Potential fields: context

As of 2002, fifty percent of the world’s seismic fleet was recording gravity and/or magnetic data; a rapid surge took place (this figure was double that of 2000). Some reasons: In the exploration industry:

• Cost effective — compared to seismics

• Complementary to seismic data in e.g. sub-salt or sub-basalt, where it is difficult to image

• New technology/interpretation techniques (e.g. GPS, gradiometers) has brought about an improvement in accuracy (position and signal)

In planetary physics

• New LEO-satellites can deliver unprecedented accuracy for under- standing planetary mass distributions, ocean flow, rebound after deglacia- tion, even mass redistribution due to global warming!

• Gravity and magnetic missions provide unique information on our sis- ter planets

Energy, Work and Potential

F

Figure 1: Force field F

Work done by field in moving particle δx is F · δx:

δW = Fxdx + Fydy + Fzdz (1)

Total work, W , in moving from x0 to x1 is

x1 W = F · dx (2) Zx0

1 In general the work done depends on the path chosen. Also in general the work done in traversing a circuit is non-zero, we write this as W = F · dx 6= 0. There areH force fields for which the work done is independent of the path. These are called conservative force fields, and for these F · dx = 0. H

Example of a non-conservative field y B B at (1,1)

1

A x Figure 2: Path 1

Let F = x2y2i + xyj Calculate work done in moving A to B along path 1.

1 1 W = Fxdx + Fydy (3) Z0 Z0 1 1 = 0+1 ydy = (4) Z0 2 Now calculate the work done in using different path 2. y 2 B B at (1,1)

A x Figure 3: Path 2

2 1 1 W = Fydy + Fxdx (5) Z0 Z0 1 2 1 = 0+1 x dx = (6) Z0 3 The work done is different — hence the force is non-conservative. It also follows that the work done in going round circuit by paths 1 and 1 1 1 2, with path 2 traversed in the negative sense, = 2 − 3 = 6 . In the other 1 direction the work = − 6 .

Example of a conservative field Let F = yi + xj For our previous paths we now find:

W1 = 1 (7) W2 = 1 (8) and the work is independent of the path in general. One can show that for the work to be independent of the path, there must exist a function such that the force F = ∇φ Depending on the convention used φ can be either plus or minus the work function. Then F =+∇W . We’ll see how this fits with gravity fields in a moment. If F =+∇W , then ∇∧ F = 0 (vector identity). Now consider the surface integral ∇∧ F · dS. We know this is zero, but: R ∇∧ F · dS = F · dl = 0 (9) Z I by Stokes’s theorem. If the integral of the work done around a closed path is zero, then F has zero curl, and is said to be irrotational.

Figure 4: Conditions associated with potentials

3 Equipotential surfaces An equipotential surface is one on which the potential remains constant:

φ(x,y,z)= const

If ˆs is a unit vector lying tangent to an equipotential surface of F, then ∂φ ˆs · F = ∂s and must vanish. Therefore field lines are always perpendicular to their equipotential surface. No work is done in moving a particle along an equipo- tential. The distance between equipotential surfaces is a measure of the density of field lines. Thus the force field F has the greatest intensity in regions where equipotential surfaces are closest.

Examples of potentials Gravity field Newton proposed the law of gravitation in 1687:

Gm1m2 |F| = r2 where m1 and m2 are two masses separated by a distance of r. m r^

r ^ r is directed from the source to the test mass

Figure 5: Definitions of m, r and ˆr

The force of one particle is directed towards the other particle; G is a constant of proportionality, the gravitational constant

−11 3 −1 −2 G = 6.67 × 10 m kg s

Let m1 be a test particle of unit mass. Then the force on it due to m2 = m is: −Gm F = ˆr (10) r2 where ˆr is a unit vector directed from the source mass to the test mass.

4 Epot r

Figure 6: Potential variation with r

Take unit mass from infinity to r, then the work done by the field is:

work done by field = + F · dr (11) Z r Gm = − 2 dr (12) Z∞ r Gm = (13) r = U (14)

This is the negative of the particle’s potential energy. So Gm E (r)= − pot r where Epot is the potential energy. We have chosen the constant of integration so that the particle has zero P.E. at ∞. Now the g field is given by: −Gm g =+∇U = ˆr (15) r2 where m is the mass causing g and ˆr is the vector from m to the observation point. NB The literature is confusing. The sign convention is arbitrary. We follow Blakely. He uses γ for G. Telford et al, amongst others, chose g = −∇U.

Other conservative potentials Any central force F = f(r)ˆr can be represented as the gradient of a potential.

5 One example arising in geophysics is the centrifugal potential, for which:

2 F = rω ˆr (16)

It has a potential φ = 1/2ω2r2

Units In S.I. g has units ms−2 (it is an acceleration). An old system used 1 Gal = 1 cms−2 = 10−2ms−2. 1 mGal is ≈ 10−6 the acceleration of gravity at the Earth’s surface. Beware: gravity unit (g.u.) is also used, 1 g.u. = 0.1 mGal.

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