# Inertial Waves - Why neglecting the advecting term?

I'm trying to derive the dispersion relation for Inertial waves.

In Cartesian coordinates:

Inviscid and incompressible fluid is rotating uniformly with Angular Velocity:

$\Omega = (0, 0, \Omega)$

The Navier Stokes equation for this problem is:

$\frac{\partial \bar{u}}{\partial t}+(\bar{u}\cdot\nabla)\bar{u} =-\frac{1}{\rho}\nabla P-2\Omega\times\bar{u}-\Omega\times(\Omega\times \bar{r})$

Where $\bar{r}=(x,y,z)$

From here,

I know that the term $(\bar{u}\cdot\nabla)\bar{u}$ is neglected compared to the Coriolis term $-2\Omega\times\bar{u}$ , But I don't understand why - and how can this be showed mathematically.

Thanks!

In general, the term cannot be neglected. It's entirely possible that the term has to stick around when advection is significant. But, since you are asking how it goes away, I'm guessing you are in a situation where it does, in fact, go away.

Like any other order of magnitude analysis, you have to non-dimensionalize your equation. This means you have to pick meaningful and relevant scales for each variable such that all of the variables become the same order of magnitude. For example, to non-dimensionalize the left-hand side, you might say that $\overline{t} = t/\tau$, $\overline{u} = u/U$, $\overline{x} = x/L$, $\overline{y} = y/L$, and $\overline{z} = z/L$ where $\tau$ and $U$ and $L$ are the relevant (constant!) scales for your problem. You would then substitute that into the equations and pull the constants out:

$$\frac{U}{\tau} \frac{\partial \overline{u}_i}{\partial \overline{t}} + \frac{U^2}{L} \frac{\partial u_i u_j}{\partial \overline{x}_j}$$

and you would need to do the same for the right hand side. Be careful, how you non-dimensionalize pressure will change how everything else falls out -- there are multiple ways to do it and different results show up.

Once you do this for all the variables in your equation, you will end up being able to combine all of your leading constants (the scales you chose) into non-dimensional numbers. This is where things like Reynolds Number and Mach Number come from.

Then, you use these non-dimensional numbers to decide which terms matter or not. So for Reynolds number for example, you would see that at high Reynolds number, the viscous term approaches zero and is negligible relative to the convective term. Conversely, at low Reynolds number, the convective term will be negligible relative to the viscous term.

So, as a final hint to help you in your process -- you should end up with a non-dimensional number called the Rossby Number, which is the ratio of convective forces to Coriolis forces. Then, you have to look at the problem you are interested in and figure out if the Rossby number is large or small and what that means for the terms in your non-dimensional equation.