notes to geof331 - tidal dynamics
TRANSCRIPT
Notes to GEOF331 - Tidal dynamicsFall 2013
Helge DrangeGeophysical Institute, University of Bergen
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November 14, 2013
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Contents
1 Equilibrium theory, the direct method 31.1 Geometry of the Earth-Moon system . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Centre of mass of the Earth-Moon system . . . . . . . . . . . . . . . . . . . . . . 41.3 Newton’s law of universal gravitation . . . . . . . . . . . . . . . . . . . . . . . . . 41.4 Gravitational forces and accelerations in the Earth and Moon system . . . . . . . 41.5 Earth’s movement around the Earth-Moon center of mass . . . . . . . . . . . . . 51.6 Net gravity on the Earth caused by the Moon . . . . . . . . . . . . . . . . . . . . 51.7 Introducing the zenith angle and simplifying . . . . . . . . . . . . . . . . . . . . . 71.8 Radial and tangential components of the tidal acceleration . . . . . . . . . . . . . 7
1.8.1 Geometric approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.8.2 Using the definition of the dot and cross products . . . . . . . . . . . . . 81.8.3 The radial component can be ignored; the tangential component is important 9
1.9 Surface elevation caused by the Moon . . . . . . . . . . . . . . . . . . . . . . . . 91.9.1 Determining the integration constant for the equilibrium tide surface ele-
vation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
2 The combined surface elevation caused by the Moon and the Sun 11
3 The Moon’s monthly and daily periods 113.1 The Moon’s sidereal and synodic (monthly) periods . . . . . . . . . . . . . . . . . 113.2 The lunar day . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
4 Introducing latitude, hour and declination angles 144.1 The three leading lunar tidal components . . . . . . . . . . . . . . . . . . . . . . 144.2 Properties of the three leading lunar components . . . . . . . . . . . . . . . . . . 16
4.2.1 Declination or nodal component, ζ1 . . . . . . . . . . . . . . . . . . . . . 174.2.2 Diurnal component, ζ2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174.2.3 Semi-diurnal component, ζ3 . . . . . . . . . . . . . . . . . . . . . . . . . . 17
5 Decomposing the solar and lunar tides into a series of simple harmonic con-stituents 175.1 Geometry of the Earth-Sun system . . . . . . . . . . . . . . . . . . . . . . . . . . 17
5.1.1 Angular speed of the Earth and the Sun . . . . . . . . . . . . . . . . . . . 195.1.2 Introducing δ, η and ω . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195.1.3 Expressing ζ2 in terms of simple harmonics . . . . . . . . . . . . . . . . . 205.1.4 Expressing ζ3 in terms of simple harmonics . . . . . . . . . . . . . . . . . 21
5.2 Moon’s leading tidal constituents . . . . . . . . . . . . . . . . . . . . . . . . . . . 225.3 Doodson’s system for labelling tidal constituents . . . . . . . . . . . . . . . . . . 22
6 Tidal forcing in the momentum equation 23
7 Laplace’s tidal equations 24
8 The tidal force expressed in terms of the tidal potential 248.1 The potential of a conservative force . . . . . . . . . . . . . . . . . . . . . . . . . 248.2 Defining the tidal potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
8.2.1 Gradient of the factor q/q3 . . . . . . . . . . . . . . . . . . . . . . . . . . 258.2.2 Gradient of the factor R/R3 . . . . . . . . . . . . . . . . . . . . . . . . . 26
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8.2.3 Resulting tidal potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . 268.3 Tidal potential expressed in terms of the zenith angle . . . . . . . . . . . . . . . 27
8.3.1 Radial and tangential tidal forces per unit mass . . . . . . . . . . . . . . . 27
9 Introducing the tidal potential in the primitive momentum equations 28
A Some key parameters of the Earth, Moon, Sun system 29
B Spherical coordinates 29B.1 Two commonly used spherical coordinate systems . . . . . . . . . . . . . . . . . . 29B.2 Volume and surface elements in spherical coordinates . . . . . . . . . . . . . . . . 30
B.2.1 Spherical volume elements . . . . . . . . . . . . . . . . . . . . . . . . . . . 30B.2.2 Spherical surface elements . . . . . . . . . . . . . . . . . . . . . . . . . . . 31B.2.3 Relationship with the Earth-Moon system . . . . . . . . . . . . . . . . . . 31B.2.4 Gradient operator in spherical coordinates . . . . . . . . . . . . . . . . . . 31
C The three lunar tidal components 32
D Binominal theorem for rational exponents 33
E Some identities 33
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1 Equilibrium theory, the direct method
A key objective in ocean tidal dynamics and analysis is to derive an analytical form of thegravitational attraction on the global ocean caused by the presence of the Moon and the Sun.Once this force is found, it can be added as a forcing term to the momentum equations which,together with the continuity equation, model the full 2- and 3-dimensional flow of the globalocean tide.
The derivation of the equilibrium tide and subsequent analysis are addressed in this note. Adescription of the basic tidal wave characteristics is given in an accompanying note. Modellingof the full 2- and 3-dimensional dynamics of the ocean tide is beyond the scope of the presentedlecture notes (but might be added at a later stage).
In the following sections, the gravitational force is derived based on what one may phrase thedirect method. An alternative approach – commonly used in tidal dynamics – derives the gravi-tational force from the tidal potential. The latter approach, which to a large extent follows thederivation of the direct method, is described in Sec. 8.
1.1 Geometry of the Earth-Moon system
The configuration of the Earth-Moon system used for deriving the properties of the tidal equi-librium is displayed in Figure 1. It follows from the figure that
r + q = R (1)
Figure 1: Illustration of the Earth-Moon system with the Earth to the left and the Moon to theright (figure greatly out of scale). O, P and L are the centre of the Earth, an arbitrary pointon Earths surface and the centers of the Moon, respectively. r is the Earth’s radius vector (frompoint O to P ), R is the position vector from the centre of the Earth to Moon’s centre (from Oto L), and q is the position vector from an arbitrary point P on Earth’s surface to L. The linebetween O and L is sometimes called the center line and the angle θ the zenith angle or thecenter angle.
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1.2 Centre of mass of the Earth-Moon system
The centre of mass of the Earth-Moon system is located along the center line OP at a distancexR (0 < x < 1) from point O (Fig. 1). We then get that
MT xR = ML (1− x)R (2)
or
x =MT
ML +MT≈ 0.012 (3)
Here ML and MT are the mass of Moon and Earth, respectively, see appendix A. With meanvalues of r and R (appendix A), we get that
x ≈ 0.73 r (4)
implying that the centre of mass of the Earth-Moon system is located about one quarter ofEarth’s radius from the surface of the Earth.
1.3 Newton’s law of universal gravitation
Newton’s law of universal gravitation states that an attractive force F is set up between any twopoint masses, varying proportional with the product of the masses (m1 and m2) and inverselyproportional with the distance l between the masses
F = Gm1m2
l2(5)
Here G = 6.674× 10−11 N m2 kg−2 is the gravitational constant.
1.4 Gravitational forces and accelerations in the Earth and Moon sys-tem
The gravitational force at the Earth’s centre because of the presence of the Moon, FTL, is
FTL = GMLMT
R2
R
R(6)
where R/R is the unit vector along the center line from Earth to Moon.
