16 Waves and instabilities

The equations of motion have wave solutions. This is obtained by inserting a complex harmonic approach $\mathbf{\psi}_0e^{-i\omega t}$ into the respective system of equations and looking for non-trivial solutions, here $\mathbf{\psi}_0$ is a complex amplitude vector. Occasionally a zonal background current is recorded due to climatological relevance. A solution is instability if $\Im\left(\omega\right) > 0$, i.e. the amplitude increases exponentially.

16.1 kinematics

Waves are periodic changes in space and time. The maximum magnitude deflection from the rest position is the amplitude. The time $T > 0$ until the movement pattern is repeated at a fixed location is called the period duration; its inverse is the frequency $f$. The angular frequency $\omega$ is defined as

\[ \begin{align} \omega \coloneqq 2\pi f = \frac{2\pi}{T}. \end{align} \]

Analogously, the wavelength $\lambda > 0$ is the length after which the movement pattern repeats itself. Wavenumber $\kappa$ and circular wavenumber $k$ are analogous to frequency and circular frequency defined by

\[ \begin{align} \kappa \coloneqq \frac{1}{\lambda}, & {} & k \coloneqq \frac{2\pi}{\lambda}. \end{align} \]

A wave of size $a$ is described by an equation of the form

\[ \begin{align} a\left(x, t\right) = Af\left(kx - \omega t + \varphi\right). \end{align} \]

Here $f$ is a real or complex-valued, $2\pi-$periodic function, i.e. h. $f\left(\xi + 2\pi\right) = f\left(\xi\right)$ for each $\xi\in \mathbb{R}$. This can, but does not have to be, a trigonometric function. $f$ is often chosen to be complex-valued, e.g. E.g. $f(x) = \exp(ix)$ if this shortens the mathematical description or makes it more clear. However, before you calculate quantities that you want to compare with measured quantities, you have to project everything onto the real axis. The argument $kx - \omega t + \varphi$ is also called the phase of the wave, in the complex plane this is the angle that the number includes with the real axis. $\varphi$ is the phase shift that allows the phase of the wave at $x, t = 0$ not to be equal to zero. You can often do without $\varphi$ in derivations by shifting the time or location coordinates accordingly. $k$ and $\omega$ can also have imaginary parts, e.g. B. with waves that propagate exponentially into a medium (so-called evanescent waves), or with instabilities in which the amplitude increases exponentially over a certain period of time. Likewise, the prefactor $A$ can also be complex and thus contain a phase shift.

In general, a wave can also propagate in a three-dimensional space, in which case we first write in general terms

\[ \begin{align} a\left(\mathbf{r}, t\right) = Af\left(\phi\left(\mathbf{r}, t\right)\right) \end{align} \]

with $\phi = \phi\left(\mathbf{r}, t\right)$ as the phase function. The phase function can, for example, have the shape

\[ \begin{align} \varphi\left(\mathbf{r}, t\right) = \mathbf{k}\cdot\mathbf{r} - \omega t\tag{16.5}\label{eq:phase_velocity_deriv} \end{align} \]

have, here $\mathbf{k} = \left(k_x, k_y, k_z\right)^T$ is called the wave vector. Its magnitude is equal to the circular wave number and points in the direction of propagation of the wave. If one introduces Eq. (16.5) after the time, follows

\[ \begin{align} \frac{d\varphi\left(\mathbf{r}, t\right)}{dt} = \mathbf{k}\cdot\frac{d\mathbf{r}}{dt} - \omega \end{align} \]

If you set the left side equal to zero, i.e. move with a point of constant phase, and choose a trajectory parallel to $\mathbf{k}$, you get the phase velocity

The dependency

\[ \begin{align} \omega = \omega\left(k\right) \end{align} \]

is called dispersion relation. Now consider the equation $f\left(x, t\right) = f_1\left(x, t\right) + f_2\left(x, t\right)$ of two waves:

\[ \begin{align} f_1\left(x, t\right) &= -\cos\left(k_1x - \omega_1t\right)\\ f_1\left(x, t\right) &= \cos\left(k_2x - \omega_2t\right)\\ f\left(x, t\right) &= -\cos\left(k_1x - \omega_1t\right) + \cos\left(k_2x - \omega_2t\right) \end{align} \]

With Eq. (A.61) follows

\[ \begin{align} -\cos\left(\alpha\right) + \cos\left(\beta\right) &= 1 - \cos\left(\alpha\right) - 1 + \cos\left(\beta\right) = 2\sin^2\left(\frac{\alpha}{2}\right) - 2\sin^2\left(\frac{\beta}{2}\right)\nonumber\\ &= 2\sin^2\left(\frac{\alpha}{2}\right) - 2\sin^2\left(\frac{\alpha}{2}\right)\sin^2\left(\frac{\beta}{2}\right) - 2\sin^2\left(\frac{\beta}{2}\right) + 2\sin^2\left(\frac{\alpha}{2}\right)\sin^2\left(\frac{\beta}{2}\right)\nonumber\\ &= 2\left[\sin^2\left(\frac{\alpha}{2}\right)\cos^2\left(\frac{\beta}{2}\right) - \cos^2\left(\frac{\alpha}{2}\right)\sin^2\left(\frac{\beta}{2}\right)\right]\nonumber\\ &= 2\left[\sin\left(\frac{\alpha}{2}\right)\cos\left(\frac{\beta}{2}\right) - \cos\left(\frac{\alpha}{2}\right)\sin\left(\frac{\beta}{2}\right)\right]\left[\sin\left(\frac{\alpha}{2}\right)\cos\left(\frac{\beta}{2}\right) + \cos\left(\frac{\alpha}{2}\right)\sin\left(\frac{\beta}{2}\right)\right]\nonumber\\ &= 2\sin\left(\frac{\alpha - \beta}{2}\right)\sin\left(\frac{\alpha + \beta}{2}\right). \end{align} \]

So it applies

\[ \begin{align} f\left(x, t\right) &= 2\sin\left(\frac{\Delta k x - \Delta\omega t}{2}\right)\sin\left(\newoverline{k}x - \newoverline{\omega}t\right) \end{align} \]

with

\[ \begin{align} \Delta k \coloneqq k_2 - k_1, & {} & \Delta\omega \coloneqq \omega_2 - \omega_1,\\ \newoverline{k} \coloneqq \frac{k_1 + k_2}{2}, & {} & \newoverline{\omega} \coloneqq \frac{\omega_1 + \omega_2}{2}. \end{align} \]

So the envelope $\sin\left(\frac{\Delta k x - \Delta\omega t}{2}\right)$ moves with the speed

\[ \begin{align} v_e = \frac{\Delta\omega}{\Delta k}. \end{align} \]

One therefore defines the group velocity $c_{\text{gr}}$ by

\[ \begin{align} c_{\text{gr}} \coloneqq \frac{\partial\omega}{\partial k}. \end{align} \]

It can therefore be derived from the dispersion relation. It is the speed at which wave packets move, i.e. at which energy is transported. Is

\[ \begin{align} c_{\text{gr}} = c \end{align} \]

regardless of $k$, the wave is called dispersion-free. You can vector the group speed

generalize. You can write with $\omega = ck$

\[ \begin{align} c_{\text{gr}} &= \frac{d\omega}{dk} = c + k\frac{dc}{dk} = c + k\frac{d\lambda}{dk}\frac{dc}{d\lambda} = c - \frac{2\pi}{k}\frac{dc}{dy} = c - \lambda\frac{dc}{d\lambda}. \end{align} \]

In the case $dc/d\lambda < 0$ one speaks of abnormal dispersion. In this case the energy propagates faster than the phase velocity. In the case

\[ \begin{align} \frac{dc}{d\lambda}\lambda > c \Leftrightarrow \frac{dc}{d\lambda} > \frac{c}{\lambda} = f \end{align} \]

or

\[ \begin{align} \frac{d\omega}{dk} < 0 \end{align} \]

propagates the energy in a direction opposite to the wave vector.

If the character of a wave allows this, it is categorized either as a longitudinal wave, in which the oscillators oscillate parallel to the wave vector, or as a transverse wave, in which they oscillate perpendicular to it. For mechanical transverse waves, the wave height $H$ as double amplitude and the steepness $S$ of a wave that passes through are introduced as further kinematic variables

\[ \begin{align} S \coloneqq \frac{H}{\lambda} \end{align} \]

is defined and is therefore dimensionless.

16.2 Categorization of waves in geofluids

Barotropic waves (see section 16.5) require boundary surfaces and have their amplitude maximum at these. Therefore they are also referred to as external waves or surface waves. The shallow water equations are usually sufficient to describe such waves.

Barocline waves (see section 16.6), however, exist inside a medium; they are therefore also referred to as internal waves.

A limiting case here are waves at a sharp interface ($N \gg 0$) inside a medium, such as. B. at an inversion or at a density jump (see, for example, the illustration at the end of section 16.5.2). They can often be described mathematically using the formulas for surface waves, but are usually classified as internal waves. Capillary waves (see section 16.4) are surface waves, but barotropy plays no role and the shallow water equations are not sufficient to describe them. Therefore they represent a special category.

16.3 Justification for linearization

If $A, B$ are two phenomena and $f\left(A + B\right)$ is, for example, a prediction, then in general it holds

\[ \begin{align} f\left(A + B\right) \not = f\left(A\right) + f\left(B\right), \end{align} \]

The total effect of $A + B$ is therefore not the sum of the individual effects $A, B$, since $f$ is not a linear map.

However, when searching for analytical solutions to a nonlinear system of equations, especially when searching for wave solutions, one is very interested in linearization. This is sometimes more sometimes less justified. It is assumed that for a quantity $A = A_0 + A'$ with $A_0$ as the background state and $A'$ as the disturbance

\[ \begin{align} \mathcal{O}\left(A_0\right) = \mathcal{O}\left(A'\right) + 1, \end{align} \]

therefore applies

\[ \begin{align} \mathcal{O}\left(A'^2\right) = \mathcal{O}\left(A_0\right) - 2. \end{align} \]

One therefore ignores all terms in which products of the perturbations occur. This leads to a linearization of the system of equations. For the derivatives, we assume a plane wave in the x-direction with angular frequency $\omega$ and angular wave number $k$, which only affects the disturbances

\[ \begin{align} \mathcal{O}\left(\frac{\partial A'}{\partial t}\right) &= \mathcal{O}\left(\omega A'\right),\\ \mathcal{O}\left(\frac{\partial A'}{\partial x}\right) &= \mathcal{O}\left(kA'\right)\\ \mathcal{O}\left(\frac{\frac{\partial A'}{\partial t}}{A\frac{\partial A'}{\partial x}}\right) &= \mathcal{O}\left(\frac{\omega}{Ak}\right) = \mathcal{O}\left(\frac{c}{A}\right).\tag{16.29}\label{eq:bed_lin_wellglg} \end{align} \]

From Eq. (16.29) it follows that one can neglect the advective terms if the phase velocity $c = \frac{\omega}{k}$ is much larger than the fluid particle velocity.

If it is possible to eliminate all nonlinear terms of a system of equations, then we know that two independent solutions to this system of equations do not affect each other. Examples are sound waves and electromagnetic waves. The linearity of these systems makes it possible to independently superimpose different sounds or signals.

16.3.1 Example: sound waves

Sound waves are compression waves, but due to adiabatic pressures and densities can be clearly mapped to one another. A wave propagating in the x direction is considered. The perturbation approach for velocity and density is

\[ \begin{align} u = U + u', & {} & \rho = \rho_0 + \rho'. \end{align} \]

The system of equations is with the abbreviated replacements

\[ \begin{align} \rho' &\to \rho,\\ u' &\to u \end{align} \]

and the approximations derived in the previous section

\[ \begin{align} \frac{\partial u}{\partial t} + U\frac{\partial u}{\partial x} &= -\frac{1}{\rho _0}\frac{\partial p}{\partial x}\tag{16.33}\label{eq:schall_1}\\ \frac{\partial\rho}{\partial t} + U\frac{\partial \rho}{\partial x} &= - \rho _0\frac{\partial u}{\partial x}\tag{16.34}\label{eq:schall_2}\\ \rho(p) &= \rho_\theta\left(\frac{p}{p_0}\right)^{c_v/c_p} \end{align} \]

The first equation is the momentum equation, the second is the continuity equation, and the third is the potential density equation Eq. (9.44). In order to be in Eq. (16.34) to express the change in density through pressure, one uses the chain rule:

\[ \begin{align} \frac{\partial\rho}{\partial t} = \frac{\partial\rho}{\partial p}\frac{\partial p}{\partial t} = \frac{\rho_\theta c_v}{p_0c_p}\left(\frac{p}{p_0}\right)^{c_v/c_p - 1}\frac{\partial p}{\partial t} = \frac{\rho_\theta c_v}{p_0c_p}\frac{\partial p}{\partial t}, \end{align} \]

analogously for $\frac{\partial}{\partial x}$. The derivation of the density according to the pressure was evaluated at the point $p = p_0$ because the pressure fluctuations are very small. Now you make an approach

\[ \begin{align} u = u_0\exp\left[i\left(kx - \omega t\right)\right], & {} & p = P\exp\left[i\left(kx - \omega t\right)\right] \end{align} \]

with possibly complex amplitudes $u_0, P$. If you insert this into the system of equations, it follows

\[ \begin{align} -i\omega u_0 + Uiku_0 &= -\frac{1}{\rho_\theta}ikP,\\ -\frac{\rho_\theta c_v}{p_0c_p}i\omega P + Uik\frac{\rho_\theta c_v}{p_0c_p}P &= -\rho_\theta iku_0\nonumber\\ \Leftrightarrow - i\omega P + UikP &= -\frac{p_0c_p}{c_v}iku_0. \end{align} \]

As a matrix equation this becomes

\[ \begin{align} \left(\begin{array}{cc} - i\omega + Uik&\frac{1}{\rho_\theta}ik\\ ik\frac{p_0c_p}{c_v}& -i\omega + Uik \end{array}\right)\left(\begin{array}{c} u_0\\ P \end{array}\right) = \mathbf{0}. \end{align} \]

Non-trivial solutions exist for

\[ \begin{align} \left(\omega - Uk\right)^2 - k^2\frac{p_0c_p}{\rho_\theta c_v} = 0\Leftrightarrow \omega = Uk\pm k\sqrt{\frac{p_0c_p}{\rho_\theta c_v}}. \end{align} \]

So for the phase velocity $c$ it follows

The equation of state was used here, $T$ is the equilibrium temperature and $\kappa = c_p/c_v>1$ is the adiabatic exponent. However, the value of $c$ also depends on the humidity. You could therefore determine the humidity by measuring the speed of sound in combination with a temperature measurement. It should be noted that sound waves are dispersion-free, but the phase velocity is space and time dependent as it depends on the temperature and humidity.

16.3.1.1 Entropy form of the pressure gradient

If you use the equation with Eq. (10.87) obtained formulation

\[ \begin{align} -\frac{1}{\rho}\nabla p = -c^{(p)}T + T\nabla s = -c^{(p)}T + c^{(p)}\frac{T}{\theta}\nabla\theta \end{align} \]

for the pressure gradient acceleration, further insights into sound waves are obtained. Since sound waves are adiabatic, the portion of the pressure gradient that is related to the gradient of entropy is not expected to contribute to sound waves. It is different with the temperature gradient. In order to obtain a closed system of equations, a prognostic equation for the temperature must therefore be found. The adiabatic form of Eq. (9.14) is

\[ \begin{align} \frac{\partial T}{\partial t} = -\nabla\cdot\left(T\mathbf{v}\right) - \left(\frac{R_d}{c^{(V)}} - 1\right)T\nabla\cdot\mathbf{v}. \end{align} \]

It is therefore recorded as a one-dimensional system of equations, neglecting advection

\[ \begin{align} \frac{\partial w}{\partial t} &= -c^{(p)}\frac{\partial T}{\partial z},\tag{16.45}\label{eq:sound_entropy_0}\\ \frac{\partial T}{\partial t} &= -\frac{R_d}{c^{(V)}}T\frac{\partial w}{\partial z}.\tag{16.46}\label{eq:sound_entropy_1} \end{align} \]

In order to verify that this actually describes sound waves, one makes the approach

\[ \begin{align} w = w_1\exp\left(ikz - i\omega t\right), & {} & T = T_0 + T_1\exp\left(ikz - i\omega t\right). \end{align} \]

The density is assumed to be homogeneous. If you insert this into the equations (16.45) - (16.46), it follows

\[ \begin{align} -i\omega w_1 = -c^{(p)}ik\frac{T_1}{\rho}, & {} & -i\omega T_1 = -\frac{R_d}{c^{(V)}}T_0ikw_1. \end{align} \]

In matrix notation you get

\[ \begin{align} \left( \begin{array}{cc} -i\omega & c^{(p)}ik\frac{1}{\rho} \\ \frac{R_d}{c^{(V)}}\newtilde{T}_0ik & -i\omega \end{array}\right) \left(\begin{array}{c} w_1 \\ \newtilde{T}_1 \end{array}\right) = \left(\begin{array}{c} 0 \\ 0 \end{array}\right). \end{align} \]

By setting the determinant to zero you get

\[ \begin{align} \omega^2 = \kappa R_dTk^2. \end{align} \]

This results in the same dispersion relation as Eq. (16.42). This verifies that the terms listed in the equations (16.45) - (16.46) do indeed describe sound waves.

16.4 Capillary waves

In the case $\eta = \hat{\eta}\exp\left(ikx - i\omega t\right)$ follows

\[ \begin{align} -\frac{1}{\rho}\frac{\partial p_k}{\partial x} &= -\frac{k^2\sigma}{\rho}\frac{\partial\eta}{\partial x}. \end{align} \]

Adding this to the right-hand side of Eq. $\href{#eq:deep_1}{(16.66)}$, is obtained

\[ \begin{align} -g\frac{\partial\eta}{\partial x} &\to -g\frac{\partial\eta}{\partial x} - \frac{\sigma k^2}{\rho}\frac{\partial\eta}{\partial x} = -g\left(1 + \frac{k^2\sigma}{\rho g}\right)\frac{\partial\eta}{\partial x}, \end{align} \]

So you have to

\[ \begin{align} g \to g\left(1 + \frac{k^2\sigma}{\rho g}\right) \end{align} \]

replace, the rest of the derivation carries over. So you get the dispersion relation of the capillary waves

In the case of very short waves, the limiting case is obtained

In this limiting case, capillary waves are abnormally dispersed.

In order to derive a boundary wall length $\lambda_K$, above which capillary effects become important, one calculates

\[ \begin{align} \frac{k_K^2\sigma}{\rho g} \stackrel{!}{\geq} \frac{1}{10} \Leftrightarrow \lambda_K \leq \sqrt{\frac{40\pi^2\sigma}{\rho g}}. \end{align} \]

The surface tension of water is approximately $\sigma \sim 75$ mN/m, from which it follows

\[ \begin{align} \lambda_K \sim 5,5\text{ cm}. \end{align} \]

16.5 Barotropic waves

16.5.1 Inertial waves

In the case of inertial waves, a vanishing horizontal pressure gradient is assumed, so that only the Coriolis force remains. The momentum equations of the SWEs Eq. (13.171) becomes, neglecting the advective terms

\[ \begin{align} \frac{du}{dt} = f_0v,& {} & \frac{dv}{dt} = -f_0u. \end{align} \]

A location-independent value $f_0$ is used for the Coriolis parameter in order to avoid the $\beta $effect. You make an approach

\[ \begin{align} u = U\exp\left(-i\omega t\right), & {} & v = V\exp\left(-i\omega t\right) \end{align} \]

with possibly complex amplitudes $U, V$. If you plug this into the system of equations, you get

\[ \begin{align} -i\omega U = f_0V, & {} & -i\omega V = -f_0 U. \end{align} \]

As a matrix equation this becomes

\[ \begin{align} \left(\begin{array}{cc} - i\omega& -f_0\\ f_0& -i\omega \end{array}\right)\left(\begin{array}{c} U\\ V \end{array}\right) = \mathbf{0}. \end{align} \]

Non-trivial solutions exist for

\[ \begin{align} -\omega^2 + f_0^2 &= 0, \end{align} \]

so for

\[ \begin{align} \omega = \pm f. \end{align} \]

The particles move in a circular path with a radius

\[ \begin{align} R_I = \frac{V}{f_0}. \end{align} \]

If you use a speed of $1$ $\frac{\text{m}}{\text{s}}$, for example $R_I = 10$ km with $f_0 = 10^{-4}\:\text{s}^{-1}$. Applies to the period

\[ \begin{align} T = \frac{2\pi}{f_0} \approx \frac{1\:\text{d}}{2\sin\left(\varphi\right)}, \end{align} \]

d = $24$ h is the duration of a day. Typical inertial periods are on the order of one day.

Since the total derivatives $\md{}$ were used, a trajectory was implicitly calculated. This trajectory is a circular path, but only assuming a location-independent Coriolis parameter. In general, however, the particle trajectories are not closed circular paths due to the beta effect. Likewise, no wind field $\mathbf{v}_h\left(\mathbf{r}, t\right)$ was determined.

