Geophysical Fluid Dynamics (Pedlosky, 1987) (6inaz/cgi-bin/Download.cgi/Pedlosky... · Geophysical...

32
Geophysical Fluid Dynamics (Pedlosky, 1987) (6 ) ( ) 19 1 11 Chapter 1 (1.4.13) (Maxwell ) ∂S ∂p T = - ∂V ∂T p (1) U V p T H S F = U - TS G = F + pV = H - TS dU = T dS - pdV (2) dH = T dS + V dp (3) dF = -SdT + V dp (4) dS = 1 T dU + p T dV (5) p V T 2 S = S (p, T ) dS = ∂S ∂p T dp + ∂S ∂T p dT (6) (3) dH = T dS + V dp = T ∂S ∂p T + V dp + T ∂S ∂T p dT (7) ∂H ∂p T = T ∂S ∂p T + V (8) ∂H ∂T p = T ∂S ∂T p (9) 1

Transcript of Geophysical Fluid Dynamics (Pedlosky, 1987) (6inaz/cgi-bin/Download.cgi/Pedlosky... · Geophysical...

Page 1: Geophysical Fluid Dynamics (Pedlosky, 1987) (6inaz/cgi-bin/Download.cgi/Pedlosky... · Geophysical Fluid Dynamics (Pedlosky, 1987) (6 ) ( ! #" $!%'&)( *) + , 19-1. 11 / Chapter 1

Geophysical Fluid Dynamics (Pedlosky, 1987)

(6 )

( ! #"$!%'&)(* )+,

19 - 1 . 11 /

Chapter 10(1.4.13) 132546#798:<;>=@?

(Maxwell:<;>=@?

)(

∂S

∂p

)

T

= −

(

∂V

∂T

)

p

(1)

ACBEDU F<GIHKJMLONKPQ D V FSR9T D p FVU 7 D T FSWEX D H FYJZE[\N]@Q D S FYJZ^C_<]@Q D F = U − TS FO`aNcbadeNf :g3h JMLYN#PQ D G = F + pV = H − TS FP@ij :g3h J#LYNKPQlknmpoIk D

dU = TdS − pdV (2)

dH = TdS + V dp (3)

dF = −SdT + V dp (4)

dS =1

TdU +

p

TdV (5)

qKr#sYt 6M7<8Mu<vYw@xKypDVD<A3z y

T: G D 2 w@c:3|eav~\E sYt ~ D JZ)^_@]@Q S = S(p, T ) k\ sE k q DY99 :Y95

dS =

(

∂S

∂p

)

T

dp+

(

∂S

∂T

)

p

dT (6)

k q@s3t A>z@D ? (3)5 D

dH = TdS + V dp =

T

(

∂S

∂p

)

T

+ V

dp+ T

(

∂S

∂T

)

p

dT (7)

qrsV D(

∂H

∂p

)

T

= T

(

∂S

∂p

)

T

+ V (8)

(

∂H

∂T

)

p

= T

(

∂S

∂T

)

p

(9)

1

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k v sYt H F 2 @ qKrKs k s k D ∂2H

∂T∂p=

∂2H

∂p∂T

v#: q D ?(8) k ?

(9)5 D

∂T

T

(

∂S

∂p

)

T

+ V

=∂

∂p

T

(

∂S

∂T

)

p

(10)

~S D ?(1) F s3t

Chapter 20(2.2.12)

C2Ω× u · dr = −2Ω

dAn

dt(11)

2.2.3

cy C

t

Chapter 3

3.6 K1 X : v⊥ v⊥ ≤ O(R−d−1)

q v@ k D T y q v@ t z>D d y "! t0(3.8.15)

u⊥ =∣

∣uH −

u‖

KK∣

∣(12)

p74 line2

σ/f > 1 F |σ/f | > 13w$#0

(3.8.17) 1&% 1 '(@")+* ^ N k = (k, l) k, :.- / K = |k| k D < F0"1 :32 4 657sOtu‖ =

1

K(k, l) · uH (13)

u⊥ =1

K(−l, k) · uH (14)98' :O? D

(u, v) F;: oIk Du =

1

K(ku‖ − lu⊥) (15)

v =1

K(lu‖ + ku⊥) (16)

k v s : q D< X ζy D

ζ =∂v

∂x−∂u

∂y=

1

K

k

l

· ∇u⊥ +

l

−k

· ∇u‖

(17)

2

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9Sq ( : (u, v) = (u0, v0) expi(kx+ ly − σt) F; s3t ? (14)5 D

l

−k

· ∇u‖ = l

(

k∂u

∂x+ l

∂v

∂x

)

− k

(

k∂u

∂y+ l

∂v

∂y

)

= l(

ik2u0 + iklv0)

− k(

iklu0 + il2v0)

= 0 (18)

3.9 3.10 K1c1

1: (a)Poincare

(:( 5: t y R : X D ) * ^ N y D t $(v#: q - $/ 9y@vE t

2: (b)Kelvin

(("!# : $ %& /98 z(' :

)t

0(3.16.1) )* 0 (3.16.6) +-,K13254 : % 2 .0/214365"789: ?

d2u

dx2+

1

x

du

dx−

(

1 +s2

x2

)

u = 0 (19)

F w)<;>= N 9: ? k@?A t

3

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3: (c)Rossby

( ts :95: D w)<; = N 9: ?5: : y D 1 w)<; = N |a

Is =∞∑

m=0

1

m!(m+ s)!

(x

2

)2m+s(20)

I−s =∞∑

m=0

1

m!(m− s)!

(x

2

)2m−s(21)

l@~S 7 8 s3t( ) : k D uE u(x) = Axs F s k D A[s(s − 1)xs−2 + sxs−2 −

xs − s2xs−2] = 0v: q D

xs−2: 5y / 8set

u(x) = Ax−s k '!" qrs>t ~S D# : k My D : 2 $cF&% 75s k I (' 2 tu1(x) =

∞∑

n=−s

anxn (22)

u2(x) =

∞∑

n=s

anxn (23)

?(22)

A>z y3?(23) F ? (19) F)# s k D

∞∑

n=−s

n(n− 1)anxn−2 =

∞∑

n=−s

nanxn−2 −

∞∑

n=−s

(

anxn + s2anx

n−2)

(24)

k v sSt ∑∞n=−s anx

n =∑∞

n=−s+2 an−2xn−2q9r9s D

n ≥ −s+2 *eO

(n2−s2)an =

an−2qrsYt

n = −s+ 1*Ea D

(−s+ 1)(−s) + (−s+ 1) + s2

a−s+1 = 0 k v sYt ~S Ds:+,l@~S # 5:5:+), - A sYt n = 2m+ s k 4 oIk D a2(m−1)+s =

1

22m(m+ s)a2m+sk v sYtc98 /.)0<? F: oVk DV :Y=@ - eD ? (20) k ?

