Why does eij*p*dij equal zero in Stokes flow uniqueness?

[davcrai]

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(e_ij*e_ij)dV

to

int(e_ij*a_ij)dV

Homework Equations

a_ij = -p*d_ij + 2u*e_ij

d_ij = 1 if i=j, 0 otherwise.

The Attempt at a Solution

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

int(e_ij*(e_ij*(p*d_ij + a_ij))dV

but why does e_ij*p*d_ij = 0 ?

 

[foxjwill]

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

 

[dextercioby]

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

 

[davcrai]

Thank you very much!