Terrain-Following Coordinate Transformation

Terrain-Following Coordinate Transformation

From the point of view of the computational model, it is highly convenient to introduce a stretched vertical coordinate system which essentially "flattens out" the variable bottom at $z=-h(x,y)$. Such "$\sigma$" coordinate systems have long been used, with slight appropriate modification, in both meteorology and oceanography [e.g., Phillips (1957) and Freeman et al. (1972)]. To proceed, we make the coordinate transformation:

$${\begin{aligned}{\hat {x}}&=x\\{\hat {y}}&=y\\\sigma &=\sigma (x,y,z)\\z&=z(x,y,\sigma )\\{\hat {t}}&=t\end{aligned}}$$

See Vertical S-coordinate for the form of $\sigma$ used here. Also, see Shchepetkin and McWilliams, 2005 for a discussion about the nature of this form of $\sigma$ and how it differs from that used in SCRUM.

In the stretched system, the vertical coordinate $\sigma$ spans the range $-1\leq \sigma \leq 0$; we are therefore left with level upper ($\sigma =0$) and lower ($\sigma =-1$) bounding surfaces. The chain rules for this transformation are:

$${\begin{aligned}\left({\partial \over \partial x}\right)_{z}&=\left({\partial \over \partial x}\right)_{\sigma }-\left({1 \over H_{z}}\right)\left({\partial z \over \partial x}\right)_{\sigma }{\partial \over \partial \sigma }\\\\\left({\partial \over \partial y}\right)_{z}&=\left({\partial \over \partial y}\right)_{\sigma }-\left({1 \over H_{z}}\right)\left({\partial z \over \partial y}\right)_{\sigma }{\partial \over \partial \sigma }\\\\{\partial \over \partial z}&=\left({\partial \sigma \over \partial z}\right){\partial \over \partial \sigma }={1 \over H_{z}}{\partial \over \partial \sigma }\end{aligned}}$$

where

$$H_{z}\equiv {\partial z \over \partial \sigma }$$

As a trade-off for this geometric simplification, the dynamic equations become somewhat more complicated. The resulting dynamic equations, after dropping the carats, are:

$${\partial u \over \partial t}-fv+{\vec {v}}\cdot \nabla u=-{\partial \phi \over \partial x}-\left({\frac {g\rho }{\rho _{o}}}\right){\frac {\partial z}{\partial x}}-g{\partial \zeta \over \partial x}+{1 \over H_{z}}{\partial \over \partial \sigma }\left[{(K_{m}+\nu ) \over H_{z}}{\partial u \over \partial \sigma }\right]+{\cal {F}}_{u}+{\cal {D}}_{u}$$

$${\frac {\partial v}{\partial t}}+fu+{\vec {v}}\cdot \nabla v=-{\frac {\partial \phi }{\partial y}}-\left({\frac {g\rho }{\rho _{o}}}\right){\frac {\partial z}{\partial y}}-g{\partial \zeta \over \partial y}+{1 \over H_{z}}{\partial \over \partial \sigma }\left[{(K_{m}+\nu ) \over H_{z}}{\partial v \over \partial \sigma }\right]+{\cal {F}}_{v}+{\cal {D}}_{v}$$

$${\frac {\partial C}{\partial t}}+{\vec {v}}\cdot \nabla C={1 \over H_{z}}{\partial \over \partial \sigma }\left[{(K_{C}+\nu ) \over H_{z}}{\partial C \over \partial \sigma }\right]+{\cal {F}}_{C}+{\cal {D}}_{C}$$

$$\rho =\rho (T,S,P)$$

$${\frac {\partial \phi }{\partial \sigma }}=\left({\frac {-gH_{z}\rho }{\rho _{o}}}\right)$$

$${\partial H_{z} \over \partial t}+{\partial (H_{z}u) \over \partial x}+{\partial (H_{z}v) \over \partial y}+{\partial (H_{z}\Omega ) \over \partial \sigma }=0$$

where

$${\vec {v}}=(u,v,\Omega )$$

$${\vec {v}}\cdot \nabla =u{\frac {\partial }{\partial x}}+v{\frac {\partial }{\partial y}}+\Omega {\frac {\partial }{\partial \sigma }}$$

The vertical velocity in $\sigma$ coordinates is

$$\Omega (x,y,\sigma ,t)={1 \over H_{z}}\left[w-{z+h \over \zeta +h}{\partial \zeta \over \partial t}-u{\partial z \over \partial x}-v{\partial z \over \partial y}\right]$$

and

$$w={\partial z \over \partial t}+u{\partial z \over \partial x}+v{\partial z \over \partial y}+\Omega H_{z}$$

Vertical Boundary Conditions

In the stretched coordinate system, the vertical boundary conditions become:

top ($\sigma =0$):

$${\begin{aligned}\left({\frac {K_{m}}{H_{z}}}\right){\frac {\partial u}{\partial \sigma }}&=\tau _{s}^{x}(x,y,t)\\\left({\frac {K_{m}}{H_{z}}}\right){\frac {\partial v}{\partial \sigma }}&=\tau _{s}^{y}(x,y,t)\\\left({\frac {K_{C}}{H_{z}}}\right){\frac {\partial C}{\partial \sigma }}&={Q_{C} \over \rho _{o}c_{P}}\\\Omega &=0\end{aligned}}$$

and bottom ($\sigma =-1$):

$${\begin{aligned}\left({\frac {K_{m}}{H_{z}}}\right){\frac {\partial u}{\partial \sigma }}&=\tau _{b}^{x}(x,y,t)\\\left({\frac {K_{m}}{H_{z}}}\right){\frac {\partial v}{\partial \sigma }}&=\tau _{b}^{y}(x,y,t)\\\left({\frac {K_{C}}{H_{z}}}\right){\frac {\partial C}{\partial \sigma }}&=0\\\Omega &=0\end{aligned}}$$

Note the simplification of the boundary conditions on vertical velocity that arises from the $\sigma$ coordinate transformation.