16.5.2 Water surface waves

At this point one assumes the absence of a zonal base current $U = 0$, o. d. A. you also align the x-direction with the wave vector, so that the linearized system of equations here

\[ \begin{align} \frac{\partial u}{\partial t} &= -g\frac{\partial\eta}{\partial x}, \tag{16.66}\label{eq:deep_1}\\ \frac{\partial u}{\partial x} + \frac{\partial w}{\partial z} &= 0\tag{16.67}\label{eq:deep_2} \end{align} \]

is. One assumes a homogeneous depth $D > 0$ and requires this as a boundary condition

\[ \begin{align} w\left(z = -D\right) \hastobe 0.\tag{16.68}\label{eq:deep_deriv_1} \end{align} \]

Now there are the approaches

\[ \begin{align} u = U\left(z\right)\exp\left(ikx - i\omega t\right), & {} & w = W\left(z\right)\exp\left(ikx - i\omega t\right). \end{align} \]

The surface deflection $\eta$ and the vertical velocity $w$ hang over

\[ \begin{align} \frac{\partial\eta}{\partial t}\left(z = 0\right) & \hastobe w\left(z = 0\right),\\ \Rightarrow -i\omega\hat{\eta} &= W\left(0\right)\tag{16.71}\label{eq:deep_deriv_2} \end{align} \]

together, where $\hat{\eta}$ is the amplitude of the surface deflection. Eq. (16.66) therefore results in $z = 0$

\[ \begin{align} -i\omega U\left(0\right) = g\frac{k}{\omega}W\left(0\right) \Rightarrow \omega^2 = igk\frac{W\left(0\right)}{U\left(0\right)}.\tag{16.72}\label{eq:deep_disp_pre} \end{align} \]

From Eq. (16.67) follows

\[ \begin{align} ikU\left(z\right) + W' = 0 \Rightarrow W' = -ikU. \end{align} \]

This and the equations (16.68) and (16.71) are through

\[ \begin{align} U\left(z\right) &= \frac{\hat{\eta}\omega}{\sinh\left(kD\right)}\cosh\left[k\left(D + z\right)\right],\tag{16.74}\label{eq:u_vert_mod_short_waves}\\ W\left(z\right) &= -\frac{i\hat{\eta}\omega}{\sinh\left(kD\right)}\sinh\left[k\left(D + z\right)\right]. \end{align} \]

fulfilled. It follows from Eq. (16.72) is the dispersion relation

In the case $\lambda \gg D$ it follows

as the dispersion relation of the so-called shallow water waves. In the opposite case $\lambda \ll D$ the dispersion relation follows the deep water waves

In the case of a density discontinuity $\rho_1 < \rho_2$ with a lighter fluid above a heavier fluid, the pressure gradient term is below the wave

\[ \begin{align} -\frac{1}{\rho_2}\frac{\partial p}{\partial x} &= -\frac{1}{\rho_2}\frac{\partial\left(g\rho_2\eta - g\rho_1\eta\right)}{\partial x} = -\frac{g}{\rho_2}\frac{\partial\left(\rho_2\eta - \rho_1\eta\right)}{\partial x} = -\frac{g}{\rho_2}\frac{\partial\left[\eta\left(\rho_2 - \rho_1\right)\right]}{\partial x}\nonumber\\ &= -\frac{g\left(\rho_2 - \rho_1\right)}{\rho_2}\frac{\partial\eta}{\partial x} = -g\frac{\Delta\rho}{\rho_2}\frac{\partial \eta}{\partial x} \end{align} \]

with $\Delta\rho \coloneqq \rho_2 - \rho_1$, where $\eta$ is to be understood as the deflection of the density discontinuity. The rest of the derivation carries over, whereby the substitution $g \to g' \coloneqq g\frac{\Delta\rho}{\rho_2}$ has to be made. For the dispersion relation one obtains

\[ \begin{align} \omega^2 &= g'k\tanh\left(k\left(z_w - z_0\right)\right), \end{align} \]

where $z_w$ denotes the position of the density discontinuity and $z_0$ denotes the depth coordinate of the bottom.

16.5.2.1 Stokes drift

Define

\[ \begin{align} \gamma \coloneqq \frac{\hat{\eta}\omega}{\sinh\left(kD\right)}, \end{align} \]

then Eq. (16.74) as

\[ \begin{align} U\left(z\right) &= \gamma\cosh\left[k\left(D + z\right)\right] \end{align} \]

note down. If you integrate this from $z = -D$ to $z = \eta$, it follows

\[ \begin{align} \int_{-D}^\eta\gamma\cosh\left[k\left(D + z\right)\right]dz = \frac{\gamma}{k}\sinh\left(k\left(D + \eta\right)\right) \stackrel{\href{ch-39-derivations-of-some-mathematical-formula.html#eq:sinh_add_1}{\text{Glg. (A.56)}}}{=} \frac{\gamma}{k}\left[\sinh\left(kD\right)\cosh\left(k\eta\right) + \cosh\left(kD\right)\sinh\left(k\eta\right)\right]. \end{align} \]

Developing this to first order in $k\eta$ (slope $\ll$ 1), it follows

\[ \begin{align} \int_{-D}^\eta\gamma\cosh\left[k\left(D + z\right)\right]dz \approx \frac{\gamma}{k}\left[\sinh\left(kD\right) + \cosh\left(kD\right)k\eta\right]. \end{align} \]

Multiplying this by the real part of $\exp\left(ikx - iwt\right)$ gives the expression

\[ \begin{align} \frac{d\newdot{v}_h}{dy} = \frac{\gamma}{k}\left[\sinh\left(kD\right) + \cosh\left(kD\right)k\eta\right]\cos\left(kx - \omega t\right) \end{align} \]

for the vertically integrated volume flow density. If you insert the real part of the surface deflection $\eta = \newhat{\eta}e^{ikx - i\omega t}$, you get

\[ \begin{align} \frac{d\newdot{v}_h}{dy} = \frac{\gamma}{k}\left[\sinh\left(kD\right) + \cosh\left(kD\right)k\newhat{\eta}\cos\left(kx - \omega t\right)\right]\cos\left(kx - \omega t\right). \end{align} \]

If you integrate this over a period it follows

\[ \begin{align} \frac{d\newdot{v}_h}{dy} = \frac{\gamma}{k}\cosh\left(kD\right)k\newhat{\eta}\frac{1}{2} = \frac{\newhat{\eta}^2\omega}{2\tanh\left(kD\right)} = \frac{\newhat{\eta}^2}{2}\sqrt{\frac{gk}{\tanh\left(kD\right)}}. \end{align} \]

This volume flow in the direction of the wave vector is called Stokes drift. Because of

\[ \begin{align} \newoverline{u}\left(z\right) = 0 \end{align} \]

for $z < -\newhat{\eta}$ this volume flow arises exclusively in the area of ​​the surface wave itself due to the increase in the amplitude of the horizontal velocity with height.

16.5.3 Poincare waves

Poincaré waves are the waves of the linearized shallow water equations on the f-plane, so the equations (13.173) - (13.174) form the underlying system of equations. These equations are in components in plane geometry

\[ \begin{align} \frac{\partial u}{\partial t} &= fv - g\frac{\partial\eta}{\partial x},\tag{16.89}\label{eq:dyn_in_grav_1}\\ \frac{\partial v}{\partial t} &= -fu - g\frac{\partial\eta}{\partial y},\tag{16.90}\label{eq:dyn_in_grav_2}\\ \frac{\partial\eta}{\partial t} &= -H\left(\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y}\right).\tag{16.91}\label{eq:dyn_in_grav_3} \end{align} \]

Here $\eta$ is the surface deflection and $H$ is the average depth. You make an approach

\[ \begin{align} u = u_0\exp\left[i\left(\mathbf{k}\cdot\mathbf{r} - \omega t\right)\right], & {} & v = v_0\exp\left[i\left(\mathbf{k}\cdot\mathbf{r} - \omega t\right)\right], & {} & \eta = \newhat{\eta}\exp\left[i\left(\mathbf{k}\cdot\mathbf{r} - \omega t\right)\right]. \end{align} \]

with a horizontal wave vector $\mathbf{k} = \left(k_x, k_y\right)^T$ and possibly complex amplitudes $u_0, v_0, \eta_0$. If you put this in, you get

\[ \begin{align} -i\omega u_0 &= fv_0 - gik_x\newhat{\eta},\\ -i\omega v_0 &= -fu_0 - gik_y\newhat{\eta},\\ -i\omega\newhat{\eta} &= -Hik_xu_0 - Hik_yv_0\\ \Leftrightarrow-i\omega u_0 - fv_0 + gik_x\newhat{\eta} &= 0,\tag{16.96}\label{eq:deriv_in_grav_1}\\ -i\omega v_0 + fu_0 + gik_y\newhat{\eta} &= 0,\tag{16.97}\label{eq:deriv_in_grav_2}\\ -i\omega\newhat{\eta} + Hik_xu_0 + Hik_yv_0 &= 0.\tag{16.98}\label{eq:deriv_in_grav_3} \end{align} \]

In matrix notation this is:

\[ \begin{align} \left(\begin{array}{ccc} - i\omega& -f&gik_x\\ f& -i\omega&gik_y\\ Hik_x&Hik_y& -i\omega \end{array}\right)\left(\begin{array}{c} u_0\\ v_0\\ \newhat{\eta} \end{array}\right) &= \left(\begin{array}{c} 0\\ 0\\ 0 \end{array}\right). \end{align} \]

Non-trivial solutions exist here if and only if

\[ \begin{align} &\Leftrightarrow\left(-i\omega\right)\left[\left(-i\omega\right)^2 - i^2Hgk_y^2\right] - f\left[-f\left(-i\omega\right) - i^2Hgk_xk_y\right] + Hik_x\left[-fgik_y - \left(-i\omega\right)gik_x\right] = 0\nonumber\\ &\Leftrightarrow\left(-i\omega\right)\left[\left(-i\omega\right)^2 + Hgk_y^2\right] - f\left[-f\left(-i\omega\right) + Hgk_xk_y\right] + Hik_x\left[-fgik_y - \left(-i\omega\right)gik_x\right] = 0\nonumber\\ &\Leftrightarrow i\omega\left[-\left(-i\omega\right)^2 - Hgk_y^2\right] + f\left[f\left(-i\omega\right) - Hgk_xk_y\right] + Hik_x\left(-fgik_y + i\omega gik_x\right) = 0\nonumber\\ &\Leftrightarrow i\omega\left[-\left(i\omega\right)^2 - Hgk_y^2\right] + f\left(-fi\omega - Hgk_xk_y\right) + Hik_x\left(-fgik_y - \omega gk_x\right) = 0\nonumber \end{align} \] \[ \begin{align} &\Leftrightarrow i\omega\left(\omega^2 - Hgk_y^2\right) + f\left(-fi\omega - Hgk_xk_y\right) + Hik_x\left(-fgik_y - \omega gk_x\right) = 0\nonumber\\ &\Leftrightarrow\omega\left(\omega^2 - Hgk_y^2\right) + f\left(-f\omega + iHgk_xk_y\right) + Hk_x\left(-fgik_y - \omega gk_x\right) = 0\nonumber\\ &\Leftrightarrow\omega\left(\omega^2 - Hgk_y^2\right) - f^2\omega + fiHgk_xk_y - Hk_xfgik_y - Hk_x\omega gk_x = 0\nonumber\\ &\Leftrightarrow\omega\left(\omega^2 - Hgk_y^2\right) - f^2\omega - Hk_x\omega gk_x = 0\nonumber\\ &\Leftrightarrow\omega\left(\omega^2 - Hgk_y^2 - f^2 - Hgk_x^2\right) = 0\nonumber\\ &\Leftrightarrow\omega\left[\omega^2 - f^2 - Hg\left(k_x^2 + k_y^2\right)\right] = 0 \end{align} \]

applies. So the dispersion relation of Poincaré waves is:

This equation has the three solutions

\[ \begin{align} \omega &= 0,\tag{16.102}\label{eq:poincare_geos_mode}\\ \omega &= \sqrt{f^2 + Hgk^2},\tag{16.103}\label{eq:poincare_grav_mode_0}\\ \omega &= -\sqrt{f^2 + Hgk^2}.\tag{16.104}\label{eq:poincare_geos_mode_1} \end{align} \]

Eq. (16.102) is the geostrophic mode, the other two equations describe Poincare waves in the true sense. Eq. (16.104) describes a wave propagating in the direction $-\mathbf{k}$ and is therefore not a physically new wave mode. For these propagating solutions $\omega^2\geq f^2$ applies, so the magnitude of the Coriolis parameter is a lower bound of the angular frequency. applies to the phase speed

\[ \begin{align} c^2 = \frac{\omega^2}{k^2} = gH + \frac{f^2}{k^2} = gH + f^2L^2\frac{1}{4\pi^2} \end{align} \]

with $L$ as the wavelength. The Poincaré waves therefore have a higher phase speed than gravity waves without the rotation of the earth. The shorter the waves become, the smaller the influence of the Coriolis force becomes, but for long waves $\omega$ goes towards $f$. The expression for $c^2$ of the Poincaré waves is higher than that of gravity waves without Coriolis force by the expression $f^2L^2\frac{1}{4\pi^2}$. To determine a cutoff wavelength $R_o$ from which the effect of the Coriolis force becomes clear, one sets

\[ \begin{align} f_0^2R_o^2\frac{1}{4\pi^2} = \frac{gH}{4\pi^2}\approx 0, 025gH. \end{align} \]

This gives the barotropic Rossby radius

\[ \begin{align} R_o = \frac{\sqrt{gH}}{\left|f\right|}.\tag{16.107}\label{eq:def_baro_ro_r} \end{align} \]

If you use typical values, $R_o \approx 2000$ km follows.

Poincaré waves are also known in English as inertia-gravity waves. This is because the restoring forces in this wave mode are gravity and the Coriolis force. The German translation „ Inertial gravity wave“ is rather unusual. Another name is Sverdrup wave.

16.5.4 Kelvin waves

One starts again from the system of equations (16.89) - (16.91), but this time one imagines the half-plane $x<0$ as a coast, so that only the remaining half-plane $x\geq 0$ can be flowed by the fluid. So the boundary condition is:

\[ \begin{align} u\left(x = 0\right) = 0. \end{align} \]

We are now looking for solutions that fulfill this globally. To do this, we start again from the f-plane. The system of equations reduces to

\[ \begin{align} 0 &= -g\frac{\partial\eta}{\partial x} + fv, \tag{16.109}\label{eq:kelvin_1}\\ \frac{\partial v}{\partial t} &= -g\frac{\partial\eta}{\partial y}, \tag{16.110}\label{eq:kelvin_2}\\ \frac{\partial\eta}{\partial t} + H \frac{\partial v}{\partial y} &= 0.\tag{16.111}\label{eq:kelvin_3} \end{align} \]

From Eq. (16.109) it follows that $v$ is geostrophically balanced, so one can eliminate $v$ from the remaining two equations:

\[ \begin{align} \frac{\partial^2\eta}{\partial t\partial x} &= -f\frac{\partial\eta}{\partial y},\\ \frac{\partial\eta}{\partial t} + H\frac{g}{f}\frac{\partial^2\eta}{\partial x\partial y} &= 0. \end{align} \]

Here you do the harmonic wave approach

\[ \begin{align} \eta = \eta_0\exp\left[i\left(k_xx + k_yy - \omega t\right)\right], \end{align} \]

from which follows

\[ \begin{align} -iwik_x &= -fik_y\Rightarrow wk_x = -ifk_y, \tag{16.115}\label{eq:kelvin_deriv_1}\\ -i\omega + H\frac{g}{f}ik_xik_y &= 0 \Rightarrow i\omega = -H\frac{g}{f}k_xk_y.\tag{16.116}\label{eq:kelvin_deriv_2} \end{align} \]

From Eq. (16.115) follows

\[ \begin{align} k_x = -\frac{i}{\omega}fk_y, \end{align} \]

what in Eq. (16.116) is used

\[ \begin{align} i\omega &= -H\frac{g}{f}k_y\left(-\frac{i}{\omega}fk_y\right) = iH\frac{g}{\omega}k_y^2 \end{align} \]

results. Define $\kappa$ by

\[ \begin{align} \kappa \coloneqq \frac{\omega}{\sqrt{gH}} > 0 \end{align} \]

with an angular frequency $\omega > 0$ assumed to be positive, follows

\[ \begin{align} k_y = \pm\kappa.\tag{16.120}\label{eq:kelvin_deriv_3} \end{align} \]

This results in transformed

With Eq. (16.115) follows

$\eta$ must not increase exponentially as a function of the distance from the coast, therefore when $f > 0$ the negative sign in Eq. (16.120), at $f < 0$ the positive. In the northern hemisphere the Kelvin wave has the coast to the right of the direction of propagation, in the southern hemisphere to the left.

The penetration lengths $l$ of the Kelvin waves are:

with that in Eq. (16.107) defined Rossby radius $R_o$.

Fig. 16.2 shows the typical penetration lengths of Kelvin waves, which are quickly in the range of $1000$ - $10000$ km, i.e. in the range of ten to several hundred percent of a quarter of the earth's circumference. This extension far exceeds the validity of the f-plane.

16.5.4.1 Equatorial Kelvin waves

At the equator you cannot set $f = f_0$, you set the $\beta $plane instead

\[ \begin{align} f = \beta y \end{align} \]

with

\[ \begin{align} \beta \coloneqq \frac{\partial f}{\partial y}\newvline_{\varphi = 0} \end{align} \]

to. The equations are therefore (16.89) - (16.91)

\[ \begin{align} \frac{\partial u}{\partial t} &= -g\frac{\partial\eta}{\partial x} + \beta yv,\\ \frac{\partial v}{\partial t} &= -g\frac{\partial\eta}{\partial y} - \beta yu,\\ \frac{\partial\eta}{\partial t} + H\left(\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y}\right) &= 0. \end{align} \]

Now you look for solutions with $v = 0$. In this case applies

\[ \begin{align} \frac{\partial u}{\partial t} &= -g\frac{\partial\eta}{\partial x},\\ 0 &= -g\frac{\partial\eta}{\partial y} - \beta yu,\\ \frac{\partial\eta}{\partial t} + H\frac{\partial u}{\partial x} &= 0. \end{align} \]

Over

\[ \begin{align} u = -\frac{g}{\beta y}\frac{\partial\eta}{\partial y} \end{align} \]

$u$ can be eliminated:

\[ \begin{align} -\frac{g}{\beta y}\frac{\partial^2\eta}{\partial t\partial y} &= -g\frac{\partial\eta}{\partial x}\\ \frac{\partial\eta}{\partial t} - \frac{gH}{\beta y}\frac{\partial^2\eta}{\partial x\partial y} &= 0 \end{align} \]

The approach

\[ \begin{align} \eta = \newhat{\eta}\exp\left[i\left(k_xx + k_yy -\omega t\right)\right] \end{align} \]

performs

\[ \begin{align} -\frac{g}{\beta y}k_y\omega &= -gik_x \Rightarrow k_y = \frac{1}{\omega}\beta yik_x,\\ -i\omega + \frac{gH}{\beta y}k_xk_y &= 0 \Rightarrow \omega^2 = gHk_x^2. \end{align} \]

So it applies

\[ \begin{align} k_x = \pm\frac{\omega}{\sqrt{gH}}. \end{align} \]

For $k_y$ follows

\[ \begin{align} k_y = \frac{1}{\omega}\beta yik_x = \pm i\frac{\beta y}{\sqrt{gH}}. \end{align} \]

Only the plus sign comes into question. So the solution is

To estimate the meridional extent $y_0$ of these waves, one sets

\[ \begin{align} 1 = \frac{2\omega y_0^2}{\sqrt{gH}} \end{align} \]

to. From this it follows

\[ \begin{align} y_0 =\sqrt{R\frac{\sqrt{gH}}{2\omega}} = 2071\text{ km} \end{align} \]

with $R$ as the equatorial radius and $H = 1$ km.

