# Meaning of the uniform ensemble

In statistical physics (and viewing this from a classical point of view) for an isolated system we can say that a system will have an energy constant in time.

People define the following uniform probability distribution of microstates ($$P$$), which is also called the uniform ensemble: $$P=0$$ if $$H(q,p)> E$$ and $$P\neq 0$$ if $$H(q,p)< E$$, where $$H$$ is the hamiltonian of a system.

But the hamiltonian of a system is also equal to the energy of a system, so how can we say that there is a probability of, in a given system, there being a microstate, a point $$(q,p)$$, where the hamiltonian is less than the energy of the system; to me this probability distribution function makes no sense. Also I don't understand the reasoning behind creating it, it just seems to come out of nowhere, but am I correct in saying that the only point of having this distribution is that from this probability density, we can find that its density is concentrated at the surface of the equienergetic surface in phase-space, which allow us to define a function of the density of the microstates $$(q,p)$$, which in turn allows us to find the expected values of observables?

Firstly, it is probably worth referring to this previous question and answer for some background information and defining equations, while making clear that your question is different from that one.

In general terms, I agree with you. The uniform ensemble, in which all states having an energy less than the prescribed value $$E$$ are equally weighted, does not have a simple physical interpretation. I don't agree with you that there is some kind of inconsistency here: $$E$$ is just the maximum possible energy, and some other symbol such as $$H(q,p)$$ represents the energy for given $$q$$ and $$p$$.

The ensemble has a small advantage: that the entropy formula $$S=k_B\ln \overline{\Omega}$$ involves taking the logarithm of a dimensionless quantity $$\overline{\Omega}$$, which we can interpret as the number of states having energy less than $$E$$ (once we have normalized the integrals over $$q$$ and $$p$$ with the appropriate factor of Planck's constant $$h$$).

For the microcanonical ensemble, one must remember that the energy shell formally has an arbitrary width $$\Delta E$$, so that $$S=k_B\ln \Omega$$ refers to the number of states with energy between $$E$$ and $$E+\Delta E$$. Then it is necessary to discuss how (for a large enough system) the choice of $$\Delta E$$ does not practically affect the calculation of thermodynamic properties, starting with $$S$$. Often, though, this is forgotten, and we just write $$S=k_B\ln \Omega$$, where $$\Omega$$ is a density of states (per unit energy); we shouldn't really do this, as $$\Omega$$ defined this way is not dimensionless. This isn't a serious deficiency, as long as it is borne in mind. Of course, the microcanonical ensemble is physically more appealing, corresponding in principle to an isolated system.

For the uniform ensemble, as you imply in your question, the $$\Delta E$$ arises "naturally" from the very rapid increase of $$\Omega(E)$$ with $$E$$, causing most of the states included in $$\overline{\Omega}(E)$$ to be concentrated near $$E$$.

All ensembles are artificial constructs, to an extent. If you are saying that the uniform ensemble is more artificial than some of the others, then yes, I think you are right.

• but how can we say that there are possible microstates of the system with an hamiltonian, which is the energy of the system, less than the energy of the system, isn't this an inconsistency? Commented Nov 11, 2018 at 23:26
• $E$ is not the energy of the system. It is a parameter defining the ensemble. All systems with an energy less than $E$ can belong to the ensemble, with equal probability. You can use a different symbol to represent the energy of any of these systems.
– user197851
Commented Nov 11, 2018 at 23:58