- #51

arildno

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Hmm..your equation is wrong (there shouldn't be a minus sign in front of the first term on the right-hand side).

Restricting myself to the case of constant density&dynamic viscosity and conservative volume forces, taking the curl of Navier-Stokes yields:

[tex]\frac{D\vec{c}}{dt}=(\vec{c}\cdot\nabla)\vec{v}+\nu\nabla^{2}\vec{c},\vec{c}\equiv\nabla\times\vec{v},\frac{D}{dt}\equiv\frac{\partial}{\partial{t}}+\vec{v}\cdot\nabla[/tex]

In the inviscid case (with volume forces being a gradient field), therefore, we have:

[tex]\frac{D\vec{c}}{dt}=(\vec{c}\cdot\nabla)\vec{v}[/tex]

(Note that this is also exact in the inviscid, barotropic case, [tex]\rho=\rho(p)[/tex])

As for the interpretation of the right-hand side, we note that if two points A,B in the fluid is separated by the vector [tex]\vec{AB}=dq\vec{s}[/tex] (where dq is an infinitesemal quantity), we have:

[tex]\vec{v}_{B}-\vec{v}_{A}=((\vec{AB}\cdot\nabla)\vec{v})\mid_{A}=dq(\vec{s}\cdot\nabla\vec{v})\mid_{A}[/tex]

This is proven as follows:

[tex]\vec{v}_{B}=\vec{v}(x_{A}+dx,y_{A}+dy,z_{A}+dz,t)=\vec{v}_{A}+dq(\vec{s}\cdot\nabla)\vec{v}\mid_{A},(dx,dy,dz)=dq\vec{s}[/tex]

If therefore A, B is fluid-particles at the same instantaneous vortex line (i.e, defined with [tex]\vec{c}[/tex] as the tangent vector at time t), we have, by the vorticity equation:

[tex]\vec{v}_{B}-\vec{v}_{A}=dq\frac{D\vec{c}}{dt}\mid_{A}[/tex]

(at time t)

That is, the vorticity change, as experienced by particle A is only affected by the perceived velocity difference along the vortex line to which A instantaneously belongs.

To proceed further, let [tex]\vec{AB}_{t+\bigtriangleup{t}}[/tex] be the vector joining A, B an instant later.

We have therefore the equality:

[tex]\vec{AB}_{t+\bigtriangleup{t}}=\vec{AB}+(\vec{v}_{B}-\vec{v}_{A})\bigtriangleup{t}=dq(\vec{c}+\frac{D\vec{c}}{dt}\bigtriangleup{t})\mid_{A}=dq\vec{c}(t+\bigtriangleup{t})\mid_{A}[/tex]

That is, particles A and B remain joined to the SAME vortex line at time [tex]t+\bigtriangleup{t}[/tex] as they were on time t.

(At time [tex]t+\bigtriangleup{t}[/tex] the vorticity experienced by particle A is [tex]\vec{c}(t+\bigtriangleup{t})[/tex]

Hence, we may conclude that in the inviscid case, vortex lines are MATERIAL curves, in that they consist of the same particles throughout time.

Restricting myself to the case of constant density&dynamic viscosity and conservative volume forces, taking the curl of Navier-Stokes yields:

[tex]\frac{D\vec{c}}{dt}=(\vec{c}\cdot\nabla)\vec{v}+\nu\nabla^{2}\vec{c},\vec{c}\equiv\nabla\times\vec{v},\frac{D}{dt}\equiv\frac{\partial}{\partial{t}}+\vec{v}\cdot\nabla[/tex]

In the inviscid case (with volume forces being a gradient field), therefore, we have:

[tex]\frac{D\vec{c}}{dt}=(\vec{c}\cdot\nabla)\vec{v}[/tex]

(Note that this is also exact in the inviscid, barotropic case, [tex]\rho=\rho(p)[/tex])

As for the interpretation of the right-hand side, we note that if two points A,B in the fluid is separated by the vector [tex]\vec{AB}=dq\vec{s}[/tex] (where dq is an infinitesemal quantity), we have:

[tex]\vec{v}_{B}-\vec{v}_{A}=((\vec{AB}\cdot\nabla)\vec{v})\mid_{A}=dq(\vec{s}\cdot\nabla\vec{v})\mid_{A}[/tex]

This is proven as follows:

[tex]\vec{v}_{B}=\vec{v}(x_{A}+dx,y_{A}+dy,z_{A}+dz,t)=\vec{v}_{A}+dq(\vec{s}\cdot\nabla)\vec{v}\mid_{A},(dx,dy,dz)=dq\vec{s}[/tex]

If therefore A, B is fluid-particles at the same instantaneous vortex line (i.e, defined with [tex]\vec{c}[/tex] as the tangent vector at time t), we have, by the vorticity equation:

[tex]\vec{v}_{B}-\vec{v}_{A}=dq\frac{D\vec{c}}{dt}\mid_{A}[/tex]

(at time t)

That is, the vorticity change, as experienced by particle A is only affected by the perceived velocity difference along the vortex line to which A instantaneously belongs.

To proceed further, let [tex]\vec{AB}_{t+\bigtriangleup{t}}[/tex] be the vector joining A, B an instant later.

We have therefore the equality:

[tex]\vec{AB}_{t+\bigtriangleup{t}}=\vec{AB}+(\vec{v}_{B}-\vec{v}_{A})\bigtriangleup{t}=dq(\vec{c}+\frac{D\vec{c}}{dt}\bigtriangleup{t})\mid_{A}=dq\vec{c}(t+\bigtriangleup{t})\mid_{A}[/tex]

That is, particles A and B remain joined to the SAME vortex line at time [tex]t+\bigtriangleup{t}[/tex] as they were on time t.

(At time [tex]t+\bigtriangleup{t}[/tex] the vorticity experienced by particle A is [tex]\vec{c}(t+\bigtriangleup{t})[/tex]

Hence, we may conclude that in the inviscid case, vortex lines are MATERIAL curves, in that they consist of the same particles throughout time.

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