6. the Gravity Field

6. The Gravity Field Ge 163 4/11/14- Outline • Spherical Harmonics • Gravitational potential in spherical harmonics • Potential in terms of moments of inertia • The Geopotential • Flattening and the excess bulge of the earth • The geoid and geoid height • Non-hydrostatic geoid • Geoid over slabs • Geoid correlation with hot-spots & Ancient Plate boundaries • GRACE: Gravity Recovery And Climate Experiment Spherical harmonics ultimately stem from the solution of Laplace’s equation. Find a solution of Laplace’s equation in a spherical coordinate system. ∇2 f = 0 The Laplacian operator can be written as 2 1 ∂ ⎛ 2 ∂f ⎞ 1 ∂ ⎛ ∂f ⎞ 1 ∂ f 2 ⎜ r ⎟ + 2 ⎜ sinθ ⎟ + 2 2 2 = 0 r ∂r ⎝ ∂r ⎠ r sinθ ∂θ ⎝ ∂θ ⎠ r sin θ ∂λ r is radius θ is colatitude λ Is longitude One way to solve the Laplacian in spherical coordinates is through separation of variables f = R(r)P(θ)L(λ) To demonstrate the form of the spherical harmonic, one can take a simple form of R(r ). This is not the only one which solves the equation, but it is sufficient to demonstrate the form of the solution. Guess R(r) = rl f = rl P(θ)L(λ) 1 ∂ ⎛ ∂P(θ)⎞ 1 ∂2 L(λ) l(l + 1) + ⎜ sinθ ⎟ + 2 2 = 0 sinθP(θ) ∂θ ⎝ ∂θ ⎠ sin θL(λ) ∂λ This is the only place where λ appears, consequently it must be equal to a constant 1 ∂2 L(λ) = constant = −m2 m is an integer L(λ) ∂λ 2 L(λ) = Am cos mλ + Bm sin mλ This becomes: 1 ∂ ⎛ ∂P(θ)⎞ ⎡ m2 ⎤ ⎜ sinθ ⎟ + ⎢l(l + 1) − 2 ⎥ P(θ) = 0 sinθ ∂θ ⎝ ∂θ ⎠ ⎣ sin θ ⎦ This equation is known as Legendre’s Associated equation The solution of P(θ) are polynomials in cosθ, involving the integers l and m. For m ≤ l, the general solution is written as (1− cos2 θ)m/2 d l + m (cos2 θ − 1)l P (cosθ) = l,m 2l l! d(cosθ)l + m These are called Associated Legendre Polynomials You will see this written as, when µ = cosθ (1− µ2 )m/2 d l + m (µ2 − 1)l P (µ) = l,m 2l l! dµl + m 2 In this case, the solution to ∇ f = 0 l f = r Pl,m (cosθ)[Am cos mλ + Bm sin mλ] −(l +1) It can also be shown that: R(r) = r −(l +1) f = r Pl,m (cosθ)[Am cos mλ + Bm sin mλ] −(l +1) f = r Pl,m (cosθ)[Am cos mλ + Bm sin mλ] l = degree m = order For m=0, there are a subset of “Associated Legendre Polynomials” referred to as “Legendre Polynomials” or Pl (cosθ) But these can only describe so-called “zonal terms” and are insufficient to fit a surface spherical harmonic. Legendre Polynomials are given by: 1 d l P (µ) = (µ2 − 1)l l l!2l dµl (i.e. n = l in previous slides) Blakely The portion of the solution which is a function of θ and λ Pl,m (cosθ)[Am cos mλ + Bm sin mλ] is known as a surface spherical harmonic, because any quantity defined over a spherical surface can be expressed as sum of surface spherical harmonics. It is important to be able to visualize the behavior of surface spherical harmonics. m = 0 No dependence on longitude and dependence in latitude is given by Legendre polynomials Zonal Harmonics Solution vanishes on l parallels of latitude 0 0 P1 (cosθ) P2 (cosθ) If 0 < m < l they vanish on 2m meridians of longitude and (l − m) parallels of latitude Tesseral Harmonics If m = l Does not change with latitude Sectorial Harmonics P 0 5 Figure from Blakely Orthogonality of surface spherical harmonic functions 2π l P ( )P ( )[cosm ,sinm ][cosm' ,sinm' ]d d 0 ∫ ∫ lm µ l'm' µ λ λ λ λ µ λ = 0 −l unless both l'= l and m'= m See Garland, Introduction to Geophysics, for demonstration Normalization Consider what would happen if we found the mean values of the squares of the surface harmonics corresponding to Pl,m(µ) over the surface of the sphere 1 1 2π 2 (l + m)! ⎡Pl,m (µ)cos mλ⎤ dµdλ = 4π ∫−1 ∫0 ⎣ ⎦ 2(2l + 1)(l − m)! The ratio of this for P4,1(µ) to P4,4 (µ) is 5! 8! : = 1: 2016 3! 1 A consequence of this is that numerical coefficients in front of the surface spherical harmonics do not themselves realistically convey the relative importance of the various terms Fully normalized surface spherical harmonics which integrate to unity over the surface of the sphere and used commonly in geodetic studies (e.g. gravity): ⎧ 1/2 [(2l + 1)] Pl,m (θ), if m = 0 m ⎪ 1/2 Pl (θ) = ⎨⎡ (l − m)!⎤ ⎪⎢2(2l + 1) ⎥ Pl,m (θ), ifm > 0 ⎩⎣ (l + m)!