A Closed-Form Solution for a Linear Viscoelastic Self-gravitating Sphere

Abstract

Following up on the classical solutions by Love for a linear-elastic self-gravitating sphere, this paper presents the corresponding extension to a linear viscoelastic body of the Kelvin–Voigt type. The solution is expressed in closed form by making use of Laplace transforms. Applications to the genesis of terrestrial planets are sought and the evolution of the Love radius and possible extensions to large deformations are discussed. As a new result, it turns out that in the early days of planet formation there is no Love radius and that it takes time for the Love radius to develop.

Appendix: Love’s Solution—Its History in Modern Form

The following passages were primarily written for the benefit of readers who do not solve linear-elastic problems on a daily basis. However, they also contain interpretations not originally provided by Love, for example intuitive explanations of the meaning of the normalizing coefficients for the displacements and for the stresses.

1.1 A.1 The Primary Assumptions of Love and Their Limitations

Love’s solution for the state of deformation in a self-gravitating sphere is a static one. Hence the balance of momentum degenerates to the following equation:

$$\begin{aligned} \begin{gathered} \nabla \cdot \varvec{\sigma }=-\rho \varvec{f}, \end{gathered}\end{aligned}$$

(23)

where $\rho$ denotes the local current mass density, and $\varvec{\sigma }$ the Cauchy stress tensor. The specific body force, $\varvec{f}$, i.e., the gravitational acceleration, is conservative and originates from self-gravity. Hence a gravitational potential $U^\mathrm {grav}(\varvec{x})$ exists, where $\varvec{x}$ denotes an arbitrary (current) position within the body, and we may write:

$$\begin{aligned} \begin{gathered} \varvec{f}(\varvec{x})=-\nabla U^\mathrm {grav}(\varvec{x}). \end{gathered}\end{aligned}$$

(24)

The gravitational potential obeys Poisson’s equation:

$$\begin{aligned} \begin{gathered} \mathrm {\Delta } U^\mathrm {grav}(\varvec{x})=4\mathrm {\pi } G\rho (\varvec{x}). \end{gathered}\end{aligned}$$

(25)

For the stress tensor we initially assume that Hooke’s law holds, so that there are no rate effects:

$$\begin{aligned} \begin{gathered} \varvec{\sigma } = \lambda \; \mathrm {Tr} \varvec{ \epsilon } \; \varvec{\mathrm {I}} + 2 \mu \; \varvec{\epsilon }, \end{gathered}\end{aligned}$$

(26)

where the linear strain tensor has been used:

$$\begin{aligned} \begin{gathered} \varvec{\epsilon }=\tfrac{1}{2}({\nabla \varvec{u}+\nabla \varvec{u}^\top )}. \end{gathered}\end{aligned}$$

(27)

$\varvec{u}$ refers to the displacement vector, i.e., to $\varvec{u}=\varvec{x}-\varvec{X}$, $\varvec{X}$ being the reference position of a material point of the sphere. $\lambda$ and $\mu$ are Lamé’s elastic constants.

At this point three remarks for putting Love’s approach in perspective are in order, which are all related somehow. They all circle around the question “What happens if the deformations prove to be large?” We shall see that they can be large, we want to point out possible remedies, we will provide some citations for further reading, but we shall not endeavor to work it all out in this paper.

The first comment concerns the nabla operator used in the aforementioned equations: Note that all nabla operators above indicate differentiation w.r.t. the current spatial position, $\varvec{x}$. This is how the theory works for “linear elasticians:” There is only one gradient in linear theory of elasticity at small deformations, namely that one. Hence for them putting an emphasis on it sounds trivial. However, there is a world outside of linear elasticity as understood by Sokolnikoff (1956) or Timoshenko and Goodier (1951), in order to quote just two references of that denomination. Indeed, it is possible to understand linear elasticity as a limit case of nonlinear materials theory. Then Hooke’s law results in a natural way written in terms of gradients with respect to the reference configuration, $\varvec{X}$, see Wang and Truesdell (1973), pp. 170, or Müller (1973), p. 72. However, in the next breath it is said, see Truesdell and Toupin (1960), Sects. 57 and 301, that it does not really matter, and these gradients can be replaced by derivatives w.r.t. the current position, since the deformations are so small. What they do not say, though, is that it does matter from a principal, didactic point-of-view.

