0
$\begingroup$

I'm learning stable fluids by reading Stam's paper.

In the paper, it says, according to Helmholtz-Hodge Decomposition, I vector field $w$ could be decomposed to a divergence-free field and a curl-free field. Thus:

$$ w = u +\nabla q \\ u = Pw = w-\nabla q$$

where $P$ is our projection P.

Given the Navier-Stokes equation: $$ \nabla \cdot u = 0\\ \frac{\partial{u}}{\partial{t}} = -(u\cdot \nabla)u - \frac{1}{\rho}\nabla p + v\nabla^2 u + f $$

Do projection $P$ to both sides of the above equation, we obtain

$$ \frac{\partial{u}}{\partial{t}} = P(-(u\cdot\nabla )u + v\nabla^2u +f)$$

I think the projection is, essentially, used to extract the divergence-free component of a vector field. So, we want to extract the divergence-free part of $-(u\cdot\nabla )u + v\nabla^2u +f$.

Later in the paper, we have $$ w_4 = w_3 - \nabla q$$ where $w_4$ is the divergence-free velocity field we want.

However, someone told me the $\nabla q$ term in this case is actually the gradient of pressure. I wonder is that correct? If it is, why?

$\endgroup$

migrated from computergraphics.stackexchange.com Nov 10 '17 at 14:05

This question came from our site for computer graphics researchers and programmers.

  • 1
    $\begingroup$ Can you provide a link or a better citation to "Stam's paper"? $\endgroup$ – rob Nov 10 '17 at 13:57
  • $\begingroup$ I think the question is on-topic here on Computer Graphics, but I don't think it quite fits our audience, and it would get better answers on Physics, so I'm sending it their way. $\endgroup$ – Dan Hulme Nov 10 '17 at 14:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy