Uniqueness of Stokes Flow: Investigating Strain and Stress Tensors

  • Thread starter Thread starter davcrai
  • Start date Start date
Click For Summary

Homework Help Overview

The discussion revolves around the uniqueness of Stokes flow in fluid mechanics, specifically focusing on the relationship between strain and stress tensors within a proof context. The original poster is exploring a mathematical transformation involving these tensors as part of their homework in a fluids class.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand the substitution of the strain tensor with the stress tensor in an integral form. They question the reasoning behind a specific term equating to zero, indicating a need for clarification on the properties of the tensors involved.

Discussion Status

Some participants have provided insights regarding the properties of the strain tensor and its traceless nature, suggesting that this is relevant to the original poster's question. The discussion appears to be progressing with some clarification being offered, though no consensus has been reached on the broader implications of the proof.

Contextual Notes

Participants are navigating the complexities of tensor properties and their implications in fluid mechanics, with specific attention to the assumptions underlying the mathematical transformations being discussed.

davcrai
Messages
13
Reaction score
0

Homework Statement


Going through a proof of uniqueness for stokes flow for a fluids class I'm taking. Part of it involves replacing the strain tensor (e) with the stress tensor (a), ie going from

(2u)*int(eij*eij)dV

to

int(eij*aij)dV


Homework Equations



aij = -p*dij + 2u*eij

dij = 1 if i=j, 0 otherwise.

The Attempt at a Solution



I can see you simply substitute for 2u*eij to the integral as follows,

int(eij*(eij*(p*dij + aij))dV

but why does eij*p*dij = 0 ?
 
Physics news on Phys.org


Please try reposting this in the physics part of the forum.
 


Because the strain tensor e is traceless. The pressure term is exactly minus the trace of the stress tensor a.
 
Last edited:


Thank you very much!
 

Similar threads

  • · Replies 18 ·
Replies
18
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 60 ·
3
Replies
60
Views
7K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 15 ·
Replies
15
Views
2K