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
$S=S_1(E_1)+S_2(E_2)=S_1(E_1)+S_2(E-E_1)$,
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
$0=\frac{\partial{S}}{\partial{E_1}}=\frac{\partial{S_1}}{\partial{E_1}}(E_1)+\frac{\partial{S_2}}{\partial{E_1}}(E-E_1)$
or thus
$\frac{\partial{S_1}}{\partial{E_1}}=\frac{\partial{S_2}}{\partial{E_2}}$.
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
$S_2(E_2)=S_2(E-E_1)=k\ln{N_2}$,
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: