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Which is the correct form of Navier Stokes stress tensor?

+ 2 like - 0 dislike
159 views

Assume that K is a diffusion coefficient for momentum diffusion.

Assume we are in two-dimensional space (horizontal divergence and vertical vorticity are relevant) on the horizontal plane. Then the friction term in the momentum equation reads either

$$\nabla\cdot(K(\nabla\mathbf{v}+\mathbf{v}\nabla)-K\nabla\cdot\mathbf{v}\mathbb{E})$$

or, after some algebra

$$\nabla(K\nabla\cdot\mathbf{v})-\nabla\times(K(\nabla\times\mathbf{v})-2(\nabla K)\nabla\cdot\mathbf{v}+2\nabla K\cdot \nabla \mathbf{v}+2\nabla K\times(\nabla\times\mathbf{v})$$

In more detail:
\begin{eqnarray}
\mathbf{F}&=&\nabla\cdot(K(\nabla\mathbf{v}+\mathbf{v}\nabla)
-K\nabla\cdot\mathbf{v}\mathbb{E})\nonumber\\
&=&\nabla K\cdot\nabla\mathbf{v}+K\nabla\cdot\nabla\mathbf{v}+\nabla K\cdot\mathbf{v}\nabla+K\nabla\cdot\mathbf{v}\nabla-\nabla(K\nabla\cdot\mathbf{v})\nonumber\\
&=&\nabla\mathbf{v}\cdot\nabla K-\nabla K\times(\nabla\times\mathbf{v})
+K\nabla(\nabla\cdot\mathbf{v})-K\nabla\times(\nabla\times\mathbf{v})
\nonumber\\&&+\nabla\mathbf{v}\cdot\nabla K +K \nabla\nabla\cdot\mathbf{v}-\nabla(K\nabla\cdot\mathbf{v})\nonumber\\
&=&-\nabla\times (K(\nabla\times\mathbf{v}))+2\nabla\mathbf{v}\cdot\nabla K
+2K\nabla\nabla\cdot\mathbf{v}-\nabla(K\nabla\cdot\mathbf{v})\nonumber\\
&=&-\nabla\times (K(\nabla\times\mathbf{v}))+2\nabla\mathbf{v}\cdot\nabla K
-\nabla(K\nabla\cdot\mathbf{v})+\\&&+2\nabla(K\nabla\cdot\mathbf{v})-2(\nabla K)\nabla\cdot\mathbf{v}\nonumber\\
&=&-\nabla\times (K(\nabla\times\mathbf{v}))+\nabla(K\nabla\cdot\mathbf{v})
+2\nabla\mathbf{v}\cdot\nabla K-2(\nabla K)\nabla\cdot\mathbf{v}\nonumber\\
&=&-\nabla\times (K(\nabla\times\mathbf{v}))+\nabla(K\nabla\cdot\mathbf{v})
-2(\nabla K)\nabla\cdot\mathbf{v}+\\&&+2\nabla K\cdot\nabla\mathbf{v}+2\nabla K\times(\nabla\times\mathbf{v})
\end{eqnarray}

This has been checked by another person meanwhile.

I have two problems with those equivalent forms.

First, when computing the energy dissipation for a constant diffusion coefficient, I get two different answers, depending on the chosen form, namely

either $K(D^2+\zeta^2)$ where $D$ is the divergence and $\zeta$ is the vorticity

or $K(E^2+F^2)$ where $F=\partial_xv+\partial_yu$ and $E=\partial_xu-\partial_yv$ are the shear and strain deformations.

Which is the right dissipation and why?

Second, the form with $D$ and $\zeta$ is 'vector invariant' or, better said, it only relies on Gauss and Stokes theorem, and is independent of the chosen coordinate system. This seems not to be the case for the form with the deformations: I personally use in my modeling not a cartesian, but a trivariate coordinate system (my grid boxes are hexagons). Then, the vector components are linearly dependent. I stumbled across the problem, that the numerical discretisation gives 2 times the expected Laplacian when I use the shear and strain deformations in the tensor, but it gives the right Laplacian if I use the divergence and vorticity in the tensor.