⎦ Note symbol Very important: This is the equation according to books by Lambeck and by Stacey. However, there are typos in the books by Garland and by Blakley Also note, in geomagnetics there is a different normalization Returning to the solution of Laplace’s equation: l −(l +1) f = ⎣⎡r ,r ⎦⎤ Pl,m (cosθ)[Am cos mλ + Bm sin mλ] (1.) We will need another constant for l, Cl; (2) if we are interested in solutions that are bounded as r∞, then we need to eliminate the rl term. Let us restrict our attention to problems that have rotational symmetry (m=0) such that the potential can be approximated solely by zonal terms (Legendre polynomials). Also, our entire solution will be made up of an infinite series of terms l, 0 to ∞ l +1 ∞ ⎛ 1⎞ f = ∑C 'l ⎜ ⎟ Pl (cosθ) l =10 ⎝ r ⎠ Since we would want our coefficients to be dimensionally equal, then this is written (a is the radius of the Earth) l +1 1 ∞ ⎛ a⎞ f = ∑Cl Pl (cosθ) a l =10 ⎝ r⎠ Now the most classical form that the gravitational potential, V, is written is: 2 GM ⎛ a ⎛ a⎞ ⎞ V = − ⎜ JoPo − J1 P1(cosθ) − J2 ⎜ ⎟ P2 (cosθ) +⎟ r ⎝ r ⎝ r ⎠ ⎠ These coefficients just describe the zonal terms and named after Harold Jeffries P1(cosθ) represents the off center potential and we choose J1=0 when we have the center of mass as our origin J2 dominates the oblate ellipsoidal form of a potential surface (geoid) J4, J6…. Are much smaller than 1000 or more and can be neglected So with P2 written out: 2 GM GMa J ⎛ 3 2 1⎞ V = − + 2 cos θ − r r 3 ⎝⎜ 2 2⎠⎟ This gives us the potential for a stationary point, r>0 i.e. not rotating An alternative means by which to formulate the potential field of a body is through placing the potential in terms of the moments of the mass distribution. Recall that some moment of inertia with respect to some axis, x', where ζ is the distance of some point from the axis, then I = ζ 2dm where m is mass and v is the ∫v volume of the entire body. Then, assuming rotational symmetry GM G ⎛ 3 2 1⎞ V = − + (A − C) cos θ − r r 3 ⎝⎜ 2 2⎠⎟ where A and C are the principal moments inertia with respect to axis passing through the equator and pole, respectively. This equation is known as MacCullagh's Formula Note the similarities between our spherical harmonic and MacCullagh's formula representations of V, and that C − A J = 2 Ma2 J2 can be measured from the precession of small satellites in orbit about the earth and is −3 J2 = 1.082626 × 10 C − A The Dynamical Ellipticity, H = , can be inferred from the C precession of the solar and lunar equinoxes and is H = 1 / 305.456 ∴ The "moment of inertia factor" is C J 2 = 2 = 0.330695 < Ma2 H 5 Now V is the gravitational potential, but it is not the entire potential that would be felt if one is fixed on a rotating earth: ω F = ω 2 (r sinθ) θ r The force can be derived from a scalar potential (via ∇Vrot ) 1 V = ω 2r2 sin2 θ rot 2 Our Geopotential, U, is U = V + Vrot Geopotential, U U = V + Vrot GM G ⎛ 3 2 1⎞ 1 2 2 2 U = − + (C − A) cos θ − − ω r sin θ r r 3 ⎝⎜ 2 2⎠⎟ 2 Geoid Let us define the geoid as the surface of constant potential, Uo, most nearly fitting the mean sea level. The geoid has equitorial and polar radii, a and c, so r = a, θ = 90 r = c, θ=0 Plug these back into the equation for the geopotential, U, just derived and with some algebra, the flattening, f , becomes a − c A − C ⎛ a2 c ⎞ 1 a2cω 2 f = = + + a Ma2 ⎝⎜ c2 2a⎠⎟ 2 GM a3ω 2 The differences between a and c are small. Note that is the ratio GM of the centrifical acceleration to gravity at the equator, a ratio known as m . So 3 1 f = J2 + m 2 2 Observed flattening A second order theory for the flattening (the one derived on the previous slide was first order) is 2 ⎛ f ⎞ m ⎛ 3 2 ⎞ J = f 1− − 1− m − f 2 3 ⎝⎜ 2 ⎠⎟ 3 ⎝⎜ 2 7 ⎠⎟ Using known values: −3 J2 = 1.08264 × 10 m = 3.461395 × 10−3 1 f = ⊕ 298.25 We refer to this as the observed flattening of the earth (it amounts to ~21km difference, a − c, between pole and equator). Hydrostatic Earth The true hydrostatic figure of a condensed rotating body can be determined. An approximate first order theory, sometimes called the Radau-Darwin approximation, to the flattening is (5 / 2)m fH = 2 ⎡ ⎛ 3 C ⎞ ⎤ 1 (5 / 2) 1 + ⎢ ⎜ − 2 ⎟ ⎥ ⎣ ⎝ 2 Ma ⎠ ⎦ A higher order treatment that resorts to numerical methods, leads to 1 f = = 3.337848 × 10−3 H 299.627 f⊕ − fH fH < f⊕ = 0.5% 0.5% × 21 km = 105 m f⊕ The earth has a significant Non-hydrostatic oblateness → the earth is bulging more at the equator than accounted for by its rotation There were as many as four ideas to explain the excess equatorial bulge of the earth: (1) the planet was spinning faster in the past; (2) mass inhomogeneities; (3) polar regions still depressed by formally larger ice caps; (4) solar and lunar tides.

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