According to Newton’s second law, this force leads to an acceleration at the center of Earth
aO =FTLMT
= GML
R2
R
R(7)
Similarly, the gravitational acceleration at point P caused by the Moon is
aP = GML
q2
q
q(8)
At point P, there is also a gravitational acceleration g towards the center of the Earth caused byEarth’s mass:
g = −G MT
r2
r
r(9)
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By inserting the numerical values of G, MT and r (appendix A), one obtains g = 9.8 m s−2, asexpected. Furthermore, (9) gives the relationship
G = gr2
MT(10)
1.5 Earth’s movement around the Earth-Moon center of mass
To maintain the Earth-Moon center of mass at 0.73 r, the Earth can either rotate around thejoint center of mass as a solid body, or the Earth can adjust it’s position around the commoncenter without rotation. The former would imply that only one side of the Earth would face theMoon at all times or, alternatively, that the Earth would make one full rotation for each fullpassing of the Moon (approximately every 28 days). This is not the case; the Moon can be viewedfrom all longitudes during the course of a lunar cycle and the Earth does not experience a lunar(near monthly) season. Therefore, the latter movement takes place as explained in Fig. 2.
This implies that every point on solid Earth describes a circular motion with radius s = 0.73 r,but with different centers, and with the rotation rate ω governed by the rotation rate of the Moonaround the Earth. Therefore, each point on Earth, whether on or below Earth’s surface, willexperience a centrifugal acceleration (i.e., outward-directed acceleration relative to the rotationof the Earth-Moon system) given by (M & P (2008), eqn. 6.28)
aω = ω2 s (11)
1.6 Net gravity on the Earth caused by the Moon
The distance between the centers of the Earth and the Moon vary slowly during the lunar month,but it can be treated as constant for the purposes considered here. If so, the magnitude of theoutward-directed centrifugal acceleration at the center of the Earth has to exactly balance themagnitude of the Moon’s gravitational pull at the Earth’s center. This means that
aω = aO (12)
or
ω2 s = GML
R2(13)
At the point P at the surface of the Earth, the gravity caused by the Moon will vary according tothe distance to the Moon, with the largest gravitational pull for the points closest to the Moon.This pull is directed along the q-vector. At the same time, the centrifugal acceleration in pointP is directed opposite to the R-vector. The net acceleration felt at P can then be expressedas
a = GML
q2
q
q− w2 s
R
R(14)
or, by means of (13),
a = GML
(q
q3− R
R3
)(15)
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Figure 2: Illustration of the movement of the Earth (Earth’s circumference in blue) aroundthe Earth-Moon center of mass (red dot), looking from the north down onto the Earth-Moonsystem. Starting with the upper, left panel, the Moon is located to the right of the Earth, in thedirection of the gray arrow. The black dots show the Earth’s center and a point on the centralline, opposite to the direction of the Moon. The red circle shows the path the Earth’s centerwould follow if it rotates around the joint center of mass. Some time later (upper, right panel),the Moon has moved counter-clockwise relative to the Earth. The Earth’s center (orange dot) islocated on the circle around the joint center of mass. Similarly, the point initially marked withthe black dot on the opposite side of the Moon describes the same circular revolution as Earth’scenter (the left-most yellow dot, with the path traced out by the movement shown by the reddotted arc). Later (lower, left panel), the Earth’s center (green dot) continue to follow the red,solid circle. The other point follows a similar circular trajectory (the second green dot and thered dotted arc). When the Moon is on the opposite side of the Earth (lower right panel), bothof the two initial points are tracing out circles with a common radius s = 0.73 r. This circularmovement will continue as the Moon rotates around the Earth. In this way, all points on andwithin the solid Earth will describe circular trajectories with radius s = 0.73 r. Note that theEarth does not rotate as a solid body around the Earth-Moon center of mass. Rather, any pointon Earth, like the left-most black dot in the upper left panel, have the same orientation withrespect to a fixed star throughout the movement. This movement is commonly described asrevolution without rotation.
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1.7 Introducing the zenith angle and simplifying
The acceleration (15) can be expressed in terms of r, R and θ. From Fig. 1, the law of cosinesgives
q2 = R2 + r2 − 2Rr cos θ (16)
or
q =(R2 + r2 − 2Rr cos θ
)1/2(17)
= R
(1− 2
r
Rcos θ +
r2
R2
)1/2
(18)
≈ R(
1− 2r
Rcos θ
)1/2
(19)
≈ R(
1− r
Rcos θ
)(20)
Here the smallness of r/R (appendix A) has been used in the second last expression, and thebinominal theorem for rational exponents (168) has been used in the last expression. Conse-quently,
1
q3=
1
R3
(1− r
Rcos θ
)−3
≈ 1
R3
(1 + 3
r
Rcos θ
)(21)
and
a =GML
R3
[(1 + 3
r
Rcos θ
)q−R
](22)
Expression (1) and the smallness of r/R give
a =GML r
R3
(3 cos θ
R
R− r
r
)(23)
Substituting G with Earth’s gravitational acceleration g from (10) yields
a = gML
MT
( rR
)3(
3 cos θR
R− r
r
)(24)
With the mass and distance ratios given in appendix A, it follows that
a ≈ 10−7 g (25)
The tidal acceleration on Earth caused by the presence of the Moon is therefore very small. Bydecomposing the tidal acceleration into one component in the direction of r and one componenttangential to the surface of the Earth, it follows that the former indeed can be ignored. Thelatter has no counterpart and it is this component that gives rise to the tides.
1.8 Radial and tangential components of the tidal acceleration
1.8.1 Geometric approach
The component of the tidal acceleration in the direction of R follows directly from (24)
aR = 3gML
MT
( rR
)3
cos θ (26)
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Figure 3: As Figure 1, but for decomposing aR (see expression 26) in the outward radial direction(ar) and in the horizontal direction pointing towards the Moon (ah).
aR can be decomposed in the radial and horizontal directions based on Figure 3.
It follows from the figure that the horizontal component of aR, in the direction towards theMoon, is
ah = 3 gML
MT
( rR
)3
sin θ cos θ (27)
Since
sin θ cos θ =1
2sin 2θ (28)
expression (27) can alternatively be expressed as
ah =3
2gML
MT
( rR
)3
sin 2θ (29)
Likewise, the radial component of aR is
3 gML
MT
( rR
)3
cos2 θ (30)
The latter, together with the radial component of (24), adds up to
ar = gML
MT
( rR
)3
(3 cos2 θ − 1) (31)
ar is directed radially outward, in the direction of r in Fig. 1.
1.8.2 Using the definition of the dot and cross products
Alternatively, ar can be obtained by taking the dot product of a and the unit vector r/r:
ar = a · rr
= gML
MT
( rR
)3(
3 cos θR · rRr
− r · rr2
)(32)
The definition of the dot product gives
R · r = Rr cos θ and r · r = r2 (33)
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so (32) becomes
ar = gML
MT
( rR
)3 (3 cos2 θ − 1
)(34)
which is identical to (31).
In a similar manner, the cross product of a with the unit vector r/r gives the horizontal compo-nent of a:
ah =∣∣∣a× r
r
∣∣∣ = gML
MT
( rR
)3(
3|R× r|Rr
cos θ − |r× r|r2
)(35)
The definition of the cross product gives
|R× r| = Rr sin θ and r× r = 0 (36)
so
ah = 3gML
MT
( rR
)3
sin θ cos θ (37)
1.8.3 The radial component can be ignored; the tangential component is impor-tant
Since the gravity on Earth’s surface caused by the Earth’s mass g is orders of magnitude largerthan the gravitational acceleration in the radial direction (see 25), ar can safely be neglectedcompared to g. Since g has no counterpart to ah, this acceleration component – aligned tangentialto the surface of the Earth – cannot be ignored. It is ah that gives rise to the tides on Earth.
