# Changeset 11331 for NEMO/trunk/doc/latex/NEMO/subfiles/annex_B.tex

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Timestamp:
2019-07-23T15:02:57+02:00 (2 years ago)
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Documentation: small corrections and tidy up of Appendix B.

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 r11151 \end{align*} Equation \autoref{apdx:B2} is obtained from \autoref{apdx:B1} without any additional assumption. \autoref{apdx:B2} is obtained from \autoref{apdx:B1} without any additional assumption. Indeed, for the special case $k=z$ and thus $e_3 =1$, we introduce an arbitrary vertical coordinate $s = s (i,j,z)$ as in \autoref{apdx:A} and & =\frac{1}{e_1\,e_2\,e_3 }\left[ {\left. {\;\;\;\frac{\partial }{\partial i}\left( {\frac{e_2\,e_3 }{e_1}\,A^{lT}\;\left. {\frac{\partial T}{\partial i}} \right|_s } \right)} \right|_s \left. -\, {e_3 \frac{\partial }{\partial i}\left( {\frac{e_2 \,\sigma_1 }{e_3 }A^{lT}\;\frac{\partial T}{\partial s}} \right)} \right|_s } \right. \\ & \qquad \qquad \quad -e_2 A^{lT}\;\frac{\partial \sigma_1 }{\partial s}\left. {\frac{\partial T}{\partial i}} \right|_s -e_1 \,\sigma_1 \frac{\partial }{\partial s}\left( {\frac{e_2 }{e_1 }A^{lT}\;\left. {\frac{\partial T}{\partial i}} \right|_s } \right) \\ & \qquad \qquad \quad\shoveright{ \left. { +e_1 \,\sigma_1 \frac{\partial }{\partial s}\left( {\frac{e_2 \,\sigma_1 }{e_3 }A^{lT}\;\frac{\partial T}{\partial s}} \right)+\frac{\partial }{\partial s}\left( {\frac{e_1 \,e_2 }{e_3 }A^{vT}\;\frac{\partial T}{\partial z}} \right)\;\;\;} \right] }\\ & \qquad \qquad \quad\shoveright{ \left. { +e_1 \,\sigma_1 \frac{\partial }{\partial s}\left( {\frac{e_2 \,\sigma_1 }{e_3 }A^{lT}\;\frac{\partial T}{\partial s}} \right)+\frac{\partial }{\partial s}\left( {\frac{e_1 \,e_2 }{e_3 }A^{vT}\;\frac{\partial T}{\partial s}} \right)\;\;\;} \right] }\\ \\ &=\frac{1}{e_1 \,e_2 \,e_3 } \left[ {\left. {\;\;\;\frac{\partial }{\partial i} \left( {\frac{e_2 \,e_3 }{e_1 }A^{lT}\;\left. {\frac{\partial T}{\partial i}} \right|_s } \right)} \right|_s \left. {-\frac{\partial }{\partial i}\left( {e_2 \,\sigma_1 A^{lT}\;\frac{\partial T}{\partial s}} \right)} \right|_s } \right. \\ \begin{array}{*{20}l} % \intertext{using the same remark as just above, it becomes:} \intertext{Using the same remark as just above, it becomes:} % &= \frac{1}{e_1 \,e_2 \,e_3 } \left[ {\left. {\;\;\;\frac{\partial }{\partial i} \left( {\frac{e_2 \,e_3 }{e_1 }A^{lT}\;\left. {\frac{\partial T}{\partial i}} \right|_s -e_2 \,\sigma_1 A^{lT}\;\frac{\partial T}{\partial s}} \right)} \right|_s } \right.\;\;\; \\ % \intertext{Since the horizontal scale factors do not depend on the vertical coordinate, the last term of the first line and the first term of the last line cancel, while the second line reduces to a single vertical derivative, so it becomes:} the two terms on the second line cancel, while the third line reduces to a single vertical derivative, so it becomes:} % & =\frac{1}{e_1 \,e_2 \,e_3 }\left[ {\left. {\;\;\;\frac{\partial }{\partial i}\left( {\frac{e_2 \,e_3 }{e_1 }A^{lT}\;\left. {\frac{\partial T}{\partial i}} \right|_s -e_2 \,\sigma_1 \,A^{lT}\;\frac{\partial T}{\partial s}} \right)} \right|_s } \right. \\ & \qquad \qquad \quad \shoveright{ \left. {+\frac{\partial }{\partial s}\left( {-e_2 \,\sigma_1 \,A^{lT}\;\left. {\frac{\partial T}{\partial i}} \right|_s +A^{lT}\frac{e_1 \,e_2 }{e_3 }\;\left( {\varepsilon +\sigma_1 ^2} \right)\frac{\partial T}{\partial s}} \right)\;\;\;} \right]} \\ % \intertext{in other words, the horizontal/vertical Laplacian operator in the ($i$,$s$) plane takes the following form:} \end{array} } \\ \intertext{In other words, the