Entropy and Thermodynamic probability
Seems like they're related....
Peter Miller
We know that an initially "highly ordered" (low entropy) state evolves spontaneously into a "less ordered" (high entropy) equilibrium state.
The fundamental assumption of statistical mechanics--that all microstates are equally probable(*)--leads to an interpretation of "equilibrium" as the most probable macrostate of a system: The one with the greatest thermodynamic probability--the largest number of microstates $w_k$.
Carter follows Boltzmann's argument to deduce the functional relationship between $S$ and $w$:
Let's venture to guess that entropy has not merely any old relationship with $w$, but is uniquely determined by $w$. Mathematically, this means that $S$ is a function of $w$, and it seems plausible that it should be an monotonically increasing function of $w$. $$S=f(w).$$
Consider the two systems pictured, labelled A and B.
How many microstates are available to the compound system? Let's assume each system is in its equilibrium state(=macrostate with the most microstates). Let's say at equilibrium that there are $w_A$ microstates for system A and $w_B$ for system B.
On the one hand that means for each microstate of system A, there are $w_B$ ways of arranging the contents of system $B$. So, since there are $w_A$ microstates for $A$, the total number of microstates for the combined system is:
$$\sum_{i=1}^{w_A}w_B=w_Aw_B$$.
On the other hand, entropy is an extensive quantity, like volume, or mass (not like temperature or pressure). We have been calculating the total entropy of two isolated systems as the sum of the entropies of the two systems, considered in isolation.
$$S_{A+B}=S_A+S_B.$$
So, we need some function that behaves like: $$f(w_Aw_B)=f(w_A)+f(w_B).$$ Does such a function exist??
YES!
If $S=k\ln w$, then...
$$k\log(w_Aw_B)=k\log w_A+k\log w_B$$
Experimental studies of simple, well understood quantum systems allows the measurement of $k$ which turns out to be $$k_B=1.38066\times 10^{-23} \text{ J/K}.$$ This is the connection between statistical mechanics and classical thermodynamics!
Thermal equilibrium between two systems
[This is not yet completely worked out... Read at your own risk.]
Now, I'm going to drop the index $k$ from $w_k$, with the understanding that $w(N,U)$ now means the equilibrium number of microstates for a given $U$ and a given $N$. That is to say, the equilibrium macrostate depends on ("is a function of...") both the internal energy and the number of particles in the system:
Consider
two spin systems, initially separate, each insulated. Schematically, each
system has its own energy, and contains a fixed number of spins, $N_A$ and
$N_B$.
They're
brought together and allowed to exchange energy, though not particles.
In the system with evenly spaced energy levels, the number of microstates for a particular macrostate.
To get the total number of microstates, we sum the product over all possible macrostates of system $A$ alone: $$w(N,s) = \sum_{s_A=-N_A/2}^{+N_A/2}w_A(N_A,s_A)w_B(N_B,s-s_A).$$
Since $U\propto s$, we could instead write this as a function of the internal energies: $$w(N,U)=\sum_{U_A}^{U_A<U}w_A(N_A,U_A)w_B(N_B,U-U_A).$$
Assuming that the product of two highly-peaked functions is also very peaky, the equilibrium state corresponds to the single, largest term in this sum, with largest joint number of microstates. That largest term where the number of microstates is a maximum) should be characterized by $dw=0$ for small variations in the partition of the energy. That is, if we write the Pfaffian of *one* term in the sum as: $$\begineq dw &= d\left[w_A(N_A,U_A)w_B(N_B,U_B)\right]_{N_A,N_B}\\ &= \left(\frac{\partial w_A}{\partial U_A}\right)_{N_A} w_B \, dU_A + w_A \left(\frac{\partial w_B}{\partial U_B}\right)_{N_B} \, dU_B,\endeq$$ then the *most likely* [equilibrium] term will be characterized by $dw=0$.
Because energy is conserved, $dU_A = -dU_B$.Also, dividing by $w_A w_B$... $$ \frac{1}{w_A}\left(\frac{\partial w_A}{\partial U_A}\right)_{N_A}=\frac{1}{w_B}\left(\frac{\partial w_B}{\partial U_B}\right)_{N_B}.$$
This is equivalent to:
$$\left(\frac{\partial \ln w_A}{\partial U_A}\right)_{N_A}=\left(\frac{\partial \ln w_B}{\partial U_B}\right)_{N_B}.$$
Classical thermodynamics: When $N$ is held constant, the second law says $$dU=T\,dS-P\,dV.$$
If we further hold the volume constant, $T\,dS = [dU]_{V,N}$, so... $$\left(\frac{\del S}{\del U}\right)_{V,N}=\frac{1}{T}.$$
