# Total multiplicity and entropy of Einstein solids

This is really a question to get a little more insight about a homework exercise.

1. In my text book, the sum of multiplicities for all possible macrostates is given to be:

$$\sum \Omega_\mathrm{total}=\binom{N_A+N_B+q_\mathrm{total}-1}{q_\mathrm{total}}$$

For two Einstein solids $A$ and $B$ of $N_A$ oscillators in solid $A$, $N_B$ oscillators in solid $B$ and $q_\mathrm{total}$ units of energy in the system. I tried deriving this myself (which the text book doesn't require) but no dice there. How can I derive this myself?

2. The problem here is that the followup question pretty much defines the long term entropy of this system to be:

$$S=k\ln\left(\sum\Omega_\mathrm{total}\right)$$

Now there's confusion. I simply have no idea what I'm doing. Why is the sum of all the multiplicities together the long term entropy? How should I reason that this is the correct answer? Some explanation would be very, very welcome.

• Are you aware that $S=k_B\ln \Omega$ is in fact the usual way to define entropy? What's throwing you off there? – Danu Aug 20 '14 at 12:13
• What's throwing me off is the difference between the entropy of the most likely macrostate and this. Why is it different? – user55789 Aug 20 '14 at 12:14

1. I will not derive this exact formula for you, but instead show you the derivation for a single solid (following Schroeder's 'Introduction to thermal physics', where I learned it). The generalization to two solids is then very simple (mostly a conceptual difference; the math hardly changes).

Consider an Einstein solid with $$N$$ oscillators and $$q$$ units of energy. The problem of calculating the multiplicity can be visualized as follows Given $$N-1$$ vertical lines - representing partitions between different oscillators - and $$q$$ dots - representing units of energy - how many ways are there of arranging these symbols? This problem is quite simple: There are a total of $$q+N-1$$ symbols, and one just has to pick the location of the $$q$$ dots (or, equivalently, the $$N-1$$ lines) to uniquely specify an arrangement. Thus, there must be $$\Omega=\binom{N+q-1}{q}$$ different arrangements corresponding to the same macrostate.

2. I will continue to consider a single solid, which is qualitatively the same thing. One considers a solid, about which is only known that it consists of $$N$$ oscillators, sharing $$q$$ units of energy. This specifies the macrostate. Given that the system is in some macrostate, the entropy is $$S=k_B\ln \Omega$$ where $$\Omega$$ is the number of microstates corresponding to that particular macrostate.

About long term and short term entropy:

Consider a system like the one you are asking about - where two subsystems share a fixed amount of energy. Then, one can say that on any practical timescales, the total system will be in the entropy- (and therefore multiplicity-)maximizing configuration. Denoting the corresponding multiplicity by $$\Omega_1$$, we can define the 'short term entropy' as $$S=k_B\ln\Omega_1$$. This is really the most practically useful concept of entropy, and therefore the prefix 'short term' is often dropped.

However, if we leave the system for extremely long times, one can say that, since all microstates are probed, we eventually lose any knowledge of the state of the system - it could be in any state. Therefore, one can define the 'long term entropy', which is defined as $$S_\text{long}=k_B\ln\left(\sum_{\text{macrostates}}\Omega\right)$$, i.e. one uses the total multiplicity, which sums over all possible states.

• In regards to (1), I believe what I was looking for is the Vandermonde identity. For (2): could you include the difference between the multiplicity over short and long time scales ? For the short time scale, the most likely probability has a corresponding combined multiplicity which is not equal to the long term total multiplicity. I don't see how that can be the case. – user55789 Aug 20 '14 at 13:10