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plasmaQ
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Does ∇⋅τ = μΔv in the Cauchy Momentum Equation?

I find two versions of the Cauchy momentum equation (1, 2): $$ \rho \frac{D\vec{v}}{Dt}=\rho\vec{g} - \nabla{p} + \mu\nabla^2\vec{v} $$ $$ \rho \frac{D\vec{v}}{Dt}=\rho\vec{g} - \nabla{p} + \nabla \cdot \bf\tau $$ where $\tau$ is the stress tensor and $\vec{v}$ is the fluid velocity.

I'm tempted to conclude that $\mu\nabla^2\vec{v} = \nabla \cdot \bf\tau$. However, when I expand and compare terms on both sides of the equation they look widely different.

Does this equality actually hold? If so, what is the physical relationship?

plasmaQ
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