Second, from the standpoint of linear elasticity, the set of Eqs. (23)–(27) serves only one purpose: It allows us to calculate the displacement $\varvec{u}(\varvec{x})$. To this end the mass density $\rho (\varvec{x})$ must be considered as known, and for linear-elasticians it is, in the simplest case, a space-independent constant, $\rho =\rho _0$. If we relate it to our problem we may consider it to be the mass density of the homogeneous sphere before gravity has been switched on. However, recall that this does not mean that the current mass density is also a constant, even if it is one in the reference state. It is dependent on deformation and it can be determined from mass conservation. In general, now turning back to nonlinear theory for a moment, it is well known that we may write:

$$\begin{aligned} \begin{gathered} \rho (\varvec{x})=\frac{\rho _0}{\mathrm {det}\varvec{F}(\varvec{x})}, \end{gathered}\end{aligned}$$

(28)

where $\varvec{F}(\varvec{x})\equiv \nabla _{\varvec{X}}\varvec{x}$ is the deformation gradient pertinent to a material point. Recall once more that Eq. (28) is the result of the physical principle of local mass conservation and geometry, i.e., nonlinear kinematics, and, as such, it holds for arbitrary deformations. If we insist on studying small deformations, we must replace Eq. (28) by:

$$\begin{aligned} \begin{gathered} \rho (\varvec{x})\approx \rho _0\left[ 1-{{\mathrm{{\textsf {Tr}}}}}\,\varvec{\epsilon }(\varvec{x})\right] . \end{gathered}\end{aligned}$$

(29)

Consequently, the argument now runs as follows: Once the linear strain $\varvec{\epsilon }(\varvec{x})$ is known from a linear-elastic analysis based on the (static) balance of momentum in combination with Hooke’s law, during which the mass density is assumed to be spatially constant, the spatial distribution of the current mass density in the strained body can be calculated from Eq. (29). In other words, one does not solve a coupled problem and does not make use of the balances of mass and momentum simultaneously. Indeed, in our problem we calculate the strains or rather the displacements from Eqs. (23)–(27) after the current mass density in the body force has been replaced by a constant reference mass density, hence mimicking homogeneous initial conditions for the mass distribution of a terrestrial planet. For conciseness of this paper, the question as to how the current mass density will look like and how it compares with today’s knowledge of the inner mass distribution of Earth will be discussed elsewhere.⁴

Third, the use of a constant reference mass density in linear elasticity turns into a very subtle point when applied to problems of self-gravity. Observe that in the current local balance of momentum the current mass density appears explicitly and linearly in three locations: (i) in the inertial term (not shown in Eq. (23), because we restrict ourselves to quasistatic conditions); (ii) in the product of the term for the body force density, and (iii) in the acceleration part of the body force density, if we consider the case of full self-gravitational interaction. The latter will be demonstrated explicitly in Subsection (A.2.1). Now recall once more that all of this is ignored in linear elasticity where the current mass density is simply replaced by a constant value $\rho _0$, everywhere. We may rephrase it in the jargon of technical mechanics by saying that the forces are applied to the undeformed structure and a first order theory is used to calculate the resulting deformation. Thus we would like to reemphasize that the model “linear elasticity” is defined by three prerequisites (also see, for example, the beginning of Kienzler and Schröder (2009), namely, first, a linear relationship between stress and strain, second, strains and displacements to be small and, third, equilibrium of an undeformed element.

However, the use of linear elasticity in self-gravitational problems remains questionable. Indeed, we shall see that for certain celestial objects, in particular the Earth, the strains we are about to predict from the linear theory of elasticity can become very large. Probably the first to notice was A.E.H. Love after applying linear elasticity in the way defined above and discovering what is known as the Love radius, a radial transition point within a self-gravitating spherical body, where radial strains switch from compression to tension and, consequently, may result in damage of the body. Love muses in sudden attacks of self-doubt about his approach, namely in Love (1892), Article 127: “There is another difficulty in the application of the result [for the strains and for the breaking stress] to the case of the Earth. The necessary limitation to the mathematical theory is that the strain found from it must always be “small”. ...” and in Love (1927), Article 75: “The Earth is an example of a body which must be regarded as being in a state of initial stress, for the stress that must exist in the interior is much too great to permit of the calculation, by the ordinary methods, of strains reckoned from the unstressed state as unstrained state.”

