Atmospheric Dynamics

Chapter 11

Atmospheric Dynamics Earth’s atmosphere is composed of a mixture of gases such as nitrogen, oxygen, carbon dioxide, water vapor, and ozone. A wide variety of fluid flows take place in the atmosphere. In this chapter, we show how the theory of fluid dynamics in Chapter 10 is applied to the atmosphere. For this purpose, we will introduce the Navier-Stokes equation, hydrostatic and geostrophic approximations, the Boussinesq approximation, potential temperature and equivalent potential temperature, quasi-geostrophic potential vorticity, buoyancy frequency, and so on. According to the variation of temperature with height, the atmosphere is conventionally divided into layers in the vertical direction. The layer from the ground up to about 15 km altitude is called the troposphere. The troposphere is bounded above by the tropopause, and in the troposphere the temperature decreases with height. The layer from the tropopause to about 50 km altitude is called the stratosphere. The stratosphere is bounded above by the stratopause, and in the stratosphere the temperature rises with height. The layer from the stratopause to about 85-90 km altitude is called the mesosphere. The mesosphere is bounded above by the mesopause, and in the mesosphere the temperature again decreases with height. The layer above the mesopause is called the thermosphere. In the thermosphere, the temperature again rises with height. Sometimes, the troposphere is also called the lower atmosphere. Most weather phenomena, such as rain, snow, thunder, and lightning, occur in the lower atmosphere. The stratosphere and mesosphere together are called the middle atmosphere. The ozone molecules stay in the lower stratosphere and form the ozone layer. The layer above the mesosphere is called the upper atmosphere.

11.1

TWO SIMPLE ATMOSPHERIC MODELS

Energy transfer in the atmosphere involves short-wave radiation emitted by the Sun and long-wave radiation emitted by Earth’s surface and atmosphere. These two wavelength ranges represent spectral regions of blackbody emission at temperatures of about 6000 and 288 K, respectively. Planck’s law states that the blackbody spectral radiance Bλ (T) at temperature T is Bλ (T) =

2hc2 hc

,

(11.1)

λ5 (e λkB T − 1)

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where h is Planck’s constant (6.626 × 10−34 J s), c is the speed of light, λ is the wavelength, and kB is Boltzmann’s constant (1.38 × 10−23 J/K). Since the blackbody radiation is isotropic, the blackbody spectral irradiance Fλ (r, n) at a point r through a surface of normal n is obtained by integration over a hemisphere on one side of the surface,  Fλ (r, n) = Bλ (T)n · s d(s), S

where S is the hemisphere and d(s) is the element of solid angle in the direction s. Let φ be the angle between s and n. Then n · s = cos φ, d = 2π sin φ dφ. Notice that Bλ (T) is independent of s and r. Then  π/2 Fλ (r, n) = πBλ (T) 2 cos φ sin φ dφ. Since

 π/2 0

0

2 cos φ sin φ dφ = 1, the blackbody spectral irradiance is given by Fλ (r, n) = πBλ (T).

Moreover, the flux density F(r, n) at the point r through the surface of normal n  ∞  ∞ F(r, n) = Fλ (r, n) dλ = π Bλ (T) dλ. (11.2) 0

By (11.1),



∞

π 0

0

 Bλ (T) dλ =

∞ 0

c1 λ5 (e

c2 λT

− 1)

dλ,

where c1 = 2hπc2 and c2 = khcB . The constant c1 is called the first radiance constant, and the constant c2 is called the second radiance constant. Take a c2 change of variable X = λT . This implies that c2 , λ = XT c2 dX dλ = − . T X2 ∞ 3 4 Since 0 eXX−1 dX = π15 ,  ∞  c1 4 ∞ X 3 c1 π 4 4 π Bλ (T) dλ = 4 T dX = T . eX − 1 c2 15c42 0 0 From this and (11.2), the Stefan-Boltzmann law for the blackbody irradiance is given by F(r, n) = σ T 4 ,

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where σ =

c1 π 4 ≈ 5.6703 × 10−8 W/(m2 K4 ). 15c42

The constant σ is called the Stefan-Boltzmann constant. Earth’s mean surface temperature is about 288 K. This observational fact can be explained by the Stefan-Boltzmann law. First we illustrate this point by the single-layer atmospheric model. Then we introduce the two-layer atmospheric model which is an extension of the single-layer atmospheric model.

11.1.1

The Single-Layer Model

In the single-layer atmospheric model, the atmosphere is taken to be a layer at a uniform temperature Ta , and the ground is assumed to emit as a blackbody at a uniform temperature Tg . We consider this simple model. The solar power per unit area at Earth’s mean distance from the Sun is called the solar constant. It is well known that the solar constant Fs = 1370 W/m2 . Assume that the Earth-atmosphere system has a planetary albedo α = 0.3. This means that an amount αFs πa2 , where a is Earth’s radius, of the solar power is reflected back to space and the remainder (1 − α)Fs πa2 is the unreflected incoming solar irradiance at the top of the atmosphere. However, the total surface area of Earth is 4πa2 . Therefore, the mean unreflected incoming solar irradiance at the top of the atmosphere is F0 =

(1 − α)Fs πa2 1 = (1 − α)Fs . 2 4πa 4

Substituting Fs = 1370 and α = 0.3 into this equality, we obtain the mean unreflected incoming solar irradiance at the top of the atmosphere is 1 (1 − 0.3) × 1370 ≈ 240 (W/m2). 4 Denote by Tsw the transmittance of any incident solar (short-wave) radiation. Then an amount Tsw F0 is absorbed by the ground. Denote by Fg the upward irradiance from the ground and denote by Tlw the transmittance of any incident thermal (long-wave) radiation. Then a upward emission Tlw Fg reaches the top of the atmosphere. Since the atmosphere is not a blackbody, the atmosphere emission Fa is both upward and downward. Assume that the whole system is in radiative equilibrium. Then the balance of irradiances on the top of the atmosphere implies F0 =

F0 = Fa + Tlw Fg and the balance of irradiances between the atmosphere and the ground is Fg = Fa + Tsw F0 .

(11.3)

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Eliminating Fa from these two equations, we get F0 − Fg = Tlw Fg − Tsw F0 , so Fg = F0

1 + Tsw . 1 + Tlw

(11.4)

Combining (11.3) with (11.4), the atmosphere emission is Fa = F0 − Tlw Fg = F0 − Tlw F0

1 + Tsw 1 − Tlw Tsw = F0 . 1 + Tlw 1 + Tlw

By the Stefan-Boltzmann law, Fa = (1 − Tlw )σ Ta4 , where σ is the Stefan-Boltzmann constant, and so (1 − Tlw )σ Ta4 = F0

1 − Tlw Tsw . 1 + Tlw

Thus, the temperature of the model atmosphere is  F0 (1 − Tlw Tsw ) Ta = 4 . σ (1 − Tlw )(1 + Tlw ) Taking transmittances Tsw = 0.9 (strong transmittance and weak absorption of solar radiation) and Tlw = 0.2 (weak transmittance and strong absorption of thermal radiation), the temperature of the model atmosphere is  (0.7 × 1370)(1 − 0.2 × 0.9) Ta = 4 ≈ 245 (K). 4 × 5.6703 × 10−8 × (1 − 0.2)(1 + 0.2) Notice that F0 ≈ 240 W/m2. By (11.4), the upward irradiance of the ground is Fg = F0

1 + Tsw 1 + 0.9 ≈ 240 × = 380 (W/m2 ). 1 + Tlw 1 + 0.2

By the Stefan-Boltzmann law, Fg = σ Tg4 . From this and (11.4), the temperature of the ground is    (0.7 × 1370)(1 + 0.9) 4 Fg 4 F0 (1 + Tsw ) Tg = = = 4 ≈ 286 (K). σ σ (1 + Tlw ) 4 × 5.6703 × 10−8 × (1 + 0.2) This temperature is close to the observed mean surface temperature of about 288 K.

11.1.2

The Two-Layer Model

The two-layer atmospheric model is an extension of the single-layer atmospheric model. This model includes two atmospheric layers, say, the upper atmosphere

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and the lower atmosphere. The upper layer mimics the stratosphere at temperature Tstrat . The lower layer mimics the troposphere at temperature Ttrop . Assume that the upper layer is transparent to solar radiation and optically thin, and its thermal absorptance is taken as   1. Kirchhoff’s law shows that the thermal emittance of the upper layer is also  and its thermal transmittance is 1 − . Assume that the lower layer has transmittances Tsw of solar radiation and Tlw of thermal radiation and that the ground emits as a blackbody at temperature Tg , and that the mean unreflected incoming solar irradiance F0 is defined as before. Under these assumptions, the emissions from the upper and lower layers, Fstrat and Ftrop , are both upward and downward, and the emission from the ground, Fg is upward. Owing to the optically thin layer, the incoming solar irradiance on the top of the lower layer is also F0 . The amounts Tsw F0 and Tlw Fstrat are absorbed by the ground. The amount Tlw Fg is absorbed by the lower layer. The amounts (1 − )Ftrop and (1 − )Tlw Fg are absorbed by the upper layer. Assume further that the whole system is in radiative equilibrium. Then the balance of irradiances above the upper atmosphere implies F0 = Fstrat + (1 − )(Ftrop + Tlw Fg ), (11.5) the balance of irradiances between the upper atmosphere and the lower atmosphere implies Ftrop + Tlw Fg = F0 + Fstrat , (11.6) and the balance of irradiances between the lower atmosphere and the ground implies Fg = Tsw F0 + Tlw Fstrat + Ftrop . (11.7) Eliminating F0 from (11.5) and (11.6), we get 2Fstrat = (Ftrop + Tlw Fg ), and so 2 Fstrat .  Substituting this into (11.5), the emission from the upper atmosphere is Ftrop + Tlw Fg =

Fstrat =

 F0 . 2−

(11.8)

The combination of (11.8) and (11.6) gives Ftrop =

2 F0 − Tlw Fg . 2−

Substituting (11.8) and (11.9) into (11.7), we get Fg =

(2 − )Tsw +  Tlw + 2 F0 − Tlw Fg , 2−

(11.9)

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and so the emission from the ground is Fg =

(2 − )Tsw +  Tlw + 2 F0 . (2 − )(1 + Tlw )

(11.10)

From this and (11.9), the emission from the lower atmosphere is Ftrop =

2 2 − (2 − )Tlw Tsw −  Tlw F0 . (2 − )(1 + Tlw )

(11.11)

However, according to the Stefan-Boltzmann law, the emissions from the upper and lower layers and the ground are, respectively, 4 Fstrat = σ Tstrat , 4 Ftrop = σ (1 − Tlw )Ttrop ,

Fg = σ Tg4 , where σ is the Stefan-Boltzmann constant. Therefore, by (11.8), (11.10), and (11.11), the temperature of the upper atmosphere is   F0 4 Fstrat Tstrat = = 4 , σ σ (2 − ) the temperature of the lower atmosphere is   2 − (2 − )T T ) F0 (2 −  Tlw Ftrop lw sw 4 Ttrop = = 4 , σ (1 − Tlw ) σ (2 − )(1 − Tlw2 ) and the temperature of the ground is   F0 ((2 − )Tsw +  Tlw + 2) 4 Fg = 4 . Tg = σ σ (2 − )(1 + Tlw )

11.2

ATMOSPHERIC COMPOSITION

We introduce the ideal gas law and some basic concepts on gases. If the mass of 1 mol is Mm and the volume of 1 mol is Vm , then the density ρ=

Mm . Vm

For an ideal gas of pressure p and temperature T, each mole of gas obeys the law pVm = RT, where R is the universal gas constant. So the ideal gas law is given by p = Ra Tρ, where Ra = R/Mm is the gas constant per unit mass of air.

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Consider a small sample of air with volume V, temperature T, and pressure p. It is composed of a mixture of gases Gi (i ∈ Z+ ). If the number of molecules of gas Gi in the sample is ni , then the total number of molecules in the sample is  n= ni . If the molecular mass of gas Gi in the sample is mi , then the total mass of the sample is  m= mi ni . If the molar mass of gas Gi in the sample is Mi , then the total molar mass is  M= Mi ni , where all the sums are taken over all the gases in the sample. Now the ideal gas law for this small sample is pV = RT, where R = nkB and kB is Boltzmann’s constant. If the molecules of gas Gi in the sample alone are to occupy the volume V at temperature T, by the ideal gas law, the pressure exerted by the molecules of Gi from the sample is kB T . V This pressure is called the partial pressure of gas Gi . The partial pressure is sometimes used to quantify chemical concentrations. If the molecules of gas Gi in the sample alone are to be held at temperature T and pressure p, then, by the ideal gas law, the volume occupied by the molecules of gas Gi from the sample is pi = ni

Vi = n i

kB T . p

This volume is called the partial volume of gas Gi .   Dalton’s Law. Let pi , p, Vi , and V be stated as above. Then pi = p and Vi = V.  In fact, notice that n = ni . It follows from the ideal gas law immediately that  k T  kB T  kB T B pi = ni = ni =n = p, V V V  k T   kB T kB T B Vi = ni = ni =n = V. p p p Define the volume mixing ratio νi of gas Gi as νi =

Vi V.

