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Next: 9.4.7 Effects on kinetic Up: 9.4 Orthogonal curvilinear coordinates Previous: 9.4.5 Horizontal tension and

9.4.6 The friction vector

The friction vector is given by the covariant divergence of the frictional stress tensor

 
$\displaystyle \rho \, F^{m}$ = $\displaystyle \tau^{mn}_{\; ; n}$  
  = $\displaystyle \tau^{mn}_{\; , n} + \Gamma^{m}_{nc} \, \tau^{nc} + \Gamma^{n}_{nc} \, \tau^{mc},$ (9.109)

where the covariant derivative of the second order stress tensor is a straightforward generalization of the result for a vector. Recall that in the quasi-hydrostatic limit, only the friction in the transverse directions is of interest, since the vertical momentum equation reduces to the inviscid hydrostatic equation. The expression (9.109) is quite general. For the purpose of representing such friction in an ocean model, it is necessary to make this result a bit more explicit. For this purpose, it is useful to start from the equivalent expression
 
$\displaystyle \rho \, F^{m}$ = $\displaystyle (\sqrt{\mathcal{G}})^{-1} \, \left( \sqrt{\mathcal{G}} \, \tau^{mn}\right)_{, n}
+ \Gamma^{m}_{ab} \, \tau^{ab}.$ (9.110)

This expression is valid for any metric. Its derivation is omitted here. To proceed, employ the expression (9.81) for the Christoffel symbol written in terms of the metric, the expression (9.100) for the stress tensor, and the diagonal form of the metric tensor. First, the contraction $\Gamma^{n}_{ab} \, \tau^{ab}$ is given by
$\displaystyle \Gamma^{m}_{ab} \, \tau^{ab}$ = $\displaystyle \frac{1}{2} \, g^{md} \, (g_{ad,b} + g_{bd,a} - g_{ab,d}) \, \tau^{ab}$  
  = $\displaystyle g^{mm} \, (g_{am, b} - \frac{1}{2} \, g_{ab,m}) \, \tau^{ab}$  
  = $\displaystyle g^{mm} \, g_{mm, b} \, \tau^{mb} -
\frac{1}{2} \, g^{mm} \, g_{ab,m} \, \tau^{ab}$ (9.111)

where there is no sum on the m label. Plugging this result into the expression (9.110) for the friction vector yields
$\displaystyle \rho \, F^{m}$ = $\displaystyle ( g_{mm} \, \sqrt{\mathcal{G}} )^{-1}
\, \left( g_{mm} \, \sqrt{\...
...}} \, \tau^{mn} \right)_{, n}
- \frac{1}{2} \, g^{mm} \, g_{ab,m} \, \tau^{ab}.$ (9.112)

Now recall that g33 = 1 and $\tau^{1}_{1} = - \tau^{2}_{2}$. Consequently, for m=1 the friction is
$\displaystyle \rho \,F^{1}$ = $\displaystyle ( g_{11} \, \sqrt{\mathcal{G}} )^{-1}
\, \left( g_{11} \, \sqrt{\...
...G}} \, \tau^{1n} \right)_{, n}
- \frac{1}{2} \, g^{11} \, g_{ab,1} \, \tau^{ab}$  
  = $\displaystyle \tau^{1n}_{,n} + \tau^{1n} \, \partial_{n} \, \ln(g_{11} \, \sqrt...
...artial_{1} \, \ln \sqrt{g_{11}}
-\frac{1}{2} \, \tau^{22} \, g^{11} \, g_{22,1}$  
  = $\displaystyle \tau^{11}_{,1} + \tau^{11} \, \partial_{1} \, \ln(g_{11} \, \sqrt...
...\sqrt{\mathcal{G}} )^{-1} ( g_{11} \, \sqrt{\mathcal{G}} \, \tau^{13} \, )_{,3}$  
  - $\displaystyle \tau^{11} \, \partial_{1} \, \ln \sqrt{g_{11}}
+\tau^{11} \, \partial_{1} \, \ln \sqrt{g_{22}}$  
  = $\displaystyle \tau^{11}_{,1} + \tau^{11} \, \partial_{1} \, \ln\mathcal{G}
+
( ...
...\sqrt{\mathcal{G}} )^{-1} ( g_{11} \, \sqrt{\mathcal{G}} \, \tau^{13} \, )_{,3}$  
  = $\displaystyle \mathcal{G}^{-1} \, (\mathcal{G} \, \tau^{11})_{,1}
+
(g_{11} \, ...
...\sqrt{\mathcal{G}} )^{-1} ( g_{11} \, \sqrt{\mathcal{G}} \, \tau^{13} \, )_{,3}$  
  = $\displaystyle \frac{\sqrt{g_{11}}}{\mathcal{G}} \, (g_{22} \, \rho \, A \, D_{T...
...\, \rho \, A \, D_{S})_{,y}
+
g_{11}^{-1/2} \, ( \rho \, \kappa \, u_{,z})_{,z}$ (9.113)

where the last step introduced the physical components, and the depth independence of the metric components has been used. Multiplying by $\sqrt{g_{11}}$ determines the physical component to the generalized zonal friction
$\displaystyle \rho \, F^{x}$ = $\displaystyle g_{22}^{-1} \, (g_{22} \, \rho \, A \, D_{T})_{,x}
+
g_{11}^{-1} \, (g_{11} \, \rho \, A \, D_{S})_{,y}
+
(\kappa \, u_{,z})_{,z}.$ (9.114)

Similar considerations lead to the second friction component
$\displaystyle \rho \,F^{2}$ = $\displaystyle \mathcal{G}^{-1} \, (\mathcal{G} \, \tau^{22})_{,2}
+
(g_{22} \, ...
...sqrt{\mathcal{G}} )^{-1} ( g_{22} \, \sqrt{\mathcal{G}} \, \tau^{31} \, )_{,3}.$ (9.115)

Multiplying by $\sqrt{g_{22}}$ leads to the generalized meridional friction component
$\displaystyle \rho \, F^{y}$ = $\displaystyle -g_{11}^{-1} \, (g_{11} \, \rho \, A \, D_{T})_{,y}
+
g_{22}^{-1} \, (g_{22} \, \rho \, A \, D_{S})_{,x}
+
( \rho \, \kappa \, v_{,z})_{,z}.$ (9.116)

Again, for Boussinesq fluids, the factors of density can be canceled on both sides, since each are formally replaced by $\rho_{o}$. For non-Boussinesq fluids, the cancelation is also often performed, since the values of the kinematic viscosities are not precisely known.


next up previous contents
Next: 9.4.7 Effects on kinetic Up: 9.4 Orthogonal curvilinear coordinates Previous: 9.4.5 Horizontal tension and
RC Pacanowski and SM Griffies, GFDL, Jan 2000