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

1r2

∂∂r

r2 ∂f∂r

⎛⎝⎜

⎞⎠⎟+

1r2 sinθ

∂∂θ

sinθ ∂f∂θ

⎛⎝⎜

⎞⎠⎟+

1r2 sin2θ

∂2 f∂λ2

= 0

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 = rlP(θ)L(λ)

l(l +1) + 1sinθP(θ)

∂∂θ

sinθ ∂P(θ)∂θ

⎛⎝⎜

⎞⎠⎟+

1sin2θL(λ)

∂2L(λ)∂λ2

= 0

This is the only place where λ appears, consequently it must be equal to a constant

1L(λ)

∂2L(λ)∂λ2

= constant = −m2

L(λ) = Am cosmλ + Bm sinmλ

m is an integer

1sinθ

∂∂θ

sinθ ∂P(θ)∂θ

⎛⎝⎜

⎞⎠⎟+ l(l +1) − m2

sin2θ⎡

⎣⎢

⎤

⎦⎥P(θ) = 0

This becomes:

This equation is known as Legendre’s Associated equation

The solution of P(θ) are polynomials in cosθ, involvingthe integers l and m. For m ≤ l, the general solution is written as

Pl ,m (cosθ) =(1− cos2θ)m /2

2l l!dl+m (cos2θ −1)l

d(cosθ)l+m

These are called Associated Legendre Polynomials

Pl ,m (µ) =(1− µ2 )m /2

2l l!dl+m (µ2 −1)l

dµ l+m

You will see this written as, when µ = cosθ

In this case, the solution to ∇2 f = 0

f = rlPl ,m (cosθ) Am cosmλ + Bm sinmλ[ ]

It can also be shown that: R(r) = r−(l+1)

f = r−(l+1)Pl ,m (cosθ) Am cosmλ + Bm sinmλ[ ]

f = r−(l+1)Pl ,m (cosθ) Am cosmλ + Bm sinmλ[ ]l = degreem = orderFor 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:

Pl (µ) =1l!2l

d l

dµ l (µ2 −1)l

(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 cosmλ + Bm sinmλ[ ]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.

No dependence on longitude and dependence in latitude is given by Legendre polynomials Zonal Harmonics

Solution vanishes on l parallels of latitude

m = 0

P10 (cosθ) P2

0 (cosθ)

If0 < m < lthey vanish on 2m meridians of longitude and (l − m) parallels of latitude

Tesseral Harmonics

Sectorial Harmonics

Ifm = lDoes not change with latitude

Figure from Blakely

P 0

5

�

Plm− l

l

∫0

2π

∫ (µ)Pl 'm ' (µ)[cosmλ,sinmλ][cosm'λ,sinm'λ]dµdλ = 0

unless both l'= l and m'= m

Orthogonality of surface spherical harmonic functions

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

14π −1

1

∫ Pl ,m (µ)cosmλ⎡⎣ ⎤⎦0

2π

∫2

dµdλ =(l + m)!

2(2l +1)(l − m)!

The ratio of this for P4,1(µ) to P4,4 (µ) is5!3!

: 8!1= 1 : 2016

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):

Plm (θ) =

(2l +1)[ ]1/2 Pl ,m (θ), if m = 0

2(2l +1) (l − m)!(l + m)!

⎡⎣⎢

⎤⎦⎥

1/2

Pl ,m (θ), ifm > 0

⎧

⎨⎪

⎩⎪

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:

f = rl ,r−(l+1)⎡⎣ ⎤⎦Pl ,m (cosθ) Am cosmλ + Bm sinmλ[ ](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 ∞

f = C 'll=1

∞

∑ 1r

⎛⎝⎜

⎞⎠⎟l+1

Pl (cosθ)0

Since we would want our coefficients to be dimensionally equal, then this is written (a is the radius of the Earth)

Now the most classical form that the gravitational potential, V, is written is:

V = −

GMr

JoPo − J1arP1(cosθ) − J2

ar

⎛⎝⎜

⎞⎠⎟2

P2 (cosθ) +⎛

⎝⎜⎞

⎠⎟

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

�

f =1a

Cll=1

∞

∑ ar

⎛ ⎝

⎞ ⎠

l+1

Pl (cosθ)0

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:

V = −GMr

+GMa2J2

r332cos2θ −

12

⎛⎝⎜

⎞⎠⎟

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 v∫ where m is mass and v is the

volume of the entire body. Then, assuming rotational symmetry

V = −GMr

+Gr3 (A − C) 3

2cos2θ −

12

⎛⎝⎜

⎞⎠⎟

where A and C are the principal moments inertia with respect to axis passing through the equator and pole, respectively. Thisequation is known as MacCullagh's Formula

Note the similarities between our spherical harmonic and MacCullagh's formula representations of V , and that

J2 =C − AMa2

J2 can be measured from the precession of small satellites inorbit about the earth and is

J2 = 1.082626 ×10−3

The Dynamical Ellipticity, H =C − AC

, can be inferred from the

precession of the solar and lunar equinoxes and is H = 1 / 305.456∴The "moment of inertia factor" is CMa2 =

J2

H= 0.330695 <

25

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

θ

F =ω 2 (r sinθ)

The force can be derived from a scalar potential (via ∇Vrot )

