Horizontal Mixing
Deviatory Stress Tensor
<wikitex>The horizontal components of the divergence of the stress tensor ( Wajsowicz, 1993) in nondimesional, orthogonal curvilinear coordinates ($\xi$, $\eta$, $s$) with dimensional, spatially-varying metric factors ($\frac{1}{m}$, $\frac{1}{n}$, $H_{z}$) and velocity components ($u$, $v$, $\omega H_{z}$) are given by:
$$ \eqalign {
F^{u} \equiv \widehat{\xi}\cdot\left(\nabla\cdot\vec{\sigma}\right) =
\frac{mn}{H_{z}} \Biggl[ & {\pder{}{\xi}} \Biggl( \frac{H_{z}{\sigma}_{\xi\xi}} {n} \Biggr) +
{\pder{}{\eta}} \Biggl( \frac{H_{z}{\sigma}_{\xi\eta}}{m} \Biggr) +
{\pder{}{s}} \Biggl( \frac{{\sigma}_{\xi s}}{mn} \Biggr) + \cr
&H_{z}{\sigma}_{\xi\eta} {\pder{}{\eta}} \left( \frac{1}{m}\right) -
H_{z}{\sigma}_{\eta\eta} {\pder{}{\xi}} \left( \frac{1}{n}\right) -
\frac{1}{n} {\sigma}_{ss}{\pder{H_{z}}{\xi}} \Biggr] \cr} \eqno{(1)} $$
$$ \eqalign {
F^{v} \equiv \widehat{\eta}\cdot\left(\nabla\cdot\vec{\sigma}\right) =
\frac{mn}{H_{z}} \Biggl[ & {\pder{}{\xi}} \Biggl( \frac{H_{z}{\sigma}_{\eta\xi}} {n} \Biggr) +
{\pder{}{\eta}} \Biggl( \frac{H_{z}{\sigma}_{\eta\eta}}{m} \Biggr) +
{\pder{}{s}} \Biggl( \frac{{\sigma}_{\eta s}}{mn} \Biggr) + \cr
&H_{z}{\sigma}_{\eta\xi} {\pder{}{\xi}} \left( \frac{1}{n} \right) -
H_{z}{\sigma}_{\xi\xi} {\pder{}{\eta}} \left( \frac{1}{m} \right) -
\frac{1}{m}{\sigma}_{ss} {\pder{H_{z}}{\eta}} \Biggr] \cr} \eqno{(2)} $$
where
$$ \eqalign {
{\sigma}_{\xi\xi} &= \left( A_{M} + \nu \right) e_{\xi\xi} + \left( \nu - A_{M}\right) e_{\eta\eta}, \cr
\noalign{\smallskip}
{\sigma}_{\eta\eta} &= \left( \nu - A_{M} \right) e_{\xi\xi} + \left( A_{M} + \nu\right) e_{\eta\eta}, \cr
\noalign{\smallskip}
{\sigma}_{ss} &= 2\,\nu\,e_{ss}, \cr
\noalign{\smallskip}
{\sigma}_{\xi\eta} &= {\sigma}_{\eta\xi} = 2\,A_{M}\,e_{\xi\eta}, \cr
\noalign{\smallskip}
{\sigma}_{\xi s} &= 2\,K_{M}\,e_{\xi s}, \cr
\noalign{\smallskip}
{\sigma}_{\eta s} &= 2\,K_{M}\,e_{\eta s}, \cr} \eqno{(3)} $$
and the strain field is:
$$ \eqalign {
e_{\xi\xi} &= m {\pder{u}{\xi}} + mnv {\pder{}{\eta}} \left( \frac{1}{m} \right), \cr
\noalign{\smallskip}
e_{\eta\eta} &= n\;{\pder{v}{\eta}} + mnu {\pder{}{\xi}} \left( \frac{1}{n} \right), \cr
\noalign{\smallskip}
e_{ss} &= \frac{1}{H_{z}} {\pder{\left( \omega H_{z} \right) }{s}} +
\frac{m}{H_{z}} u {\pder{H_{z}}{\xi}} +
\frac{n}{H_{z}} v {\pder{H_{z}}{\eta}}, \cr
\noalign{\smallskip}
2\,e_{\xi\eta} &= \frac{m}{n} {\pder{\left( nv \right) }{\xi}} +
\frac{n}{m} {\pder{\left( mu \right) }{\eta}}, \cr
\noalign{\smallskip}
2\,e_{\xi s} &= \frac{1}{mH_{z}} {\pder{\left( mu \right) }{s}} +
m H_{z} {\pder{\omega}{\xi}}, \cr
\noalign{\smallskip}