Consequently, we could come to the conclusion to abandon linear elasticity completely and to use a deformation-wise nonlinear theory instead. Indeed, this was done in a series of papers from the school of Seth, who was one of the first to study and use nonlinear deformation measures for elasticity problems, see, e.g., Chattarji (1953) or Bose and Chattarji (1963). In principle, this requires solving the coupled problem, namely the balances of mass and momentum, unless an empirical expression for the (current) mass density distribution is assumed. The latter was the case in the aforementioned papers from the school of Seth. One of their main conclusions was that the position of the Love radius as predicted by linear elasticity at small deformations does not change much when switching to large deformation theory.

This in mind, do we now feel completely reconciled by using the linear theory elasticity in context with problems of self-gravity? The answer is unfortunately “no.” Indeed, a detailed study of the nonlinear problem shows that there are many open questions, ranging from numerical issues to the use of the proper nonlinear stress-strain relationship.⁵ Nevertheless, this paper is not the place to explore this issue completely. We must and will assume the position of Galileo’s Simplicio, who would use the linear-elastic solution anyway, no matter how large the self-gravitating mass really is. To quote Churchill: “Now this is not the end. It is not even the beginning of the end. But it is, perhaps, the end of the beginning” or shall we say the dawn of awareness?

In the next section we shall briefly summarize Love’s linear elasticity results at small deformations and provide some additional comments for better explanation and clarification of the problem. For example we shall introduce and interpret various parameters in terms of their physical meaning, which can be used for normalization of the solution. Moreover, we will show that for certain celestial objects, such as Mercury, the linear elasticity solution with small deformations can be considered as valid. It will also serve as a starting point as well as for comparison with the results from linear viscoelasticity in Sect. 3.

1.2 A.2 Review of Love’s Linear-Elastic Model of Self-gravitation

1.2.1 A.2.1 Analysis of the Strictly Radially Symmetric Case

We start from the Poisson equation describing Newtonian gravity as shown in Eq. (24) and assume purely radial dependencies:

$$\begin{aligned} \begin{gathered} \frac{1}{r^2}\frac{\mathrm {d}}{\mathrm {d} r}\left( r^2\frac{\mathrm {d} U^\mathrm {grav}(r)}{\mathrm {d} r}\right) =4\mathrm {\pi } G\rho (r)\,\,\qquad \Rightarrow \qquad \,\, \frac{\mathrm {d} U^\mathrm {grav}}{\mathrm {d} r}= G\frac{m(r)}{r^2}, \end{gathered}\end{aligned}$$

(30)

where m(r) denotes the total mass within a spherical region of radial extension r:

$$\begin{aligned} \begin{gathered} m(r)=4\mathrm {\pi }\int \limits _{\tilde{r}=0}^{\tilde{r}=r}\rho (\tilde{r})\tilde{r}^2\mathrm {d}\tilde{r}\, , \quad 0\le r\le r_\mathrm {o}, \end{gathered}\end{aligned}$$

(31)

and $r_\mathrm {o}$ stands for the current outer radius of the spherical body. Consequently, according to Eq. (24), under these circumstances the volume density of body force is given by:

$$\begin{aligned} \begin{gathered} \rho (r)\varvec{f}(r)=-G\frac{\rho (r)m(r)}{r^2}\, \varvec{e}_r. \end{gathered}\end{aligned}$$

(32)

This includes the well-known high school result according to which the gravitational force at a distance r within a homogeneous sphere is given by Newton’s law of gravity for point masses: The attracting mass is given by all the matter below the position r, i.e., m(r), and can be thought of as being concentrated in the origin of the sphere, i.e., $r=0$. The to-be-attracted mass at the radial position r is given by $\mathrm {d}m=\rho (r)\, \mathrm {d}V$, $\mathrm {d}V$ being the corresponding volume element to be used for multiplication in Eq. (24). Moreover, the gravitational force is attractive, as indicated by the negative direction of the current radial unit vector, $\varvec{e}_r$. It should be pointed out that in this equation the mass density within the sphere does not necessarily have to be homogeneous. Rather it can be a function of the current radius, $\rho (r)$, and the high-school result still holds. This is very often not clearly stated in textbooks, especially if use of the Poisson equation is avoided for mathematical simplicity.

