Using Gibbs entropy to sum entropy of subsystems. Why does this not equal the Gibbs entropy of the total system?

I have found the Gibbs entropy of a simple system and the summed Gibbs entropy of its subsystems. I believed that these values should be equal. They are not. I'd be grateful for any ideas regarding where I have went wrong.

Consider an isolated system, $$S$$, which can be split into two subsystems: $$L$$ and $$R$$.

There are ten microstates of system $$S$$. Subsystem $$L$$ can be in one of three microstates, $$A_1^L$$, $$B^L_1$$ and $$B^L_2$$. Subsystem $$R$$ can be in one of seven microstates, $$A_1^R$$, $$A_2^R$$, $$A_3^R$$, $$A_4^R$$, $$B^R_1$$, $$B^R_2$$ and $$B^R_3$$.

If subystem $$L$$ is in a microstate of type $$A$$ then so is $$R$$ (same for $$B$$ microstates). This is due to conservation of energy. Let $$E(A^L)$$ be the energy of the subsystem $$L$$ in an $$A$$ state. Likewise for other terms of this form. Let $$E(A^L)+E(A^R)=E(B^L)+E(B^R)=U$$, where $$U$$ is the total energy of $$S$$. Let $$E(A^L) \neq E(B^L)$$.

The ten possible microstates of the overall system $$S$$ are thus:

$$A_1^L$$ $$A_1^R$$,

$$A_1^L$$ $$A_2^R$$,

$$A_1^L$$ $$A_3^R$$,

$$A_1^L$$ $$A_4^R$$,

$$B_1^L$$ $$B_1^R$$,

$$B_1^L$$ $$B_2^R$$,

$$B_1^L$$ $$B_3^R$$,

$$B_2^L$$ $$B_1^R$$,

$$B_2^L$$ $$B_2^R$$,

$$B_2^L$$ $$B_3^R$$

Each microstate of $$S$$ is equally likely. Thus the probability $$S$$ is in any one of these ten microstates is $$\frac{1}{10}$$.

The Gibbs total entropy of system $$S$$, given by $$-\sum_i P_i \log P_i$$, is simply $$\log 10$$. (Let $$k_B=1$$)

Let us find the Gibbs entropy of subsystem $$L$$. The probability it is in state $$A_1^L$$ is $$\frac{4}{10}$$. This is because there are four overall system microstates where $$L$$ is in this state. Likewise $$L$$ is in state $$B_1^L$$ with probability $$\frac{3}{10}$$, equivalently for state $$B_2^L$$. Thus the entropy of subsystem $$L$$ is

$$-[\frac{4}{10}\log\frac{4}{10}+\frac{3}{10}\log\frac{3}{10}+\frac{3}{10}\log\frac{3}{10}]=-[\frac{4}{10}\log\frac{4}{10}+\frac{6}{10}\log\frac{3}{10}]$$

Likewise the entropy of subsystem $$R$$ is

$$-[\frac{1}{10}\log\frac{1}{10}+\frac{1}{10}\log\frac{1}{10}+\frac{1}{10}\log\frac{1}{10}+\frac{1}{10}\log\frac{1}{10} + \frac{2}{10}\log\frac{2}{10}+\frac{2}{10}\log\frac{2}{10}+\frac{2}{10}\log\frac{2}{10}]$$

$$=-[\frac{4}{10}\log\frac{1}{10}+\frac{6}{10}\log\frac{2}{10}]$$

The sum of subsystem $$R$$ and subsystem $$L$$ entropies is,

$$2\log10 -\frac{4}{10}\log4 -\frac{6}{10}\log 6$$

$$=2\log10 -\frac{1}{10}\log 256-\frac{1}{10}\log 46656$$

$$=2\log10 -\log5.1$$

$$\neq \log 10$$

So, the total entropy is not equal to the summed entropy of the system's subsystems. What has went wrong? Why can Gibbs entropy not be manipulated in this way?

• If $L$ has 1 $A$-type and 2 $B$-types, $R$ has 4 $A$-types and 1 $B$-type, you have 6 states, not 10. So $R$ had 3 $B$-type states. Suggested reading: en.wikipedia.org/wiki/Conditional_entropy – Sean E. Lake Nov 11 '19 at 15:23
• Sorry, that was a typo. R has 3 B-type states. Typo now corrected. – safcphysics Nov 11 '19 at 15:33

Long story short, the correct equation is: \begin{align} H(X,Y) & = H(X|Y) + H(Y) \end{align} because the definition of how joint probability $$P(X,Y)$$ and conditional probability $$P(X|Y)$$ is \begin{align} P(X,Y) & = P(X|Y) P(Y). \end{align} When $$X$$ and $$Y$$ are independent, $$P(X|Y) = P(X)$$.
For completeness, conditional entropy is defined as \begin{align} H(X|Y) & = - \sum_{X,Y} P(X,Y) \ln P(X|Y) \end{align} and, if $$P(X|Y) = P(X)$$ then you can show that $$H(X|Y) = H(X)$$.