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Proposition 11.1. Let ni , n, pi , p, and νi be stated as above. Then ni pi = . νi = n p In fact, by the definition of the partial volume, νi =

ni kBpT Vi ni = k T = . B V n n p

Similarly, by the definition of the partial pressure, ni kBVT pi ni = k T = . B p n n V pi ni = . n p Define the mean molecular mass of the sample as m = mn . Similarly, define the mean molar mass of the sample as M = Mn . Proposition 11.2. Let νi , mi , Mi , and m, M be stated as above. Then  m = mi νi ,  M = Mi νi .

Thus, νi =

In fact, by the definitions of m, M, and Proposition 11.1,   ni  mi ni m m = = = mi = mi νi , n n n  Mi ni  ni M M = = = Mi = Mi νi . n n n Define the mass mixing ratio μi of gas Gi as mi pi . μi = m p Mass and volume mixing ratios are more convenient measures of the concentration of an atmospheric gas when the transport of chemicals is being studied. Proposition 11.3. Let μi , νi , pi , p, mi , and m be stated as above. Then mi μi = νi . m

11.3

HYDROSTATIC BALANCE EQUATION

We will give several forms of the hydrostatic balance equation by using the ideal gas law and then derive the basic properties involving the atmospheric pressure and density. Consider a small cylinder of air with height z and horizontal cross-sectional area A. The mass of the cylinder of air is m = ρAz, where ρ is the density. Let g = |g|, where g be the gravitational acceleration. There are three

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vertical forces acting on the small cylinder of air: the downward gravitational force gm, the upward pressure force p(z)A on the bottom of the cylinder, and the downward pressure force p(z + z)A on the top of the cylinder. Since the small cylinder of air is in hydrostatic equilibrium, the balance of the three forces implies gρAz = p(z)A − p(z + z)A. Using Taylor’s expansion, dp z, dz and then canceling Az, we obtain the hydrostatic balance equation: p(z + z) ≈ p(z) +

dp = −gρ. dz Assume that the air is an ideal gas and is in hydrostatic balance. Then the air obeys the ideal gas law: p = ρRa T, where p, ρ, T, and Ra are stated as in Section 11.2, and so the hydrostatic balance equation takes the form dp gp =− dz Ra T or g dp =− dz. p Ra T If the temperature T is a function of height z, and the pressure at the ground is p0 , integrating both sides with respect to z from the ground z = 0 upward, we get  z g dz ln p − ln p0 = − . Ra 0 T(z ) Taking exponentials on the both sides, − Rga

p = p0 e

z

dz 0 T(z )

.

Therefore, the pressure p is a function of the height z. The simplest case is that of an isothermal temperature profile, i.e., T = T0 =constant. Then − RagzT

p = p0 e

= p0 e− H , z

0

where H = RagT0 is called the pressure scale height. Thus, in the isothermal case, the pressure decays exponentially with height.

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Similarly, eliminating the pressure p from the ideal gas law and the hydrostatic balance equation, the alternative form of the hydrostatic balance equation is dρ gρ =− dz Ra T or dρ g =− dz. ρ Ra T Integrating both sides with respect to z from the ground z = 0 upward,  z dz g ln ρ − ln ρ0 = − , Ra 0 T(z ) where ρ0 is the density at the ground. Taking exponentials on both sides, we get − Rga

ρ = ρ0 e

z

dz 0 T(z )

.

Therefore, the density ρ is also a function of the height z. In the isothermal case, T = T0 = constant, and so − RagzT

ρ = ρ0 e

= ρ0 e− H , z

0

where H = RagT0 . Therefore, in the isothermal case, the density also decays exponentially with height.

11.4

POTENTIAL TEMPERATURE

Consider a small air parcel with volume V, pressure p, and temperature T. Assume that the air in the parcel is of unit mass and is in hydrostatic balance. Let w be the enthalpy and  be the internal energy per unit mass. Since the enthalpy w =  + ρp and the thermodynamic relation d = T ds + ρp2 dρ, where s is the entropy, a small change of enthalpy dw = d +

1 p 1 dp − 2 dρ = T ds + dp. ρ ρ ρ

According the ideal gas law p = Ra Tρ, the small change of enthalpy becomes dw = T ds +

Ra T dp. p

On the other hand, for unit mass of ideal gas,  = cν T, where cν is the specific heat capacity at constant volume and is independent of T. Using the ideal gas law p = Ra Tρ, we find the enthalpy is p w =  + = cν T + Ra T = cp T, ρ

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where cp = cν + Ra . So the small change of enthalpy dw = cp dT. Therefore, T ds +

Ra T dp = cp dT. p

(11.12)

Dividing both sides of (11.12) by T, ds +

cp Ra dp = dT, p T

which is equivalent to ds = cp d(ln T) − Ra d(ln p). Integrating both sides and then letting κ =

Ra cp ,

(11.13)

we find the entropy is

s = cp ln T − Ra ln p + s0 = cp (ln T − κ ln p) + s0 = cp ln(Tp−κ ) + s0 , where s0 is a constant. When the air in the parcel is compressed adiabatically from pressure p and temperature T to pressure p0 and temperature θ, where p0 is usually taken to be 1000 hPa, we find the temperature θ. Since the process is adiabatic, the entropy satisfies the condition ds = 0. From this and (11.13), cp d(ln T) = Ra d(ln p). Integrating both sides and using the end conditions T = θ and p = p0 , we get



 θ p0 cp ln = Ra ln , (11.14) T p and so

θ =T

p0 p

κ ,

where κ =

Ra Ra = . cp cν + Ra

The quantity θ is called the potential temperature. It depends on temperature T  κ p0 and pressure p. The expression θ = T p is equivalent to ln θ = ln(Tp−κ ) + ln pκ0 . So the specific entropy s = cp ln(Tp−κ ) + s0 = cp ln θ + s1 ,

(11.15)

where s1 = s0 − cp ln pκ0 = s0 − cp κ ln p0 = s0 − Ra ln p0 . This means that the specific entropy relates to the potential temperature. When a mass of air is subject to an adiabatic change, since the entropy is constant in the adiabatic process,

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it follows from (11.15) that its potential temperature is constant. Conversely, when the mass is subject to a nonadiabatic change, its potential temperature will change.  κ Differentiating θ = T pp0 with respect to t, dθ = dt Notice that κ = d dt



Ra cp .

p0 p

κ



p0 p

κ

dT d +T dt dt



p0 p

κ .

Then

=κ

p0 p

and so 1 dθ = dt cp

κ−1



p0 p

−

p0 p2

κ

cp



dp Ra =− dt cp p



p0 p

κ

dp , dt

 dT TRa dp − . dt p dt

Applying the ideal gas law p = Ra Tρ gives



 1 p0 κ dT 1 dp dθ = cp − . dt cp p dt ρ dt By (11.12), the part in brackets cp

dT 1 dp ds − =T . dt ρ dt dt

Therefore, 1 dθ = dt cp Here the product of T and by Q, i.e., T ds dt = Q. So

ds dt



p0 p

κ T

ds . dt

is the adiabatic heating rate per unit mass, denoted dθ Q = dt cp



p0 p

κ .

This equation is called the thermodynamic energy equation.

11.5

LAPSE RATE

Denote by T the air temperature. The quantity (z) = −

dT dz

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is called the lapse rate of temperature with height. It represents the rate of decrease of temperature with height. It is clear that (i) if the temperature decreases with height in some region, then  > 0 in that region; (ii) if the temperature increases with height in some region, then  < 0 in that region. In the troposphere  > 0, and in the stratosphere  < 0.

11.5.1

Adiabatic Lapse Rate

As an air parcel rises adiabatically, the rate of decrease of temperature with height, following the adiabatic parcel, is called the adiabatic lapse rate, denoted by a . Now we find the adiabatic lapse rate. Consider an adiabatically rising air parcel with pressure p and temperature T. By (11.12), ds +

cp Ra dp = dT, p T

(11.16)

where Ra and cp are stated as above. Notice that ds = 0 in an adiabatic process. By (11.16), it follows that the vertical derivatives of the temperature and pressure of the parcel satisfy cp dT Ra dp = p dz T dz or Ra T dp dT = . dz pcp dz gp From this and the hydrostatic balance equation dp dz = − Ra T , the adiabatic lapse rate is dT g a = − = , dz cp

where g = |g| and cp = cν + Ra . Here g is the gravitational acceleration, Ra is the gas constant per unit mass of air, and cν is the specific heat capacity at constant volume. The adiabatic lapse rate for dry air is called the dry adiabatic lapse rate, denoted by d . It is approximately 9.8 K/km. The actual lapse rate in the atmosphere generally differs from the dry adiabatic lapse rate. To investigate this, consider a dry air parcel that is originally at an equilibrium position at height z0 with temperature T0 , pressure p0 , and density ρ0 , all equal to the values for the surroundings. Suppose that an instantaneous upward force is applied to the

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parcel so that the parcel rises adiabatically through a small height z from its equilibrium position at height z0 to the height z = z0 + z without influencing its surroundings. Then the parcel temperature has increased from T0 to Tp and dTp . dz On the other hand, the environment temperature has increased from T0 to Te , and dTe Te = T0 − e z, where e = − . dz If e = d , then there is a difference between the temperature of the parcel and the temperature of its surrounding environment: Tp = T0 − d z,

where d = −

Tp − Te = (e − d )z.

(11.17)

Since the pressures inside and outside the parcel at height z + z are same, these pressures are both equal to dpe z. dz According to the ideal gas law p = Ra Tρ, the densities inside and outside the parcel are, respectively, p ρp = , Ra Tp p ρe = . Ra Te p = p0 +

From this and (11.17), ρe − ρp Tp − Te (e − d )z = = . ρp Te Te

(11.18)

If e < d , then Tp < Te and ρe < ρp . So the parcel temperature is less than environment temperature and the parcel is heavier than its environment. In this case the atmosphere is said to be statically stable. If e > d , then Tp > Te and ρe > ρp . So the parcel temperature is higher than the environment temperature, and the parcel is lighter than its environment. In this case the atmosphere is said to be statically unstable. If e = d , then Tp = Te and ρe = ρp . In this case the atmosphere is said to have neutral stability.

11.5.2

Buoyancy Frequency

Suppose that V is the volume of the air parcel at height z, and ρp and ρe are the densities inside and outside the parcel, respectively. Then the masses inside and outside the parcel are ρp V and ρe V, respectively. So the upward buoyancy force

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on the parcel at height z is equal to g times the difference between these two masses: g(ρe V − ρp V) = gV(ρe − ρp ). On the other hand, since the acceleration of the parcel is d dt(z) 2 , according to Newton’s second law, the upward buoyancy force is equal to 2

d2 (z) , dt2 i.e., the mass inside the parcel times the acceleration of the parcel. Therefore, ρp V

gV(ρe − ρp ) = ρp V

d2 (z) . dt2

Dividing both sides by ρp V,

 ρe − ρp d2 (z) =g . ρp dt2

Combining this with (11.18), we get d2 (z) g(e − a ) = z 2 Te dt which is equivalent to d2 (z) + N 2 z = 0 dt2



 g(a − e ) N = . Te 2

The quantity N is called the buoyancy frequency or the Brunt-V¨ais¨al¨a frequency. The buoyancy frequency is a useful measure of atmosphere stratification. It relates to the potential temperature of the environment as follows. Let θe be the potential temperature of the environment at height z. By (11.14), it follows that Ra Ra ln θe = ln Te + ln p0 − ln p. cp cp Differentiating both sides with respect to z, and then using the hydrostatic gp balance equation dp dz = − Ra Te , we get 1 dθe 1 dTe Ra dp 1 dTe g = − = + . θe dz Te dz cp p dz Te dz Te cp Notice that lapse rates a =

g cp

e and e = − dT dz . Then

a − e 1 dθe = . θe dz Te

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e If the potential temperature θe increases with height, dθ dz > 0, and so e < a . In this case, the atmosphere is statically stable. If the potential temperature θe e decreases with height, dθ dz < 0, and so e > a . In this case, the atmosphere is

statically unstable. Notice that N 2 =

g(a −e ) . Te

N2 =

Then

g dθe . θe dz

This shows that the buoyancy frequency closely relates to the potential temperature of the environment.

11.6

CLAUSIUS-CLAPEYRON EQUATION

The relationship between the temperature of a liquid and its vapor pressure is not a straight line in a temperature-pressure diagram. The vapor pressure of water, for example, increases significantly more rapidly than the temperature. This behavior can be explained with the Clausius-Clapeyron equation: Lp dp , = dT Rv T 2 where p and T are the pressure and temperature at the phase transition, L is the latent heat of vaporization per unit mass, and Rv is the specific gas constant for the vapor. Consider a parcel of moist air. Denote by pv and psv the partial pressures of water vapor and the saturation water vapor at the phase transition, respectively. Replacing p by pv or psv in the Clausius-Clapeyron equation, we get dpv Lpv = , dT Rv T 2 dpsv Lpsv = dT Rv T 2 or dpv L dT = , pv Rv T 2 L dT dpsv = . psv Rv T 2

(11.19)

Notice that if L is a constant, integrating both sides of (11.19), we get the following for the partial pressures of water vapor and saturation water vapor: L



pv (T) = pv (T0 )e Rv psv (T) = psv (T0 )e

L Rv

where T0 is a constant reference temperature.

