I'm trying to gain some intuition on the Boltzmann distribution (also called Gibbs canonical distribution). The definition is very broad: it's an exponential probability distribution that is a function of an energy and temperature, $$ P(x) \propto e^{-\epsilon(x)/kT(x)} $$ where $\epsilon$ and $T$ are the energy and temperature of the state $x$, and $k$ is the Boltzmann constant.
An example is the 1D Gaussian distribution, which has pdf $$P(x) \propto e^{-\frac{(x-\mu)^2}{2\sigma^2}}.$$ Apparently this is the Boltzmann distribution corresponding to the potential energy $$\epsilon(x) = \frac{(x-\mu)^2}{2\sigma^2}.$$
How do I interpret this? I can see that if a point $x$ is far from the mean, then it has high potential energy.
In statistics, the idea of "entropy" is related to the spread of a distribution. Is this saying that a high entropy distribution has high potential energy?
The potential energy ${1\over2}k(x-\mu)^2$ is exactly the well known potential energy of a harmonic spring with offset $\mu$. A spring in a thermal bath where the kinetic energy can be neglected, or can be integrated over will have a position distribution proportional to $e^{-{1\over 2}\beta k (x-\mu)^2}$ where $\beta = {1\over \kappa T}$