General Circulation of the Atmosphere

Abstract

The general circulation of the atmosphere and ocean is what drives the transport of heat, chemical compounds, and whatever. It is a very complicate subject so that we will start softly just to see how much energy need to be transported and what are the characteristic times. We will threat separately transport in the atmosphere and in the ocean (in the next chapter) but the general constraints for the transport will be defined in this chapter.

Appendices

Appendix

Latitude Dependence of Solar Radiation

First of all we need to consider the geometry as shown in Fig. 6.13. To understand this figure, the observer must imagine himself to be at the point P at latitude $\phi$. In this case, the local vertical is given by the direction OZ, and then the zenith angle of the sun (the angular distance between the vertical and the direction of the sun) is $\theta$. As we have seen already, the incident flux is given by

$$\begin{aligned} F=S_0\left( \frac{d_m}{d}\right) ^2 \end{aligned}$$

(6.48)

where d is the instantaneous distance from the sun and $d_m$ the average distance. What is of interest to us is the average value of the flux $\langle F\rangle$ in some regions at latitude $\phi$, so that

$$\begin{aligned} \langle F \rangle =S_0\left( \frac{d_m}{d}\right) ^2\int _{t_{sunrise}}^{t_{sunset}}\frac{F(t)}{T} dt \end{aligned}$$

(6.49)

where T is the length of the day. From Fig. 6.13, we should find a relationship between the solar zenith angle and the hour angle h, that is, the angle at which the Earth should rotate so that the meridian at P is just below the sun. From the spherical triangle PDN, we have

$$\begin{aligned} \cos {\theta }=\sin {\phi }\sin {\delta }+\cos {\phi }\cos {\delta }\cos {h} \end{aligned}$$

(6.50)

Fig. 6.13

The geometry of the solar radiation im**ing on the atmosphere

Fig. 6.14

The annual distribution of solar radiation in $w\ m^{-2}$ and its average approximated with the Legendre expansion

where $\delta$ is the solar declination, that is, the angular distance of the sun with respect to the equatorial plane, and $\phi$ is the latitude. At this point, it is simple to find a relationship between the hour angle and the time because if $\Omega$ is the angular velocity of the Earth, then $dh=\Omega dt$ and substituting in Eq. (6.50) we obtain

$$\begin{aligned} \langle F \rangle =\frac{S_0}{2\pi }\left( \frac{d_m}{d}\right) ^2\int _{-H}^{H}(\sin {\phi }\sin {\delta }+\cos {\phi }\cos {\delta }\cos {h})dh \end{aligned}$$

(6.51)

where H is the hour of sunset/sunrise obtained from (6.51) by putting $\theta =0$, that is,

$$\begin{aligned} \cosh {H}=-\tanh {\phi }\tan {\delta } \end{aligned}$$

(6.52)

When performing the integration we have

$$\begin{aligned} \langle F \rangle =\frac{S_0}{\pi }\left( \frac{d_m}{d}\right) ^2(H\sin {\phi }\sin {\delta }+\cos {\phi }\cos {\delta }\sin {H}) \end{aligned}$$

(6.53)

Now if we want the annual average of solar radiation at some latitude we need to make the average of (6.53) over $\theta$ the conventional polar angle describing a planetary orbit. Let $\theta = 0$ at the vernal equinox. The declination $\delta$ as a function of orbital position (with $\epsilon$ the obliquity)

$$\begin{aligned} \delta =\epsilon \sin {\theta } \end{aligned}$$

(6.54)

Figure 6.14 on one side shows the radiation distribution over a year and the comparison with the Legendre polynomial approximation. We assume a constant distance from the Sun.

The Jet Production by Eddies

We write the two-dimensional equation of motion in the x (zonal direction)

$$\begin{aligned} \frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}-fv=-\frac{1}{\rho }\frac{\partial p}{\partial x}-D \end{aligned}$$

(6.55)

where D is some generic drag force. This equation can be manipulated and becomes

$$\begin{aligned} \frac{\partial u}{\partial t}&+\frac{\partial u^2}{\partial x}+\frac{\partial uv}{\partial y}-u\left( \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right) -fv=\nonumber \\&\frac{\partial u}{\partial t}+\frac{\partial u^2}{\partial x}+\frac{\partial uv}{\partial y}-fv=-\frac{1}{\rho }\frac{\partial p}{\partial x}-D \end{aligned}$$

(6.56)

where we have used the non-divergence for the velocity. Now we assume that the velocity can be thought as the sum of the zonal mean ${\bar{u}}, {\bar{v}}$ and their deviation $u={\bar{u}}+u', v={\bar{v}}+v'$ so that we have

$$\begin{aligned} u^2={\bar{u}}^2+2{\bar{u}} u'+u \prime ^2,\,\,\,\,\,\,\,\,\,uv={\bar{u}} {\bar{v}}+{\bar{u}} v'+{\bar{v}} u' \end{aligned}$$

(6.57)

The zonal average for each variable A is defined as

$$\begin{aligned} {\bar{A}}=\int _0^L A dx \end{aligned}$$

(6.58)

where L is the length of the latitude circle $L=2\pi a \cos \phi$ with $\phi$ latitude and a radius of the Earth. The definition (6.52) implies that the average of a derivative with respect to x is always zero and the zonal average of (6.50) becomes

$$\begin{aligned} \frac{\partial {\bar{u}}}{\partial t}+\frac{\partial {\bar{u}} {\bar{v}}}{\partial y}+\frac{\partial {\overline{u'v'}}}{\partial y}-f{\bar{v}}=-D \end{aligned}$$