(21) F 2 s3t ( 1 )2 w) ; = N |a

Ks =π

2

I−s(x)− Is(x)

sin(sπ)(25)

4

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' @ 9 : ?(19)

: : q#rKsYt c s c:c $ a ' sYtM : k D _]#[N :O Ks =

(−1)s

2

(

dI−sds−dIsds

)

(26)q 7 8 s3t9Sq D Z |a5: @ 9 ψ(x) =dΓ(s)

ds/Γ(s) F 7lsYt H T l5 D

ψ(x)Γ(x) =d

ds

∫ ∞

0e−sxs−1 dx

=

∫ ∞

0e−s lnxxs−1 dx = Γ(s− 1) + (s− 1)ψ(s− 1) = · · ·

= Γ(s)

(

1

s− 1+

1

s− 2+ · · ·+

1

2+ 1 + ψ(1)

)

k v s : q D Q γ = −ψ(1) = 0.5772· k ~S Dψ(s) = −γ + 1 +

1

2+ · · ·+

1

s− 1 :ψ F ~S D

K0(x) = − ln(x

2

)

I0(x) +∞∑

m=0

ψ(m+ 1)

(m!)2

(z

2

)2m= − lnx− γ + ln 2 + · · · (27)

k cs3t( ) d

ds

1

(s+m)!=

d

ds

1

Γ(s+m+ 1)= −

ψ(s+m+ 1)

Γ(s+m+ 1)

q#r#sYt ~ D ?(20) k ?

(21)5 DdIsds

=

∞∑

m=0

−ψ(s+m+ 1)

m!(m+ s)!

(x

2

)2m+s+

∞∑

m=0

ln(x

2

) 1

m!(m+ s)!

(x

2

)2m+s(28)

dIsds

=

∞∑

m=0

ψ(m− s+ 1)

m!(m− s)!

(x

2

)2m+s−

∞∑

m=0

ln(x

2

) 1

m!(m− s)!

(x

2

)2m−s(29)

9Sq Ds = 0 F )# s k D ? (27) F s3t ( 1 )!&" x→∞

* csK0(x)

:.9y DK0(x) =

π

2xe−x

1−1

8z+

9

128z2+ · · ·

(30)

3.18 p109 3 '3.13 y D 3.14

3.20 !"$# ,c4%"&(')*%+-,/.0 71 D2<9 9: ?5:435 768d2x

dt2+ x = sin t (31)

x(0) = 0 (32)

dx

dt(0) = 1 (33)

5

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F:po t ASBED j w 5 F 9 s3tL[f ](s) =

∫ ∞

0e−stf(t) dt (34)

k\ s3t : k D T : ( ) 5L[f ′](s) = sL[f ](s)− f(0) (35)

L[sin t] =1

s2 + 1(36)

L[eat] =1

s− a(a < s) (37)

L[teat] =1

(s− a)2(38)

qrs3t : - j w 5 F 9 9: ? (31) !k D

L[f ′′] = sL[f ′]− f(0) = s2L[f ]− sf(0)− f ′(0) (39)qrsV D 35(32) (33) F/ ~V D

L[x](s) =s2 + 2

(s2 + 1)2=−1/4

(s+ i)2+−1/4

(s− i)2+

3i/4

s+ i+−3i/4

s− i(40)

k v s : q D ~S Dx(t) = −

1

4t(eit + e−it) +

3

4(eit + e−it)i = −

1

2t cos t+

3

2sin t (41)

0(3.23.12) 0 (3.23.13)

ui = −<(ilAei(kix+ly−σt)) (42)

vi = <(ikiAei(kix+ly−σt)) (43)

ur = <(ilAei(krx+ly−σt)) (44)

ur = −<(ikrAei(krx+ly−σt)) (45)

(46)

0(3.23.14)?(3.23.14)

:A2/4

:1/4 M s D A yKv#: q A F A/2 k 4 57&8(1I t0

(3.24.37) 0 (3.24.38a)k =

∂θ

∂x= ks +

(

x− t∂σ(ks)

∂k

)

∂k

∂x= ks(x, t) (47)

σ = −∂θ

∂t= −

(

x− t∂σ(ks)

∂k

)

∂k

∂x+ σ(ks) (48)

6

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0(3.24.42)

E(xs, t)∆x = ... (49)0(3.24.54)

l =

2

πAne

i(kmx−σ(km)t) ×

∫ km+∆

km−∆dk exp i

[

(x− Cgt)(k − km)−C ′′g

6(k − km)

3t

]

(50)

0(3.24.55) 1

∫ ∞

0cos(

ak3 + bk)

dk =π

(3a)1/3Ai((3a)1/3b) (51)

( ) ξ = b

(3a)1/3k *Ea D 98 F ? (51)

)# s k D∫ ∞

0cos

(

b3ξ3

3ξ3+ bk

)

dk =π

bξAi(ξ) (52)

k v s3t /# D ξ′ = bk

ξk * oIk D∫ ∞

0cos

(

ξk +k3

3

)

dk = πAi(ξ) (53)

98 F D ξ q 2 9 s k D−

∫ ∞

0cos

(

ξk +k3

3

)

k2 dk = πdAi

dξ(54)

?(53) k ? (54)

5 D−π

(

dAi

dξ− ξAi

)

=

∫ ∞

0cos

(

ξk +k3

3

)

(ξ + k2) dk (55)

θ = ξk +ξ3

3

5 Ddθ = (ξ + k2)dk

5 D T y D∫ ∞

0cos θdθ

?= 0

k v s3tdAi

dx− xAi = 0 (56)98 5 ?

