Tuesday, January 10, 2012

Where does the ln come from in S = k ln(W) ?

The relationship between entropy ($S$) and degeneracy ($W$ or $g$ depending on the book) $$S = k\ln(W)$$ is one of the fundamental equations of statistical mechanics.  But where does it come from?  Or more precisely, why $\ln(W)$ and not, say, $\sin(W)$?  And why does $k$ have units of J/K?

I believe it goes back the second law of thermodynamics, i.e. it is based on observation.  The second law can be stated in many ways and one is that heat ($q$) spontaneously flows only from hot ($T_h$) to cold ($T_c$) bodies.

Mathematically this can be stated as $$\frac{q}{T_c}-\frac{q}{T_h}>0$$ or $$\Delta S >0$$ where $S$ is defined* as $$dS=\frac{dq_{rev}}{T}$$ This establishes the units of $S$ and, hence, $k$.

Furthermore, since $q$ (like all energy) is additive, $S$ must be additive: $S_{total}=S_A+S_B$.  However, $W_{total}=W_AW_B$, so $S=kW$ won't work.  However, $S = k\ln(W)$ will.

The final question is now whether the logarithm is the only mathematical function for which $f(xy)=f(x)+f(y)$.  It turns out that it is** if we require the function to be continuous (thanks to Niels Grønbæk for help here), which we do since $S$ as defined by the second law is non-discrete like the energy.

Another important property of $\ln(W)$ is that is has a maximum value when $W$ is largest.  So the most probable state will have the largest entropy.

* This definition begs the question: if heat is transferred, what $T$ do you use, the one before or after the heat transfer?  The answer is that $dq$ has to be so small that $T$ is not affected.  Since $T$ is not affected this is a $rev$ersible process.  Furthermore, notice that this law also introduces the concept of temperature.

** Se also http://www.physicsforums.com/showthread.php?t=566358.  Not that I understand all of it!

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