1 1 T0 − T





1 1 T0 − T

,



,

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11.6.1

363

Saturation Mass Mixing Radio

By mv denote the molecular mass of water vapor and by m denote the mean molecular mass of moist air. By Proposition 11.3, the mass mixing ratio of water vapor mv pv μ= , m p where mmv is constant and is approximately 0.622. Define the saturation mass mixing ratio as mv psv μs = , m p where psv is the partial pressure of the saturation water vapor and p is the pressure at the phase transition. If μ < μs , then the air is said to be unsaturated, and if μ = μs , then the air is said to be saturated. In this case, the corresponding adiabatic lapse rate is called the saturation adiabatic lapse rate, denoted by s , dT , dz where T is the temperature at the phase transition. If μ > μs , then the air is said to be supersaturated. s = −

11.6.2

Saturation Adiabatic Lapse Rate

Consider a saturated air parcel of unit mass. Assume that the parcel rises a small distance z and its temperature increases by a small amount T. Then the saturation mass mixing ratio has an increase −μs , and so the small amount of latent heat given to the parcel is −Lμs , where L is the latent heat of vaporization per unit mass and μs is the saturation mass mixing ratio. On the other hand, a small heat input is given by Ts, where ds is an increase of the entropy into the parcel. By (11.12) and the hydrostatic balance equation, the small heat input Ts = cp T −

Ra T p = cp T + gz, p

where cp = cν + Ra and g = |g| and g is the gravitational acceleration. The small latent heat is equal to the small heat input at saturation, so −Lμs = cp T + gz. This is equivalent to dμs = −

cp g dT − dz. L L

(11.20)

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The increase of the saturation mass mixing ratio is directly calculated as follows. By the definition, mv psv μs = , m p where psv depends only on T, taking logarithms and then differentiating both sides, since mmv is constant, we get dμs dpsv dp = − . μs psv p From this, (11.19), and the hydrostatic balance equation, it follows that dμs L g = dT + dz, μs Ra T Rv T 2 and so the increase of the saturation mass mixing ratio is dμs =

μs L μs g dT + dz. 2 Ra T Rv T

(11.21)

From (11.20) and (11.21) for dμs , it follows that − or

cp g μs L μs g dT + dT − dz = dz L L Rv T 2 Ra T

−cp

This is equivalent to

L2 μs +1 cp Rv T 2



dT = g

Lμs +1 Ra T

 dz.

 Lμs Ra T + 1 g dT  , = − dz cp L2 μs + 1 cp Rv T 2

i.e., the saturated adiabatic lapse rate s =



g  cp

Lμs Ra T

+1

L2 μs cp Rv T 2

Notice that the adiabatic lapse rate a = 



+1

.

g cp .

Therefore, Lμs + 1 Ra T . s = a  2 L μs + 1 c R T2 p v

This shows that the saturated adiabatic lapse rate and the adiabatic lapse rate have a close relation.

Atmospheric Dynamics Chapter | 11

11.6.3

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Equivalent Potential Temperature

Dividing both sides of (11.20) by T, cp dT dμs g dz =− − . T LT LT From this and the hydrostatic balance equation

dp p

= − Rga T dz, it follows that

dT dp Ra L dμs =− + κ , where κ = , cp T T p cp which is equivalent to



Lμs d cp T

 = −d(ln T) + κ d(ln p).

Hence, Lμs + ln T − κ ln p = c, cp T where c is a constant. Taking exponentials on both sides, T = ec . pκ

Lμs

e cp T This is equivalent to

T

p0 p

κ

Let

θe = T

Lμs

e cp T = ec pκ0 .

p0 p

κ

Lμs

e cp T .

The quantity θe is called the equivalent potential temperature. It depends on temperature T and pressure p. The potential temperature given in Section 11.4 is

κ p0 θ =T . p Therefore, Lμs

θe = θe cp T , where μs is the saturation mass mixing ratio, T is the temperature, L is the latent heat of vaporization per unit mass, and cp = cν + Ra . This equality shows a close relation between the equivalent potential temperature and the potential temperature. When the air in a rising parcel remains unsaturated, the latent heat is not released. So L = 0, and so θe = θ, i.e., the equivalent potential temperature is

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equal to the potential temperature. Once saturation takes place, the latent heat is released. So L = 0, and so θe = θ, i.e., the equivalent potential temperature is not equal to the potential temperature.

11.7

MATERIAL DERIVATIVES

Consider a blob of the atmosphere fluid and denote by r(t) the position of the blob at time t. The Eulerian velocity of the blob at time t is equal to the current time rate of change of the blob’s position: v(r(t), t) =

dr , dt

(11.22)

where v(r(t), t) = (v1 (x(t), y(t), z(t), t), v2 (x(t), y(t), z(t), t), v3 (x(t), y(t), z(t), t)), r = (x(t), y(t), z(t)). Notice that dr = dt



 dx(t) dy(t) dz(t) , , . dt dt dt

Then (v1 (x(t), y(t), z(t), t), v2 (x(t), y(t), z(t), t), v3 (x(t), y(t), z(t), t))

 dx(t) dy(t) dz(t) = , , . dt dt dt Comparing both sides of this equality, we get dx = v1 (x(t), y(t), z(t), t), dt dy = v2 (x(t), y(t), z(t), t), dt dz = v3 (x(t), y(t), z(t), t). (11.23) dt Similarly, the Eulerian acceleration of the blob at time t is equal to the second derivative of the position vector: a(r(t), t) = Notice that d2 r = dt2



d2 r . dt2

 d2 x d2 y d2 z , , . dt2 dt2 dt2

Atmospheric Dynamics Chapter | 11

Then

a(r(t), t) =

367

 d2 x d2 y d2 z , , . dt2 dt2 dt2

Differentiating (11.23) once again, applying the chain rule, we find the three components on the right-hand side are d2 x ∂v1 dx ∂v1 dy ∂v1 dz ∂v1 = + + + = v · grad v1 + 2 ∂x dt ∂y dt ∂z dt ∂t dt d2 y ∂v2 dx ∂v2 dy ∂v2 dz ∂v2 = + + + = v · grad v2 + 2 ∂x dt ∂y dt ∂z dt ∂t dt d2 z ∂v3 dx ∂v3 dy ∂v3 dz ∂v3 = + + + = v · grad v3 + ∂x dt ∂y dt ∂z dt ∂t dt2

∂v1 , ∂t ∂v2 , ∂t ∂v3 . ∂t

Therefore, the Eulerian acceleration a = (v · ∇)v +

∂v . ∂t

Define the material derivative as ∂ D = + (v · ∇) Dt ∂t or D ∂ ∂ ∂ ∂ = + v1 + v2 + v3 . Dt ∂t ∂x ∂y ∂z Especially, define the material derivatives of the position vector r as dr Dr = , Dt dt Dr dr = (r = |r|). Dt dt So the Eulerian acceleration is written simply as Dv . (11.24) Dt The material derivatives differ from the partial derivatives. The material D derivative Dt represents the rate of change with respect to time following the moving fluid blob, while the partial derivative ∂t∂ represents the rate of change with respect to time at a fixed point. Property. The material derivative has the following simple properties. a=

(i) Let c be a constant. Then Dc Dt = 0. (ii) Let f and g be functions of time t. Then D(f ± g) Df Dg = ± , Dt Dt Dt

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D(fg) Df Dg = g+f , Dt Dt Dt 

 D gf 1 Df Dg = 2 g−f . Dt g Dt Dt (iii) Let F and G be vectors with respect to t. Then D(F · G) DF DG = ·G+F· , Dt Dt Dt D(F × G) DF DG = ×G+F× . Dt Dt Dt Proof. We only prove (iii) here. Assume that F and G are both the three-dimensional vectors and F = (f1 , f2 , f3 ) and G = (g1 , g2 , g3 ). Then the scalar product of F and G is F · G = f1 g1 + f2 g2 + f3 g3 . So ∂f1 ∂f2 ∂f3 ∂(F · G) ∂g1 ∂g2 ∂g3 = g1 + f1 + g2 + f2 + g3 + f3 . ∂t ∂t ∂t ∂t ∂t ∂t ∂t On the other hand, by

 ∂F ∂f1 ∂f2 ∂f3 = , , , ∂t ∂t ∂t ∂t

 ∂g1 ∂g2 ∂g3 ∂G = , , , ∂t ∂t ∂t ∂t it follows that the scalar product of

∂F ∂t

(11.25)

and G is

∂F ∂f1 ∂f2 ∂f3 ·G= g1 + g2 + g3 , ∂t ∂t ∂t ∂t and the scalar product of F and F·

∂G ∂t

is

∂G ∂g1 ∂g2 ∂g3 = f1 + f2 + f3 . ∂t ∂t ∂t ∂t

Adding these two equations together gives ∂F ∂G ∂f1 ∂f2 ∂f3 ∂g1 ∂g2 ∂g3 ·G+F· = g1 + g2 + g3 + f1 + f2 + f3 . ∂t ∂t ∂t ∂t ∂t ∂t ∂t ∂t From this and (11.25), it follows that ∂(F · G) ∂F ∂G = ·G+F· . ∂t ∂t ∂t

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369

Similarly, v1

∂G ∂(F · G) ∂F = v1 · G + v1 F · , ∂x ∂x ∂x

v2

∂(F · G) ∂F ∂G = v2 · G + v2 F · , ∂y ∂y ∂y

∂(F · G) ∂F ∂G = v3 · G + v3 F · . ∂z ∂z ∂z Adding these four equations together, by the definition of material derivative, we get v3

D(F · G) DF DG = ·G+F· , Dt Dt Dt i.e., the first equality of (iii). Similarly, the second equality of (iii) can be derived. In Section 10.4, the Eulerian form of the continuity equation was given. By the definition of the material derivative, the continuity equation takes the form Dρ + ρ div v = 0. Dt In fact, by the definition of the material derivative, Dρ ∂ρ + ρ div v = + (v · ∇)ρ + ρ div v. Dt ∂t Let v = (v1 , v2 , v3 ). Then the sum of the last two terms on the right-hand side is 



∂ρ ∂ρ ∂ρ ∂v1 ∂v2 ∂v3 + v2 + v3 +ρ +ρ (v · ∇)ρ + ρ div v = v1 + ρ ∂x ∂y ∂z ∂x ∂y ∂z =

∂(ρv1 ) ∂ρv2 ∂ρv3 + + = div (ρv), ∂x ∂y ∂z

and so

Dρ ∂ρ + ρ div v = + div (ρv). Dt ∂t From this and the Eulerian form of the continuity equation: we get

∂ρ ∂t

+ div (ρv) = 0,

Dρ + ρ div v = 0. Dt This is called the Lagrangian form of the continuity equation. Euler’s equation in a gravitational field given in Section 10.4 is ∂v ∇p + (v · ∇)v = − − gk. ∂t ρ

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By the definition of the material derivative, it is rewritten in the form Dv ∇p =− − gk. Dt ρ

11.8

VORTICITY AND POTENTIAL VORTICITY

Vorticity and potential vorticity are two important concepts in the atmosphere. The curl of the atmospheric fluid velocity v is called the vorticity, denoted by  :  = ∇ × v = curl v. For a three-dimensional flow, by the definition of the curl, the vorticity is given by





 ∂v3 ∂v2 ∂v1 ∂v3 ∂v2 ∂v1  = curl v = − i+ − j+ − k, ∂y ∂z ∂z ∂x ∂x ∂y where v = (v1 , v2 , v3 ) is the fluid velocity. For a two-dimensional flow in which the direction of the fluid velocity is parallel to the xy-plane and the fluid velocity is independent of z, the vorticity is given by

 ∂v2 ∂v1  = curl v = − k. ∂x ∂y where v = (v1 , v2 ). In order to investigate the relationship between vorticities in an inertial frame and in a rotating frame, the definition of vorticity shows that we need to know only the relationship between the velocities in these two frames. Suppose that the rotating frame R rotates at a constant angular velocity  with respect to the inertial frame I and that the z-axes of both frames are in the direction of . If the position vector r(t) is viewed in the rotating frame, then the change of the position vector between t and t + t is (r)R = r(t + t) − r(t). If the position vector is viewed in the inertial frame, then the rotation gives an extra change  × rt in r. So the change of the position vector at times t and t + t is (r)I = (r)R +  × rt. This is equivalent to



 r r = +  × r. t I t R Let t → 0. Then



 dr dr = +  × r. dt I dt R

Atmospheric Dynamics Chapter | 11

 By (11.22), satisfy

dr dt I

= vI and



dr dt R

371

= vR . So the velocities in these two frames

vI = vR +  × r, i.e., the velocity in the inertial frame is equal to the velocity in the rotating frame plus the vector product of the angular velocity and the position vector. Taking the curl on both sides, and then using Property 10.1, we get curl vI = curl vR + curl ( × r), i.e., I = R + curl ( × r). (11.26) This shows that the vorticity in the inertial frame is equal to the vorticity in the rotating frame plus the curl of the vector product of the angular vector and the position vector. We compute the term curl ( × r). By the definition of the curl, curl ( × r) = ∇ × ( × r). Replacing F and G by  and r, respectively, in Property 10.7, we get ∇ × ( × r) = (r · ∇) − ( · ∇)r + (∇ · r) − r(∇ · ). Therefore, curl ( × r) = (r · ∇) − ( · ∇)r + (∇ · r) − r(∇ · ). (11.27) We compute each term on the right-hand side of (11.27). Let  = (1 , 2 , 3 ), r = (x, y, z). Notice that the angular velocity  is a constant. Then ∂ ∂ ∂ +y +z = 0, (r · ∇) = x ∂x ∂y ∂z

 ∂ ∂ ∂ ( · ∇)r = 1 + 2 + 3 (x, y, z) ∂x ∂y ∂z = (1 , 2 , 3 ) = . Notice that ∂x ∂y ∂z + + = 3, ∂x ∂y ∂z ∂1 ∂2 ∂3 ∇ · = + + = 0. ∂x ∂y ∂z ∇ ·r =

Then (∇ · r) = 3, r(∇ · ) = 0.