Note that ah only varies with θ, the zenith or central angle in Fig. 1. θ depends on the latitudeand longitude of the position P , as well as the declination angle of the Moon (and the Sun).Introduction of these geometric factors are presented in Sec. 4.
The horizontal component of the tidal acceleration or tidal force – which is the actual tidalacceleration or force – is often named the tractive acceleration or force.
1.9 Surface elevation caused by the Moon
The tidal acceleration ah will give rise to changes in the sea level. This again leads to a pressuregradient or, alternatively, a pressure force. The pressure force per unit mass (i.e., acceleration)is, per definition (M & P (2008), eqn. 6.6)
−1
ρ∇p (38)
Here p is pressure and ρ is density.
For a fluid with uniform (constant) density, the hydrostatic approximation gives
p = ρgξ (39)
Here ξ is the elevation compared to a flat ocean. This means that the pressure force per unitmass can be expressed as
−g∇ξ (40)
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It follows then from Newton’s second law that
−g∇hξ = ah (41)
where ∇h denotes the gradient operator tangential to the Earth’s surface.
It is convenient to express the differential equation (41) in terms of the spherical coordinatesystem to the right in Fig. 9. In this case the z axis of the spherical coordinate system is orientedin the direction of R in Fig. 1. In this system, only the eθ component (see 157) has a contribution,so (41) becomes
−gr
∂ξ
∂θ= ah (42)
or, by means of (29),∂ξ
∂θ+
3
2
ML
MT
r4
R3sin 2θ = 0 (43)
Integration over θ, using∫
sin 2θ dθ = −(1/2) cos 2θ, gives the equilibrium tide’s surface eleva-tion
ξ =3
4
ML
MT
r4
R3cos 2θ + C (44)
where C is the integration constant.
1.9.1 Determining the integration constant for the equilibrium tide surface eleva-tion
Conservation of water volume requires that the surface elevation ξ must vanish when integratedover the sphere. This constraint can be used to determine the integration constant C by inte-grating (44) over a sphere with constant radius r by means of the zenith-angle coordinate systemin the right panel of Fig. 9.
The first term on the right-hand-side of (44) is a constant multiplied with the factor cos 2θ.Integration over the sphere of the given term therefore corresponds to solving the double-integral(similarly to 155)
r2
∫ 2π
Ψ=0
dΨ
∫ π
θ=0
cos 2θ sin θ dθ (45)
The integral involving θ can be solved by using the identity∫cos 2x sinx dx =
cosx
2− cos 3x
6+ C∗ (46)
where C∗ is an integration constant. Expression (45) then becomes
r22π
[cos θ
2− cos 3θ
6
]π0
= −r2 4π
3(47)
Thus, integration of (44) over a sphere with constant radius r leads to
3
4
ML
MT
r4
R3
(−r2 4π
3
)+ 4r2πC = 0 (48)
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Consequently,
C =1
4
ML
MT
r4
R3(49)
The equilibrium tide is therefore governed by the expression
ξ =1
4
ML
MT
r4
R3(3 cos 2θ + 1) (50)
Alternatively, sincecos 2θ = 2 cos2 θ − 1 (51)
the equilibrium tide may be put in the form
ξ =3
2
ML
MT
r4
R3
(cos2 θ − 1
3
)(52)
2 The combined surface elevation caused by the Moon andthe Sun
Expression (52) is also valid for the Sun, but then with the solar quantities. If the subscripts Land S denote the Moon and the Sun, respectively, we get that the total surface tidal elevationξt becomes
ξt =3
2
ML
MT
r4
R3L
(cos2 θL −
1
3
)+
3
2
MS
MT
r4
R3S
(cos2 θS −
1
3
)
=3
2
r4
MT
[ML
R3L
(cos2 θL −
1
3
)+MS
R3S
(cos2 θS −
1
3
)](53)
3 The Moon’s monthly and daily periods
3.1 The Moon’s sidereal and synodic (monthly) periods
The lunar orbital period depends on the frame of reference. Seen from a fixed star (far away),the lunar period is known as the sidereal period (from “sidus”, or “star” in Latin). Seen from theEarth, the lunar period is known as the synodic period (from “synodus”, Greek from “comingtogether”).
A full rotation of the Moon around the Earth as seen from a distant star gives a sidereal periodTL = 27.32 d. The synodic period can be found by considering the Sun-Earth-Moon system seenfrom a fixed star from above (north), as illustrated in Fig. 4.
If the Earth were not rotating around the Sun, the Moon’s rotation rate as seen from the Earthwould equal
ωL TL = 2π (54)
Since the Earth does rotate around the Sun with the rotation rate ωTS and rotation periodTTS = 365.25 d, the Moon will not be back at the initial position – seen from the Earth – after
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Figure 4: Illustration of the Sun (black circle in the center), the Earth (solid blue dot, encirclingthe Sun along the blue circle), and the Moon (solid red dot, encircling the Earth along the redcircle), viewed from the north. ωL and TL are the lunar rotation rate and period seen from adistant star, respectively. Similarly, ωTS and TTS are the Earth’s rotation rate and rotationperiod around the Sun. Point A denotes the initial position of the Moon, in this case exactly onthe center line between the Earth and the Sun. Point B refers to the position of the Moon afterthe sidereal period TL. Point C denotes the full lunar rotation as observed from the Earth, inthis case when the Moon is back on the center line between the Earth and the Sun. The rotationperiod for the latter is the synodic period.
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the sidereal period TL. Actually, the Moon needs to move from point B to point C in Fig. 4 inorder to be back at the initial position A, seen from the Earth. We are therefore searching for aperiod T ∗ > TL in order for the Moon to get to position C in Fig. 4.
Based on (54), T ∗ can be derived from the expression
ωL T∗ = 2π + ωTS T
∗ (55)
where the last term corresponds to the angle between the points A, B and the center of the Sunor, which is equivalent, the angle between the points B, C and the center of the Earth in Fig. 4.Division by the factor 2π T ∗ gives
1
TL=
1
T ∗ +1
TTS(56)
For TL = 27.32 d and TTS = 365.25 d one obtains that the synodic period
T ∗ = 29.53 d (57)
3.2 The lunar day
Following the logic from section 3.1, the lunar day T ∗ seen from Earth – due to the Earth’srotation around it’s own axis – will be somewhat longer than that seen from a fixed star. Thesituation is illustrated in Fig. 5.
Figure 5: Illustration of the lunar day. The Moon is the filled red dot encircling the Earth alongthe red circle, viewed from above (north). The Earth is the blue circle. If the Moon was fixedin space and if it was located above point A initially, the Moon would be located in the sameposition after one full rotation of the Earth (TT = 1 d). However, during the period TT , theMoon has moved with the rotation rate ωL. The Earth must therefore do an additional rotation,corresponding to move point A to position B, before the Moon is again located in zenith relativeto the initial point A.
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One full rotation of the Earth can be expressed as
ωT TT = 2π (58)
where TT = 1 d.
Taking into account that simultaneous rotation of the Moon around the Earth, gives
ωT T∗ = 2π + ωL T
∗ (59)
where T ∗ is the lunar day seen from the Earth.
Division by 2π T ∗ gives1
TT=
1
T ∗ +1
TL(60)
With TL = 27.32 d and TT = 1 d, we obtain
T ∗ = 24hr50min (61)
The lunar day observed from the Earth is therefore 50 min longer than the day on Earth.