(51) /68 z t ( 1 )0(3.24.57)

φn(x, t) = 21/2π1/2An(km)

|C ′′g (km)(t/2)|

1/3cos [kmx− σ(km)t]× ... (57)

7

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0(3.25.21)

σmn = −β

2knm(58)0

(3.25.22) : F M s3t ψ(r = 1) = 00(3.25.25) cuE 1 : 9: ? F q % 75s3t

∇2φ+ λ2φ = 0 (59)

8 aD∇2 =

∂2

∂φ2+

1

r

∂r+

1

r2∂2

∂θ2F) aD w9 φ(r, θ) = R(r)Θ(θ) F&%

75s k Dd2R

dr2+1

r

dR

drR

r

+ λ2r2 = −

dθΘ

= m2 (my ) (60)

RM; s 96: ? F: o3k D R(r) = Jm(λr) k v sat ly φ(1, θ) = 0

DJm(λ) = 0

qrs3t $ A D Jm : n : knm eD

Jm(knm) = 0qrs3t

( ) r′ = λr k\ s k Dr2R′′ + rR′ + (r2 −m2)R = 0 (61) aD

R(r) = rk∑∞

n=0 anrn k 2 # :F s3t98 F ? (61)

) rn:

5:3=@ 0@v s z @y D

k(k − 1)a0 + ka0 −m2a0 = 0

v#: q Dk = m

qrs3t Az9Drm+1

:3=@cy(m + 1)ma1 + (m + 1)a1 −m

2a1 = 0v#: q D

my qrsV D

a1 = 0 k v s3t /# D rm+2 :Y=@5: D a0 = −4(m+ 1)a2 k v s3t"!# u9 Dan = −

1

n(2m+ n)an−2

qrs3tY ~S D

R(r) = a0

∞∑

l=0

(−1)lm!

22ll!(m+ l)!r2l+m (62)

~S DJm(r) =

∞∑

l=0

(−1)l

22ll!(m+ l)!r2l+m (63)

$: k v s3t ( 1 )0(3.25.30): Jacobi-Anger $&%

eiz cos θ =∞∑

n=−∞

inJn(z)einθ (64)

Jn(z) =i−n

π

∫ ∞

0eiz cos θ cos(nθ) dθ (65)

8

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( ) n F +9 k s3t ASBEDeiz cos θ = 1−

1

2!z2 cos2 θ +

1

4!z4 cos4 θ −+i

(

z cos θ −1

3!z3 cos3 θ +

1

5!z5 cos5 θ−

)

(66)

2l ≥ n: k D

∫ π

0cos2l θ cosnθ dθ =

1

22l+1

∫ ∞

0

2(2l)!

(l + n)!n!dθ =

π

22l(2l)!

(l + n)!n!qr D2l < n

: k y T y 0qrs3t

n ,9: '!" t ( 1 )0(3.26.12)

B(Km,Kn) = ... (67)

p157 line 2

For (3.26.8)0(3.26.28)

ψ = ψ0(x, y, t, t) +1

βψ1(x, y, t, t) + · · · (68)

0(3.26.37)

Vj = (K2j + F )2a2j4

= (K2j + F )Ej (69)0(3.26.38)

d

dt(V1 + V2 + V3) = 0 (70)0

(3.26.40) 1the radius of gyration( ! ) k y D Q3Z^3k x: tK98 y 6 # s RE D , : :: ! q M sV k 2 t0(3.26.41) 9 y 99 t0(3.26.43) 1$C1 0

dα1dt

= −B(K2,K3)

(K21 + F )A2α3 (71)

0(3.26.44)

d2α1dt2

=B(K2,K3)B(K1,K2)

(K23 + F )(K21 + F )A22α1 (72)

0(3.26.47) 1$C1 0

9

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(K21 + F )a24+ · · · (73)0

(3.26.48)

da1dt

= C1[(C2 − a21)(C3 − a

21)]1/2 (74)0

(3.27.14) 91K1Fourier amplitude ψ(k, l) as

E =

∫ ∫

Adx dy

[

|∇ψ|2 + F |ψ|2

2

]

= · · · (75)

0(3.27.19)

K22 =

∫∞−∞

∫∞−∞K2E(k, l) dk dl

∫∞−∞

∫∞−∞ E(k, l) dk dl

(76)

0(3.28.18) If (3.28.8) is used in (3.28.16) it follows that for the similarity solution ε = tg(Kt)

F = −Kε(K)

t(77)

( ) g 1w@M|> qKrKs9 k 6 sYt ε(K, t) = tg(Kt) F t q < D k q T s k D

∫ K

0

∂ε

∂tdK =

∫ K

0

[

g(Kt) +Ktg′(Kt)]

dK

=1

t

[

∫ α/t

0g(α) + αg′(α) dα

]

t2g(α

t

)

=Kε(K)

t ~S D ?(77) /98 z t ( 1 )

p176 line 1

In Section 3.26 it ... 3.28.3

upper panely

left paneltlower panel

yright panel

t

Chapter 40(4.3.31) 0 (4.3.32)

ME = ρ

(

i

∫ ∞

0u dz + j

∫ ∞

0v dz

)

= δEρU

2−i+ j (78)

ME =τ × k

f(79)

10

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0(4.4.14) 13254

u = 1−cos[(1− z)k] cosh[(1 + z)k] + cosh[(1− z)k] cos[(1 + z)k]

cos 2k + cosh 2k

≈k4

6(1− z2)(5− z2) (80)

F < s3tcos[(1− z)k] = 1−

1

2(1− z)2k2 +

1

24(1− z)4k4 − · · ·

cosh[(1 + z)k] = 1 +1

2(1 + z)2k2 +

1

24(1 + z)4k4 + · · ·

:2?: F s k D

cos[(1− z)k] cosh[(1 + z)k]

≈ 1 +k2

2[(1 + z)2 − (1− z)2]−

k4

4[(1− z)2(1 + z)2] +

k4

24[(1− z)4 + (1 + z)4]

= 1 + 2k2 +k4

6

(

−1 + 6z2 − z4)

!&" Dcos[(1− z)k] cosh[(1 + z)k] ≈ 1− 2k2 +

k4

6(−1 + 6z2 − z4)

~S D ?(80)

: y2 +

k4

3(−1 + 6z2 + z4)

qrs3t 9 DY y(

1−1

2(2k)2 +

1

24(2k)4 + · · · 1 +

1

2(2k)2 +

1

24(2k)4

)−1

=1

2

(

1 +2

3k4 + · · ·

)−1

≈1

2

(

1−2k4

3

)

~S D1−

(

)

(

)= 1−

[

1 +k4

6(−1 + 6z2 + z4)

](

1−2k4

3

)

=k4

6(5− z2)(1− z2)

0(4.10.12)

uE = u− u0 =α

2[τ − (k × τ )] (81)

4.10.2 :

k × τy

τ × kt0

(4.13.25)

11

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0 = −E1/2V

∂2ψ

∂η2−l

r

∂ψ

∂η

+1

Rel2

∂2

∂η2−l

r

∂η

ψ (82)

0(4.13.35) )* 0 (4.13.37)ACBED ?