16.5.5 Rossby-Haurwitz fashions

Rossby-Haurwitz modes are the eigenmodes of the barotropic vorticity equation Eq. (15.92) on the sphere. Therefore, the notation from section 15.1.3.6 is adopted in this section. Spherical surface functions $Y_{l, m}\left(\phi, \lambda\right)$ with $l \geq 0$ and $-l \leq m \leq l$ satisfy

\[ \begin{align} \Delta Y_{l, m} \stackrel{\href{ch-41-orthogonal-function-systems.html#eq:spherical_harm_prop_1}{\text{Glg. (C.166)}}}{=} -\frac{l\left(l + 1\right)}{a^2}Y_{l, m}. \end{align} \]

Do the approach for the current function $\psi = \psi\left(\phi, \lambda, t\right)$

\[ \begin{align} \psi = Y_{l, m}e^{-i\omega t},\tag{16.144}\label{eq:rossby-haurwitz_ansatz} \end{align} \]

follows from this

\[ \begin{align} J\left(\zeta, \psi\right) = J\left(\Delta\psi, \psi\right) \propto \frac{\partial\psi}{\partial\lambda}\frac{\partial\psi}{\partial\phi} - \frac{\partial\psi}{\partial\phi}\frac{\partial\psi}{\partial\lambda} = 0. \end{align} \]

If you put Eq. (16.144) into the barotropic vorticity equation in the form Eq. (15.101), you get

\[ \begin{align} \Delta\frac{\partial\psi}{\partial t} &= -\frac{\partial\psi}{a\cos\left(\phi\right)\partial\lambda}\beta + \frac{1}{a^2\cos\left(\phi\right)}J\left(\Delta\psi, \psi\right) = -\frac{\partial\psi}{a\cos\left(\phi\right)\partial\lambda}\beta\nonumber\\ \Delta\frac{\partial\psi}{\partial t} &= -\frac{\partial\psi}{a^2\cos\left(\phi\right)\partial\lambda}2\omega\cos\left(\phi\right) = -\frac{\partial\psi}{a^2\partial\lambda}2\Omega\nonumber\\ \Rightarrow-i\omega l\frac{l\left(l + 1\right)}{a^2} &= -\frac{2\Omega}{a^2}im\nonumber\\ \Rightarrow\omega\frac{l\left(l + 1\right)}{a^2} &= \frac{2\Omega}{a^2}m\tag{16.146}\label{eq:rossby-haurwitz_disprel_deriv} \end{align} \]

From this follows the dispersion relation of the Rossby-Haurwitz modes:

In the case $l = 0$, $m = 0$ and Eq. (16.146) is satisfied for every $\omega$. From Eq. (16.147) follows

\[ \begin{align} -\frac{1}{\left(l + 1\right)}2\Omega \leq \omega\left(l, m\right) \leq \frac{1}{\left(l + 1\right)}2\Omega. \end{align} \]

For the phase velocity (expressed as zonal angular velocity) one obtains

\[ \begin{align} \frac{\omega}{\frac{2\pi}{\Delta\lambda}} = \frac{\omega\Delta\lambda}{2\pi} = \frac{\omega 2\pi}{2\pi m} = \frac{\omega}{m} = \frac{1}{l\left(l + 1\right)}2\Omega > 0. \end{align} \]

Rossby-Haurwitz fashions always propagate to the East. Using Eq. (15.139), which is

\[ \begin{align} \Delta\Phi = \beta\frac{\partial\psi}{\partial y} + f\Delta\psi, \end{align} \]

The geopotential $\Phi$ can be determined from $\psi$. First you set a Rossby-Haurwitz fashion

\[ \begin{align} \psi = \psi_0Y_{l, m}e^{-i\omega t} \end{align} \]

one, this implies

\[ \begin{align} \Delta\Phi &= \frac{2\Omega\cos\left(\phi\right)}{a^2\cos\left(\phi\right)}\frac{\partial\psi}{\partial\phi} - \frac{2\Omega\sin\left(\phi\right)}{a^2}l\left(l + 1\right)\psi_0Y_{l, m}e^{-i\omega t}\nonumber\\ \Rightarrow\Delta\Phi &= \frac{2\Omega}{a^2}\frac{\partial\psi}{\partial\phi} - \frac{2\Omega\sin\left(\phi\right)}{a^2}l\left(l + 1\right)\psi_0Y_{l, m}e^{-i\omega t}. \end{align} \]

The approach is now made for the geopotential $\Phi$

\[ \begin{align} \Phi = \sum_{l' = 0}^\infty\sum_{m' = -l'}^{l'}\Phi_{l', m'}Y_{l', m'}. \end{align} \]

If you put this in, you get

\[ \begin{align} \sum_{l' = 0}^\infty\sum_{m' = -l'}^{l'}-\frac{l'\left(l' + 1\right)}{a^2}\Phi_{l', m'}Y_{l', m'} &= \frac{2\Omega}{a^2}\frac{\partial\psi}{\partial\phi} - \frac{2\Omega\sin\left(\phi\right)}{a^2}l\left(l + 1\right)\psi_0Y_{l, m}e^{-i\omega t}\nonumber\\ \Rightarrow\sum_{l' = 0}^\infty\sum_{m' = -l'}^{l'}-l'\left(l' + 1\right)\Phi_{l', m'}Y_{l', m'} &= 2\Omega\frac{\partial\psi}{\partial\phi} - 2\Omega\sin\left(\phi\right)l\left(l + 1\right)\psi_0Y_{l, m}e^{-i\omega t}. \end{align} \]

Eq. (C.180) is

\[ \begin{align} \sin\left(\phi\right)Y_{l, m} = \sqrt{\frac{l^2 - m^2}{4l^2 - 1}}Y_{l - 1, m} + \sqrt{\frac{\left(l + 1\right)^2 - m^2}{4\left(l + 1\right)^2 - 1}}Y_{l + 1, m}. \end{align} \]

This implies

\[ \begin{align} \sum_{l' = 0}^\infty\sum_{m' = -l'}^{l'}-l'\left(l' + 1\right)\Phi_{l', m'}Y_{l', m'} &= 2\Omega\frac{\partial\psi}{\partial\phi} - 2\Omega l\left(l + 1\right)\psi_0\left[\sqrt{\frac{l^2 - m^2}{4l^2 - 1}}Y_{l - 1, m} + \sqrt{\frac{\left(l + 1\right)^2 - m^2}{4\left(l + 1\right)^2 - 1}}Y_{l + 1, m}\right]e^{-i\omega t}. \end{align} \]

16.5.6 Barotropic Rossby waves

In section 16.5.3 $\beta = 0$ was assumed. In this case, the Poincaré waves derived there are the most general wave solutions. A new type of waves arises when the width dependence of $f$ is taken into account.

One assumes a homogeneous zonal base current $U$. One assumes $u_0 = 0$, and one also assumes $k_y = 0$, the disturbances should only be x-dependent. You already know:

• Since $U$ is homogeneous and $u_0 = 0$, $v = v\left(x, t\right)$, the solutions are divergence-free.

Therefore one can use the barotropic vorticity equation. You do the approach

\[ \begin{align} v = v_0\exp\left[i\left(kx - \omega t\right)\right] \end{align} \]

then apply

\[ \begin{align} \zeta = ikv, & {} & \frac{\partial\zeta}{\partial t} = k\omega v, & {} & \frac{\partial\zeta}{\partial x} = -k^2v. \end{align} \]

Plugging this into the barotropic vorticity equation you get

\[ \begin{align} k\omega v_0 &= Uk^2v_0 - v_0\beta\Leftrightarrow\omega = Uk - \frac{\beta}{k}.\tag{16.159}\label{eq:rossby_welle_barotrop_disp} \end{align} \]

Eq. (16.159) is the dispersion relation of barotropic Rossby waves. If they span a complete circle of latitude, they are also referred to as planetary waves.

The following applies to the phase velocity of barotropic Rossby waves

In the hypothetical case $U\to0$, $L = 2\pi a$ and $\varphi = 0$ holds

\[ \begin{align} \left|c\right| = 2\omega\frac{1}{a}\frac{4\pi^2 a^2}{4\pi^2} = 2\omega a = 930\:\frac{\text{m}}{\text{s}}. \end{align} \]

This is an amount limit of $c$. No wave can propagate faster than the speed $2\omega a$, and the system of governing equations cannot transmit information faster.

16.5.6.1 Clear understanding

The y component of the momentum equation in the shallow water model is

\[ \begin{align} \md{v} = -fu - g\frac{\partial\eta}{\partial y}.\tag{16.162}\label{eq:momentum_swe_y} \end{align} \]

Do a linear Taylor expansion for $f$

\[ \begin{align} f = f_0 + \beta y. \end{align} \]

A homogeneous zonal base current $U$ can be geostrophically balanced by surface deflection

\[ \begin{align} 0 = -f_0U - g\frac{\partial\eta}{\partial y}. \end{align} \]

In this case, Eq. (16.162) to

\[ \begin{align} \md{^2y}{t^2} = -\beta U y. \end{align} \]

This corresponds to a harmonic oscillator with the natural frequency

\[ \begin{align} \omega_{\text{ind}} = \sqrt{\beta U}. \end{align} \]

Therefore $U>0$ must be western. The natural frequency was marked with an index $\text{ind}$ because it is the individual oscillation frequency of the particles and not the frequency $\omega$ of the wave.

Another clear explanation comes from the barotropic vorticity equation, which is:

\[ \begin{align} f + \zeta = \text{const}. \end{align} \]

If a particle is deflected northward from its resting position in the northern hemisphere when there is a western background current, then $f$ increases. Therefore $\zeta$ must decrease and the particle acquires anticyclonic relative vorticity. It then oscillates southward beyond its rest position, with $f$ decreasing. Therefore, the particle acquires cyclonic relative vorticity and veers northward again. The mechanism of barotropic Rossby waves is the conservation of absolute vorticity, see also Fig. 16.3.

Barotropic Rossby waves are relevant in the atmosphere and ocean despite the extensive simplifications (barotropy and freedom from divergence). In the atmosphere, due to their freedom from divergence, they are most applicable to the middle troposphere and provide a simple view of the meandering polar jet. The term wave number is often used here; a wave of wave number $n$ has a zonal spatial period of $\frac{2\pi}{n}$ as an angle. Rossby waves always propagate westward in the ocean due to the small easterly (east-facing) base current (it is $\beta\geq0$ over the entire Earth).

16.5.7 Oceanic tide

The oceanic tide is a global phenomenon, so advection ($\propto 1/\lambda$) can be neglected. Barotropy can also be assumed. Therefore, the linear shallow water equations Equations (13.173) - (13.174) are used as an equation system. These are in components

\[ \begin{align} \frac{\partial u}{\partial t} &= -\frac{\partial\left(gd + U\right)}{a\cos\left(\phi\right)\partial\lambda} + 2\Omega\sin\left(\phi\right)v,\\ \frac{\partial v}{\partial t} &= -\frac{\partial\left(gd + U\right)}{a\partial\phi} - 2\Omega\sin\left(\phi\right)u,\\ \frac{\partial d}{\partial t} &= -D\left(\frac{\partial u}{a\cos\left(\phi\right)\partial\lambda} + \frac{\partial v}{a\partial\phi} - v\frac{\tan\left(\phi\right)}{a}\right). \end{align} \]

Here $U = U\left(\phi, \lambda, t\right)$ is the tidal potential. $\Omega$ is the angular velocity of the Earth's rotation. These equations are also known as Laplace's tidal equations (Laplace's tidal equations (LTEs)). The approach is taken here

\[ \begin{align} u = \sum_{l = 0}^\infty\sum_{m = -l}^l\newtilde{u}_{l, m}Y_{l, m}\left(\phi, \lambda\right)e^{-i\omega t}, & {} & v = \sum_{l = 0}^\infty\sum_{m = -l}^l\newtilde{v}_{l, m}Y_{l, m}\left(\phi, \lambda\right)e^{-i\omega t},\\ d = \sum_{l = 0}^\infty\sum_{m = -l}^l\newtilde{d}_{l, m}Y_{l, m}\left(\phi, \lambda\right)e^{-i\omega t}, & {} & U = \sum_{l = 0}^\infty\sum_{m = -l}^l\newtilde{U}_{l, m}Y_{l, m}\left(\phi, \lambda\right)e^{-i\omega t}, \end{align} \]

Here $Y_{l, m}\left(\phi, \lambda\right)$ are spherical surface functions, see section C.5. They apply

\[ \begin{align} \frac{\partial Y_{l, m}}{\partial\phi} &\stackrel{\href{ch-41-orthogonal-function-systems.html#eq:spherical_harmonic_deriv_theta}{\text{Glg. (C.167)}}}{=} -m\tan\left(\phi\right) Y_{l, m} - \sqrt{l^2 - m^2 + l - m}Y_{l, m + 1}\exp\left(-i\lambda\right),\\ \frac{\partial Y_{l, m}}{\partial\lambda} &\stackrel{\href{ch-41-orthogonal-function-systems.html#eq:spherical_harmonic_deriv_phi}{\text{Glg. (C.168)}}}{=} imY_{l, m},\\ \sin\left(\phi\right)Y_{l, m} &\stackrel{\href{ch-41-orthogonal-function-systems.html#eq:spherical_harm_prop_2}{\text{Glg. (C.171)}}}{=} \sqrt{\frac{l^2 - m^2}{4l^2 - 1}}Y_{l - 1, m} + \sqrt{\frac{\left(l + 1\right)^2 - m^2}{4\left(l + 1\right)^2 - 1}}Y_{l + 1, m}. \end{align} \]

With the abbreviations

\[ \begin{align} f_{l, m} = \sqrt{l^2 - m^2 + l - m},& {} & g_{l, m} = \sqrt{\frac{l^2 - m^2}{4l^2 - 1}},& {} & h_{l, m} = \sqrt{\frac{\left(l + 1\right)^2 - m^2}{4\left(l + 1\right)^2 - 1}} \end{align} \]

you can do this in the form

\[ \begin{align} \frac{\partial Y_{l, m}}{\partial\phi} &\stackrel{\href{ch-41-orthogonal-function-systems.html#eq:spherical_harmonic_deriv_theta_geo}{\text{Glg. (C.178)}}}{=} -m\tan\left(\phi\right) Y_{l, m} - f_{l, m}Y_{l, m + 1}\exp\left(-i\lambda\right),\\ \frac{\partial Y_{l, m}}{\partial\lambda} &\stackrel{\href{ch-41-orthogonal-function-systems.html#eq:spherical_harmonic_deriv_phi_geo}{\text{Glg. (C.179)}}}{=} imY_{l, m},\\ \sin\left(\phi\right)Y_{l, m} &\stackrel{\href{ch-41-orthogonal-function-systems.html#eq:spherical_harm_prop_2_geo}{\text{Glg. (C.180)}}}{=} g_{l, m}Y_{l - 1, m} + h_{l, m}Y_{l + 1, m} \end{align} \]

note down. Put all of this into the LTEs and you get

\[ \begin{align} &-i\omega\sum_{l = 0}^\infty\sum_{m = -l}^l\newtilde{u}_{l, m}Y_{l, m}\nonumber\\ &= \sum_{l = 0}^\infty\sum_{m = -l}^l-im\frac{g\newtilde{d}_{l, m} + \newtilde{U}_{l, m}}{a\cos\left(\phi\right)}Y_{l, m} + 2\Omega\newtilde{v}_{l, m}\left(g_{l, m}Y_{l - 1, m} + h_{l, m}Y_{l + 1, m}\right),\\ &-i\omega\sum_{l = 0}^\infty\sum_{m = -l}^l\newtilde{v}_{l, m}Y_{l, m}\nonumber\\ &= \sum_{l = 0}^\infty\sum_{m = -l}^l\frac{g\newtilde{d}_{l, m} + \newtilde{U}_{l, m}}{a}\left(m\tan\left(\phi\right) Y_{l, m} + f_{l, m}Y_{l, m + 1}\exp\left(-i\lambda\right)\right) - 2\Omega\newtilde{u}_{l, m}\left(g_{l, m}Y_{l - 1, m} + h_{l, m}Y_{l + 1, m}\right),\\ &-i\omega\sum_{l = 0}^\infty\sum_{m = -l}^l\newtilde{d}_{l, m}Y_{l, m}\nonumber\\ &= -\frac{D}{a}\sum_{l = 0}^\infty\sum_{m = -l}^lim\frac{\newtilde{u}_{l, m}}{\cos\left(\phi\right)}Y_{l, m} - \newtilde{v}_{l, m}\left(m\tan\left(\phi\right) Y_{l, m} + f_{l, m}Y_{l, m + 1}\exp\left(-i\lambda\right)\right) - \newtilde{v}_{l, m}\tan\left(\phi\right)Y_{l, m}\nonumber\\ &= -\frac{D}{a}\sum_{l = 0}^\infty\sum_{m = -l}^lim\frac{\newtilde{u}_{l, m}}{\cos\left(\phi\right)}Y_{l, m} - \newtilde{v}_{l, m}\left[\left(m + 1\right)\tan\left(\phi\right) Y_{l, m} + f_{l, m}Y_{l, m + 1}\exp\left(-i\lambda\right)\right]. \end{align} \]

Coastlines, bathymetry and ground friction modify this significantly.

16.5.7.1 Clear understanding

In contrast to many clear illustrations, it is less the vertical but predominantly the horizontal component of $-\nabla U$ that is relevant for the oceanic tide. However, what is somewhat counterintuitive is the fact that in most places it is not the period $\sim$ 25 hours that predominates, but rather the first harmonic of the lunar tide with a period of $\sim$ 12.5 hours. To understand this clearly, place a KS at the center of the earth and align the negative x-axis with the moon. Lunar quantities (relative to the moon) are denoted by the index $l$ and terrestrial quantities by the index $t$. For the x-coordinate $x_S$ of the center of gravity one obtains

\[ \begin{align} x_S = \frac{m_lx_l + m_tx_t}{m_l + m_t} = \frac{m_lx_l}{m_l + m_t} = \frac{1}{1 + \frac{m_t}{m_l}}x_l = -0,73 a. \end{align} \]

The center of gravity of the moon and earth is therefore within the earth. At point $x = -a$, the tidal force $\mathbf{f}_T$ is directed towards the moon. At $x = a$ applies

\[ \begin{align} f_{T, x} = \omega_l^2\left(a - x_S\right) - \frac{Gm_l}{\left(a - x_l\right)^2} = 1,73\omega_l^2a - \frac{Gm_l}{\left(a - x_l\right)^2}. \end{align} \]

$\omega_l$ is the angular velocity of the moon's rotation around the earth. It is easy to verify that

\[ \begin{align} f_{T, x} > 0 \end{align} \]

applies. At the two points considered so far, the tidal force has no horizontal component. If you move away from these points, the horizontal tidal force is directed towards these points. This implies that there must be another point on every semicircle between $x = -a$ and $x = a$ where $\mathbf{f}_T = -\nabla U$ has no horizontal component. This suggests that the period $\sim$ 12.5 hours is more dominant than the lunar fundamental period.

16.6 Baroclinic waves

16.6.1 Vertical modes

Here we assume a flat subsurface at a depth of $z = -D < 0$ and demand the boundary conditions

\[ \begin{align} w\left(z = 0\right) &= 0,\\ w\left(z = -D\right) &= 0. \end{align} \]

The following system of equations is used, whereby one o. B. d. A. starts from a wave propagating in the x direction:

\[ \begin{align} \frac{\partial u}{\partial t} &= -\frac{1}{\rho_0}\frac{\partial p'}{\partial x},\\ \frac{\partial w}{\partial t} &= -\frac{1}{\rho_0}\frac{\partial p'}{\partial z},\\ \frac{\partial\rho'}{\partial t} &= -\rho_0\frac{\partial u}{\partial x} - \rho_0\frac{\partial w}{\partial z}. \end{align} \]

Here is

\[ \begin{align} \rho = \rho_0 + \rho' \end{align} \]

with a homogeneous background density $\rho_0$ and a fluctuation $\rho'$. Analogously, one decomposes the pressure $p$, where should apply

\[ \begin{align} \frac{\partial p_0}{\partial z} = -g\rho_0. \end{align} \]

You now make the approaches

\[ \begin{align} u = U\left(z\right)\exp\left(ikx-i\omega t\right), & {} & v = V\left(z\right)\exp\left(ikx-i\omega t\right),\\ p' = P\left(z\right)\exp\left(ikx-i\omega t\right), & {} & \rho' = \newtilde{\rho}\left(z\right)\exp\left(ikx-i\omega t\right). \end{align} \]

If you insert this into the applicable system of equations, it follows

\[ \begin{align} -i\omega U = -\frac{ikP}{\rho_0} \Rightarrow \omega U &= \frac{kP}{\rho_0}, \tag{16.195}\label{eq:v_mode_deriv_2}\\ -i\omega W = -\frac{P'}{\rho_0} \Rightarrow \omega W &= -i\frac{P'}{\rho_0}, \tag{16.196}\label{eq:v_mode_deriv_1}\\ -i\omega\newtilde{\rho} = -\rho_0ikU - \rho_0W' \Rightarrow \omega\newtilde{\rho} &= \rho_0kU - i\rho_0W'. \end{align} \]

The third equation is always satisfiable, $\newtilde{\rho}\left(z\right)$ can be chosen accordingly. For $W$ you do the approach

\[ \begin{align} W\left(z\right) = \newhat{w}\sin\left(n\pi\frac{z}{D}\right) \end{align} \]

with $n\geq 1$. From Eq. (16.196) follows

\[ \begin{align} P' = \rho_0i \omega\newhat{w}\sin\left(n\pi\frac{z}{D}\right). \end{align} \]

Therefore you bet for $P$

\[ \begin{align} P\left(z\right) = -\frac{D\rho_0i\omega\newhat{w}}{n\pi}\cos\left(n\pi\frac{z}{D}\right) \end{align} \]

to. Putting this into Eq. (16.196) follows

\[ \begin{align} U\left(z\right) = \frac{kP}{\omega\rho_0} = -\frac{ki\newhat{w}D}{n\pi}\cos\left(n\pi\frac{z}{D}\right). \end{align} \]

So it applies

\[ \begin{align} u\left(x, z, t\right) &= \frac{Dk\newhat{w}}{n\pi}\cos\left(n\pi\frac{z}{D}\right)\exp\left[i\left(kx - \omega t - \frac{\pi}{2}\right)\right]\nonumber\\ \Rightarrow u\left(x, z, t\right) &= -i\frac{Dk\newhat{w}}{n\pi}\cos\left(n\pi\frac{z}{D}\right)\exp\left[i\left(kx - \omega t\right)\right]\nonumber\\ \Rightarrow u\left(x, z, t\right) &= -i\frac{Dk\newhat{w}}{n\pi}\cos\left(n\pi\frac{z}{D}\right)\left[\cos\left(kx - \omega t\right) + i\sin\left(kx - \omega t\right)\right]\nonumber\\ \Rightarrow u\left(x, z, t\right) &= \frac{Dk\newhat{w}}{n\pi}\cos\left(n\pi\frac{z}{D}\right)\left[-i\cos\left(kx - \omega t\right) + \sin\left(kx - \omega t\right)\right]\nonumber\\ \Rightarrow\Re\left(u\left(x, z, t\right)\right) &= \frac{Dk\newhat{w}}{n\pi}\cos\left(n\pi\frac{z}{D}\right)\sin\left(kx - \omega t\right). \end{align} \]

The horizontal divergence $\delta$ at the surface results in

\[ \begin{align} \delta\left(z = 0\right) = \frac{\partial\Re\left(u\left(x, z, t\right)\right)}{\partial x} = \frac{Dk^2\newhat{w}}{n\pi}\cos\left(n\pi\frac{z}{D}\right)\cos\left(kx - \omega t\right). \end{align} \]

Vertical modes therefore lead to alternating stripes of divergence and convergence at the surface. In the area of ​​convergence, the surface waves are compressed horizontally, which destabilizes the wave field. Conversely, the wave field is stabilized in the area of ​​divergence, which also contributes to the fact that in this area the water comes from the depths to the surface and this water has not yet been exposed to the effects of wind.