What is going on here?

asked May 1 in Computational Physics by almi (10 points) [ revision history ]
edited May 10 by Arnold Neumaier

@ArnoldNeumaier

According to what you write, the claimed equivalence should not be a true equivalence. Please give the details for ''after some algebra'' so that one can point to the mistake in the algebra.

Sorry for not reacting promply.

Please note that I do not deal with a three-dimensional problem, but I only consider a horizontal plane. Therefore the deformation matrix is 2 x 2. The algebra is

\begin{eqnarray}
\mathbf{F}&=&\nabla\cdot(K(\nabla\mathbf{v}+\mathbf{v}\nabla)
-K\nabla\cdot\mathbf{v}\mathbb{E})\nonumber\\
&=&\nabla K\cdot\nabla\mathbf{v}+K\nabla\cdot\nabla\mathbf{v}+\nabla K\cdot\mathbf{v}\nabla+K\nabla\cdot\mathbf{v}\nabla-\nabla(K\nabla\cdot\mathbf{v})\nonumber\\
&=&\nabla\mathbf{v}\cdot\nabla K-\nabla K\times(\nabla\times\mathbf{v})
+K\nabla(\nabla\cdot\mathbf{v})-K\nabla\times(\nabla\times\mathbf{v})
\nonumber\\&&+\nabla\mathbf{v}\cdot\nabla K +K \nabla\nabla\cdot\mathbf{v}-\nabla(K\nabla\cdot\mathbf{v})\nonumber\\
&=&-\nabla\times (K(\nabla\times\mathbf{v}))+2\nabla\mathbf{v}\cdot\nabla K
+2K\nabla\nabla\cdot\mathbf{v}-\nabla(K\nabla\cdot\mathbf{v})\nonumber\\
&=&-\nabla\times (K(\nabla\times\mathbf{v}))+2\nabla\mathbf{v}\cdot\nabla K
-\nabla(K\nabla\cdot\mathbf{v})+\\&&+2\nabla(K\nabla\cdot\mathbf{v})-2(\nabla K)\nabla\cdot\mathbf{v}\nonumber\\
&=&-\nabla\times (K(\nabla\times\mathbf{v}))+\nabla(K\nabla\cdot\mathbf{v})
+2\nabla\mathbf{v}\cdot\nabla K-2(\nabla K)\nabla\cdot\mathbf{v}\nonumber\\
&=&-\nabla\times (K(\nabla\times\mathbf{v}))+\nabla(K\nabla\cdot\mathbf{v})
-2(\nabla K)\nabla\cdot\mathbf{v}+\\&&+2\nabla K\cdot\nabla\mathbf{v}+2\nabla K\times(\nabla\times\mathbf{v})
\end{eqnarray}

This has been checked by another person meanwhile. 

I ask myself, why physicists prefer the deformation form over the vector invariant form. Computing deformations on grid boxes, which are not rectangular, makes headaches. I use a trivariate grid, in which it seems that the entity which is formed out of two equilateral triangles, hence a rhombus ( this looks like "<|>") replace the usual rectangular grid. Velocity components are given tangentially on the edges. If I compute the shear deformation for rhombus which consists of an upper tip and an lower tip triangle using the Gauss-Green theorem, I get half of the vorticity. But the vorticity is clearly not a shear deformation!

So my question is, why the vector invariant form is discarded, but the deformation form is 'right', but does only work for rectangular grids. I thought that physics should not depend on a man chosen coordinate system.

1 Answer

+ 1 like - 0 dislike

Your two expression $K(D^2+\zeta^2)$ or $K(E^2+F^2)$ cannot be correct since they are quadratics in first order derivatives of $v$ while the expression $F$ is linear in second order derivatives of $v$. Indeed, for constant $K$, the two expressions at the top of your post are equivalent and reduce to $K$ times the Laplacian of $v$. 

Thus it seems that not the formulas you started with are incorrect but the way you use them in special cases. 

answered May 15 by Arnold Neumaier (12,355 points) [ no revision ]

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