State Counting in Thermodynamics

Let’s say you have a box full of particles, and each of those particles is independently in one of N states. Each state has a probability of occuring. Let us further suppose that none of these probabilities are the same – if so, then there is one state with the highest probability. If there are then properties E and V which are diagonal and have a 1-1 correspondence to each other amongst all the states, then the most common E and most common V definitely are related to each other by E=MV^2/2.

Examine the Speed and Energy distributions of the ideal gas. The modal values of these two distributions do not correspond to each other via the E = MV^2/2 relation. Paradox?


The modal value of P(v) is the value of v for which the integral from v to v+dv is maximal for any sufficiently small dv. Imagine sliding a small window of width dv over P(v) and finding the interval which contained the most probability density – this interval always includes the modal value of P(v), and that value can be found by taking dv to zero. An identical process applies to the modal energy.

The resolution of the paradox lies in the fact that dv and dE are both the wrong measures to use. The correct measure is number of quantum states, which classically corresponds to phase space. The correct procedure would be to find P in terms of phase space, then to find the values of E and v associated with the modal state. E definitely is the same as mv^2/2 in that case.


Explicitly prove this by showing how the phase space approach derives the correct values of E and V corresponding to using the quantum partition function with the classical limit of particle in a box.

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