= Solution
Linearize the inviscid momentum equation in the uniformly rotating frame. The <Coriolis acceleration> is $2\boldsymbol\Omega\times\delta\mathbf u$, and the equilibrium pressure gradient cancels gravity. For perturbations proportional to $e^{ik_xx-i\omega t}$, the horizontal components are
$$
-i\omega\rho\,\delta u_x-2\Omega\rho\,\delta u_y
=-ik_x\delta p,
$$
$$
-i\omega\rho\,\delta u_y+2\Omega\rho\,\delta u_x=0.
$$
The vertical component is
$$
-i\omega\rho\,\delta u_z
=-g\delta\rho-\frac{d\delta p}{dz}.
$$
Linearizing mass conservation,
$$
\partial_t\delta\rho+\nabla\mathbin\cdot(\rho\delta\mathbf u)=0,
$$
gives
$$
-i\omega\delta\rho+\delta u_z\frac{d\rho}{dz}
=-\rho\left(ik_x\delta u_x+\frac{d\delta u_z}{dz}\right).
$$
Finally, linearizing the adiabatic pressure equation gives
$$
-i\omega\delta p+\delta u_z\frac{dp}{dz}
=-\gamma p\left(ik_x\delta u_x+\frac{d\delta u_z}{dz}\right).
$$
These are the stated five equations. Self-gravity contributes no perturbation because it is neglected, and the equilibrium centrifugal term has already been absorbed or omitted.
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