4.6.1 Differential operators

In MOM, the radial coordinate is taken as

r = a + z (4.58)

where a is the earth's radius. z=0 is assumed to be the position of the resting ocean, which is defined parallel to the geoid. $z=-H(\lambda,\phi)$ is the position of the ocean bottom. As mentioned earlier, although the geoid is not a perfect sphere, the relatively mild deviations from a sphere are ignored in MOM, which allows for spherical coordinates. Finally, since z=r-a, the unit vector $\hat{z}$ points in the radial direction $\hat{r}$

$$\hat{z} = \hat{r}.$$ (4.59)

Consistent with the Traditional Approximation (see Marshall et al. 1997 for a review), the differential operators used in the model take on the following form (see Appendix A of Washington and Parkinson (1986) for derivations). The gradient operator is given by

 

$$\nabla \Psi \hat{\lambda} \, \left( \frac{\Psi_{\lambda}}{a \, \cos \phi} \right) + \hat{\phi} \, \left( \frac{\Psi_{\phi}}{a} \right) + \hat{z} \, \Psi_{z}$$ =

$$\nabla_{h} \Psi + \hat{z} \, \Psi_{z}.$$ = (4.60)

The three-dimensional divergence operator acting on a vector ${\bf u} = ({\bf u}_{h},w)$ is given by

 

$$\nabla \cdot {\bf u} \left( \frac{1}{a \, \cos \phi }\right) [u_{\lambda} + (v \cos \phi)_{\phi} ] + w_{z}$$ =

$$\nabla_{h} \cdot {\bf u}_{h} + w_{z}.$$ = (4.61)

If ${\bf u}$ is the velocity field, then its three dimensional divergence vanishes since the fluid is always assumed incompressible in MOM. The three-dimensional curl operator acting on the velocity is given by

$${\bf\omega} \hat{\lambda} \left( \frac{1}{a} \frac{\partial w}{\partial \phi}... ...\frac{1}{a \, \cos\phi} \frac{\partial (u \, \cos\phi)}{\partial \phi} \right),$$ = (4.62)

where ${\bf\omega} = \nabla \wedge {\bf u}$ is the three dimensional vorticity vector. Often, the vertical component of the vorticity will be written

$$\zeta = \hat{z} \cdot {\bf\omega} = \left( \frac{1}{a \, \cos\phi} \right) \left[ v_{\lambda} - (u \, \cos\phi)_{\phi} \right].$$ (4.63)

The three-dimensional Laplacian is given by

 

$$\nabla \cdot (\nabla \Psi) \left(\frac{1}{a^2 \, \cos\phi} \right) \left( \frac{1}{\cos\phi} \, \Psi_{\lambda\lambda} + (\cos\phi\cdot\Psi_\phi)_\phi \right) + \Psi_{zz}$$ =

$$\nabla_{h} \cdot (\nabla_{h} \Psi) + \Psi_{zz}.$$ = (4.64)