Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

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

IT is often stated the flow with high Re is not isotropic, meaning there is no uniform or dominant direction of the flow. But this seems wrong to me - -while there's certainly cases where no dominant direction of flow can be discerned in highly turbulent situations, there's for example the flow through pipes or along fast moving object where there's clearly a strong anisotropy. Even an eddy in a stirred coffee cup has an easily discernible direction of flow locally.

The Reynolds number as such tells me the relation between inertial and viscous forces - is there a number or metric that tells me how non-isotropic a flow is?

share|cite|improve this question
I guess you are confusing the two terms "isentropic" and "isotropic". "Isotropic" means independent from the direction whereas "isentropic" means that the entropy is constant (there are no irreversible processes going on that would increase it). – Dilaton Jan 21 '13 at 15:13
yes. I corrected this, thanks. – mart Jan 21 '13 at 15:17

All flows (high Re flows or not) are non-isotropic. Flow is a response to a non-equilibrium and occurs because of a gradient of some scalar quantity which has a preferred direction (because a gradient is a vector). So by that note, does one need a number of any sort to quantify it?

From a more deeper perspective, non-equilibrium flows themselves may be associated with some global entropy-like metric that enables the study of pattern formation in nature, that could be applicable a wide category of flows from convection rolls in the atmosphere to crystal formation in solids. But it is not clear if one exists---a debate on the nature of time and the existence of the arrow of time in thermodynamics But that is not the purpose of this question.

The OP raises some questions about the isotropic nature at the Kolmogorov scale: Local isotropy in high Reynolds number flow is Kolmogorov's hypothesis. At the scale that local isotropy prevails---the Kolmogorov scale---kinetic energy is dissipated by viscosity into thermal energy. In essence the momentum in the fluid packet is equivalent to random motion of its molecules, and effectively there is no bulk convective flow. Furthermore the flow is statistically isotropic. All the kinetic energy in a turbulent flow is in its large scale where the flow is non-isotropic.

share|cite|improve this answer
"All flows are non-isotropic" depends on the volume in consideration: If you look at an eddy or whirl, the net direction is practically zero, as is - trivial example - the flow inside a closed container if you look at the whole container. Low Re implies strong viscous forces implying a lot of particles moving in the same direction = non isotropic - but the reverse does not hold true! – mart Jan 22 '13 at 8:36
Yes, but if you enclose a flow in a such a volume that you cannot "see" a net direction you also don't "see" a flow. The idea of observing something in control volume is always done to ignore the details of the flow in that control volume, but to study what goes in and what goes out. If you are then looking at the details of your enclosed eddy, you are effectively looking at a local control volume which has net directional flow. So you cannot ask questions both ways. – Sankaran Jan 22 '13 at 16:26
contd... Another way to think about this is how you would write an equation for that flow. If you writing a local equation you see both flow and direction. If you are writing a global one you integrate over both the flow field details as well as the directionality. – Sankaran Jan 22 '13 at 16:28
But you can have conditions with a lot of movement and not a "lot of direction" - wikipedia describes turbulence as locally isotropic - but is Re a metric for this isotropy? – mart Jan 23 '13 at 8:40
I have edited my answer for the question you've raised – Sankaran Jan 23 '13 at 17:11

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.