Elastic response theory

The interior of the planet is modeled as a uniform, incompressible, elastic solid possessing the isotropic stress-strain relation

$$\sigma_{ij} = -p\,\delta_{ij} + \mu\left(\frac{\partial \xi_i}{\partial x_j}+\frac{\partial \xi_j}{\partial x_i}\right),$$ (C.21)

and subject to the incompressibility constraint

$$\nabla\cdot \mbox{\boldmath$\xi$} = 0.$$ (C.22)

Here, $\sigma_{ij}({\bf r})$ is the stress tensor, $\delta_{ij}$ the identity tensor, $\xi$$({\bf r})$ the elastic displacement, $p({\bf r})$ the pressure, and $\mu$ the (uniform) rigidity of the material making up the planet (Riley 1974e).

Force balance inside the planet yields (Love 2011)

$$\frac{\partial\sigma_{ij}}{\partial x_j}=\rho\,\frac{\partial{\mit\Phi}}{\partial x_i}.$$ (C.23)

It follows from Equations (C.21) and (C.22) that

$$\mu\,\nabla^{\,2} \mbox{\boldmath$\xi$} - \nabla(p+\rho\,{\mit\Phi}) = {\bf0}.$$ (C.24)

Writing

$$p(r,\theta,\phi) = p_0(r) -\rho\,\mit\Phi_2(r,\theta,\phi) + p_2(r,\theta,\phi),$$ (C.25)

Equation (C.24) yields

$$\frac{d}{dr}\left(p_0+\rho\,{\mit\Phi}_0\right) = 0,$$ (C.26)

$$\mu\,\nabla^{\,2} \mbox{\boldmath$\xi$} -\nabla p_2 = 0.$$ (C.27)

Taking the divergence of the previous equation, and making use of Equation (C.22), we find that $\nabla^{\,2} p_2=0$, which implies that $p_2({\bf r})$ is a solid harmonic (of degree 2). Incidentally, $p_2$ would be zero were the planet in hydrostatic equilibrium.

It is helpful to define the radial component of the elastic displacement,

$$\xi_r = \frac{x_i\,\xi_i}{r},$$ (C.28)

as well as the stress acting (outward) across a constant $r$ surface,

$$X_i= -\frac{x_j\,\sigma_{ij}}{r} = p\,\frac{x_i}{r} - \frac{\mu}{... ...ial \xi_i}{\partial r}-\xi_i + \frac{\partial (r\,\xi_r)}{\partial x_i}\right],$$ (C.29)

where use has been made of Equation (C.21). Of course, the radial displacement at $r=a$ is equivalent to the displacement of the planet's surface:

$$\xi_r(a,\theta,\phi)=\delta_2(\theta,\phi).$$ (C.30)

The stress at any point on the surface $r=a$ must be entirely radial (because it would be impossible to balance a tangential surface stress), and such as to balance the weight of the column of displaced material directly above the point in question. In other words,

$$X_i(a,\theta,\phi) = g\,\rho\,\delta_2\left(\frac{x_i}{r}\right)_{r=a}.$$ (C.31)

It follows from Equations (C.18), (C.25), and (C.29) that

$$p_0(a) = 0,$$ (C.32)

$$\left(p_2\,\frac{x_i}{r} - \frac{\mu}{r}\left[r\,\frac{\partial \... ...rtial r} - \xi_i +\frac{\partial (r\,\xi_r)}{\partial x_i}\right] \right)_{r=a} = \rho\,g \left(\frac{2}{5}\,\delta_2-\zeta_2\right)\left(\frac{x_i}{r}\right)_{r=a},$$ (C.33)

where

$$\zeta_2(\theta,\phi) = - g^{\,-1}\chi_2(a,\theta,\phi).$$ (C.34)

Equations (C.16), (C.26), and (C.32) yield

$$p_0(r) = \frac{\rho\,g}{2\,a}\,(3\,a^{\,2}-r^{\,2}).$$ (C.35)

It remains to solve Equations (C.22) and (C.27), subject to the boundary conditions (C.30) and (C.33).

