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2 votes
1 answer
49 views

Proving $σ_{ji,j} + f_i = ρ(Dv_i/Dt) \implies (σ_{ij} - ρv_iv_j) + f_i = \partial(ρv_i)/ \partial t$

Exercise : Prove that $$σ_{ji,j} + f_i = ρ(Dv_i/Dt) \implies (σ_{ij} - ρv_iv_j)_{,j} + f_i = \partial(ρv_i)/ \partial t$$ where $σ_{ij} = σ_{ji}$ is the stress tensor. Attempt : It is : $$\...
Rebellos's user avatar
  • 21.4k
6 votes
1 answer
467 views

Net torque on a surface in Stokes flow

I'm having difficulty seeing how the net torque on a surface in Stokes flow is non-zero. For Stokes flow, we have $\nabla\cdot\sigma = 0$, where $\sigma$ is a symmetric stress tensor. The net torque ...
Lukas Bystricky's user avatar
1 vote
0 answers
358 views

Can you recover a stress tensor from a velocity field?

Consider the Cauchy momentum equation for a steady flow: $$ \rho\mathbf{u}\cdot\nabla\mathbf{u} = \rho\mathbf{g} + \nabla\cdot\mathbf{\sigma} $$ Here, $\mathbf{\sigma}$ is the stress tensor and $\...
Jonathan Michael Foonlan Tsang's user avatar
2 votes
2 answers
174 views

Why is this true: $\nabla \cdot (S\cdot \vec v )=S:(\nabla \otimes \vec v)+\vec v \cdot (\nabla \cdot S)$?

I am doing fluid mechanics and I don't understand a particular step that is being used. It is the following step which I don't understand: $\nabla \cdot (S\cdot \vec v )=S:(\nabla \otimes \vec v)+\...
user104662's user avatar