Vrot =12ω 2r2 sin2θ

Our Geopotential, U, isU = V +Vrot

r

Geopotential, U

U = V +Vrot

U = −GMr

+Gr3 C − A( ) 3

2cos2θ −

12

⎛⎝⎜

⎞⎠⎟−

12ω 2r2 sin2θ

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, sor = a, θ = 90

r = c, θ=0

Plug these back into the equation for the geopotential, U, just derivedand with some algebra, the flattening, f , becomes

f =a − ca

=A − CMa2

a2

c2 +c

2a⎛⎝⎜

⎞⎠⎟+

12a2cω 2

GM

The differences between a and c are small. Note that a3ω 2

GM is the ratio

of the centrifical acceleration to gravity at the equator, a ratio knownas m . So

f =32J2 +

12m

Observed flattening

A second order theory for the flattening (the one derived on theprevious slide was first order) is

J2 =23f 1− f

2⎛⎝⎜

⎞⎠⎟−m3

1− 32m −

27f⎛

⎝⎜⎞⎠⎟

Using known values:

J2 = 1.08264 ×10−3

m = 3.461395 ×10−3

f⊕ =1

298.25We refer to this as the observed flattening of the earth (it amountsto ~21km difference, a − c, between pole and equator).

Hydrostatic Earth The true hydrostatic figure of a condensed rotating body canbe determined.

An approximate first order theory, sometimes called the Radau-Darwin approximation, to the flattening is

fH =(5 / 2)m

1+ (5 / 2) 1− 32

CMa2

⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

2

A higher order treatment that resorts to numerical methods, leads to

fH =1

299.627= 3.337848 ×10−3

fH < f⊕ f⊕ − fHf⊕

= 0.5% 0.5% × 21 km = 105 m

The earth has a significant Non-hydrostatic oblateness → the earth isbulging 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. (3) & (4) can be easily dismissed.

1.  The first serious idea was that the fossil bulge was a “relic” from when the Earth was spinning faster. The earth had a more significant bulge in the past, but because of viscous delay it takes time to relax.

McKenzie [1966] had concluded that the viscosity of the lower mantle must be > 2X1026 Pa-s. This idea was falsified by more direct inference of the present-day mantle viscosity from post-glacial rebound.

2. Mass (density) inhomogeneities inside the Earth. Soon after McKenzie’s estimate, Goldreich & Toomre [1969] show that the C’-B’ = 6.9x10-6 Ma2 B’-A’ = 7.2x10-6 Ma2 [The primes denote the “non-hydrostatic” contributions to the 3 principal moments of inertia. In other words, mass inhomogeneities between 2 axis within the equatorial plane (B’-A’) are as large (in fact larger) than that related to the rotation axis (C’-B’).

Equipotential Surfaces and the geoid Obviously, a true equipotential surface of the earth may be different from a simple. Ellipsoidal hydrostatic earth. Let N be the geoid height, or small displacement if the actual geopotential from our ellipsoidal geopotential (a reference surface U(r ), not yet specified).

Uo = const. = U(r + N )

=U(r) + ∂U∂r

N

≅U(r) + ∂Uo

∂rN

=U(r) − gN(where g is the local acceleration of gravity, or g = −∂Uo / ∂r)

∴N =U −Uo

gSometimes, you see this written asgN = ΔU

Reference Geoid (ellipsoidal Geopotential)

N

Geoid

Mass Excess δρ>0

�

A density anomaly at degree l at a depth b falls off as :4πG2l + 1

⋅bl+ 2

rl+1

L=2 L=4 surface 1 0.2 mid mantle 0.46 0.043 CMB 0.16 0.0055

Amplitudes ~ l-1

�

Vl = Clm( )2 + Sl

m( )2[ ]m=0

l

∑⎧ ⎨ ⎩

⎫ ⎬ ⎭

1/ 2

Free air gravity

Gravity and Geoid over subduction zones

Hager

Non-hydrostatic Geoid with Hotspot locations

Correlation of ‘Residual Geoid’ & HS with Continents in Jurassic

Anderson [1982]

Present Day Plate Boundaries & Non-hydrostatic Geoid

Chase & Sprowl [1983]

Mesozoic Plate Boundaries & Non-hydrostatic Geoid

Chase & Sprowl [1983]

Aside: The density anomaly within slabs in light of the arguments in the Chase [1979] reading.

~80 km <ΔT>~600°C

δρ = ρoα vΔT

= (4,000 kg m-3)(3×10-5 K-1)(600 iC)=78 kg m-3

= 0.08 g cm-3

~2x assumed in geoid models

Additional Background for the Chase [1979] reading

McKenzie [1969]

GRACE: Gravity Recovery And Climate Experiment • Launched in March of 2002 • Accurately maps variations in the Earth's gravity field • Two identical spacecrafts flying about 220 kilometers apart in a polar orbit 500 km. • Maps the gravity fields by making accurate measurements of the distance between the two satellites, using GPS and a microwave ranging system • A ‘new’ gravity model every 30 days • Precision should increase as the orbit decays and GRACE moves down to a lower elevation

Gravity anomalies from decades of tracking Earth-orbiting satellites

Gravity anomalies from 111 days of GRACE data (GGM01S)

Gravity anomalies from 363 days of GRACE data (GGM02S)

GRACE Change in the geoid in 2003

Change in gravity associated with the Dec. 2004 Sumatra Earthquake (Mw~9.2)

de Linage et al. [2009]

Some comments on an up-coming homework

Application of results from HW2

What is the geoid anomaly associated with isostatic compensation? Ideas help to better understand gravity in both isostatic as well as dynamic earth models.