= Viscous stress tensor
{title2=$\boldsymbol\sigma_{\mathrm{visc}}=2\mu D$}
= Viscous stress
{synonym}
The <viscous stress tensor> is the part of the <Cauchy stress tensor> associated with deformation rates. For an isotropic <Newtonian fluid> in <incompressible flow>,
$$
\boldsymbol\sigma=-pI+\boldsymbol\sigma_{\mathrm{visc}},\qquad
\boldsymbol\sigma_{\mathrm{visc}}=2\mu D,\qquad
D=\frac12\left[\nabla\mathbf u+(\nabla\mathbf u)^T\right],\qquad
\operatorname{tr}D=\nabla\cdot\mathbf u=0.
$$
Here $p$ is <pressure>, $I$ is the <identity matrix>, $\mu$ is <dynamic viscosity>, and $D$ is the <strain-rate tensor>. The convention is $(\nabla\mathbf u)_{ij}=\partial_j u_i$; $T$ denotes <transpose> and $\operatorname{tr}$ denotes <trace>. Across a plane with unit <normal vector> $\mathbf n$, the viscous <traction> is $\boldsymbol\sigma_{\mathrm{visc}}\mathbf n$. The <tensor> makes this rule work for any plane orientation. https://www.damtp.cam.ac.uk/user/tong/fluids/fluids.pdf[David Tong, Fluid Mechanics, section 3.1.1, equations (3.5)–(3.6)]{ref}
In the worked <Couette flow>, $\mathbf u=(Uy/h,0,0)$, so
$$
D=\frac{U}{2h}\begin{pmatrix}0&1&0\\1&0&0\\0&0&0\end{pmatrix},\qquad
\boldsymbol\sigma_{\mathrm{visc}}
=\begin{pmatrix}0&2&0\\2&0&0\\0&0&0\end{pmatrix}\,\mathrm{Pa}.
$$
In particular, $D_{xy}=10\,\mathrm{s^{-1}}$ while the <shear rate> is $20\,\mathrm{s^{-1}}$; the factor $2$ in $2\mu D$ recovers $\tau_{xy}=2\,\mathrm{Pa}$. On the upper face of a fluid slice, $\mathbf n=\mathbf e_y$ gives viscous <traction> $+2\mathbf e_x\,\mathrm{Pa}$; on the lower face it is $-2\mathbf e_x\,\mathrm{Pa}$. Constant <stress> can therefore transmit a <force> between the plates even though its <divergence>, and the net viscous <force> on each slice, vanish.
Compare planar extension $\mathbf u=(ax,-ay,0)$ with <solid-body rotation> $\mathbf u=(-\Omega y,\Omega x,0)$:
$$
D_{\mathrm{extension}}=\begin{pmatrix}a&0&0\\0&-a&0\\0&0&0\end{pmatrix},\qquad
\nabla\mathbf u_{\mathrm{rotation}}=\begin{pmatrix}0&-\Omega&0\\\Omega&0&0\\0&0&0\end{pmatrix},\qquad
D_{\mathrm{rotation}}=0.
$$
Extension changes lengths and produces viscous <normal stresses> $\pm2\mu a$. Rotation preserves distances and gives zero <viscous stress>, despite its nonzero <velocity gradient> and <vorticity> $2\Omega\mathbf e_z$. For $a=\Omega=1\,\mathrm{s^{-1}}$ and $\mu=0.10\,\mathrm{Pa\,s}$, extension gives $\sigma_{\mathrm{visc},xx}=0.20\,\mathrm{Pa}$ and $\sigma_{\mathrm{visc},yy}=-0.20\,\mathrm{Pa}$; rotation gives zero. A rotating fluid may still have a <pressure gradient> supplying its centripetal <acceleration>.
\Image[/viscosity-strain-versus-rotation.png]
{title=Deformation produces Newtonian viscous stress; rigid rotation does not}
{description=Blue arrows show instantaneous velocity. The dashed square is a material patch at $t=0$; its orange outline at $t=0.65\,\mathrm s$ is obtained from the exact trajectories of planar extension and solid-body rotation, with $a=\Omega=1\,\mathrm{s^{-1}}$. The marked corner makes rotation visible. Both motions preserve area, but only extension changes the patch's shape.}
For an isotropic <Newtonian fluid> in three-dimensional <compressible flow>, separate shape change from <volume> change:
$$
\boldsymbol\sigma_{\mathrm{visc}}
=2\mu\left(D-\frac{\nabla\cdot\mathbf u}{3}I\right)
+\zeta(\nabla\cdot\mathbf u)I.
$$
The first term is <deviatoric stress>, with zero <trace>; $\zeta$ is <volume viscosity>, also called <bulk viscosity>. For uniform expansion $\mathbf u=b\mathbf x$, $D=bI$ and $\nabla\cdot\mathbf u=3b$, so the shear contribution vanishes and the viscous <stress> is $3\zeta bI$. Thus <dynamic viscosity> and <volume viscosity> describe distinct responses, and <incompressible flow> removes only the volume-change term. https://notes.shreyasmandre.com/fluid-dynamics-notes/Ch2.html[Shreyas Mandre, Fluid Dynamics Lecture Notes, section 2.11.3, equation (2.49)]{ref}
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