The mode $n = 0$ is also called external mode because it has no internal vertical structure.

16.6.2 Gravity waves

Baroclinic gravity waves are the baroclinic analogue of Poincaré waves. First you make a perturbation of the shape

\[ \begin{align} u = U + u', & {} & v = v', & {} & w = w',\tag{16.204}\label{eq:gw_disp_deriv_0}\\ p = p_0\left(z\right) + p', & {} & \theta = \theta_0\left(z\right) + \theta', & {} & f = f_0.\tag{16.205}\label{eq:gw_disp_deriv_1} \end{align} \]

Because of Eq. (16.205) is Eq. (16.204) no restriction of the background current to the x-direction; by rotating the coordinate system around the vertical axis, the background wind can assume any direction. In particular, all curvature terms are neglected and one can use Cartesian coordinates $\left(x, y, z\right)$. The background state is supposed to be hydrostatic, so

\[ \begin{align} \frac{\partial p_0}{\partial z} = -g\rho_0\left(z\right)\tag{16.206}\label{eq:hydrostatic_gravity_wave} \end{align} \]

fulfill. A dry adiabatic system is assumed; in such a system the first law of thermodynamics becomes true

\[ \begin{align} \md{\theta} = 0 & {} & \Leftrightarrow & {} & \md{\theta} + \md{\theta'} = 0 & {} & \Leftrightarrow \md{\theta'} + w'\frac{d\theta_0}{dz} = 0. \end{align} \]

One now introduces an approximate material derivation $\mdtilde{}$ in which disturbing products are neglected. Equal signs are noted below for this purpose. This gives you

\[ \begin{align} \mdtilde{\theta'} + w'\frac{d\theta_0}{dz} &= 0. \end{align} \]

If you multiply this equation by $\frac{g}{\theta_0}$, it follows

\[ \begin{align} \mdtilde{b'} + w'\frac{g}{\theta_0}\frac{d\theta_0}{dz} = 0, \end{align} \]

where the definition of the so-called buoyancy

\[ \begin{align} b' \coloneqq g\frac{\theta'}{\theta_0} \end{align} \]

was used. From Eq. (16.253) can be seen

\[ \begin{align} \frac{g}{\theta_0}\frac{d\theta_0}{dz} = N^2, \end{align} \]

what on

\[ \begin{align} \mdtilde{b'} + w'N^2 = 0 \end{align} \]

leads. With Eq. (9.69) is obtained

\[ \begin{align} \md{p} + \frac{c^{(p)}}{c^{(V)}}p\nabla\cdot\mathbf{v} &= 0\nonumber\\ \Leftrightarrow \frac{1}{\kappa p}\md{p} + \nabla\cdot\mathbf{v} &= 0\nonumber\\ \Leftrightarrow \frac{1}{\kappa\rho R_dT}\md{p} + \nabla\cdot\mathbf{v} &= 0. \end{align} \]

With Eq. (16.42) this can be expressed in terms of the speed of sound:

\[ \begin{align} \frac{1}{c_s^2\rho}\md{p} + \nabla\cdot\mathbf{v} &= 0 \end{align} \]

If you linearize this equation, you get

\[ \begin{align} \frac{1}{c_s^2\rho}\mdtilde{p} + \nabla\cdot\mathbf{v}' &= 0\nonumber\\ \Leftrightarrow \frac{1}{c_s^2\rho_0}\mdtilde{p}' + \frac{1}{c_s^2\rho_0}w'\frac{dp_0}{dz} + \nabla\cdot\mathbf{v}' &= 0\nonumber\\ \Leftrightarrow \frac{1}{c_s^2\rho_0}\mdtilde{p}' - \frac{g}{c_s^2}w' + \nabla\cdot\mathbf{v}' &= 0. \end{align} \]

For the vertical equation of motion one obtains using Eq. (13.151)

\[ \begin{align} \mdtilde{w'} = -g - \frac{1}{\rho}\frac{\partial p}{\partial z} = -\frac{1}{\rho}\frac{\partial p'}{\partial z} - \left(\frac{1}{\rho}\right)'\frac{\partial p_0}{\partial z} \approx - \frac{1}{\rho_0}\frac{\partial p'}{\partial z} + \frac{\rho'}{\rho_0^2}\frac{\partial p_0}{\partial z} = -\frac{1}{\rho_0}\frac{\partial p'}{\partial z} - \frac{g\rho'}{\rho_0}. \end{align} \]

To express the perturbation in the density through perturbations in the potential temperature and pressure, write down Eq. (9.65) in the form

\[ \begin{align} \rho = \frac{p_\text{ref}^{R_d/c^{(p)}}}{R_d}\frac{p^{1/\kappa}}{\theta}. \end{align} \]

In the first order, this applies

\[ \begin{align} \rho' &= \frac{1}{\kappa}\frac{\rho_0}{p_0}p' - \frac{\rho_0}{\theta}\theta' \approx \frac{p'}{c_s^2} - \frac{\rho_0}{\theta_0}\theta' \end{align} \]

From this it follows

\[ \begin{align} \mdtilde{w'} = -\frac{1}{\rho_0}\frac{\partial p'}{\partial z} - \frac{gp'}{\rho_0c_s^2} + \frac{g}{\theta_0}\theta' = -\frac{1}{\rho_0}\frac{\partial p'}{\partial z} - \frac{gp'}{\rho_0c_s^2} + b'. \end{align} \]

In summary, the linearized system of equations is:

Terms that would disappear with the hydrostatic approximation were marked blue, and all terms that would disappear if sound waves were ignored were marked red. The vertical dependency on $\rho_0$ is disturbing. Therefore one defines

\[ \begin{align} u'' \coloneqq \sqrt{\frac{\rho_0}{\rho_\text{SFC}}}u', & {} & v'' \coloneqq \sqrt{\frac{\rho_0}{\rho_\text{SFC}}}v', & {} & w'' \coloneqq \sqrt{\frac{\rho_0}{\rho_\text{SFC}}}w',\tag{16.225}\label{eq:bretherton_0}\\ b'' \coloneqq \sqrt{\frac{\rho_0}{\rho_\text{SFC}}}b', & {} & p'' \coloneqq \sqrt{\frac{\rho_\text{SFC}}{\rho_0}}p'.\tag{16.226}\label{eq:bretherton_1} \end{align} \]

which is called Bretherton transformation. The index SFC denotes the values ​​on the earth's surface. The transformation is essentially done by multiplying the system of equations by $\sqrt{\frac{\rho_0}{\rho_\text{SFC}}}$. The vertical leads require special attention:

\[ \begin{align} \frac{dw'}{dz} &= \sqrt{\frac{\rho_\text{SFC}}{\rho_0}}\frac{dw''}{dz} - \frac{1}{2}\sqrt{\frac{\rho_\text{SFC}}{\rho_0}}w''\frac{1}{\rho_0}\frac{d\rho_0}{dz} = \sqrt{\frac{\rho_\text{SFC}}{\rho_0}}\frac{dw''}{dz} + \frac{H}{2}w'\\ \frac{dp'}{dz} &= \sqrt{\frac{\rho_0}{\rho_\text{SFC}}}\frac{dp''}{dz} + \frac{1}{2}\sqrt{\frac{\rho_0}{\rho_\text{SFC}}}p''\frac{1}{\rho_0}\frac{d\rho_0}{dz} = \sqrt{\frac{\rho_0}{\rho_\text{SFC}}}\frac{dp''}{dz} + \frac{1}{2}p'\frac{1}{\rho_0}\frac{d\rho_0}{dz} = \sqrt{\frac{\rho_0}{\rho_\text{SFC}}}\frac{dp''}{dz} - \frac{H}{2}p' \end{align} \]

This was done

\[ \begin{align} \frac{1}{\rho_0}\frac{d\rho_0}{dz} = -\frac{1}{H}\tag{16.229}\label{eq:scale_h_density} \end{align} \]

with the scale height $H$ used. From this it follows

\[ \begin{align} -\frac{1}{\rho_0}\frac{\partial p'}{\partial z} &= -\sqrt{\frac{1}{\rho_0\rho_\text{SFC}}}\frac{dp''}{dz} + \frac{H}{2\rho_0}p' = -\sqrt{\frac{1}{\rho_0\rho_\text{SFC}}}\frac{dp''}{dz} + \frac{H}{2\sqrt{\rho_0\rho_\text{SFC}}}p'',\\ -\frac{\partial w'}{\partial z} &= -\sqrt{\frac{\rho_\text{SFC}}{\rho_0}}\frac{dw''}{dz} - \frac{H}{2}w' = -\sqrt{\frac{\rho_\text{SFC}}{\rho_0}}\frac{dw''}{dz} - \frac{H}{2}\sqrt{\frac{\rho_\text{SFC}}{\rho_0}}w''. \end{align} \]

In summary, this leads to the following system of equations:

An approach is now made for each of the five occurring variables

\[ \begin{align} \psi'' = A_\psi\exp\left[i\left(kx + ly + mz - \omega t\right)\right], \end{align} \]

where $A_\psi$ can be complex. It applies

\[ \begin{align} \mdtilde{A_\psi} = i\left(Uk - \omega\right)A_\psi = -i\omega_IA_\psi, \end{align} \]

where the definition of the so-called intrinsic frequency

\[ \begin{align} \omega_I \coloneqq \omega - kU \end{align} \]

was used. This is the frequency that would be measured in the coordinate system moving at speed $U$. If you write down the derived system of equations as a matrix $A$ and neglect the sound wave terms (red terms), you get

\[ \begin{align} A\cdot\left(\begin{array}{c} A_u\\ A_v\\ A_w\\ A_b\\ A_{p/\rho_\text{SFC}} \end{array}\right) = \left(\begin{array}{ccccc} -i\omega_I & -f & 0 & 0 & ik \\ f & -i\omega_I & 0 & 0 & il \\ 0 & 0 & \textcolor{blue}{-i\omega_I} & -1 & im - \frac{1}{2H}\\ 0 & 0 & N^2 & -i\omega_I & 0\\ ik & il & im + \frac{1}{2H} & 0 & 0\\ \end{array}\right)\cdot\left(\begin{array}{c} A_u\\ A_v\\ A_w\\ A_b\\ A_{p/\rho_\text{SFC}} \end{array}\right) = 0. \end{align} \]

Nontrivial solutions exist if the determinant $A$ of $A$ vanishes:

\[ \begin{align} A &\hastobe 0\nonumber\\ \Leftrightarrow 0 &= -i\omega_I\left\{-i\omega_I\left[i\omega_I\left(-m^2 - \frac{1}{4H^2}\right)\right] - il\left[\textcolor{blue}{-i\omega_I^2l} + N^2il\right]\right\}\nonumber\\ &+f\left\{f\left[i\omega_I\left(-m^2 - \frac{1}{4H^2}\right)\right] \textcolor{magenta}{- il\left[-\omega_I^2ik + ikN^2\right]}\right\}\nonumber\\ &+ik\left[\textcolor{magenta}{f\left(-\omega_I^2il + ilN^2\right)} - ik\left(\textcolor{blue}{i\omega_I^3} - N^2i\omega_I\right)\right]\nonumber\\ \Leftrightarrow 0 &= \left(f^2 - \omega_I^2\right)i\omega_I\left(-m^2 - \frac{1}{4H^2}\right) - \omega_Il\left[\textcolor{blue}{-i\omega_I^2l} + N^2il\right] + k^2\left(\textcolor{blue}{i\omega_I^3} - N^2i\omega_I\right)\nonumber\\ \end{align} \]

The magenta colored terms cancel each other out. The geostrophic mode $\omega_I = 0$ is of no further interest here, which is why division by $i\omega_I$ can be done:

\[ \begin{align} 0 &= \left(f^2 - \omega_I^2\right)\left(-m^2 - \frac{1}{4H^2}\right) + \textcolor{blue}{\omega_I^2l^2} - N^2l^2 + k^2\left(\textcolor{blue}{\omega_I^2} - N^2\right)\nonumber\\ \Leftrightarrow \omega_I^2\left(\textcolor{blue}{k^2 + l^2} + m^2 + \frac{1}{4H^2}\right) &= f^2\left(m^2 + \frac{1}{4H^2}\right) + N^2\left(l^2 + k^2\right)\nonumber \end{align} \]

This is the Dispersion relation of gravity waves, this is also visualized in Fig. 16.4. $\omega_I^2$ is a weighted average of the shares $f^2$ and $N^2$ with the weighting factors $m^2 + \frac{1}{4H^2}$ and $k^2 + l^2$. So $\omega_I^2$ is always between $f^2$ and $N^2$.

Often one wants to know the period of the expected oscillation as a function of the atmospheric background state $\left(U, N\right)$. The period assuming that the wave vector is parallel to the background wind is shown in Fig. 16.5.

16.6.2.1 Borderline cases

For $k^2 + l^2 \to 0$ (waves that do not propagate horizontally relative to the base current) one obtains from Eq. (16.242) Inertial waves:

For $m^2 + \frac{1}{4H^2} \to \infty$ one gets from Eq. (16.242) oscillations with frequency $N$:

$H \to \infty$ can be interpreted as neglecting vertical density fluctuations. $m \to 0$ means that the wave vector is oriented purely horizontally. These waves arise in a continuous, thermally stable medium when a particle is deflected vertically. A particle of density $\rho _0 = \rho (z_0)$ has its rest position at the height $z_0$, which is shown here above. A. is set to $z_0 = 0$. If you deflect the particle vertically, the equation of motion is:

\[ \begin{align} \frac{d^2z}{dt^2} = -\frac{1}{\rho_0}\frac{\partial p}{\partial z} - g = g\frac{\rho}{\rho_0} - g = \frac{g}{\rho_0}\left(\rho - \rho_0\right), \tag{16.245}\label{eq:momentum_schichtung} \end{align} \]

Let $z(t)$ be the deflection. It was assumed that the particle does not change its density during the deflection. If one approximates the deviation $\rho (z) - \rho _0$ using a Taylor expansion

\[ \begin{align} \rho\left(t\right) - \rho_0 = \frac{\partial\rho}{\partial z}(z = 0)z, \end{align} \]

this follows for the oscillation equation

\[ \begin{align} \frac{d^2z}{dt^2}(t) = \frac{g}{\rho_0}\frac{\partial\rho}{\partial z}z(t)\tag{16.247}\label{eq:schichtungs_dg}. \end{align} \]

If you set $z = Z_1\exp\left(iNt\right) + Z_2\exp\left(-iNt\right)$, this follows for the angular frequency

\[ \begin{align} N^2 = -\frac{g}{\rho}\frac{\partial\rho}{\partial z}, \end{align} \]

$N$ is the Brunt-Väisälä frequency, it is a measure of stability. If you insert $N$ into (16.247), you get

\[ \begin{align} \frac{d^2z}{dt^2} = -N^2z. \end{align} \]

This shows that in the case $N^2 > 0$ a sine wave follows as the solution, while in the case $N^2 < 0$ a real exponential function is the solution because the particle is accelerated away from its origin. In the case $N^2> 0$ the stratification is stable, while in the case $N^2<0$ it is unstable. Stable stratification is also referred to as strong stratification. In the case $N^2 = 0$ the stratification is neutral.

The above derivation is not yet completely complete because incompressibility was assumed. Strictly speaking, it is not the density that is retained during the deflection, but rather the potential density relative to the reference level. If $\newtilde{\rho}(z)$ denotes the density of the particle at a deflection $z$, then equation (16.245) becomes

\[ \begin{align} \frac{d^2z}{dt^2} = \frac{1}{\newtilde{\rho}\left(z\right)}g\rho\left(z\right) - g = \frac{g}{\newtilde{\rho}\left(z\right)} \left(\rho\left(z\right) - \newtilde{\rho}\left(z\right)\right). \end{align} \]

The expression $\rho\left(z\right) - \newtilde{\rho}\left(z\right)$ is the density difference between the surrounding fluid and the particle under consideration. If you bring both systems adiabatic to $z = 0$, then according to Eq. (9.44) for their density difference

\[ \begin{align} \Delta\rho = \left(\rho\left(z\right) - \newtilde{\rho}\left(z\right)\right)\left(\frac{p_0}{p}\right)^{1/\kappa}. \end{align} \]

Here $p \coloneqq p\left(z\right)$ and $p_0 \coloneqq p\left(0\right)$. In the first order $p = p_0$, $\Delta\rho$ is then the difference of the potential densities with respect to the reference level $z = 0$. Therefore you can also use the potential densities $\rho_\theta$ (related to $z = 0$). One obtains in the first order

\[ \begin{align} \frac{d^2z}{dt^2} = \frac{g}{\newtilde{\rho}(z)}\left(\rho_\theta(z) - \rho_\theta\left(0\right)\right) = \frac{g}{\newtilde{\rho}(z)}\frac{\partial\rho_\theta}{\partial z}z. \end{align} \]

For the Brunt-Väisälä frequency follows

This is the general expression for the Brunt-Väisälä frequency in a compressible medium. $N^2$ is a field that can depend on all three coordinates and time. The potential density must always be related to the level at which you are. In a dry atmosphere $N^2$ can also be expressed in terms of the potential temperature $\theta$:

\[ \begin{align} \rho_\theta = \frac{p_0}{R_d\theta} \end{align} \]

Thus follows

\[ \begin{align} \frac{\partial\rho_\theta}{\partial z} = -\frac{p_0}{R_d\theta^2}\frac{\partial\theta}{\partial z}. \end{align} \]

For the Brunt-Väisälä frequency this means

\[ \begin{align} N^2 &= \frac{g}{\theta}\frac{\partial\theta}{\partial z}.\tag{16.256}\label{eq:bv_freq_theta} \end{align} \]

Such waves, which are also referred to in English as buoyancy oscillations (the German translation „ stratification wave“ is rather unusual) arise, for example, as lee waves behind orography, in this case sinusoidal trajectories $(ut, z_0\sin\left(Nt\right))^T$ result. When saturation is reached, clouds form in the wave crests.

