We derive a general form of the viscous tensor and show that in two spatial dimensions it can have an antisymmetric part. For electron transport, the Hall viscosity emerges.
Properties of the viscous tensor
A central component of the Navier-Stokes equation (including the relaxation term)
\[\rho\frac{\mathrm d}{\mathrm dt}v\equiv
\rho
\left[
\partial_t
+ \left(v\cdot\nabla\right)
\right] v
=
\nabla\cdot\Pi-\nabla P+\rho a
+\left.\partial_t(\rho v)\right|_\text{Streu}\]
is the viscous tensor $\Pi$. (As usual, $P$ is the pressure, $\rho a$ represents external forces like gravity or the Lorentz force.) This tensor parametrizes internal shear forces and compression due to particle velocity fluctuations. In the absence of momentum-relaxing scattering, it is the single source of dissipation in the system being described by the equation above. We assign the following properties to it:
-
Compression and shear forces only: viscosity is unrelated to local rotations.
-
Small changes, i.e., linear response: $\Pi=\Pi(J(v))$ is a linear function of the spatial derivatives forming the Jacobi matrix $J(v)=(\partial v_i/\partial x_j)$ of the velocity field $v=v(x,t)$.
-
Homogeneous fluid: $\Pi$ must be invariant under global rotations.
Property 1 tells us that $\Pi$ must be symmetric, $\Pi_{ij}=\Pi_{ji}$. One might be tempted to conclude that, because there is no antisymmetric part in $\Pi$, there can’t be something like a “Hall viscosity” by symmetry. More than one textbook does or at least suggests that, however we shall see below that this is not strictly true.
Property 2 allows us to express $\Pi$ as
\[\Pi_{ij}
=\eta_{ij}\cdot J(v)
=\sum_{kl}\eta_{ijkl}J_{kl}(v)
=
\sum_{kl}\eta_{ijkl}\frac{\partial v_k}{\partial x_l},\]
introducing the viscosity $\eta=(\eta_{ijkl})$ as a tensor of rank 4. At first sight it seems that this beast has $2\times2\times2\times2=16$ components, however it is symmetric under $(ij)\leftrightarrow(ji)$ (from property 1) and $(kl)\leftrightarrow(lk)$ (again, no local rotations, therefore antisymetric combinations of velocity gradients vanish), reducing the number of independent components to nine. Still a lot, but we can reduce this further.
Property 3 implies
$
U^T\Pi\left(J(v)\right)U
=
\Pi\left(U^TJ(v)U\right)
$
for an arbitrary orthogonal matrix $U$.
A unitary basis for rank-2 tensors in two dimensions are the Pauli matrices $\sigma^\alpha$, $\alpha=0\ldots3$. For rank-4 tensors, the Kronecker product basis $\{\sigma^\alpha\otimes\sigma^\beta:\alpha,\beta=0\ldots3\}$ would be appropriate. In this basis, the viscosity can be decomposed into
\[\eta=\sum_{\alpha\beta}\eta_{\alpha\beta}\sigma^\alpha\otimes\sigma^\beta.\]
A rotation in two dimensions by an angle $\pi/2$ can be parametrized by the matrix ${\rm i}\sigma^2$. From property 3 above we then obtain with $U(\pi/2)=\sigma^2\otimes\sigma^2$
\[\begin{eqnarray*}
U^T\eta U
&\stackrel!=&
\eta
\\
\Rightarrow\left(\sigma^2\otimes\sigma^2\right)^T\eta\left(\sigma^2\otimes\sigma^2\right)
&=&
\sum_{\alpha\beta}\eta_{\alpha\beta}
\left(\sigma^2\otimes\sigma^2\right)^T
\left(\sigma^\alpha\otimes\sigma^\beta\right)
\left(\sigma^2\otimes\sigma^2\right)
\\
&=&
\sum_{\alpha\beta}\eta_{\alpha\beta}
\left((\sigma^2)^T\sigma^\alpha\sigma^2\right)\otimes
\left((\sigma^2)^T\sigma^\beta\sigma^2\right)
\\
&\stackrel!=&
\sum_{\alpha\beta}\eta_{\alpha\beta}
\sigma^\alpha\otimes\sigma^\beta.
