# Can a macrostate of a system be defined upon distributional quantities of the energies?

Is it valid to define for a system, a macrostate based upon statistical quantities of the particle energy distributions? Eg. that a macrostate is based upon the variance of the kinetic energies of the particles? Would another potentially valid macrostate value be the skew of the distribution of the energies? Those macrostate values are to be used in order to calculate the entropy changes in a system over time as the number (size) of microstates that satisfy the macrostate.

Let's do the math first. We have a collection of particles $$i=1,2,\cdots$$ with energies $$E_i$$ and all we know is their variance. By the maximum entropy principle we seek the probability distribution $$p_i$$ that maximizes the entropy functional, $$-\sum_i p_i \log p_i$$ under the constraints $$\sum_i p_i = 1,\quad \sum_i p_i E_i^2 = \sigma^2$$ This variational problem has a known solution: $$p_i$$ is a Gaussian function with variance $$\sigma^2$$ regardless of its mean (see wikipedia; on the same web page there is also the case of a distribution with known variance and skewness).
Does it make physical sense? A physical macrostate is a macroscopic state in equilibrium that can be prepared so as to satisfy a number of specifications. For example, a $$(U,V,N)$$ system is prepared by putting $$N$$ particles in a rigid insulated box with volume $$V$$ (we may have to play with the temperature to hit the exact value of $$U$$ we want). Can we prepare a system of particles such that the variance or skewness of the distribution is exactly what we want, independently of the mean of the distribution? I think it's safe to say that answer is no.
• It seems to me that on the first OP question ("Is it valid to define for a system, a macrostate based upon statistical quantities of the particle energy distributions?") we can still answer "yes".In fact, what is the canonical distribution? it is the probability distribution that maximises the entropy subject to the constraint that the average (or mean) energy $\sum_i p_i E_i = <E>$ is fixed. – Aleksey Druggist Aug 22 '19 at 17:16