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the In case of seismic data, we actually obtain two velocity profiles: vP(r) and vS(r), respectively the velocities of P-waves and S-waves. These are related to the material properties by KS + 4 3 = , 2 vS = 2 vP Here KS is the adiabatic modulus of incompressibility (or the bulk modulus; we already defined it for zero temperature), and is the shear modulus. Note K S = -(P / ln V ) S (adiabatic bulk modulus) Both of these coefficients have the units of pressure (force per unit area), and they represent the force per unit area that arises as a result of a relative distortion of the medium, either a pure relative change of volume in the case of the incompressibility modulus, or a pure shear (no change in volume, just a change in shape). Obviously, for a liquid or a gas (no long-range order), the P-wave velocity just becomes the sound velocity: P 2 vP = , S and the S-wave velocity vS drops to zero because =0 (there are no S-waves). With seismic data available for a planet, we can compute the socalled seismic parameter: Feb. 28 p. 1 2 2 ( r ) = v P - 4 vS = 3 K S P = S Now if we assume that within the planet dP P d S we get the Adams-Williamson equation: 1 d 2 d r = -4 G r 2 dr dr Temperature corrections to the A-W eqn.: dP M (r ) = - g = -G 2 dr r . Now P CV T = CV . = T T S Differentiate the equation of state P = P ( , T ) : dP P d P dT = + dr dr T dr T 1 d dT = KT + CV . dr dr Now Feb. 28 p. 2 T P KT = K S - CV T so, including temperature corrections, we get C T d dP 1 d dT = KS + CV - 2 V . dr dr dr dr If the temperature follows an adiabat, no corrections are needed. The actual temperature distribution is not far from adiabatic, so the corrections are small. For a planet without any interior discontinuities, the 2nd-order A-W D.E. can just be integrated to solve for the planet's structure. But if we have a major discontinuity, as is the case for the Earth's coremantle boundary, we need an additional boundary condition. Otherwise, the value of the density discontinuity cannot be determined from the existing boundary conditions: 2 Feb. 28 p. 3 Traditionally, the additional boundary condition for the Earth is the Earth's axial moment of inertia, C, defined by C = 2 d 0 a a 2 -2 2 - a 2 -2 dz (, z ) , where is the distance from the planet's rotation axis in a cylindrical coordinate system, and z is the distance along the rotation axis, measured from the center of gravity: Feb. 28 p. 4 For a uniform-density sphere, C=0.4 Ma2. The Earth's C can be independently determined by two different techniques (to be discussed later), with the result CEarth=0.33 Ma2. Here is the profile of the Earth's core, superimposed on the Fe calculations presented earlier: Feb. 28 p. 5 So far, seismology is only available for the Earth (and, in a very limited way, for the Moon). Feb. 28 p. 6 For determining interior structure of a planet, we thus have the following approaches: (a) M=M(a) (very crude) (b) Use the seismic parameter (r) to get the density profile (r) (Earth only) (c) Use free-oscillation frequencies to constrain interior structure (will not be discussed in this class; so far only available for the Earth and the Sun). (d) Measure the response of the planet to perturbations: hydrostatic-equilibrium response, and departures from hydrostatic equilibrium. We will now go on to discuss (d): Gravitational Potentials: We will be working in a spherical polar coordinate system defined as follows: Feb. 28 p. 7 The differential-equation form for the gravitational potential is (see Jackson): 2 V = -4 G (Poisson's equation). Outside the planet, where the mass density is zero, this becomes Laplace's equation: 2V = 0 In the coordinate system that we are using, the general solution to Laplace's equation is most conveniently given in the form of an expansion in spherical harmonic functions. These functions have the property that they are complete and orthonormal. This will be Feb. 28 p. 8 very convenient for analysis later on. As discussed Jackson, by the general