Geophysical Fluid Dynamics (Pedlosky, 1987) (6inaz/cgi-bin/Download.cgi/Pedlosky... · Geophysical...
Transcript of Geophysical Fluid Dynamics (Pedlosky, 1987) (6inaz/cgi-bin/Download.cgi/Pedlosky... · Geophysical...
Geophysical Fluid Dynamics (Pedlosky, 1987)
(6 )
( ! #"$!%'&)(* )+,
19 - 1 . 11 /
Chapter 10(1.4.13) 132546#798:<;>=@?
(Maxwell:<;>=@?
)(
∂S
∂p
)
T
= −
(
∂V
∂T
)
p
(1)
ACBEDU F<GIHKJMLONKPQ D V FSR9T D p FVU 7 D T FSWEX D H FYJZE[\N]@Q D S FYJZ^C_<]@Q D F = U − TS FO`aNcbadeNf :g3h JMLYN#PQ D G = F + pV = H − TS FP@ij :g3h J#LYNKPQlknmpoIk D
dU = TdS − pdV (2)
dH = TdS + V dp (3)
dF = −SdT + V dp (4)
dS =1
TdU +
p
TdV (5)
qKr#sYt 6M7<8Mu<vYw@xKypDVD<A3z y
T: G D 2 w@c:3|eav~\E sYt ~ D JZ)^_@]@Q S = S(p, T ) k\ sE k q DY99 :Y95
dS =
(
∂S
∂p
)
T
dp+
(
∂S
∂T
)
p
dT (6)
k q@s3t A>z@D ? (3)5 D
dH = TdS + V dp =
T
(
∂S
∂p
)
T
+ V
dp+ T
(
∂S
∂T
)
p
dT (7)
qrsV D(
∂H
∂p
)
T
= T
(
∂S
∂p
)
T
+ V (8)
(
∂H
∂T
)
p
= T
(
∂S
∂T
)
p
(9)
1
k v sYt H F 2 @ qKrKs k s k D ∂2H
∂T∂p=
∂2H
∂p∂T
v#: q D ?(8) k ?
(9)5 D
∂
∂T
T
(
∂S
∂p
)
T
+ V
=∂
∂p
T
(
∂S
∂T
)
p
(10)
~S D ?(1) F s3t
Chapter 20(2.2.12)
−
∮
C2Ω× u · dr = −2Ω
dAn
dt(11)
2.2.3
cy C
t
Chapter 3
3.6 K1 X : v⊥ v⊥ ≤ O(R−d−1)
q v@ k D T y q v@ t z>D d y "! t0(3.8.15)
u⊥ =∣
∣
∣uH −
u‖
KK∣
∣
∣(12)
p74 line2
σ/f > 1 F |σ/f | > 13w$#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
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
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
' @ 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
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
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
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 = −
dΘ
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 $&%
eiz cos θ =∞∑
n=−∞
inJn(z)einθ (64)
Jn(z) =i−n
π
∫ ∞
0eiz cos θ cos(nθ) dθ (65)
8
( ) n F +9 k s3t ASBEDeiz cos θ = 1−
1
2!z2 cos2 θ +
1
4!z4 cos4 θ −+i
(
z cos θ −1
3!z3 cos3 θ +
1
5!z5 cos5 θ−
)
(66)
2l ≥ n: k D
∫ π
0cos2l θ cosnθ dθ =
1
22l+1
∫ ∞
0
2(2l)!
(l + n)!n!dθ =
π
22l(2l)!
(l + n)!n!qr D2l < n
: k y T y 0qrs3t
n ,9: '!" t ( 1 )0(3.26.12)
B(Km,Kn) = ... (67)
p157 line 2
For (3.26.8)0(3.26.28)
ψ = ψ0(x, y, t, t) +1
βψ1(x, y, t, t) + · · · (68)
0(3.26.37)
Vj = (K2j + F )2a2j4
= (K2j + F )Ej (69)0(3.26.38)
d
dt(V1 + V2 + V3) = 0 (70)0
(3.26.40) 1the radius of gyration( ! ) k y D Q3Z^3k x: tK98 y 6 # s RE D , : :: ! q M sV k 2 t0(3.26.41) 9 y 99 t0(3.26.43) 1$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
(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
]
=α
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
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
0 = −E1/2V
2ε
∂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
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
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 $C1 0δ = O(4× 10−2) = O(ε) (96)
p352 T N <6Ky k :gYh X R/2 F z ' : q#rKsYt ez ~ D @ q y3/2R
D2 9 q y
5/2RD 9 q y
3R k v sYtp352
δ k ∆ 0 t0
(6.5.1)
ln θ∗ =1
γln p∗ − ln ρ∗ +
R
cpln p0∗ − lnR (97)
0(6.5.2)
...− εFρ+O(ε2F 2) (98)
p356 1 0 (6.5.10) 1M1CpT∗ = O(gD)
y D -Kv O(105m2/s2)
q D v O(104m2/s2)
t y DMKS
8(1 D H∗ ≤ O(U2f0) = O(10−2m2/s3)
t0(6.5.16)
...− S−1
(
∂u0∂z
∂θ0∂x
+∂v0∂z
∂θ0∂y
)
(99)
0(6.5.25) 0 (6.5.26)θ0y
(6.5.26)
Π∗ = ...+ εFθs
εξ∂θ
∂x+ εη
∂θ
∂y+ εδ cot θ0
∂θ
∂y
(100)
14
(6.5.29)
d0dtζ0 + βy+
F
(1/θs)(∂θs/∂z)
d0dt
∂θ0∂z
+ ... (101)
(6.9.1) ... given by (4.10.20), i.e.,(6.9.13) ... is on a scale L = O(100km) rather than La = O(103km), so that
1
gDε
d0dt
(
p∗aρs
)
= ... (102)
! !#"$%'&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
... = ...−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
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
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
4π
∫ 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)
dΦ
dz= 0 z = 1 (120)
(6.13.47)
H = ... (121)
18
(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
∂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
(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
(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
∆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
)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
)
(ρ
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
(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
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
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=
2Ω
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
∂
∂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
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
(7.10.7)
...
(
k2 +π2
4
)
. (210)
(7.10.12)7 X (U2 − U1)
(U1 − U2)
(7.14.5b)
dχ
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
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
: 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)