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Combining these results with (11.27), we get curl ( × r) = 2. From this and (11.26), I = R + 2, where  is the constant rotation vector. Sometimes, I is called the absolute vorticity and R is called the relative vorticity. Thus, the absolute vorticity is equal to the relative vorticity plus twice the rotation vector. Let I be the absolute vorticity, θ the potential temperature, and ρ the fluid density. The quantity I · grad θ P= ρ is called the potential vorticity.

11.9

NAVIER-STOKES EQUATION

The Navier-Stokes equation is an equation of motion involving viscous fluids. Here Newton’s second law is applied to a small moving blob of a viscous fluid, and then the Navier-Stokes equation is derived.

11.9.1

Navier-Stokes Equation in an Inertial Frame

Consider a blob of cuboidal shape instantaneously with sides x, y, and z. Its volume V = xyz and its mass m = ρV, where ρ is the density. Applying Newton’s second law gives F = (ρV)a,

(11.28)

where a is the acceleration of the blob and F is the vector sum of the pressure force, the gravitational force, and the frictional force acting on the blob, i.e., F = Fp + Fg + Fv . First, we compute the pressure force acting on the blob. The pressure force at position x is p(x, y, z)yz in the positive x-direction and the pressure force at position x + x is p(x + x, y, z)yz in the negative x-direction, where yz is the area of the relevant wall of the blob. Using Taylor’s theorem gives

 ∂p p(x + x, y, z)yz = p(x, y, z) + (x, y, z)x + · · · yz. ∂x Therefore, the net pressure force in the positive x-direction is ∂p V. ∂x Similarly, the net pressure forces in the positive y-direction and the positive z-direction are, respectively, (p(x, y, z) − p(x + x, y, z))yz ≈ −

Atmospheric Dynamics Chapter | 11

(p(x, y, z) − p(x, y + y, z))xz ≈ −

373

∂p V, ∂y

∂p V. ∂z Summarizing, the net pressure force in three directions is 

∂p ∂p ∂p Fp = − i+ j+ k V = −(∇p)V, ∂x ∂y ∂z (p(x, y, z) − p(x, y, z + z))xy ≈ −

where i = (1, 0, 0), j = (0, 1, 0), and k = (0, 0, 1). Next, we compute the gravitational force acting on the blob. The gravitational force is the mass m of the blob times g acting downward. Since m = ρV, the gravitational force Fg = −mgk = −(ρV)gk, where g = |g| and k = (0, 0, 1). Finally, we compute the frictional force acting on the blob. For simplicity, we consider a special case where the frictional force acting on the blob is provided by a horizontal force τ = (τx , τy , 0) alone. The net frictional force acting on the blob is the difference between the stresses on the top and the bottom, (τx (z + z) − τx (z))xy ≈

∂τx V, ∂z

∂τy V, ∂z where xy is the area of the relevant wall of the blob. Therefore, the net frictional force is

 ∂τy ∂τx Fv = i+ j V, (11.29) ∂z ∂z (τy (z + z) − τy (z))xy ≈

where i = (1, 0, 0) and j = (0, 1, 0). In general, the net frictional force acting on the blob is  Vη ∇ 2 v + 13 ∇(∇ · v) if the fluid is compressible, Fv = Vη∇ 2 v if the fluid is incompressible, where η is the dynamic viscosity and ∇ 2 = ∇ · ∇. Combining the above results with (11.28), we get (ρV)a = −(∇p)V − (ρV)gk + Fv . Canceling V, and then combining the result with (11.24), we get Dv ∇p =− − gk + Fv , Dt ρ

(11.30)

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where Fv =

Fv . ρV

If the frictional force acting on blob is provided by the horizontal force τ = (τx , τy , 0) alone, then, by (11.29), Fv = In general, Fv =

⎧ ⎨ ⎩



η 1 2 ρ ∇ v + 3 ∇(∇ η 2 ρ∇ v

1 ∂τy 1 ∂τx i+ j. ρ ∂z ρ ∂z

(11.31)

· v) if the fluid is compressible, if the fluid is incompressible.

Equation (11.30) is called the Navier-Stokes equation or the momentum equation in an inertial frame. Comparing the Navier-Stokes equation with Euler’s equation given in the end of Section 11.7, we see that the frictional term Fv is added to Euler’s equation. Therefore, the Navier-Stokes equation is a generalization of Euler’s equation.

11.9.2

Navier-Stokes Equation in a Rotating Frame

For the large-scale atmospheric flows, the rotation of Earth cannot be ignored. So Earth’s rotation must be incorporated into the Navier-Stokes equation, and this will modify the Navier-Stokes equation. Consider an inertial frame I and a rotating frame R. Suppose that the frame R rotates at a constant angular velocity  with respect to the frame I, and that the z-axes of the two frames are both in the direction of . In Section 11.8 we obtained the following relationship between the velocities in both frames,



 dr dr = +  × r. dt I dt R A double application of this equation gives

2 





 d r d dr d dr dr = = +× dt2 I dt dt I I dt dt I R dt I



 

  d dr dr = +×r +× +×r . dt dt R dt R R This is equivalent to

2 

 



 d r d dr d dr = + ( × r) +  × dt dt R R dt dt R dt2 I R +  × ( × r).

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Clearly, the first term on the right-hand side

 

2  d dr d r = . dt dt R R d2 t R Let  = (1 , 2 , 3 ) and r = (x, y, z). Notice that

 dx dy dz dr = , , . dt dt dt dt Then

       i j k   i j k     d d dr     ( × r) =  1 2 3  =  1 2 3  =  × , dt dt  x y z   dx dy dz  dt   dt dt dt i.e., the second term is equal to the third term on the right-hand side. Therefore,

2 

2 

 d r d r dr = + 2 × +  × ( × r). dt R dt2 I dt2 R However, by (11.24), the accelerations in the inertial frame and the rotating frame are, respectively,

2  d r , aI = dt2 I

2  d r aR = , dt2 R and by (11.22), the velocity in the rotating frame is

 dr vR = . dt R Therefore, aI = aR + 2 × vR +  × ( × r), where the term 2 × vR is the Coriolis acceleration and the term  × ( × r) is the centripetal acceleration. On the other hand, by (11.30), the Navier-Stokes equation in an inertial frame takes the form aI = −

∇p − gk + Fv , ρ

where k = (0, 0, 1). Therefore, aR + 2 × vR +  × ( × r) = −

∇p − gk + Fv . ρ

 Dv  By (11.24), aR = DT . If we drop the subscript R, the Navier-Stokes equation R in the rotating frame is

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Dv ∇p =− − 2 × v −  × ( × r) − gk + Fv , Dt ρ

(11.32)

where k = (0, 0, 1).

11.9.3

Component Form of the Navier-Stokes Equation

By using Cartesian coordinates, we take unit vectors i pointing eastward and j pointing northward, and k pointing upward at a point on Earth’s surface. Consider small incremental distances: dx = r cos φ dλ dy = r dφ dz = dr

in the eastward (zonal) direction,

in the northward (meridional) direction, in the vertical direction,

where φ, λ, and z are latitude, longitude, and the vertical distance from Earth’s surface, respectively, and r = a + z, where a is Earth’s radius. Let the velocity v = v1 i + v2 j + v3 k. Notice that i, j, and k change with time. Then Dv Dv1 Di Dv2 Dj Dv3 Dk = i + v1 + j + v2 + k + v3 . Dt Dt Dt Dt Dt Dt Dt First, we compute the material derivatives of the unit vector k. Notice that k = rr , where r is the magnitude of position vector r. Then

 Dk D  r 1 Dr r Dr 1 Dr Dr = = − 2 = −k . Dt Dt r r Dt r Dt Dt r Dt By the definition of the material derivative of the position vector given in Section 11.7,

 1 dr dr Dk = −k . Dt r dt dt By (11.22), (11.23) and dz = dr, Dk 1 v1 i + v2 j = (v − v3 k) = . Dt r r Next, we compute the material derivative of the unit vector j. Let the constant angular velocity  = (j cos φ + k sin φ). It follows from D Dt = 0 that cos φ

Dj D cos φ Dk D sin φ +j + sin φ +k = 0. Dt Dt Dt Dt

This is equivalent to Dj Dφ Dk Dφ Dφ v1 i + v2 j = j tan φ − tan φ −k = (j tan φ − k) − tan φ . Dt Dt Dt Dt Dt r

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By the definition of the material derivative, (11.23), and dy = rdφ, we get ∂φ ∂φ ∂φ Dφ ∂φ = + v1 + v2 + v3 Dt ∂t ∂x ∂y ∂z ∂φ ∂φ dx ∂φ dy ∂φ dz dφ 1 dy v2 = + + + = = = . ∂t ∂x dt ∂y dt ∂z dt dt r dt r Therefore, Dj v2 v1 i + v2 j v1 tan φ v2 = (j tan φ − k) − tan φ =− i − k. Dt r r r r Finally, we compute the material derivative of the unit vector i. Notice that i = j × k. Applying property (iii) of the material derivative given in Section 11.7, we get Di D(j × k) Dj Dk −v1 tan φi − v2 k v1 i + v2 j = = ×k+j× = ×k+j× . Dt Dt Dt Dt r r By the orthogonality of the unit vectors, i × k = −j,

k × k = 0,

j × i = −k,

j × j = 0,

and so v1 tan φj − v1 k Di = . Dt r Summarizing these results, we get Dv Dv1 v1 tan φj − v1 k Dv2 −v1 tan φi − v2 k Dv3 = i + v1 + j + v2 + k Dt Dt r Dt r Dt v1 i + v2 j . + v3 r The Coriolis term, the vector product of 2 and v, is given by 2 × v = 2(v3 cos φ − v2 sin φ)i + 2v1 sin φj − 2v1 cos φk, and the pressure gradient term is given by

 ∇p 1 ∂p ∂p ∂p − =− i+ j+ k . ρ ρ ∂x ∂y ∂z Disregard the centripetal acceleration  × ( × r). Let Fv = Fx i + Fy j + Fz k. Then the Navier-Stokes equation (11.32) is written in the form Dv1 v1 tan φj − v1 k Dv2 −v1 tan φi − v2 k Dv3 i + v1 + j + v2 + k Dt r Dt Dt

r v1 i + v2 j 1 ∂p ∂p ∂p +v3 =− i+ j + k − 2(v3 cos φ − v2 sin φ)i r ρ ∂x ∂y ∂z −2v1 sin φj + 2v1 cos φk − gk + Fx i + Fy j + Fz k.

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Collecting the terms in i, j, and k, we can write the Navier-Stokes equation (11.32) in the component form:

 ⎧ 1 ∂p v1 Dv1 ⎪ ⎪ = − + 2 + (v2 sin φ − v3 cos φ) + Fx , ⎪ ⎪ Dt ρ ∂x r cos φ ⎪ ⎪ ⎪

 ⎪ ⎨ Dv 1 ∂p v2 v3 v1 2 =− − − 2 + v1 sin φ + Fy , ⎪ Dt ρ ∂y r r cos φ ⎪ ⎪ ⎪ ⎪ ⎪ Dv3 1 ∂p v12 + v22 ⎪ ⎪ ⎩ =− + + 2v1 cos φ − g + Fz , Dt ρ ∂z r where the first two equations are called the horizontal momentum equations and the third equation is called the vertical momentum equation.

