It appears to be a well-known fact that a probability distribution on phase space will tend towards the distribution that maximizes entropy. The Wikipedia article on maximum entropy states:
"The motivation is twofold: first, maximizing entropy minimizes the amount of prior information built into the distribution; second, many physical systems tend to move towards maximal entropy configurations over time."
There is the MaxEnt principle of Jaynes, and perhaps even the second law of thermodynamics.
Suppose we have classical phase space with $2n$ variables $x_i$, $p_i$. Given an energy function $u(x,p) = p^2/2m$, and a probability distribution $D$, one has average energy $U(D) = \frac{1}{h}\int dxdp Du$ and entropy $S(D)=-\frac{k}{h}\int dxdpD\log(D)$, and normalize by setting $\frac{1}{h}\int dxdp D = 1$. One can use Lagrange multipliers to find the distribution that maximizes $S$, and it turns out to be $\frac{1}{Z}e^{-\beta u}$, the Boltzmann-Gibbs distribution, a Gaussian, where $\beta=\frac{1}{k_BT}$ and $Z=\frac{1}{h}\int e^{-\beta u}$ is the partition function.
My question is practically speaking, how does this work? So if I have an ideal gas, and I let it settle into equilibrium, I should have the distribution $D=\frac{1}{Z}e^{-\beta u}$ on phase space. Then I add some information: move this particle to the left, that particle to the right, I squeeze $D$ a little, or move it to the right. Now I've added, say, 100 $bit$ of info, i.e. the new distribution $D'$ has $\Delta S=S(D)-S(D') = k*100 \,bit$.
Will $D'$ just bounce back to the distribution $D$, perhaps releasing some energy?
