Statistical Mechanics – Additivity of the Gibbs Entropy

by

0

Today we’ll prove an incidental result, with a view to interpreting \beta in the next post.

Consider two closed systems in loose thermal contact. Such a pair of systems can exchange energy with one another, but the microstate one occupies is in no way correlated with the microstate occupied by the other. That is to say, when an otherwise isolated system with state occupation probabilities \{p_\alpha\} comes into loose thermal contact with a second system, the set \{p_\alpha\} stays exactly the same.

Let’s consider the two systems together as a single, composite system with microstates \{ij\} and corresponding occupation probabilities \{p_{ij}\}. Each microstate of the joint system is uniquely determined by the microstate occupied by each of the two composite systems.

So, if system one can occupy states \{i\} with probability \{p_i^1\} and system two can occupy states \{j\} with probability \{p_j^2\}, the combined microstate ij means system one is in state i and system two is in state j with probability p_{ij}=p_i^1 p_j^2.

Now consider the Gibbs entropy of the combined system. By definition,

\displaystyle S_G=-\sum_{i,j} p_{ij}\ln p_{ij}

\displaystyle S_G=-\sum_{i,j} p_i^1 p_j^2\ln p_i^1 p_j^2

Using the fact that \ln xy=\ln x + \ln y, the sum can be split into two parts:

\displaystyle S_G=-\sum_j p_j^2\sum_i p_i^1 \ln p_i^1 - \sum_i p_i^1 \sum_j p_j^2 \ln p_j^2

The sum over all microstates of the sets \{p_i^1\} and \{p_j^2\} are each equal to one, so

\displaystyle S_G=-\sum_i p_i^1 \ln p_i^1-\sum_j p_j^2 \ln p_j^2

S_G=S_G^1+S_G^2

Hence the Gibbs entropy of the composite system is the sum of the Gibbs entropies of the individual systems it comprises. With this result, we can consider the behaviour of two closed systems brought into contact with one another and find the role of thermodynamic beta.

Return to top

Leave a comment