11.10

GEOSTROPHIC BALANCE EQUATIONS

The component form of the Navier-Stokes equation is complicated. We need to simplify it for motions associated with large-scale weather systems. Since the depth of the atmosphere is much less than Earth’s radius, in the component form of the Navier-Stokes equation, we first can replace the distance r by Earth’s radius a with negligible error,

 ⎧ 1 ∂p v1 Dv1 ⎪ ⎪ = − + 2 + (v2 sin φ − v3 cos φ) + Fx , ⎪ ⎪ ⎪ Dt ρ ∂x a cos φ ⎪ ⎪

 ⎪ ⎨ Dv 1 ∂p v2 v3 v1 2 =− − − 2 + v1 sin φ + Fy , (11.33) ⎪ Dt ρ ∂y a a cos φ ⎪ ⎪ ⎪ ⎪ ⎪ Dv3 1 ∂p v12 + v22 ⎪ ⎪ ⎩ =− + + 2v1 cos φ − g + Fz . Dt ρ ∂z a 1| Assume that |v3 cos φ|  |v2 sin φ| and a |v cos φ  2, and the Coriolis parameter f = 2 sin φ. This is a common assumption for weather systems. Then the first equation in (11.33) reduces to

Dv1 1 ∂p =− + f v 2 + Fx . Dt ρ ∂x Similarly, for weather systems, we usually assume that |v2av3 |  2|v1 sin φ| 1| and a |v cos φ  2 and the Coriolis parameter f = 2 sin φ. Then the second equation reduces to Dv2 1 ∂p =− − f v 1 + Fy . Dt ρ ∂y Also, we usually assume that equation reduces to

v12 +v22 a

 g and 2v1 cos φ  g. Then the third

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Dv3 1 ∂p =− − g + Fz . Dt ρ ∂z Combining these with (11.33), the component form of the Navier-Stokes equation is simplified further to ⎧ 1 ∂p Dv1 ⎪ ⎪ =− + f v 2 + Fx , ⎪ ⎪ Dt ρ ∂x ⎪ ⎪ ⎪ ⎨ Dv 1 ∂p 2 =− − f v 1 + Fy , (11.34) ⎪ Dt ρ ∂y ⎪ ⎪ ⎪ ⎪ ⎪ Dv3 1 ∂p ⎪ ⎩ =− − g + Fz , Dt ρ ∂z where the Coriolis parameter f = 2 sin φ. The f -plane and the β-plane are two Cartesian coordinate systems. In the f plane the Coriolis parameter is assumed to have constant value f0 = 2 sin φ0 , where φ0 is a constant latitude. In the β-plane the Coriolis parameter is assumed φ0 to vary linearly with latitude f = f0 + βy, where β = 2 cos is the north-south a variation of the Coriolis force and a is Earth’s radius. ∂p 1 Assume further that Dv Dt  f v2 and that ∂x is the only term that can balance the large term f v2 . This is also a common assumption for weather systems. If we disregard frictional force, the first equation in (11.34) reduces to f v2 =

1 ∂p . ρ ∂x

∂p 2 Similarly, assume further that Dv Dt  f v1 and that ∂y is the only term that can balance the large term f v1 . If we disregard frictional force, the second equation in (11.34) reduces to

f v1 = −

1 ∂p . ρ ∂y

These two equations are called the geostrophic balance equations. The geostrophic balance equations are a good approximation for the horizontal momentum equations. 3 Usually, we assume further that Dv Dt  g and that the vertical pressure ∂p gradient ∂z is the only term that can balance the large g term. If we disregard frictional force, the third equation in (11.34) reduces to ∂p = −gρ. ∂z This equation is just the hydrostatic balance equation in Section 11.3. The hydrostatic balance equation is a good approximation for the vertical momentum equation.

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11.11 BOUSSINESQ APPROXIMATION AND ENERGY EQUATION In the Boussinesq approximation, the atmospheric flows are always assumed to satisfy div v = 0, where v is the fluid velocity. This assumption implies that the continuity equation decouples into two equations: div v = 0, Dρ = 0. Dt The first equation is equivalent to ∂v1 ∂v2 ∂v3 + + = 0, ∂x ∂y ∂z where v = (v1 , v2 , v3 ). The second equation is called the density equation. It states that for atmospheric flow, the density is constant on following a moving fluid blob. But this does not imply that the density is uniform everywhere. Therefore, we must still allow for vertical density stratification. We separate the density into a background ρ depending only on height z and a deviation ρ  as follows: ρ(x, y, z, t) = ρ  (x, y, z, t) + ρ(z). Substituting this equation into the density equation leads to Dρ  Dρ + = 0. Dt Dt Since ρ depends only on z, ∂ρ ∂ρ ∂ρ = = = 0, ∂t ∂x ∂y and so Dρ ∂ρ ∂ρ ∂ρ ∂ρ dρ = + v1 + v2 + v3 = v3 . Dt ∂t ∂x ∂y ∂z dz Therefore, Dρ  dρ + v3 = 0. Dt dz g dρ Denote ρ0 = ρ(0) and NB2 = − . This equation is equivalent to ρ0 dz g Dρ  − v3 NB2 = 0. ρ0 Dt The quantity NB is called the Boussinesq buoyancy frequency for the stratified atmospheric fluid with div v = 0.

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Similarly, the pressure is separated into a background p depending only on height z and a deviation p : p(x, y, z, t) = p (x, y, z, t) + p(z), and so ∂p ∂p dp = + . ∂z ∂z dz  However, by the hydrostatic balance equation ∂p ∂z = −gρ and ρ = ρ + ρ, it follows that ∂p = −gρ  − gρ. ∂z Therefore,

dp ∂p + = −gρ  − gρ. ∂z dz From this equation, we see that if the background pressure satisfies the hydrostatic balance equation: dp = −gρ, dz then the deviation also satisfies the hydrostatic balance equation: ∂p = −gρ  . ∂z Since p depends only on z, ∂p ∂p = = 0. ∂x ∂y Notice that p = p + p. Then ∂p ∂p ∂p ∂p = + = , ∂x ∂x ∂x ∂x ∂p ∂p ∂p ∂p = + = . ∂y ∂y ∂y ∂y Assume that ρ = ρ0 . Then the first two equations of (11.34) reduce to Dv1 1 ∂p = − + f v 2 + Fx , Dt ρ0 ∂x Dv2 1 ∂p = − − f v 1 + Fy . Dt ρ0 ∂y Summarizing up all results, a system of five equations with respect to p and ρ  is obtained as follows:

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⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨

Dv1 1 ∂p =− + f v 2 + Fx , Dt ρ0 ∂x Dv2 1 ∂p =− − f v 1 + Fy , Dt ρ0 ∂y

∂p = −gρ  , ⎪ ∂z ⎪ ⎪ ⎪ ⎪ ⎪ g Dρ  ⎪ 2 ⎪ ⎪ ⎪ ρ Dt − NB v3 = 0, ⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ∂v1 + ∂v2 + ∂v3 = 0, ⎩ ∂x ∂y ∂z where ρ0 = ρ(0), NB2 = − ρg0

dρ dz ,

(11.35)

v = (v1 , v2 , v3 ), and the material derivative

D ∂ ∂ ∂ ∂ = + v1 + v2 + v3 . Dt ∂t ∂x ∂y ∂z This system of equations is called the Boussinesq equation. Because of the presence of the quadratic terms in the material derivative, this system of equations is sometimes called the nonlinear Boussinesq equation. We make the further approximation linearizing these equations. Dropping the quadratic terms in the material derivatives and disregarding frictional force, the system of equations (11.35) can be approximated by ⎧ 1 ∂p ∂v1 ⎪ ⎪ − = f v + Fx , ⎪ 2 ⎪ ⎪ ∂t ρ0 ∂x ⎪ ⎪ ⎪ ∂v2 1 ∂p ⎪ ⎪ = −f v − + Fy , ⎪ 1 ⎪ ⎪ ρ0 ∂y ⎪ ⎨ ∂t ∂p (11.36) = −gρ  , ⎪ ∂z ⎪ ⎪  ⎪ g ∂ρ ⎪ 2 ⎪ ⎪ ⎪ ρ ∂t − NB v3 = 0, ⎪ ⎪ 0 ⎪ ⎪ ∂v1 ∂v2 ∂v3 ⎪ ⎪ + + = 0. ⎩ ∂x ∂y ∂z This system of equations is called the linearized Boussinesq equation. In (11.36), disregarding friction and then multiplying the first equation by ρ0 v1 , the second equation by ρ0 v2 , the third equation by v3 , the fourth equation  by gρ2 , and the fifth equation by p , we get NB

∂v1 ∂p − ρ0 v1 f v2 + v1 = 0, ∂t ∂x ∂v2 ∂p ρ0 v2 + ρ0 v2 f v1 + v2 = 0, ∂t ∂y ρ0 v1

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∂p + v3 gρ  = 0, ∂z g2 ρ  ∂ρ  − gρ  v3 = 0, NB2 ρ0 ∂t ∂v1 ∂v2 ∂v3 p + p + p = 0. ∂x ∂y ∂z v3

Adding these five equations together gives ρ0 v1

∂v1 ∂v2 ∂p ∂p ∂p ∂v1 ∂v2 + ρ0 v2 + v1 + v2 + v3 + p + p ∂t ∂t ∂x ∂y ∂z ∂x ∂y 2   g ρ ∂ρ ∂v3 + p + 2 = 0. ∂z NB ρ0 ∂t

A short calculation shows that this equation is equivalent to 

  ρ0 ∂ gρ  2 2 2 v + v2 + + ∇ · (vp ) = 0, 2 ∂t 1 NB ρ0 which is called the energy equation, where v = (v1 , v2 , v3 ). In the energy equation, the term ρ20 (v12 + v22 ) is clearly the kinetic energy per unit volume of the horizontal motion, the term involving (ρ  )2 can be interpreted as the available potential energy, and the term vp can be interpreted as an energy flux. Overall, the equation states that the kinetic energy and available potential energy within a volume increases if there is an energy flux into the volume, and the kinetic energy and available potential energy within a volume decreases if there is an energy flux out of the volume.

11.12

QUASI-GEOSTROPHIC POTENTIAL VORTICITY g

g

Consider a geostrophic flow vg = (v1 , v2 , 0). Start from Boussinesq equag Dv1

g

Dv

g

g

tion (11.35). Assume further that Dt  f0 v2 and Dt2  f0 v1 , where f0 is the Coriolis parameter (f0 = 2 sin φ0 ) and φ0 is a constant latitude. This is a common assumption for weather systems. Then the first two equations in (11.35) reduce to geostrophic balance equations, i.e., the geostrophic flow satisfies the following geostrophic balance equations: g

1 ∂p , f0 ρ0 ∂x 1 ∂p = − . f0 ρ0 ∂y

v2 = g

v1 

∂ψ Let ψ = f0pρ0 . Then v2 = ∂ψ ∂x and v1 = − ∂y . The function ψ is called the geostrophic streamfunction. g

g

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g

The geostrophic flow (v1 , v2 , 0) differs from the true velocity. The differences of components between the true velocity (v1 , v2 , v3 ) and the geostrophic g g flow (v1 , v2 , 0) are denoted by g

v1a = v1 − v1 , g

v2a = v2 − v2 , v3a = v3 . The vector va = (v1a , v2a , v3a ) is called the ageostrophic velocity. Another approximation beyond geostrophic balance on a β-plane is given by the quasi-geostrophic equations: ⎧ g Dg v1 ⎪ g ⎪ − f0 v2a − βyv2 = 0, ⎪ ⎪ ⎪ Dt ⎪ ⎪ g ⎪ ⎪ Dg v2 g ⎪ ⎪ + f0 v1a + βyv1 = 0, ⎪ ⎪ Dt ⎪ ⎨ a ∂v3a ∂v1 ∂v2a (11.37) + + = 0, ⎪ ⎪ ∂x ∂y ∂z ⎪ ⎪  ⎪ ⎪ ⎪ g Dg ρ = N 2 v a , ⎪ B 3 ⎪ ⎪ ρ0 Dt ⎪ ⎪  ⎪ ⎪ ⎩ ∂p = −gρ  , ∂z where Dg ∂ g ∂ g ∂ = + v1 + v2 Dt ∂t ∂x ∂y is the material derivative following the geostrophic flow. In the quasi-geostrophic approximation, frictional force is disregarded and the Coriolis parameter 2 cos φ0 y, a where φ0 is a constant latitude and a is Earth’s radius. The quasi-geostrophic equations hold in general for large-scale, low-frequency motions. The quasi-geostrophic equations can conveniently be combined as follows. Differentiating the first equation in (11.37) with respect to y, and then multiplying both sides by −1, we get   g g ∂v1 ∂v a ∂v Dg g − + f0 2 + βy 2 + βv2 = 0. Dt ∂y ∂y ∂y f = f0 + βy = 2 sin φ0 +

Differentiating the second equation in (11.37) with respect to x, we get  g g ∂v a ∂v Dg ∂v2 + f0 1 + βy 1 = 0. Dt ∂x ∂x ∂x