4 Introducing latitude, hour and declination angles
Up to now, no specific geographic reference has been given beyond the zenith angle shown inthe left panel of Fig. 6. To determine the tide at any point P on the Earth’s surface relativethe the Moon or the Sun (or other celestial bodies), it is convenient to introduce three angles:The latitude ϕ of P , the hour angle (playing the role of longitude) Ψ of P , and the declinationangle δ of the celestial body, see the right panel in Fig. 6 for the exact definition of the threeangles.
4.1 The three leading lunar tidal components
We start by considering the two-body Earth-Moon system as illustrated in Fig. 6.
The two position vectors r and R can be expressed in terms of ϕ, Ψ and δ. The projection of Ronto the x- and z-axes give R cos δ and R sin δ, respectively, see left panel of Fig. 7. Thus,
R = R (cos δ, 0, sin δ) (62)
Likewise, the projection of r onto the equatorial plane gives a vector of length r cosϕ (rightpanel of Fig. 7). Decomposing this vector onto the x and y axes gives
r = r (cosϕ cos Ψ, cosϕ sin Ψ, sinϕ) (63)
Expression (52) states that the surface elevation of the equilibrium tide ξ is proportional to thefactor cos2 θ − 1/3. The task is therefore to express cos θ in terms of ϕ, Ψ and δ. This can bereadily done by means of the dot product
r ·R = rR cos θ
= rR (cosϕ cos Ψ cos δ + sinϕ sin δ) (64)
15
Figure 6: The tidal equilibrium configuration defined by R, r and θ (left) and the new anglesϕ, Ψ of and δ (right). x, y, z denotes the Cartesian coordinates with x, y spanning put Earth’sequatorial plane and z pointing northward. The x-axis is aligned along the projection of thecenter line (or vector R) onto the equatorial plane. O is the center of the Earth, L is the Moon,and P is any arbitrary point on the surface of the Earth with latitude ϕ measured from theequatorial plane. Point V is where the center line crosses the surface of the Earth (or the pointon the surface of the Earth under the Moon) with the declination angle δ. The hour angle Ψ isthe angle on the equatorial plane between the x-axis and the longitude of point P .
Figure 7: As Fig. 6, illustrating of the decomposition of R (left panel) and r (right panel) ontothe Cartesian x, y, z-system.
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From the above relationship,
cos θ = cosϕ cos Ψ cos δ + sinϕ sin δ (65)
and
cos2 θ − 1
3= cos2 ϕ cos2 Ψ cos2 δ + 2 cosϕ cos Ψ cos δ sinϕ sin δ + sin2 ϕ sin2 δ − 1
3(66)
The identitysin(2a) = 2 sin a cos a (67)
can be applied to the ϕ- and δ-factors in the second term on the right hand side of (66), resultingin
cos2 θ − 1
3= cos2 ϕ cos2 Ψ cos2 δ +
1
2sin(2ϕ) cos Ψ sin(2δ) + sin2 ϕ sin2 δ − 1
3(68)
The right hand side of (68) contains terms with variations on three, main time scales. Keepingthe latitude of the position P fixed so ϕ = const., the first term on the right hand side of (68)varies mainly because of cos2 Ψ. This term gives rise of the semi-diurnal lunar tide. The secondterm on the right hand side of (68) varies mainly because of cos Ψ, describing the diurnal lunartide. And finally, the third term vary with sin2 δ, describing the tidal response to the nearbiweekly variations in the lunar declination.
The split into the three time scales can be facilitated by expressing
cos2 θ − 1
3= ζ1 + ζ2 + ζ3 (69)
where
ζ1 =3
2
(sin2 ϕ− 1
3
)(sin2 δ − 1
3
)(70)
ζ2 =1
2sin(2ϕ) cos Ψ sin(2δ) (71)
ζ3 =1
2cos2 ϕ cos(2Ψ) cos2 δ (72)
Here, ζ1 describes the biweekly variations, ζ2 describes the diurnal variations, and ζ3 describes thesemi-diurnal variations. The transformation of (68) into (69)–(72) is given in appendix C.
4.2 Properties of the three leading lunar components
From (70)–(72), it follows that all components depend on the latitude ϕ. Secondly, the identi-ties
cos2 δ =1
2(1 + cos 2δ) (73)
and
sin2 δ =1
2(1− cos 2δ) (74)
show that all contributions depend on cos 2δ, implying that all contributions have a periodicityof TN ∼ 14 d. And thirdly, only ζ2 and ζ3 depend on the hour angle Ψ.
Other characteristics are listed below.
17
4.2.1 Declination or nodal component, ζ1
• ζ1 = 0 for sin2 ϕ = 1/2, or for ϕ = ±35◦
• Since the Moon’s declination angle δ varies between 18.5–28.5◦, sin2 δ−1/3 does not vanish
4.2.2 Diurnal component, ζ2
• Vary with Ψ with the lunar day with period 24h50min
• sin 2δ = 0 twice a month, so ζ2 vanishes twice a month
• δ-modulation largest when the Moon is high on sky; 18.5◦ < δ < 28.5◦ with a period of18.6 yr
• Largest at ϕ = ±45◦, vanishes at equator and the poles
4.2.3 Semi-diurnal component, ζ3
• Vary with Ψ with half of the lunar day with period 12h25min
• δ-modulation decreases with increasing δ; min(cos2 δ) ≈ 0.77 for δ = 28.5◦
• Largest for ϕ = 0 (equator), vanishes at the poles
5 Decomposing the solar and lunar tides into a series ofsimple harmonic constituents
The total tide can be expressed as a sum of many harmonic contributions. Each of these con-tributions – called tidal component, tidal constituent or harmonic constituent – is representedby a simple, harmonic cosine function. A capital letter (sometimes a combination of capitaland small letters), plus a numerical subscript, is used to designate each constituent. For in-stance, the semi-diurnal tidal contribution from the Moon and the Sun are named M2 and S2,respectively.
Each constituent is typically described by its speed (or frequency) expressed as degrees per solarhour (hereafter deg/hr). The speed of a constituent is 360◦/T , where T (hr) is the period. ForS2, the speed is then 360◦/12 hr = 30◦/ hr.
5.1 Geometry of the Earth-Sun system
The Earth-Sun system, viewed from the Earth (i.e., a geocentric view), can be presented bymeans of the declination angle δ, the ecliptic plane angle η and the longitude angle ω as shownin Fig. 8. From this figure, the following relationships are obtained
h = R sin δ (75)
h = d sin η (76)
d = R sinω (77)
18
Figure 8: A geocentric view of the Earth-Solar system. R is the solar position vector, describinga declination angle δ above Earth’s equatorial plane. The Sun’s path on sky follows the ecliptic,with η describing the angle between Earth’s equatorial plane and the ecliptic. The third angleω describes the Sun’s longitudinal angle with respect to the vernal equinox � (the point wherethe ascending Sun crosses the equatorial plane). The vernal equinox � is also known as the firstpoint of Aries.
19
The first two expressions give
sin δ =d
Rsin η (78)
Combined with the third expression, one obtains
sin δ = sinω sin η (79)
5.1.1 Angular speed of the Earth and the Sun
For the geocentric view in Fig. 8, the longitude angle ω has two contributions:
• The angular speed of the Earth relative to the Sun (as seen from the Earth):
360◦
24 hr= 15
deg
hr(80)
Note that this speed includes the slow, annual rotation of the Sun around the Earth.
