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The Boltzmann Distribution

In defining the concept of temperature, we have examined the behaviour of systems in thermal contact (i.e. able to exchange energy back and forth) when the total energy  U  is fixed. In the real world, however, it is not often that we know the total energy of an arbitrary system; there is no ``energometer'' that we can stick into a system and read off its energy! What we often do know about a system it its temperature. To find this out, all we have to do is stick a calibrated thermometer into the system and wait until equilibrium is established between the thermometer and the system. Then we read its temperature off the thermometer. So what can we say about a small system<sup>15.23</sup>  ${\cal S}$  (like a single molecule) in thermal equilibrium with a large system (which we usually call a ``heat reservoir''  ${\cal R}$)  at temperature   $\tau = k_{\rm B} T$?

Well, the small system can be in any one of a large number of fully-specified states. It is convenient to be invent an abstract label for a given fully-specified state so that we can talk about its properties and probability. Let's call such a state   $\ket{\alpha}$  where  $\alpha$  is a ``full label'' - i.e.  $\alpha$  includes all the information there is about the state of  ${\cal S}$.  It is like a complete list of which car is parked in which space, or exactly which coins came up heads or tails in which order, or whatever. For something simple like a single particle's spin,  $\alpha$  may only specify whether the spin is up or down. Now consider some particular fully-specified state   $\ket{\alpha}$  whose energy is   $\varepsilon_\alpha$. As long as  ${\cal R}$  is very big and  ${\cal S}$ is very small,  ${\cal S}$  can - and sometimes will - absorb from  ${\cal R}$  the energy   $\varepsilon_\alpha$  required to be in the state   $\ket{\alpha}$,  no matter how large   $\varepsilon_\alpha$  may be. However, you might expect that states with really big   $\varepsilon_\alpha$  would be excited somewhat less often than states with small   $\varepsilon_\alpha$,  because the extra energy has to come from  ${\cal R}$,  and every time we take energy out of  ${\cal R}$  we decrease its entropy and make the resultant configuration that much less probable. You would be right. Can we be quantitative about this?

Well, the combined system   $\{ {\cal S} + {\cal R} \}$ has a multiplicity function  $\Omega$  which is the product of the multiplicity function   $\Omega_{\cal S} = 1$  for  ${\cal S}$ [which equals 1 because we have already postulated that  ${\cal S}$  is in a specific fully specified state   $\ket{\alpha}$] and the multiplicity function   $\Omega_{\cal R} = e^{\sigma_{_{\cal R}}}$  for  ${\cal R}$:

$$\Omega \; = \; \Omega_{\cal S} \times \Omega_{\cal R} \; = \; 1 \times e^{\sigma_{_{\cal R}}}$$

Moreover, the probability   ${\cal P}_\alpha$  of finding  ${\cal S}$  in state   $\ket{\alpha}$  with energy   $\varepsilon_\alpha$  will be proportional to this net multiplicity:

$${\cal P}_\alpha \; \propto \; e^{\sigma_{_{\cal R}}}$$

We must now take into account the effect on this probability of removing the energy   $\varepsilon_\alpha$  from  ${\cal R}$  to excite the state   $\ket{\alpha}$.

The energy of the reservoir  ${\cal R}$  before we brought  ${\cal S}$  into contact with it was  U.  We don't need to know the value of  U,  only that it was a fixed starting point. The entropy of  ${\cal R}$  was then   $\sigma_{_{\cal R}}(U)$.  Once contact is made and an energy   $\varepsilon_\alpha$  has been ``drained off'' into  ${\cal S}$,  the energy of  ${\cal R}$  is   $(U - \varepsilon_\alpha)$  and its entropy is   $\sigma_{_{\cal R}}(U - \varepsilon_\alpha)$.

Because   $\varepsilon_\alpha$  is so tiny compared to  U,  we can treat it as a ``differential'' of  U  (like ``dU'')  and estimate the resultant change in   $\sigma_{_{\cal R}}$  [relative to its old value   $\sigma_{_{\cal R}}(U)$]  in terms of the derivative of   $\sigma_{_{\cal R}}$  with respect to energy:

$$\sigma_{_{\cal R}}(U +dU) \; = \; \sigma_{_{\cal R}}(U) \; . . . . . . partial \sigma_{_{\cal R}}\over \partial U \right) \cdot dU$$

or in this case (with   $dU \equiv - \varepsilon_\alpha$)

$$\sigma_{_{\cal R}}(U - \varepsilon_\alpha) \; = \; \sigma_{ . . . . . . _{\cal R}}\over \partial U \right) \cdot \varepsilon_\alpha$$

But this derivative is by definition the inverse temperature of  ${\cal R}$:   ${\displaystyle {\partial \sigma_{_{\cal R}}\over \partial U} \equiv {1 \over \tau} }$.  Thus

$$\sigma_{_{\cal R}}(U - \varepsilon_\alpha) \; = \; \sigma_{_{\cal R}}(U) \; - \, {\varepsilon_\alpha \over \tau}$$

and thus the probability of finding  ${\cal S}$  in the state   $\ket{\alpha}$  obeys

$${\cal P}_\alpha \; \propto \; e^{\sigma_{_{\cal R}}(U - \va . . . . . . \sigma_{_{\cal R}}(U) - {\varepsilon_\alpha \over \tau}\right]$$

$$\hbox{\rm or} \quad {\cal P}_\alpha \; \propto \; e^{\sig . . . . . . } \cdot \exp \left( -{\varepsilon_\alpha \over \tau} \right)$$

Since   $e^{\sigma_{_{\cal R}}(U)}$  is a constant independent of either   $\varepsilon_\alpha$  or  $\tau$, that term will be the same for any state   $\ket{\alpha}$  so we may ignore it and write simply

$${\cal P}_\alpha \; \propto \; \exp \left( - {\varepsilon_\alpha \over \tau} \right)$$ (15.16)

This is the famous BOLTZMANN FACTOR that describes exactly how to calculate the relative probabilities of different states   $\ket{\alpha}$  of a system in thermal contact with a heat reservoir at temperature  $\tau$. It is probably the single most useful rule of thumb in all of thermal physics.