The economic combinatorial problem
I mentioned in this post that an economy is a combinatorial problem (along with some hints at entropy); let me sketch out the mathematical side of the argument.
I connected $\log M$ (where M is the currency supply) with $\beta = 1 / T$ in the partition function awhile ago (setting $k_{B} = 1$). If we take the thermodynamic definition of temperature:
$$
\frac{1}{T} = \frac{dS}{dE}
$$
as an analogy (where $S$ is entropy and $E$ is energy), we can write (putting in a constant $c_{0}$ that corresponds to the constant $\gamma M_{0}$ in the information transfer model):
$$
\log M/c_{0} = \frac{dS}{dN}
$$
where we've used the correspondence of the demand (NGDP, or $N$ -- i.e. aggregate demand AD) with the energy of the system. We don't know what the economic entropy is at this point. However, if we take (using the solution shown here):
$$
N = M^{1/k}
$$
Then we can write down
$$
k \log N/c_{0} = \frac{dS}{dN}
$$
So that, integrating both sides,
$$
S = k \; (N/c_{0}) (\log N/c_{0} - 1) + C
$$
Which, via Stirling's approximation, gives us (dropping the integration constant $C$)
$$
S \simeq k \log \; (N/c_{0})!
$$
If we compare this equation with Boltzmann's grave:
$$
S = k \log W
$$
We can identify $(N/c_{0})!$ with the number of microstates in the economy. The factorial $N!$ counts the number of permutations of $N$ objects and we can see that $c_{0}$ adjusts for the distinguishability of given permutations -- all the permutations where dollars are moved around in the same company or industry are likely indistinguishable. This could lend itself to an interpretation of the constant $\gamma$ mentioned above: large economies are diverse and likely have the same relative size manufacturing sectors and service sectors, for example -- once you set the scale of the money supply $M_{0}$, the relative industry sizes (approximately the same in advanced economies) are set by $\gamma$.
This picture provides the analogy that a larger economy ($N$) has larger entropy (economic growth produces entropy) and lower temperature ($1/\log M$).