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Adding these two equations gives    g  

a g g g ∂v1 ∂v2 ∂v2a ∂v1 ∂v2 ∂v1 Dg − + + f0 + + βy + Dt ∂y ∂x ∂x ∂y ∂x ∂y g

+ βv2 = 0. We compute each term on the left-hand side of (11.38). g g ∂ψ Since v1 = − ∂ψ ∂y and v2 = ∂x , the first term is  

 g g ∂v1 ∂v2 Dg Dg ∂ 2 ψ ∂ 2ψ − + = + . Dt ∂y ∂x Dt ∂x2 ∂y2 From ψ =

p f0 ρ0

and the fifth equation in (11.37), it follows that gρ  ∂ψ =− . ∂z f0 ρ0

Combining this and the fourth equation in (11.37), it follows that     Dg Dg gρ  f0 ∂ψ a v3 = = − 2 . Dt ρ0 NB2 Dt NB ∂z Again, by the third equation of (11.37), the second term is   

a  ∂v3a ∂v1 ∂v2a f02 ∂ψ Dg ∂ f0 + = −f0 = . ∂x ∂y ∂z Dt ∂z NB2 ∂z g

Since v2 =

∂ψ ∂x

and v1 = − ∂ψ ∂y , the third term is  g 

 g ∂v1 ∂v2 ∂ 2ψ ∂ 2ψ βy + = βy − + = 0. ∂x ∂y ∂y∂x ∂x∂y g

Notice that ∂(f0 + βy) = 0, ∂t g ∂(f0

+ βy) = 0, ∂x

g ∂(f0

+ βy) g = βv2 . ∂y

v1 v2

Adding these three equations together gives g

βv2 =

∂(f0 + βy) g ∂(f0 + βy) g ∂(f0 + βy) + v1 + v2 . ∂t ∂x ∂y

(11.38)

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From this and

Dg Dt

=

∂ ∂t

∂ ∂ + v1 ∂x + v2 ∂y , the last term is g

g

Dg (f0 + βy) . Dt Summarizing all the results, by (11.38), the quasi-geostrophic equations are combined as follows Dg ζ = 0, Dt where Dg ∂ ∂ψ ∂ ∂ψ ∂ = − + Dt ∂t ∂y ∂x ∂x ∂y g

βv2 =

and ∂ 2ψ ∂ 2ψ ∂ ζ = f0 + βy + + + ∂x2 ∂y2 ∂z



f02 ∂ψ NB2 ∂z

 =: f0 + βy + Lψ.

D ζ

The equation Dtg = 0 is called the quasi-geostrophic potential vorticity equation. The quantity ζ is called the quasi-geostrophic potential vorticity associated with the geostrophic flow. The operator   f02 ∂ ∂2 ∂2 ∂ L= 2 + 2 + ∂z NB2 ∂z ∂x ∂y is called the elliptic operator, and    ∂v3a f02 ∂ψ Dg ∂ − = f . 0 Dt ∂z NB2 ∂z ∂z The term on the left-hand side, Dg − Dt



∂ ∂z



f02 ∂ψ NB2 ∂z

 ,

is often called the stretching term. It can generate vorticity by differential vertical motions. If a cylindrical blob of air enters a region where the vertical velocity increases with height, it is stretched vertically. Conversely, if the cylindrical blob of air enters a region where the vertical velocity decreases with height, it is shrunken vertically.

11.13

GRAVITY WAVES

Gravity waves are a class of atmospheric waves and are frequently observed in the atmosphere. The famous lee waves manifested as parallel bands of cloud downstream of mountain ranges are gravity waves. The linearized Boussinesq equation is a good tool to use to develop models of gravity waves.

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387

Internal Gravity Waves

We concentrate on small-scale gravity waves. These waves are called internal gravity waves. The horizontal scale of internal gravity waves is so small that Earth’s rotation can be negligible. Therefore, this can be used to simplify the linearized Boussinesq equation by disregarding the Coriolis terms. In this case, the linearized Boussinesq equation (11.36) becomes a system of linear partial differential equations: ⎧ 1 ∂p ∂v1 ⎪ ⎪ + = 0, ⎪ ⎪ ⎪ ∂t ρ0 ∂x ⎪ ⎪  ⎪ 1 ∂p ∂v2 ⎪ ⎪ + = 0, ⎪ ⎪ ⎪ ρ0 ∂y ⎪ ⎨ ∂t ∂p (11.39) = −gρ  , ⎪ ∂z ⎪ ⎪ ⎪ g ∂ρ  ⎪ ⎪ ⎪ − NB2 v3 = 0, ⎪ ⎪ ρ ∂t ⎪ 0 ⎪ ⎪ ∂v2 ∂v3 ⎪ ∂v1 ⎪ + + = 0, ⎩ ∂x ∂y ∂z where the velocity v = (v1 , v2 , v3 ), ρ0 is the constant, and NB is the Boussinesq buoyancy frequency. We look for linear plane-wave solutions, propagating in the xz-plane and independent of y, of the form   {v1 , v2 , v3 , p , ρ  } = Re ( v1 , v2 , v3 , p, ρ )ei(kx+mz−ωt) , where  v1 ,  v2 ,  v3 are complex amplitudes. Substituting this expression into (11.39), we obtain a system of algebraic equations as follows: ⎧ −iω v1 + ik ρp0 = 0, ⎪ ⎪ ⎪ ⎪ ⎪ v2 = 0, ⎪ ⎨ −iω im p + g ρ = 0, ⎪ ⎪ ρ  ⎪ ⎪ iωg ρ0 + NB2  v3 = 0, ⎪ ⎪ ⎩ ik v1 + im v3 = 0. This system of algebraic equations is equivalent to ⎧ ω v1 = k ρp0 , ⎪ ⎪ ⎪ ⎪ ⎪ v2 = 0, ⎪ ⎨ ω im p = −g ρ, ⎪ ⎪ ρ  ⎪ N 2 ⎪ ⎪ B v3 = −iωg ρ0 , ⎪ ⎩ k v1 = −m v3

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or

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨

 v1 k = ,  p ωρ0 k  v3 =− ,  v1 m

iNB2 ρ0 ρ  = , ⎪ ⎪ ⎪  v3 ωg ⎪ ⎪ ⎪ ⎪  p ig ⎪ ⎪ ⎪ = , ⎪ ⎪ ρ  m ⎪ ⎪ ⎩  v2 = 0.

(11.40)

Multiplying the first four equations in (11.40), we get



 2   iNB ρ0  v1  v3 ρ  p k k ig 1= = − ,  p v1  v3 ρ  ωρ0 m ωg m and so k2 NB2 . m2 This is the dispersion relation for internal gravity waves. It relates the angular frequency ω to the components k, m of the wave vector k = (k, 0, m). On the other hand, solving the system of algebraic equations (11.40), we obtain a nontrivial solution  v1 ,  v2 ,  v3 , and  ρ in favor  p: ⎧ k p  v1 = ρ0 ω , ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ v2 = 0, ω2 =

k p ⎪ ,  v3 = − mρ ⎪ ⎪ 0ω ⎪ ⎪ ⎩ p ρ  = − im g . 2

Arbitrarily choosing  p to be real, we find the linear plane-wave solution of (11.39) is  i(kx+mz−ωt)  ⎧  = p cos(kx + mz − ωt), pe p = Re  ⎪ ⎪ ⎪  i(kx+mz−ωt)  ⎪ ⎪ ⎪ v1 = Re  v1 e = kp cos(kx + mz − ωt), ⎪ ⎪  i(kx+mz−ωt)  ρ0 ω ⎨ = 0, v2 = Re  v2 e ⎪   ⎪ k2 p ⎪ ⎪ v3 = Re  v3 ei(kx+mz−ωt) = − mρ cos(kx + mz − ωt), ⎪ 0ω ⎪ ⎪   ⎪ ⎩ ρ  = Re ρ p ei(kx+mz−ωt) = m g sin(kx + mz − ωt). These are called the polarization relations for internal gravity waves. The polarization relations may verify the precise phase relations between the velocity, density, and pressure disturbances.

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k2 N 2

By the dispersion relation ω2 = m2B , the angular frequency has two possible solutions ω = ± NmB k . Define a group velocity vector

 ∂ω ∂ω (z) cg = (c(x) , 0, c ) = , 0, . g g ∂k ∂m Its vertical component is c(z) g =

NB k ∂ω =∓ 2 . ∂m m

The signs are chosen as follows. For an atmospheric internal gravity wave generated near the ground and (z) propagating information upward, cg > 0. By convention, k > 0. So the vertical (z) NB k velocity cg = m2 and the angular frequency ω = − NmB k . For this choice of signs, in the xz-plane, the phase surfaces kx + mz − ωt = constant move obliquely downward in the direction of the wave vector k = (k, 0, m). However, the propagation of information represented by the group velocity vector is obliquely upward. The velocity vector (v1 , 0, v3) is parallel to the slanting phase surfaces and the fluid blobs oscillate up and down these surfaces. Secondly, we look for plane-wave solutions, now allowing variations in y as well, of the form   v1 , v2 , v3 , p, ρ )ei(kx+ly+mz−ωt) , {v1 , v2 , v3 , p , ρ  } = Re ( where  v1 ,  v2 ,  v3 are complex amplitudes. Substituting this into (11.39), we obtain a system of algebraic equations: ⎧  p ⎪ ⎪ = 0, −iω v1 + ik ⎪ ⎪ ρ 0 ⎪ ⎪ ⎪ ⎪  p ⎪ ⎪ −iω v2 + il = 0, ⎪ ⎪ ⎨ ρ0 ik v + il v + im v3 = 0, 1 2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ρ  ⎪ ⎪ iωg + NB2  v3 = 0, ⎪ ⎪ ρ ⎪ 0 ⎪ ⎩ im p + g ρ = 0. It follows from the first two equations that  v2 = kl  v1 . Combining this with the third equation, we get equations is equivalent to

 v3  v1

+l = − k mk . Therefore, this system of algebraic 2

2

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⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

k  v1 = ,  p ωρ0 k2 + l2  v3 =− ,  v1 mk iN 2 ρ0 ρ  = B ,  v3 ωg  p ig = , ρ  m l  v2 = .  p ωρ0

Multiplying the first four equations, we get

 2    2 iNB ρ0  v1  v3 ρ k  p k + l2 ig 1= = , −  p v1  v3 ρ  ωρ0 mk ωg m and so (k2 + l2 )NB2 . m2 This is the dispersion relation for linear internal gravity waves. It relates the angular frequency ω to the components of the wave vector k = (k, l, m). On the other hand, solving the system of algebraic equations, we obtain a nontrivial solution  v1 ,  v2 ,  v3 , and  ρ in favor  p: ⎧ p  v1 = ρk , ⎪ ⎪ 0ω ⎪ ⎪ ⎪ ⎪ p ⎪ v2 = ρl , ⎨ 0ω ω2 =

2 2 ⎪ ⎪  v3 = − (kmρ+l0 ω)p , ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ p ρ  = − im g .

Arbitrarily choose  p to be real. The polarization relations for internal gravity waves are ⎧   i(kx+mz−ωt)  pe = p cos(kx + mz − ωt), p = Re  ⎪ ⎪ ⎪ ⎪ ⎪ ⎪  k p ⎪ ⎪ v = Re  ⎪ v1 ei(kx+mz−ωt) = cos(kx + mz − ωt), 1 ⎪ ⎪ ρ ⎪ 0ω ⎪ ⎪  i(kx+mz−ωt)  ⎨ l p v2 = Re  v2 e = cos(kx + mz − ωt), ρ0 ω ⎪ ⎪ ⎪ ⎪  i(kx+mz−ωt)  (k2 + l2 ) p ⎪ ⎪ v = Re  v e = − cos(kx + mz − ωt), ⎪ 3 3 ⎪ ⎪ mρ ω 0 ⎪ ⎪ ⎪  i(kx+mz−ωt)  m p ⎪ ⎪ ⎩ ρ  = Re ρ e = sin(kx + mz − ωt). g