16.6.2.2 Hydrostatic gravity waves

Neglecting the blue terms in Eq. (16.242) provides the dispersion relation of hydrostatic gravity waves:

16.6.2.3 Group speed

From this equation it follows

\[ \begin{align} \nabla_\mathbf{k}\omega_I^2 = \left(\begin{array}{c} \frac{2N^2k}{m^2 + \frac{1}{4H^2}}\\ \frac{2N^2l}{m^2 + \frac{1}{4H^2}}\\ -2m\frac{N^2\left(k^2 + l^2\right)}{\left(m^2 + \frac{1}{4H^2}\right)^2} \end{array}\right). \end{align} \]

With $\omega' = \sign\left(\omega\right)\sqrt{\omega^2}' = \frac{1}{2\omega}\left(\omega^2\right)'$ one obtains for the group velocity of hydrostatic baroclinic gravity waves

\[ \begin{align} \mathbf{c}_\text{gr} = \frac{1}{\omega_I}\left(\begin{array}{c} \frac{N^2k}{m^2 + \frac{1}{4H^2}}\\ \frac{N^2l}{m^2 + \frac{1}{4H^2}}\\ -m\frac{N^2\left(k^2 + l^2\right)}{\left(m^2 + \frac{1}{4H^2}\right)^2} \end{array}\right) = \frac{1}{\omega_I}\frac{N^2}{m^2 + \frac{1}{4H^2}}\left(\begin{array}{c} k\\ l\\ -m\frac{k^2 + l^2}{m^2 + \frac{1}{4H^2}} \end{array}\right). \end{align} \]

Horizontally, the group velocity points in the same direction as the phase velocity. Vertically, the two velocities have opposite signs. For the vertical wavelength $l_z$ applies

\[ \begin{align} m^2 = \frac{4\pi^2}{l_z^2} \Rightarrow \frac{m^2}{\frac{1}{4H^2}} = \frac{16\pi^2H^2}{l_z^2} \approx 158\frac{H^2}{l_z^2}. \end{align} \]

Often applies

\[ \begin{align} \frac{m^2}{\frac{1}{4H^2}} \gg 1. \end{align} \]

Under this condition applies

\[ \begin{align} \mathbf{c}_\text{gr} &= \frac{\sign\left(\omega_I\right)}{\sqrt{f^2 + \frac{N^2}{m^2}\left(k^2 + l^2\right)}}\frac{N^2}{m^2}\left(\begin{array}{c} k\\ l\\ -\frac{k^2 + l^2}{m} \end{array}\right) = \frac{\sign\left(\omega_I\right)}{\sqrt{f^2m^2 + N^2\left(k^2 + l^2\right)}}\frac{N^2}{m}\left(\begin{array}{c} k\\ l\\ -\frac{k^2 + l^2}{m} \end{array}\right)\\ \Rightarrow\mathbf{c}_\text{gr}\cdot\mathbf{k} &\propto k^2 + l^2 - \left(k^2 + l^2\right) = 0. \end{align} \]

Phase velocity and group velocity are therefore perpendicular to each other.

16.6.2.4 Amplitude behavior

From the Bretherton transformation equations (16.225) - (16.226) and Eq. (16.229) follows

\[ \begin{align} u' \coloneqq \sqrt{\frac{\rho_\text{SFC}}{\rho_0}}u'' = \exp\left(\frac{z}{2H}\right)u'', & {} & v' \coloneqq \sqrt{\frac{\rho_\text{SFC}}{\rho_0}}v'' = \exp\left(\frac{z}{2H}\right)v'',\\ w' \coloneqq \sqrt{\frac{\rho_\text{SFC}}{\rho_0}}w'' = \exp\left(\frac{z}{2H}\right)w'', & {} & p' \coloneqq \sqrt{\frac{\rho_0}{\rho_\text{SFC}}}p'' = \exp\left(-\frac{z}{2H}\right)p''. \end{align} \]

The amplitude of the movements increases exponentially with height, and the height of the pressure fluctuations decreases exponentially.

16.6.2.5 Lee waves

Lee waves are stationary, i.e. h. it applies $\omega = 0$. From this it follows

\[ \begin{align} \omega_I = \omega - Uk = -Uk\Rightarrow \omega_I^2 = U^2k^2. \end{align} \]

If one assumes hydrostatic gravity waves, this can be done using Eq. Equate (16.257), from this it follows

\[ \begin{align} \omega_I^2 = U^2k^2 &= f^2 + \frac{N^2\left(k^2 + l^2\right)}{m^2 + \frac{1}{4H^2}}. \end{align} \]

Assuming $m^2 \gg \frac{1}{4H^2}$, $f = 0$ and $l = 0$, it follows

\[ \begin{align} \omega_I^2 = U^2k^2 = \frac{N^2k^2}{m^2}\Rightarrow U^2 &= \frac{N^2}{m^2} \Rightarrow m^2 = \frac{N^2}{U^2}. \end{align} \]

This follows for the intrinsic phase velocity

\[ \begin{align} \mathbf{c}_{I, \text{ph}} = \frac{\omega_I}{\sqrt{k^2 + m^2}}\frac{\mathbf{k}}{\left|\mathbf{k}\right|} = -\frac{Uk}{k^2 + m^2}\left( \begin{array}{c} k\\ 0\\ m \end{array} \right) \end{align} \]

as well as for the intrinsic group speed

\[ \begin{align} \mathbf{c}_{I, \text{gr}} = \nabla_\mathbf{k}\omega_I = \left( \begin{array}{c} -\frac{N}{m}\\ 0\\ \frac{Nk}{m^2} \end{array} \right). \end{align} \]

The intrinsic phase velocity is directed downwards and against the wind direction, the intrinsic group velocity is also negative in the x direction, but directed upwards.

In the case of non-hydrostatic gravity waves, Eq. (16.242), also under the assumptions $f = 0$ and $l = 0$,

\[ \begin{align} U^2k^2 &= \frac{N^2k^2}{m^2 + \frac{1}{4H^2} + k^2} \Leftrightarrow U^2\left(k^2 + m^2 + \frac{1}{4H^2}\right) = N^2\nonumber\\ \Leftrightarrow m^2 &= \frac{N^2}{U^2} - k^2 - \frac{1}{4H^2}. \end{align} \]

In the case $m^2 < 0$ the waves are vertically evanescent, so they cannot propagate vertically. This is the case for

\[ \begin{align} \frac{N^2}{U^2} - k^2 - \frac{1}{4H^2} < 0 & {} & \Leftrightarrow k^2 > \frac{N^2}{U^2} - \frac{1}{4H^2}\nonumber\\ \Leftrightarrow \frac{4\pi^2}{l_x^2} > \frac{N^2}{U^2} - \frac{1}{4H^2} & {} & \Leftrightarrow \frac{l_x^2}{4\pi^2} < \frac{1}{\frac{N^2}{U^2} - \frac{1}{4H^2}}\nonumber\\ \Leftrightarrow l_x < \frac{2\pi}{\sqrt{\frac{N^2}{U^2} - \frac{1}{4H^2}}}.& {} & \end{align} \]

Those spectral components of the orography that fulfill this inequality cannot generate upward propagating waves. Substituting typical values ​​of $N = 0.02$ Hz, $U = 8$ m/s and $H = 8$ km gives $l_x \approx 2.5$ km. With higher wind speed, $l_x$ increases.

16.6.3 Rossby waves

In order to understand so-called Rossby waves, one uses the quasi-geostrophic equation system derived in Sect. 13.11, consisting of the trend equation Eq. (13.226) and $\omega-$equation Eq. (13.231), but here in discretized form. The prognostic (state-determining) variables here are two current functions $\psi_1$, $\psi_3$. The resulting five layers are summarized in Tab. 16.1. The boundary conditions $\omega_0 = \omega_4 = 0$ are used for calculation.

Overview of the quasi-geostrophic two-layer model. See [26].

Index	Druck / hPa	definierte Funktion
$0$	$0$	$\omega_0 = 0$ (Randbedingung)
$1$	$250$	$\psi_1$
$2$	$500$	$\omega_2$
$3$	$750$	$\psi_3$
$4$	$1000$	$\omega_4 = 0$ (Randbedingung)

This approach is called quasigeostrophic two-layer model.

For the quasi-geostrophic potential vorticities applies

\[ \begin{align} q_1 &= f + \Delta\psi_1 + \frac{f_0^2}{\sigma}\frac{\frac{\partial\psi}{\partial p}\newvline_{_2} - \frac{\partial\psi}{\partial p}\newvline_{_0}}{\Delta p} = f + \Delta\psi_1 + \frac{f_0^2}{\sigma\Delta p^2}\left(\psi_3 - \psi_1\right) - \frac{f_0^2\frac{\partial\psi}{\partial p}\newvline_{_0}}{\sigma\Delta p},\\ q_3 &= f + \Delta\psi_3 + \frac{f_0^2}{\sigma}\frac{\frac{\partial\psi}{\partial p}\newvline_{_4} - \frac{\partial\psi}{\partial p}\newvline_{_2}}{\Delta p} = f + \Delta\psi_3 - \frac{f_0^2}{\sigma\Delta p^2}\left(\psi_3 - \psi_1\right) + \frac{f_0^2\frac{\partial\psi}{\partial p}\newvline_{_4}}{\sigma\Delta p}. \end{align} \]

$\Delta p \coloneqq500$ hPa is defined. No statement can be made about the last terms within the framework of this model, so they are assumed to be constant and are therefore ignored. One further defines a barotropic and a baroclinic component of the current function:

\[ \begin{align} \psi_M \coloneqq&\frac{\psi_1 + \psi_3}{2}, & {} & \psi_T \coloneqq \frac{\psi_1 - \psi_3}{2}. \end{align} \]

In addition, a baroclinic potential vorticity is defined:

\[ \begin{align} q_T \coloneqq\frac{q_1 - q_3}{2} = \frac{\zeta_1 - \zeta_3}{2} - \frac{f_0^2}{\sigma\Delta p^2}\left(\psi_1 - \psi_3\right) \end{align} \]

Furthermore, the stability wave number $K$ is given by

\[ \begin{align} K^2 \coloneqq\frac{2f_0^2}{\sigma\Delta p^2} \end{align} \]

defined. This allows you to take notes

\[ \begin{align} q_1 = f + \zeta_1 - K^2\psi_T, & {} & q_3 = f + \zeta_3 + K^2\psi_T, & {} & q_T = \zeta_T - K^2\psi_T. \end{align} \]

Now you make a disruption approach

\[ \begin{align} \psi = \newoverline{\psi} + \psi' = -\newoverline{u}y + \psi' \end{align} \]

with a mean homogeneous zonal flow vector $\newoverline{u}$. Now we assume a special case: we assume that the background wind is at level 1

\[ \begin{align} \newoverline{u}_1 = \newoverline{u}_T \end{align} \]

blowing and around level 3 the background wind

\[ \begin{align} \newoverline{u}_3 = -\newoverline{u}_T. \end{align} \]

Then apply

\[ \begin{align} \newoverline{\psi}_1 = -\newoverline{u}_Ty, & {} & \newoverline{\psi}_3 = \newoverline{u}_Ty. \end{align} \]

From this it follows

\[ \begin{align} \newoverline{\psi}_T &= -\newoverline{u}_Ty. \end{align} \]

For the ground state $\newoverline{\mathbf{v}_h}$ one can therefore note

\[ \begin{align} \newoverline{\mathbf{v}_h}_1 &= \left(\begin{array}{c} \newoverline{u}_T\\ 0 \end{array}\right),\\ \newoverline{\mathbf{v}_h}_3 &= \left(\begin{array}{c} - \newoverline{u}_T\\ 0 \end{array}\right). \end{align} \]

Let the field $\newoverline{u}_T$ be independent of $y$, then the following applies

\[ \begin{align} \newoverline{\zeta}_1 = \newoverline{\zeta}_3 = 0. \end{align} \]

For the potential vorticities applies

\[ \begin{align} \newoverline{q}_1 &= f_0 + \beta y - K^2\newoverline{\psi}_T = f_0 + \left(\beta + K^2\newoverline{u}_T\right)y,\\ \newoverline{q}_3 &= f_0 + \beta y + K^2\newoverline{\psi}_T = f_0 + \left(\beta - K^2\newoverline{u}_T\right)y. \end{align} \]

This field is stationary:

\[ \begin{align} \left(\frac{\partial}{\partial t} + \newoverline{\mathbf{v}_h}_1\cdot\nabla\right)\newoverline{q}_1 &= 0\\ \left(\frac{\partial}{\partial t} + \newoverline{\mathbf{v}_h}_3\cdot\nabla\right)\newoverline{q}_3 &= 0 \end{align} \]

If one also applies the perturbation approach to the form of material derivation used here and to the potential vorticity, it follows

\[ \begin{align} \md{_h^{(g)}} &= \frac{\partial}{\partial t} + \newoverline{\mathbf{v}_h}\cdot\nabla + \mathbf{v}_h'\cdot\nabla,\\ q &= \newoverline{q} + q'. \end{align} \]

Now one assumes again that the disturbances only have meridional components, i.e

\[ \begin{align} \mathbf{v}_{h, g, 1}' &= \left(\begin{array}{c} 0\\ v_1' \end{array}\right),\\ \mathbf{v}_{h, g, 3}' &= \left(\begin{array}{c} 0\\ v_3' \end{array}\right). \end{align} \]

Apply to the power functions

\[ \begin{align} v_1' = \frac{\partial\psi_1'}{\partial x}, & {} & v_3' = \frac{\partial\psi_3'}{\partial x}. \end{align} \]

This follows for the potential vorticites

\[ \begin{align} q_1' = \frac{\partial^2\psi_1'}{\partial x^2} - K^2\psi_T', & {} & q_3' = \frac{\partial^2\psi_3'}{\partial x^2} + K^2\psi_T'. \end{align} \]

To examine the tendency equations, write down

\[ \begin{align} \frac{\partial\newoverline{q}}{\partial t} + \newoverline{\mathbf{v}_h}\cdot\nabla\newoverline{q} + \frac{\partial q'}{\partial t} + \newoverline{\mathbf{v}_h}\cdot\nabla q' + \mathbf{v}_h'\cdot\nabla\newoverline{q} + \mathbf{v}_h'\cdot\nabla q' = 0. \end{align} \]

The first two terms result in zero, as already noted. The final dot product also disappears due to the orthogonality of the vectors involved. Therefore apply

\[ \begin{align} \left(\frac{\partial}{\partial t} + \newoverline{u}_T\frac{\partial}{\partial x}\right)q_1' + \left(\beta + K^2\newoverline{u}_T\right)\frac{\partial\psi_1'}{\partial x} &= 0, \tag{16.298}\label{eq:dgl_baroklin_1}\\ \left(\frac{\partial}{\partial t} - \newoverline{u}_T\frac{\partial}{\partial x}\right)q_3' + \left(\beta - K^2\newoverline{u}_T\right)\frac{\partial\psi_3'}{\partial x} &= 0.\tag{16.299}\label{eq:dgl_baroklin_2} \end{align} \]

Since the equations (16.298) - (16.299) are linear, one can understand the solutions as superpositions of plane waves and therefore makes an approach

\[ \begin{align} \psi_1' &= A_1\exp\left[i\left(kx - \omega t\right)\right] = :A_1f\left(x, t\right),\\ \psi_3' &= A_3\exp\left[i\left(kx - \omega t\right)\right] = :A_3f\left(x, t\right). \end{align} \]

This gives you

\[ \begin{align} q_1' &= -k^2A_1f - K^2\frac{A_1 - A_3}{2}f,\\ q_3' &= -k^2A_3f + K^2\frac{A_1 - A_3}{2}f. \end{align} \]

So the system of equations to be solved is

\[ \begin{align} \left(\omega - k\newoverline{u}_T\right)\left(k^2A_1 + K^2\frac{A_1 - A_3}{2}\right) + \left(\beta + K^2\newoverline{u}_T\right)kA_1 &= 0,\\ \left(\omega + k\newoverline{u}_T\right)\left(k^2A_3 - K^2\frac{A_1 - A_3}{2}\right) + \left(\beta - K^2\newoverline{u}_T\right)kA_3 &= 0. \end{align} \]

In matrix notation this is:

\[ \begin{align} \left(\begin{array}{cc} M_{1, 1}&M_{1, 2}\\ M_{2, 1}&M_{2, 2} \end{array}\right)\left(\begin{array}{c} A_1\\ A_3 \end{array}\right) = \mathbf{0} \end{align} \]

with the matrix coefficients

\[ \begin{align} M_{1, 1} &= \left(\omega - k\newoverline{u}_T\right)\left(k^2 + \frac{K^2}{2}\right) + \left(\beta + K^2\newoverline{u}_T\right)k,\\ M_{1, 2} &= -\frac{K^2}{2}\left(\omega - k\newoverline{u}_T\right),\\ M_{2, 1} &= -\frac{K^2}{2}\left(\omega + k\newoverline{u}_T\right),\\ M_{2, 2} &= \left(\omega + k\newoverline{u}_T\right)\left(k^2 + \frac{K^2}{2}\right) + \left(\beta - K^2\newoverline{u}_T\right)k. \end{align} \]

Setting the determinant to zero results in

\[ \begin{align} & M_{1, 1}M_{2, 2} - M_{2, 1}M_{1, 2}\hastobe0\nonumber\\ &\Rightarrow\left(\omega^2 - k^2\newoverline{u}_T^2\right)\left(k^2 + \frac{K^2}{2}\right)^2 + 2k\left(k^2 + \frac{K^2}{2}\right)\left(\omega\beta + kK^2\newoverline{u}_T^2\right) + \left(\beta^2 - K^4\newoverline{u}_T^2\right)k^2 - \frac{K^4}{4}\left(\omega^2 - k^2\newoverline{u}_T^2\right) = 0\nonumber\\ &\Rightarrow\left(\omega^2 - k^2\newoverline{u}_T^2\right)\left(k^4 + k^2K^2\right) + 2k\left(k^2 + \frac{K^2}{2}\right)\left(\omega\beta + kK^2\newoverline{u}_T^2\right) + \left(\beta^2 - K^4\newoverline{u}_T^2\right)k^2 = 0\nonumber\\ &\Rightarrow\left(\omega^2 - k^2\newoverline{u}_T^2\right)\left(k^2 + K^2\right) + \left(2k + \frac{K^2}{k}\right)\left(\omega\beta + kK^2\newoverline{u}_T^2\right) + \beta^2 - K^4\newoverline{u}_T^2 = 0\nonumber\\ &\Rightarrow\left(\omega^2 - k^2\newoverline{u}_T^2\right)\left(k^2 + K^2\right) + 2k\left(\omega\beta + kK^2\newoverline{u}_T^2\right) + \frac{K^2\omega\beta}{k} + \beta^2 = 0\nonumber\\ &\Rightarrow\omega^2k^2 + \omega^2K^2 - k^4\newoverline{u}_T^2 + k^2K^2\newoverline{u}_T^2 + 2k\omega\beta + \frac{K^2\omega\beta}{k} + \beta^2 = 0\nonumber \end{align} \] \[ \begin{align} &\Rightarrow\omega^2\left[k^2 + K^2\right] + \omega\left[2k\beta + \frac{K^2\beta}{k}\right] - k^4\newoverline{u}_T^2 + k^2K^2\newoverline{u}_T^2 + \beta^2 = 0\nonumber\\ &\Rightarrow\omega^2 + \omega\frac{\beta}{k}\frac{2k^2 + K^2}{k^2 + K^2} + \frac{k^2K^2\newoverline{u}_T^2 - k^4\newoverline{u}_T^2 + \beta^2}{k^2 + K^2} = 0\nonumber\\ &\Rightarrow\omega_{1, 2} = -\frac{\beta}{2k}\frac{2k^2 + K^2}{k^2 + K^2}\pm\sqrt{\frac{\beta^2}{4k^2}\frac{\left(2k^2 + K^2\right)^2}{\left(k^2 + K^2\right)^2} - \frac{k^2K^2\newoverline{u}_T^2 - k^4\newoverline{u}_T^2 + \beta^2}{k^2 + K^2}}\nonumber\\ &\Rightarrow\omega_{1, 2} = -\frac{\beta}{2k}\frac{2k^2 + K^2}{k^2 + K^2}\pm\frac{1}{k^2 + K^2}\sqrt{\frac{\beta^2}{4k^2}\left(4k^4 + K^4 + 4k^2K^2\right) - \left(k^2 + K^2\right)\left(k^2K^2\newoverline{u}_T^2 - k^4\newoverline{u}_T^2 + \beta^2\right)}\nonumber\\ &\Rightarrow\omega_{1, 2} = -\frac{\beta}{2k}\frac{2k^2 + K^2}{k^2 + K^2}\nonumber\\ &\pm\frac{1}{k^2 + K^2}\sqrt{\beta^2k^2 + \frac{\beta^2K^4}{4k^2} + \beta^2K^2 - k^4K^2\newoverline{u}_T^2 - k^2\beta^2 - K^4k^2\newoverline{u}_T^2 - K^2\beta^2 + k^6\newoverline{u}_T^2 + K^2k^4\newoverline{u}_T^2}\nonumber\\ &\Rightarrow\omega_{1, 2} = -\frac{\beta}{k}\frac{k^2 + K^2/2}{k^2 + K^2}\pm\frac{1}{k^2 + K^2}\sqrt{\frac{\beta^2K^4}{4k^2} + \newoverline{u}_T^2k^2\left(k^4 - K^4\right)}.\tag{16.311}\label{eq:disp_rel_baroklin} \end{align} \]

Assuming $k < K$, one maximizes the expression at the root by setting $\newoverline{u}_T = 0$, from which it follows

Baroclinic Rossby waves always propagate westward. Baroclinic Rossby waves are also called planetary waves when they span a complete circle of latitude.