(6.59)

Since ${\bar{v}}$ is small or zero Eq. (6.53) becomes

$$\begin{aligned} \frac{\partial {\bar{u}}}{\partial t}=-\frac{\partial {\overline{u'v'}}}{\partial y}+f{\bar{v}}-D \end{aligned}$$

(6.60)

The first term of the right represents the eddy flux in the direction x transported along y. If this flux increases with y the zonal flow decelerates while the opposite happens if the flux decreases with y,$\partial {\overline{u'v'}}/\partial y<0$. It can be shown that this correlation can be expressed as a function of the meridional flux of vorticity $v\zeta$

$$\begin{aligned} v\zeta =v\left( \frac{\partial v}{\partial x}-\frac{\partial u}{\partial y }\right) =\frac{1}{2}\frac{\partial }{\partial x}(v^2-u^2)-\frac{\partial }{\partial y}(uv) \end{aligned}$$

(6.61)

where we have used the non-divergence for the velocities. After the decomposition $v\zeta =({\bar{v}}+v')({\bar{\zeta }} +\zeta ')$ we can average zonally to get

$$\begin{aligned} \overline{v' \zeta '} =-\frac{\partial }{\partial y}\overline{u'v'} \end{aligned}$$

(6.62)

And (6.54) becomes

$$\begin{aligned} \frac{\partial {\bar{u}}}{\partial t}=-\overline{v'\zeta '}-r{\bar{u}} \end{aligned}$$

(6.63)

Now we can think of the presence of Rossby waves of the form

$$\begin{aligned} \psi =A\cos (kx+ly-\omega t) \end{aligned}$$

(6.64)

With a phase velocity $c=\omega /k$ and with a dispersion relation

$$\begin{aligned} \omega =ck={\bar{u}} k-\frac{\beta k}{k^2+l^2} \end{aligned}$$

(6.65)

The momentum flux can be evaluated considering that $u'=-\partial \psi \partial y$ and $v'=\partial \psi \partial x$ so that we have

$$\begin{aligned} \overline{u'v'}=-A^2\frac{kl}{2} \end{aligned}$$

(6.66)

And the group velocity in the meridional direction

$$\begin{aligned} G_y=\frac{\partial \omega }{\partial l}=\frac{2\beta kl}{(k^2+l^2)^2} \end{aligned}$$

(6.67)

We see that if $kl<0$ the momentum flux is northward ($\overline{u'v'}$ is positive) while the group velocity is southward. The opposite happens if $kl>0$ . In the first case we are south of the source of Rossby waves while we are north of the source in the second case. This is shown in Fig. 6.11. It is also clear that the group velocity has the opposite sign of the phase velocity. The conclusion is that energy propagates away from the source region with the group velocity while the vorticity flux converges toward the source accelerating the flow.

Pseudomomentum

Things can be seen from another point of view. Consider the vorticity equation

$$\begin{aligned} \frac{\partial \zeta }{\partial t}+u\frac{\partial \zeta }{\partial x}+v\frac{\partial \zeta }{\partial y}+\beta v=S-D \end{aligned}$$

(6.68)

And its linearized version with $\zeta ={\bar{\zeta }}+\zeta '$

$$\begin{aligned} \frac{\partial \zeta '}{\partial t}=-{\bar{u}}\frac{\partial \zeta '}{\partial x}-\gamma v'+S'-D' \end{aligned}$$

(6.69)

where

$$\begin{aligned} \gamma =\beta -\frac{\partial ^2 {\bar{u}}}{\partial y^2} \end{aligned}$$

(6.70)

We now multiply (6.63) by $\zeta '/\gamma$ and zonally average to obtain

$$\begin{aligned} \frac{\partial P}{\partial t}=-\overline{v'\zeta '}-\frac{\overline{\zeta ' S'}-\overline{\zeta ' D'}}{\gamma } \end{aligned}$$

(6.71)

where pseudomomentum P is defined as

$$\begin{aligned} P=\frac{1}{2 \gamma }\overline{{\zeta '}^2}=\frac{\gamma }{2}\overline{\eta ^2} \end{aligned}$$

(6.72)

Now $\eta =\zeta '/\gamma$ and has dimension of length and can be thought as the meridional displacement that perturbs the vorticity. From (6.65) and (6.57), we have

$$\begin{aligned} \frac{\partial {\bar{u}}}{\partial t}+\frac{\partial P}{\partial t}=\frac{\partial {\bar{u}}}{\partial t}-\overline{v'\zeta '}=0 \end{aligned}$$

(6.73)

This means the quantity $P+{\bar{u}}$ is a constant if the dissipation is zero. Now we can relate the flux of momentum $\overline{u'v'}$ to P simply noting that from (6.56), (6.67) can be written as

$$\begin{aligned} \frac{\partial {\bar{u}}}{\partial t}-\frac{\partial }{\partial y}\overline{u'v'}=0 \end{aligned}$$

(6.74)

