Statistical physics and momentum conservation

Question

In statistical physics one usually looks at energy as a conserved quantity and e.g. in the canonical ensemble assumes a constant average energy of the ensemble. Now why don't we usually do this for other conserved quantities like momentum? Why not do a 'canonical' ensemble with momentum exchange? Is it more complicated or simply never useful?

Answer 1

Momentum, unlike energy, is a vector quantity. This means that if you have a large number of particle moving at random their momenta tend to cancel out, rather than adding the way a scalar quantity like energy would.

If the total momentum of the component parts of your system does not add to 0, then this simply means that your entire system is moving in some direction and you can always move to a reference frame where the total momentum is 0. This means you can't really have 'internal momentum' in the same way as internal energy.

Similarly if the total angular momentum of your system is non-zero that simply means that the system as a whole is rotating and does not tell you much about the microstructure.

Answer 2

Indeed they are conserved quantities so there are canonical ensembles with linear momentum exchange and angular momentum exchange with a reservoir. In fact Gibbs used angular momentum as an example to demonstrate the canonical ensemble's use beyond simply energy (Gibbs 1902, see around eqn 98).

Considering linear and angular momentum together, the canonical probability expression for state $i$ is $$ p_i = \exp( [\Omega - E_i + \vec v \cdot \vec P_i + \vec \omega \cdot \vec L_i]/kT) $$ where the state $i$ has total energy $E_i$, total linear momentum $\vec P_i$, total angular momentum $\vec L_i$. What's neat is that the conjugate variables $\vec v$ and $\vec \omega$ are basically linear velocity and angular velocity of the reservoir. $\Omega$ is the applicable thermodynamic potential, I'm not sure what to call it though, "inertial free energy"?

To be honest this ensemble does not get used very much. We do not often deal with systems where we are uncertain about total orientation (spherically symmetric knowledge) or about total position (linearly symmetric knowledge). Usually our thermodynamic samples are in a box sitting upright in a particular drawer, all momentum conservation becoming irrelevant.

However there are at least a few cool situations where it appears:

• Nonzero $\vec \omega$ can be used to analyse gases in rotating containers --- in particular gas centrifuges. (See e.g., Balian's From Microphysics to Macrophysics vol 1, p. 339) Or how about Rotating black holes?
• Did you know that if you rapidly rotate cold objects, their nuclei develop a net spin (and a net magnetisation)? The effect is weak but easy to calculate using this ensemble with nonzero $\vec \omega$.(Jaynes 1957)
• Nonzero $\vec v$ can be used to consider what is a gas in a moving environment. For example consider a gas (e.g. a blackbody radiation, or photon, gas) inside a relativistically moving environment. There is some disagreement over how temperature behaves under Lorentz transformations, though. (e.g., Eberly and Kujawski 1968 but many other papers before and after).

Answer 3

As the previous answers have noted, momentum conservation often isn't useful. However, there are exceptions -- especially in transport problems. Although you're not necessarily directly using the canonical ensemble for these problems, they are decidedly statistical mechanics problems. For example, (quasi)momentum conservation in phonon normal scattering processes leads to a modified distribution function. (See equation 6.8 in this paper -- often regarded as a classic.)

Answer 4

I think it wouldn't be useful. In statistical mechanics you want to model the microscopic behaviour of a thermodynamical system. In the laws of thermodynamics there's no mention to momentum. But I believe nothing forbids you from talking about the average momentum of a statistical mechanical system.