From Satellite Gradiometry Data to Subcrustal Stress Due to Mantle Convection

Abstract

Subcrustal stress induced by mantle convection can be determined by the Earth’s gravitational potential. In this study, the spherical harmonic expansion of the simplified Navier–Stokes equation is developed further so satellite gradiometry data (SGD) can be used to determine the subcrustal stress. To do so, we present two methods for producing the stress components or an equivalent function thereof, the so-called S function, from which the stress components can be computed numerically. First, some integral estimators are presented to integrate the SGD and deliver the stress components and/or the S function. Second, integral equations are constructed for inversion of the SGD to the aforementioned quantities. The kernel functions of the integrals of both approaches are plotted and interpreted. The behaviour of the integral kernels is dependent on the signal and noise spectra in the first approach whilst it does not depend on extra information in the second method. It is shown that recovering the stress from the vertical–vertical gradients, using the integral estimators presented, is suitable, but when using the integral equations the vertical–vertical gradients are recommended for recovering the S function and the vertical–horizontal gradients for the stress components. This study is theoretical and numerical results using synthetic or real data are not given.

Appendices

Appendix A: Proofs of Closed-Form Formulae

All of the closed formulae can be derived based on the well-known formula of Legendre expansion of distance (cf. Heiskanen and Moritz 1967, p. 33):

$$ \frac{1}{l} = \sum\limits_{n = 0}^{\infty } {s^{n} P_{n} \left( {\cos \psi } \right)} $$

(15)

In order to obtain the closed formula for the kernel presented in Eq. (11c) take the derivative of both sides of (15) and multiply the results by:

$$ s^{3} \frac{\partial }{\partial s}\left( \frac{1}{l} \right) = s^{3} \left( {\frac{\cos \psi - s}{{l^{3} }}} \right) = \sum\limits_{n = 0}^{\infty } {ns^{n + 2} P_{n} \left( {\cos \psi } \right)} $$

(16)

Now if another derivative with respect to s is taken, the results will be

$$ \frac{\partial }{\partial s}\left[ {s^{3} \frac{\partial }{\partial s}\left( \frac{1}{l} \right)} \right] = 2s^{2} \left( {\cos \psi - 2s} \right) + \frac{{s^{3} }}{{l^{2} }}\left( {s - \cos \psi } \right)^{2} = \sum\limits_{n = 0}^{\infty } {n\left( {n + 2} \right)s^{n + 1} P_{n} \left( {\cos \psi } \right)} $$

(17)

when the contributions of zero- and first-degrees are removed from Eq. (17) the closed-form formula of the kernel of Eq. (11c) is derived.

Similarly, it will not be difficult to show that Eq. (13c) is

$$ s\frac{{\partial^{2} }}{{\partial s^{2} }}\left( {\frac{{s^{2} }}{l}} \right) = \frac{2s}{l} - \frac{{s^{2} }}{{l^{3} }}\left( {3s - 2\cos \psi } \right) + \frac{{s^{2} }}{{l^{3} }}\left( {s - \cos \psi } \right)\left( { - 2 + 3s\frac{s - \cos \psi }{{l^{2} }}} \right) $$

(18)

and by removing the contributions of the first- and second-degree Eq. (10d) we have:

$$ \begin{aligned} & s\frac{\partial }{{\partial s}}\left[ {s\frac{{\partial ^{2} }}{{\partial s^{2} }}\left( {\frac{{s^{2} }}{l}} \right)} \right] \\ & = \frac{{2s^{2} }}{l} + \frac{s}{{l^{3} }}\left( {s - \cos \psi } \right)\left[ { - 2 + 3\frac{s}{{l^{3} }}\left( {3s - 2\cos \psi } \right)} \right] \\ & \quad - \frac{{2s}}{{l^{3} }}\left( {3s - 2\cos \psi } \right) + \frac{{s^{2} }}{{l^{5} }}\left( {s - \cos \psi } \right)^{2} \left( {2 - 15\frac{s}{{l^{2} }}\left( {s - \cos \psi } \right)} \right) \\ & \quad + \frac{{s^{2} }}{{l^{5} }}\left[ {15s^{2} + 3\cos \psi \left( {3\cos \psi - 8s} \right)} \right] \\ \end{aligned} $$

(19)

Again, when the contributions of the zero- and first-degree are removed the kernel function will be obtained.

The proof of Eq. (12c) is simply followed based on the following relation:

$$ \sum\limits_{n = 0}^{\infty } {\left( {2n + 1} \right)s^{n} P_{n} \left( {\cos \psi } \right)} = \frac{{2s\cos \psi + l^{2} }}{{l^{3} }}. $$

(20)

The following closed-form formula has been derived by Martinec (2003):

$$ \sum\limits_{n = 1}^{\infty } \frac{( 2n + 1 )s^{n + 1} }{n( n + 1 )( n + 2 )} \frac{dP_{n} ( \cos \psi )}{d\psi } = \sin \psi \left[ \frac{3}{2l} + \frac{s^{2} ( l + 1 )}{2l( l + 1 - s\cos \psi )} + \left( 1 - \frac{3\cos \psi }{2s} \right) \times \left( \frac{1}{1 - \cos \psi } - \frac{l + s}{l\left( {l + s - \cos \psi } \right)} \right) - \frac{3}{2s}\ln \left( \frac{l + s - \cos \psi }{1 - \cos \psi } \right) \right] $$

(21)

To derive the kernel of Eq. (11f) it is enough to consider s = 1 and removing the first degree contribution.