This equation can be interpreted as an equation of continuity where the acceleration is given by the divergence of the momentum flux that can also be written as

$$\begin{aligned} F_y=-\overline{u'v'}=G_yP \end{aligned}$$

(6.75)

This can be shown for the case of the Rossby wave and for ${\bar{u}}={\text{ c }onstant}$. From (6.58) we have

$$\begin{aligned} u'=Al&\sin {(kx+ly-\omega t) },\,\,\,\,\,\,\,\ v'=-Ak\sin {(kx+ly-\omega t)} \end{aligned}$$

(6.76)

$$\begin{aligned}&\quad \zeta '=-A(k^2+l^2)\cos {(kx+ly-\omega t)} \end{aligned}$$

(6.77)

And now we can use (6.61) and (6.66) to demonstrate (6.69). Now we can use (6.65) with (6.61) to show that the steady state

$$\begin{aligned} r{\bar{u}}=\frac{1}{\gamma }\left( \overline{\zeta ' S'}-\overline{\zeta ' D'}\right) \end{aligned}$$

(6.78)

In the forcing region the forcing term (first on the right) will dominate and an eastward flow will result while far from the forcing region the dissipation term will dominate with a resulting westward flow.

The Hadley Circulation as a Shallow Water Case

The problem of the Hadley circulation is particularly simple if we consider it in the approximation of shallow water. To do this we write the equation of motion

$$\begin{aligned}&\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}-fv=-\frac{1}{\rho }\frac{\partial p}{\partial x} \nonumber \\&\frac{\partial v}{\partial t}+u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}+fu=-\frac{1}{\rho }\frac{\partial p}{\partial y} \end{aligned}$$

(6.79)

The other equation we need is the continuity. We assume the total depth of the fluid to be made of an unperturbed depth H and a perturbation depth h such that in (6.73) $dp=-\rho g dh$. The continuity then reads

$$\begin{aligned} \begin{aligned} \frac{\partial }{\partial t}(h+H)*u\frac{\partial }{\partial x}(h+H)+v\frac{\partial }{\partial y}(h+H)\\ =\frac{1}{H}\frac{\partial }{\partial t}h+\frac{\partial }{\partial x}\left[ \left( {1+\frac{h}{H}}\right) u\right] +\frac{\partial }{\partial y}\left[ \left( {1+\frac{h}{H}}\right) v\right] \end{aligned} \end{aligned}$$

(6.80)

Equations (6.73) and (6.74) are made non-dimensional using the Rossby radius $\sqrt{g}/f$ as the length scale and the inverse of the Coriolis parameter as time scale. We then obtain after zonally averaging

$$\begin{aligned} \begin{aligned} \frac{\partial u }{\partial t}+v\frac{\partial u}{\partial y}-v&=-\alpha u\\ \frac{\partial v }{\partial t}+v\frac{\partial v}{\partial y}+u&=-\frac{\partial h}{\partial y}-\alpha v \end{aligned} \end{aligned}$$

(6.81)

where we have introduced also a viscous drag with coefficient $\alpha$. As for the continuity we introduce a normalized thickness $\eta =h/H$ and obtain

$$\begin{aligned} \frac{\partial \eta }{\partial t}+\frac{\partial }{\partial x}[(1+\eta )u]+ \frac{\partial }{\partial y}[(1+\eta )v] \end{aligned}$$

(6.82)

The depth of the fluid can be assimilated as geopotential depth so we have

$$\begin{aligned} h+H=\frac{RT}{g}=\frac{RT_0}{g}\left( 1+\frac{T}{T_0}\right) \end{aligned}$$

(6.83)

Equation (6.76) can be averaged zonally and also we can introduce a forcing Q on temperature such that

$$\begin{aligned} \frac{\partial \eta }{\partial t}+ \frac{\partial }{\partial y}(hv)=\frac{\partial \eta }{\partial t}+ \frac{\partial }{\partial y}[(1+\eta )v]=Q \end{aligned}$$

(6.84)

where Q as the form of a Newtonian cooling

$$\begin{aligned} Q=\frac{\eta _E-\eta }{\tau } \end{aligned}$$

(6.85)

We can solve for the variables $u,v,\eta$ in the case of steady state. Then we have

$$\begin{aligned} \frac{\partial u}{\partial y}&=1-\alpha \frac{u}{v} \nonumber \\ \frac{\partial v}{\partial y}&=-[uv+\alpha v^2+Q][v^2-(1+\eta )]^{-1} \nonumber \\ \frac{\partial \eta }{\partial y}&=[(1+\eta )(\alpha v+u)+Qv][v^2-(1+\eta )]^{-1} \end{aligned}$$

(6.86)

Before going to the numerical resolution of such equations is rather instructive to consider a simple case in which there is non-viscous drag $\alpha =0$ and we consider a stationary situation with all the accelerations to be zero. In that case the system reduces to

$$\begin{aligned} \frac{\partial u}{\partial y}=1;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,v\frac{\partial v}{\partial y}+u=-\frac{\partial h}{\partial y};\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{\partial (hv) }{\partial y}=Q \end{aligned}$$

(6.87)

In the second equation we can neglect the term $vv_y$ if we assume that the zonal flow (u) is geostrophic. The first and second equations can be easily integrated giving

$$\begin{aligned} u=y C_1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\eta =\frac{1}{2}y^2+c_2 \end{aligned}$$

