Elasticity is founded on displacement fields, Stokes flows on velocity fields, and tides on acceleration fields. Denote g(x,t) as the acceleration field. In Newtonian theory it is a gradient,

g= -grad Phi, (1)

where Phi is the Newtonian gravitational potential.

Now, the gradient of the gravitational acceleration (1),

E_(ij)= (1/2)[d_i g_j+ d_j g_i], (2)

where d_i denotes the partial differential in the i direction. In Newtonian theory, the strain (2)

is given by

E = - grad grad Phi, (3)

which is symmetric, but not traceless. To make it so, so that it corresponds to a strain, we subtract off the diagonal component

E= - grad grad Phi + (1/3) div grad Phi I (4)

where I is the identity tensor.

If the elastic tensor, C, is constant and strongly elliptic then

C_(ijkl)x_i x_j y_k y_l >0,

for any pair of vectors x,y in a region R. The equilibrium conditions are

C_(ijkl) g_(k,lj) =0, (5)

where the comma denotes differentiation, and suppose that g vanishes on the boundary dR. In terms of the Lame coefficient lambda and shear stress, mu, both of which we assume are constant, the elasticity tensor is

C_(ijkl)= lambda d_(ij) d_(kl) + mu {d_(ik) d_(jl) + d_(il)d_(jk)},

where d_(ij) denotes the Kroneckar delta function. Strong ellipicity is sure when

mu >0, lambda _2 mu > 0.

Since g=g(r) is only radial, the stress field,

s_(ij) = C_(ijkl) g_(k,l),

has components

s_(rr)= [lambda+2mu] g'(r) + 2 lambda g(r)/r (6a)

s_(theta theta) =s_(phi phi) = lambda g'(r) + 2[lambda+mu]g(r)/r (6b)

and all other components are zero, where the prime denotes differentiation with respect to r. The equilibrium equations are

{(lambda+2 mu)g' +2 lambda g/r}' + 4 mu {g/r}' =0.

The terms (lambda+2 mu) factor out leaving

g" + 2g'/r -2 g/r^2 =0,

whose general solution is

g(r) = c_1 r + c_2/r^2, (7)

and c_1 and c_2 are integration constants.

If we consider two spherical shells at r=a and r=b, with b>a, we can consider several cases.

The most important one is "mixed" one where (7) vanishes on a and (6a) vanishing at b. The former leads to the identification of the constants in (7), c_1= w^2, the square of the angular velocity, and c_2=m, the gravitational parameter. Then the vanishing of (7),

w^2 a^3 = m =constant, (8)

which is none other than Kepler's III. Then the vanishing of (6a) at r=b leads to

3 lambda+ 2 mu =4 mu (a/b)^3,

where

m/w^2 =(a/b)^3.

As a->0, we have pure centrifugal acceleration, and the limiting condition

lambda= - (2/3) mu, (9)

at the point where the bulk modulus vanishes. The bulk modulus is a measure of how resistant a material is to compression.

The limiting case (9) occurs when the radial stresses vanish at both r=a and r=b, i.e,

(3 lambda +2 mu)/4 mu = m/a^3 = m/b^3 . (10)

Since a and b are distinct, (10) can only be satisfied it (9) holds and m=0, i.e., there is no gravitational acceleration so that the acceleration is purely centripetal.

Now, what appears as a little confusing is that we are thought to think of centripetal acceleration as being synonymous with Newton's inverse square law. If alpha is the angle between the force F directed to the focus and the radius r directed to the origin of the instantaneous circle tangent to the trajectory at the point in question, then we have

F sin alpha = v^2/p = h^2/r^2 p sin^2 alpha, (10)

where p is the radius of curvature, and h is the angular momentum per unit mass. (The second equality in (10) is just the definition of angular momentum h= r^2 w sin alpha since v=wr. )

It turns out that

p sin^3 alpha =constant

for all conics, ellipses, circles, hyperbolas and parabolas, and the central force in (9) becomes

F r^2 = h^2= constant,

which is Newton's inverse-square law. So we are inclined to think that Kepler's third will always be obeyed. This is not true, and it will hold usually at one point; in the present case where the gravitational acceleration vanishes at r=a.

The surface that describes a geoid contains both centripetal acceleration and gravitational acceleration. In a non-rotating earth, the surface of the earth r=a (mean radius) has only a gravitational potential acting on it from the outside where g=m/a^2 is the surface gravity. But since the earth is rotating it won't remain spherical because it will be deformed by centrifugal forces. The new terms in the potential will contain a second-order Legendre polynomial, P_2(cos theta), at lowest order.

The potential outside the earth will be

V= (m/a)[1-(r/a)^2 P_2(cos theta)] + (1/3)w^2r^2[1-P_2(cos theta)],

where the P_2 terms take into account the deformations due to tidal forces (1/r^3) and to the centrifugal forces. The values of the angles, theta =0, +/- pi/2 will represent the principal axes of an ellipsoid. If we consider only the deformation terms in P_2

Delta V = -{m/a^3 - (1/3)w^2}r^2 P_2(cos theta),

and apply Kepler's third then two are comparable, the rotational case being -1/3 less than the gravitational case. Consider the surfaces

r=a[ 1+ X P_2(cos theta)] and r=a[1-(1/3)X P_2(cos theta)],

where X is a small constant, the principles axes for gravitational defomation is

a = (1+ X) for theta=0, and b = c =(1-X/2) for theta =+/- pi/2. (11)

For rotational deformation

a=b =(1+X/6) for theta=+/- pi/2 and c =a (1- X/3) for theta=0. (12)

The gravitational deformation (11) yields a prolate ellipsoid a>b=c, while the rotational deformation, (12) an oblate ellipsoid a=b>c. Note that the sum of the principal axes always remains 3a. [cf Murray & Dermott, "Solar System Dynamics" p. 156]

Thus, all accelerations are not equivalent, and can be distinguished by the deformations they cause. We have contrasted these types of deformations with the sectional curvatures they cause in the previous blog. It is also important to realize that so long as the strain tensor (4) remains symmetric there can be no Faraday effect that would define a Hodge dual.

## Opmerkingen