in contact with a heat bath (system 2): a system large enough that any heat exchanges will not significantly change its temperature, *T*_{2}; namely, that *E*_{1}≪*E*=*E*_{1}+*E*_{2}, and *S*_{2}(*E*_{2}) is a smooth function.

Now, the entropy of our heat bath is

,

where *N*_{2} is the number of microstates. From our above assumptions, we can approximate *S*_{2}(*E*–*E*_{1}) by the Taylor series to first order about *E*:

. However, we note from our previous work that the derivative in the linear term is simply , and so we get

. Solving this for the number of microstates, we get:

.

Thus, the probability of any microstate of system 1 with energy *E*_{1} will be proportional to the above, and thus, as the first exponential is a constant for all possible states of our system, the probability is proportional to ; this is the Boltzmann factor, which is key to deriving a number of statistics used in both classical and quantum statistical mechanics.

Tags: Boltzmann Constant, Boltzmann Factor, Energy, Entropy, Friday Physics, physics, Temperature, Thermodynamics

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May 11, 2013 at 2:49 pm |

I have also seen the boltzmann distribution in terms of probability, would a “microstate”, N, be synonomus with probability, P?

Its really confusing when different texts and resources use different variables.

May 21, 2013 at 1:24 pm |

Not exactly. For more on microstate versus macrostate, see here: Microstate (statistical mechanics). In particular:

May 11, 2013 at 3:29 pm |