2\,e_{\eta s} &= \frac{1}{nH_{z}} \; {\pder{\left(nv\right)}{s}} \;+
n\;H_{z} {\pder{\omega}{\eta}}. \cr} \eqno{(4)} $$
Here, $A_{M}(\xi,\eta)$ and $K_{M}(\xi,\eta,s)$ are the spatially varying horizontal and vertical viscosity coefficients, respectively, and $\nu$ is another (very small, often neglected) horizontal viscosity coefficient. Notice that because of the generalized terrain-following vertical coordinates of ROMS, we need to transform the horizontal partial derivatives from constant z-surfaces to constant s-surfaces. And the vertical metric or level thickness is the Jacobian of the transformation, $H_{z}={\pder{z}{s}}$. Also in these models, the vertical velocity is computed as $\frac{\omega H_{z}}{mn}$ and has units of $\hbox{m}^3/\hbox{s}$.</wikitex>
Transverse Stress Tensor
<wikitex>Assuming transverse isotropy, as in Sadourny and Maynard (1997) and Griffies and Hallberg (2000), the deviatoric stress tensor can be split into vertical and horizontal sub-tensors. The horizontal (or transverse) sub-tensor is symmetric, it has a null trace, and it possesses axial symmetry in the local vertical direction. Then, transverse stress tensor can be derived from (1) and (2) yielding
$$ \eqalign {
H_{z}F^{u} &= {n^2}m {\partial\over\partial\xi}\left(\frac{H_{z}F^{u\xi}}{n}\right) +
{m^2}n {\partial\over\partial\eta}\left(\frac{H_{z}F^{u\eta}}{m}\right) \cr
\noalign{\smallskip}
H_{z}F^{v} &= {n^2}m {\partial\over\partial\xi}\left(\frac{H_{z}F^{v\xi}}{n}\right) +
{m^2}n {\partial\over\partial\eta}\left(\frac{H_{z}F^{v\eta}}{m}\right) \cr}
\eqno{(5)} $$
where
$$ \eqalign {
F^{u\xi} &= \frac{1}{n} \;A_{M}\left[
\frac{m}{n} {\pder{\left( nu \right) }{\xi}} \;-
\frac{n}{m} {\pder{\left( mv \right) }{\eta}} \right], \cr
\noalign{\smallskip}
F^{u\eta} &= \frac{1}{m} A_{M}\left[
\frac{n}{m} {\pder{\left( mu \right) }{\eta}} +
\frac{m}{n} {\pder{\left( nv \right) }{\xi}} \;\right], \cr
\noalign{\medskip}
F^{v\xi} &= \frac{1}{n} \;A_{M}\left[
\frac{m}{n} {\pder{\left( nv \right) }{\xi}} \;+
\frac{n}{m} {\pder{\left( mu \right) }{\eta}} \right], \cr
\noalign{\smallskip}
F^{v\eta} &= \frac{1}{m} A_{M}\left[
\frac{n}{m} {\pder{\left( mv \right) }{\eta}} -
\frac{m}{n} {\pder{\left( nu \right) }{\xi}} \;\right]. \cr}
\eqno{(6)} $$
Notice the flux form of (5) and the symmetry between the $F^{u\xi}$ and $F^{v\eta}$ terms which are defined at density points on a C-grid. Similarly, the $F^{u\eta}$ and $F^{v\xi}$ terms are symmetric and defined at vorticity points. These staggering positions are optimal for the discretization of the tensor; it has no computational modes and satisfy first-moment conservation.
