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

Question

I find two versions of the Cauchy momentum equation (1, 2):
$$
rho frac{Dvec{v}}{Dt}=rhovec{g} - nabla{p} + munabla^2vec{v}
$$
$$
rho frac{Dvec{v}}{Dt}=rhovec{g} - nabla{p} + nabla cdot bftau
$$
where $tau$ is the stress tensor and $vec{v}$ is the fluid velocity.
I'm tempted to conclude that $munabla^2vec{v} = nabla cdot bftau$. 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?

Amey Joshi · Accepted Answer

The stress tensor is related to velocity in a Newtonian fluid as
begin{equation}tag{1}
tau_{ij} = muleft(frac{partial v_i}{partial x_j} + frac{partial v_j}{partial x_i}right)
end{equation}
If we assume that the viscosity of the fluid is independent of its position then the divergence of stress is
begin{equation}tag{2}
frac{partialtau_{ij}}{partial x_j} = mufrac{partial^2 v_i}{partial x_j^2} +
mufrac{partial}{partial x_i}left(frac{partial v_j}{partial x_j}right).
end{equation}
If we further assume that the fluid is incompressible then $nablacdotvec{v} = 0$ so that equation (2) simplifies to
begin{equation}tag{3}
frac{partialtau_{ij}}{partial x_j} = mufrac{partial^2 v_i}{partial x_j^2}.
end{equation}
It is under these assumptions (marked bold) that $nablacdottau = munabla^2vec{v}$.
You may find the third chapter of Batchelor's book interesting if you wish to know more about this topic.

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

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