# The Canonical Ensemble

I was reading Blundell and I came with the topic of the Canonical Ensemble.

And as long as I did understand, the system when is in contact with the large reservoir and it’s left in this condition for a sufficiently amount of time so they both come into thermal equilibrium, the temperature is fixed (and the temperature of the reservoir is equal to the temperature of the system).

Although, in thermal equilibrium we expect the energy content of the system to remains constant, but the the the canonical ensemble says we rather have some probability distributuion of the different (but possible) values of energy the system can have.

This is given by the Boltzmann Distribution.

But why is the energy not fixed? That is a requirement for being in the thermal equilibrium state, right?

(I think I might be missing something in the middle of this but I’m not sure of what. Thank you for your time)

• "This is given by the Boltzmann Distribution" It seems to me that in relation to the canonical ensemble, the corresponding distribution is called the Gibbs distribution – Aleksey Druggist Dec 11 '19 at 1:00
• @AlekseyDruggist I have search in many books and all of them demonstrate a relation between the Canonical Ensemble and the Boltzmann Distribution – user249212 Dec 11 '19 at 22:04
• If you focus on classical textbooks, then the Gibbs distribution unambiguously. See, for example, Landau, Lifshits, "Theoretical Physics in Ten Volumes, Volume V: Statistical Physics. Part 1." , Chapter III: "Gibbs distribution", Section 28: "Gibbs distribution" – Aleksey Druggist Dec 11 '19 at 22:24
• @AlekseyDruggist Thank you for suggestion! I am going to search on that books and see what I can get from them :) – user249212 Dec 12 '19 at 11:20

• I am by no means an expert in this area but to my understanding you are correct. The heat bath experiences a tiny change in its energy (relative to its size) while the change in the system's energy is more significant. This is why the probability of finding the system at some energy is is given by $P_r = \alpha \times \Omega_{bath}(E_{total}-E_{system})$, the bath is so much larger that it's number of possible microstates is enormous compared to the system. – Charlie Dec 9 '19 at 14:21