(4.13.35)y

ψ = ψI(r)−E1/2H

E1/4V

∂ψI∂r

r=1

e−η (83)

qKr ~C Dτ @ q v 8(1 vEY? ψI(r = 1) =

∂ψI∂r

r=1

a /68 s3tp248 line7

geostrophic velocities are not divergent, ...0(4.14.6)

∂K21∂t

= −

∂t

∫ ∫

(K −K1)2ε(K) dK

∫∞0 ε(K) dK

− · · · (84)

p249 line−18

that the eddies become statistically ...

p251 line−9

able to continuously accept ...

Chapter 5

chapter 5 )curlτ k k · curlτ t t

5.3 K1circulation (e.g. the western intensification) does not seem ...0(5.3.3) 1'

= k · curlτ∗ρ+ τ

(x)∗

β0ρf

(85)

0(5.3.5) 0 (5.3.8) 0 (5.3.10)?(5.3.5)

: 9 y 99 t ? (5.3.8)D ?

(5.3.10) $ <My D XE k XW

; s 9 : 99 tp268 11 0 31 0

ρ

∫ XE

XW

curlτ(x′, y) dx′ (86)

12

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0(5.4.2)

δMy

Munk:MD!&"

δSy

Stommel:St0

(5.4.25) 0 (5.4.27)ζyξ3w$# t0

(5.6.20)" c: 5y ξ → ∞ F D(' 2 X vK k 2 t c8u#Eyf(s(0, y)) = g(s(0, y))

v 1f(s(x, y)) = g(s(x, y)) k s z D f k g y k ' :u q K: 9 E rcsYt0

(5.10.3)z eDA > 0

t0(5.10.10)

CsA2

= yr (87)

p307 1 1)* 2 ... walls at x = 0 or l.0(5.12.25b)

∫ ∞

0wc(ξ, y,∞)dξ = +

βL

f0

τ (y)

f(XE , y) (88)

0(5.12.30)

vBβ0L/f0(EH/2)(∂3vB/∂x3)

≤ O(E1/2H )¿ 1 (89)

0(5.12.33)

fuB = −∂pB∂y

+1

2

∂2vB∂ξ2

(90)

0(5.13.16)

∂2ψB∂η2

+∂ψB∂η

∂B

∂x(x, 1) = 0 (91)

0(5.13.20)

ψ = −β

απcosπy +Φ

(

y +αx

β

)

(92)

5.13.3 1 ... for curlτ = − sinπy and ...

13

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Chapter 60(6.2.28c)

tan θ = tan θ0 +L

r0y cos−2 θ0 +

(

L

r0

)2

y2tan θ0cos2 θ0

+ · · · . (93)

6.3 K1 (6.3.2) 11 0 91K1 0

ε = O(10−1), β−1 = O(1) (94)

ε = O(5× 10−3), β−1 = O(0.5) (95)

p346 1 2 $C1 0δ = O(4× 10−2) = O(ε) (96)

p352 T N <6Ky k :gYh X R/2 F z ' : q#rKsYt ez ~ D @ q y3/2R

D2 9 q y

5/2RD 9 q y

3R k v sYtp352

δ k ∆ 0 t0

(6.5.1)

ln θ∗ =1

γln p∗ − ln ρ∗ +

R

cpln p0∗ − lnR (97)

0(6.5.2)

...− εFρ+O(ε2F 2) (98)

p356 1 0 (6.5.10) 1M1CpT∗ = O(gD)

y D -Kv O(105m2/s2)

q D v O(104m2/s2)

t y DMKS

8(1 D H∗ ≤ O(U2f0) = O(10−2m2/s3)

t0(6.5.16)

...− S−1

(

∂u0∂z

∂θ0∂x

+∂v0∂z

∂θ0∂y

)

(99)

0(6.5.25) 0 (6.5.26)θ0y

(6.5.26)

Π∗ = ...+ εFθs

εξ∂θ

∂x+ εη

∂θ

∂y+ εδ cot θ0

∂θ

∂y

(100)

14

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(6.5.29)

d0dtζ0 + βy+

F

(1/θs)(∂θs/∂z)

d0dt

∂θ0∂z

+ ... (101)

(6.9.1) ... given by (4.10.20), i.e.,(6.9.13) ... is on a scale L = O(100km) rather than La = O(103km), so that

1

gDε

d0dt

(

p∗aρs

)

= ... (102)

! !#"$%'&1/10

( ) * + p367 ,-.0/ 21 3 CGS 4 ) MKS 4 256 7( 105dyne = 1N 8:9<; 107erg = 1J

)ε =

U

f0L=

5× 10−3* += 9

(6.9.14) ... condition for (6.9.11) becomes(6.10.1) ?>7@A(6.5.12) 90B A (6.10.3) CDE 0 F 2G!%H%I JK LM NO P( ) ?*+Q0 )

∂Π0∂t

+∂

∂x(u0Π0) +

∂y(v0Π0) + β

∂ψ

∂x= 0 (103)

SR −ρsψ CTVU )W 2,3,4 X

−x(ρsu0ψΠ0)−

y(ρsv0ψΠ0)−

x

(

ρsβψ2

2

)

)ZY W1 X Π0

[ \]_^

−ρsψ∂

∂t

(

∂2ψ

∂x2+∂2ψ

∂y2+

1

ρs

∂z

ρsS

∂ψ

∂z

)

= −∂

∂x

(

ρsψ∂2ψ

∂x∂t

)

−∂

∂y

(

ρsψ∂2ψ

∂y∂t

)

−∂

∂z

(

ρsψ

S

∂2ψ

∂z∂t

)

+ρs∂ψ

∂x

∂t

(

∂ψ

∂x

)

+ ρs∂ψ

∂y

∂t

(

∂ψ

∂y

)