16.6.4 Natural oscillations of a hydrostatic atmosphere

Write down the equations summarized in Section 13.7 in their adiabatic form:

\[ \begin{align} \frac{\partial u}{\partial t} + u\frac{\partial u}{\partial x} + v\frac{\partial u}{\partial y} + \omega\frac{\partial u}{\partial p} &= -\frac{\partial\Phi}{\partial x} + fv,\\ \frac{\partial v}{\partial t} + u\frac{\partial v}{\partial x} + v\frac{\partial v}{\partial y} + \omega\frac{\partial v}{\partial p} &= -\frac{\partial\Phi}{\partial y} - fu,\\ \frac{\partial^2\Phi}{\partial p\partial t} + \left(u\frac{\partial}{\partial x} + v\frac{\partial}{\partial y}\right)\frac{\partial\Phi}{\partial p} + \frac{\alpha^2}{g^2}N^2\omega &= 0,\\ \frac{\partial\omega}{\partial p} &= -\nabla\cdot\mathbf{v}_h. \end{align} \]

If you neglect advection, you get

\[ \begin{align} \frac{\partial u}{\partial t} &= -\frac{\partial\Phi}{\partial x} + fv,\\ \frac{\partial v}{\partial t} &= -\frac{\partial\Phi}{\partial y} - fu,\\ \frac{\partial^2\Phi}{\partial p\partial t} + \sigma\omega &= 0,\\ \frac{\partial\omega}{\partial p} &= -\nabla\cdot\mathbf{v}_h. \end{align} \]

with the static stability parameter $\sigma = \frac{\alpha^2}{g^2}N^2$. This is the underlying system of equations for this section. If you write down the derivatives and differential operators in geographical coordinates, you get.

\[ \begin{align} \frac{\partial u}{\partial t} &= -\frac{1}{a\cos\left(\phi\right)}\frac{\partial\Phi}{\partial\lambda} + fv,\tag{16.321}\label{eq:hydrostatic_normal_0}\\ \frac{\partial v}{\partial t} &= -\frac{1}{a}\frac{\partial\Phi}{\partial\phi} - fu,\tag{16.322}\label{eq:hydrostatic_normal_1}\\ \frac{\partial^2\Phi}{\partial p\partial t} + \sigma\omega &= 0,\tag{16.323}\label{eq:hydrostatic_normal_2}\\ \frac{\partial\omega}{\partial p} &= -\frac{1}{a\cos\left(\phi\right)}\frac{\partial u}{\partial\lambda} - \frac{1}{a}\frac{\partial v}{\partial\phi} - \frac{v}{a}\tan\left(\phi\right).\tag{16.324}\label{eq:hydrostatic_normal_3} \end{align} \]

Now we divide it into a background state $\newoverline{\mathbf{\psi}}$ and deviations $\mathbf{\psi}'$, where $\sigma$ is a property of the background state. One assumes $\newoverline{u} = \newoverline{v} = \newoverline{\omega} = 0$. As approaches for fields $\psi$ are used

\[ \begin{align} \psi' = \psi_0\left(\frac{p}{p_0}\right)^\chi h\left(\phi\right)\exp\left(im\lambda - i\gamma t\right) \end{align} \]

with a complex amplitude $\psi_0$, a zonal wavenumber $m \in \mathbb{N}$ and a continuously differentiable function $h\left(\phi\right)$. $p_0$ is a reference pressure and $\chi > 0$ is a parameter. If you insert this and the definition of $f$ into the equations (16.321) - (16.324), you get

\[ \begin{align} -i\gamma u_0 &= -\frac{im\Phi_0}{a\cos\left(\phi\right)} + 2\Omega\sin\left(\phi\right)v_0,\\ -i\gamma hv_0 &= -\frac{\Phi_0 h'}{a} - 2\Omega\sin\left(\phi\right)u_0h,\\ -i\gamma\frac{\chi}{p}\Phi_0 &= -\sigma\omega_0,\\ \frac{\chi}{p}h\omega_0 &= -\frac{imh}{a\cos\left(\phi\right)}u_0 - \frac{v_0h'}{a} - \frac{v_0}{a}h\tan\left(\phi\right). \end{align} \]

The fractions are multiplied away:

\[ \begin{align} -i\gamma a\cos\left(\phi\right)u_0 &= -im\Phi_0 + 2\Omega a\cos\left(\phi\right)\sin\left(\phi\right)v_0 = -im\Phi_0 + \Omega a\sin\left(2\phi\right)v_0,\\ -i\gamma ahv_0 &= -h'\Phi_0 - 2\Omega a\sin\left(\phi\right)hu_0,\\ -i\gamma\chi\Phi_0 &= -\sigma p\omega_0,\\ \chi ha\cos\left(\phi\right)\omega_0 &= -imhpu_0 - ph'\cos\left(\phi\right)v_0 - ph\sin\left(\phi\right)v_0 \end{align} \]

If you bring everything to the left side, you get

\[ \begin{align} -i\gamma a\cos\left(\phi\right)u_0 + im\Phi_0 - \Omega a\sin\left(2\phi\right)v_0 &= 0,\\ -i\gamma ahv_0 + h'\Phi_0 + 2\Omega a\sin\left(\phi\right)hu_0 &= 0,\\ -i\gamma\chi\Phi_0 + \sigma p\omega_0 &= 0,\\ \chi ha\cos\left(\phi\right)\omega_0 + imhpu_0 + ph'\cos\left(\phi\right)v_0 + ph\sin\left(\phi\right)v_0 &= 0. \end{align} \]

The terms are now sorted in the order $u_0$, $v_0$, $\Phi_0$, $\omega_0$:

\[ \begin{align} \begin{array}{ccccc} -i\gamma a\cos\left(\phi\right)u_0 &- \Omega a\sin\left(2\phi\right)v_0 &+ im\Phi_0 &+ 0 &= 0,\\ 2\Omega a\sin\left(\phi\right)hu_0 &- i\gamma ahv_0 &+ h'\Phi_0 &+ 0 &= 0,\\ 0 &+ 0 &-i\gamma\chi\Phi_0 &+ \sigma p\omega_0 &= 0,\\ imhpu_0 &+ \left[ph'\cos\left(\phi\right) + ph\sin\left(\phi\right)\right]v_0 &+ 0 &+ \chi ha\cos\left(\phi\right)\omega_0 &= 0. \end{array} \end{align} \]

This leads directly to matrix notation

\[ \begin{align} M\cdot\mathbf{x} = \mathbf{0} \end{align} \]

with

\[ \begin{align} M \coloneqq \left( \begin{array}{cccc} -i\gamma a\cos\left(\phi\right) & -\Omega a\sin\left(2\phi\right) & im & 0,\\ 2\Omega a\sin\left(\phi\right)h & -i\gamma ah & h' & 0,\\ 0 & 0 & -i\gamma\chi & \sigma p,\\ imhp & ph'\cos\left(\phi\right) + ph\sin\left(\phi\right) & 0 & \chi ha\cos\left(\phi\right) \end{array}\right) \end{align} \]

and

\[ \begin{align} \mathbf{v} \coloneqq \left(\begin{array}{c} u_0\\ v_0\\ \Phi_0\\ \omega_0\end{array}\right). \end{align} \]

16.6.4.1 Atmospheric tide

Important differences between the atmospheric and the oceanic tide (see section 16.5.7) are as follows:

• The oceanic tide can be viewed as barotropic, effects of the ocean's baroclinicity play a minor role.
• In the atmosphere, baroclinity is much more relevant; the barotropic approximation would filter out essential structural elements of the atmospheric tide.
• The oceanic tide is generated primarily by fluctuations in gravity caused by the moon. This leads to a base period of just under 25 hours.
• The atmospheric tide is driven primarily by radiation and by processes triggered by radiation (chemical reactions and convection).
• Due to barotropy, the oceanic tide is virtually independent of height, while the amplitude of the atmospheric tide increases exponentially with height. This means that the tide plays a role in the atmosphere, especially in the mesosphere (50 - 80 km), where it can reach large amplitudes. In the troposphere their amplitude is small.

16.7 Instabilities

If one finds angular frequencies with a positive imaginary part when a plane wave is introduced as a disturbance into a system of equations and subsequent linearization (justified for small amplitudes), then a instability is present. The amplitude of the wave grows exponentially in this initial linear time before nonlinear effects become dominant and cause the wave to break with subsequent dissipation of the kinetic energy.

16.7.1 Stratification instability

The atmosphere over a given point on the earth's surface has the stratification $T = T\left(z\right)$. Usually is

\[ \begin{align} \frac{\partial T}{\partial z} < 0, \end{align} \]

at $\frac{\partial T}{\partial z}>0$ one speaks of an inversion. The dry adiabatic temperature gradient is according to Eq. (9.47)

\[ \begin{align} \Gamma_d = \frac{g}{c^{(p)}} \end{align} \]

defined. In the case

\[ \begin{align} \frac{\partial T}{\partial z} < -\Gamma_d \end{align} \]

The layering is unstable, and even unconditionally unstable, since both a saturated and an unsaturated air particle that is deflected from its rest position does not return to it.

However, if the air is saturated, the cooling that occurs as a result of the uplift causes condensation or resublimation to occur, releasing phase transition heat. If you deflect the particle vertically at $z$, the lifted condensation level (LCL) $z_\text{LCL}$ is the height at which it would be saturated, so it satisfies the equation

\[ \begin{align} p_v^{(S)}\left(T\left(z\right) - \Gamma_d\left(z_{\text{LCL}} - z\right)\right) \equiv p_v\left(z\right)\left(1 - \frac{\Gamma_d\left(z_{\text{LCL}} - z\right)}{T\left(z\right)}\right)^{\frac{g}{R_d\Gamma_d}}. \end{align} \]

During the subsequent further uplift, it is initially assumed that all the water remains in the cloud, which is referred to as reversible wet adiabates; Occasionally the word pseudoadiabatic is used instead of wet adiabatic, since the question of whether this process is adiabatic or not depends on which system you are looking at: if you look at the air including the condensates as a system, the process is adiabatic because it does not exchange heat with its surroundings; However, if you only look at the gas phase, the process is diabatic. In the case of irreversible wet adiabates, which will be discussed in more detail in section 16.7.1.3, it is assumed that all the water rains out. This will now be considered. So assume that the air in a particle with volume $V$ and constant mass $m$ consists of a mixture of moist air and condensation products of a phase, with all components having the same temperature $T$. With the first law of thermodynamics Eq. (9.10) is obtained

\[ \begin{align} &c^{(V)}\frac{dT}{dz} + p\frac{d}{dz}\frac{1}{\rho} = \frac{1}{\rho}\frac{dq}{dz}\nonumber\\ \Rightarrow\frac{dT}{dz} &= \frac{1}{c^{(V)}}\left[\frac{1}{\rho}\frac{dq}{dz} - p\frac{d}{dz}\frac{1}{\rho}\right] = \frac{1}{\rho c^{(V)}}\left[\frac{dq}{dz} + \frac{p}{\rho}\frac{d\rho}{dz}\right] = -\frac{g}{c^{(V)}}\left[\frac{dq}{dp} + \frac{p}{\rho}\frac{d\rho}{dp}\right].\tag{16.346}\label{eq:feuchtad_deriv_1} \end{align} \]

The basic hydrostatic equation was used here. For the heat output per volume due to the phase transition $dq$ applies

\[ \begin{align} dq = \frac{c_{c}}{v}_hdm_c \end{align} \]

with $m_c$ as the condensed or resublimated mass and $c_{c}$ as the phase transition heat. With the equation of state for water vapor applies

\[ \begin{align} p_v &= \rho_vR_vT\Rightarrow m_v = \frac{V_vp_v}{R_vT} = \frac{V_hp_v}{R_vT} = \frac{m_hR_hp_v^{(S)}\left(T\right)}{pR_v}\nonumber\\ \Rightarrow dm_c &= -dm_v = \frac{m_hR_hp_v^{(S)}\left(T\right)}{pR_v}\left(\frac{dp}{p} - \frac{dp_v^{(S)}}{p_v^{(S)}} + \frac{dm_c}{m_h}\right)\nonumber\\ \Rightarrow dm_c\left(1 - \frac{R_hp_v^{(S)}\left(T\right)}{pR_v}\right) &= \frac{m_hR_hp_v^{(S)}\left(T\right)}{pR_v}\left(\frac{dp}{p} - \frac{dp_v^{(S)}}{p_v^{(S)}}\right)\nonumber\\ \Rightarrow dm_c &= m_h\frac{\frac{dp}{p} - \frac{dp_v^{(S)}}{p_v^{(S)}}}{\frac{pR_v}{R_hp_v^{(S)}} - 1}. \end{align} \]

Here $p_v^{(S)} = p_v^{(S)}\left(T\right)$ is the saturation vapor pressure. It follows

\[ \begin{align} dq &= \frac{c_c}{T}\frac{dp - \frac{p}{p_v^{(S)}}dp_v^{(S)}}{\frac{pR_v}{p_v^{(S)}} - R_h}\nonumber\\ \Rightarrow\frac{dq}{dp} &= \frac{c_c}{\frac{pR_vT}{p_v^{(S)}} - R_hT}\left(1 - \frac{p}{p_v^{(S)}}\frac{dp_v^{(S)}}{dp}\right) = \frac{c_c}{\frac{pR_vT}{p_v^{(S)}} - TR_h}\left(1 + \frac{1}{g\rho}\frac{dT}{dz}\frac{p}{p_v^{(S)}}\frac{dp_v^{(S)}}{dT}\right)\nonumber\\ &= \frac{c_c}{TpR_v - Tp_v^{(S)}R_h}\left(p_v^{(S)} + \frac{p}{g\rho}\frac{dT}{dz}\frac{dp_v^{(S)}}{dT}\right)\nonumber\\ &= \frac{c_cp_v^{(S)}}{TpR_v - Tp_v^{(S)}R_h} + \frac{p}{Tg\rho}\frac{c_c}{pR_v - p_v^{(S)}R_h}\frac{dp_v^{(S)}}{dT}\frac{dT}{dz}. \end{align} \]

Putting this into Eq. (16.346) and applying the chain rule again, you get

\[ \begin{align} \frac{dT}{dz} &= -\frac{g}{c^{(V)}}\Bigg[\frac{c_cp_v^{(S)}}{TpR_v - Tp_v^{(S)}R_h}\nonumber\\ & + \frac{p}{Tg\rho}\frac{c_c}{pR_v - p_v^{(S)}R_h}\frac{dp_v^{(S)}}{dT}\frac{dT}{dz} + \frac{p}{\rho}\frac{\partial\rho}{\partial p} - \frac{p}{g\rho^2}\frac{\partial\rho}{\partial T}\frac{dT}{dz}\Bigg]. \end{align} \]

Now you need an equation of state. applies to the density

\[ \begin{align} \rho = \rho_h + \rho_c = \frac{p}{R_hT} + \rho_c. \end{align} \]

This is what you get

\[ \begin{align} \frac{dT}{dz} &= \frac{g}{c^{(V)}}\Bigg[ - \frac{c_cp_v^{(S)}}{TpR_v - Tp_v^{(S)}R_h}\nonumber\\ & - \frac{p}{Tg\rho}\frac{c_c}{pR_v - p_v^{(S)}R_h}\frac{dp_v^{(S)}}{dT}\frac{dT}{dz} - \frac{p}{\rho R_hT} - \frac{p^2}{R_hg\rho^2T^2}\frac{dT}{dz}\Bigg]\nonumber\\ &= \frac{g}{c^{(V)}}\Bigg[ - \frac{c_cp_v^{(S)}}{TpR_v - Tp_v^{(S)}R_h} - \frac{p}{Tg\rho}\frac{c_c}{pR_v - p_v^{(S)}R_h}\frac{dp_v^{(S)}}{dT}\frac{dT}{dz}\nonumber\\ & - 1 + \frac{\rho _c}{\rho} - \frac{1}{g}\left(1 - \frac{\rho_c}{\rho}\right)\frac{p}{\rho T}\frac{dT}{dz}\Bigg]. \end{align} \]

This is now converted to $\frac{dT}{dz}$:

\[ \begin{align} & \frac{dT}{dz}\left(1 + \frac{p}{Tc^{(V)}\rho}\frac{c_c}{pR_v - p_v^{(S)}R_h}\frac{dp_v^{(S)}}{dT} + \frac{1}{c^{(V)}}\left(1 - \frac{\rho_c}{\rho}\right)\frac{p}{\rho T}\right)\nonumber\\ &= \frac{g}{c^{(V)}}\left(\frac{\rho_c}{\rho} - 1\right) - \frac{gc_cp_v^{(S)}\left(T\right)}{c^{(V)}T\left(R_vp - p_v^{(S)}R_h\right)}\nonumber\\ \Rightarrow\frac{dT}{dz} &= -\frac{g}{c^{(p)}}\frac{1 - \frac{\rho_c}{\rho} + \frac{c_cp_v^{(S)}}{T\left(R_vp - p_v^{(S)}R_h\right)}}{\frac{c^{(V)}}{c^{(p)}} + \frac{p}{T\rho c^{(p)}}\frac{c_c}{pR_v - p_v^{(S)}R_h}\frac{dp_v^{(S)}}{dT} + \left(1 - \frac{\rho_c}{\rho}\right)\frac{p}{c^{(p)}T\rho}}\nonumber\\ &= -\Gamma_d\beta \end{align} \]

Here is

The Clausius-Clapeyron equation can be used for the derivative $\frac{dp_v^{(S)}}{dT}$. The so-called moist adiabatic temperature gradient is now given by

\[ \begin{align} \Gamma_h\coloneqq\Gamma_d\beta \end{align} \]

defined. If you put in Eq. (16.354) $\rho_c = 0$ and $p_v \ll p$, but keeps terms with $\frac{dp_v^{(S)}}{dT}$, follows

\[ \begin{align} \frac{dT}{dz} &= -\frac{g}{c^{(p)}}\frac{1 + \frac{c_cp_v^{(S)}}{TR_vp}}{\frac{c^{(V)}}{c^{(p)}} + \frac{p}{T\rho c^{(p)}}\frac{c_c}{pR_v}\frac{dp_v^{(S)}}{dT} + \frac{p}{c^{(p)}T\rho}}. \end{align} \]

If you insert the equation of state of ideal gases $p = \rho R_dT$ here, you get the expression that is often found in the literature

\[ \begin{align} \frac{dT}{dz} &= -\frac{g}{c^{(p)}}\frac{1 + \frac{c_cp_v^{(S)}}{TR_vp}}{\frac{c^{(V)}}{c^{(p)}} + \frac{R_d}{c^{(p)}}\frac{c_c}{pR_v}\frac{dp_v^{(S)}}{dT} + \frac{c^{(p)} - c^{(V)}}{c^{(p)}}}\nonumber\\ \Leftrightarrow\frac{dT}{dz} &= -\frac{g}{c^{(p)}}\frac{1 + \frac{c_cp_v^{(S)}}{TR_vp}}{1 + \frac{R_dc_c}{R_vc^{(p)}p}\frac{dp_v^{(S)}}{dT}} \equiv -\Gamma_d\beta' \end{align} \]

with a simplified factor

\[ \begin{align} \beta'&= \beta'\left(p, T\right) \coloneqq\frac{1 + \frac{c_cp_v^{(S)}}{TR_vp}}{1 + \frac{\frac{M_v}{M_d}c_c}{pc^{(p)}}\frac{dp_v^{(S)}}{dT}}\stackrel{\href{ch-04-statistical-physics.html#eq:clausius-clapeyron_vereinfacht}{\text{Glg. (5.237)}}}{\approx}\frac{1 + \frac{c_cp_v^{(S)}}{TR_vp}}{1 + \frac{\frac{M_v}{M_d}c_c^2p_v^{(S)}}{R_vc^{(p)}T^2p}}\nonumber \end{align} \]

which no longer depends on the density of the water that has already condensed out. In the case $\frac{dp_v^{(S)}}{dT} = p_v^{(S)} = 0$ it follows

\[ \begin{align} \Gamma_h = \Gamma_d. \end{align} \]

In the case of saturated air and

\[ \begin{align} \frac{\partial T}{\partial z} < -\Gamma_h \end{align} \]

the layering is also labile. In the case

\[ \begin{align} -\Gamma_d \leq \frac{\partial T}{\partial z} < -\Gamma_h \end{align} \]

the layering is conditionally unstable, since only a saturated particle does not return to its rest position when it is deflected, so the instability depends on the humidity. In the case of unsaturated air and

\[ \begin{align} \frac{\partial T}{\partial z} = -\Gamma_d \end{align} \]

or saturated air and

\[ \begin{align} \frac{\partial T}{\partial z} = -\Gamma_h \end{align} \]

one speaks of neutral stratification.

16.7.1.1 Equivalent potential temperature

To create the so-called equivalent potential temperature $\theta_e$ (also: pseudoequivalent potential temperature or, more rarely, pseudopotential temperature) of a particle, two steps are necessary:

1. Raise the particle dry adiabatically to its LCL.
2. Raise the particle irreversibly moist adiabatically to an infinite height. The potential temperature then obtained is $\theta_e$.

This size is not only adjusted for the pressure difference, but also for the existing moisture, which could lead to an increase in temperature in the form of phase transition enthalpy. It is therefore well suited to comparing the temperatures of different air masses.