• The angular speed of the Sun relative to the vernal equinox �:
360◦
365.24 d= 0.041
deg
hr(81)
From a geocentric view, ω increases with time t according to
ω = 0.041deg
hr· t def
= ε t (82)
where ε = 0.041 deg/hr.
In a similar manner, the longitudinal angle Ψ in the right panel of Fig. 7 increases with time:
Ψ = 15deg
hr· t def
=(γ − ε) t (83)
where γ = 15.041 deg/hr. γ represents the rotation speed of the Earth seen from a distantstar.
5.1.2 Introducing δ, η and ω
Both ζ2 (71) and ζ3 (72) include a harmonic term with argument 2δ. This term can be expressedby means of sin δ;
sin 2δ = 2 sin δ cos δ = 2 sin δ√
1− sin2 δ (84)
where the identities (170) and (171) have been used.
The declination angle |δ| < 23.5◦, so | sin δ| is a rather small factor. The first order approximationof the binominal theorem (168) can then be applied to (84)
sin 2δ ≈ 2 sin δ
(1− 1
2sin2 δ
)(85)
20
5.1.3 Expressing ζ2 in terms of simple harmonics
With the above definitions and simplifications, (71) can be expressed by means of ϕ, δ, η, ω andt:
ζ2 =1
2sin(2ϕ) sin 2δ cos Ψ
(85)= sin(2ϕ) sin δ
(1− 1
2sin2 δ
)cos Ψ
(79)= sin(2ϕ) sin η
(sinω − 1
2sin2 η sin3 ω
)cos Ψ
(172)= sin(2ϕ) sin η
[(1− 3
8sin2 η
)sinω +
1
8sin2 η sin 3ω
]cos Ψ
= sin(2ϕ) sin η
[(1− 3
8sin2 η
)sin εt+
1
8sin2 η sin 3εt
]cos(γ − ε)t (86)
The numbers above the equal sign show the expressions used. For the last equality, the angularspeeds from section 5.1.1 have been introduced.
For η = 23.5◦, the 1/8-term is small (about 2 percent contribution), so (86) can be expressedas
ζ2 = sin(2ϕ) sin η
(1− 3
8sin2 η
)sin εt cos(γ − ε)t
= C2 sin εt cos(γ − ε)t (87)
where C2 includes the time-independent contribution to ζ2. (87) states that the temporal vari-ation of ζ2 is governed by the basic speed γ − ε = 15 deg/hr, modulated by the (slow) speedε = 0.041 deg/hr.
The identity
sin a− sin b = 2 sin
(1
2(a− b)
)cos
(1
2(a+ b)
)(88)
can be used to split (87) into two simple harmonics. With ε = (a− b)/2 and γ − ε = (a+ b)/2,we get that a = γ and b = γ − 2ε. Thus,
ζ2 =C2
2(sin γt− sin(γ − 2ε)t) (89)
Expression (89) can alternatively be put in the form
ζ2 =C2
2(sin((γ − ε) + ε)t− sin((γ − ε)− ε)t) (90)
The Sun’s diurnal contribution has therefore speeds γ and γ − 2ε (from 89) symmetrically dis-tributed about the basic speed γ − ε (expressions 87 and 90).
In summary, ζ2 gives rise to two tidal constituents, originating from the Sun’s half-yearly variationof the declination angle. The two constituents are called the solar declination constituents:
K1 Declinational diurnal constituent (will also get a contribution from the Moon, see Sec. ??);it’s speed is γ = 15.041 deg/hr and period is 23.93 hr
P1 Principal (main) solar declinational diurnal constituent; it’s speed is γ − 2ε = 14.959 deg/hrand period is 24.07 hr
21
5.1.4 Expressing ζ3 in terms of simple harmonics
ζ3 can rewritten as
ζ3 =1
2cos2 ϕ cos2 δ cos 2Ψ
(171)=
1
2cos2 ϕ (1− sin2 δ) cos 2Ψ
(79)=
1
2cos2 ϕ (1− sin2 η sin2 ω) cos 2Ψ
(175)=
1
2cos2 ϕ
[1− 1
2sin2 η (1− cos 2ω)
]cos 2Ψ (91)
If (91) is split in one part with and one part without ω-dependency, one obtains
ζ3 =1
2cos2 ϕ
[(1− 1
2sin2 η
)cos 2Ψ +
1
2sin2 η cos 2ω cos 2Ψ
]=
1
2cos2 ϕ
[(1− 1
2sin2 η
)cos(2(γ − ε)t) +
1
2sin2 η cos 2εt cos(2(γ − ε)t)
](92)
For the last equality, the angular speeds from section 5.1.1 have been introduced.
Similarly to ζ2, the cos 2εt cos(2(γ− ε)t) factor consists of a temporal variation with basic speed2(γ − ε), slowly modulated by variation with speed 2ε. These two temporal variations can besplit into two simple harmonics by means of the identity
cos a+ cos b = 2 cos
(1
2(a+ b)
)cos
(1
2(a− b)
)(93)
For (a − b)/2 = 2ε and (a + b)/2 = 2(γ − ε), one obtains a = 2γ and b = 2(γ − 2ε). Conse-quently,
ζ3 =1
2cos2 ϕ
[(1− 1
2sin2 η
)cos(2(γ − ε)t)
+1
4sin2 η cos 2γt
+1
4sin2 η cos(2(γ − 2ε)t)
](94)
In summary, ζ3 gives rise to three semi-diurnal tidal constituents, originating from Earth’s dailyrotation:
S2 is the principal (main) solar semi-diurnal tide with speed 2(γ − ε) = 30 deg/hr and a periodof 12 hr.
K2 is the semi-diurnal declination tide from the Sun and Moon (the latter contribution is givenin Sec. 5.2), often called the lunisolar semi-diurnal constituent. It’s peed is 2γ = 30.082deg/hr and the period is 11.97 hr
The third constituent is the second semi-diurnal solar declination tide. It’s speed is 2(γ −2ε) = 29.918 deg/hr and the period is 12.03 hr
Since |η| < 23.5◦, | sin2 η| < 0.16, so the 2γ and 2(γ−2ε) constituents in (94) are small comparedto S2. But since K2 gets a similar contribution from the Moon, this constituent is larger thanthe third constituent. One may therefore neglect the third component, but not K2.
22
5.2 Moon’s leading tidal constituents
Moon’s declination varies from 18.4◦ to 28.6◦ in 18.6 years, but can be considered constant forone month. Moon’s tidal constituents follow then those of the Sun.
A full rotation of the Moon around the Earth as seen from a distant star gives the Moon’s siderealperiod TL = 27.32 d (see Sec. 3.1). The Moon’s angular speed s is then
s =360◦
27.32 d= 0.549
deg
hr(95)
The relative hour angle follows that of the Sun by substituting s for ε in (83)
γ − s = 14.492deg
hr(96)
where γ = 15.041 deg/hr is the sidereal rotation speed of the Earth. Furthermore, the dailylunar period is 24h50min, and the half-daily lunar period is 12h25min (see 61).
Transferring the results from the Sun, one obtains
Table 1: Overview of the major lunar tidal constituents, excluding variations in the Moon’s orbitaround the Earth.
