The entropy

The discussion in the previous section brings us naturally to the introduction of the Entropy of the system, which we define as:

$$S(E,V,N) = k_{\rm B}\ln \Omega(E,V,N).$$ (3.10)

This is an extensive function, i.e. proportional to the size of the system, and a function of state, because $\Omega$ is a function of state. Since $S$ is a monotonic function of the statistical weight $\Omega$, the principle of being maximum w.r.t. any internal constraint extends to the entropy too. This makes the connection with the second law of thermodynamics.

Figure: Volume of gas partitioned in two sub-volumes.

Let us see now what are the implications of this principle. Imagine to divide our system by introducing a partitioning wall in a way that does not change the total energy of the system, so that we have ^3.4 :

The last equality arises because of the above definition of the entropy, which is an extensive function under the conditions that surface effects can be neglected. This partitioning acts as a constraint and so it may decrease the entropy of the system $S$ by reducing the number of available microstates. It is, in fact, a triple constraint, as it imposes partitions on $E$, $V$ and $N$ and each of these constraints decreases the entropy, if it causes a distribution of those variables that are different from what they would be in absence of the constraint.

Now we ask: what would happen if we lifted one of the constraints, for example by allowing transfer of energy across the partitioning wall? Since the system is isolated from the outside and since we are not allowing the partitioning wall to move, no work is being performed, then from the first law we have that any energy transfer between the two subsystems is due to heat flow. If the constraint causes the energies to be only infinitesimally different from what they would be without it, say by an amount $dE_1$ in system 1 and $dE_2 = -dE_1$ in system 2 (because of a), then upon its removal the change of entropy would also be infinitesimal, but positive (because removal of a constraint increases the number of microstates available to the system and so it increases its entropy):

$$dS = dS_1 + dS_2 = \left ( \frac{\partial S_1}{\partial E_1}\righ... ...1} dE_1 + \left ( \frac{\partial S_2}{\partial E_2}\right )_{V_2,N_2} dE_2 > 0,$$ (3.12)

where the partial derivatives are taken in the system with the constraint on. Let us assume that we are in a system for which the number of microstates increases as the energy increases, ^3.5 then the partial derivatives in Eq.  are positive. This means that if heat flows from system 1 to system 2, we have $dE_1 < 0$ and so $dS_1 < 0$, i.e the entropy decreases in a system that loses heat at constant volume and constant number of particles. The viceversa is also true, i.e. we have $dE_2 > 0$ and so $dS_2 > 0$. Overall the entropy must increase and so the increase in system 2 must overcompensate the decrease in system 1, which means that we must have $\left ( \frac{\partial S_2}{\partial E_2}\right )_{V_2,N_2} > \left ( \frac{\partial S_1}{\partial E_1}\right )_{V_1,N_1}$.