(6.88)

The constant $c_1=0$ while $c_2$ can be found with the requirement that when $y=Y_H$ (the limit of the Hadley cell) the perturbation $\eta =0$ so we have

$$\begin{aligned} u=y;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \eta _a(y)=\frac{1}{2}\left( Y_H^2-y^2\right) \end{aligned}$$

(6.89)

Then v can be obtained integrating the third of (6.86) so that

$$\begin{aligned} v(y)=\frac{1}{1+\eta _a(y)}\int _0^y \frac{\eta _E(y')-\eta _a(y')}{\tau }dy' \end{aligned}$$

(6.90)

The equilibrium temperature is parameterized in a simple form

$$\begin{aligned} \eta _E(y)={\left\{ \begin{array}{ll} H_E, &{} \text {if}\, |y |<Y_E \\ 0, &{} \text {if}\, |y |>Y_E \end{array}\right. } \end{aligned}$$

The values of $H_E$ and $Y_E$ must be assigned and they have to do with the intensity and the extension of the forcing while $Y_H$ can be determined. In this simple case we can solve the integral (6.84) for an $y \le Y_E$ with the result

$$\begin{aligned} v(y)=\left[ H_E y-\frac{1}{2}\left( Y_H^2y-\frac{y^3}{3}\right) \right] [1+\eta _a(y)]^{-1} \end{aligned}$$

(6.91)

For $y=Y_E$ we have $v(Y_E)=0$ and then

$$\begin{aligned} H_EY_E-\frac{1}{3}Y_H^3=0,\,\,\,\,\,\Rightarrow {Y_H=(3H_EY_E)^{1/3}} \end{aligned}$$

(6.92)

When $y\ge Y_E$ the result for the same integral can be obtained from the rule

$$\begin{aligned} v(y)=\int _0^{Y_E}+\int _{Y_E}^y=\left[ H_EY_E-0.5\left( Y_H^2y-\frac{Y_E^3}{3}\right) \right] [1+\eta _a(y)]^{-1} \end{aligned}$$

(6.93)

These solutions are represented in Fig. 6.7 together with the numerical solutions for the case with $\alpha \ne 0$. A critical parameter in this case is the determination of the width of the Hadley cell, $Y_H$. The interested reader can use the original paper by Polvani and Sobel.

Baroclinic Instability

We have given a qualitative explanation for the baroclinic instability and now we should give a more analytical explanation. But this is more than an excuse to introduce some interesting properties of vorticity and waves. We consider an atmospheric slab where we suppose a wind shear exist in the zonal wind produced by a meridional temperature gradient. The mean zonal wind ${\bar{u}}(z,y)$ is in geostrophic equilibrium with the pressure gradient

$$\begin{aligned} {\bar{u}}(z,y)=-\frac{1}{f}\frac{\partial \Phi }{\partial y} \end{aligned}$$

(6.94)

where $\Phi (z,y)$ is the geopotential which is in hydrostatic equilibrium

$$\begin{aligned} \frac{\partial \Phi }{\partial z}+\frac{R{\bar{T}}}{H}=0 \end{aligned}$$

(6.95)

As promised we now use the definition of potential vorticity (5.126) as

$$\begin{aligned} q=\nabla ^2 \Phi +f+\frac{1}{\rho }\frac{\partial }{\partial z}\left( \frac{f}{N^2}\rho \frac{\partial \Phi }{\partial z}\right) \end{aligned}$$

(6.96)

where we have used the definition of vorticity

$$\begin{aligned} \zeta \approx \frac{\partial v_g}{\partial x}-\frac{\partial u_g}{\partial y}=\frac{\partial }{\partial x}\frac{\partial \Phi }{\partial x}+\frac{\partial }{\partial y}\frac{\partial \Phi }{\partial y} \end{aligned}$$

(6.97)

We now proceed with the perturbation approach so that

$$\begin{aligned} \Phi ={\bar{\Phi }}+\Phi ',\,\,\,\,\,\,\ q={\bar{q}}+q',\,\,\,\,\,\,\,u={\bar{u}}_g+u',\,\,\,\,\,\,\,v'=v_g,\,\,\,\,\,\,\,w'=w \end{aligned}$$

(6.98)

And write the perturbation potential vorticity

$$\begin{aligned} q'=\frac{1}{f}\nabla ^2 \Phi '+\frac{1}{\rho }\frac{\partial }{\partial z}\left( \frac{f}{N^2}\rho \frac{\partial \Phi '}{\partial z}\right) \end{aligned}$$

(6.99)

And then the variation of potential vorticity due to the advection of meridional vorticity

$$\begin{aligned} \left( \frac{\partial }{\partial t} +{\bar{u}}_g \frac{\partial }{\partial x} \right) q'+\frac{1}{f}\frac{\partial \Phi ' }{\partial x}\frac{\partial {\bar{q}}}{\partial y}=0 \end{aligned}$$

(6.100)

From (6.90) we have

$$\begin{aligned} \frac{\partial {\bar{q}}}{\partial y}=\frac{\partial ^2 {\bar{u}}_g}{\partial y^2}+\beta -\frac{1}{\rho }\frac{\partial }{\partial z}\left( \frac{f}{N^2}\rho \frac{\partial {\bar{u}}_g}{\partial z}\right) \end{aligned}$$

