Uniqueness of Stokes Flow: Investigating Strain and Stress Tensors

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SUMMARY

The discussion focuses on the proof of uniqueness for Stokes flow in fluid mechanics, specifically addressing the relationship between the strain tensor (e) and the stress tensor (a). The transformation from the integral involving the strain tensor to one involving the stress tensor is highlighted, with the equation aij = -p*dij + 2u*eij being central to the discussion. The traceless nature of the strain tensor is emphasized, explaining why the term eij*p*dij equals zero due to the pressure term being the negative trace of the stress tensor.

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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 ?
 
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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!
 

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