Solving N-S Eq: Fourier Space, Pressure, Interpretation

  • Thread starter Thread starter dakold
  • Start date Start date
  • Tags Tags
    Navier-stokes
dakold
Messages
13
Reaction score
0
As show below the Fourier Transform of Naiver-Stokes equation. I wonder if the pressure should be in the Fourier transform? In the below transformation there is no pressure.

N-S
\frac{\partial\vec{u}}{\partial t}\+(\vec{u}\bullet\nabla)\vec{u}=-\frac{\nabla P}{\rho}+\nu\nabla^{2}

N-S in Fourier space \frac{\partial u_{\alpha}}{\partial t}=-i\int(k_{\beta}-p_{\beta}u_{\beta}(\vec{p})u_{\alpha}(\vec{k}-\vec{p})d^{3}p+i\frac{k_{\alpha}}{k^{2}}\int(p_{\gamma}(k_{\beta}-p_{\beta})u_{\beta}(\vec{p})u_{\gamma}(\vec{k}-\vec{p})<br /> d^{3}p-\nu k^{2} u_{\alpha}(\vec{k})

I got another question also, how can one interpretate the second term on the left hand side?
(\vec{u}\bullet\nabla)\vec{u}
Is this acceleration?
 
Physics news on Phys.org
I don't know about the pressure terms in Fourier space. What do the 'p' terms represent if they do not represent pressures?

The (\vec{u}\bullet\nabla)\vec{u} term does indeed represent an acceleration. Both terms on the left hand side of the N-S equations represent acceleration.

The first term represents accelerations arising if the flow is unsteady, and the second term represents the acceleration of a fluid particle arising from its movement in the velocity field.
 
the small 'p' is the momentum.
Thanks that what I thought
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
Back
Top