16.7.1.2 Reversible equivalent potential temperature

The reversible equivalent potential temperature is defined analogously to the equivalent potential temperature, except that a reversible wet adiabatic is pursued in step 2 from section 16.7.1.1.

16.7.1.3 CAPE

The derivations carried out in this section refer to a vertical column of air, where dynamic, i.e. h. Effects caused by the velocity field are not taken into account. One defines a function $f\left(z, z'\right)$ as the specific force, i.e. the acceleration, that acts on the particle at $z$ if it is brought adiabatically to the level $z'$ and assumes a hydrostatic atmosphere, i.e

\[ \begin{align} f\left(z, z'\right) \coloneqq -\frac{1}{\rho'\left(z, z'\right)}\frac{\partial p\left(z'\right)}{\partial z} - g \stackrel{\href{ch-12-important-approximations.html#eq:hydrostatic}{\text{Glg. (13.122)}}}{=} \frac{g\rho\left(z'\right)}{\rho'\left(z, z'\right)} - g = g\frac{\rho\left(z'\right) - \rho'\left(z, z'\right)}{\rho'\left(z, z'\right)}, \end{align} \]

where $\rho'\left(z, z'\right)$ is the density of the particle at $z$ after moving it adiabatically to $z'$. One also defines a function $\newtilde{F}\left(z, z'\right)$ by

\[ \begin{align} \newtilde{F}\left(z, z'\right) \coloneqq g\int_{z}^{z'}\frac{\rho\left(z''\right) - \rho'\left(z, z''\right)}{\rho'\left(z, z''\right)}dz'' \end{align} \]

as the work done by the layering on a particle that was deflected adiabatically from $z$ to $z'$. Usually you choose $z = z_{\mathrm{2m}} \coloneqq z_{\mathrm{SFC}} + 2$ m, you define

\[ \begin{align} F\left(z\right) &\coloneqq \newtilde{F}\left(z_{\text{2m}}, z\right) = g\int_{z_{\text{2m}}}^{z}\frac{\rho\left(z''\right) - \rho'\left(z_{\text{2m}}, z''\right)}{\rho'\left(z_{\text{2m}}, z''\right)}dz'' = g\int_{z_{\text{2m}}}^{z}\frac{\rho\left(z'\right) - \rho'\left(z_{\text{2m}}, z'\right)}{\rho'\left(z_{\text{2m}}, z'\right)}dz'\nonumber\\ &= g\int_{z_{\text{2m}}}^{z}\frac{\rho\left(z'\right) - \rho'\left(z'\right)}{\rho'\left(z'\right)}dz' \end{align} \]

with the misuse of the term

\[ \begin{align} \rho'\left(z'\right) \coloneqq \rho'\left(z_{\text{2m}}, z'\right). \end{align} \]

The same applies to the integrand

\[ \begin{align} f\left(z'\right) \coloneqq g\frac{\rho\left(z'\right) - \rho'\left(z'\right)}{\rho'\left(z'\right)}\tag{16.368}\label{eq:uplift} \end{align} \]

defined. Many of the quantities commonly used in connection with convection can be defined in terms of the functions $F$ and $f$. The so-called convectively available potential energy CAPE is defined by

\[ \begin{align} \CAPE \coloneqq \int_{z_{\text{2m}}}^{\infty}\Theta\left(f\left(z\right)\right)dz,\tag{16.369}\label{eq:def_cape} \end{align} \]

where $\Theta$ denotes the $\Theta $function. Realistically, of course, you don't integrate to infinity, but only up to a height $z_T$, where the tropopause is typically located, since there is no longer any instability above that anyway. In the case of dry air one can write for $f$

\[ \begin{align} f\left(z'\right) = g\frac{\frac{p}{R_dT} - \frac{p}{R_dT'}}{\frac{p}{R_dT'}} = g\frac{\frac{1}{T} - \frac{1}{T'}}{\frac{1}{T'}} = g\frac{T' - T}{T} = g\frac{\theta'\left(\frac{p}{p_0}\right)^{R_d/c_d^{(p)}} - \theta\left(\frac{p}{p_0}\right)^{R_d/c_d^{(p)}}}{\theta\left(\frac{p}{p_0}\right)^{R_d/c_d^{(p)}}} = g\frac{\theta' - \theta}{\theta}. \end{align} \]

In the case of moist air one can replace $T \to T_v$, here $T_v$ is the value in Eq. (5.164) defined virtual temperature, since at this temperature the dry air would have the same density as the existing moist air. The level of free convection $z_{\text{LFC}}$ is the height at which the adiabatic rising particle is lighter than the surroundings for the first time

\[ \begin{align} z_{\text{LFC}} \coloneqq\inf\{z\newvline f\left(z\right) > 0\}. \end{align} \]

When the particle reaches this height, it will continue to rise, which is called free convection, i.e. convection that would also take place without dynamic forcing. It is called analogous

\[ \begin{align} z_{\text{EL}} \coloneqq\sup\{z\newvline f\left(z\right) > 0, z < 2z_T\} \end{align} \]

as equilibrium level (equilibrium level). To be on the safe side, the deep stratosphere was excluded. This is the height at which a freely rising particle would be heavier than its surroundings. Often in Eq. (16.369) omitted the $\Theta $function and only integrated from $z_{\mathrm{LFC}}$ to $z_{\mathrm{EL}}$, which is correct if there is thermal lability in exactly one height interval, which is often the case. The maximum cloud top is defined as the height at which such a particle would have lost its kinetic energy again, i.e

\[ \begin{align} z_{\text{MCT}} \coloneqq\inf\{z\newvline \newtilde{F}\left(z_{\text{LFC}}, z\right) < 0\}. \end{align} \]

The phenomenon $z_{\mathrm{MCT}} > z_{\mathrm{EL}}$ leads to the development of gravity waves at the tropopause (overshooting tops) in the case of long-range convection, which is generally the case wherever a stable height interval follows an unstablely layered height interval.

If a particle has reached the so-called release temperature at a height of two meters, it can rise into the LFC even without dynamic drive, i.e

\[ \begin{align} 0 \equiv g\int_{z_{\text{2m}}}^{z_{\text{LFC}}}\frac{\rho\left(z'\right) - \newtilde{\rho}'\left(z'\right)}{\newtilde{\rho}'\left(z'\right)}, \end{align} \]

where $\newtilde{\rho}'\left(z'\right)$ is the density of the particle lifted adiabatically from $z_{\mathrm{2m}}$ to $z'$, which would be the trigger temperature at a height of two meters. So the energy $\CIN$ that has to be overcome

\[ \begin{align} \CIN \coloneqq -\newtilde{F}\left(z_{\text{2m}}, z_{\text{LFC}}\right). \end{align} \]

is called convective barrier or convective inhibition. The height at which a particle heated to the trigger temperature would be saturated during adiabatic ascent is called convective condensation level (CCL). This is at least as big as the LCL, so

\[ \begin{align} z_{\text{CCL}} \geq z_{\text{LCL}}. \end{align} \]

16.7.2 Kelvin-Helmholtz instability

The Kelvin-Helmholtz instability is the instability of a vertically sheared horizontal flow.

16.7.2.1 In discrete layering

The movement is y-symmetrical, so the xz plane is considered. The xy plane coincides with the equilibrium position of a phase boundary. The medium above the boundary layer ($z > 0$) is given the index 1, the medium below it ($z < 0$) is given the index 2. The media are barotropic and incompressible. At time $t = 0$ the flow field will be through

\[ \begin{align} U_1 > 0, & {} & U_2 > 0, & {} & V_1 = V_2 = W_1 = W_2 = 0 \end{align} \]

described. The Coriolis force is neglected. It applies

\[ \begin{align} \nabla \times \mathbf{v}_1 = \nabla \times \mathbf{v}_2 = \mathbf{0}. \end{align} \]

Due to the circulation law in the form Eq. (15.39) this also applies to $t > 0$ as long as the media does not mix. Therefore the velocity fields are described by two current functions

\[ \begin{align} \psi_1 = U_1x + \psi_1', & {} & \psi_2 = U_2 x + \psi_2',\tag{16.379}\label{eq:khi_discrete_stream} \end{align} \]

where the deleted variables stand for the disturbances. Because of the lack of divergence,

\[ \begin{align} \Delta\psi_1' = \Delta\psi_2' = 0\tag{16.380}\label{eq:khi_deriv_laplace}. \end{align} \]

The deflection of the phase boundary is denoted by $\zeta$. With $\mathbf{n}$ as the normal vector of the surface and $\zeta$ as the surface deflection, the kinematic boundary condition applies

\[ \begin{align} \mathbf{n}\cdot\nabla\psi_1 = \mathbf{n}\cdot\left(\frac{\partial\zeta}{\partial t}\mathbf{e}_z\right) = \mathbf{n}\cdot\nabla\psi_2\text{ bei }z = \zeta.\tag{16.381}\label{eq:khi_discrete_kbc} \end{align} \]

Neglecting the surface tension, the dynamic boundary condition follows

\[ \begin{align} p_1 = p_2\text{ bei }z = \zeta.\tag{16.382}\label{eq:khi_discrete_dbc} \end{align} \]

For the normal vector $\mathbf{n}$ applies

\[ \begin{align} \mathbf{n} = \frac{1}{\sqrt{1 + \left(\frac{\partial\zeta}{\partial x}\right)^2}}\left(\begin{array}{c} -\frac{\partial\zeta}{\partial x}\\ 1\end{array}\right). \end{align} \]

The kinematic boundary condition Eq. (16.381) becomes

\[ \begin{align} -\left(U_1 + \frac{\partial\psi_1'}{\partial x}\right)\frac{\partial\zeta}{\partial x} + \frac{\partial\psi_1'}{\partial z} = \frac{\partial\zeta}{\partial t} = -\left(U_2 + \frac{\partial\psi_2'}{\partial x}\right)\frac{\partial\zeta}{\partial x} + \frac{\partial\psi_2'}{\partial z}\text{ bei }z = \zeta, \end{align} \]

where the root expression was multiplied out. To linearize this expression, apply it at $z = 0$ and neglect quadratic terms:

\[ \begin{align} -U_1\frac{\partial\zeta}{\partial x} + \frac{\partial\psi_1'}{\partial z} = \frac{\partial\zeta}{\partial t} = -U_2\frac{\partial\zeta}{\partial x} + \frac{\partial\psi_2'}{\partial z}\text{ bei }z = 0\tag{16.385}\label{eq:khi_discrete_deriv_kbc} \end{align} \]

Since the flow field is rotation-free and one also assumes an ideal fluid, one can use the time-dependent Bernoulli equations (see equation (15.122)) of the two layers

\[ \begin{align} \frac{\partial\psi_1}{\partial t} + \frac{1}{2}\left|\nabla\psi_1\right|^2 + \frac{p_1}{\rho_1} + gz &= C,\\ \frac{\partial\psi_2}{\partial t} + \frac{1}{2}\left|\nabla\psi_2\right|^2 + \frac{p_2}{\rho_2} + gz &= C_2 \end{align} \]

with real constants $C, C_2$. The dynamic boundary condition Eq. (16.382) implies identity at $z = \zeta$

\[ \begin{align} \rho_1\left(C - \frac{\partial\psi_1}{\partial t} - \frac{1}{2}\left|\nabla\psi_1\right|^2 - gz\right) = \rho_2\left(C_2 - \frac{\partial\psi_2}{\partial t} - \frac{1}{2}\left|\nabla\psi_2\right|^2 - gz\right).\tag{16.388}\label{eq:khi_discrete_deriv_1} \end{align} \]

Without disruption applies

\[ \begin{align} \rho_1\left(C - \frac{1}{2}U_1^2\right) = \rho_2\left(C_2 - \frac{1}{2}U_2^2\right).\tag{16.389}\label{eq:khi_discrete_deriv_2} \end{align} \]

Subtracting Eq. (16.389) from Eq. (16.388) leads to

\[ \begin{align} \rho_1\left(-\frac{\partial\psi_1}{\partial t} + \frac{1}{2}U_1^2 - \frac{1}{2}\left|\nabla\psi_1\right|^2 - gz\right) = \rho_2\left(-\frac{\partial\psi_2}{\partial t} + \frac{1}{2}U_2^2 - \frac{1}{2}\left|\nabla\psi_2\right|^2 - gz\right),\tag{16.390}\label{eq:khi_discrete_deriv_3} \end{align} \]

which applies at $z = \zeta$. With the equations (16.379) - (16.379) follows

\[ \begin{align} \frac{1}{2}\left|\nabla\psi_j\right|^2 &= \frac{1}{2}\left(U_j^2 + \left(\frac{\partial\psi_j'}{\partial x}\right)^2 + 2U_j\frac{\partial\psi_j'}{\partial x} + \left(\frac{\partial\psi_j'}{\partial z}\right)^2\right) \end{align} \]

for $j = 1, 2$. Putting this into Eq. (16.390), you get

\[ \begin{align} & \rho_1\left(-\frac{\partial\psi_1}{\partial t} - \frac{1}{2}\left(\frac{\partial\psi_1'}{\partial x}\right)^2 - U_1\frac{\partial\psi_1'}{\partial x} - \frac{1}{2}\left(\frac{\partial\psi_1'}{\partial z}\right)^2 - gz\right)\nonumber\\ &= \rho_2\left(-\frac{\partial\psi_2}{\partial t} - \frac{1}{2}\left(\frac{\partial\psi_1'}{\partial x}\right)^2 - U_2\frac{\partial\psi_1'}{\partial x} - \frac{1}{2}\left(\frac{\partial\psi_1'}{\partial z}\right)^2 - gz\right),\tag{16.392}\label{eq:khi_discrete_deriv_4} \end{align} \]

at $z = \zeta.$ This equation is linearized by evaluating it at $z = 0$ and neglecting the quadratic terms:

\[ \begin{align} \rho_1\left(\frac{\partial\psi_1}{\partial t} + U_1\frac{\partial\psi_1'}{\partial x} + g\zeta\right) = \rho_2\left(\frac{\partial\psi_2}{\partial t} + U_2\frac{\partial\psi_1'}{\partial x} + g\zeta\right)\tag{16.393}\label{eq:khi_discrete_deriv_5} \end{align} \]

Now you do the approach for $j = 1, 2$

\[ \begin{align} \psi_j' = A_j\left(z\right)\exp\left(ikx - i\omega t\right). \end{align} \]

With the Laplace equations Eq. (16.380) follows

\[ \begin{align} -k^2A_j + \frac{d^2A_j}{dz^2} &= 0. \end{align} \]

The solutions are of the form

\[ \begin{align} A_j\left(z\right) = A_\pm\exp\left(\pm kz\right). \end{align} \]

With the specifications

\[ \begin{align} \lim_{z \to\infty} \psi_1' = 0, & {} & \lim_{z \to -\infty} \psi_2' = 0, \end{align} \]

which mean that the disturbance should disappear into infinity, follow

\[ \begin{align} \psi_1' = A_-\exp\left[ikx - i\omega t - kz\right], & {} & \psi_2' = A_+\exp\left[ikx - i\omega t + kz\right]. \end{align} \]

The surface deflection is taken accordingly

\[ \begin{align} \zeta = \zeta_0\exp\left(ikx - i\omega t\right) \end{align} \]

to. If you insert this into the equations (16.385) and (16.393), you get

\[ \begin{align} -U_1ik\zeta_0 - kA_- &= -i\omega\zeta_0 = -U_2ik\zeta_0 + kA_+,\tag{16.400}\label{eq:khi_discrete_deriv_6}\\ \rho_1\left(-i\omega A_- + ikU_1A_- + g\zeta_0\right) &= \rho_2\left(-i\omega A_+ + ikU_2A_+ + g\zeta_0\right).\tag{16.401}\label{eq:khi_discrete_deriv_7} \end{align} \]

Eq. (16.400) implied

\[ \begin{align} A_- = \frac{i}{k}\zeta_0\left(\omega - U_1k\right), & {} & A_+ = \frac{i}{k}\zeta_0\left(U_2k - \omega\right). \end{align} \]

Putting this into Eq. (16.401), follows

\[ \begin{align} \rho_1\left(-i\omega\frac{i}{k}\zeta_0\left(\omega - U_1k\right) + ikU_1\frac{i}{k}\zeta_0\left(\omega - U_1k\right) + g\zeta_0\right) &= \rho_2\left(-i\omega\frac{i}{k}\zeta_0\left(U_2k - \omega\right) + ikU_2\frac{i}{k}\zeta_0\left(U_2k - \omega\right) + g\zeta_0\right)\nonumber\\ \Leftrightarrow\rho_1\left(\frac{\omega}{k}\left(\omega - U_1k\right) - U_1\left(\omega - U_1k\right) + g\right) &= \rho_2\left(\frac{\omega}{k}\left(U_2k - \omega\right) - U_2\left(U_2k - \omega\right) + g\right)\nonumber\\ \Leftrightarrow\rho_1\left(\omega\left(\omega - U_1k\right) - kU_1\left(\omega - U_1k\right) + kg\right) &= \rho_2\left(\omega\left(U_2k - \omega\right) - U_2k\left(U_2k - \omega\right) + kg\right). \end{align} \]

This is a quadratic equation for $\omega$, i.e. a dispersion relation. Further algebraic transformation results

\[ \begin{align} \omega^2 - 2k\omega\frac{\rho_1U_1 + \rho_2U_2}{\rho_1 + \rho_2} - kg\frac{\rho_2 - \rho_1}{\rho_1 + \rho_2} + k^2\frac{U_1^2\rho_1 + U_2^2\rho_2}{\rho_1 + \rho_2} = 0. \end{align} \]

This gives two solutions for the angular frequency:

\[ \begin{align} \omega_\pm &= \frac{\rho_1U_1k + \rho_2U_2k}{\rho_1 + \rho_2} \pm \left[kg\frac{\rho_2 - \rho_1}{\rho_1 + \rho_2} - k^2\frac{U_1^2\rho_1\rho_2 + U_2^2\rho_1\rho_2 - 2\rho_1\rho_2U_1U_2}{\left(\rho_1 + \rho_2\right)^2}\right]^{1/2}\nonumber\\ \Leftrightarrow\omega_\pm &= \frac{\rho_1U_1k + \rho_2U_2k}{\rho_1 + \rho_2} \pm \left[kg\frac{\rho_2 - \rho_1}{\rho_1 + \rho_2} - k^2\rho_1\rho_2\frac{\left(U_1 - U_2\right)^2}{\left(\rho_1 + \rho_2\right)^2}\right]^{1/2} \end{align} \]

Instability exists in the case

\[ \begin{align} kg\frac{\rho_2 - \rho_1}{\rho_1 + \rho_2} &- k^2\rho_1\rho_2\frac{\left(U_1 - U_2\right)^2}{\left(\rho_1 + \rho_2\right)^2} < 0\nonumber\\ \Leftrightarrow kg\frac{\rho_2 - \rho_1}{\rho_1 + \rho_2} & < k^2\rho_1\rho_2\frac{\left(U_1 - U_2\right)^2}{\left(\rho_1 + \rho_2\right)^2}\nonumber \end{align} \]

so if

• the density difference is small enough,
• the wave is short enough or the
• speed difference is large enough.

An example is the formation of water surface waves under the influence of wind. One can assume that there are almost always spectral components in the wind field that satisfy the condition Eq. (16.406). The further development of the system over time can look different, e.g. B. this depends on whether the media can mix or not.

