The microcanonical ensemble is given by $\rho = \frac{\delta(H-E)}{\int dxdp\;\delta(H-E)}$. This corresponds to a uniform distribution on the energy shell. The interpretation of this is that we don't know the microstate of the system, we only know the total energy so we average over all possibilities.

However we also know other conserved quantities of the system. For example the momentum, the angular momentum or the center of gravity.

For example if I have a (perfectly isolating) balloon filled with an ideal gas that is floating in the air, I can clearly say that the gas inside has no total momentum, no total angular momentum and I also know its center of gravity. However when I use the micro canonial ensemble to describe it I only fix the energy to a specific value. This means I also allow for states where all particles move in the same direction (which is clearly not the case).

Shouldn't we change the micro canonical ensemble to something like $$\rho = \frac{\delta(H-E)\delta(\sum_i p_i - P)\delta(\sum_i x_i \times p_i - L)\delta(\sum_i x_i - R)}{\int dxdp\; \delta(H-E)\delta(\sum_i p_i - P)\delta(\sum_i x_i \times p_i - L)\delta(\sum_i x_i - R)}?$$

  • $\begingroup$ You are confusing internal energy of the system given by its components and macroscopic energy of the whole system. The momentum, angular momentum and center of gravity are dependent on an external referential which is outside of the system. The micro canonical ensemble defines the Hamiltonian in the reference of the system in which P, L and R can be set at zero. Else they would be constants which would make no difference as we are only interested in the fact that the quantity is conserved. $\endgroup$
    – G.Clavier
    Commented May 1, 2020 at 16:32

1 Answer 1


Indeed, statistical physics describes a state of the systems as a function of its integrals of motion: the total energy, the three components of the total momentum, and the three components of the total angular momentum. Energy is somewhat special, since the other integrals determine the motion of the system as a whole. Nevertheless, it is implicit in most books on statistical mechanics that we are working in the reference frame where the net momentum and angular momentum are zero.

I suggest to consult chapter 4 of Landau&Lifshitz' Statistical physics, which is explicitly devoted to this subject (in fact, it proposes the very same equation that is given in the question, apart from the average position, which is not an integral of motion).

  • $\begingroup$ If I choose a reference frame in which the total momentum and angular momentum vanish I should at least have a $\delta(\sum_i p_i)\delta(\sum_i x_i \times p_i)$. Otherwise I also consider microstates with all particles moving in one direction (which I don't want to consider, do I?). PS: The center of gravity is an integral of motion (In the sense that $R(t)-Pt = const$). I know its trivial but formally it is an independent conserved quantity. $\endgroup$
    – toaster
    Commented May 1, 2020 at 21:11
  • 1
    $\begingroup$ The trick is that we are talking about a gas confined to a reservoir: we set to zero the momentum of the whole (gas + reservoir), but then consider the gas only. Thus all the configurations of the gas are possible, since its molecules change their momentum when hitting the walls of the reservoir. The fluctuations like the one that you suggest should cause the reservoir vibrate. $\endgroup$
    – Roger V.
    Commented May 2, 2020 at 8:01
  • $\begingroup$ Ok I understand your point. So for example if I try to describe the gas in my room, then it is clear. But if I try to describe gas for example in a balloon then I should formally add $\delta(\sum_i p_i)$ (do you agree with that?). However I guess it will not change the behavior of the macroscopic state significantly, so I can probably omit it. $\endgroup$
    – toaster
    Commented May 2, 2020 at 15:12
  • $\begingroup$ Balloon case is tricky, since there is also the gas outside, to balance the pressure. What is certain is that such fluctuations are as improbable, as all the molecules gathering in one half of the reservoir. $\endgroup$
    – Roger V.
    Commented May 2, 2020 at 16:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.