+ρsS

∂ψ

∂z

∂t

(

∂ψ

∂z

)

SR 97Ba`b A Cc (6.10.4)

... = −∂

∂xρsu0ψ [...] −

∂yρsv0ψ [...] − ... (104)

(6.10.6) d2efV?>7@gh

15

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... = ...−Evε

∫ ∫

Adx dy... (105)

Y A ( B Y E y = y1)y = y2

8#E v0 = v1 = 0Y

x A ∂u0∂t

+ u0∂u0∂x

= −1

ρs

∂p

∂x

( SR x Q0 )

∂t

xu0 dx = 0

) Y u0(x, y1/y2, 0, t = 0) = 0 C [Q )S Q t ! u0(x, y1/y2, 0, t) =

0)<Y (6.10.10)

O(ε) = O(F )) E

(6.10.14)

E∗ =ρs2

(

u2∗ + v2∗ +g2

N2s

(

δθ∗θs

)2)

(106)

(6.10.16) ?>7@

1

S

∂2ψ

∂z∂t=

1

S

∂θ0∂t

=1

S

d0θ0dt

= −w1

)ZY 2 ∫ ∫

dx dyρsp0u0 · ∇θ0 =

∫ ∫

dx dy∇ · [ρsp0u0θ0]−

∫ ∫

dx dyρsθ0u0 · ∇p0 = 0

( ]_^ (6.10.17)

... = ...− δEf0

∫ ∫

Adx∗dy∗

(

u2∗ + v2∗2

)

z=0

(107)

(6.10.21)

S = i [...]

+ j

[

...+ ...−ρsψLy

εr0tan θ0

∂p0∂x

]

θ0 "

(6.10.22) p1ψ

(p1ψ)

(6.10.22) #!

16

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J = i [...]

+ j

[

v0E + ρs

[

ψv1 + p1∂ψ

∂x−Ly

εr0tan θ0ψv0

]

+ ρsp0v0/ε

]

+ k...

(6.10.23)

cos θ

ε

cos θ

ε cos θ0

(6.11.11) ?>7@

w1 = −1

S

d0θ0dt

= −1

S

(

∂t+ u0

∂x+ v0

∂y

)

∂ψ

∂z]_^ ` b

ψ = ψ1(z)ei(φ−σt) + ψ2(z)e

−i(φ−σt)

) N X I 0 Y ) C Q ) #( ^Z O((mH)−1)

Q ∂2ψ

∂z∂t= Aez/2Hσ

(

m cosφ+1

2Hsinφ

)

→ Aez/2Hmσ cosφ

)ZY ) * +Q0 (6.11.12) (6.11.13)d

dt

d0dt

(6.11.15)

σ/m

Cgz= −

k2 + l2 +m2/S

2m2/S< 0 (108)

(6.11.19) ?>7@A

(6.11.21)' ∂ψ

∂z∼ −mAez/2H sinφ CZDE ^ ( Z]:^ d0θ0

dt→

∂θ0∂t

) P A(6.11.6) CD

(6.11.21)

u0 = −∂ψ

∂y= Alez/2H sinφ (109)

v0 =∂ψ

∂x= −Akez/2H sinφ (110)

(111)

(6.12.14)

17

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Φ = cos[(λS)1/2z] (112)

(6.13.14)

ρsp0w1x,y = −ρs

φ

Su0

∂2φ

∂x∂z

x,y

≥ 0 (113)

(6.13.15)

( )x,y≡

k

∫ 2π/k

0dx

∫ 1

−1dy (114)

(6.13.29) (6.13.30)

ρsp0w1x,y =

|A|2kmρs(0)

4Su0 (115)

ρsp0w1x,y =

−|B|2kmρs(0)

4Su0 (116)

(6.13.38) 1 U0 u0 A 2

lim

K2→K2s+α2

S(1+µ)

φ &!'[ C<D

(6.13.41) (6.13.42)

ρsθ0v0x,y

=ρsη

20S

K2s −K2

km

4ez/H (117)

ρsp0w1x,y

= u0ρsη

20

K2s −K2

km

4ez/H (118)

(6.13.43)

p0

(

∂η

∂x

)

z=0

x,y

= ... (119)

(6.13.45) Multiplication by p0 = φ and ...(6.13.46)

dz= 0 z = 1 (120)

(6.13.47)

H = ... (121)

18

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(6.13.53)

K2s >> K2 +n2π2

S(122)

(6.13.54)

βv0 =1

S

∂H

∂z(123)

(6.14.8)

∂θ0x

∂t= −w1

xS −∂

∂y(θ′0v

′0

x) +H

x(124)

(6.14.7’)

∂u0x

∂t= v1

x −∂

∂y(u′0v

′0

x) + Fx

x(125)

(6.14.9)

Fx =F∗φ

ρ∗εUf0(126)

(6.14.18) ... and hence u0

x and θ0xcan be ...

(6.14.20a)

∂u0x

∂t= v1∗

x +1

ρs∇ · (ρsF ) + Fx

x(127)

(6.14.24)

...−∂Fx

x

∂z(128)

p401 1 ... have, from (6.6.10) and ...(6.14.27)

v′0ηBx= −p′0

∂ηB∂x

x

(129)

(6.14.29)

...+∂F ′

y

∂x−∂F ′

x

∂y(130)

(6.14.32)

19

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∂θ′0∂t

+ u0x∂θ

′0

∂x+ v′0

∂θ0x

∂y+ w′

1S = H′ (131)

(6.14.33)

v′0Π′0

x= −

∂t

(

Π′20 /2

x

∂Π0x/∂y

)

+D (132)

(6.14.34)

D = Π′0

1

ρS

∂z

ρsH

S+∂F ′

y

∂x−∂F ′

x

∂y

/∂Π0

x

∂y

x

(133)

(6.14.34) If ∂Π0

x/∂y should vanish ... by multiplying (6.14.29) by ψ′ ... expression for v′0Π

′0

x.

Should both u0x and ∂Π0

x/∂y vanish for ...