(6.101)

we now consider the thermodynamic equation

$$\begin{aligned} \frac{d}{dt}\left( \frac{\partial \Phi }{\partial z}\right) +w N^2=0 \end{aligned}$$

(6.102)

And its linearized form

$$\begin{aligned} \left( \frac{\partial }{\partial t}+{\bar{u}}_g\frac{\partial }{\partial x}\right) \frac{\partial \Phi '}{\partial z}+v_g\frac{\partial }{\partial y}\left( \frac{\partial {\bar{\Phi }}}{\partial z}\right) +wN^2=0 \end{aligned}$$

(6.103)

Substituting (6.88) and remembering that $v_g=(\partial \Phi '/\partial x$)

$$\begin{aligned} \left( \frac{\partial }{\partial t}+{\bar{u}}_g\frac{\partial }{\partial x}\right) \frac{\partial \Phi '}{\partial z}-\frac{\partial \Phi '}{\partial x}\frac{\partial {\bar{u}}_g}{\partial z}+wN^2=0 \end{aligned}$$

(6.104)

This equation is applied to the top and lower boundaries where $w=0$ gives the boundary condition

$$\begin{aligned} \left( \frac{\partial }{\partial t}+{\bar{u}}_g\frac{\partial }{\partial x}\right) \frac{\partial \Phi '}{\partial z}-\frac{\partial \Phi '}{\partial x}\frac{\partial {\bar{u}}_g}{\partial z}=0 \end{aligned}$$

(6.105)

A possible solution to (6.94) is assumed of the form

$$\begin{aligned} \Phi '={\tilde{\Phi }}(y,z)exp[ik(x-ct)] \end{aligned}$$

(6.106)

That once substituted in (6.94) would give

$$\begin{aligned} ({\bar{u}}_g-c)\left[ \frac{\partial ^2 {\tilde{\Phi }} }{\partial y^2}-k^2 {\tilde{\Phi }}+\frac{1}{\rho }\frac{\partial }{\partial z}\left( \frac{f^2}{N^2}\rho \frac{\partial {\tilde{\Phi }}}{\partial z}\right) \right] +{\tilde{\Phi }}\frac{\partial {\bar{q}}}{\partial y}=0 \end{aligned}$$

(6.107)

In the same way substitution in Eq. (6.99) gives

$$\begin{aligned} ({\bar{u}}_g-c)\frac{\partial {\tilde{\phi }}}{\partial z}-{\tilde{\Phi }}\frac{\partial {\bar{u}}_g}{\partial z}=0 \end{aligned}$$

(6.108)

At this point to solve for Eqs. (6.101) and (6.102) we assume that the meridional gradient of potential vorticity is given by

$$\begin{aligned} \frac{\partial q}{\partial y}=-\frac{f^2}{N^2}\frac{\partial {\bar{u}}_g}{\partial z}\delta (z=z_{bot})-\frac{\partial q}{\partial y}+\frac{f^2}{N^2}\frac{\partial {\bar{u}}_g}{\partial z}\delta (z=z_{top}) \end{aligned}$$

(6.109)

In practice we attribute the meridional gradient of the vorticity to the changes happening in the delta Dirac layer at the top and bottom of the atmospheric slab. Further we assume the zonal velocity has a constant shear $\Lambda$ such that $\Lambda =\partial u_g/\partial z$. With these conditions in mind (6.101) and (6.102) reduce to

$$\begin{aligned} \begin{aligned} (\Lambda z-c)\left( \frac{\partial ^2{\tilde{\Phi }}}{\partial y^2}-k^2{\tilde{\Phi }}+\frac{f^2}{N^2}\frac{\partial ^2{\tilde{\Phi }}}{\partial z^2}\right)&=0\\ (\Lambda z-c)\frac{\partial {\tilde{\Phi }}}{\partial z}-{\tilde{\Phi }}\Lambda&=0 \end{aligned} \end{aligned}$$

(6.110)

Because the equations are homogeneous in y and z we may separate the variables according to

$$\begin{aligned} {\tilde{\Phi }}(y,z) = \sin (l y) {\tilde{\Psi }} (z) \end{aligned}$$

(6.111)

That one substituted in (6.104) gives

$$\begin{aligned} \begin{aligned} (\Lambda z-c)\frac{\partial ^2 {\tilde{\Psi }}}{\partial z^2}-\alpha ^2 {\tilde{\Psi }}=0\\ (\Lambda z-c)\frac{\partial {\tilde{\Psi }}}{\partial z}-\Lambda {\tilde{\Psi }}=0 \end{aligned} \end{aligned}$$

(6.112)

where

$$\begin{aligned} \alpha ^2=\frac{N^2}{f^2} (k^2+l^2) \end{aligned}$$

(6.113)

Assume a solution of the form

$$\begin{aligned} {\tilde{\Psi }}(z)=A\sinh (\alpha z)+B\cosh (\alpha z) \end{aligned}$$

(6.114)

which is substituted in the second of (6.106) with the boundary condition at $z=0$ and $z=H$. We obtain

$$\begin{aligned} \begin{aligned} c\alpha A-\Lambda B=0\\ [\alpha (\Lambda H-c)\cosh \alpha H-\Lambda \sinh \Lambda H]A+[\alpha (\Lambda H-c)\sinh \alpha H-\Lambda \cosh \alpha H]B =0 \end{aligned} \end{aligned}$$