solution to Laplace's equation can be written as V (r , , ) = l =0 m = - l [ lm r l + lm r -(l +1) ] Ylm ( , ) +l For solutions of interest to us, the terms that blow up at infinite r should be discarded. So we set all the alphas equal to zero. The problem is now to relate the beta coefficients, which can be measured by spacecraft, to something of interest about the planet. First, here are some mathematical relations from the relevant sections in Jackson: Feb. 28 p. 9 Feb. 28 p. 10 Feb. 28 p. 11 Here is a convenient table: Feb. 28 p. 12 Normally, one generates the Ylm for high l,m with a numerical code. Here is an example for you to play with. Note, however, that the NORMALIZED associated Legendre polynomials calculated by this code, are different by a normalizing factor from the associated Legendre polynomials defined by Jackson. Lots of different normalizations are used in the planetary literature, so you have to be careful. (normalized) Plm = 2(l - m)!(2l + 1) m Pl (as defined by Jackson), for m 0 (l + m)! (normalized) Plm = (2l + 1) Pl m (as defined by Jackson), for m = 0 C 100 C 200 300 340 C C ------------------------------------------------C C generates NORMALIZED associated legendre polynomials C as a function of x = cos(theta) C C Note: P(1,1) = p_(00), etc. C subroutine nasleg(x,p,nulim) implicit double precision(a-h,o-z) sample program to show how NASLEG.F works implicit double precision(a-h,o-z) dimension p(61,61) common/lps2/pun(30) nulim=21 i=-1 i=i+1 if (i.ge.21) go to 300 x=(i-10)*0.1 call nasleg(x,p,nulim) analytic forms for P_2 and P_2^1: p20=1.5*x**2-0.5 p21=-3.*x*dsqrt(1.d0-x**2) znorm20=dsqrt(5.0d0) znorm21=dsqrt(5.d0/3.d0) write(*,340) x,p(3,1),p20*znorm20,p(3,2),p21*znorm21 go to 100 write(*,*) 'done' format(f6.3,4f12.6) end Feb. 28 p. 13 C C C C double precision n,m dimension p(61,61) as a bonus, the array pun(30) contains the UNNORMALIZED (i.e., usual) even Legendre polynomials P_2, P_4, P_6, ..., P_60 common/lps2/pun(30) mulim=nulim do 100 nu = 1,nulim do 100 mu = 1,mulim p(nu,mu) = 0.d0 continue a = dsqrt(3.d0) xp = dsqrt(1.d0-x*x) p(1,1)=1.d0 p(2,1)=a*x p(2,2)=a*xp do 300 mu=1,mulim m = dble(mu-1) if(mu.le.2) go to 200 a = dsqrt((2.d0*m+1.d0)/(2.d0*m)) p(mu,mu)=a*xp*p(mu-1,mu-1) continue do 300 nu=3,nulim n=dble(nu-1) if(nu.le.mu) go to 300 a = dsqrt((2.d0*n-1.d0)/(n-m)*(2.d0*n+1.d0)/(n+m)) b = dsqrt((2.d0*n+1.d0)/(2.d0*n-3.d0)* (n-m-1.d0)/(n-m)*(n+m-1.d0)/(n+m)) p(nu,mu) = a*x*p(nu-1,mu)-b*p(nu-2,mu) continue do 400 nu2=1,(nulim-1)/2 nu = 2*nu2 + 1 pun(nu2)=p(nu,1)/dsqrt(dble(2*(nu-1)+1)) continue return end 100 C C 200 C 1 300 C 400 C C With all these numerical preliminaries over with, let's try to look at the expression for the potential in a more intuitive way. We suppose that we are at an external point of observation at position r from the planet's center of mass, such that any mass element in the planet is always at a distance r' < r from the center of mass. Feb. 28 p. 14 Suppose that the angle between the mass element vector r' and the observer vector r is gamma: Now we can write r - r = r 2 + r 2 - 2r r = r 1 + , where r 2 r = 2 - 2 cos r r Now by hypothesis, < 1 , so expand in powers of : 1 1 1 1 = (1 - 1 + ...) , 2 r - r r 1 + r Feb. 28 p. 15 And, in particular (see Jackson), 1 r l = l +1 Pl (cos ) . r - r l = 0 r Here Pl ( x) = Pl m = 0 ( x) = ordinary Legendre polynomial. This allows us to carry out a series expansion for V(r): V (r , , ) = +l G 3 d r (r)[1 - 1 + ...] = 2 r l = 0 m = -l lm r -(l +1) Ylm ( , ) So we can use this expansion, comparing term by term, to figure out what the beta-coefficients represent in terms of integrals over the planet's mass distribution. [Hint: this is also a good approach to solving Homework #5]. In working out the integrals, we will need to use the expansion, derived in Jackson, 4 + l * Pl (cos ) = Ylm ( , ) Ylm ( , ) , 2l + 1 m = -l where ( , ) are the angular coordinates of vector r and ( , ) are the angular coordinates of vector r. Feb. 28 p. 16

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