The biharmonic friction operator can be computed by applying twice the tensor operator (5), but with the squared root of the biharmonic viscosity coefficient (Griffies and Hallberg, 2000). For simplicity and momentum balance, the thickness $H_{z}$ appears only when computing the second harmonic operator as in Griffies and Hallberg (2000).</wikitex>
Rotated Transverse Stress Tensor
<wikitex>In some applications with tall and steep topography, it will be advantageous to reduce substantially the contribution of the stress tensor (5) to the vertical mixing when operating along constant $s$-surfaces. The transverse stress tensor rotated along geopotentials (constant depth) is, then, given by
$$ \eqalign {
H_{z}R^{u} &= {n^2}m {\pder{}{\xi}} \Biggl( \frac{H_{z}R^{u\xi}} {n} \Biggr) +
{m^2}n {\pder{}{\eta}} \Biggl( \frac{H_{z}R^{u\eta}}{m} \Biggr) +
{\pder{}{s}} \Biggl( R^{us} \Biggr) \cr
\noalign{\smallskip}
H_{z}R^{v} &= {n^2}m {\pder{}{\xi}} \Biggl( \frac{H_{z}R^{v\xi}} {n} \Biggr) +
{m^2}n {\pder{}{\eta}} \Biggl( \frac{H_{z}R^{v\eta}}{m} \Biggr) +
{\pder{}{s}} \Biggl( R^{vs} \Biggr) \cr}
\eqno{(7)} $$
where
$$ \eqalign {
R^{u\xi} = &\frac{1}{n}\; A_{M} \left[
\frac{1}{n}\;\left( m {\pder{\left(nu\right)}{\xi}} -
m {\pder{z}{\xi}} \frac{1}{H_{z}}
{\pder{\left( nu \right) }{s}} \right) -
\frac{1}{m} \left( n {\pder{\left( mv \right) }{\eta}} -
n {\pder{z}{\eta}} \frac{1}{H_{z}}
{\pder{\left( mv \right) }{s}}\right)
\right], \cr
\noalign{\medskip}
R^{u\eta} = &\frac{1}{m} A_{M} \left[
\frac{1}{m} \left( n {\pder{\left( mu \right) }{\eta}} -
n {\pder{z}{\eta}} \frac{1}{H_{z}}
{\pder{\left( mu \right) }{s}} \right) +
\frac{1}{n}\;\left( m {\pder{\left( nv \right) }{\xi}} -
m {\pder{z}{\xi}} \frac{1}{H_{z}}
{\pder{\left( nv \right) }{s}} \right)
\right], \cr
\noalign{\medskip}
R^{us} = &m {\pder{z}{\xi}} A_{M} \left[
\frac{1}{n}\;\left( m {\pder{z}{\xi}} \frac{1}{H_{z}}
{\pder{\left( nu \right) }{s}} -
m {\pder{\left( nu \right) }{\xi}} \right) -
\frac{1}{m} \left( n {\pder{z}{\eta}} \frac{1}{H_{z}}
{\pder{\left( mv \right) }{s}} -
n {\pder{\left( mv \right) }{\eta}} \right)
\right] +\cr
&n\; {\pder{z}{\eta}} A_{M} \left[
\frac{1}{m} \left( n {\pder{z}{\eta}} \frac{1}{H_{z}}
{\pder{\left( mu \right) }{s}} -
n {\pder{\left( mu \right) }{\eta}} \right) +
\frac{1}{n}\;\left( m {\pder{z}{\xi}} \frac{1}{H_{z}}
{\pder{\left( nv \right) }{s}} -
m {\pder{\left( nv \right) }{\xi}} \right)
\right], \cr
\noalign{\bigskip}
R^{v\xi} = &\frac{1}{n}\;A_{M} \left[
\frac{1}{n}\;\left( m {\pder{\left( nv \right) }{\xi}} -
m {\pder{z}{\xi}} \frac{1}{H_{z}}
{\pder{\left( nv \right) }{s}} \right) +
\frac{1}{m} \left( n {\pder{\left( mu \right) }{\eta}}-
n {\pder{z}{\eta}} \frac{1}{H_{z}}
{\pder{\left( mu \right) }{s}} \right)
\right], \cr
\noalign{\medskip}
R^{v\eta} = &\frac{1}{m}A_{M} \left[