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391

Inertia Gravity Waves

We concentrate on large-scale gravity waves. These waves are called inertia gravity waves. They have a horizontal scale of hundreds of kilometers and periods of several hours. Inertia gravity waves are the generalization of internal gravity waves to the case that the Coriolis parameter is not equal to zero, i.e., f0 = 0. In this case, the linearized Boussinesq equation (11.36) on an f -plane becomes ⎧ 1 ∂p ∂v1 ⎪ ⎪ − f0 v2 + = 0, ⎪ ⎪ ∂t ρ0 ∂x ⎪ ⎪ ⎪ ⎪ ⎪ ∂v2 1 ∂p ⎪ ⎪ ⎪ + f0 v1 + = 0, ⎪ ⎪ ∂t ρ0 ∂y ⎪ ⎪ ⎪ ⎨ ∂v ∂v2 ∂v3 1 (11.41) + + = 0, ⎪ ∂x ∂y ∂z ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ g ∂ρ  ⎪ ⎪ − NB2 v3 = 0, ⎪ ⎪ ρ0 ∂t ⎪ ⎪ ⎪ ⎪  ⎪ ⎪ ⎩ ∂p = −gρ  . ∂z We look for linear plane-wave solutions, propagating in the xz-plane and independent of y, of the form   v1 , v2 , v3 , p, ρ )ei(kx+mz−ωt) . {v1 , v2 , v3 , p , ρ  } = Re ( Substituting this expression into (11.41), we obtain a system of algebraic equations: ⎧ ik ⎪ ⎪ v2 +  p = 0, v1 − f0 ⎪ −iω ⎪ ρ0 ⎪ ⎪ ⎪ ⎪ ⎪ −iω v2 + f0 v1 = 0, ⎪ ⎪ ⎪ ⎨ ik v1 + im v3 = 0, ⎪ ⎪ ⎪ ⎪ ⎪ iωg ⎪ ⎪  + NB2 v3 = 0, ⎪ ⎪− ρ ρ ⎪ 0 ⎪ ⎪ ⎩ im p + g ρ = 0. By the second equation,  v1 = get

iω v2 . f0 

Combining this with the first equation, we

iρ0 (ω2 − f02 )  p = .  v2 kf0 Therefore, this system of algebraic equations is equivalent to

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⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

iρ0 (ω2 − f02 )  p = ,  v2 kf0 f0  v2 = ,  v1 iω m  v1 =− ,  v3 k ωg  v3 = 2 , ρ  iNB ρ0 im ρ  =− .  p g

Multiplying these equations, we get   

   iρ0 (ω2 − f02 )  p v2  v1  v3 ρ f0  m im  ωg − = − ,  v2  v1  v3 ρ  p kf0 iω k g iNB2 ρ0 i.e., 1 =

m2 (ω2 −f02 ) . k2 NB2

So

k2 NB2 . m2 This is the dispersion relation for inertia gravity waves. On the other hand, solving the system of algebraic equations, we obtain a nontrivial solution  v1 ,  v2 ,  v3 , and  ρ in favor  p: ⎧ kω p ⎪  v1 = , ⎪ ⎪ ⎪ ρ0 (ω2 − f02 ) ⎪ ⎪ ⎪ kf0 p ⎪ ⎪ ⎪ , v2 = −i ⎨ ρ0 (ω2 − f02 ) k2 ω p ⎪ ⎪ ⎪  v3 = − , ⎪ 2 − f 2) ⎪ mρ (ω ⎪ 0 ⎪ 0 ⎪ ⎪ m p ⎪ ⎩ρ  = −i . g ω2 = f02 +

Arbitrarily choosing  p to be real, we find the plane-wave solution is  i(kx+mz−ωt)  ⎧  pe = p cos(kx + mz − ωt), p = Re  ⎪ ⎪ ⎪  i(kx+mz−ωt)  kω p ⎪ ⎪ v1 = Re  v1 e = cos(kx + mz − ωt), ⎪ ⎪ 2 ⎪ ρ0 (ω − f02 ) ⎪ ⎪ ⎪  kf0 p ⎪ ⎨ v = Re  v2 ei(kx+mz−ωt) = sin(kx + mz − ωt), 2 ρ0 (ω2 − f02 ) ⎪ ⎪  i(kx+mz−ωt)  ⎪ k2 ω p ⎪ ⎪ v = Re  v e = − cos(kx + mz − ωt), 3 3 ⎪ ⎪ 2 ⎪ mρ0 (ω − f02 ) ⎪ ⎪ ⎪  i(kx+mz−ωt)  m p ⎪ ⎩ ρ  = Re ρ e = sin(kx + mz − ωt). g These are the polarization relations for inertia gravity waves.

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393

ROSSBY WAVES

Rossby waves are another class of atmospheric waves. These waves are also called planetary waves. They have horizontal scales of thousands of kilometers and periods of several days. Rossby waves are associated with many observed large-scale disturbances in the troposphere and the stratosphere, so they are very important for understanding many large-scale atmospheric phenomena. The quasi-geostrophic potential vorticity equation allows us to set up a simple model for the large-scale Rossby waves. Consider a uniform zonal background flow (U, 0, 0), where U is a constant. Denote by ψ ∗ its geostrophic stream function. The geostrophic stream function satisfies ∂ψ ∗ U = − , ∂y ∂ψ ∗ 0 = . ∂x The second equation implies that ψ ∗ is independent of x. Integrating the first equation with respect to y, we obtain a geostrophic stream function given by ψ ∗ = −Uy. If we have a zonal background flow plus a small disturbance, the geostrophic stream function of the total flow is ψ = ψ ∗ + ψ  = −Uy + ψ  , where ψ  is due to a small disturbance. From Section 11.12, the quasigeostrophic potential vorticity equation of the total flow is Dg ζ = 0, Dt where Dg ∂ ∂ψ ∂ ∂ψ ∂ = − + , Dt ∂t ∂y ∂x ∂x ∂y and the quasi-geostrophic potential vorticity of the total flow is   f02 ∂ψ ∂ 2ψ ∂ 2ψ ∂ ζ = f0 + βy + 2 + + . ∂x ∂y2 ∂z NB2 ∂z It is clear that ∂ 2ψ ∂ 2ψ ∂ + + ∂z ∂x2 ∂y2



f02 ∂ψ NB2 ∂z

 =

∂ 2 (−Uy + ψ  ) ∂ 2 (−Uy + ψ  ) + ∂x2 ∂y2   f02 ∂(−Uy + ψ  ) ∂ + . ∂z NB2 ∂z

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Notice that U is a constant and ∂ 2 (−Uy + ψ  ) ∂ 2ψ  = , ∂x2 ∂x2 ∂ 2 (−Uy + ψ  ) ∂ 2ψ  = , ∂y2 ∂y2     f02 ∂(−Uy + ψ  ) f02 ∂ψ  ∂ ∂ = . ∂z NB2 ∂z ∂z NB2 ∂z Then ∂ 2ψ ∂ 2ψ ∂ + + 2 2 ∂x ∂y ∂z



f02 ∂ψ NB2 ∂z



∂ 2ψ  ∂ 2ψ  ∂ = + + 2 2 ∂x ∂y ∂z



f02 ∂ψ  NB2 ∂z

 .

So the quasi-geostrophic potential vorticity for the total flow becomes ζ = f0 + βy + Lψ  , where ∂2 ∂2 ∂ L= 2 + 2 + ∂x ∂y ∂z



 f02 ∂ , NB2 ∂z

and the quasi-geostrophic potential vorticity equation becomes Dg (f0 + βy) Dg (Lψ  ) + = 0, Dt Dt

(11.42)

where Dg ∂ ∂ψ ∂ ∂ψ ∂ = − + . Dt ∂t ∂y ∂x ∂x ∂y Notice that

∂ψ ∗ ∂x

= 0 and

∂ψ ∗ ∂y

= −U. It is clear from ψ = ψ ∗ + ψ  that ∂ψ ∂ψ  = , ∂x ∂x ∂ψ ∂ψ  = −U + , ∂y ∂y

and so the two terms on the left-hand side of (11.42) are, respectively,

 Dg (f0 + βy) ∂ ∂ψ ∂ ∂ψ ∂ ∂ψ ∂ψ  = − + (f0 + βy) = β =β , Dt ∂t ∂y ∂x ∂x ∂y ∂x ∂x

  Dg (Lψ  ) ∂ψ ∂ψ ∂ψ  ∂ψ ∂ψ  = L − + Dt ∂t ∂y ∂x ∂x ∂y

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  ∂ψ  ∂ψ  ∂ψ  ∂ψ  ∂ψ  − −U + + ∂t ∂y ∂x ∂x ∂y

 

  ∂ψ ∂ψ ∂ ∂ = L +U = +U Lψ  . ∂t ∂x ∂t ∂x = L

From this and (11.42), the quasi-geostrophic potential vorticity equation for the total flow is

 ∂ ∂ ∂ψ  +U Lψ  + β = 0. ∂t ∂x ∂x Now we look for a plane-wave solution of the form    ei(kx+ly+mz−ωt) , ψ  = Re ψ where ω is the angular frequency, and k, l, and m are the components of the wave vector k = (k, l, m). Substituting this into the quasi-geostrophic potential vorticity equation, we obtain an algebraic equation:

    ∂ ∂  i(kx+ly+mz−ωt) L Re ψe +U ∂t ∂x   i(kx+ly+mz−ωt)   ∂ Re ψe +β =0 ∂x or 

 i(kx+ly+mz−ωt)       ∂ ψe ∂ ∂ ei(kx+ly+mz−ωt) + β Re L ψ +U = 0. ∂t ∂x ∂x This implies that  i(kx+ly+mz−ωt) 

    ∂ ψe ∂ ∂ i(kx+ly+mz−ωt)  L ψe +β +U = 0. (11.43) ∂t ∂x ∂x Assume that NB is a constant. Then   f02 ∂ 2 f02 ∂ ∂2 ∂2 ∂ ∂2 ∂2 L= 2 + 2 + = + + , ∂x ∂y ∂z NB2 ∂z ∂x2 ∂y2 NB2 ∂z2 and so

  i(kx+ly+mz−ωt) L ψe   f02 ∂ 2  i(kx+ly+mz−ωt) ∂2 ∂2 e = + 2+ 2 2 ψ ∂x2 ∂y NB ∂z  = − k +l + 2

2

f2 m2 02 NB

 ei(kx+ly+mz−ωt) . ψ

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From this, the first term of (11.43) is

  ∂ ∂ ei(kx+ly+mz−ωt) +U L ψ ∂t ∂x  

 2 ∂ ∂ 2 2 2 f0 ei(kx+ly+mz−ωt) =− k +l +m 2 +U ψ ∂t ∂x NB   2 2 2 2 f0  i(kx+ly+mz−ωt) , = i(ω − Uk) k + l + m 2 ψe NB and the second term of (11.43) is  i(kx+ly+mz−ωt)  e ∂ ψ  i(kx+ly+mz−ωt) . β = ikβ ψe ∂x By (11.43), we get   f02 2 2 2 ei(kx+ly+mz−ωt) i(ω − Uk) k + l + m 2 ψ NB ei(kx+ly+mz−ωt) = 0. + ikβ ψ This is simplified as



(ω − Uk) k + l 2

2

f2 + m2 02 NB

 + kβ = 0,

and so ω = Uk −

kβ f02 NB2

k2 + l2 + m2

.

This is the dispersion relation for Rossby waves. Let β = 0. Then ω = kU. The waves are merely carried along with the background flow. Therefore, β is crucial to the existence of Rossby waves. The zonal phase speed of the waves is c=

ω β =U− k k2 + l2 + m2

f02 NB2

.

Assume that k, l, and m are real and that m is nonzero. Notice that β=

2 cos φ0 > 0. a

Then 0
β f2 k2 + l2 + m2 02 NB

<

β . k2 + l2

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Since the crests and troughs of stationary waves do not move with respect to the ground, ω = 0, and so c = 0. This implies that there will be a vertical propagation and 0
k2

β . + l2

This inequality is called the Charney-Drazin criterion. Let β . Uc = 2 k + l2 Then Uc increases with increasing horizontal wavelength. In the Northern Hemisphere stratosphere, it has been observed that the winter background winds are eastward and stationary Rossby waves have large horizontal scales, while the summer background winds are westward and stationary Rossby waves do not exist. Define a group velocity vector

 ∂ω ∂ω ∂ω (y) (z) cg = (c(x) , c , c ) = , , . g g g ∂k ∂l ∂m Its vertical component is c(z) g =

∂ω = ∂m

2f02 βkm

NB2

k2 + l2 + m2

f02 NB2

2 .

By convention, k > 0. If m > 0, then the vertical component of the group velocity is positive and the waves propagate information upward. For upwardpropagating waves, the phase surfaces kx + ly + mz − ωt = constant slope westward with height. We use moving fluid blobs A, B, C, etc., lying along a line of latitude to describe the behavior of Rossby waves. Owing to the Coriolis term in the quasigeostrophic potential vorticity, according to conservation of potential vorticity, a northward-moving blob loses some disturbance vorticity, and then a southwardmoving blob must gain some disturbance vorticity. When blob A moves southward, the increase of the disturbance vorticity associated with blob A causes blob B to move southward, and then the increase of the disturbance vorticity associated with blob B will make blob C move southward and make blob A move northward again. After a short time, it is observed that the sinusoidal pattern of the blobs has moved westward, although each individual blob oscillates only north-south.