(6.115)

Being a homogeneous system we can put the determinant to zero to obtain the phase velocity

$$\begin{aligned} c=\frac{\Lambda H}{2}\left[ 1\pm \left( 1-\frac{4\cosh \alpha H}{\alpha H\sinh \alpha H}+\frac{4}{\alpha ^2 H^2}\right) ^{1/2}\right] \end{aligned}$$

(6.116)

When the content of the square root becomes negative the wave will grow. The critical wavelength at which this happens can be found with the condition

$$\begin{aligned} 1-\frac{4\cosh \alpha H}{\alpha H\sinh \alpha H}+\frac{4}{\alpha ^2 H^2}=\frac{\alpha ^2 H^2}{4}-\frac{\alpha H}{\tanh \alpha H}+1=0 \end{aligned}$$

(6.117)

This equation can be solved for the critical value $\alpha _c$ giving

$$\begin{aligned} \frac{\alpha _cH}{2}=\coth \left( \frac{\alpha _cH}{2}\right) \end{aligned}$$

(6.118)

That is

$$\begin{aligned} \alpha _c H\approx 2.4 \end{aligned}$$

(6.119)

If we assume $k=l$ from (6.107) we get

$$\begin{aligned} 2\left( \frac{kNH}{f}\right) ^2=5.76\Rightarrow kL_{\rho }\approx 1.7\Rightarrow \frac{2\pi }{L}\approx 1.7L_{\rho }\Rightarrow L\approx 3700 \end{aligned}$$

(6.120)

With $L_{\rho }\approx 1000$ km. L represent the threshold for the wavelength of the most unstable wave. Notice that this is much longer than the result we got in the qualitative treatment.

Fig. 6.15

The meridional movement of parcel along the isentropes

Baroclinic Instability: The Physical Approach

It is rather interesting to report the original physical approach to the baroclinic instability as reported by John Green in his seminal 1960 paper. He considered a meridional section of the atmosphere Fig. 6.15 where the inclination of the constant potential temperature lines is given by

$$\begin{aligned} \alpha _{\theta }=\frac{\partial z}{\partial y} =\frac{\partial \theta }{\partial y} \Big / \frac{\partial \theta }{\partial z}=-\frac{f}{N^2}\frac{\partial {\bar{u}}}{\partial z} \end{aligned}$$

(6.121)

where we have used the relations

$$\begin{aligned} N^2=\frac{g}{\theta _0}\frac{\partial \theta }{\partial z},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,f\frac{\partial {\bar{u}}}{\partial z}=-\frac{g}{\theta _0}\frac{\partial \theta }{\partial y} \end{aligned}$$

(6.122)

Now we assume as shown in Fig. 6.15 to move a parcel of air of mass $m_o$ from point A to point B along a path $\delta L$ and return the displaced mass $m_1$ to the initial point A and the change in potential energy is then

$$\begin{aligned} \Delta P=g\delta L\sin \alpha (m_0-m_1) \end{aligned}$$

(6.123)

If the interchange is carried out adiabatically we can associate at the $m_0$ the entropy $s_0$ and the relation between entropy and mass is the same as between entropy and potential temperature

$$\begin{aligned} \frac{m_1}{m_0}=\exp \left\{ \frac{s_0-s_1}{C_p}\right\} \end{aligned}$$

(6.124)

Because of the smallness of $(s_0-s_1)$ we have

$$\begin{aligned} \frac{m_1-m_0}{m}\approx \frac{s_0-s_1}{C_p} \end{aligned}$$

(6.125)

where $m=(m_0+m_1)/2$. This can be substituted in (6.117) to get

$$\begin{aligned} dP=g\delta L m\frac{(s_0-s_1)}{C_p} \end{aligned}$$

(6.126)

where

$$\begin{aligned} \Delta s=s_0-s_1=\frac{\partial s}{\partial y}\delta y+\frac{\partial s}{\partial z}\delta z=\frac{C_p \delta L}{\theta }\left( -\frac{\partial \theta }{\partial y}\cos \alpha +\frac{\partial \theta }{\partial z}\sin \alpha \right) \end{aligned}$$

(6.127)

where we have used the relation $ds=C_p d\theta /\theta$. We now would like to consider the definition of $\alpha _{\theta }$ so that (6.121) is written as

$$\begin{aligned} \begin{aligned} \Delta s=C_p \frac{\delta L}{\theta }\frac{\partial \theta }{\partial z}\left( \alpha _{\theta }\cos \alpha +\sin \alpha \right) =C_p\frac{\delta L}{\theta }\frac{\partial \theta }{\partial y} \left( \frac{\alpha _{\theta }\cos \alpha -\sin \alpha }{\alpha _{\theta }}\right) \\ \approx C_p\delta L\frac{f}{g}\frac{\partial u}{\partial z}\frac{\alpha _{\theta }-\alpha }{\alpha _{\theta }} \end{aligned} \end{aligned}$$

(6.128)

