Dynamics of an isolated, viscoelastic, self-gravitating body

Abstract

This paper is devoted to an alternative model for a rotating, isolated, self-gravitating, viscoelastic body. The initial approach is quite similar to the classical one, present in the works of Dirichlet, Riemann, Chandrasekhar, among others. Our main contribution is to present a simplified model for the motion of an almost spherical body. The Lagrangian function ${\fancyscript{L}}$ and the dissipation function ${\fancyscript{D}}$ of the simplified model are:

$$\begin{aligned} {\fancyscript{L}}=\frac{\omega \cdot \mathrm{I}\omega }{2}+ \frac{1}{36\, \mathrm{I}_\circ }(\Vert \dot{Q}\Vert ^2-\gamma \Vert Q\Vert ^2) \end{aligned}$$

and

$$\begin{aligned} {\fancyscript{D}}=\frac{\nu }{36\, \mathrm{I}_\circ }\Vert \dot{Q}\Vert ^2 \end{aligned}$$

where $\omega$ is the angular velocity vector, $Q$ is the quadrupole moment tensor, $\mathrm{I}=\mathrm{I}_\circ \mathrm{\!\ \mathbb {I}d\!\ } -Q/3$ is the usual moment of inertia tensor with $\mathrm{I}_\circ$ equal to the moment of inertia of the spherical body at rest, $\gamma$ is an elastic constant, and $\nu$ is a damping coefficient. The angular momentum $\mathrm{I}\omega$ transformed to an inertial reference frame is conserved. The constants $\gamma$ and $\nu$ must be determined experimentally. We believe this to be the simplest model one can get without loosing the symmetries and the conserved quantities of the original problem. This model can be used as a building block for the study of many-body planetary systems.

Appendices

Appendix 1

We perform the Taylor expansion of (32) up to order $\varepsilon ^2$. If $A=\exp (\varepsilon B) \in \mathrm{\!\ SSym_{+}(3)\!\ }$ then:

$$\begin{aligned} \phi (\varepsilon B)= & {} -\frac{G\rho ^2}{2} \int _{\fancyscript{B}}\int _{\fancyscript{B}}\frac{1}{\Vert \exp (\varepsilon B)(x-y)\Vert }dxdy+C\\= & {} -\gamma \int _{0}^{\infty }\frac{1}{\sqrt{\det (\exp (2\varepsilon B)+\lambda \mathrm{\!\ \mathbb {I}d\!\ } )}}d\lambda +C. \end{aligned}$$

Thus, (choosing $\phi (0)=0$),

$$\begin{aligned} \phi (\varepsilon B) = \varepsilon D\phi (0) B+\frac{1}{2}\varepsilon ^2 D^2\phi (0)B^2+\fancyscript{O}(\varepsilon ^3). \end{aligned}$$

Since the derivatives of the integrand are continuous and bounded, we may perform the following calculations

$$\begin{aligned}&D\phi (\varepsilon B)B=\frac{d}{d\varepsilon }\phi (\varepsilon B)=\gamma \int _{0}^{\infty }\frac{\mathrm{\!\ Tr\!\ } \left( (\exp (2\varepsilon B)+\lambda \mathrm{\!\ \mathbb {I}d\!\ } )^{-1}\exp (2\varepsilon B)B \right) }{\sqrt{\det (\exp (2\varepsilon B)+\lambda \mathrm{\!\ \mathbb {I}d\!\ } )}}d\lambda ,\\&D^2\phi (\varepsilon B)B^2=\frac{d}{d\varepsilon }D\phi (\varepsilon B)B=-\gamma \int _{0}^{\infty }\frac{\mathrm{\!\ Tr\!\ } \left( (\exp (2\varepsilon B)+\lambda \mathrm{\!\ \mathbb {I}d\!\ } )^{-1}\exp (2\varepsilon B)B \right) ^2}{4\sqrt{\det (\exp (2\varepsilon B)+\lambda \mathrm{\!\ \mathbb {I}d\!\ } )}} \\&\quad -\frac{\mathrm{\!\ Tr\!\ } \left( -4(\exp (2\varepsilon B)+\lambda \mathrm{\!\ \mathbb {I}d\!\ } )^{-2}\exp (4\varepsilon B)B^2 + 4 (\exp (2\varepsilon B)+\lambda \mathrm{\!\ \mathbb {I}d\!\ } )^{-1} \exp (2\varepsilon B)B^2 \right) }{2\sqrt{\det (\exp (2\varepsilon B)+\lambda \mathrm{\!\ \mathbb {I}d\!\ } )}}d\lambda . \end{aligned}$$