Angular speed Period Tidal constituent(deg/hr) (hr)
γ − 2s 13.943 25.82 O1 Principal diurnal lunar tideγ 15.041 23.93 K1 Daily declination tide
2(γ − s) 28.984 12.42 M2 Principal semi-diurnal lunar tide2(γ − 2s) 27.886 12.91 Semi-diurnal lunar declination tide
2γ 30.082 11.97 K2 Semi-diurnal lunar declination tide
5.3 Doodson’s system for labelling tidal constituents
To facilitate a tabular overview of the (many) tidal constituents, Doodson invented a 6-digitsystem characterising the speed of the constituents. The speed of the individual contributionscan be described by means of 6 basic speeds f1...f6,
n1f1 + n2f2 + ...+ n6f6 (97)
where n1 is a positive integer and n2...n6 are integers between −5 and +5.
A convenient choice of speeds are
n1 = γ − s γ = 15.041 deg/hrn2 = s s = 0.549 deg/hrn3 = ε ε = 0.041 deg/hrn4...n6 Frequencies related to the distance between
the planets (not part of the theory here)
23
Table 2 summarises the Doodson coeffisients for four of the leading tidal constituents. As anexample, the speed of the S2 tide 2(γ − ε) can be expressed as
2n1 + 2n2 − 2n3 + 0n4 + 0n5 + 0n6 = 2 (γ − s) + 2 s− 2 ε = 2 (γ − ε) (98)
To circumvent negative coefficients, it is common practice to add 5 to the coefficients n2...n6,see Table 2.
Table 2: Overview of the major solar and lunar tides expressed in terms of the Doodson coeffi-cients.
Tide Period Speed Doodson coefficients Non-negative coefficients(hr) (deg/hr) n1n2n3.n4n5n6 (add 5 to n2...n6)
M2 12.42 28.98 = 2(γ − s) 200.000 255.555S2 12.00 30.00 = 2(γ − ε) 22-2.000 273.555K1 23.93 15.041 = γ 110.000 165.555O1 25.82 13.94 = γ − 2s 1-10.000 145.555
6 Tidal forcing in the momentum equation
The full 3-dimensional tidal flow, taking into account bathymetry and coastlines, can be describedby the primitive equations. The 3-dimensional momentum equation can be expressed in the form(M & P (2008), Chap. 6)
Du
Dt+
1
ρ∇p+ f z× u = −g z + F (99)
In (99), u is the 3-dimensional velocity field, Du/Dt is the total derivative of u, p is pressure, ρis density, f is the Coriolis parameter, z is the radial, outward-directed unit vector on a sphere,and F is friction.
The tidal forcing from the Moon and the Sun at the surface of the Earth can be added to (99)by introducing the pressure force (40) caused by the tidal elevation ξt (53):
Du
Dt+
1
ρ∇p+ f z× u = −g z− g∇ξt + F (100)
where
ξt =3
2
r4
MT
[ML
R3L
(cos2 θL −
1
3
)+MS
R3S
(cos2 θS −
1
3
)](101)
In the expression for ξt, cos2 θ − 1/3 is written in terms of the latitude ϕ and the hour angle Ψof a position P on Earth’s surface, as well as the declination angle δ of the celestial body (seeSec. 4 and expression 68):
cos2 θ − 1
3= cos2 ϕ cos2 Ψ cos2 δ +
1
2sin(2ϕ) cos Ψ sin(2δ) + sin2 ϕ sin2 δ − 1
3(102)
The hour (“longitude”) angle Ψ needs to be referenced to a fixed geographic point. A commonreference point is the point of the vernal equinox � (also known as the first point of Aries), seeFig. 8.
24
The angles ϕ, Ψ and δ can be computed in Fortran, C and Python, for any time, from e.g. theNaval Observatory Vector Astrometry Software (NOVAS, see Kaplan et al., 2011,http://aa.usno.navy.mil/software/novas/novas_info.php).
The momentum equation (100), with expressions (101) and (102), together with the continuityequation, models the full 3-dimensional dynamics of the tide. The set of equations, incorporatingrealistic bathymetry, can only be solved numerically.
7 Laplace’s tidal equations
When tidal forcing is introduced to the (quasi-)linearised version of the shallow water equations,the obtained equations are known as Laplace’s tidal equations (LTE). Tidal flow is then describedas the flow of a barotropic fluid, forced by the tidal pull from the Moon and the Sun. The phrase“shallow water equations” reflects that the wavelength of the resulting motion is large comparedto the thickness of the fluid.
The horizontal components of the momentum equation and the continuity equation can then beexpressed as
∂u
∂t− fv = −g ∂
∂x(η + ξt) (103)
∂v
∂t+ fu = −g ∂
∂y(η + ξt) (104)
∂η
∂t+
∂
∂x(uh) +
∂
∂y(vh) = 0 (105)
In the above equations, ξt is the (prescribed) tidal forcing and η is the resulting surface elevation,h is the ocean depth, and ξt is obtained from (101) and (102).
The horizontal momentum equations are linear, but inclusion of a friction term will typically turnthe equations non-linear. Likewise, the divergence terms in the continuity equation are nonlinearbecause of the product uh and vh. Solution of LTE requires discretisation and subsequentnumerical solution.
8 The tidal force expressed in terms of the tidal poten-tial
The direct method described in Sec. 1 can be conveniently carried out by introducing the tidalpotential as described below.
8.1 The potential of a conservative force
The gravitational force is an example of a conservative force, i.e., a force with the property thatthe work done in moving a particle from position a to position b is independent of the pathtaken (think of changes in the potential energy when an object in Earth’s gravitational field ismoved from one position to a new position; the path taken is irrelevant, only the height differencebetween the start and end positions matter).
25
It then follows that the gravitational force can be expressed as the negative gradient of a potentialΦ
F = −∇Φ (106)
Once Φ is found, the resulting force can be readily obtained by taking the gradient of Φ. Thissimplification is a main motivation of introducing the potential.
8.2 Defining the tidal potential
From (15), we have that the tidal acceleration can be expressed as
a = GML
(q
q3− R
R3
)(107)
Let Ft denote the tidal force per unit mass (or acceleration). It follows then from the aboveexpression that
Ft = GML
(q
q3− R
R3
)(108)
The task is therefore to determine the scalar function Φ, the tidal potential, satisfying
Ft = −∇Φ (109)
8.2.1 Gradient of the factor q/q3
We start by searching for a scalar with the property that the gradient of the scalar equals q/q3.It follows from Fig. (1) that
q = R− r (110)
In the above expression, R is a constant (fixed) vector whereas r varies spatially as it can orientedat any location on Earth’s surface.
In a Cartesian coordinate system with time-invariant unit vectors ex, ey and ez, we may ex-press
r = x ex + y ey + z ez (111)
andR = Rx ex +Ry ey +Rz ez (112)
Therefore,q = (Rx − x) ex + (Ry − y) ey + (Rz − z) ez (113)
The magnitude of q becomes
q = |q| = |R− r| =[(Rx − x)2 + (Ry − y)2 + (Rz − z)2
]1/2(114)
soq3 =
[(Rx − x)2 + (Ry − y)2 + (Rz − z)2
]3/2(115)
Furthermore,
∇(
1
q
)=
∂
∂x
(1
q
)ex +
∂
∂y
(1
q
)ey +
∂
∂z
(1
q
)ez (116)
26
In the above expression
∂
∂x
(1
q
)=
∂
∂x
(1
[(Rx − x)2 + (Ry − y)2 + (Rz − z)2]1/2
)
= −1
2
−2(Rx − x)
[(Rx − x)2 + (Ry − y)2 + (Rz − z)2]3/2
(117)
Similar results are obtained for the y- and z-derivatives in (116). Consequently,
∇(
1
q
)=
(Rx − x) ex + (Ry − y) ey + (Rz − z) ezq3
=q
q3(118)
The q/q3-term in (108) can therefore be expressed as ∇(1/q). This is a classical result from bothelectrodynamics and analysis involving gravity.