horizontal/vertical Laplacian operator in the ($i$,$s$) plane takes the following form:} \end{array} } \\ % {\frac{1}{e_1\,e_2\,e_3}} \left[ {{ \begin{array}{*{20}c} {1+a_1 ^2} \hfill & {-a_1 a_2 } \hfill & {-a_1 } \hfill \\ {-a_1 a_2 } \hfill & {1+a_2 ^2} \hfill & {-a_2 } \hfill \\ {-a_1 } \hfill & {-a_2 } \hfill & {\varepsilon +a_1 ^2+a_2 ^2} \hfill \\ {1+a_2 ^2 +\varepsilon a_1 ^2} \hfill & {-a_1 a_2 (1-\varepsilon)} \hfill & {-a_1 (1-\varepsilon) } \hfill \\ {-a_1 a_2 (1-\varepsilon) } \hfill & {1+a_1 ^2 +\varepsilon a_2 ^2} \hfill & {-a_2 (1-\varepsilon)} \hfill \\ {-a_1 (1-\varepsilon)} \hfill & {-a_2 (1-\varepsilon)} \hfill & {\varepsilon +a_1 ^2+a_2 ^2} \hfill \\ \end{array} }} \right] \right)\left( {\frac{\partial \rho }{\partial k}} \right)^{-1} \] In practice, isopycnal slopes are generally less than $10^{-2}$ in the ocean, so $\textbf {A}_{\textbf I}$ can be simplified appreciably \citep{cox_OM87}: and, as before, $\epsilon = A^{vT} / A^{lT}$. In practice, $\epsilon$ is small and isopycnal slopes are generally less than $10^{-2}$ in the ocean, so $\textbf {A}_{\textbf I}$ can be simplified appreciably \citep{cox_OM87}. Keeping leading order terms\footnote{Apart from the (1,0) and (0,1) elements which are set to zero. See \citet{griffies_bk04}, section 14.1.4.1 for a discussion of this point.}: \begin{subequations} \label{apdx:B4} \] To prove \autoref{apdx:B5} by direct re-expression of \autoref{apdx:B_ldfiso} is straightforward, but laborious. To prove \autoref{apdx:B_ldfiso_s} by direct re-expression of \autoref{apdx:B_ldfiso} is straightforward, but laborious. An easier way is first to note (by reversing the derivation of \autoref{sec:B_2} from \autoref{sec:B_1} ) that the weak-slope operator may be \emph{exactly} reexpressed in non-orthogonal $i,j,\rho$-coordinates as the (rotated,orthogonal) isoneutral axes to the non-orthogonal $i,j,\rho$-coordinates. The further transformation into $i,j,s$-coordinates is exact, whatever the steepness of the $s$-surfaces, in the same way as the transformation of horizontal/vertical Laplacian diffusion in $z$-coordinates, in the same way as the transformation of horizontal/vertical Laplacian diffusion in $z$-coordinates in \autoref{sec:B_1} onto $s$-coordinates is exact, however steep the $s$-surfaces. \label{sec:B_3} The second order momentum diffusion operator (Laplacian) in the $z$-coordinate is found by The second order momentum diffusion operator (Laplacian) in $z$-coordinates is found by applying \autoref{eq:PE_lap_vector}, the expression for the Laplacian of a vector, to the horizontal velocity vector: \end{align*} Using \autoref{eq:PE_div}, the definition of the horizontal divergence, the third componant of the second vector is obviously zero and thus : the third component of the second vector is obviously zero and thus : \[ \Delta {\textbf{U}}_h = \nabla _h \left( \chi \right) - \nabla _h \times \left( \zeta \right) + \frac {1}{e_3 } \frac {\partial }{\partial k} \left( {\frac {1}{e_3 } \frac{\partial {\textbf{ U}}_h }{\partial k}} \right) & = \frac{1}{e_1} \frac{\partial \left( {A^{lm}\chi   } \right)}{\partial i} -\frac{1}{e_2} \frac{\partial \left( {A^{lm}\zeta } \right)}{\partial j} +\frac{1}{e_3} \frac{\partial u}{\partial k}      \\ +\frac{1}{e_3} \frac{\partial}{\partial k} \left( \frac{A^{vm}}{e_3} \frac{\partial u}{\partial k} \right)      \\ D^{\textbf{U}}_v & = \frac{1}{e_2 }\frac{\partial \left( {A^{lm}\chi   } \right)}{\partial j} +\frac{1}{e_1 }\frac{\partial \left( {A^{lm}\zeta } \right)}{\partial i} +\frac{1}{e_3} \frac{\partial v}{\partial k} +\frac{1}{e_3} \frac{\partial}{\partial k} \left( \frac{A^{vm}}{e_3} \frac{\partial v}{\partial k} \right) \end{align*}