Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If an ideal gas is flowing with a velocity $v$, how is the equipartition theorem applied.

Normally, we can say that $\frac{1}{2}mv_{x,rms}^2=\frac{1}{2}k_BT$. We can do the same thing for $v_y$ &c. But, when a gas is flowing in the x direction, I don't think that $v_{x,rms}=v_{y,rms}$. I'm not too sure of this, the distribution may be such that the rms velocities are preserved. If the rms velocities aren't preserved, obviously we cannot use $\frac{1}{2}k_BT$ as the temperature is the same (or is it?). So how does one analyse such a situation with the equipartition theorem?

I'm not very good at Hamiltonian mechanics, so Wikipedia isn't helping.

Question sparked off by Air velocity in a double-skin facade

share|cite|improve this question
up vote 1 down vote accepted

Do the work in the rest frame of the (bulk) gas. It's really that easy.

In other words for a gas whose bulk flow is along the $x$ axis the $y$ and $z$ velocity distributions are given by equipartition as is the deviation of the $x$ velocity from the bulk velocity.

share|cite|improve this answer
Makes sense; thanks! In retrospect it feels obvious.. All inertial frames are equivalent. – Manishearth Feb 23 '12 at 5:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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