# A simple question on finding the number of macro states for a system of two Einstein solids

Consider two identical Einstein solids each with $$N$$ oscillators in thermal contact with each other and suppose that $$q_{\mathrm{total}}=q_A+q_B=2N$$

How many different macrostates are there ( i.e. possible values for a total value of A)?

My attempt and reasoning:

I imagined that if we had two boxes A and B. I started by saying that micros-states corresponding first box is $$q_A=2N, q_B=0$$, the next is $$q_A=2N-1, q_B=1$$ and so on until.......$$q_A=2N-2N=0, qB=2N$$. Therefore the total no of micro-states is $$2N+1$$.

Question: Is this correct? and if it is not, is there any mathematically rigorous way to describe it?

• You seem to be assuming that each oscillator can have only one quantum. This is surely incorrect. Sep 5, 2018 at 21:34
• @PhilipWood I didn't think I implied that at all. Rather I am saying that the total number of energy units in A is $q_A$. ?Now as to whether or not that means that all the energy units exist in one of A's N oscillators or all N of them I haven'y specified at all. Sep 7, 2018 at 19:45

This is more a question of how you define a macrostate in your model. In the conventional Einstein solid model, we consider the energy observable. The reasoning for this is that we are using the model to study thermal energy (specifically heat capacity), and so all internal energy is taken to be heat, and changes in heat are directly observable.

From that conventional perspective, with two Einstein solids in thermal contact, and the total energy fixed, we cannot define separate macrostates for each of the subsystems. If this were an actual physical system, as it evolved the two subsystems would constantly exchange heat energy and thus there would be no distinguishable macrostates with the same total energy.

Consistent with that view, the correct way of looking at your model is to consider both Einstein solids together as a single solid. In which case for a fixed total energy, there is only one macrostate.

2N+1 is correct.

A mathematical answer is more easily found if we consider the more general case of many Einstein solids. The number of macro sates will be the number of solids, times the average of each solid, minus the half the degrees of freedom.

$$\text{#Einstein solids} \cdot \frac{1}{N}\sum_{i}^{N}(\text{directions of movement})n_i +\text{#Einstein solids}-\frac{f}{2}$$

For each oscillator we have 2 directions of movement as they can move in + or - whatever direction we are alligned in, we think of each particle as 3 oscillators so we do not need to worry about an oscillator moving in more than 2 directions. For a solid we have f=2. Using that we have two einstien solids we get: $$2 \cdot \frac{1}{N}\sum_{i}^{N}(2)n_i +2 -\frac{6}{2}$$ $$\frac{2}{N} 2\frac{N(N+1)}{2} +2 -3$$ $$2N+1$$