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

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The discussion focuses on the Fourier Transform of the Navier-Stokes (N-S) equation, specifically questioning the absence of pressure in the Fourier space representation. The N-S equation is presented as \(\frac{\partial\vec{u}}{\partial t}+(\vec{u}\bullet\nabla)\vec{u}=-\frac{\nabla P}{\rho}+\nu\nabla^{2}\), with its Fourier space transformation detailed. Participants confirm that the term \((\vec{u}\bullet\nabla)\vec{u}\) represents acceleration, clarifying that both terms on the left side of the N-S equations denote acceleration, while the small 'p' refers to momentum rather than pressure.

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dakold
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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 interpret the second term on the left hand side?
(\vec{u}\bullet\nabla)\vec{u}
Is this acceleration?
 
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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
 

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