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Next: Darwin-Radau equation Up: Yielding of an elastic Previous: Gravitational potential theory


Elastic response theory

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

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

and subject to the incompressibility constraint

$\displaystyle \nabla\cdot$$\displaystyle \mbox{\boldmath$\xi$}$$\displaystyle = 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)

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

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

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

Writing

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

Equation (C.24) yields

    $\displaystyle \frac{d}{dr}\left(p_0+\rho\,{\mit\Phi}_0\right)$ $\displaystyle = 0,$ (C.26)
and   $\displaystyle \mu\,\nabla^{\,2}$$\displaystyle \mbox{\boldmath$\xi$}$$\displaystyle -\nabla p_2$ $\displaystyle = 0.$ (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,

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

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

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

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

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

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

    $\displaystyle p_0(a)$ $\displaystyle = 0,$ (C.32)
and   $\displaystyle \left(p_2\,\frac{x_i}{r} - \frac{\mu}{r}\left[r\,\frac{\partial \...
...rtial r} - \xi_i +\frac{\partial (r\,\xi_r)}{\partial x_i}\right] \right)_{r=a}$ $\displaystyle = \rho\,g \left(\frac{2}{5}\,\delta_2-\zeta_2\right)\left(\frac{x_i}{r}\right)_{r=a},$ (C.33)

where

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

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

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

$\displaystyle \xi_i = A\,r^{\,2}\,\frac{\partial p_2}{\partial x_i} + B\,p_2\,x_i + \frac{\partial\phi_2}{\partial x_i},$ (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

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

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

    $\displaystyle \frac{\partial}{\partial x_i}(r\,\xi_r)$ $\displaystyle = (2\,A+B)\,r^{\,2}\,\frac{\partial p_2}{\partial x_i} + 2\,(2\,A+B)\,p_2\,x_i + 2\,\frac{\partial\phi_2}{\partial x_i},$ (C.38)
and   $\displaystyle r\,\frac{\partial\xi_i}{\partial r}-\xi_i$ $\displaystyle = 2\,A\,r^{\,2}\,\frac{\partial p_2}{\partial x_i} + 2\,B\,p_2\,x_i,$ (C.39)

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

    $\displaystyle \left[(2\,A+B)\,r^{\,2}\,p_2 + 2\,\phi_2\right]_{r=a}$ $\displaystyle = a\,\delta_2,$ (C.40)
and   $\displaystyle \left[1-4\,(A+B)\,\mu\right]\left(p_2\,x_i\right)_{r=a}$      
    $\displaystyle -\mu\left[(4\,A+B)\,r^{\,2}\,\frac{\partial p_2}{\partial x_i}+2\,\frac{\partial \phi_2}{\partial x_i}\right]_{r=a}$ $\displaystyle = \rho\,g\left(\frac{2}{5}\,\delta_2-\zeta_2\right) (x_i)_{r=a},$ (C.41)

respectively. The previous equation implies that

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

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

    $\displaystyle -2\,A\,a\left.p_2\right\vert _{r=a}$ $\displaystyle = \delta_2,$ (C.43)
and   $\displaystyle [1-4\,(A+B)\,\mu]\left.p_2\right\vert _{r=a}$ $\displaystyle = \rho\,g\left( \frac{2}{5}\,\delta_2-\zeta_2 \right),$ (C.44)

respectively.

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

    $\displaystyle 4\,A\,\mu + 5\,B\,\mu$ $\displaystyle = 0,$ (C.45)
and   $\displaystyle 10\,A\,\mu + 2\,B\,\mu$ $\displaystyle = 1,$     (C.46)

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

    $\displaystyle A\,\mu$ $\displaystyle = \frac{5}{42},$ (C.47)
and   $\displaystyle B\,\mu$ $\displaystyle = -\frac{4}{42}.$     (C.48)

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

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

where

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

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

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

As a specific example, suppose that

    $\displaystyle \zeta_2(\theta,\phi)$ $\displaystyle = -\zeta\,a\,P_2(\cos\theta),$ (C.53)
    $\displaystyle \delta_2(\theta,\phi)$ $\displaystyle = -\frac{2}{3}\,\epsilon\,a\,P_2(\cos\theta),$ (C.54)
and   $\displaystyle X_2(\theta,\phi)$ $\displaystyle = X\,P_2(\cos\theta),$     (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

    $\displaystyle X$ $\displaystyle = \rho\,g\,a\left(\zeta-\frac{4}{15}\,\epsilon\right),$ (C.56)
and   $\displaystyle \epsilon$ $\displaystyle = \frac{15}{38}\,\frac{X}{\mu}= \frac{15}{4}\,\frac{\zeta}{1+\tilde{\mu}},$ (C.57)

where

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

is the planet's effective rigidity.
next up previous
Next: Darwin-Radau equation Up: Yielding of an elastic Previous: Gravitational potential theory
Richard Fitzpatrick 2016-03-31