Substitution in (6.120) would give

$$\begin{aligned} dP=g\delta L^2 m\sin \alpha \left( \frac{\alpha _{\theta }\cos \alpha -\sin \alpha }{\alpha _{\theta }}\right) \frac{f}{g}\frac{\partial u}{\partial z}\approx \delta L^2 m\alpha \left( \frac{\alpha _{\theta }-\alpha }{\alpha _{\theta }}\right) f\frac{\partial u}{\partial z} \end{aligned}$$

(6.129)

The maximum value is obtained for $\alpha =\tfrac{1}{2}\alpha _{\theta }$ so that the change in potential energy per unit mass is given by

$$\begin{aligned} dPE=\frac{dP}{\delta m}=\delta L^2\frac{1}{4}\alpha _{\theta }f\frac{\partial u}{\partial z} \end{aligned}$$

(6.130)

Formally this can be written as (to mimic the energy of an oscillator) with $\delta y \approx \delta L$

$$\begin{aligned} \frac{1}{2}(\delta y)^2_{max}=\frac{1}{2}\delta y^2\left( \frac{1}{4}\alpha _{\theta }f\frac{\partial u}{\partial z}\right) =\frac{1}{2}\delta y^2\left( \frac{1}{2}\frac{f}{N}\frac{\partial u}{\partial z}\right) ^2 \end{aligned}$$

(6.131)

In analogy with the harmonic oscillator we associate the frequency of the term in parenthesis to the growth rate $\sigma$

$$\begin{aligned} \sigma =\frac{1}{2}\frac{f}{N}\frac{\partial u}{\partial z} \end{aligned}$$

(6.132)

with parameters

$$\begin{aligned} f=10^{-4}\ s^{-1},\,\,\,N=10^{-2}\ s^{-1},\,\,\,\frac{\partial u}{\partial z}=3\ \text {ms}^{-1}/\text {km},\,\,\, \end{aligned}$$

(6.133)

We obtain a growth rate of roughly 1 day. John Green in his book gives again another way to evaluate the growth rate. First of all the available potential energy (APE) is evaluated with the interchange of two particles A and B with densities $\rho _A$ and $\rho _B$ (with $\rho _A$ > $\rho _B$) initially at altitudes $z_A$ and $z_B$. Then the position is switched so that the resulting APE is given by the difference

$$\begin{aligned} APE=g(\rho _A z_A+\rho _B z_b-\rho _A z_B-\rho _Bz_A)= g(\rho _A-\rho _B)(z_A-z_B)=g\Delta \rho \Delta z \end{aligned}$$

Now if we refer to Fig. 6.15 the change in density can be written as

$$\begin{aligned} \Delta \rho =\left( \frac{\partial \rho }{\partial z}\delta z+\frac{\partial \rho }{\partial y}\delta y\right) =\delta L\left( \frac{\partial \rho }{\partial z}\alpha +\frac{\partial \rho }{\partial y}\right) \end{aligned}$$

Again if we use $\alpha _{\theta }=-(\partial \theta /\partial y)/(\partial \theta /\partial z)$ we have

$$\begin{aligned} APE=g\delta L^2\alpha \left( 1-\frac{\alpha }{\alpha _{\theta }}\right) \frac{\partial \rho }{\partial y}=\frac{1}{2}g\delta L^2 \frac{\partial \rho }{\partial y}\alpha =\frac{1}{2}\rho v^2 \end{aligned}$$

We can obtain an estimate of the growth rate observing that

$$\begin{aligned} \frac{1}{\delta L}\frac{d \delta L}{dt}=\left( \frac{g\alpha }{\rho }\frac{\partial \rho }{\partial y}\right) ^{1/2} \end{aligned}$$

Considering that $g\alpha \approx 0.01$ and $(1/\rho )\partial \rho /\partial y \approx 10^{-8}$ we obtain a growth rate of roughly 1 day.

Baroclinic Instability: The Charney Problem

The problem solved by Eric Eady assumed the atmosphere to be a slab of finite thickness limited by rigid lid. A zonal wind shear was present as a consequence of temperature meridional gradient. The solution of the problem was published by Eady in 1949, that is, 2 years after Jule Charney had solved a much more complex problem that did assume a variable Coriolis parameter, f, and used a rigid lid only at the bottom. The solution of the Charney problem requires then a vertical structure that can be simulated with a two-level model. However is much more instructive to understand the physics underlying the solution. First step is to specify that we assume a density and zonal velocity changing with eight in such a way that

$$\begin{aligned} \begin{aligned} {\bar{u}}&=\Lambda z\\ -\frac{1}{\rho }\frac{\partial \rho }{\partial z}&=\frac{1}{H}=\text{ c}onst \end{aligned} \end{aligned}$$

(6.134)

Then we can write Eq. (6.101) once we execute the derivation

$$\begin{aligned} \begin{aligned}&({\bar{u}} -c)\left[ \frac{f^2}{N^2}\left( \frac{\partial ^2 \Phi }{\partial z^2}-\frac{1}{H}\frac{\partial \Phi }{\partial z}\right) +\frac{\partial ^2 \Phi }{\partial y^2}-k^2\Phi \right] + \left[ \beta +\frac{f^2}{N^2}\frac{\Lambda }{H}\right] \Phi =0\\&-c\frac{\partial \Phi }{\partial z}-\Lambda \Phi =0 \end{aligned} \end{aligned}$$