(6.14.35)

−1

ρs

∂χ

∂y=

∂y

[

ψ′

S

∂θ′0∂t

x

+E1/2v

2εψ′ζ ′0

x−ψ′H′x

S

]

/u0x +

E1/2v

2εζ0x(134)

−1

ρs

∂χ

∂y=

∂y

[

v′0ηBx−

∂t

θ′20x

2S+ u0

xθ′0∂ηB∂x

x

+E1/2v

2εθ′0ζ

′0

x−θ′0H

′x

S

/∂θ0

x

∂y

]

+E1/2v

2εζ0x(135)

v′0ηBx− u0

xθ′0∂ηB∂x

x

/∂θ0

x

∂y= ηB

−∂θ′0∂t− S

E1/2v

2εζ ′0 +H

x

/∂θ0

x

∂y(136)

p403 l5

condition at z = 0 contains no ... propotional to ηB∂θ′0/∂tx.

p403 13

is simply χ = const.(6.14.37) ' #!There is particularly intriguing connection between the Eliassen ... we define wave

activity (density), Aw,

Aw =ρsΠ′2

0

x/2

∂Π0x/∂y

. (137)

... by the Eliassen–Palm flux, ρsF , while the

p403 !p403

( B ! N [ )R Z[ ( A (6.14.39)

9 8C WKB(J) )

20

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(6.14.41)

v′0Π′0

x= ∇ · F = G0e

−(αr)2 (138)

(6.14.42) _ (6.14.43)that if χ is a solution ...

χ = ρs∂χ

∂z(139)

(6.14.44) (6.14.45)ζ ) 7 Y E

χ = −G02α2

∫ r

0

(

1− e−α2ζ2

ζ

)

dζ (140)

χ = ρs∂χ

∂z= ρs

z

r

∂χ

∂r(141)

χ = −ρsG02α2

z

r2(1− e−α

2r2) (142)

(6.14.46)

v1∗x = −

G02α2

[

(1− e−α2r2)

r2

(

1−2z2

r2

)

+2α2z2

r2e−α

2r2

]

(143)

w1∗x =

G02α2

[

2(1− e−α2r2)

r2+ 2α2e−α

2r2

]

(144)

6.14.2

&v1∗

x/

(

G02

)

8V9S; ∇ · F x/

(

G02

)

(6.14.47) y = 0 (6.14.20a), (6.14.41), and (6.14.46a) imply that ...(6.14.48) 1

∂Π′0

∂t= −c

∂Π′0

∂x(145)

(6.14.48)

(u0x − c)

∂x

∂2ψ′

∂x2+∂2ψ′

∂y2+

1

ρs

∂z

ρsS

∂ψ′

∂z

+∂Π0

x

∂y

∂ψ′

∂x= D′ (146)

p407 21 #! unless fortuitously, ∂Π0

x/∂y vanishes there.

p411 l7

large KS1/2 ...

21

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(6.15.34)-(6.15.36)

j = 0, 1, 2, · · ·(6.16.1) as in section 6.2.(6.16.25b) (6.16.26)d0dt0

d0dt.

(6.16.38)

S = i

[

2∑

K=1

−ψK∂2ψK∂x∂t

− u(0)K ψKΠ

(0)K − β

ψ2K2

DK

D−r22ψ2∂ψ2∂x

D2D

]

+ · · · (147)

(6.18.2) The derivative in (6.18.1) may be ...

... =ρs(hn)w1(hn)− ρs(hn+1)w1(hn+1)

dn+O(dn), (148)

(6.18.9)

w1(hN+1) = u0(ZN ) · ∇ηB +E1/2v

2εζ0(x, y, ZN ), (149)

(6.18.10)

d0dt

[

ζ0(ZN ) + βy −Fθs(hN )

dNρs(ZN )

p0(ZN )− p0(ZN−1)

[θs(ZN−1)− θs(ZN )]ρs(hN )

+ηBdN

]

= −E1/2V

2εdNζ0(ZN ).(150)

A ( ρs(hN ) ∼= ρs(ZN ) C [ E (6.18.11) ... compared with (6.16.25b), (6.16.29), and (6.16.30) respectively, ...(6.18.13b)

[

∂t+∂ψ(Z2)

∂x

∂y− ...

] [

...+ βy +ηBd2

]

= ... (151)

p430 l1

... for the Fn in (6.18.14) may be ...(6.18.18)

F1 =f20L

2

(∆θs/θs)g(Z1 − Z2)D= F2 (152)

(6.18.19)

22

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∆θsθs

=θs(Z1)− θs(Z2)

θs(h2)(153)

(6.19.12)

w∗(x, y, 1) = ... = k · curl∗τ∗ρsf

=τ0

ρsLfcurlτ + ... (154)

(6.19.13)

w1(x, y, 1) =

τ0ρsUDβ0L

(155)

(6.19.14)

M (y)s = ... =

τ0ρsUDβ0L

curlτ, (156)

(6.20.2a)

... +

(

r0r∗

)

1

cos θ

∂u

∂φ

]

= 0, (157)

(6.20.2d)

... = ...+Fr∗

ρ0U2Ωδ (158)

(6.20.24) ?>7@#

α = α(φ, θ) C [ E B α z Y E A

(6.20.23)∇Π C Q0 ) A ]_^

sin θ∂u

∂z· ∇Π = −k ×∇ρ · ∇Π+ α cos θk(j · [∇ρ×∇Π]) · ∇Π

= k · (∇ρ×∇Π) + α cos θ∂Π

∂z(j · [∇ρ×∇Π]) (159)

F (6.20.21) C z ∇Π C Q0 )

∂z(∇ρ×∇Πsin θ) · ∇Π = sin θ∇

∂ρ

∂z×∇Π · ∇Π+ sin θ∇ρ×∇

∂Π

∂z· ∇Π

= −∇ρ×∇Π · ∇

(

∂Π

∂zsin θ

)

+ (cos θj · [∇ρ×∇Π])∂Π

∂z(160)

SR ^ C Q0 ) A Cc (6.20.26) ?>7@(6.20.21)

j)k C Q7 )

w = α(∇ρ×∇Π) · k (161)

v = α(∇ρ×∇Π) · j (162)

23

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)ZY 2 α C Q0

(∇ρ×∇Π) · j = tan θ

(

∂ρ

∂z

∂2ρ

∂φ∂z−∂ρ

∂φ

∂2ρ

∂z2

)

= tan θ∂

∂z

(

∂ρ/∂φ

∂ρ/∂z

)

z

)2

= − tan θ∂

∂z

(

∂z

∂φ

)

ρ

(

∂ρ

∂z

)2

(163)

