matsuno gill

Slow-wave MG

Gill's paper nondimensionalizes the equations using the length scale corresponding to the equatorial rossby radius on a $\beta$ plane ($\beta = \frac{2\Omega \cos(0)}{a}$) Letting $c = \frac{ND}{\pi}$, with $z_{\textrm{top}} = D$ and $N$ the B-V frequency, then the equatorial Rossby radius is $\sqrt{\frac{c}{2\beta}}$ and the time scale $\sqrt{\frac{1}{2\beta c}}$ . The resulting equations of motion are $$\begin{align*} \pder{u}{t} - \frac{1}{2} y v &= -\pder{p}{x} \\ \pder{v}{t} + \frac{1}{2} y u &= -\pder{p}{y} \\ \pder{p}{t} + \pder{u}{x} + \pder{v}{y} &= -Q \end{align*}$$ with $Q$ our forced heating. We then add ODE damping to heat and wind, under which the steady state equations look like

$$\begin{align*} \varepsilon u - \frac{1}{2} y v &= -\pder{p}{x} \\ \varepsilon v+ \frac{1}{2} y u &= -\pder{p}{y} \\ \varepsilon p + \pder{u}{x} + \pder{v}{y} &= -Q \\ (w &= \varepsilon p + Q) \\ \end{align*}$$

Under the cosine-x-gaussian heating profile, the steady-state kelvin wave takes the form $$\begin{align*} u &= p = \frac{1}{2} q_0(x) \exp(-\frac{1}{4} y^2) \\ v &= 0 \\ w &= \frac{1}{2}(\varepsilon q_0(x) + F(x)) \exp(-\frac{1}{4}y^2) \\ \varepsilon q_0 + \der{q_0}{x} \stackrel{?}{=}& -F(x) \propto \cos(x) \\ \end{align*}$$ From the last equation we see that the Kelvin wave has spatial decay rate $\varepsilon$. The paper states that in this nondimensional form it moves with unit speed (verify this?), so it exhibits spatial decay rate $\varepsilon$.

Implications for large-earth simulation:

The wave moves at unit speed when $1 \sim \tilde{x} = x\sqrt{\frac{c}{2\beta}}$, $1 \sim \tilde{t} = t \sqrt{\frac{1}{2\beta c}}$. Since we won't be varying $N$ or $D$, we can assume $c = 1$. If $\frac{\Delta \tilde{x}}{\Delta \tilde{t}} = 1$, then $\frac{\Delta x}{ \Delta t} = \frac{\sqrt{\frac{c}{2\beta}}}{\sqrt{\frac{1}{2\beta c}}} = \sqrt{\frac{2c^2 \beta}{2\beta}} = c.$

Therefore, if I keep $\beta$ constant on a larger earth, then the wave will travel at a speed that depends only on the depth of the atmosphere and the B-V frequency.

Since $\beta = \frac{2\Omega \cos(\phi_0)}{a}$, then if we set $a = 10a_0$ and $\Omega = \frac{\Omega_0}{10}$, then $\beta = \frac{2 \cdot \Omega_0}{10 \cdot 10 a} = \frac{\beta_0}{100}$. The conclusion is that Rossby radius of deformation $\frac{NH}{\pi f}$ should be kept constant by keeping $f$ constant!

BV frequency derivation (slightly wrong, there's a typo)

Assume $p(z=0) = p_0, T(z=0) = T_0$. A constant BV frequency gives $\frac{g}{\theta}\pder{\theta}{z} = N^2$ so $$\begin{align*} & \pder{}{z} \log(\theta) = \frac{N^2}{g} \\ \implies& \log(\theta) - \log(\theta_0) = \frac{N^2}{g}(z-z_0) \\ \implies& \log\left( \frac{T \left(\frac{p_0}{p} \right)^{\kappa}}{T_0} \right) = \frac{N^2}{g} z \\ \implies& T = T_0 \left(\frac{p}{p_0} \right)^\kappa \exp\left(\frac{N^2}{g} z\right) \end{align*}$$ and using hydrostatic balance we get $$\begin{align*} &\pder{p}{z} = -\rho g \\ \implies& \pder{p}{z} = -\frac{pg}{R_d T} \\ \implies& \pder{p}{z} = -\frac{pg}{R_d T_0 \left(\frac{p}{p_0} \right)^\kappa \exp\left(\frac{N^2}{g} z\right)} \\ \implies& \pder{p}{z} = -\frac{pg p_0^\kappa}{R_d T_0 p^\kappa \exp\left(\frac{N^2}{g} z\right)} \\ \implies& \pder{p}{z} = -\frac{p^{1-\kappa}g p_0^\kappa}{R_d T_0} \exp\left(-\frac{N^2}{g} z\right) \\ \implies& \int_{0}^{z'} p^{\kappa-1} \pder{p}{z} \intd{z} = -\frac{g p_0^\kappa}{R_d T_0} \int_0^{z'} \exp\left(-\frac{N^2}{g} z\right) \intd{z} \\ \implies& \int_{p_0}^{p'} p^{\kappa-1} \intd{p} = -\frac{g p_0^\kappa}{R_d T_0} \int_0^{z'} \exp\left(-\frac{N^2}{g} z\right) \intd{z} \\ \implies& p^\kappa - p_0^\kappa = -\frac{g p_0^\kappa}{R_d T_0} \frac{g}{N^2} \left(1-\exp\left(-\frac{N^2}{g}z\right)\right) \\ \implies& p^\kappa =p_0^\kappa \left[1 -\frac{g^2}{R_d T_0 N^2} \left(1-\exp\left(-\frac{N^2}{g}z\right)\right) \right] \\ \end{align*}$$

and based on $p(z=\infty) = 0$ we see $T_0 = \frac{g^2}{R_d N^2}$

$$\begin{align*} &p = p_0 \left[1 -\frac{g^2}{R_d T_0 N^2} \left(1-\exp\left(-\frac{N^2}{g}z\right)\right) \right]^{1/\kappa} \\ \implies& \left(\frac{p}{p_0}\right)^\kappa = \left[1 -\frac{g^2}{R_d T_0 N^2} \left(1-\exp\left(-\frac{N^2}{g}z\right)\right) \right]\\ \implies& 1-\left(\frac{p}{p_0}\right)^\kappa = \frac{g^2}{R_d T_0 N^2} \left(1-\exp\left(-\frac{N^2}{g}z\right)\right) \\ \implies& 1-\frac{R_d T_0 N^2}{g^2}\left[1-\left(\frac{p}{p_0}\right)^\kappa\right] = \exp\left(-\frac{N^2}{g}z\right) \\ \implies& \log\left[1-\frac{R_d T_0 N^2}{g^2}\left(1-\left(\frac{p}{p_0}\right)^\kappa\right)\right] = \left(-\frac{N^2}{g}z\right) \\ \implies& -\frac{g}{N^2}\log\left[1-\frac{R_d T_0 N^2}{g^2}\left(1-\left(\frac{p}{p_0}\right)^\kappa\right)\right] = z \end{align*}$$