(6.135)

where we have assumed a streamfunction of the form $\Phi (y,z)\exp [ik(x-ct)]$ and the second equation is the thermodynamic equation written at $z=0$ where the vertical velocity is zero. We now assume that

$$\begin{aligned} \Phi (z,y)=A(z)\sin {ly} \end{aligned}$$

(6.136)

So that (6.129) becomes

$$\begin{aligned} \begin{aligned}&(\Lambda z-c)\left[ \frac{f^2}{N^2}\left( A_{zz}-\frac{A_{z}}{H}\right) -K^2A\right] +\left[ \beta +\frac{f^2}{N^2}\frac{\Lambda }{H}\right] A=0\\&-cA_z-\Lambda A=0 \end{aligned} \end{aligned}$$

(6.137)

where $K^{2}=k^2+l^2$. We now follow the consideration of Pedlovski and note that the derivative of the potential vorticity $\partial q/ \partial y$ (last term in (6.129)) is not zero and this means that the equation becomes singular each time $z=c/\Lambda$. If c is complex as should be for a growing wave the singularity will lie on the z-plane while if it is real a positive will lie in the domain of the problem $z \ge 0$. Such points correspond to the so-called it critical levels. The structure of the solution must contain the balance between the second derivative of A and the term involving $q_y$. Without the second derivative the condition at the surface in (6.131) could not be satisfied. Based on that we can get a rough estimate of the depth scale of the perturbation. From the first (6.131), we have the balance between the first derivative and the gradient of potential vorticity

$$\begin{aligned} \Lambda d \frac{f^2}{N^2d^2}A \approx \left[ \beta +\frac{f^2}{N^2}\frac{\Lambda }{H}\right] A \end{aligned}$$

(6.138)

From which we get the vertical scale $d_{\beta }=d$

$$\begin{aligned} d=d_{\beta }=\frac{f^2}{N^2}\frac{\Lambda }{\beta } \end{aligned}$$

(6.139)

If we assume the shear $\Delta z \approx H\Lambda$ we have

$$\begin{aligned} \Delta {\bar{u}}_{\beta } \approx \Lambda d_{\beta }=\frac{f^2}{N^2}\frac{\Lambda ^2}{\beta } \end{aligned}$$

(6.140)

The larger the $\beta$ the smaller is the shear which is confined near the ground. We can also get an estimate of the horizontal scale L based on the same argument of the deformation radius

$$\begin{aligned} L=L_{\beta }=\frac{Nd_{\beta }}{f}=\frac{f\Lambda }{N\beta } \end{aligned}$$

(6.141)

which say that the horizontal scale is proportional to the vertical scale. With $f=10^{-4}\, \text {s}^{-1}$, $N=10^{-2}\, \text {s}^{-1}$, $\beta =1.5\cdot 10^{-11} \, \text {m}^{-1}\, \text {s}^{-1}$, and $\Lambda =3 \cdot 10^{-3}\, \text {s}^{-1}$ we have $d_{beta}\approx 20\ \, \text {km}$ $d_{beta}\approx 2000 \, \text {km}$. Before discussing these numbers we can get an estimate of the growth rate $\sigma$ by observing that the imaginary component of the phase velocity $c_i$ will be of the order of the shear $\Delta u =\Lambda d$ so that we have

$$\begin{aligned} \sigma =kc_i\approx \frac{\Lambda d}{L}= \Lambda d\frac{f}{Nd}=\frac{f}{N} \Lambda \approx 3 \cdot 10^{-5}s^{-1} < \tfrac{1}{2} day \end{aligned}$$

(6.142)

And it is independent from d. All the previous considerations are based on the fact that in the last term of (6.131) the last term is dominated by beta, that is,

$$\begin{aligned} H>\frac{f^2\Lambda }{N^2\beta }=d_{\beta } \end{aligned}$$

(6.143)

On the other hand, if $\beta$ is negligible with respect to the other term in (6.132) we have simply

$$\begin{aligned} d=H \end{aligned}$$

(6.144)

And in this case the horizontal scale is simply the deformation radius NH/f. It is clear that if $d_{beta}<H$ the scale height is given by (6.133) because it is the smaller term which determìnes the dominant term in the right-hand side of (6.132). There is another way to look at the same problem and that is the motion should be within the hedge of instability, that is,

$$\begin{aligned} \frac{w}{v}=\mathcal {O}\left( \frac{\partial \theta /\partial y}{\partial \theta /\partial z}\right) =\frac{f\Lambda }{N^2} \end{aligned}$$

(6.145)

For large $\beta$ the vorticity equation

$$\begin{aligned} \frac{D\zeta }{Dt}+\beta v\approx f\frac{\partial w}{\partial z} \end{aligned}$$

(6.146)

reduces to

$$\begin{aligned} \beta v\approx f\frac{\partial w}{\partial z}\Rightarrow {\frac{w}{v}}=\frac{d \beta }{f} \end{aligned}$$

(6.147)

which when compared with (6.139) gives (6.133). Finally we can get another interesting relation from (6.133) that can be rewritten as

$$\begin{aligned} \Lambda H=\Delta u=\beta L\frac{NH}{f}\approx \beta L^2 \end{aligned}$$

(6.148)

where we have assumed $L_{\rho }\approx L$. This relation is similar to Eq. (6.31) which we found discussing the Hadley cell.