Physics Friday 67

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:
S=k\ln{N}, 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 E1 and E2, these energies may change, but their sum, the total energy E=E1+E2, 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 N1 and N2 microstates, respectively, then the combination system has N=N1N2 microstates, so the total entropy is S=k\ln{N}=k(\ln{N_1}+\ln{N_2})=S_1+S_2; this should be maximized. Now, taking these as a function of energy, and noting that as E=E1+E2 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 \frac{\partial{S}}{\partial{E_1}}=0. This, in turn, tells us that
or thus
Thus, we expect that \frac{\partial{S}}{\partial{E}} is equal to some property which must be the same for two systems in equilibrium. Considering the older thermodynamic definition of entropy given by dS=\frac{dQ_{rev}}{T}, we then see that \frac{\partial{S}}{\partial{E}}=\frac{1}{T}, 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, T2; namely, that E1E=E1+E2, and S2(E2) is a smooth function.

Now, the entropy of our heat bath is
where N2 is the number of microstates. From our above assumptions, we can approximate S2(EE1) by the Taylor series to first order about E:
S_2(E-E_1)\approx{S_2}(E)-\frac{\partial{S_2}(E)}{\partial{E_1}}E_1. However, we note from our previous work that the derivative in the linear term is simply \frac{\partial{S_2}(E)}{\partial{E_1}}=\frac{1}{T}, and so we get
k\ln{N_2}\approx{S_2}(E)-\frac{E_1}{T}. Solving this for the number of microstates, we get:
N_2\approx e^{\frac{S_2(E)}{k}}e^{-\frac{E_1}{kT}}.
Thus, the probability of any microstate of system 1 with energy E1 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 e^{-\frac{E_1}{kT}}; this is the Boltzmann factor, which is key to deriving a number of statistics used in both classical and quantum statistical mechanics.


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6 Responses to “Physics Friday 67”

  1. Physics Friday 71: The Isothermal Atmosphere « Twisted One 151’s Weblog Says:

    […] molecule at a height z. Thus, our exponential proportionality we just found is, in fact, simply the Boltzmann factor. (I vaguely recall reading somewhere, though I don’t recall where, that reasoning similar to […]

  2. Physics Friday 73 « Twisted One 151’s Weblog Says:

    […] (to define a particular path through the phase space of the system): Now, recalling that we can define temperature via , we can define heat capacity as . For gases, the two conditions we usually consider are […]

  3. Physics Friday 82 « Twisted One 151’s Weblog Says:

    […] Friday 82 By twistedone151 One might recall my previous post where I considered a system in contact with a thermal reservoir with which it could exchange […]

  4. Ron Pearson Says:

    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.

    • twistedone151 Says:

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

      A particular set of values of energy, number of particles and volume of an isolated thermodynamic system is said to specify a particular macrostate of it. In this description, microstates appear as different possible ways the system can achieve a particular macrostate.

  5. Ron Pearson Says:

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