Re: deriving the Navier-Stokes viscosity term

Dr Will · June 4, 2023, 14:25

I have 3 references -
Sonin - Equations of Motion for Viscous Fluids - https://web.mit.edu/2.25/www/pdf/viscous_flow_eqn.pdf
Chakraborty - Navier Stokes continued
https://www.youtube.com/watch?v=0ndp5yAOklc - starting at 8:00
Fitzpatrick - Isotropic tensors
https://farside.ph.utexas.edu/teachi...d/node252.html

I'm trying to find an understandable derivation of the viscosity term of the N-S eqs. Sonin has a first principles analysis which is straightforward but a little too complicated ...

Chakraborty has what looks like a simpler approach using properties of an isometric rank 4 tensor. It's only simple, however, if the proof that an isometric rank 4 tensor has the required form, which he sort of proves, but ... does he?

Fitzpatrick has rigorous proof that an isometric rank 4 tensor has the required form. It's complicated.

So, my questions are, is Chakraborty's method rigorous or heuristic, and, what is the easiest way to derive the N-S viscosity term?

FMDenaro · June 4, 2023, 15:00

Quote:

Originally Posted by Dr Will

I have 3 references -
Sonin - Equations of Motion for Viscous Fluids - https://web.mit.edu/2.25/www/pdf/viscous_flow_eqn.pdf
Chakraborty - Navier Stokes continued
https://www.youtube.com/watch?v=0ndp5yAOklc - starting at 8:00
Fitzpatrick - Isotropic tensors
https://farside.ph.utexas.edu/teachi...d/node252.html

I'm trying to find an understandable derivation of the viscosity term of the N-S eqs. Sonin has a first principles analysis which is straightforward but a little too complicated ...

Chakraborty has what looks like a simpler approach using properties of an isometric rank 4 tensor. It's only simple, however, if the proof that an isometric rank 4 tensor has the required form, which he sort of proves, but ... does he?

Fitzpatrick has rigorous proof that an isometric rank 4 tensor has the required form. It's complicated.

So, my questions are, is Chakraborty's method rigorous or heuristic, and, what is the easiest way to derive the N-S viscosity term?

Why the isotropic tensor, are you looking for the bulk viscosity?
All is concerned to the viscosity is in the Newtonian model.

Dr Will · June 4, 2023, 15:14

Quote:

Originally Posted by FMDenaro

Why the isotropic tensor, are you looking for the bulk viscosity?
All is concerned to the viscosity is in the Newtonian model.

The relationship of the stress tensor to the strain rate tensor is specified by an isotropic rank 4 tensor.

FMDenaro · June 4, 2023, 15:26

Why don’t you start by Batchelor textbook?

Dr Will · June 4, 2023, 15:59

Quote:

Originally Posted by FMDenaro

Why don’t you start by Batchelor textbook?

I downloaded it (from pdfcoffee). It quotes the isotropic tensor theorem and uses that. Then there is an alternate derivation which is opaque to me.

FMDenaro · June 5, 2023, 05:38

Sec 3.3

T = A: grad V

Accepting that the fourth-order viscosity tensor A is isotropic for statistical reasons, you have only entries in the main diagonal.
The simmetry is applied for i-j indices.

Dr Will · June 5, 2023, 13:04

Unless I'm mistaken, there is an easy proof that a rank 4 isotropic tensor has the required form ....

Let V = R2 to make it simple.

By rotating the axes 90 deg., f1 = e2, and f2 = -e1 it's easy to show the non-zero components of an isotropic rank 4 isotropic tensor T are [t1111 0 0 t1122; 0 t1212 t1221 0; 0 t1221 t1212 0; t1122 0 0 t111].

Why? t1111 corresponds to e1xe1xe1xe1 in e basis coordinates and f1xf1xf1xf1 (= e2xe2xe2xe2) in f basis coordinates, so t1111 = t2222, and similarly for components with 2 1s and 2 2s. If a component has 1 2 and 3 1s, for example t2111, it corresponds to e2xe1xe1xe2 in e basis coordinates and f2xf1xf1xf1 (= (-e1)xe2xe2xe2) in f basis coordinates and is hence 0. Similarly for a component with 1 1 and 3 2s.

LuckyTran · June 6, 2023, 07:58

The point of introducing the isotropic rank 4 tensor is to show that the tensor can be no more complicated than such. That is, stress-strain relations and 4th order isotropic tensors are an extremely constrained class, there's only one form they can take. It's more a case of: what is the worst result possible?

It's not about deriving N-S from first principles. Newton's law is a law, stress is proportional to strain via an isotropic scalar.

June 4, 2023, 14:25	Re: deriving the Navier-Stokes viscosity term	#1
Dr Will New Member William Flannery Join Date: Jun 2023 Posts: 6 Rep Power: 3	I have 3 references - Sonin - Equations of Motion for Viscous Fluids - https://web.mit.edu/2.25/www/pdf/viscous_flow_eqn.pdf Chakraborty - Navier Stokes continued https://www.youtube.com/watch?v=0ndp5yAOklc - starting at 8:00 Fitzpatrick - Isotropic tensors https://farside.ph.utexas.edu/teachi...d/node252.html I'm trying to find an understandable derivation of the viscosity term of the N-S eqs. Sonin has a first principles analysis which is straightforward but a little too complicated ... Chakraborty has what looks like a simpler approach using properties of an isometric rank 4 tensor. It's only simple, however, if the proof that an isometric rank 4 tensor has the required form, which he sort of proves, but ... does he? Fitzpatrick has rigorous proof that an isometric rank 4 tensor has the required form. It's complicated. So, my questions are, is Chakraborty's method rigorous or heuristic, and, what is the easiest way to derive the N-S viscosity term?

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June 4, 2023, 15:26		#4
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,882 Rep Power: 73	Why don’t you start by Batchelor textbook?

June 5, 2023, 05:38		#6
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,882 Rep Power: 73	Sec 3.3 T = A: grad V Accepting that the fourth-order viscosity tensor A is isotropic for statistical reasons, you have only entries in the main diagonal. The simmetry is applied for i-j indices.

June 5, 2023, 13:04		#7
Dr Will New Member William Flannery Join Date: Jun 2023 Posts: 6 Rep Power: 3	Unless I'm mistaken, there is an easy proof that a rank 4 isotropic tensor has the required form .... Let V = R2 to make it simple. By rotating the axes 90 deg., f1 = e2, and f2 = -e1 it's easy to show the non-zero components of an isotropic rank 4 isotropic tensor T are [t1111 0 0 t1122; 0 t1212 t1221 0; 0 t1221 t1212 0; t1122 0 0 t111]. Why? t1111 corresponds to e1xe1xe1xe1 in e basis coordinates and f1xf1xf1xf1 (= e2xe2xe2xe2) in f basis coordinates, so t1111 = t2222, and similarly for components with 2 1s and 2 2s. If a component has 1 2 and 3 1s, for example t2111, it corresponds to e2xe1xe1xe2 in e basis coordinates and f2xf1xf1xf1 (= (-e1)xe2xe2xe2) in f basis coordinates and is hence 0. Similarly for a component with 1 1 and 3 2s.

June 6, 2023, 07:58		#8
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,753 Rep Power: 66	The point of introducing the isotropic rank 4 tensor is to show that the tensor can be no more complicated than such. That is, stress-strain relations and 4th order isotropic tensors are an extremely constrained class, there's only one form they can take. It's more a case of: what is the worst result possible? It's not about deriving N-S from first principles. Newton's law is a law, stress is proportional to strain via an isotropic scalar.