8.2.2 Gradient of the factor R/R3
The second term on the right-hand side of (108) can also be turned into the gradient of a scalar.For this we use that
r ·R = xRx + y Ry + z Rz (119)
Consequently,
∇(r ·R) =∂
∂x(r ·R) ex +
∂
∂y(r ·R) ey +
∂
∂z(r ·R) ex
= Rx ex +Ry ey +Rz ez
= R (120)
The R/R3-term in (108) is therefore equivalent to ∇(r ·R).
8.2.3 Resulting tidal potential
By inserting (118) and (120) into (108), we obtain
Ft = −GML∇(−1
q+
r ·RR3
+ C
)(121)
where C is a constant.
C can be determined by noting that Ft vanishes at Earth’s center, see the discussion in Sec. 1.6.When r = 0, q equals R, and we get from (121) that C = 1/R. Therefore
Ft = −GML∇(−1
q+
r ·RR3
+1
R
)= −∇Φ (122)
where the tidal potential Φ has been introduced from (109). The tidal potential can thereforebe put in the form
Φ = GML
(−1
q+
r ·RR3
+1
R
)(123)
27
8.3 Tidal potential expressed in terms of the zenith angle
Next step is to express (123) by means of the zenith angle θ (see Fig. 1). For the 1/q-term theprocedure follows that of Sec. 1.7, starting with the law of cosines (16),
q2 = R2 + r2 − 2Rr cos θ = R2[1 +
r
R
( rR− 2 cos θ
)](124)
Consequently,
q = R[1 +
r
R
( rR− 2 cos θ
)]1/2(125)
The factor r/R has been introduced in the above expressions since it is a small number (ap-pendix A), allowing for series expansion. The quantity 1/q can then be expended by applyingthe binomial theorem (168). We neglect terms proportional to and smaller than (r/R)3, giv-ing
1
q=
1
R
[1 +
r
R
( rR− 2 cos θ
)]−1/2
=1
R
[1− 1
2
r
R
( rR− 2 cos θ
)+
(− 12 )(− 3
2 )
2
( rR
)2 ( rR− 2 cos θ
)2
+ . . .
]≈ 1
R
[1− 1
2
( rR
)2
+r
Rcos θ +
3
2
( rR
)2
cos2 θ
]=
1
R
[1 +
r
Rcos θ +
1
2
( rR
)2 (3 cos2 θ − 1
)](126)
For the r ·R-term, the scalar product gives
r ·R = r R cos θ
sor ·RR3
=r
R2cos θ (127)
By combining (123), (126) and (127), one obtains the tidal potential expressed in terms of thezenith angle θ,
Φ = −3
2GML
r2
R3
(cos2 θ − 1
3
)(128)
8.3.1 Radial and tangential tidal forces per unit mass
Once the tidal potential Φ is determined, the radial and tangential tidal forces per unit masscan be determined from (109). For the gradient operator, we use the spherical (zenith) coordi-nate system shown in the right panel of Fig. 9. In this system, the gradient operator is givenby (157).
The radial component Ft,r of the tidal force per unit mass is then
Ft,r = −∂Φ
∂r= GML
r
R3
(3 cos2 θ − 1
)(129)
28
or, by introducing the gravitational acceleration g from (10),
Ft,r = gML
MT
( rR
)3 (3 cos2 θ − 1
)(130)
Ft,r is directed outward, i.e., in the er-direction.
Similarly, the tangential (horizontal) component Ft,h of the tidal force per unit mass is
Ft,h = −1
r
∂Φ
∂r= 3GML
r
R3cos θ sin θ (131)
or, with sin 2θ = 2 sin θ cos θ and by introducing the gravitational acceleration g from (10),
Ft,h =3
2gML
MT
( rR
)3
sin 2θ (132)
Ft,h is oriented in the eθ direction, i.e., towards the center line connecting the centers of theEarth and the Moon (see Fig. 1).
Ft,r and Ft,h are identical to the tidal acceleration in (31) and (29), demonstrating the consistencyof the direct method presented in Sec. 1 and the method of tidal potential of this section.
9 Introducing the tidal potential in the primitive momen-tum equations
The tidal forcing of the ocean can be introduced by adding the tidal forcing per unit mass (108)to the 3-dimensional momentum equation (99).
The tidal forcing on the Earth’s surface Φt caused by the Moon (ΦL) and the Sun (ΦS) is
Φt = ΦL + ΦS (133)
ΦL and ΦS are both on the from of (128), so
Φt = −3
2gr4
MT
[ML
R3L
(cos2 θL −
1
3
)+MS
R3S
(cos2 θS −
1
3
)](134)
where G = g r2/MT from (10) has been used.
The resulting momentum equation becomes
Du
Dt+
1
ρ∇p+ f z× u = −g z−∇(Φt) + F (135)
Alternatively, the gravity term can be expressed as a gradient of a potential
g z = −∇Φg (136)
where Φg = g z. It is therefore common to put (135) in the form
Du
Dt+
1
ρ∇p+ f z× u = −∇(Φg + Φt) + F (137)
Note that Φt is a function of position and time whereas Φg can be considered time-independent.
Equation (135) or (137), together with the continuity equation, is used to simulate the full 3-dimensional flow of the ocean tide. The linearised shallow water equivalent of the above equationsis given by (103) and (104).
29
A Some key parameters of the Earth, Moon, Sun sys-tem
The mass of the Earth MT , Moon ML and Sun MS are
MT = 5.974 · 1024 kg (138)
ML = 7.347 · 1022 kg (139)
MS = 1.991 · 1930 kg (140)
The following mass ratios are then obtained
MT
ML≈ 80 ,
ML
MT≈ 0.012 (141)
MT
MS≈ 3× 10−6 ,
ML
MT≈ 3× 105 (142)
(143)
Earth’s mean radius isr = 6.37 · 106 m (144)
Often a, in stead of r, is used to denote Earth’s mean radius.
The mean distance between Earth and Moon, and Earth and Sun, are
RTL,mean = 3.844 · 108 m (145)
RTS,mean = 1.496 · 1011 m (146)
The following distance ratios are obtained
r
RTL,mean≈ 0.017 ,
RTL,mean
r≈ 60 (147)
r
RTS,mean≈ 4× 10−5 ,
RTS,mean
r≈ 2× 104 (148)
(149)
B Spherical coordinates
B.1 Two commonly used spherical coordinate systems
Spherical coordinates are, by construction, convenient for any type of spherical problems. Fig-ure 9 illustrates two commonly used variants of spherical coordinates.
The system to the left is described by Ψ (angle in longitudinal direction, 0 ≥ Ψ ≥ 2π), ϕ(angle in the latitudinal direction, −π/2 ≥ ϕ ≥ π/2) and r (length in the radial direction,r ≥ 0). This coordinate system is commonly used in oceanography and meteorology because ofthe correspondence with Earth’s geographical latitude and longitude position.
30
Figure 9: Illustration of two spherical coordinate systems.
The system to the right is described by Ψ (azimuth angle, 0 ≥ Ψ ≥ 2π), θ (zenith angle,0 ≥ θ ≥ π) and r (length in the radial direction, r ≥ 0). This coordinate system is commonlyused in problems involving gravitational or electric potential as will become clear shortly.