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11.15

ATMOSPHERIC BOUNDARY LAYER

Since the frictional force has only a small impact on gravity waves and Rossby waves in the atmosphere, friction effects are usually ignored. However, friction effects are sometimes very important in the region near Earth’s surface, especially in the lowest several kilometers of the atmosphere. The region near Earth’s surface is called the atmospheric boundary layer. By (11.36), noticing that ∂p ∂p = , ∂x ∂x ∂p ∂p = , ∂y ∂y the first two equations of the linearized Boussinesq equations on an f -plane reduce to ∂v1 1 ∂p = − + f0 v2 + Fx , ∂t ρ0 ∂x 1 ∂p ∂v2 = − − f0 v1 + Fy , ∂t ρ0 ∂y where v1 , v2 are the horizontal components of the velocity, f0 is the Coriolis parameter, and ρ0 = ρ(0). Assume that frictional stress is a horizontal force τ = (τx , τy , 0). Replacing ρ by ρ0 in (11.31), we get Fv =

1 ∂τy 1 ∂τx i+ j. ρ0 ∂z ρ0 ∂z

So 1 ∂τx , ρ0 ∂z 1 ∂τy Fy = , ρ0 ∂z Fx =

and so 1 ∂p ∂v1 = − + f0 v2 + ∂t ρ0 ∂x ∂v2 1 ∂p = − − f0 v1 + ∂t ρ0 ∂y

1 ∂τx , ρ0 ∂z 1 ∂τy . ρ0 ∂z

These two equations are both linear. In order to separate these two equations, the atmospheric flow is separated into a pressure-driven flow and a frictional-stress-driven flow. Corresponding to this separation, the horizontal velocity components for the atmospheric flow are separated into sums of those of the velocities for the pressure-driven flow and the frictional-stress-driven flow, i.e.,

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v1 = v1 + v1τ , p v2 = v2 + v2τ , p

where the subscripts p and τ denote the pressure-driven flow and the frictional stress-driven flow, respectively. Then the above two linear equations are separated into two systems of equations, i.e., the pressure-driven flow satisfies the system of equations p

∂v1 1 ∂p p − f0 v2 = − , ∂t ρ0 ∂x p ∂v2 1 ∂p p + f0 v1 = − , ∂t ρ0 ∂y

(11.44)

and the frictional-stress-driven flow satisfies the system of equations ∂v1τ 1 ∂τx − f0 v2τ = , ∂t ρ0 ∂z ∂v2τ 1 ∂τy + f0 v1τ = . ∂t ρ0 ∂z

(11.45)

If the frictional-stress-driven flow is steady, then ∂v1τ ∂v τ = 2 = 0. ∂t ∂t From this and (11.45), it follows that 1 ∂τx , f0 ρ0 ∂z 1 ∂τy . = f0 ρ0 ∂z

v2τ = − v1τ

Assume that the frictional stress exists significantly only in a boundary layer of depth d above the flat ground at z = 0. This layer is called a frictional boundary layer. Then τx and τy are nonzero for 0 ≤ z < d but vanish for z ≥ d. Denote by τx0 and τy0 the surface stresses exerted by the ground on the lowest layer of the atmosphere. Integrating these equations through the depth of this boundary layer, we get  d τ0 v2τ dz = x , f0 ρ0 0  d τy0 v1τ dz = − . f0 ρ0 0 Denote V1τ



d

= 0

v1τ dz,

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V2τ =

0

d

v2τ dz.

The quantities V1τ and V2τ are called the Ekman volume transports, and represent the horizontal fluxes of volume within this boundary layer. Then τy0 , V1τ = − f0 ρ0 τ0 V2τ = x , f0 ρ0 and so (V1τ , V2τ , 0) =

1 (−τy0 , τx0 , 0). f0 ρ0

Let the surface stress τ 0 = (τx0 , τy0 , 0). Then the scalar product of these two vectors (V1τ , V2τ , 0) · τ 0 = −τy0 τx0 + τx0 τy0 = 0, i.e., the Ekman volume transport in the frictional boundary layer is perpendicular to the surface stress. If the pressure-driven flow is steady, then p

p

∂v1 ∂v = 2 = 0. ∂t ∂t From this and (11.44), it follows that p

1 ∂p , f0 ρ0 ∂x 1 ∂p = − , f0 ρ0 ∂y

v2 = p

v1 and so p

p

∂v1 ∂v 1 ∂ 2p 1 ∂ 2p + 2 =− + = 0. ∂x ∂y f0 ρ0 ∂y∂x f0 ρ0 ∂x∂y Assume that the atmospheric flow satisfies div v = 0, i.e., ∂v1 ∂v2 ∂v3 + + = 0. ∂x ∂y ∂z Notice that v1 = v1 + v1τ and v2 = v2 + v2τ . Then  p 

  p ∂v1 ∂v2 ∂v1τ ∂v τ ∂v3 ∂v1 ∂v2 =− + = − + − + 2 ∂z ∂x ∂y ∂x ∂y ∂x ∂y

τ τ ∂v1 ∂v = − + 2 . ∂x ∂y p

p

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Integrating both sides through the depth of the boundary layer, we get 

 d 

τ  d ∂V1 ∂V τ ∂ ∂ + 2 , v3 |d0 = − v1τ dz + v2τ dz = − ∂x 0 ∂y 0 ∂x ∂y where V1τ and V2τ are the Ekman volume transports. Since the ground is flat, v3 = 0 at z = 0. Denote by v3d the value of the vertical velocity at z = d. Then

τ  ∂V1 ∂V2τ d v3 = − + . ∂x ∂y In this equality, the left-hand side is a upward flow out of the top of the boundary layer and the right-hand side is the horizontal convergence of the Ekman volume transports, i.e., the horizontal convergence of the Ekman volume transports must be balanced by a upward flow out of the top of the boundary layer. Conversely, horizontal divergence of the Ekman volume transports is balanced by a downward flow into the top of the boundary layer. The velocity v3d is called the Ekman pumping velocity. It depends only on the frictional-stress-driven flow. From 1 (V1τ , V2τ , 0) = (−τy0 , τx0 , 0), f0 ρ0 it follows that ∂V1τ 1 ∂τy0 = − , ∂x f0 ρ0 ∂x ∂V2τ 1 ∂τx0 = , ∂y f0 ρ0 ∂x and so the Ekman pumping velocity is  

τ  ∂τy0 ∂V1 ∂V2τ 1 ∂τx0 d + − = . v3 = − ∂x ∂y f0 ρ0 ∂x ∂y

(11.46)

Notice that the surface stress τ 0 = (τx0 , τy0 , 0). By the definition of the curl, the curl of the surface stress is   ∂τy0 ∂τy0 ∂τx0 ∂τx0 0 curl τ = − i+ j+ − k. ∂z ∂z ∂x ∂y Therefore, the vertical component of the curl of the surface stress is curlz τ 0 =

∂τy0 ∂x

−

∂τx0 , ∂y

and so v3d =

1 curlz τ 0 . f0 ρ0

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This shows that the Ekman pumping velocity is proportional to the vertical component of the curl of the surface stress. Imitating the kinetic theory in fluid dynamics, the stress components and the vertical derivatives of the horizontal velocity are assumed to satisfy ∂v1τ , ∂z ∂v τ τy = ρ0 ν 2 , (11.47) ∂z where the quantity ν is called the kinematic eddy viscosity. Assume further p p that the flow is steady, ν is a constant, and the pressure-driven flow (v1 , v2 ) is independent of z within the boundary layer and equal to the large-scale purely zonal flow satisfying τx = ρ0 ν

p

v1 = V(y), p v2 = 0. Since there can be no flow at the ground with friction, the boundary conditions on the total flow are (v1 , v2 ) → (0, 0) as z → 0, (v1 , v2 ) → (V(y), 0) (z  d). In terms of the stress-driven flow, by v1 = v1 + v1τ and v2 = v2 + v2τ , these become p

p

(v1τ , v2τ ) → (−V(y), 0) as z → 0, (v1τ , v2τ ) → (0, 0) (z  d). By (11.45) and (11.47), the steady stress-driven flow satisfies ∂ 2 v1τ , ∂z2 ∂ 2vτ f0 v1τ = ν 22 . ∂z τ τ Let λτ = v1 + iv2 . Then it follows from the coupled differential equations that f0 v2τ = −ν

∂ 2 v1τ ∂ 2 v2τ f0 v2τ f0 v1τ f0 ∂ 2 λτ f0 = + i = − + i = i (v1τ + iv2τ ) = i λτ , ∂z2 ∂z2 ∂z2 ν ν ν ν i.e., λτ satisfies the second-order equation ∂ 2 λτ f0 = i λτ . ∂z2 ν A direct check shows that the second-order equation has two solutions: λτ = Ae

±(1+i)



f0 2ν z

,

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where A is a constant. Since (v1τ , v2τ ) → (0, 0) (z  d), the minus sign must be chosen and the plus sign is deleted. So the solution of the second-order equation is −(1+i)



f0

2ν z . λτ = Ae Taking the real and imaginary parts and noticing that λτ = v1τ + iv2τ , we get    f f0 − 2ν0 z τ v1 = Ae cos z , 2ν    f f0 − 2ν0 z τ v2 = −Ae sin z . 2ν

Applying the boundary condition v1τ → −V(y) as z → 0 gives A = −V(y), and so v1τ v2τ

 f − 2ν0 z

= −V(y)e

 f − 2ν0 z

= V(y)e

 cos 

sin

 f0 z , 2ν 

f0 z . 2ν

Since the pressure-driven flow is p

v1 = V(y), p v2 = 0, the full solution is

 z z p v1 = v1 + v1τ = V(y) 1 − e− h cos , h z z p v2 = v2 + v2τ = V(y)e− h sin , h

 where h = 2ν f0 . This solution is called Ekman’s solution, and the corresponding boundary layer is called the Ekman layer. From Ekman’s solution, we see that for any fixed y, the horizontal velocity vector (v1 , v2 ), as a function of hz , represents a spiral which is called Ekman’s spiral. Ekman’s spiral shows that the deflection of the wind in the boundary layer is mostly to the low-pressure side of the geostrophic, large-scale flow. By (11.46) and (11.47),

 ν ∂ 2 v2 ∂ 2 v1  d v3 = − . f0 ∂z∂x ∂z∂y z=0

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From the full solution, it follows that  z ∂v1 z dV = 1 − e− h cos , ∂y h dy ∂v2 = 0, ∂x and so   z z  1 dV 1 −z  dV ∂ 2 v1  = e h cos + sin  = ,  ∂z∂y z=0 h h h z=0 dy h dy  ∂ 2 v2  = 0. ∂z∂x z=0  Notice that h = 2ν f0 . Then v3d = −

1 ν dV = hξ , f0 h dy 2

where ξ = − dV dy is the relative vorticity of the free-atmosphere flow. If the freeatmosphere flow is not a purely zonal flow, this relationship also holds. This relationship shows that the Ekman pumping velocity is upward under a cyclone in the free atmosphere for ξ > 0 and downward under an anticyclone for ξ < 0.

PROBLEMS 11.1 Show that



∞ 0

X3 π4 dX = −1 15

eX

which is used to derive the Stefan-Boltzmann law for the black-body irradiance. 11.2 Show that if the solar constant Fs = 1370 W/m2 and the planetary albedo α = 0.3, then Earth’s effective emitting temperature T ≈ 255 K. 11.3 Show that if the solar luminosity Fs = 2619 W/m2 and the planetary albedo α = 0.7, then the effective temperature T ≈ 242 K. These are values appropriate to the planet Venus. 11.4 Let v be the atmospheric fluid velocity. Show that the material acceleration may be expressed by

2 Dv ∂v |v| = + grad + (curl v) × v. Dt ∂t 2 11.5 Consider an air parcel with local speed v satisfying v = |v|s, where s is a unit vector. Show that its material acceleration can be described by Dv D|v| Ds = s + |v| . Dt Dt Dt

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11.6 Let ρsv be the density of water vapor at saturation, L be the latent heat of vaporization per unit mass, and Rv be the specific gas constant for the vapor. Show that the density of water vapor at saturation is a function of temperature given by c d ρsv = e− T , T where c is a constant and d = RLv . 11.7 Suppose that the Rossby waves are independent of height and suppose also that there is no background flow. For these Rossby waves, show their quasi-geostrophic potential vorticity and their dispersion relation between the angular frequency and the wave-vector components.

BIBLIOGRAPHY Boas, M.L., 1983. Mathematical Methods in the Physical Sciences. Wiley, New York. Bolton, D., 1980. The computation of equivalent potential temperature. Mon. Weather Rev. 108, 1046-1053. Durran, D.R., 1993. Is the Coriolis force really responsible for the inertial oscillation? Bull. Am. Meteorol. Soc. 74, 2179-2184. Gill, A.E., 1982. Atmosphere-Ocean Dynamics. Academic Press, London. Holton, J.R., 2004. An Introduction to Dynamic Meteorology, fourth ed. Academic Press, Burlington, MA. Landau, L.D., Lifshitz, E.M., 1975. Statistical Physics, Part I, third ed. Pergamon, Oxford. Lorenz, E.N., 1955. Available potential energy and the maintenance of the general circulation. Tellus 7, 157-167. Marshall, J., Plumb, R., 2008. Atmosphere, Ocean and Climate Dynamics, An Introductory Text. Academic Press, New York. Pedlosky, J., 1987. Geophysical Fluid Dynamics, second ed. Springer-Verlag, New York. Thuburn, J., Craig, G.C., 2000. Stratospheric influence on tropopause height: the radiative constraint. J. Atmos. Sci. 57, 17-28.