# LES question: Why must the eddy viscosity be rotation-invariant?

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 April 12, 2019, 13:44 LES question: Why must the eddy viscosity be rotation-invariant? #1 New Member   Join Date: Apr 2019 Posts: 5 Rep Power: 6 hello all, I am a new beginner in LES and CFD. I am reading all I can on LES modelling, but I think I am missing something. All the papers and books state that for eddy viscosity models, the eddy viscosity must be rotation invariant. That means that we have to use the same eddy viscosity for the u,v, and w equations, correct? Why is that the case? I understand that the physical viscosity is a scalar, so rotation invariant. However, the eddy viscosity is a pure model term, not physical. Why can we not use different eddy viscosities in different equations? For example, in the wall normal direction, why can it not be different from the wall parallel directions? This has me very confused! Does anybody know? Are there any models with different eddy viscosities? thank you all greeting ulfu

April 13, 2019, 19:52
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Paolo Lampitella
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Quote:
 Originally Posted by ulfu hello all, I am a new beginner in LES and CFD. I am reading all I can on LES modelling, but I think I am missing something. All the papers and books state that for eddy viscosity models, the eddy viscosity must be rotation invariant. That means that we have to use the same eddy viscosity for the u,v, and w equations, correct? Why is that the case? I understand that the physical viscosity is a scalar, so rotation invariant. However, the eddy viscosity is a pure model term, not physical. Why can we not use different eddy viscosities in different equations? For example, in the wall normal direction, why can it not be different from the wall parallel directions? This has me very confused! Does anybody know? Are there any models with different eddy viscosities? thank you all greeting ulfu
There are indeed anisotropic eddy viscosity models, where the eddy viscosity is a 4th order tensor in general. Check the Sagaut book for an account of such models.

What rotation invariance refers to is the fact that the eddy viscosity, as a function of the resolved velocity field, cannot depend from the frame of reference, including its rotation.

 April 14, 2019, 17:19 #3 Senior Member   Join Date: Oct 2017 Location: United States Posts: 233 Blog Entries: 1 Rep Power: 9 Historically, eddy-viscosity stems from Boussinesq hypothesis, which is analogous to Newton's linear stress-rate-of-strain relation, where the shear stress is proportional to the strain rate with the physical viscosity, and it is isotropic as you point out since it's a scalar. The eddy-viscosity pretty much follow the same way.

 April 14, 2019, 17:40 #4 Senior Member   Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,737 Rep Power: 71 I think that there is a motivation in the hystorical assumption of the fully developed turbulence. For transitional flows there were DNS studies that showed that anisotropy exists. However, there is not a definite assessment that anisotropic SGS eddy viscosity models perform better than classical models

April 15, 2019, 12:50
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 Originally Posted by sbaffini What rotation invariance refers to is the fact that the eddy viscosity, as a function of the resolved velocity field, cannot depend from the frame of reference, including its rotation.

Thank you for your answer, I will check the book and learn about the models. I am still not sure I get what rotational invariance means, though. Is it the same as isotropy, or do these concepts differ?

Also, I am trying to combine the two issues you mentioned in my head but I fail:

1) Anisotropic models exist
2) the eddy visocity cannot depend on the frame of reference.

Is 1 not a contradiction of 2?

Let us say we are considering a boundary layer, where x is the streamwise direction and y is normal to the wall. A reasonable eddy viscosity could now be constant in x, but e.g. have y ^ 3 towards the wall - is this not a a contradiction of rotational invariance? I am sorry if this is stupid, but I cannot find the solution.

Thanks a lot
ulfu

April 15, 2019, 12:52
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 Originally Posted by FMDenaro I think that there is a motivation in the hystorical assumption of the fully developed turbulence. For transitional flows there were DNS studies that showed that anisotropy exists. However, there is not a definite assessment that anisotropic SGS eddy viscosity models perform better than classical models

so Professor does this mean that we could have 3 different eddy viscosities, for each direction? Or does the viscosity have to be a tensor?

April 15, 2019, 19:04
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Lucky
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 Originally Posted by ulfu Let us say we are considering a boundary layer, where x is the streamwise direction and y is normal to the wall. A reasonable eddy viscosity could now be constant in x, but e.g. have y ^ 3 towards the wall - is this not a a contradiction of rotational invariance? I am sorry if this is stupid, but I cannot find the solution.
This is a non-homogeneous mapping, but is almost guaranteed to be an isotropic eddy viscosity. A function of x,y,z that spits out a single scalar value for the eddy viscosity means that it's still a rank 0 tensor which is always isotropic and always has rotational invariance (a scalar is a scalar).

Isotropic tensors have rotational invariance by definition.

Quote:
 Originally Posted by ulfu so Professor does this mean that we could have 3 different eddy viscosities, for each direction? Or does the viscosity have to be a tensor?
In general the anisotropic eddy viscosity is a 3x3x3x3 fourth order tensor. The eddy viscosity maps a tensor (the velocity gradient) onto a tensor (the Reynolds stresses). There aren't 3 directions but actually a bunch of dyadic directions. An anisotropic physical diffusion coefficient is often considered as having 3 directions because it maps a vector (the scalar gradient) onto a vector (the flux vector). But you are on the right track.... If there is directionality to this mapping, then it could be anisotropic.

Actually physical viscosity can also be generalized to an anisotropic tensor. The requirement that the eddy viscosity is rotationally invariant is historical, where it is normally treated as a scalar. But no matter how complicated your function, if your physical or eddy viscosity comes out to be a scalar then it is rotationally invariant.

 April 15, 2019, 21:50 #8 Senior Member     Paolo Lampitella Join Date: Mar 2009 Location: Italy Posts: 2,146 Blog Entries: 29 Rep Power: 38 They are two completely different concepts. Assume you have a certain reference frame, a velocity field and the resulting anisotropic eddy viscosity model, which has a certain action on it. Now, rotation invariance means that, if you just change the reference frame, all else being equal, your model action can't change. For scalar eddy viscosities it is, I guess, obvious. For anisotropic ones it means, for example, that the action on the wall normal direction is independent from it being along x, y, z or any combination of them.

 April 16, 2019, 04:09 #9 Senior Member   Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,737 Rep Power: 71 The key is that the eddy viscosity should be seen as a tensor M_eddy. This way, in matrix notation, you could suppose to find the eigenvalue and express the tensor in the main system as a diagonal matrix having the eigenvalues along the main entries. Invariance to rotation is the assumption that the matrix is diagonal in any system of reference and reduce to nothing else that eddy_viscosity*I. Therefore a single value along each direction is supposed to exist.