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

I'm considering to study some high-dimensional Navier-Stokes equations. One problem is to do write the viscous equation for vorticity, helicity and other conserved quantities. I think it might be better if it is possible to work with differential form and exterior calculus? Is there any reference that I may find somewhere?

share|cite|improve this question
You mean in a form other than $\rho (\partial_t + v \cdot \nabla) v= f+ \dot{\overline \sigma}(\dot \nabla)$? Or formulas for the quantities you mention? – Muphrid Dec 25 '12 at 22:09

MS Mohamed et al begin with a "standard vector calculus formulation of the NS equations",

$$ \frac{∂ \bf u}{∂ t} - \mu ∆ {\bf u} + ({\bf u} \cdot \nabla) {\bf u} + \nabla p = 0 $$ $$ \nabla \cdot {\bf u} = 0 $$

and use the rotational form,

$$ \frac{∂ \bf{u}}{∂ t} + \mu \nabla \times \nabla \times {\bf u} - {\bf u}\times(\nabla \times {\bf u}) + \nabla p^d = 0 $$

to eventually derive

$$ \frac{∂ {\bf u}^\flat}{∂ t} + (-1)^{N+1} \mu \star d \star d {\bf u}^\flat + (-1)^{N+2}\star ({\bf u}^\flat \wedge \star d{\bf u}^\flat) + d p^d = 0, $$ $$ \star d \star {\bf u}^\flat = 0 $$

But they also later derive it as

$$ \frac{∂ d {\bf u}^\flat}{∂ t} + (-1)^{N+1} \mu d \star d \star d {\bf u}^\flat + (-1)^{N}d\star({\bf u}^\flat \wedge \star d{\bf u}^\flat) = 0, $$

Judging from the stars, -1s, and flats, these results could probably be simplified by substitution of the codifferential operator $d^\dagger$. Both are also the rotational form; I would rather provide a convective form, but I'm unaware of such a derivation in the literature.

I urge Troy to actually post an answer rather than referring us to his book, but lacking the points I cannot leave a comment to that effect.

share|cite|improve this answer

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.