(6.21.1b)

u

cos θ

∂ρ

∂φ+ v

∂ρ

∂θ+ w

∂ρ

∂z= λ

∂2ρ

∂z2(164)

p444(6.21.5) V

W =We

p445 l13-l14

... in Chapter where it is ...(6.21.9) 2 ... by (6.21.1a,b), (6.21.5), and (6.21.7) ...

p444 l-3

... nothing in (6.21.1b) could balance ...(6.21.11b)

(u, v) =

(

D

δa

)2

(u, v) (165)

(6.21.12)

(

D

δa

)2 [ u

cos θ

∂ρ

∂φ+ v

∂ρ

∂θ+ w

∂ρ

∂z

]

=δDD

∂2ρ

∂z2, (166)

(6.21.14) ... independent of we and that

p447 l6

Since then λ = δD/δa ¿ 1, the obvious approximation to (6.21.1b) is(6.21.23)

w = (sin θ)−2[

k−1∂m

∂φ(ekz − 1) + · · · (167)

(6.21.25)

ekz [· · · ] + e2kz [· · · ] = 0 (168)

24

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(6.21.25) In order that (6.21.20) may truly be ...

p449 1 by (6.21.1a)]. Thus if we ...

p450 7 ..., ρ must equal ρs, hence ...(6.21.32)

C =

(

∂ρs/∂φ

λsin θ

)1/3

(169)

p450 9 C versus ∂ρs/∂φ is shown ...(6.21.33)

−4

27

w3eλ2

sin2 θ <∂ρs∂φ

< 0. (170)

(6.21.34)

C ∼we sin θ

λ(171)

(6.21.35) 3 <

u =

[

∂θ

(ρsk

)

+ z∂k

∂θ

ρsk

]

ekz

sin θ, (172)

p452 2 anomaly is connected to ρ ...(6.21.40)

Π = aρ+ b(p+ ρz) = sin θ∂ρ

∂z(173)

(6.21.41)

∂2ρ

∂z2=

1

sin θ

(

a∂ρ

∂z+ bz

∂ρ

∂z

)

= ... (174)

(6.21.44) ' 2 ..., or ρ→ 0 as z → −∞, but ...

C = −ρs...

(175)

(6.21.47)

25

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w = we =

−u

cos θ

∂ρs∂φ− v

∂ρs∂θ

∂ρs∂z

(176)

(6.21.48)

we =

∂ρs∂φ

∂θC sin θ − ...

...(177)

p456 1 ... nonlinear equations (6.20.7) make it ...

p457 6 ... it is W = UD/r0 whose magnitude ...(6.22.8)

(zn+1 − zn)∇HunH + ... (178)

where unH is the ...(6.22.25)

γn ≡ρn+1 − ρn

∆ρ(179)

(6.22.27)

∂φh2 = 2

sin2 θ

γ2we (180)

(6.22.39)

∂φ

(

h2 +γ1γ2h21

)

= ... (181)

(6.22.41)

h =

D20 +H22

1 +γ1γ2

(

1−f

f1

)2

1/2

(182)

p464 21 3 2 , ∂h1∂θ

= ... (183)

p467 7 26

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contours of h+ γ1/γ2h1) comes directly ...(6.23.6) ... a function of θ, then ...

... = −γ2

2 sin2 θwe(θ)... (184)

(6.23.7a) (6.23.7b) '

φE − φp = ... (185)

... The longitude φp marks the ...

p473 l11

... is so, (6.23.10) becomes ...

p476 l2 p476 l4... (6.23.23) may ... when used in (6.23.23) recovers ...

p476 l-10

... smaller and φp moves westward ...(6.24.5a)

... = −ρgDz + ... (186)

(6.24.12)

γ

ε=

r0

L2

U, (187)

p484 l7

... p(0) of ξ and η.(6.24.15)

(

∂τ+ ... + ... = −

γ

ε cos θ[...] (188)

(6.24.26)

... = −γ

ε

[

vTcos θ

sin θ+ ... (189)

(6.24.27)

A = ...+γ

ε

p(1)

cos θ sin θ... (190)

B = ...+ vTΠ−γ

ε

p(1)

cos θ sin θ... (191)

(6.24.28)

27

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∂τ

∫ ∫

A0

... (192)

(6.24.31)

wT = −

[

∂ρ(0)

∂T+ ... (193)

(6.24.32)

[

∂τ+ ... (194)

(6.24.33)

...+

(

uT∂

∂x+ vT

∂y

)[

∂2ψ

∂x2+∂2ψ

∂y2+ ...

]

(195)

1 Chapter 7

p506 l6

Now by (7.2.6), (6.5.3), and (6.5.13)(7.3.41)

... = 2∂E(φ)

∂t(196)

(7.4.14)

∂z

∫ 1

−1ρsφw1 dy = ... (197)

(7.5.3)

... = β

∫ zT

0

∫ 1

−1...dz = ... (198)

(7.5.6)

∫ zT

0

∫ 1

−1...dz = ...+

∫ 1

−1dyU0J(y, 0)

∂ηB∂y

(199)

(7.5.22)

...+

[

S∂ηB∂y−1

2

∂U0∂z

]

χ = 0 (200)

...−1

2

∂U0∂z

χ = 0 (201)

28

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p520(7.6.3)-(7.6.6) ) %2& C Q0 b A

Y E I E E (7.6.20)

ρs Sρs2

(7.7.32)

... = ρs|Φ(z)|2

2k∂α

∂zcos2 lnye

2kcit. (202)

(7.7.35)

... = ρskci2|c|2

e2kcit cos2 lny > 0 (203)

p532(7.8.3)

With (7.8.2) the normal ...(7.8.7)

... =β0

∂U∗/∂z∗

N2sf20D + ... (204)

(7.8.27)

limci→0

∫ zc+

zc−...dz = ... (205)

(7.8.28)

limci→0

∫ zc+

zc−...dz = ... (206)

(7.8.56)

ξ20

[

ξ02−

(

1 +2α1

2α1 − 1

)]

= 0 (207)

(7.10.4) ... (7.10.2a,b) by [Φ∗

1/(U1 − c)]d1 and [Φ∗2/(U2 − c)]d1 respectively. ...

2∑

n=1

∫ 1

−1dy

dΦn

dy+ ...

... (208)

(7.10.6)

... =2∑

n=1

dn

∫ 1

−1dy

[

dΦn

dy+ ...