However, as outlined before, it is customary in linear elasticity to use the body force of an undeformed structure in Eq. (23). More specifically, we pretend everything is initially homogeneous and use a constant mass density, $\rho _0$, such that:

$$\begin{aligned} \begin{gathered} \rho (r)\varvec{f}(r)\approx -G\frac{\rho _0 m(r)}{r^2}\, \varvec{e}_r\approx -\frac{4\pi G\rho _0^2}{3} r\, \varvec{e}_r. \end{gathered}\end{aligned}$$

(33)

Note that a two-step approximation was involved here. First, the current mass density, $\rho (r)$, in Eq. (23) or in (32) was replaced by the reference mass density, $\rho _0$. Second, no distinction is made between the current and the reference radius on the right hand side of Eq. (32). We will now use the approximation (33) in Eq. (23), which reads in spherical coordinates as follows:

$$\begin{aligned} \frac{\partial \sigma _{rr} }{\partial r}+{\frac{1}{r}} \frac{\partial \sigma _{r\vartheta } }{\partial \vartheta }+{\frac{1}{r\mathrm {sin}\vartheta }} \frac{\partial \sigma _{ r\varphi } }{\partial \varphi }+{\frac{2\sigma _{rr}-\sigma _{\vartheta \vartheta }-\sigma _{\varphi \varphi }+\sigma _{r\vartheta } \cot \vartheta }{r}}&=\frac{4\pi G\rho _0^2}{3} r, \nonumber \\ \frac{\partial \sigma _{r\vartheta } }{\partial r} +{\frac{1}{r}} \frac{\partial \sigma _{\vartheta \vartheta } }{\partial \vartheta }+{\frac{1}{r\mathrm {sin}\vartheta }} \frac{\partial \sigma _{ \vartheta \varphi } }{\partial \varphi } +{\frac{3\sigma _{r\vartheta } +\left( \sigma _{\vartheta \vartheta } -\sigma _{\varphi \varphi } \right) \; \mathrm{cot}\vartheta }{r}}&=0 , \\ \frac{\partial \sigma _{r\varphi } }{\partial r} +{\frac{1}{r}} \frac{\partial \sigma _{\vartheta \varphi } }{\partial \vartheta }+{\frac{1}{r\mathrm {sin}\vartheta }} \frac{\partial \sigma _{ \varphi \varphi } }{\partial \varphi } +{\frac{3\sigma _{r\varphi } +2\sigma _{\vartheta \varphi } \cot \vartheta }{r}}&=0. \nonumber \end{aligned}$$

(34)

Moreover, Hooke’s law reads in spherical coordinates as follows:

$$\begin{aligned} \sigma _{rr}&=\lambda (\epsilon _{\vartheta \vartheta }+\epsilon _{\varphi \varphi })+(\lambda +2\mu )\epsilon _{rr}\; ,\;\sigma _{\vartheta \vartheta }=\lambda (\epsilon _{rr}+\epsilon _{\varphi \varphi })+(\lambda +2\mu )\epsilon _{\vartheta \vartheta }, \nonumber \\ \sigma _{\varphi \varphi }&=\lambda (\epsilon _{rr}+\epsilon _{\vartheta \vartheta })+(\lambda +2\mu )\epsilon _{\varphi \varphi }\; , \\ \sigma _{r\vartheta }&=2\mu \,\epsilon _{r\vartheta }\; ,\sigma _{r\varphi }=2\mu \,\epsilon _{r\varphi }\; ,\sigma _{\vartheta \varphi }=2\mu \,\epsilon _{\vartheta \varphi }. \nonumber \end{aligned}$$

(35)

And finally the linear strain tensor is linked with spatial derivatives of the displacements by:

$$\epsilon _{rr}=\frac{\partial u_r}{\partial r}\;,\; \epsilon _{\vartheta \vartheta }=\frac{1}{r}\frac{\partial u_\vartheta }{\partial \vartheta }+\frac{u_r}{r}\;,\; \epsilon _{\varphi \varphi }=\frac{1}{r\mathrm {sin}\vartheta }\frac{\partial u_\varphi }{\partial \varphi }+\frac{u_r}{r}+\frac{\mathrm {cot}\vartheta }{r}u_\vartheta , $$

$$\begin{aligned} \begin{gathered} \epsilon _{r\varphi }=\tfrac{1}{2}\left( \frac{1}{r\mathrm {sin}\vartheta }\frac{\partial u_r}{\partial \varphi }+\frac{\partial u_\varphi }{\partial r}-\frac{u_\varphi }{r}\right) , \end{gathered}\end{aligned}$$

(36)

$$\epsilon _{r\vartheta }=\tfrac{1}{2}\left( \frac{1}{r}\frac{\partial u_r}{\partial \vartheta }+\frac{\partial u_\vartheta }{\partial r}-\frac{u_\vartheta }{r}\right) \;,\; \epsilon _{\vartheta \varphi }=\tfrac{1}{2}\left( \frac{1}{r\mathrm {sin}\vartheta }\frac{\partial u_\vartheta }{\partial \varphi }+\frac{1}{r}\frac{\partial u_\varphi }{\partial \vartheta }-\frac{\mathrm {cot}\vartheta }{r}u_\varphi \right) . $$

We now proceed to solve these equations. To this end we make use of the semi-inverse method. Because of symmetry it seems reasonable to seek for solutions with the following ansatz:

$$\begin{aligned} \begin{gathered} u_{r}=u_{r}(r)\; , \quad u_{\vartheta }=0\; , \quad u_{\varphi }=0. \end{gathered}\end{aligned}$$

(37)

Consequently, we find for the linear strains:

$$\begin{aligned} \begin{gathered} \epsilon _{rr}=u'_{r}(r)\; ,\qquad \epsilon _{\vartheta \vartheta }\equiv \epsilon _{\varphi \varphi }=\frac{u_{r}}{r}\; ,\qquad \epsilon _{r\vartheta }=\epsilon _{r\varphi }=\epsilon _{\vartheta \varphi }\equiv 0\;, \end{gathered}\end{aligned}$$

(38)

where the dash refers to a differentiation w.r.t. r. Because of that Hooke’s law (35) reduces to:

$$\begin{aligned} \begin{gathered} \sigma _{rr}=(\lambda +2\mu )u'_{r}+2\lambda \frac{u_{r}}{r} ,\qquad \sigma _{\vartheta \vartheta }\equiv \sigma _{\varphi \varphi }=\lambda u'_{r}+2(\lambda +\mu )\frac{u_{r}}{r},\; \end{gathered}\end{aligned}$$

(39)

$$\sigma _{r\vartheta }=\sigma _{r\varphi }=\sigma _{\vartheta \varphi }\equiv 0\; . $$

Thus, the angular components of the balance of momentum shown in Eq. (34) are identically satisfied and the first one results in an ordinary differential equation of second order (a dash refers to differentiation with respect to the radius, r):

$$\begin{aligned} \begin{gathered} u''_{r}+2\frac{u'_{r}}{r}-2\frac{u_{r}}{r^2}=\frac{4\pi \rho ^2_0 G}{3(\lambda +2\mu )}r. \end{gathered}\end{aligned}$$

(40)

The general solution consists of the full solution to the homogeneous part and one particular solution of the inhomogeneous case. It reads with two constants of integration, A and B, respectively, as follows:

$$\begin{aligned} \begin{gathered} u_{r}=Ar+\frac{B}{r^2}+\frac{4\pi \rho ^2_0 G}{30(\lambda +2\mu )}r^3. \end{gathered}\end{aligned}$$

(41)

Two conditions are needed in order to determine the two constants of integration. First, we require that the solution does not become singular at $r=0$ and, second, the traction must be continuous at the outer radius, $r_\mathrm {o}$, of the sphere. Hence, $\sigma _{rr}\big \vert _{r=r_\mathrm {o}}=0$, because the influence of the outer atmospheric pressure of roughly $1 \mathrm {bar}$ is negligibly small when it comes to the deformation of a solid. With Eq. (39)$_1$ we find that:

$$\begin{aligned} \begin{gathered} B=0\;,\qquad \qquad A=-\frac{4\pi G \rho ^2_0 }{30(\lambda +2\mu )}\frac{3-\nu }{1+\nu }r^2_\mathrm{o}\equiv -\frac{4\pi G \rho ^2_o}{90\,k}\frac{3-\nu }{1-\nu }r^2_\mathrm{o}, \end{gathered}\end{aligned}$$