\frac{1}{m} \left( n {\pder{\left( mv \right) }{\eta}} -
n {\pder{z}{\eta}} \frac{1}{H_{z}}
{\pder{\left( mv \right) }{s}} \right) -
\frac{1}{n}\;\left( m {\pder{\left( nu \right) }{\xi}} -
m {\pder{z}{\xi}} \frac{1}{H_{z}}
{\pder{\left( nu \right )}{s}} \right)
\right], \cr
\noalign{\medskip}
R^{vs} = &m {\pder{z}{\xi}} A_{M} \left[
\frac{1}{n}\;\left( m {\pder{z}{\xi}} \frac{1}{H_{z}}
{\pder{\left( nv \right) }{s}} -
m {\pder{\left( nv \right) }{\xi}} \right) +
\frac{1}{m} \left( n {\pder{z}{\eta}} \frac{1}{H_{z}}
{\pder{\left( mu \right) }{s}} -
n {\pder{\left( mu \right) }{\eta}} \right)
\right] +\cr
&n\; {\pder{z}{\eta}} A_{M} \left[
\frac{1}{m} \left( n {\pder{z}{\eta}} \frac{1}{H_{z}}
{\pder{\left( mv \right) }{s}} -
n {\pder{\left( mv \right) }{\eta}} \right) -
\frac{1}{n}\;\left( m {\pder{z}{\xi}} \frac{1}{H_{z}}
{\pder{\left( nu \right) }{s}} -
m {\pder{\left( nu \right) }{\xi}} \right)
\right]. \cr} \eqno{(8)} $$
Notice that transverse stress tensor remains invariant under coordinate transformation. The rotated tensor (7) retains the same properties as the unrotated tensor (5). The additional terms that arise from the slopes of $s$-surfaces along geopotentials are discretized using a modified version of the triad approach of Griffies et al. (1998).</wikitex>
Radiation Stresses
Guidelines for Coefficient Values
<wikitex> The horizontal viscosity and diffusion coefficients are scalars which are read in from ocean.in. Several factors to consider when choosing these values are:
- spindown time The spindown time on wavenumber $k$ is ${1 \over k^2 \nu_2}$ for the Laplacian operator and ${1 \over k^4 \nu_4}$ for the biharmonic operator. The smallest wavenumber corresponds to the length $2 \Delta x$ and is $k = {\pi \over \Delta x}$, leading to
$$
\Delta t < t_{damp} = {\Delta x^2 \over \pi^2 \nu_2}
\mbox{\quad or \quad} {\Delta x^4 \over \pi^4 \nu_4}
$$ This time should be short enough to damp out the numerical noise which is being generated but long enough on the larger scales to retain the features you are interested in. This time should also be resolved by the model timestep.
- boundary layer thickness The western boundary layer has a thickness proportional to:
$$
\Delta x < L_{BL} = \left( {\nu_2 \over \beta} \right)^{1 \over 3}
\mbox{\quad and \quad}
\left( {\nu_4 \over \beta} \right)^{1 \over 5}
$$ for the Laplacian and biharmonic viscosity, respectively. We have found that the model typically requires the boundary layer to be resolved with at least one grid cell. This leads to coarse grids requiring large values of $\nu$.
Horizontal Diffusion
We have chosen anything from zero to the value of the horizontal viscosity for the horizontal diffusion coefficient. One common choice is an order of magnitude smaller than the viscosity.
</wikitex>