\end{eqnarray*}\]
Including the $(ij)\leftrightarrow(ji)$ and $(kl)\leftrightarrow(lk)$ symmetries of $\eta$ this leave us with the pairs $(\alpha,\beta)=(0,0),(1,1),(1,3),(3,1)$, and $(3,3)$ in the summation above.
While $\eta$ must be symmetric in pairs, we can now split $\eta=\eta^{\rm s}+\eta^{\rm a}$ into a symmetric part with \(\eta^{\rm s}_{ijkl}=\eta^{\rm s}_{klij}\) and an antisymmetric part with \(\eta^{\rm a}_{ijkl}=-\eta^{\rm a}_{klij}\) under $(ijkl)\leftrightarrow(klij)$. This eventually reduces the independent parameters of $\eta$ to three, namely $\eta_{11}$ and $\eta_{00}$ for the symmetric viscosity, and $\eta_{13}$ for the antisymmetric viscosity. We obtain
\[\begin{eqnarray*}
\eta^{\rm s}
&=&
\eta_{00}\sigma^0\otimes\sigma^0
+\eta_{11}\left(
\sigma^1\otimes\sigma^1+\sigma^3\otimes\sigma^3
\right),
\\
\eta^{\rm a}
&=&
\eta_{13}\left(
\sigma^3\otimes\sigma^1-\sigma^1\otimes\sigma^3
\right).
\end{eqnarray*}\]
Associating the three parameters of the viscosity tensor with common literature notation, we have
\[\begin{eqnarray*}
\Pi_{ij}&=&\Pi_{ij}^{\rm s}+\Pi_{ij}^{\rm a},
\\
\Pi_{ij}^{\rm s}
&=&
\eta\left(
\frac{\partial v_i}{\partial x_j}
+\frac{\partial v_j}{\partial x_i}
\right)
+\eta'\delta_{ij}\nabla\cdot v,
\\
\Pi_{ij}^{\rm a}
&=&
\frac{\eta_{\rm H}}2\sum_k
\left[
\epsilon_{ik}\left(\frac{\partial v_k}{\partial x_j}
+\frac{\partial v_j}{\partial x_k}\right)
+\epsilon_{jk}\left(\frac{\partial v_k}{\partial x_i}
+\frac{\partial v_i}{\partial x_k}\right)
\right],
\end{eqnarray*}\]
more explicitly
\[\begin{eqnarray*}
\Pi^\text s
&=&
\eta
\left(
\begin{array}{cc}
2\frac{\partial v_1}{\partial x_1}
&
\frac{\partial v_2}{\partial x_1}
+\frac{\partial v_1}{\partial x_2}
\\
\frac{\partial v_2}{\partial x_1}
+\frac{\partial v_1}{\partial x_2}
&
2\frac{\partial v_2}{\partial x_2}
\end{array}
\right)
+
\eta'
\left(
\begin{array}{cc}
\frac{\partial v_1}{\partial x_1}+\frac{\partial v_2}{\partial x_2}
&
0
\\
0
&
\frac{\partial v_1}{\partial x_1}+\frac{\partial v_2}{\partial x_2}
\end{array}
\right),
\\
\Pi^\text a
&=&
\eta_\text H
\left(
\begin{array}{cc}
\frac{\partial v_2}{\partial x_1}
+\frac{\partial v_1}{\partial x_2}
&
\frac{\partial v_2}{\partial x_2}
-\frac{\partial v_1}{\partial x_1}
\\
\frac{\partial v_2}{\partial x_2}
-\frac{\partial v_1}{\partial x_1}
&
-\left(
\frac{\partial v_2}{\partial x_1}
+\frac{\partial v_1}{\partial x_2}
\right)
\end{array}
\right)
\end{eqnarray*}\]
with the standard viscosity $\eta=\eta_{11}$, the second viscosity $\eta’=\eta_{00}-\eta_{11}$, and the Hall viscosity $\eta_{\rm H}=\eta_{13}$.
We note that this works in two spatial dimensions only. In three dimensions, one is tempted to assume that the 81 components of $\eta_{ijkl}$ (36 parameters left after symmetry reduction from criterion 1) allow for even more choice. However one can show that in three dimensions isotropy implies that $\eta^{\rm a}=0$.
Readings: