Fluid Dynamics

Abstract

As the reader may know (and as for all other chapters) a topic so huge as fluid dynamics would require books, treatises, etc. However we need to introduces only the basis so that we could understand what comes later in terms of transport of energy and chemicals around the atmosphere and the ocean. As usual the treatment will be at two different levels, qualitative in the main text and much more detailed in the appendix. We will not go into the philosophy of fluid dynamic because we will use the most practical tools it produced for our purposes that we will use in the book. Consider the smoke that exit from a chimney and suppose to set a thermometer just above the chimney at a fixed point. The measurement will produce a series of value that will give the temperature at that point as a function of time T(t) so that the derivative $\partial T/\partial t$ will give the rate of change of temperature at the specific point of measurement. Now suppose to follow a parcel of air coming out of the chimney.

Appendices

Appendix

Diffusion Equation

There is an interesting application of the equation of mass balance for methane ($CH_4$) which is one of the so-called greenhouse gases. This application is given as an exercise in the beautiful Jacob’s book. Methane has a constant mixing ratio in the troposphere of about $1.9 \times 10^{-6}$ that is $1.9\,\text {ppm}$ while present in the stratosphere a constant scale height of about $60\,\text {km}$. Assuming that there is no production in the stratosphere we can write Eq. (5.10) at steady state ($dn/dt=0$) as

$$\begin{aligned} \frac{d}{dz}\left( n_a K\frac{dC}{dz}\right) =ln_M \end{aligned}$$

(5.64)

is the number density of methane. Here l is the local loss of methane (in $s^{-1}$ and $n_m$. We can integrate (5.64) from the tropopause ($z_T$) to the top of the atmosphere to get

$$\begin{aligned} -n_a K\frac{dC}{dz}\Bigg |_{z_t} =\int _{z_t}^\infty ln_Mdz=L \end{aligned}$$

(5.65)

Now assuming the exponential distribution for the mixing ratio we have

$$\begin{aligned} C(z)=C_T\exp \{(z-z_T)/h\},\,\,\,\,\,\,\,\Longrightarrow \frac{dC}{dz}\Bigg |_{z_t}=-\frac{C_T}{h} \end{aligned}$$

(5.66)

So we obtain a relation between the total loss in the stratosphere with the mixing ratio at the tropopause.

$$\begin{aligned} L=\frac{n_a K C_T}{h} \end{aligned}$$

(5.67)

Using the known values for $n_a=5\times 10^{24} \, \text {molecules}\,\text {cm}^{-3}$, $K=0.7\,\text {m}^2\,\text {s}^{-1} s$ we obtain $L\approx 1.2\times 10^{14}\,\text {molecules}\, \text {m}^{-2}\text {s}^{-1}$. Equation (5.65) also constitutes an explanation for the distribution of methane given by (5.66). As a matter of fact we can write

$$\begin{aligned} \frac{dC}{dz}=\frac{d}{dz}\left( \frac{n_M}{n_a}\right) \end{aligned}$$

where $n_M$ and $n_a$ are the number density for methane and the atmosphere, respectively. After some rearranging it is easy to show that

$$\begin{aligned} -Kn_a\frac{dC}{dz}=-Kn_M\left( \frac{1}{H_a}-\frac{1}{h}\right) \end{aligned}$$

(5.68)

where $H_a$ and h are the scale height of the atmosphere and methane defined as

$$\begin{aligned} \frac{1}{H_a}=-\frac{1}{n_a}\frac{dn_a}{dz},\,\,\,\,\,\,\,\,\,\,\frac{1}{h}=-\frac{1}{n_M}\frac{dn_M}{dz} \end{aligned}$$

Now we use (5.64) to write

$$\begin{aligned} \frac{d}{dz}\left( n_a K\frac{dC}{dz}\right) =\frac{d}{dz}Kn_M\left( \frac{1}{H_a}-\frac{1}{h}\right) =ln_M \end{aligned}$$

Assuming a constant scale height for methane the integration gives

$$\begin{aligned} K\left( \frac{1}{H_a}-\frac{1}{h}\right) =l h \end{aligned}$$

That can be solved for h giving

$$\begin{aligned} \frac{2}{h}=\frac{1}{H_a}+\left( \frac{1}{H_a^2}+\frac{4l}{K}\right) ^{1/2} \end{aligned}$$

(5.69)

We see that for a very large K methane is completely mixed ($h=H_a$).

Geostrophic Motion in the Ocean

Some interesting application of the geostrophic motion can be made for the ocean (following, for example, Marshall and Plumb). As a first approximation for the ocean we can consider a constant density so that the pressure is proportional to height $\eta$ above some reference level $z_0$ (see Fig. 5.9) and we can write

$$\begin{aligned} p(z_0)=p_s+g\rho _{ref}(\eta -z_0) \end{aligned}$$

(5.70)

Fig. 5.9

The setting for explaining the stratification with depth for a geostrophic current. On the left the single layer ocean is shown while on the right two layers of different densities are shown with the lower layer at rest

where $p_s$ is the pressure near the surface and $\rho _{ref}$ a constant value for the density. Using Eq. (5.24) we get in vectorial form

$$\begin{aligned} \boldsymbol{u}=\frac{g}{f}\boldsymbol{k}\times \boldsymbol{\nabla }\eta \end{aligned}$$

(5.71)

It is interesting to use (5.71) to compare the height, $\eta$, for the atmosphere and the ocean. What makes the difference is the velocity that in the atmosphere may be of the order of $10 \,\text {ms}^{-1}$ while in the ocean is a hundred times less. If the L is the horizontal length and $\Delta z$ the vertical displacement we have

$$\begin{aligned} u\approx \frac{g\Delta z}{fL} \end{aligned}$$

(5.72)

That gives $100\,\text {m}$ over $1000\,\text {km}$ for the atmosphere and $1\,\text {m}$ for the ocean over the same distance. These conclusions are valid near the surface while at some depth Eq. (5.70) must be generalized by substituting the $\rho _{ref}$ with some average density $\langle \rho \rangle$ in the layer between the generic z and $\eta$, that is, $p(z)=p_s+g\langle \rho \rangle (\eta -z)$. We have then

$$\begin{aligned} \boldsymbol{k}\times \boldsymbol{\nabla }p=g\langle \rho \rangle \boldsymbol{k}\times \boldsymbol{\nabla }\eta +g(\eta -z)\boldsymbol{k}\times \boldsymbol{\nabla }\langle \rho \rangle \end{aligned}$$

(5.73)

The geostrophic balance gives

$$\begin{aligned} \begin{aligned} \boldsymbol{u}=\frac{1}{f\rho _{ref}}\boldsymbol{k}\times \boldsymbol{\nabla }p=\frac{g}{f\rho _{ref}}\left[ \langle \rho \rangle \boldsymbol{k}\times \boldsymbol{\nabla }\eta +g(\eta -z)\boldsymbol{k}\times \boldsymbol{\nabla }\langle \rho \rangle \right] \\ \approx \frac{g}{f}\boldsymbol{k}\times \boldsymbol{\nabla }\eta +\frac{g(\eta -z)}{f\rho _{ref}}\boldsymbol{k}\times \boldsymbol{\nabla }\langle \rho \rangle \end{aligned} \end{aligned}$$

(5.74)

We see that the geostrophic currents are determined by two terms, the first one due to surface height variations and the other to horizontal density changes. For uniform density, the second term in (5.74) can be neglected. However it is observed that at depth the currents are much slower indicating some compensation between the two. For $\boldsymbol{u}\rightarrow 0$ and for $z=-H$ with $H<<\eta$ (5.74) indicates that

$$\begin{aligned} |\nabla \eta | \approx \frac{H}{\rho }|\nabla \langle \rho \rangle | \end{aligned}$$

(5.75)

Now remember that buoyancy frequency

$$\begin{aligned} N^2=-\frac{g}{\rho }\frac{d\rho }{dz}\approx -\frac{g}{\rho }\frac{\Delta \rho }{\Delta z} \end{aligned}$$

(5.76)

While (5.75) is approximately

$$\begin{aligned} \frac{\Delta \eta }{\Delta x}\approx \frac{H}{\rho }\frac{\Delta \rho }{\Delta x} \end{aligned}$$

And so dividing by $N^2$ we have

$$\begin{aligned} \frac{1}{N^2}\frac{\Delta \eta }{\Delta x}\approx \frac{H}{\rho }\frac{\Delta \rho }{\Delta x}\frac{\rho }{g}\frac{\Delta z}{\Delta \rho }\Longrightarrow \frac{g}{N^2 H}\frac{\Delta \eta }{\Delta x}\approx \frac{\Delta z}{\Delta x} \end{aligned}$$

(5.77)

We see that the isopycnal slope $\Delta z/\Delta x$ is amplified by a factor $g/N^"H$ with respect to the free surface slope $\Delta \eta /\Delta x$. The amplification factor can be evaluated for $N\approx 10^{-5}\, \text {s}^{-1}$ and $H=1\,\text {km}$ that gives a factor 400. This means that for a tilt of 1 m at the surface there is a tilt of $400\,\text {m}$ at depth.

There is another way to obtain the same result as given by Pond and Pickard. We assume a two-layer fluid with the upper layer with density $\rho _1$ and is moving while the lower layer is at rest with density $\rho _2$. The pressure at some depth z in the lower layer is (see Fig. 5.9 right)

$$\begin{aligned} p=-\int _{\eta }^z \rho gdz=-\int _{\eta }^d \rho _1 gdz-\int _{d}^z \rho _2 gdz \end{aligned}$$

(5.78)

where $\eta$ is the height of the surface and d is the level of the interface between the two layers measured from $z=0$. Integrals in (5.78) can be solved to give

$$\begin{aligned} p=\rho _1g(\eta -d)+\rho _2g(d-z)=\rho _1g\eta +(\rho _2-\rho _1)gd-\rho _2gz \end{aligned}$$

(5.79)

The pressure gradient is then

$$\begin{aligned} \frac{\partial p}{\partial x}=\rho _1g\frac{\partial \eta }{\partial x}+(\rho _2-\rho _1)g\frac{\partial d}{\partial x} \end{aligned}$$

(5.80)

That must be zero because the layer is at rest so we get

$$\begin{aligned} \frac{\partial d}{\partial x}=-\frac{\rho _1}{ \rho _2-\rho _1}\frac{\partial \eta }{\partial x} \end{aligned}$$

(5.81)

The same derivative could be made for the thickness of the upper layer $D=\eta -d$ (see Fig. 5.9) so we have

$$\begin{aligned} \frac{\partial D}{\partial x}=\frac{\partial \eta }{\partial x}-\frac{\partial d}{\partial x}=\left( \frac{\rho _2}{\rho _2-\rho _1}\right) \frac{\partial \eta }{\partial x} \end{aligned}$$

(5.82)

Now $(\rho _2-\rho _1)<<\rho _2$ and again the slope of the interface is much larger than the slope of the surface.

Thermal Wind and Taylor–Proudman Theorem

In the case of the ocean, the pressure can be related to the water density by the hydrostatic equilibrium and as consequence the change of geostrophic velocities with depth can be written as

$$\begin{aligned} \begin{aligned} f\frac{\partial }{\partial z}(\rho u)=-\frac{\partial }{\partial y}\frac{\partial p}{\partial z}=g\frac{\partial \rho }{\partial y}\\ f\frac{\partial }{\partial z}(\rho v)=\frac{\partial }{\partial x}\frac{\partial p}{\partial z}=-g\frac{\partial \rho }{\partial x} \end{aligned} \end{aligned}$$

(5.83)

Considering the change in density is small with respect to the change in velocity more simply we have the change with z given by

$$\begin{aligned} \frac{\partial u}{\partial z}=\frac{g}{\rho f}\frac{\partial \rho }{\partial y},\,\,\,\,\,\,\,\,\,\, \frac{\partial v}{\partial z}=-\frac{g}{\rho f}\frac{\partial \rho }{\partial x} \end{aligned}$$

(5.84)

If we consider a situation in which density decreases toward east (increasing x) then $\partial \rho /\partial x < 0$ and v increase with z. The component u will also increase with height in the water column if the density decreases to the south ($\partial \rho /\partial y < 0$). The ocean in the upper layer can be considered a barotropic fluid when pressure is only a function of density. For such a fluid we can write the geostrophic equilibrium in vectorial form as

$$\begin{aligned} 2\boldsymbol{\Omega }\times \boldsymbol{u}+\frac{1}{\rho }\boldsymbol{\nabla }p+\boldsymbol{\nabla }\phi =0 \end{aligned}$$

(5.85)

where $\phi$ is the gravitational potential.

We can apply the operator $\boldsymbol{\nabla \times }$ to get

$$\begin{aligned} \boldsymbol{\nabla }\times \left( 2\boldsymbol{\Omega }\times \boldsymbol{u}+\frac{1}{\rho }\boldsymbol{\nabla }p+\boldsymbol{\nabla }\phi \right) =\boldsymbol{\nabla \times }(2\boldsymbol{\Omega }\times \boldsymbol{u})-\frac{\boldsymbol{\nabla }p \times \boldsymbol{\nabla }\rho }{\rho ^2}=0 \end{aligned}$$

(5.86)

The cross product between the pressure and density gradient is zero in a barotropic fluid and once the first term is developed we have

$$\begin{aligned} (2\boldsymbol{\Omega }\cdot \boldsymbol{\nabla })\boldsymbol{u}=0 \end{aligned}$$

(5.87)

Fig. 5.10

The Taylor–Proudman theorem states that the circulation stays constant in the direction parallel to the angular velocity (left). An obstacle at the bottom (the dashed cylinder) creates a vertical column (right)

If $\omega$ is in the z-direction this relation implies

$$\begin{aligned} \frac{\partial \boldsymbol{u}}{\partial z}=0 \end{aligned}$$

(5.88)

This expresses the Taylor–Proudman theorem that in barotropic fluid the velocity remains constant with height. In particular (5.88) also applies to the vertical component w so that $\Omega \partial w/\partial z =0$. This means that if $w=0$ somewhere it must be zero everywhere and so rotation suppresses vertical motions and a flow assume the shape seen in Fig. 5.10. The column which develops above a possible obstacle is called a Taylor column. A last notation is about the occurrence of Taylor columns in the atmosphere. Equation (5.88) has been obtained in the hypothesis of constant density and such thing just does not exist.

At this point we would like to illustrate another simple way to obtain the thermal wind equation. This is done by evaluating the pressure at the same height z over two columns at different temperatures and with different scale heights, H and $H+\Delta H$. We have

$$\begin{aligned} p=p_0\exp \{-z/H\},\,\,\,\,\,\,\,\, p'=p_0\exp \{-z/(H+\Delta H)\} \end{aligned}$$

(5.89)

Now the exponential for $p'$ can be written as (assuming $\Delta H<<H$)

$$\begin{aligned} -\frac{z}{H+\Delta H}=-\frac{z}{H\left( 1+\frac{\Delta H}{H}\right) } \approx -\frac{z}{H}\left( 1-\frac{\Delta H}{H}\right) \approx -\frac{z}{H}-\frac{z \Delta H}{H^2} \end{aligned}$$

The pressure difference $\Delta p=p-p'$ results

$$\begin{aligned} \Delta p=p-p'=p_0\exp (-z/H)\left( 1-\exp \{-z\Delta H/H^2\}\right) =p[1-\exp (-z\Delta H/H^2)] \end{aligned}$$

Assuming $z\Delta H/H<<1$ we have

$$\begin{aligned} \Delta p\approx p\frac{z\Delta H}{H^2}\approx p\frac{g}{R}\frac{\Delta T}{T^2}z \approx \frac{g\rho \Delta T }{T}z \end{aligned}$$

We get then

$$\begin{aligned} \frac{1}{f \rho }\frac{\partial p}{\partial x}=\frac{g}{fT}\frac{\partial T}{\partial x}z=-v_g \end{aligned}$$

(5.90)

Driving with respect to z we get one component of the thermal wind equation.

Finally we want to demonstrate the result given by (5.31). We start by writing according to (5.24) that

$$\begin{aligned} u_g=\frac{1}{\rho f}\frac{\partial }{\partial y}(\rho g dz)=-\frac{1}{f}\frac{\partial \phi }{\partial y} \end{aligned}$$

(5.91)

where we have used the definition of geopotential $d\phi =-gdz$. The difference between $u_g(p_1)$ at pressure $p_1$ and $u_g(p_0)$ with $p_1<p_0$ is given by

$$\begin{aligned} u_g(p_1)-u_g(p_0)=-\frac{1}{f}\frac{\partial \delta \phi }{\partial y} \end{aligned}$$

(5.92)

where $\delta \phi$ is the geopotential difference between $p_1$ and $p_0$, that is,

$$\begin{aligned} \delta \phi =\int _{p_0}^{p_1}gdz=\int _{p_0}^{p_1}\frac{dp}{\rho }=R{\bar{ T}}\ln {\frac{p_0}{p_1}} \end{aligned}$$

(5.93)

where we have used $\rho =p/RT$ and ${\bar{ T}}$ is the average temperature of the layer between $p_0$ and $p_1$. Using (5.89) and the corresponding for $v_g$ we have (5.31).

Thermodynamic Equation: Second Time Around

We would like to prepare the terrain for what we are going to do next in particular with the thermodynamic equation. We could apply to this equation the total time derivative d/dt and obtain the thermodynamic equation. We just write the derivative of Eq. (2.25) for adiabatic motion ($ds=0$) so that we have

$$\begin{aligned} \frac{\partial \ln \theta }{\partial t}+u\frac{\partial \ln \theta }{\partial x}+v\frac{\partial \ln \theta }{\partial y}+w\frac{\partial \ln \theta }{\partial z} \end{aligned}$$

(5.94)

At this point we assume that the field $\theta (x,y,z,t)$ can be written as

$$\begin{aligned} \theta (x,y,z,t)=\theta _0(z)+\theta '(x,y,z,t) \end{aligned}$$

(5.95)

where $\theta _0$ depends only on z and $\theta '<<\theta _0$. The derivative (5.94) reads

$$\begin{aligned} \frac{1}{\theta _0}\left( \frac{\partial \theta '}{\partial t}+u\frac{\partial \theta '}{\partial x}+v\frac{\partial \theta '}{\partial y}\right) +w\frac{d \ln \theta _0}{d z}=0 \end{aligned}$$

(5.96)

We now remember that the buoyancy frequency $N^2$ can be written as

$$\begin{aligned} N^2=\frac{g}{\theta _0}\frac{d\theta _0}{dz}=\frac{g}{T}(\Gamma _d-\Gamma ) \end{aligned}$$

(5.97)

And drop** the prime Eq. (5.96) becomes

$$\begin{aligned} \frac{g}{\theta _0}\left( \frac{\partial \theta }{\partial t}+u\frac{\partial \theta }{\partial x}+v\frac{\partial \theta }{\partial y}\right) +w N^2=0 \end{aligned}$$

(5.98)

We may now assume the same approximation for T so that $T(x,y,z,t)=T_=(z)+T'(x,y,z,t)$ and also observe that $\theta /\theta _0\approx T/T_0$ and (9.58) becomes

$$\begin{aligned} \frac{g}{T_0}\left( \frac{\partial T}{\partial t}+u\frac{\partial T}{\partial x}+v\frac{\partial T}{\partial y}\right) +w N^2=0 \end{aligned}$$

(5.99)

We now introduce the notation for the geopotential $\Phi$ such that

$$\begin{aligned} \frac{d \Phi }{dz}=\frac{RT}{H} \end{aligned}$$

(5.100)

where H is the scale height. Then Eq. (5.99) becomes

$$\begin{aligned} \frac{H}{R}\frac{g}{T_0}\left( \frac{\partial \Phi }{\partial t}+u\frac{\partial \Phi }{\partial x}+v\frac{\partial \Phi }{\partial y}\right) +w N^2=\left( \frac{\partial \Phi }{\partial t}+u\frac{\partial \Phi }{\partial x}+v\frac{\partial \Phi }{\partial y}\right) +w N^2=0 \end{aligned}$$

(5.101)

where we have substituted for $H=RT_0/g$. In this form we will use the thermodynamic equation later on.

Vorticity Equation

Meteorology books most of the time are translation of fluid dynamics into applied physics. The equation of vorticity falls into the same fashion and you find very cumbersome derivation of it while it could be done by any second year physics students reading the Feynman books. So we start out by writing the equation of motion

$$\begin{aligned} \frac{\partial \boldsymbol{u}}{\partial t}+(\boldsymbol{u}\cdot \boldsymbol{\nabla })\boldsymbol{u}=-\frac{\boldsymbol{\nabla }p}{\rho }-\boldsymbol{\nabla }\phi \end{aligned}$$

(5.102)

We now observe the vector identity

$$\begin{aligned} (\boldsymbol{\nabla }\times \boldsymbol{u})\times \boldsymbol{u}=(\boldsymbol{u}\cdot \boldsymbol{\nabla })\boldsymbol{u}-\tfrac{1}{2}\boldsymbol{\nabla }(\boldsymbol{u}\cdot \boldsymbol{u}) \end{aligned}$$

(5.103)

If we now define the vector vorticity

$$\begin{aligned} \boldsymbol{\omega }=\boldsymbol{\nabla }\times \boldsymbol{u} \end{aligned}$$

(5.104)

Then Eq. (5.105) becomes

$$\begin{aligned} \frac{\partial \boldsymbol{u}}{\partial t}+\boldsymbol{\omega }\times \boldsymbol{u}+\tfrac{1}{2}\boldsymbol{\nabla }(\boldsymbol{u}\cdot \boldsymbol{u})=\frac{\boldsymbol{\nabla }p}{\rho }-\boldsymbol{\nabla }\phi \end{aligned}$$

(5.105)

Operating on this equation with ${\boldsymbol{\nabla }}\times$ and remembering that the curl of any gradient is zero we have

$$\begin{aligned} \frac{\partial \boldsymbol{\omega }}{\partial t}+\boldsymbol{\nabla }\times (\boldsymbol{\omega }\times \boldsymbol{u})=0 \end{aligned}$$

(5.106)

Now this equation does not look like (5.55) but we can do better. As matter of fact the curl can be written as

$$\begin{aligned} \boldsymbol{\nabla }\times (\boldsymbol{\omega }\times \boldsymbol{u})=\boldsymbol{\omega }(\boldsymbol{\nabla }\cdot \boldsymbol{u})-\boldsymbol{u}(\boldsymbol{\nabla }\cdot \omega )+(\boldsymbol{u}\cdot \boldsymbol{\nabla })\omega )-(\omega \cdot \boldsymbol{\nabla })\boldsymbol{u} \end{aligned}$$

We see that the first and second terms on the r.h.s. of this equation are zero and we remain with

$$\begin{aligned} \frac{\partial \boldsymbol{\omega }}{\partial t}+(\boldsymbol{u}\cdot \boldsymbol{\nabla })\omega -(\omega \cdot \boldsymbol{\nabla })\boldsymbol{u}=0 \end{aligned}$$

That is

$$\begin{aligned} \frac{d\boldsymbol{\omega }}{dt}=(\omega \cdot \boldsymbol{\nabla })\boldsymbol{u} \end{aligned}$$

(5.107)

