[LES] Dynamic Smagorinsky coefficient: scalar or a tensor?

Moreza7 · January 24, 2021, 12:56

Hello,
First of all, Sorry if the question is amateur I'm am very noob in Einstien notation!

As you know, SGS stresses tensor is $\tau_{ij}$ .
Also, Germano identity is a tensor $\mathcal{L}_{ij}$ .

Then the coefficient of dynamic Smagorinsky is calculated by 2 tensors (with or without Lily's correction):

$C=-\frac{1}{2\widetilde{\Delta}^2}\frac{\langle\mathcal{L}_{ij}\mathcal{M}_{ij}\rangle}{\langle\mathcal{M}_{ij}\mathcal{M}_{ij}\rangle}$

While

does not have an index, it seems that it's a scalar! I really don't understand how can it be possible that the divide of two tensors is a scalar!

This means: (suppose for example we have a 2d case)

$C=-\frac{1}{2\widetilde{\Delta}^2}\frac{\langle\mathcal{L}_{11}\mathcal{M}_{11}+\mathcal{L}_{12}\mathcal{M}_{12}+\mathcal{L}_{21}\mathcal{M}_{21}+\mathcal{L}_{22}\mathcal{M}_{22}\rangle}{\langle\mathcal{M}_{11}\mathcal{M}_{11}+\mathcal{M}_{12}\mathcal{M}_{12}+\mathcal{M}_{21}\mathcal{M}_{21}+\mathcal{M}_{22}\mathcal{M}_{22}\rangle}$

My question is, why it is not calculated like this:
$C_{x}=-\frac{1}{2\widetilde{\Delta}^2}\frac{\langle\mathcal{L}_{11}\mathcal{M}_{11}+\mathcal{L}_{12}\mathcal{M}_{12}\rangle}{\langle\mathcal{M}_{11}\mathcal{M}_{11}+\mathcal{M}_{12}\mathcal{M}_{12}\rangle}$

$C_{y}=-\frac{1}{2\widetilde{\Delta}^2}\frac{\langle\mathcal{L}_{21}\mathcal{M}_{21}+\mathcal{L}_{22}\mathcal{M}_{22}\rangle}{\langle\mathcal{M}_{21}\mathcal{M}_{21}+\mathcal{M}_{22}\mathcal{M}_{22}\rangle}$

Then we can have $\nu_{t_{x}}$ and $\nu_{t_{y}}$ which can be used in

and

momentum equations.

FMDenaro · January 24, 2021, 16:00

The fact that the eddy viscosity is a scalar function or a tensor function depends on the chosen functional relation you use for the model. It is not a result of the Germano identity and the consequent contraction.

Indeed, there are anisotropic models in tensor form, see Sagaut.

Moreza7 · January 24, 2021, 16:26

Quote:

Originally Posted by FMDenaro

The fact that the eddy viscosity is a scalar function or a tensor function depends on the chosen functional relation you use for the model. It is not a result of the Germano identity and the consequent contraction.

Indeed, there are anisotropic models in tensor form, see Sagaut.

Thanks for your reply.

My question was a little mathematical: How can it be possible to divide two tensors and the result is a scalar?
$C=-\frac{1}{2\widetilde{\Delta}^2}\frac{\langle\mathcal{L}_{ij}\mathcal{M}_{ij}\rangle}{\langle\mathcal{M}_{ij}\mathcal{M}_{ij}\rangle}$

The numerator is the product of two tensors and the denomerator is the same. But the left hand side is a scalar.

FMDenaro · January 24, 2021, 16:43

You have to consider the double scalar product between the two tensors, it is a scalar function

Moreza7 · January 24, 2021, 18:31

Quote:

Originally Posted by FMDenaro

You have to consider the double scalar product between the two tensors, it is a scalar function

Thanks a lot dear professor.

January 24, 2021, 12:56	[LES] Dynamic Smagorinsky coefficient: scalar or a tensor?	#1
Moreza7 Senior Member Join Date: Jan 2018 Posts: 121 Rep Power: 8	Hello, First of all, Sorry if the question is amateur I'm am very noob in Einstien notation! As you know, SGS stresses tensor is $\tau_{ij}$ . Also, Germano identity is a tensor $\mathcal{L}_{ij}$ . Then the coefficient of dynamic Smagorinsky is calculated by 2 tensors (with or without Lily's correction): $C=-\frac{1}{2\widetilde{\Delta}^2}\frac{\langle\mathcal{L}_{ij}\mathcal{M}_{ij}\rangle}{\langle\mathcal{M}_{ij}\mathcal{M}_{ij}\rangle}$ While does not have an index, it seems that it's a scalar! I really don't understand how can it be possible that the divide of two tensors is a scalar! This means: (suppose for example we have a 2d case) $C=-\frac{1}{2\widetilde{\Delta}^2}\frac{\langle\mathcal{L}_{11}\mathcal{M}_{11}+\mathcal{L}_{12}\mathcal{M}_{12}+\mathcal{L}_{21}\mathcal{M}_{21}+\mathcal{L}_{22}\mathcal{M}_{22}\rangle}{\langle\mathcal{M}_{11}\mathcal{M}_{11}+\mathcal{M}_{12}\mathcal{M}_{12}+\mathcal{M}_{21}\mathcal{M}_{21}+\mathcal{M}_{22}\mathcal{M}_{22}\rangle}$ My question is, why it is not calculated like this: $C_{x}=-\frac{1}{2\widetilde{\Delta}^2}\frac{\langle\mathcal{L}_{11}\mathcal{M}_{11}+\mathcal{L}_{12}\mathcal{M}_{12}\rangle}{\langle\mathcal{M}_{11}\mathcal{M}_{11}+\mathcal{M}_{12}\mathcal{M}_{12}\rangle}$ $C_{y}=-\frac{1}{2\widetilde{\Delta}^2}\frac{\langle\mathcal{L}_{21}\mathcal{M}_{21}+\mathcal{L}_{22}\mathcal{M}_{22}\rangle}{\langle\mathcal{M}_{21}\mathcal{M}_{21}+\mathcal{M}_{22}\mathcal{M}_{22}\rangle}$ Then we can have $\nu_{t_{x}}$ and $\nu_{t_{y}}$ which can be used in and momentum equations.

January 24, 2021, 16:00		#2
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,896 Rep Power: 73	The fact that the eddy viscosity is a scalar function or a tensor function depends on the chosen functional relation you use for the model. It is not a result of the Germano identity and the consequent contraction. Indeed, there are anisotropic models in tensor form, see Sagaut. Moreza7 likes this.

January 24, 2021, 16:43		#4
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,896 Rep Power: 73	You have to consider the double scalar product between the two tensors, it is a scalar function Moreza7 likes this.

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