- #1

Apashanka

- 429

- 15

If the velocity gradient decomposition is done by symmetric and antisymmetric parts then ##\frac{\partial v^i}{\partial x^j}=\sigma_{ij}+\omega_{ij}## where ##\sigma _{ij}=\frac{1}{2}(\frac{\partial v^i}{\partial x^j}+\frac{\partial v^j}{\partial x^i})## and ##\omega_{ij}=\frac{1}{2}(\frac{\partial v^i}{\partial x^j}-\frac{\partial v^j}{\partial x^i})## where ##\omega_{ij}## is the component of vorticity e.g ##\omega=\frac{1}{2}(\nabla × v),\omega_{ij}=-\epsilon_{ijk}\omega_k## and ##\sigma_{ij}## is the shear part as it doesn't include rotation (e.g if ##v=\Omega × x=\epsilon_{pqr}\Omega_q x_r## putting this in ##\sigma_{ij}## gives 0 and ##\omega_{ij}## gives ##\Omega_k## where ##\Omega ## is the angular velocity. here we have used that ##\epsilon_{ijk}=1## for ##i,j,k## placed in antisymmetric order.

But in the paper quoted in eq 2(a) the velocity gradient contains an extra term ##\frac{1}{3}\delta_{ij}\theta## where ##\theta## is the trace of ##\sigma_{ij}## can anyone please explain why this extra term is added??

But in the paper quoted in eq 2(a) the velocity gradient contains an extra term ##\frac{1}{3}\delta_{ij}\theta## where ##\theta## is the trace of ##\sigma_{ij}## can anyone please explain why this extra term is added??