]

... (209)

29

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(7.10.7)

...

(

k2 +π2

4

)

. (210)

(7.10.12)7 X (U2 − U1)

(U1 − U2)

(7.14.5b)

dz= 0, z = 0, zT (211)

(7.14.17)

χ2 CTVU (7.14.32)

... = ...+∆c

U0 − c0[...] = ... (212)

(7.14.46)

ρs CTVU (7.14.49) Now (7.14.48) implies ...

p591 3 Z3... K defined by (7.12.6), ... given by (7.12.9) which ...

p593 l-1

In deriving (7.16.7a,b) ...(7.16.8)

... = ...+r2

2

K2

k2U2s. (213)

!% CVU !E !! #"$ #%& +C' Q )(+*, ( -./0412354 O67)89 J: <; =?>A@B?67 = %

)DCEGFIH

( -./04123J4 O67)89! J:KL<; =I>@MBI67 = % ) N!OQPR( ST UV<W#" >JXY67 = % ) [Z\^] ( S!T UV_W#" >JXY6!7 = %

)a`b \c

( ST UVdW" >!X[67 = % ) efhgi ( ST UVdW">X67 = %)

j<kmln( ST UVdW " >X67 = % ) Iop+qr (

s16 t uJvxw yz

))|) ~

(!s

16 t! uv)w yz ) o ( S!T UV<W#" >JXY6!7 = %)

( ST UV_W " >X67 = % ) Y ( ST UV_W ">X67 = %)Y c F

( ST UV_W " >X67 = % ) [ 3 ( ST

30

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UVdW" >!X[67 = %) E

( ST UV5W " >!X[67 = % ) ( ST U)V5W" >X[67 = % ) ( ST UV5W" >X?67 = % ) k *H

( ST UV5W" >!X?67 = % ) ( 2 UU !U 67 )

!

31

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: ε =

U

fL(1.2.2)

: δ =

D

L(1.3.4)

: S =

N2D

f2L2=

LD

L(1.3.1)or(1.3.3)

: F =

f2L2

gD=

L2

R2(3.12.9)

: EV =

2AV

fD2; EH =

2AH

fD2(4.5.7)

!"$#&%(' : Re =

UL

AH

(4.6.2)

)*+, : N2 =

g

θs

dθs

dzor

g

ρs

dρs

dz(6.5.14)or(6.8.5)

-.0/1 (2: β = β0

L2

U(3.17.9)

3 *4.0/: p = ps(z) + ρsUf0Lp

′ (6.2.18)564.0/: ρ = ρs(z)(1 + εFρ′) (6.2.21)

7(8:

uvwρ′p′

=

u0v00ρ0p0

+ ε

u1v1w1ρ1p1

+ · · ·

9:;<:: v0 = −

∂p0

∂x; −u0 = −

∂p0

∂y(6.3.6)

=*>?<:: θ0 =

∂p0

∂zor ρ0 = −

∂p0

∂z(6.5.8)or(6.8.6)

: ε =

U

fL(1.2.2)

@ABC: δ =

D

L(1.3.4)

: S =

N2D

f2L2=

LD

L(1.3.1)or(1.3.3)

: F =

f2L2

gD=

L2

R2(3.12.9)

(D: EV =

2AV

fD2; EH =

2AH

fD2(4.5.7)

!"0#E%$' : Re =

UL

AH

(4.6.2)

)0*+, : N2 =

g

θs

dθs

dzor

g

ρs

dρs

dz(6.5.14)or(6.8.5)

-.0/1 $2: β = β0

L2

U(3.17.9)

3 *4.0/: p = ps(z) + ρsUf0Lp

′ (6.2.18)564.0/: ρ = ρs(z)(1 + εFρ′) (6.2.21)

7(8:

uvwρ′p′

=

u0v00ρ0p0

+ ε

u1v1w1ρ1p1

+ · · ·

9:;<:: v0 = −

∂p0

∂x; −u0 = −

∂p0

∂y(6.3.6)

=0*>?<:: θ0 =

∂p0

∂zor ρ0 = −

∂p0

∂z(6.5.8)or(6.8.6)

: ε =

U

fL(1.2.2)

: δ =

D

L(1.3.4)

: S =

N2D

f2L2=

LD

L(1.3.1)or(1.3.3)

: F =

f2L2

gD=

L2

R2(3.12.9)

: EV =

2AV

fD2; EH =

2AH

fD2(4.5.7)

!"$#&%(' : Re =

UL

AH

(4.6.2)

)*+, : N2 =

g

θs

dθs

dzor

g

ρs

dρs

dz(6.5.14)or(6.8.5)

-.0/1 (2: β = β0

L2

U(3.17.9)

3 *4.0/: p = ps(z) + ρsUf0Lp

′ (6.2.18)564.0/: ρ = ρs(z)(1 + εFρ′) (6.2.21)

7(8:

uvwρ′p′

=

u0v00ρ0p0

+ ε

u1v1w1ρ1p1

+ · · ·

9:;<:: v0 = −

∂p0

∂x; −u0 = −

∂p0

∂y(6.3.6)

=*>?<:: θ0 =

∂p0

∂zor ρ0 = −

∂p0

∂z(6.5.8)or(6.8.6)

: ε =

U

fL(1.2.2)

@ABC: δ =

D

L(1.3.4)

: S =

N2D

f2L2=

LD

L(1.3.1)or(1.3.3)

: F =

f2L2

gD=

L2

R2(3.12.9)

(D: EV =

2AV

fD2; EH =

2AH

fD2(4.5.7)

!"0#E%$' : Re =

UL

AH

(4.6.2)

)0*+, : N2 =

g

θs

dθs

dzor

g

ρs

dρs

dz(6.5.14)or(6.8.5)

-.0/1 $2: β = β0

L2

U(3.17.9)

3 *4.0/: p = ps(z) + ρsUf0Lp

′ (6.2.18)564.0/: ρ = ρs(z)(1 + εFρ′) (6.2.21)

7(8:

uvwρ′p′

=

u0v00ρ0p0

+ ε

u1v1w1ρ1p1

+ · · ·

9:;<:: v0 = −

∂p0

∂x; −u0 = −

∂p0

∂y(6.3.6)

=0*>?<:: θ0 =

∂p0

∂zor ρ0 = −

∂p0

∂z(6.5.8)or(6.8.6)