Hence,

$$\begin{aligned} D\phi (0)B= & {} 0,\\ D^2\phi (0)B^2= & {} -4\gamma \int _{0}^{\infty }\frac{1}{2(1+\lambda )^{\frac{3}{2}}}\left( (1+\lambda )^{-2}-(1+\lambda )^{-1} \right) d \lambda \mathrm{\!\ Tr\!\ } (B^2)=\frac{8}{15}\gamma \mathrm{\!\ Tr\!\ } (B^2). \end{aligned}$$

Recalling that $\gamma =3 M^2 G/(10R)$, we get

$$\begin{aligned} \phi (\varepsilon B)= \frac{2}{25} \frac{M^2 G}{R} \varepsilon ^2 \mathrm{\!\ Tr\!\ } (B^2) +\fancyscript{O}(\varepsilon ^3). \end{aligned}$$

(61)

Appendix 2

Love (1944) (chapter XI) studied the relative equilibria of Eq. (20) for an elastic, incompressible, homogeneous, and isotropic body. Love assumed, as we did, that the body angular velocity was small in such a way that he could linearize the problem around the spherical shape of equilibrium. Then he used spherical coordinates $(r,\theta ,\phi )$ to write the deformed surface of the body as $r=R+\varepsilon S$, where $S=S(\theta ,\phi )$ and $R$ is the radius of the sphere of reference. Finally, he supposed $\varepsilon S$ to be expanded in a series $\sum \varepsilon _n S_n$ of surface spherical harmonics and following this expansion he also expanded all other quantities in the problem: pressure, deformation, and boundary conditions, in spherical harmonics and powers of $r$. Equating powers of $r$ he obtained recurrence relations for the coefficients of the spherical harmonics and solved the linear problem. Up to second order spherical harmonics the parametrization for the deformed surface becomes

$$\begin{aligned} r=R\left( 1-\frac{2}{3}\varepsilon \left( \frac{3}{2}\cos ^2\theta -\frac{1}{2} \right) \right) , \end{aligned}$$

(62)

where $\varepsilon$ is the flattening of the body up to first order in $\varepsilon$. The deformation $u$ of the body (here we follow the notation of Love and writes $u(x)$ for what we had previously written as $u(x)-x$) up to the second order, is

$$\begin{aligned} \mathbf {u}(x_1,x_2,x_3)= -A_2 r^2 \nabla p_2-B_2 p_2 \mathbf {r}- \nabla \phi _2 \end{aligned}$$

where $p_2, \phi _2$ are spherical solid harmonics:

$$\begin{aligned} p_2=\frac{2}{3} \left( \frac{r}{R}\right) ^2 \left( \frac{3}{2}\cos ^2\theta -\frac{1}{2} \right) = \frac{1}{3R^2}\left( 2x_3^2-x_1^2-x_2^2 \right) \end{aligned}$$

and $\phi _2$ is proportional to $p_2$ (we are following Love’s notation). The equations of motion, boundary conditions and $\mathrm{\!\ div\!\ } \, \mathbf {u}=0$ imply (Love (1944), chapter XI, paragraph 177, equations (22) to (28)) $A_2=-5 B_2/4$ and $2\phi _2=-R^2(B_2+4A_2)p_2$ . So,

$$\begin{aligned} \mathbf {u}= & {} - \frac{B_2}{6R^2}\left( 3x_1^3+3x_1x_2^2+ 9x_1x_3^2-8R^2 x_1,3x_2^3+3x_2x_1^2+9x_2x_3^2-8R^2x_2,\right. \\&\left. -6x_3^3-12x_3x_1^2- 12x_3x_2^2+16R^2x_3 \right) . \end{aligned}$$

Using the definition of flattening in Eq. (2), Eq. (62), and that the surface of the deformed body is also given by $x+u(x)$ with $\Vert x\Vert =R$ we get a relation between the flattening $\varepsilon$ and the coefficient $B_2$: $\varepsilon =5B_2/(2(1+5B_2/6))\approx 5B_2/2$. Now, using the elastic energy formula (21) and the elastic strain tensor in Eq. (23) we obtain that the elastic energy of Love’s deformation as a function of $\varepsilon$ is:

$$\begin{aligned} E_{Love}=\mu \int _{{\fancyscript{B}}} \mathrm{\!\ Tr\!\ } (D\mathbf {u}^2)(\mathbf {x}) d\mathbf {x} = \left( \frac{4\pi R^3}{3}\right) \left( \frac{2\cdot 19}{3\cdot 25}\mu \right) \varepsilon ^2= \frac{k_{Love}}{2}\varepsilon ^2. \end{aligned}$$

(63)

In our approach, the deformation of the body is a priori imposed as $e^{\varepsilon B}x -x \approx \varepsilon B x$ where $\varepsilon$ is the flattening of the ellipsoid up to first order in $\varepsilon$, and

$$\begin{aligned} B=\frac{1}{3}\left( \begin{matrix} 1 &{} 0 &{} 0 \\ 0 &{} 1 &{} 0 \\ 0 &{} 0 &{} -2 \end{matrix}\right) . \end{aligned}$$

The elastic energy associated to this deformation is

$$\begin{aligned} E=\mu \int _{{\fancyscript{B}}} \mathrm{\!\ Tr\!\ } ((\varepsilon B)^2) d\mathbf {x}= \left( \frac{4\pi R^3}{3}\right) \left( \frac{2}{3}\mu \right) \varepsilon ^2 =\frac{k}{2}\varepsilon ^2. \end{aligned}$$

(64)

So, the elastic rigidity with respect to ellipticity $\varepsilon$-deformations found by Love and ours have a ratio $k_{Love}/k=19/25$. This is exactly the ratio between the elastic contribution to the Love numbers of each model as it should be. Notice that Love solves the Euler–Lagrange equations associated to the elastic energy functional up to first order in $\varepsilon$. Indeed he is minimizing this energy functional under the boundary conditions of linear elasticity. Our deformation neither satisfies the elastic boundary conditions nor minimizes the elastic functional energy. So, our a priori imposed deformation overestimates both the elastic energy and the stress while Love’s deformation gives the correct value up to order $\varepsilon$. Since both models lead to a deformed surface represented by the same ellipsoid up to first order in $\varepsilon$, the gravitational energy of both models coincide.

Appendix 3

In Appendix 2 we presented a quantitative comparison between our results and some of those obtained by Love for the equilibrium of a rotating elastic body. The computations of Love are involved which makes the comparison not easy. In this appendix we use a simple example to explain the qualitative difference between our approach and that by Love.

Consider a homogeneous rod of natural length $\ell$, elastic modulus $\lambda$, and linear density $\rho$ rotating with constant angular velocity $\omega$ around its center point. Let $s\in [-\ell /2,\ell /2]$ be a point in the undeformed rod and $p(s)$ be the position of this point after deformation (Lagrangian description). If $u(s)=p(s)-s$ denotes the deformation of the point originally at $s$ then the equilibrium equation is:

$$\begin{aligned} -\lambda u^{\prime \prime }(s) =\rho \omega ^2s, \qquad u^\prime (-\ell /2)=u^\prime (\ell /2)=0 \end{aligned}$$

(65)

This equation is analogous to the equation solved by Love in the sense that both are linear equations that were obtained under the assumption of small deformations and linear constitutive relations. The solution to problem (65) is:

$$\begin{aligned} u^\prime (s)=-\frac{c s^2}{2}+ \frac{c \ell ^2}{8}, \quad u(s)=-\frac{c s^3}{6}+ \frac{c \ell ^2 s}{8}, \quad \mathrm{where}\quad c=\frac{\rho \omega ^2}{\lambda } \end{aligned}$$

Since the problem treated by Love is three-dimensional, he was not able to obtain a simple solution as this one. He obtained a solution in the form of an infinite series in powers of a scalar parameter $\varepsilon$ that essentially measures the flattening of the deformed body. His solution satisfies the boundary conditions accordingly. Notice that the solution to equation (65), which is linear, is a nonlinear function (a cubic polynomial). The same happens for the solution found by Love. Now, the solution to problem (65) is the critical point of the functional

$$\begin{aligned} u\rightarrow E(u)=\int _{-\ell /2}^{\ell /2} \left[ \frac{\lambda }{2}u^{\prime 2}-\rho \omega ^2s u \right] ds, \quad \mathrm{with}\quad u^\prime (-\ell /2)=u^\prime (\ell /2)=0 \end{aligned}$$

(66)

The first term in $E$

$$\begin{aligned} E_{e}(u)=\int _{-\ell /2}^{\ell /2} \frac{\lambda }{2}u^{\prime 2}ds \end{aligned}$$

is the elastic energy of the rod. The second term

$$\begin{aligned} E_c(u)=-\int _{-\ell /2}^{\ell /2} \rho \omega ^2s u ds \end{aligned}$$

is the centrifugal energy of the rod in the following sense. The energy spent to move a point particle of mass $m$ from the origin to a point $s$ in the centrifugal field $\omega ^2 s$ is $-m\omega ^2s^2/2$. So the centrifugal energy of the rod after deformation minus the centrifugal energy before deformation is

$$\begin{aligned} -\int _{-\ell /2}^{\ell /2} \rho \frac{\omega ^2}{2} {\underbrace{[s+u]}_{=p(s)}}^2 ds+ \int _{-\ell /2}^{\ell /2} \rho \frac{\omega ^2 s^2}{2} ds= -\int _{-\ell /2}^{\ell /2} \rho \omega ^2s u ds-\int _{-\ell /2}^{\ell /2} \rho \frac{\omega ^2}{2} u^2 ds \end{aligned}$$

that is equal to $E_c(u)$ except for the negligible term which is quadratic in $u$. It is important to say that we are disregarding variations of density after the deformation of the rod. This can be done when the elastic constant is large compared to the centripetal force, namely, if

$$\begin{aligned} \frac{\rho \omega ^2\ell ^2}{\lambda }\ll 1 \end{aligned}$$

The length of the deformed rod minus its natural length is

$$\begin{aligned} \delta =2p(\ell /2)-\ell =2u(\ell /2)= 2\left( \frac{\ell }{2}+\frac{c\ell ^3}{24}\right) -\ell =\frac{c\ell ^3}{12} \end{aligned}$$

(67)

The elastic energy stored in the rod is

$$\begin{aligned} E_e=\int _{-\ell /2}^{\ell /2} \frac{\lambda }{2}u^{\prime 2}ds= \frac{1}{240}\lambda \ell ^5 c^2=\frac{1}{2}\frac{144}{120} \frac{\lambda }{\ell }\delta ^2 \end{aligned}$$

(68)

The centrifugal energy of the rod is

$$\begin{aligned} E_c(u)=-\int _{-\ell /2}^{\ell /2} \rho \omega ^2s u ds=-\frac{\rho \omega ^2\ell ^5c}{120}= -\frac{\rho \omega ^2\ell ^2}{10}\delta \end{aligned}$$

(69)

The length of the rod as a function of $\omega ^2$ is given by the critical point of the total energy of the rod as a function of $\delta$:

$$\begin{aligned} 0=\frac{d}{d\delta }E=\frac{d}{d\delta } \left\{ \frac{1}{2}\frac{144}{120}\frac{\lambda }{\ell }\delta ^2-\frac{\rho \omega ^2\ell ^2}{10}\delta \right\} \Longrightarrow \delta = \frac{\rho \ell ^3}{12\lambda }\omega ^2 \end{aligned}$$

that gives exactly the value in Eq. (67), as expected.

Our approach to problem (65) is the following. We a priori assume that the deformation is linear, $u(s)=bs$, where $b$ is a free parameter (Love obtains a nonlinear deformation as a solution of a linear equation and we propose a linear deformation as an “approximation” to the real solution). The linear deformation neither solves the equation nor the boundary conditions of problem (65). In order to determine $b$ we use the variational characterization (66) of problem (65) to obtain the function:

$$\begin{aligned} b\rightarrow \int _{-\ell /2}^{\ell /2} \left[ \frac{\lambda }{2}b^{2}-\rho \omega ^2s^2b\right] ds =\frac{\lambda b^2 \ell }{2}-\frac{\rho \omega ^2\ell ^3b}{12}. \end{aligned}$$

The critical point of this function is

$$\begin{aligned} b=\frac{\rho \omega ^2\ell ^2}{12\lambda }=\frac{c\ell ^2}{12} \end{aligned}$$

In this case the length of the deformed rod minus its natural length is

$$\begin{aligned} \delta =2p(\ell /2)-\ell =2u(\ell /2)=b\ell =\frac{c\ell ^3}{12}, \end{aligned}$$

(70)

the elastic energy stored in the rod is

$$\begin{aligned} E_e=\int _{-\ell /2}^{\ell /2} \frac{\lambda }{2}u^{\prime 2}ds= \frac{\lambda b^2\ell }{2}=\frac{1}{2}\frac{\lambda }{\ell }\delta ^2 \end{aligned}$$

and the centrifugal energy of the rod is

$$\begin{aligned} E_c(u)=-\int _{-\ell /2}^{\ell /2} \rho \omega ^2s u ds=-\frac{\rho \omega ^2\ell ^3b}{12}= -\frac{\rho \omega ^2\ell ^2}{12}\delta \end{aligned}$$

As above, the length of the rod as a function of $\omega ^2$ is given by the critical point of the total energy of the rod as a function of $\delta$:

$$\begin{aligned} 0=\frac{d}{d\delta }E=\frac{d}{d\delta } \left\{ \frac{1}{2}\frac{\lambda }{\ell }\delta ^2 -\frac{\rho \omega ^2\ell ^2}{12}\delta \right\} \Longrightarrow \delta = \frac{\rho \ell ^3}{12\lambda }\omega ^2 \end{aligned}$$

that gives exactly the value in Eq. (70), as expected, and also the same value in Eq. (67), as not expected. Therefore in this case the approximated method, which is similar to that used in the paper, gives exactly the same rigidity with respect to variations of $\omega ^2$ as the exact solution. In general this does not happen. For instance, the elastic rigidity we obtain in our paper is close to (but different from) that obtained by Love with a rigorous analysis.