In our case we consider on the z-component of the vector $\boldsymbol{\omega }$ that is $\zeta +f$ so that Eq. (5.111) becomes (5.55)

$$\begin{aligned} \frac{d}{dt}(\zeta +f)=(\zeta +f)\frac{\partial w}{\partial z}=-(\zeta +f)\left( \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right) \end{aligned}$$

(5.108)

Actually we should include also the term with the divergence of $\omega$ while it can be easily shown from scaling arguments that can be neglected. If we call $U\approx 10 \, \text {ms}^{-1}$ a typical velocity and $L \approx 1000\,\text {km}$ a typical length we have the following order of magnitude:

$$\begin{aligned} \zeta =\frac{U}{L}\approx 10^{-5} s^{-1},\,\,\,\,\,\,f\left( \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right) \approx 10^{-9}s^{-2},\,\,\,\,\,\boldsymbol{u}(\boldsymbol{\nabla }\cdot \omega )\approx \frac{U\zeta }{L}\approx 10^{-10} \end{aligned}$$

(5.109)

Notice that in the calculation of the divergence of $\omega$ only $\zeta$ enters, while for mid-latitude $\zeta<<f$.

Rossby Waves

When the velocity divergence is zero we have the conservation of the absolute vorticity

$$\begin{aligned} \frac{d}{dt}(\zeta +f)=0 \end{aligned}$$

(5.110)

That can be written as

$$\begin{aligned} \frac{\partial \zeta }{\partial t}+u\frac{\partial \zeta }{\partial x}+v\frac{\partial \zeta }{\partial y}+\beta v=0 \end{aligned}$$

(5.111)

Now we consider a situation where there is a flow mainly in the x-direction with the velocity given by $u={\bar{ u}}+u'$. The velocity in y-direction is just $v'$ and we have that both $u'$ and $v'$ are negligible with respect to ${\bar{ u}}$. We these conditions are inserted in (5.115) we have

$$\begin{aligned} \frac{\partial \zeta '}{\partial t}+{\bar{ u}}\frac{\partial \zeta '}{\partial x}+u'\frac{\partial \zeta '}{\partial x}+v'\frac{\partial \zeta '}{\partial y}+\beta v'=0 \end{aligned}$$

The terms which include the product of two primed quantities can be neglected because are much smaller than the other terms and the equation reduces to

$$\begin{aligned} \frac{\partial \zeta }{\partial t}+{\bar{ u}}\frac{\partial \zeta }{\partial x}+\beta v=0 \end{aligned}$$

(5.112)

We then assume that the velocities can be obtained from a streamfunction $\psi$ such that

$$\begin{aligned} u=-\frac{\partial \psi }{\partial y},\,\,\,\,\,\,\,\,\,\,v=\frac{\partial \psi }{\partial x},\,\,\,\,\,\,\,\,\,\,\zeta =\frac{\partial ^2 \psi }{\partial x^2}+\frac{\partial ^2 \psi }{\partial y^2}=\nabla ^2 \psi \end{aligned}$$

(5.113)

And (5.112) becomes

$$\begin{aligned} \left( \frac{\partial }{\partial t}+{\bar{ u}}\frac{\partial }{\partial x}\right) \nabla ^2\psi +\beta \frac{\partial \psi }{\partial x} =0 \end{aligned}$$

(5.114)

Then assume a solution of the form of a two-dimensional wave

$$\begin{aligned} \psi (x,y,t)=\psi _0\cos (\omega t+ky+ly) \end{aligned}$$

(5.115)

So that

$$\begin{aligned} \nabla ^2 \psi =-(k^2+l^2)\psi ,\,\,\,\,\,\,\,\,\frac{\partial \psi }{\partial x}=-k\psi _0 \sin (\omega t+kx+ly) \end{aligned}$$

(5.116)

Substituting in (5.115) we have the dispersion relation

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

(5.117)

