Why is the stress tensor simplifying like this

$$\begin{eqnarray} \nabla \cdot \boldsymbol \tau &=& 2 \mu \nabla \cdot \boldsymbol \varepsilon\\ &=& \mu \nabla \cdot \left( \nabla\mathbf{u} + (\nabla\mathbf{u}) ^\mathrm{T} \right)\\ &=& \mu \, \nabla^2 \mathbf{u} \end{eqnarray}$$

Given $$\nabla \cdot \boldsymbol u =0 \$$ How do you go from step 3 to step 4 and get rid of the transpose? Isn't the divergence operator distributive

• Where did you find this? Is there any more context? – probably_someone Oct 10 at 14:37
• What is ${\bf u}$? Is it a matrix or a vector or a matrix of vectors? when you say $\nabla {\bf u}$ do you mean curl or divergence? Can you please give more details? – Prahar Oct 10 at 14:46
• @Prahar I'm fairly certain the OP means the "gradient of a vector", which gives you a 3x3 matrix that's the transpose of the Jacobian (en.wikipedia.org/wiki/Gradient#Gradient_of_a_vector). Any other interpretation wouldn't make sense, as you can't add a vector and its transpose, but you can add two 3x3 matrices. – probably_someone Oct 10 at 14:47
• Sorry for not providing context. This is fluid Dynamics. U is the velocity field, Tau is the viscous stress tensor, mu is the viscousity en.wikipedia.org/wiki/Navier%E2%80%93Stokes_equations this simplification is for incompressible flows – ChemEng Oct 10 at 14:50

The divergence of the transpose of a vector gradient is equal to the gradient of the divergence of the vector. You can prove it quickly using some index notation:

$$\nabla \cdot (\nabla \vec{u})^T = \frac{\partial^2 u_j}{\partial x_i \partial x_j} = \frac{\partial}{\partial x_i}\frac{\partial u_j}{\partial x_j} = \nabla(\nabla \cdot \vec{u})$$

I assumed everything is in orthonormal coordinates so no need to raise or lower indices. Since the flow is incompressible by your assumption, this term cancels out, and you're only left with the divergence of the vector gradient (which gives you the vector Laplacian).

• Thanks now it makes sense – ChemEng Oct 10 at 15:44
• Happy to help! Be sure to tick one of the answers with the green checkmark so people know your question was answered. – aghostinthefigures Oct 10 at 15:49

I'm going to guess that the RHS really means $$\nabla^j ( \nabla_i u_j + \nabla_j u_i ) = \nabla_j \nabla_i u^j + \nabla^2 u_i .$$ Now, $$\nabla_j \nabla_i u^j = [ \nabla_j , \nabla_i ] u^j + \nabla_i (\nabla_j u^j) = R_{ij} u^j + \nabla_i (\nabla_j u^j) .$$ You're probably working in a Ricci flat background and the velocity vector is divergenceless so $$\nabla_j \nabla_i u^j=0$$ and therefore $$\nabla^j ( \nabla_i u_j + \nabla_j u_i ) = \nabla^2 u_i .$$

• Coming from a (classical) fluid dynamics background, this is going to be confusing. We don't use contravariant (raised indices) pretty much ever. So while this is correct, it's using notation that is outside of what fluids folks are typically trained to use. – tpg2114 Oct 10 at 15:00
• @tpg2114 - That's exactly what the answer has also done though. – Prahar Oct 10 at 16:30
• What I mean is -- few with a classical fluids background will know what most of the notation you're using is saying. In this approach, we write the right-hand side as $\partial x_j ( \partial x_j u_i + \partial x_i u_j)$. It doesn't negate what you wrote, but the traditional notation is different and we don't know what a "Ricci flat background" is or what $R_{ij}$ is or what $|\nabla_j, \nabla_i|$ means, for example. Just seeing a $\nabla$ with indices is something that wouldn't show up in a traditional fluids text. I'm not saying your wrong, just saying this is going to confuse some people. – tpg2114 Oct 10 at 16:38
• The traditional notation is what the other answer here used and it is much more recognizable for people from the fluid dynamics background. – tpg2114 Oct 10 at 16:39
• I must admit that Ricci-anything is not something I recall ever seeing in a fluids course (In GR context, yeah obviously). I think this answer is missing the target audience entirely. – Kyle Kanos Oct 10 at 16:50