(42)

because $\lambda =\tfrac{E\nu }{(1-2\nu )(1+\nu )}$ and $\mu =\tfrac{E}{2(1+\nu )}$, E being Young’s modulus and $\nu$ Poisson’s ratio, respectively. Hence the radial displacement reads:

$$\begin{aligned} u_{r}=-\frac{2\pi G\rho _0^2r_\mathrm {o}^2}{15(\lambda +2\mu )}\left( \frac{3-\nu }{1+\nu }-\frac{r^2}{r_\mathrm {o}^2}\right) r\equiv -\frac{2\pi G\rho _0^2r_\mathrm {o}^2}{45\;k}\frac{1+\nu }{1-\nu }\left( \frac{3-\nu }{1+\nu }-\frac{r^2}{r_\mathrm {o}^2}\right) r. \end{aligned}$$

(43)

For completeness, the nonvanishing stresses then follow from Eq. (39) as:

$$\begin{aligned} \sigma _{rr }&=-\frac{2\pi G\rho _0^2r_\mathrm {o}^2}{15}\frac{3-\nu }{1-\nu } \left( 1-\frac{r^2}{r_\mathrm {o}^2}\right) , \\ \sigma _{\vartheta \vartheta }&\equiv \sigma _{\varphi \varphi }=- \frac{2\pi G\rho _0^2r_\mathrm {o}^2}{15}\frac{3-\nu }{1-\nu } \left( 1-\frac{1+3\nu }{3-\nu }\frac{r^2}{r_\mathrm {o}^2}\right) . \nonumber \end{aligned}$$

(44)

Note the common factor $\tfrac{2\pi G\rho _0^2r_\mathrm {o}^2}{15}$ in front of all these expressions. On first glance it does not allow for an easy intuitive interpretation. However, on second thought, note that (within the approximations made) the total mass of the gravitating sphere, $m_0$, the gravitational acceleration on its surface, g, and its surface area, $A_\mathrm {o}$, are given by:

$$\begin{aligned} \begin{gathered} m_0=\frac{4\pi }{3} \rho _0 r_\mathrm {o}^3\;,\quad g=\frac{Gm_0}{r_\mathrm {o}^2}\;,\quad A_\mathrm {o}=4\pi r_\mathrm {o}^2. \end{gathered}\end{aligned}$$

(45)

Hence, we may write the ominous factor as:

$$\begin{aligned} \begin{gathered} \frac{2\pi G\rho _0^2r_\mathrm {o}^2}{15}\equiv \frac{3m_0g}{10A_\mathrm {o}}, \end{gathered}\end{aligned}$$

(46)

and interpret it, with the exception of the fraction $\tfrac{3}{10}$, as an average “pressure,” namely the ratio between the “total gravitational force,” $m_0g$, distributed over the total surface area, $A_\mathrm {o}$.

1.2.2 A.2.2 Dimensionless Formulation

For a numerical analysis it is best to work with dimensionless quantities. Since the outer radius, $r_\mathrm {o}$, is the only length parameter in the problem, there is no other choice for a dimensionless distance and a dimensionless displacement but to define:

$$\begin{aligned} \begin{gathered} x=\frac{r}{r_\mathrm {o}}\;,\quad u=\frac{u_r}{r_\mathrm {o}}. \end{gathered}\end{aligned}$$

(47)

This allows us to rewrite Eq. (43) as follows:

$$\begin{aligned} \begin{gathered} u=-\frac{\alpha }{20}\left( \frac{3-\nu }{1+\nu }-x^2\right) x\equiv -\frac{\alpha _k}{30}\frac{1+\nu }{1-\nu }\left( \frac{3-\nu }{1+\nu }-x^2\right) x, \end{gathered}\end{aligned}$$

(48)

with two dimensionless factors:

$$\begin{aligned} \begin{gathered} \alpha =\frac{8\pi G \rho ^2_0 r^2_\mathrm {o}}{3(\lambda +2\mu )}\;,\quad \alpha _{k}=\frac{4\pi G \rho ^2_0 r^2_\mathrm {o}}{3\,k}, \end{gathered}\end{aligned}$$

(49)

because the bulk modulus is given by $k=\frac{E}{3(1-2\nu )}$.