Note that the set of coordinates in the order (Ψ, ϕ, r) for the system to the left in Figure 9,and (r, θ, Ψ) for the system to the right in the figure, are both right-hand coordinate systems.Also note that the actual naming of the various angles may vary depending on the problem; theazimuth angle Ψ is, as an example, commonly labelled λ in meteorology and oceanography.
B.2 Volume and surface elements in spherical coordinates
B.2.1 Spherical volume elements
The volume element formed by perturbing each of the three coordinates, denoted by δ in thefollowing, can be obtained as follows:
For the system to the left in Figure 9, small changes in r form a line segment of length δr inthe direction of r; small changes in ϕ form an upward directed arc of length r δϕ; and smallchanges in Ψ form an arc in the xy-plane of length r cosϕ δΨ. The resulting volume element isthe product of the three length contributions
δVΨϕr = r2 cosϕ δr δϕ δΨ (150)
Correspondingly, for the system to the right in the figure, small changes in r form a line segmentof length δr in the direction of r, small changes in θ form an downward directed arc of lengthr δθ, and small changes in Ψ form an arc in the xy-plane of length r sin θ δΨ. The resultingvolume element is
δVrθΨ = r2 sin θ δr δθ δΨ (151)
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B.2.2 Spherical surface elements
If r is kept constant at r = a, the surface area S of the sphere spanned out by the two pairsof angles (Ψ, ϕ) and (θ, Ψ) follows from the procedure with the volume elements above whenchanges in r are ignored
δSΨϕ = a2 cosϕ δϕ δΨ (152)
δSθΨ = a2 sin θ δθ δΨ (153)
Integration over the sphere according to∫ 2π
Ψ=0
∫ π/2
ϕ=−π/2δSΨϕ dϕ dΨ (154)
∫ 2π
Ψ=0
∫ π
θ=0
δSθΨ dθ dΨ (155)
result in, as expected, 4πa2, the surface area of a sphere with radius a.
B.2.3 Relationship with the Earth-Moon system
Note that in the system to the right in Figure 9, the zenith angle θ is always non-negative. Forr = const, θ = const and 0 ≤ Ψ < 2π, the resulting geometry forms a cone with angle 0 ≤ θ ≤ πcentered around the positive z-axis. A cone is also spanned out for r = const, ϕ = const and0 ≥ Ψ ≥ 2π in the system to the left in the figure. In the latter case, however, the latitudinalangle −π/2 ≤ ϕ ≤ π/2 takes both signs.
It is the non-negative zenith-angle θ in the coordinate system to the right in Figure 9, similarlyto the angular zenith distance to the Moon θ in tidal theory (see Figure 1), that makes the (θ,Ψ)-system particularly suited for problems involving gravitational potential (or problems involvingelectric potentials).
B.2.4 Gradient operator in spherical coordinates
The gradient operator expressed in terms of the two spherical coordinate systems in Fig. 9 canbe expressed in terms of the two system’s scale factors.
For the system to the left in Fig. 9, the scale factors in the Ψ, ϕ, r-directions are r cosϕ, r and1, respectively. The resulting gradient operator is therefore
∇ =
(1
r cosϕ
∂
∂Ψ,
1
r
∂
∂ϕ,∂
∂r
)=
1
r cosϕ
∂
∂ΨeΨ +
1
r
∂
∂ϕeϕ +
∂
∂rer (156)
In the above expression, eΨ, eϕ and er are the unit vectors in the Ψ, ϕ, r-directions, respec-tively.
Similarly, for the right system in Fig. 9, the scale factors in the r, θ,Ψ-directions are 1, r andr sin θ, respectively. The resulting gradient operator becomes
∇ =
(∂
∂r,
1
r
∂
∂θ,
1
r sin θ
∂
∂Ψ
)=
∂
∂rer +
1
r
∂
∂θeθ +
1
r sin θ
∂
∂ΨeΨ (157)
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C The three lunar tidal components
To show the identity between the right hand sides of (68) and (69), one can first note that ζ2 iscommon to both expressions and can therefor be ignored in the following.
Expansion and reordering of ζ1 gives
ζ1 =3
2
(sin2 ϕ− 1
3
)(sin2 δ − 1
3
)=
3
2sin2 ϕ sin2 δ − 1
2sin2 ϕ− 1
2sin2 δ +
1
6
= sin2 ϕ sin2 δ +1
2sin2 ϕ sin2 δ − 1
2sin2 ϕ− 1
2sin2 δ +
1
6
= sin2 ϕ sin2 δ +1
2sin2 ϕ (sin2 δ − 1)− 1
2sin2 δ +
1
6
= sin2 ϕ sin2 δ − 1
2sin2 ϕ cos2 δ − 1
2sin2 δ +
1
6(158)
In the second last line, the identity
sin2 a+ cos2 a = 1 (159)
has been used.
Furthermore
ζ3 =1
2cos2 ϕ cos(2Ψ) cos2 δ
= cos2 ϕ cos2 Ψ cos2 δ − 1
2cos2 ϕ cos2 δ (160)
Here the identitycos 2a = cos2 a− sin2 a = 2 cos2 a− 1 (161)
has been used (with the help of identity (159) in the last equality).
With ζ1 from (158) and ζ3 from (160), one obtains
ζ1 + ζ3 = sin2 ϕ sin2 δ − 1
2sin2 ϕ cos2 δ − 1
2sin2 δ +
1
6
+ cos2 ϕ cos2 Ψ cos2 δ − 1
2cos2 ϕ cos2 δ (162)
The second and last term on the right hand side can be combined into
−1
2cos2 δ (163)
so
ζ1 + ζ3 = sin2 ϕ sin2 δ − 1
2cos2 δ − 1
2sin2 δ +
1
6+ cos2 ϕ cos2 Ψ cos2 δ (164)
Comparison between the right hand sides of (68) and (164), and remembering that ζ2 has alreadybeen taken care of, shows that the two last terms on the right hand side of (164) have theircounterparts on the right hand side of (68). For the remaining terms on the right hand sideof (164), one obtaines
sin2 ϕ sin2 δ − 1
2cos2 δ − 1
2sin2 δ +
1
6= sin2 ϕ sin2 δ − 1
2+
1
6= sin2 ϕ sin2 δ − 1
3(165)
which is identical to the remaining terms on the right hand side of (68).
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D Binominal theorem for rational exponents
For all rational1 numbers r, the following identity exists
(1 + x)r = 1 + r x+ rr − 1
2x2 + r
r − 1
2
r − 2
3x3 + ... (166)
For a derivation, see e.g.http://www.trans4mind.com/personal_development/mathematics/series/binomialProofAllAlgebra.htm.
For |x| � 1, the linear and quadric approximation of (166) becomes
(1 + x)r = 1 + r x (167)
and
(1 + x)r = 1 + r x+r(r − 1)
2x2 (168)
E Some identities∫cos 2x sinx dx =
cosx
2− cos 3x
6+ C (169)
sin 2a = 2 sin a cos a (170)
sin2 a+ cos2 a = 1 (171)
sin3 a =1
4(3 sin a− sin 3a) (172)
cos 2a = cos2 a− sin2 a (173)
cos 2a = 2 cos2 a− 1 (174)
= 1− 2 sin2 a (175)
sin a± sin b = 2 sin
(1
2(a± b)
)cos
(1
2(a∓ b)
)(176)
cos a+ cos b = 2 cos
(1
2(a+ b)
)cos
(1
2(a− b)
)(177)
1A rational number is any number that can be expressed as the quotient or fraction p/q of two integers, withthe denominator q not equal to zero.
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