That is an equation we have already used. It is interesting to cast the conservation of vorticity in a slight different form. Just write Eq. (5.110) as

$$\begin{aligned} \frac{d}{dt}(\zeta +f)=\frac{\partial \zeta }{\partial t}+\boldsymbol{v}\cdot \nabla ( \zeta +f)=0 \,\,\,\,\Rightarrow \frac{\partial \zeta }{\partial t}=-\boldsymbol{v}\cdot \boldsymbol{\nabla } \zeta -\boldsymbol{v}\cdot \boldsymbol{\nabla } f \end{aligned}$$

(5.118)

The terms on the right are the advection of relative vorticity and the advection of planetary vorticity. Now if f would be constant the only contribution would come from the relative vorticity. On the other hand, referring at Fig. 5.7 we see that in the sector where the waves moves southward we have $-\boldsymbol{v}\cdot \boldsymbol{\nabla } \zeta <0$ while $-\boldsymbol{v}\cdot \boldsymbol{\nabla } f>0$. This means that in this sector the advection of planetary vorticity moves the waves westward while the advection of relative vorticity moves the waves eastward. The relative contribution of the two effects can be evaluated from the ratio

$$\begin{aligned} \frac{|-v\cdot \nabla f|}{|-v\cdot \nabla \zeta |}\approx \frac{U(f/L)}{U(U/L^2)}\approx \frac{fL}{U} \end{aligned}$$

(5.119)

We see that the contribution of the long waves to the planetary vorticity advection increases with the wavelength and so long waves move slower than short waves. It is interesting to discuss stationary waves ($c=0$). The number of waves n and wavelength L contained in a latitude circle at latitude $\phi$ is given by

$$\begin{aligned} n=\frac{2\pi \cos \phi }{L} \end{aligned}$$

(5.120)

That substituted in (5.117) with $c=0$ gives

$$\begin{aligned} n_s=a \cos \phi \sqrt{\frac{\beta }{U}} \end{aligned}$$

(5.121)

At mid-latitude for a $10 \,\text {ms}^{-1}$ we have between five and six waves. Then we would have progressive waves $c>0$ for $n>6$ and retrograde waves for $n<5$.

An Application of Potential Vorticity

We consider the conservation of potential vorticity

$$\begin{aligned} \frac{\zeta +f}{\Delta z}=\text {const} \end{aligned}$$

(5.122)

And assume that $\Delta z$ is the distance between two surfaces at potential temperature $\Delta \theta$. We also have

$$\begin{aligned} (\zeta +f)\frac{\Delta \theta }{\Delta z}=\text {const} \end{aligned}$$

(5.123)

which is a particular form of Ertel potential vorticity. We now consider the potential temperature as given by $\theta ={\bar{\theta }}+\theta '$ and indicate $\Delta \theta /\Delta z\approx \theta _z$ with the subscript indicating the derivative. Then we can write

$$\begin{aligned} (\zeta +f)\theta _z\approx {\bar{ \theta }}_z(\zeta +f)\left( 1+\frac{\theta '_z}{{\bar{\theta }}_z}\right) \end{aligned}$$

(5.124)

We then observe that

$$\begin{aligned} \theta '_z=\frac{N^2{\bar{ \theta }}}{g},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \frac{\theta '_z}{{\bar{\theta }}_z}\approx -\frac{\rho '_z}{\bar{\rho }} \end{aligned}$$

And substitution (5.124) gives

$$\begin{aligned} {\bar{\theta }}_z(\zeta +f)\left( 1-\frac{g}{N^2} \frac{\rho '_z}{\bar{\rho }}\right) ={\bar{\theta }}_z(\zeta +f)\left( 1+\frac{1}{N^2{\bar{ \rho }}} \frac{\partial ^2 p'}{\partial z^2}\right) \end{aligned}$$

(5.125)

We now use the definition of streamfunction $\psi =p/\rho f$ and the fact that at middle latitude $\zeta<<f$ to find a different form of Ertel potential vorticity q

$$\begin{aligned} q=\left( \zeta +f+\frac{f^2}{N^2}\frac{\partial ^2\psi }{\partial z^2}\right) \end{aligned}$$

(5.126)

that we will use later.