Let us try a solution to Equations (C.22) and (C.27) of the form

$$\xi_i = A\,r^{\,2}\,\frac{\partial p_2}{\partial x_i} + B\,p_2\,x_i + \frac{\partial\phi_2}{\partial x_i},$$ (C.36)

where $A$ and $B$ are spatial constants, and $\phi_2({\bf r})$ is a solid harmonic of degree 2 (Love 2011). It follows that

$$r\,\xi_r = (2\,A+B)\,r^{\,2}\,p_2 + 2\,\phi_2,$$ (C.37)

where use has been made of Equation (C.5). Moreover,

$$\frac{\partial}{\partial x_i}(r\,\xi_r) = (2\,A+B)\,r^{\,2}\,\frac{\partial p_2}{\partial x_i} + 2\,(2\,A+B)\,p_2\,x_i + 2\,\frac{\partial\phi_2}{\partial x_i},$$ (C.38)

$$r\,\frac{\partial\xi_i}{\partial r}-\xi_i = 2\,A\,r^{\,2}\,\frac{\partial p_2}{\partial x_i} + 2\,B\,p_2\,x_i,$$ (C.39)

where use has been made of Equations (C.5)–(C.7). Thus, the boundary conditions (C.30) and (C.33) become

$$\left[(2\,A+B)\,r^{\,2}\,p_2 + 2\,\phi_2\right]_{r=a} = a\,\delta_2,$$ (C.40)

$$\left[1-4\,(A+B)\,\mu\right]\left(p_2\,x_i\right)_{r=a}$$

$$-\mu\left[(4\,A+B)\,r^{\,2}\,\frac{\partial p_2}{\partial x_i}+2\,\frac{\partial \phi_2}{\partial x_i}\right]_{r=a} = \rho\,g\left(\frac{2}{5}\,\delta_2-\zeta_2\right) (x_i)_{r=a},$$ (C.41)

respectively. The previous equation implies that

$$\left[(4\,A+B)\,a^{\,2}\,p_2 + 2\,\phi_2\right]_{r=a} = 0.$$ (C.42)

Hence, the boundary conditions (C.40) and (C.41) reduce to

$$-2\,A\,a\left.p_2\right\vert _{r=a} = \delta_2,$$ (C.43)

$$[1-4\,(A+B)\,\mu]\left.p_2\right\vert _{r=a} = \rho\,g\left( \frac{2}{5}\,\delta_2-\zeta_2 \right),$$ (C.44)

respectively.

The expression for $\xi_i$ given in Equation (C.36) satisfies Equations (C.22) and (C.27) provided that

$$4\,A\,\mu + 5\,B\,\mu = 0,$$ (C.45)

$$10\,A\,\mu + 2\,B\,\mu = 1,$$ (C.46)

respectively, where use has been made of Equations (C.5)–(C.7). It follows that

$$A\,\mu = \frac{5}{42},$$ (C.47)

$$B\,\mu = -\frac{4}{42}.$$ (C.48)

Hence, the boundary conditions (C.43) and (C.44) yield

$$\delta_2 = h_2\,\zeta_2,$$ (C.49)

where

$$h_2 = \frac{5/2}{1+(19/2)\, (\mu/\rho\,g\,a)}.$$ (C.50)

The dimensionless quantity $h_2$ is termed a Love number of degree 2 (Love 2011).

The radial component of the elastic (i.e., non-hydrostatic) stress acting (outward) across the surface $r=a$ takes the form

$$X_2= \left(p_2\,\frac{x_i}{r} - \frac{\mu}{r}\left[r\,\frac{\part... ...rac{x_i}{r}\right)_{r=a} = -\rho\,g \left(\zeta_2-\frac{2}{5}\,\delta_2\right),$$ (C.51)

where use has been made of Equation (C.33). Equations (C.49) and (C.50) imply that this stress is related to the radial strain at the surface of the planet according to

$$\frac{\delta_2}{a} = -\frac{5}{19}\,\frac{X_2}{\mu}.$$ (C.52)

As a specific example, suppose that

$$\zeta_2(\theta,\phi) = -\zeta\,a\,P_2(\cos\theta),$$ (C.53)

$$\delta_2(\theta,\phi) = -\frac{2}{3}\,\epsilon\,a\,P_2(\cos\theta),$$ (C.54)

$$X_2(\theta,\phi) = X\,P_2(\cos\theta),$$ (C.55)

where $\zeta$ is a dimensionless measure of the strength of the tidal field, and $\epsilon$ is the tidally induced planetary ellipticity. It follows that

$$X = \rho\,g\,a\left(\zeta-\frac{4}{15}\,\epsilon\right),$$ (C.56)

$$\epsilon = \frac{15}{38}\,\frac{X}{\mu}= \frac{15}{4}\,\frac{\zeta}{1+\tilde{\mu}},$$ (C.57)

where

$$\tilde{\mu} = \frac{19}{2}\,\frac{\mu}{\rho\,g\,a}$$ (C.58)

is the planet's effective rigidity.