Whilst the appearance of $\alpha$ is a straightforward consequence of Eq. (43) the need for $\alpha _k$ must be explained. To this end note that the dimensionless expressions in the parentheses of Eq. (48) still contain Poisson’s ratio. However, Poisson’s ratio of a terrestrial planet is not an immediately accessible parameter. A homogenization technique has to be applied in order to find out which effective elastic parameters such an object has. On the other hand, if we evaluate the parentheses in this equation at the surface of the planet, i.e., at $x=1$, we obtain twice the fraction $\tfrac{1-\nu }{1+\nu }$. Hence Poisson’s ratio disappears completely in the expression for the normalized displacement if we use the dimensionless factor $\alpha _k$. In this case one only needs to know the effective compressibility of the planet, a parameter that can vary within certain physically reasonable bounds. And what is more, since $u(x=1)$ can be interpreted as an average strain characterizing the state of deformation of the planet, which we wish to access numerically, it is very useful to have one elastic parameter less to worry about. A final comment is in order in context with Eq. (48). It obviously provides a restriction to the size of $\alpha$ so that small strain theory applies. However, as it was pointed out above, we will discuss this issue in detail in Müller and Weiss (2016) and not here.

Later we shall be interested in a strain-based failure criterion. Hence it is useful to know the strains explicitly:

$$\begin{aligned} \begin{gathered} \epsilon _{rr}=-\frac{\alpha }{20}\left( \frac{3-\nu }{1+\nu }-3x^2\right) \,,\quad \epsilon _{\vartheta \vartheta }\equiv \epsilon _{\varphi \varphi }=-\frac{\alpha }{20}\left( \frac{3-\nu }{1+\nu }-x^2\right) . \end{gathered}\end{aligned}$$

(50)

We now turn to the stresses given by Eq. (44). Differently from the case of length related quantities there are various possibilities for normalization. First, as Eq. (44) suggestively seems to indicate, we can use the factor $\tfrac{2\pi G\rho _0^2r_\mathrm {o}^2}{15}$, which we have interpreted as an “average gravitational pressure” before. However, alternatively, we may use combinations of (effective) elastic constants. There is $\lambda +2\mu$, which is related to the velocity of P-waves $\left( \mathrm {v}_\mathrm {P}=\sqrt{\tfrac{\lambda +2\mu }{\rho _0}}\right)$, and, hence, a physically accessible quantity. Moreover, we may turn to the compressibility k, which is a direct measure of the resistance of a planet’s response to its own self-gravity. However, in this paper we restrict ourselves for simplicity to the choice:

$$\begin{aligned} \begin{gathered} \tilde{\varvec{\sigma }}=\frac{\varvec{\sigma }}{\lambda +2\mu } \end{gathered}\end{aligned}$$

(51)

and obtain:

$$\begin{aligned} \begin{gathered} \tilde{\sigma } _{rr } =-\frac{\alpha }{20}\frac{3-\nu }{1-\nu } \left( 1-x^2\right) ,\qquad \tilde{\sigma }_{\vartheta \vartheta }\equiv \tilde{\sigma } _{\varphi \varphi }=- \frac{\alpha }{20}\frac{3-\nu }{1-\nu } \left( 1-\frac{1+3\nu }{3-\nu }x^2\right) . \end{gathered}\end{aligned}$$

(52)

1.2.3 A.2.3 Numerical Evaluation and Graphical Representation

Figure 6 (left) illustrates the dependence of the normalized displacement, $u\equiv \tfrac{u_r}{r_\mathrm {o}}$ per $\alpha _k$, on $x\equiv \tfrac{r}{r_\mathrm {o}}$ for three different choices of Poisson’s ratio, $\nu =0$ (red), $\nu =0.3$ (green), and $\nu =0.5$ (blue). Note that, as it should be, the radial displacement is negative and that the curves show a minimum. Because of Eq. (36)$_1$ we may interpret the slope of the curves as radial strain multiplied by $\alpha _k$. Hence the only positive radial strains can be found to the right of the minimum. The transition point between positive and negative strains (see Fig. 6, right), identifiable by locating the minimum of the displacement, is a.k.a. the Love radius and given by:

$$\begin{aligned} \begin{gathered} r_\mathrm{{Love}}=r_0\sqrt{\frac{3-\nu }{3(1+\nu )}}. \end{gathered}\end{aligned}$$

(53)

Fig. 6

Normalized displacement versus dimensionless radius (see text)

