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

Let's imagine a section of a pipe through which a fluid, gas for example, flows. When there is no pressure gradient, there is no flow. However, that does not mean that the molecules are at rest. They are "Maxwell-Boltzmann" distributed but the net macroscopic velocity is zero. When the pressure gradient is turned on, the velocity prefers to be (opposite) to the direction of pressure gradient. Do you agree that this is a more ordered state then a gas at rest and therefore has a lower entropy? Is a lower entropy (one of) the cause of flows getting unstable?

share|cite|improve this question
btw, there might be a flow without pressure gradient --- you can just move to another reference frame. So just the fact that fluid is flowing does not make it more ordered, but you might have already came up with that argument. – Yrogirg Apr 29 '13 at 13:14

The connection between disorder and entropy is more complicated than that; it is made in statistical physics on the level of a microscopic model of the system, not macroscopic model.

Entropy is a more general concept, applicable even if there is no microscopic model at hand, but then the connection to disorder is not applicable.

Flowing fluid is not in a state of equilibrium, so strictly speaking it does not have thermodynamic entropy.

In continuum mechanics, one can extend the notion of entropy to such non-equilibrium flows if locally the fluid is close to equilibrium. Then the entropy of unit mass depends on its internal state, described by e.g. pressure and temperature. In this approximate theory, the velocity of the fluid does not have direct connection to its entropy.

share|cite|improve this answer
Even I had thought so until I came across this example of Landau and Lifshitz's volume 5 (Problem 3 in section 42). They calculate the Helmholtz free energy of a rotating gas as $F = F_0 - NTlog\left[\frac{2T}{m\Omega^2R^2}\exp\left(\frac{m\Omega^2R^2}{2T} - 1\right)\right]$. Entropy is just the $-\partial F/\partial T$. I am wondering, if such a formula can be derived for a gas flowing through a cylinder. – Amey Joshi Apr 28 '13 at 11:12

I'll explain how the entropy is calculated in fluid dynamics. For ordinary fluid flows the assumption of local equilibrium is valid. That it is assumed that the fluid can be diveded into many small volumes each of which being at equilibrium with say temperature $T$ and density $\rho$. This model also suggests that the specific entropy $s$ [J/(K kg)] is given by the equilibrium relation $s = s(\rho, T)$. Thus the entropy of the overall flow is

$$ S = \int_V s\left[\rho(\boldsymbol r ), T(\boldsymbol r ) \right]\, d \boldsymbol r $$

As you see the difference in entropy in the two cases is not due to the flow (velocity). It is due just to the fact that the thermodynamic quantities upon which the entropy depends in two cases are different. There is no ordering coming from the fact that flowing fluid is somehow more ordered.

However this is true only if the assumption of local equilibrium is valid and it is valid for particularly all flows you see in everyday life. But there are indeed formalism that go beyond the assumption and suggest entropy is the function of fluxes --- heat flux, diffusion flux and stress tensor. One such approach was introduced by Jou, Lebon and Casas-Vâazquez.

share|cite|improve this answer
You're right, in fact, with your reply I now understand Jan Lalinsky's response as well. The formula I wrote in my comment is actually for a gas rotating as a whole and not the one that is having a vorticity. Thank you both for clarifying the point. – Amey Joshi Apr 28 '13 at 15:59

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.