16.7.2.2 In continuous layering

Consider the xz plane again and ignore the Coriolis acceleration. You also do the Boussinesq approximation (see section 13.6). The momentum, continuity equations and the first law of thermodynamics read with $\sigma$ as the potential density

\[ \begin{align} \frac{\partial u}{\partial t} + u\frac{\partial u}{\partial x} + w\frac{\partial w}{\partial z} &= -\frac{1}{\rho}\frac{\partial p}{\partial x},\\ \frac{\partial w}{\partial t} + u\frac{\partial w}{\partial x} + w\frac{\partial w}{\partial z} &= -\frac{1}{\rho}\frac{\partial p}{\partial z} - g,\\ \frac{\partial u}{\partial x} + \frac{\partial w}{\partial z} &= 0,\\ \frac{\partial\sigma}{\partial t} + u\frac{\partial\sigma}{\partial x} + w\frac{\partial\sigma}{\partial z} &= 0. \end{align} \]

Now one assumes a vertically sheared background wind field $\newoverline{u}\left(z\right) = u\left(x, z, t\right) - u'\left(x, z, t\right)$, the vertical background wind $\newoverline{w}$ disappears. The background density $\newoverline{\rho}\left(z\right) = \rho\left(x, z, t\right) - \rho'\left(x, z, t\right)$ is in hydrostatic balance with the background pressure field $\newoverline{p}\left(z\right) = p\left(x, z, t\right) - p'\left(x, z, t\right)$, i.e

\[ \begin{align} \frac{\partial\newoverline{p}}{\partial z} &= -g\newoverline{\rho}.\tag{16.411}\label{eq:khi_deriv_1} \end{align} \]

Plugging this into the equations and linearizing them, you get

\[ \begin{align} \frac{\partial u'}{\partial t} + \newoverline{u}\frac{\partial u'}{\partial x} + w\frac{\partial\newoverline{u}}{\partial z} &= -\frac{1}{\rho}\frac{\partial p'}{\partial x} \approx -\frac{1}{\newtilde{\rho}}\frac{\partial p'}{\partial x},\\ \frac{\partial w}{\partial t} + \newoverline{u}\frac{\partial w}{\partial x} &= -\frac{1}{\rho}\frac{\partial p}{\partial z} - g \approx -\frac{1}{\newtilde{\rho}}\frac{\partial p}{\partial z} - \frac{g\rho}{\newtilde{\rho}} = -\frac{1}{\newtilde{\rho}}\frac{\partial p'}{\partial z} - \frac{g\rho'}{\newtilde{\rho}},\\ \frac{\partial u'}{\partial x} + \frac{\partial w}{\partial z} &= 0,\\ \frac{\partial\sigma'}{\partial t} + \newoverline{u}\frac{\partial\sigma'}{\partial x} + w\frac{\partial\newoverline{\sigma}}{\partial z} &= 0. \end{align} \]

In the x-component of the momentum equation, $\rho \to \newtilde{\rho}$ was replaced in the denominator of the pressure gradient term, where $\newtilde{\rho}$ denotes an average density assumed to be homogeneous; in the z-component, the right-hand side was multiplied by $\frac{\rho}{\newtilde{\rho}}$ before Eq. (16.411) was used. The disturbance of the flow is divergence-free, so a current function $\psi = \psi\left(x, z, t\right)$ is used

\[ \begin{align} u' = \frac{\partial\psi}{\partial z}, & {} &w = -\frac{\partial\psi}{\partial x} \end{align} \]

to. This means that the continuity equation is automatically fulfilled. The remaining three equations are

\[ \begin{align} \frac{\partial^2\psi}{\partial t\partial z} + \newoverline{u}\frac{\partial^2\psi}{\partial z\partial x} - \frac{\partial\psi}{\partial x}\frac{\partial\newoverline{u}}{\partial z} &= -\frac{1}{\newtilde{\rho}}\frac{\partial p'}{\partial x},\\ -\frac{\partial^2\psi}{\partial t\partial x} - \newoverline{u}\frac{\partial^2\psi}{\partial x^2} &= -\frac{1}{\newtilde{\rho}}\frac{\partial p'}{\partial z} - \frac{g\rho'}{\newtilde{\rho}},\\ \frac{\partial\sigma'}{\partial t} + \newoverline{u}\frac{\partial\sigma'}{\partial x} - \frac{\partial\psi}{\partial x}\frac{\partial\newoverline{\sigma}}{\partial z} &= 0. \end{align} \]

Now you continue to do the approaches

\[ \begin{align} \psi\left(x, z, t\right) = \psi_0\left(z\right)\exp\left[ik\left(x - ct\right)\right], & {} & p'\left(x, z, t\right) = p_0\left(z\right)\exp\left[ik\left(x - ct\right)\right],\\ \rho'\left(x, z, t\right) = \rho_0\left(z\right)\exp\left[ik\left(x - ct\right)\right], & {} & \sigma'\left(x, z, t\right) = \sigma_0\left(z\right)\exp\left[ik\left(x - ct\right)\right]. \end{align} \]

It follows

\[ \begin{align} -ikc\frac{d\psi_0}{dz} + \newoverline{u}ik\frac{d\psi_0}{dz} - ik\psi_0\frac{\partial\newoverline{u}}{\partial z} &= -ik\frac{p_0}{\newtilde{\rho}},\\ -k^2c\psi_0 + \newoverline{u}k^2\psi_0 &= -\frac{1}{\newtilde{\rho}}\frac{dp_0}{dz} -g\frac{\rho_0}{\newtilde{\rho}},\\ -ikc\sigma_0 + \newoverline{u}ik\sigma_0 -ik\psi_0\frac{d\newoverline{\sigma}}{dz} &= 0. \end{align} \]

This results in simplification

\[ \begin{align} -c\frac{d\psi_0}{dz} + \newoverline{u}\frac{d\psi_0}{dz} - \psi_0\frac{\partial\newoverline{u}}{\partial z} &= -\frac{p_0}{\newtilde{\rho}},\\ \psi_0k^2\left(\newoverline{u} - c\right) &= -\frac{1}{\newtilde{\rho}}\frac{dp_0}{dz} - g\frac{\rho_0}{\newtilde{\rho}}, \tag{16.426}\label{eq:khi_deriv_2}\\ \sigma_0\left(\newoverline{u} - c\right) - \psi_0\frac{d\newoverline{\sigma}}{dz} &= 0.\tag{16.427}\label{eq:khi_deriv_3} \end{align} \]

If you differentiate the first equation according to $z$, you get

\[ \begin{align} \frac{d^2\psi_0}{dz^2}\left(\newoverline{u} - c\right) - \psi_0\frac{d^2\newoverline{u}}{dz^2} &= -\frac{1}{\newtilde{\rho}}\frac{dp_0}{dz}. \end{align} \]

If you put Eq. (16.426), follows

\[ \begin{align} \left(\newoverline{u} - c\right)\left[\frac{d^2\psi_0}{dz^2} - k^2\psi_0\right] - \psi_0\frac{d^2\newoverline{u}}{dz^2} - g\frac{\rho_0}{\newtilde{\rho}} &= 0. \end{align} \]

To eliminate $\rho_0$, one first sets Eq. (16.427) to $\sigma_0$ and gets

\[ \begin{align} \sigma_0 &= \frac{\psi_0}{\newoverline{u} - c}\frac{d\newoverline{\sigma}}{dz}. \end{align} \]

O.B.d. A. you can place the reference level in the level you are currently considering and use $\rho_0 = \sigma_0$. This results in used

\[ \begin{align} \left(\newoverline{u} - c\right)\left[\frac{d^2\psi_0}{dz^2} - k^2\psi_0\right] -\psi_0\frac{d^2\newoverline{u}}{dz^2} - \frac{\psi_0}{\newoverline{u} - c}\frac{g}{\newtilde{\rho}}\frac{d\newoverline{\sigma}}{dz} &= 0. \end{align} \]

Here you set the Brunt-Väisälä frequency

\[ \begin{align} N^2 \coloneqq -\frac{g}{\newtilde{\rho}}\frac{d\newoverline{\sigma}}{dz} \end{align} \]

one and receives

\[ \begin{align} \left(\newoverline{u} - c\right)\left(\frac{d^2}{dz^2} - k^2\right)\psi + \left(\frac{N^2}{\newoverline{u} - c} - \frac{d^2\newoverline{u}}{dz^2}\right)\psi = 0.\tag{16.433}\label{eq:taylor-goldstein} \end{align} \]

This equation is called Taylor-Goldstein equation. It is an eigenvalue problem $\left\{\psi, c\right\}$ for the function $\psi$ with the eigenvalue $c$ depending on $\newoverline{u}\left(z\right)$. If $\left\{\psi, c\right\}$ is a solution, then $\left\{\psi^\star, c^\star\right\}$ is also a solution. A positive imaginary part of $c$ means that $\psi$ is an unstable solution. A shear $\newoverline{u} = \newoverline{u}\left(z\right)$ is stable if and only if all solutions of Eq. (16.433) have real eigenvalues. It is now further assumed that at $z = 0, H$ the boundary conditions $w = 0$ apply, from which it follows

\[ \begin{align} \psi\left(0\right) = \psi\left(H\right) = 0. \end{align} \]

Now define $\phi$ by

\[ \begin{align} \psi = \sqrt{\newoverline{u} - c}\phi. \end{align} \]

Differentiating according to $z$ gives

\[ \begin{align} \frac{d\psi}{dz} &= \frac{1}{2}\frac{1}{\sqrt{\newoverline{u} - c}}\phi\frac{d\newoverline{u}}{dz} + \sqrt{\newoverline{u} - c}\frac{d\phi}{dz},\\ \Rightarrow\frac{d^2\psi}{dz^2} &= \frac{1}{\sqrt{\newoverline{u} - c}}\frac{d\newoverline{u}}{dz}\frac{d\phi}{dz} - \frac{1}{4}\frac{1}{\sqrt{\newoverline{u} - c}^3}\phi\left(\frac{d\newoverline{u}}{dz}\right)^2 + \frac{1}{2}\frac{1}{\sqrt{\newoverline{u} - c}}\frac{d^2\newoverline{u}}{dz^2}\phi + \sqrt{\newoverline{u} - c}\frac{d^2\phi}{dz^2}. \end{align} \]

Putting this into Eq. (16.433) and dividing by $\sqrt{\newoverline{u} - c}$, you get

\[ \begin{align} \frac{d}{dz}\left[\left(\newoverline{u} - c\right)\frac{d\phi}{dz}\right] - \left[k^2\left(\newoverline{u} - c\right) + \frac{1}{2}\frac{d^2\newoverline{u}}{dz^2} + \frac{1}{\newoverline{u} - c}\left(\frac{1}{4}\left(\frac{d\newoverline{u}}{dz}\right)^2 - N^2\right)\right]\phi = 0.\tag{16.438}\label{eq:khi_deriv_4} \end{align} \]

The boundary conditions remain under the assumption $c \not= \newoverline{u}$

\[ \begin{align} \phi\left(0\right) = \phi\left(H\right) = 0. \end{align} \]

The Eq. (16.438) is now multiplied by $\phi^\star$:

\[ \begin{align} \phi^\star\frac{d}{dz}\left[\left(\newoverline{u} - c\right)\frac{d\phi}{dz}\right] - \left[k^2\left(\newoverline{u} - c\right) + \frac{1}{2}\frac{d^2\newoverline{u}}{dz^2} + \frac{1}{\newoverline{u} - c}\left(\frac{1}{4}\left(\frac{d\newoverline{u}}{dz}\right)^2 - N^2\right)\right]\left|\phi\right|^2 = 0.\tag{16.440}\label{eq:khi_deriv_5} \end{align} \]

If you integrate the first term from $0$ to $H$, it follows using partial integration

\[ \begin{align} \int_0^H\phi^\star\frac{d}{dz}\left[\left(\newoverline{u} - c\right)\frac{d\phi}{dz}\right]dz &= -\int_0^H\left(\newoverline{u} - c\right)\left|\frac{d\phi}{dz}\right|^2dz. \end{align} \]

Eq. (16.440) is now integrated from $0$ to $H$:

\[ \begin{align} \int_0^H\left[N^2 - \frac{1}{4}\left(\frac{d\newoverline{u}}{dz}\right)^2\right]\frac{\left|\phi\right|^2}{\newoverline{u} - c}dz &= \int_0^H\left(\newoverline{u} - c\right)\left(\left|\frac{d\phi}{dz}\right|^2 + k^2\left|\phi\right|^2\right) + \frac{1}{2}\frac{d^2\newoverline{u}}{dz^2}\left|\phi\right|^2dz.\tag{16.442}\label{eq:khi_deriv_6} \end{align} \]

Now write for the phase velocity

\[ \begin{align} c = c_r + ic_i \end{align} \]

with $c_r, c_i\in \mathbb{R}$. The imaginary part of Eq. (16.442) is written as

\[ \begin{align} c_i\int_0^H\left[N^2 - \frac{1}{4}\left(\frac{d\newoverline{u}}{dz^2}\right)^2\right]\frac{\left|\phi\right|^2}{\left|\newoverline{u} - c\right|^2}dz &= -c_i\int_0^H\left(\left|\frac{d\phi}{dz}\right|^2 + k^2\left|\phi\right|^2\right)dz\nonumber\\ \Rightarrow c_i\int_0^H\left[N^2 - \frac{1}{4}\left(\frac{d\newoverline{u}}{dz^2}\right)^2\right]\frac{\left|\phi\right|^2}{\left|\newoverline{u} - c\right|^2} + \left|\frac{d\phi}{dz}\right|^2 + k^2\left|\phi\right|^2dz &= 0. \end{align} \]

If $N^2 - \frac{1}{4}\left(\frac{d\newoverline{u}}{dz^2}\right)^2 > 0$ everywhere in the interval $\left(0, H\right), $ then $c_i = 0$. One defines the Richardson number $R_i$ by

The frequency

is called Prandtl frequency. The condition

is sufficient for stability.

16.7.3 Barotropic instability

Assume a western base current $U>0$ that depends only on $y$

\[ \begin{align} U = U\left(y\right) \end{align} \]

and a disorder

\[ \begin{align} \mathbf{v}_h' = \left(u, v\right)^T = \mathbf{k}\times\nabla\psi \end{align} \]

with a current function $\psi$. Consider a zonal channel of extent $2b$ with $b>0$, the line $y = 0$ lies in the middle of this channel. The boundary conditions are $\omega = 0$ at $p = 0$ and $p = p_0$ and $v = 0$ at $y = \pm b$. Then the horizontal wind is divergence-free and the barotropic vorticity equation can be applied,

\[ \begin{align} \frac{\partial\zeta}{\partial t} + \beta v + u\frac{\partial\zeta}{\partial x} + v\frac{\partial\zeta}{\partial y} = 0. \end{align} \]

Let $\beta$ be homogeneous, see section 13.9.2. If you do a perturbation approach according to Sect. 16.3, you get

\[ \begin{align} \frac{\partial\zeta'}{\partial t} + \beta v + U\frac{\partial\zeta'}{\partial x} + v\frac{\partial\newoverline{\zeta}}{\partial y} = 0. \end{align} \]

Here the vorticity was written $\zeta = \newoverline{\zeta} + \zeta'$ with $\newoverline{\zeta}$ as the vorticity of $U$ and $\zeta'$ as the vorticity of $\mathbf{v}_h'$. Set $\psi$ for the perturbation

\[ \begin{align} \psi\left(x, y, t\right) = Y\left(y\right)\psi_0e^{i\left(kx - \omega t\right)} \end{align} \]

to. With $Y\left(\pm b\right)\hastobe0$ the boundary conditions on $v$ are satisfied, one also obtains

\[ \begin{align} v &= \frac{\partial\psi}{\partial x} = ik\psi,\\ \zeta' &= \Delta \psi = -k^2\psi + Y''\psi_0e^{i(kx - \omega t)},\\ \frac{\partial\zeta'}{\partial x} &= -ik^3\psi + Y''\psi_0ike^{i\left(kx - \omega t\right)}.\\ \end{align} \]

Plugging this into the barotropic vorticity equation and subtracting $\psi_0e^{i\left(kx - \omega t\right)}$, you get

\[ \begin{align} i\omega k^2Y - i\omega Y'' + \beta ikY - Uik^3Y + Y''ikU - ikY\frac{d^2U}{dy^2} &= 0\nonumber\\ \Leftrightarrow \omega k^2Y - \omega Y'' + \beta kY - Uk^3Y + Y''kU - kY\frac{d^2U}{dy^2} &= 0\tag{16.457}\label{eq:baro_inst_deriv_1}. \end{align} \]

Instability means that $\omega$ is not purely real,

\[ \begin{align} \omega = \omega_r + i\omega _c, \end{align} \]

with $\omega_r, \omega_c\in \mathbb{R}\setminus\{0\}$. Inserting into Eq. (16.457) returns

\[ \begin{align} k^2Y\left(\omega_r + i\omega_c\right) - Y''\left(\omega_r + i\omega_c\right) + \beta kY - Uk^3Y + Y''kU = 0. \end{align} \]

The imaginary part of this is

\[ \begin{align} k^2Y\omega_c - Y''\omega_c = 0\Leftrightarrow Y'' = k^2Y. \end{align} \]

Substituting this into the real part, you get

\[ \begin{align} k^2Y\omega_r& - k^2Y\omega_r + \beta kY - Uk^3Y + Uk^3Y - kY\frac{d^2U}{dy^2} = 0\Rightarrow\frac{d^2U}{dy^2} = \beta.\tag{16.461}\label{eq:notw_baro_inst} \end{align} \]

this is a necessary condition. If you put Eq. (16.461), Eq. (16.457) (we don't yet know whether $\omega$ is real)

\[ \begin{align} \omega k^2Y - \omega Y'' - Uk^3Y + Y''kU = 0.\tag{16.462}\label{eq:baro_inst_deriv_2} \end{align} \]

Assuming a jet-shaped background wind

\[ \begin{align} U\left(y\right) = U_{\text{min}} + \frac{1}{2}\left(U_{\text{max}} - U_{\text{min}}\right)\left(1 + e^{i \frac{y\pi}{b}}\right) \end{align} \]

with $U_{\mathrm{min}}, U_{\mathrm{max}}>0$, one obtains in the imaginary part of Eq. (16.462)

\[ \begin{align} k^3Y\frac{1}{2}\left(U_{\text{max}} - U_{\text{min}}\right)\sin\left(\frac{y\pi}{b}\right) = Y''k\frac{1}{2}\left(U_{\text{max}} - U_{\text{min}}\right)\sin\left(\frac{y\pi}{b}\right), \end{align} \]

so is

\[ \begin{align} Y'' = k^2Y. \end{align} \]

The fact that a state is unstable does not necessarily mean that it will collapse. However, an arbitrarily small initial deflection $v$ is sufficient to generate an exponentially increasing disturbance. In reality one can assume the existence of such a very small initial disturbance, so that instability in reality always leads to the collapse of the basic state.

16.7.4 Baroclinic instability

For the expression under the root in Eq. (16.311) now becomes an abbreviation

\[ \begin{align} g = g\left(\newoverline{u}_T, k\right) \coloneqq\frac{\beta^2K^4}{4k^2} + \newoverline{u}_T^2k^2\left(k^4 - K^4\right) \end{align} \]

defined. In the case $g<0$ applies

\[ \begin{align} \omega = \omega_r + i\frac{1}{k^2 + K^2}\omega_i' \end{align} \]

with

\[ \begin{align} \omega_i' \coloneqq\pm\sqrt{\left|g\left(\newoverline{u}_T, k\right)\right|}. \end{align} \]

This means that in the case of the negative sign the amplitude increases exponentially, so there is instability. Define

\[ \begin{align} N \coloneqq\frac{\omega_i'}{k^2 + K^2} = \frac{\sqrt{\left|g\left(\newoverline{u}_T, k\right)\right|}}{k^2 + K^2}, \end{align} \]

this is the growth constant of the disorder. So it applies in the unstable case

\[ \begin{align} N^2 + \frac{g}{\left(k^2 + K^2\right)} &= N^2 + \frac{\frac{\beta^2K^4}{4k^2} + \newoverline{u}_T^2k^2\left(k^4 - K^4\right)}{\left(k^2 + K^2\right)^2} = 0. \end{align} \]

At this point a dimensionless circular wave number is defined

\[ \begin{align} \kappa \coloneqq\frac{k}{K}, \end{align} \]

that follows

\[ \begin{align} N^2 + \frac{\frac{\beta^2}{4k^2} + \newoverline{u}_T^2k^2\left(\kappa^2 + 1\right)\left(\kappa^2 - 1\right)}{\left(\kappa^2 + 1\right)^2} &= N^2 + \frac{\beta^2}{4k^2\left(\kappa^2 + 1\right)^2} + \newoverline{u}_T^2k^2\frac{\kappa^2 - 1}{\kappa^2 + 1} = 0\nonumber\\ \Rightarrow N^2 + \frac{\beta^2}{4k^2\left(\kappa^2 + 1\right)^2} + \frac{\beta^2}{K^2}\newoverline{u}^{\star2}_T\kappa^2\frac{\kappa^2 - 1}{\kappa^2 + 1} &= 0 \end{align} \]

with a dimensionless thermal wind

\[ \begin{align} \newoverline{u}_T^\star \coloneqq\frac{\newoverline{u}_TK^2}{\beta}. \end{align} \]

From this it follows

\[ \begin{align} \frac{N^2K^2}{\kappa^2\beta^2} + \frac{1}{4\kappa^4\left(\kappa^2 + 1\right)^2} + \newoverline{u}^{\star2}_T\frac{\kappa^2 - 1}{\kappa^2 + 1} &= 0 \end{align} \]

With the dimensionless growth rate

\[ \begin{align} n \coloneqq\frac{NK}{\kappa\beta} \end{align} \]

you can do this as

\[ \begin{align} n^2 + \frac{1}{4\kappa^4\left(\kappa^2 + 1\right)^2} + \newoverline{u}^{\star2}_T\frac{\kappa^2 - 1}{\kappa^2 + 1} &= 0 \end{align} \]

note down. If you convert this to $\newoverline{u}^{\star2}_T$, it follows

\[ \begin{align} \newoverline{u}^{\star2}_T = n^2\frac{1 + \kappa^2}{1 - \kappa^2} + \frac{1}{4\kappa^4\left(1 - \kappa^4\right)}. \end{align} \]

Fig. 16.7 shows the thermal wind shear required for different wave numbers and growth rates. From the definitions of $\kappa$ and $\newoverline{u}^\star_T$ it is clear that as thermal stability decreases, one moves to the top left in this representation, i.e. towards baroclinic lability.

In baroclinic instability, potential energy of the stratification, which is continually generated by the radiation field, is converted into kinetic energy of the horizontal wind before it is dissipated.