This result was first mentioned by Love in his books on linear elasticity, namely in Article 127 of Love (1892) and later in Article 98 of Love (1906) or Love (1927). An intuitive explanation for the necessity of its occurrence is as follows: Unlike a homogeneous, isotropic sphere subjected to a constant external pressure, the state of strain in our case is not homogeneous and not isotropic. We face a nonconstant “external hydrostatic pressure,” so-to-speak, given by an effective gravitational force depending linearly on the distance from the center. This in combination with Poisson’s effect, i.e., the ability of a radial strain “making up” for the lateral contractions, $\epsilon _{\vartheta \vartheta }$ and $\epsilon _{\varphi \varphi }$, which are purely compressive in nature everywhere: The contractive force is proportional to r, the stretching stress (compensation of the lateral contraction) is proportional to $r^2$. Hence from some r onward the force is not strong enough for compression, resulting in a transition from the negative to the positive, in other words in the existence of a Love radius. In his later editions Love does no longer comment on the physical significance of his radius. However, the first edition makes it perfectly clear that he was aiming at a failure criterion, namely specifically at what is known today as maximum principal strain theory. Love realized that the angular strains are always negative, whereas the radial strains may become positive above the Love radius, cf. Eq. (50) and Fig. 6 (right), and he provided an expression for the maximum radial strain, which is the one at the surface:

$$\begin{aligned} \begin{gathered} \epsilon _{rr}\big \vert _{x=1}=\frac{\alpha }{5}{\frac{\nu }{1+\nu }}. \end{gathered}\end{aligned}$$

(54)

According to Love the corresponding tensile breaking stress is then given by:

$$\begin{aligned} \begin{gathered} T_{0}=E\epsilon _{rr}\big \vert _{x=1}=\frac{E\alpha }{5}{\frac{\nu }{1+\nu }}. \end{gathered}\end{aligned}$$

(55)

Fig. 7

Normalized displacement versus dimensionless radius for Mercury and Earth (see text)

This is quite a daring concept, because a planet like Earth is heterogeneous, and surely not perfectly linear elastic, and the materials it is made of might not be susceptible to strain-based failure, and so on, and so on. But even if we accept his idea in principle, what is the proper Young’s modulus to be used for a heterogeneous object like Earth? On second glance, however, note that the factor $\alpha$ also contains Young’s modulus in its denominator (cf. Eq. (49)$_1$) since $\lambda +2\mu =\tfrac{(1-\nu )E}{(1-2\nu )(1+\nu )}$. Thus we do not need to know E but only Poisson’s ratio, $\nu$, which runs within well-known bounds, namely $0\le \nu \le 0.5$. Hence, we may rewrite Love’s result as follows:

$$\begin{aligned} \begin{gathered} T_0=\frac{\nu (1-2\nu )}{1-\nu }\frac{8\pi G\rho _0^2r_\mathrm {o}^2}{15}. \end{gathered}\end{aligned}$$

(56)

If we assume Poisson’s ratio to be that of iron, $\nu =0.3$, and the mean mass density of Earth $\rho _0=5500\,\tfrac{\mathrm {kg}}{\mathrm {m}^3}$ together with its (average) outer radius $r_\mathrm {o}=\mathrm {6370\,km}$, we obtain $T_0>30{,}000\,\mathrm {MPa}$, which is a value for a breaking stress far beyond physical credibility. It may be for that reason that Love did not present his original idea in later editions of his book any more. Nevertheless, it is a fact that the quality of the radial strain changes when passing the Love radius and it remains to be seen if the breaking stress can be brought to physically reasonable values if a deformation-wise nonlinear theory is used.

Figure 7 shows a plot of Eq. (48)$_1$ when data of Mercury (index M) and Earth (index E) were used for evaluation. Specifically we have values for the average mass densities of $\rho _0^{\mathrm {E}}=5500\tfrac{\mathrm {kg}}{\mathrm {m}^3}$ and $\rho _0^{\mathrm {M}}=5400\tfrac{\mathrm {kg}}{\mathrm {m}^3}$ and (average) outer radii of $r_\mathrm {o}^{\mathrm {E}}=\mathrm {6370\,km}$ and $r_\mathrm {o}^{\mathrm {M}}=\mathrm {2440\,km}$, respectively. For the elastic data we assume in both cases the values of iron, i.e., $E=210$ GPa and $\nu =0.3$. This leads to the red curve for Mercury and to the blue one for Earth. Consequently, the strains for Mercury are below two percent but the ones for Earth are huge and amount to a maximum of fourteen percent. Hence, one may question the validity of the use of linear elasticity in case of very large self-gravitating masses and turn to a nonlinear formulation instead. Indeed, this has been done by the Indian school of Seth, who was one of the pioneers in large strain measures. We will not discuss this in detail here and refer the interested reader to Chap. 3 of the upcoming publication by Müller and Weiss (2016).