Appendix B: Azimuths α and $\alpha^{\prime}$

Based on the well-known rule of cosine in spherical trigonometry one can write (cf. Heiskanen and Moritz 1967, p. 113):

$$ \cos \psi = \cos \theta \cos \theta^{\prime} + \sin \theta \sin \theta^{\prime}\cos \left( {\lambda^{\prime} - \lambda } \right) $$

(22)

and cos α will be:

$$ \cos \alpha = \frac{\cos \theta - \cos \theta \cos \psi }{\sin \theta \sin \psi }. $$

(23)

Substitution of Eq. (23) into Eq. (22) and simplification of the results yields:

$$ \cos \alpha = \frac{{\cos \theta^{\prime}\sin \theta - \sin \theta^{\prime}\cos \theta \cos \left( {\lambda^{\prime} - \lambda } \right)}}{\sin \psi } $$

(24)

According to sine rule, one can simply obtain:

$$ \sin \alpha = \frac{{\sin \theta \sin \left( {\lambda^{\prime} - \lambda } \right)}}{\sin \psi }. $$

(25)

The associated Legendre function of degree and order 1 has the following relation with the derivative of Legendre polynomials (cf. Heiskanen and Moritz 1967, p. 22):

$$ P_{n1} \left( {\cos \psi } \right) = \sin \psi \frac{{{\text{d}}P_{n} \left( {\cos \psi } \right)}}{{{\text{d}}\left( {\cos \psi } \right)}} $$

(26)

According to Eq. (26) the derivatives of Legendre polynomials with respect to θ and λ will be (Heiskanen and Moritz 1967, p. 22):

$$ \frac{{\text{d}P_{n} \left( {\cos \psi } \right)}}{{\text{d}\theta }} = \frac{{\text{d}P_{n} \left( {\cos \psi } \right)}}{{\text{d}\left( {\cos \psi } \right)}}\left( { - \sin \psi } \right)\frac{{\text{d}\psi }}{{\text{d}\theta }} = - P_{n1} \left( {\cos \psi } \right)\frac{{\text{d}\psi }}{{\text{d}\theta }} $$

(27)

$$ \frac{{\text{d}P_{n} \left( {\cos \psi } \right)}}{{\sin \theta \text{d}\theta }} = \frac{{\text{d}P_{n} \left( {\cos \psi } \right)}}{{\sin \theta d\left( {\cos \psi } \right)}}\left( { - \sin \psi } \right)\frac{{\text{d}\psi }}{{\text{d}\lambda }} = - \frac{{P_{n1} \left( {\cos \psi } \right)}}{\sin \theta }\frac{{\text{d}\psi }}{{\text{d}\lambda }} $$

(28)

By taking the derivatives of Eq. (22) with respect to θ and λ we obtain:

$$ \frac{{\text{d}\psi }}{{\text{d}\theta }} = \frac{{\sin \theta \cos \theta^{\prime} - \cos \theta \sin \theta^{\prime}\cos \left( {\lambda^{\prime} - \lambda } \right)}}{\sin \psi } = \cos \alpha $$

(29)

$$ \frac{{\text{d}\psi }}{{\text{d}\lambda }} = \frac{{\sin \theta^{\prime}\sin \left( {\lambda^{\prime} - \lambda } \right)}}{\sin \psi } = \sin \alpha $$

(30)

Finally, we can write:

$$ \begin{aligned} \frac{{{\text{d}}P_{n} \left( {\cos \psi } \right)}}{{{\text{d}}\theta }} & = - P_{{n1}} \left( {\cos \psi } \right)\cos \alpha \quad {\text{and}} \\ \frac{{{\text{d}}P_{n} \left( {\cos \psi } \right)}}{{\sin \theta {\text{d}}\lambda }} & = P_{{n1}} \left( {\cos \psi } \right)\sin \alpha . \\ \end{aligned} $$

(31)

Similarly, we can write the following relation for the azimuth from the integration point to the computation point:

$$ \cos \left( {360 - \alpha^{\prime}} \right) = \cos \alpha^{\prime} = \frac{{\cos \theta^{\prime} - \cos \theta^{\prime}\cos \psi }}{{\sin \theta^{\prime}\sin \psi }} $$

(32)

In such a case the derivatives of Eq. (22) should be taken now with respect to θ ^′ and λ ^′. After simplification of the results we have:

$$ \frac{{\text{d}\psi }}{{\text{d}\theta^{\prime}}} = \frac{{\sin \theta^{\prime}\cos \theta - \cos \theta^{\prime}\sin \theta \cos \left( {\lambda^{\prime} - \lambda } \right)}}{\sin \psi } = \cos \alpha^{\prime} $$

(33)

$$ \frac{{\text{d}\psi }}{{\text{d}\lambda }} = \frac{{\sin \theta \sin \left( {\lambda^{\prime} - \lambda } \right)}}{\sin \psi } = - \sin \alpha^{\prime} $$

(34)

Now according to Eq. (26) and similar to Eqs. (27) and (28) we obtain:

$$ \begin{aligned} \frac{{{\text{d}}P_{n} \left( {\cos \psi } \right)}}{{{\text{d}}\theta ^{\prime } }} & = - P_{{n1}} \left( {\cos \psi } \right)\cos \alpha ^{\prime } \quad {\text{and}} \\ \frac{{{\text{d}}P_{n} \left( {\cos \psi } \right)}}{{\sin \theta ^{\prime } {\text{d}}\lambda ^{\prime } }} & = P_{{n1}} \left( {\cos \psi } \right)\sin \alpha ^{\prime } . \\ \end{aligned} $$

(35)