Viscoelastic deformations of a large, rotating and gravitating body: the basic theory

Hans-Peter Plag, Nevada Bureau of Mines and Geology, University of Nevada, Reno, Nevada, USA Email: hpplag@unr.edu

April 5, 2011

0.1 Linearized differential equations for deformations

0.1.1 General setting of the model

The linearized differential equation system for the tidal deformation u and the incremental potential φ is given by

ρü = ρ {∇φ + (u · ∇)∇(Φ + Ψ) − 2Ω × u̇} +
∇(u · ∇P) − (∇· u)∇P − (u · ∇)∇P + ∇· σ ,  (1)
∇^{2}φ = 4πG ∇· (ρu) ,

where ρ = ρ(r) denotes the density, Φ is the (negative) gravitational potential

Φ(r) =^{def} G ∫_{V} \frac{ρ(r′)}{|r − r′|} d^{3}r′ ,  (2)

with gravitational constant G,

Ψ(r) =^{def} \frac{1}{2}(Ω^{2}r^{2} − (Ω · r)^{2}) = \frac{1}{2}[Ω × r]^{2}  (3)

with Ω = |Ω|, r = |r| denotes the (negative) rotational potential for angular velocity vector Ω, P the hydrostatic pressure and σ is the incremental pseudo-stress tensor with components

σ_{ij} = λ∂_{k}u_{k}δ_{ij} + μ(∂_{i}u_{j} + ∂_{j}u_{i}) ,  (4)

when the material is assumed to be isotropic. Here, λ and μ denote the Lamé parameters.

Since in hydrostatic equilibrium the pressure P must satisfy

∇P = ρ∇(Φ + Ψ) ,  (5)

it follows ∇ρ × ∇(Φ + Ψ) = 0 and the first equation (1) may be simplified to give

ρü = ρ {∇(φ + u · ∇(Φ + Ψ)) − (∇· u)∇(Φ + Ψ) − 2Ω × u̇} + ∇· σ .  (6)

Equation system (6) has to be suplemented by interior boundary conditions

[n · σ]_{−}^{+} = 0 ,  (0.7a)

\{ \begin{matrix}[u]_{−}^{+} = 0 & if welded \\ [n · u]_{−}^{+} = 0 & if frictionless \end{matrix}\} ,  (0.7b)
[φ]_{−}^{+} = 0 ,  (0.7c)
[n · (∇φ − 4πGρu)]_{−}^{+} = 0 ,  (0.7d)

and surface boundary conditions

n · σ|_{−} = F ,  (0.8a)
φ|_{−} = φ|_{+} ,  (0.8b)
n · (∇φ − 4πGρu)|_{−} = n · ∇φ|_{+} + 4πG τ ,  (0.8c)

where n denotes the exterior normal to the boundary, F is the external force per area acting on the surface, and τ is the surface mass density. If one is studying body tides, one has F = 0 and τ = 0. However, for loading tides F and τ are the sources, which generate the tides.

The last equation in (1) together with the boundary conditions (0.8b) and (0.8c) may be solved by the methods of potential theory to give

φ(r) = −G ∫_{V} \frac{∇· (ρ(r′)u(r′))}{|r − r′|} d^{3}r′ + G ∫_{∂V} \frac{τ(r′) + ρ(r′) (u(r′) · n′)}{|r − r′|}d^{2}r′ + φ^{P}(r)  (9)

where φ^{P} denotes the tide generating primary potential which is smooth and satisfies the Laplace equation ∇^{2}φ^{P} = 0 inside and at the surface of the earth.

0.1.2 First order differential equation system

We take the undisturbed system to be spherically symmetric (ρ = ρ(r)) and non-rotating. Then equation (1) simplifies further to

\begin{matrix}ρü = & ρ\{∇[φ − g(u · e_{r})] + g(∇· u)e_{r}\} + ∇· σ , \\ ∇^{2}φ = & 4πG ∇· (ρu) , \end{matrix}  (10)

where g = −dΦ/dr is the absolut value of the gravitational acceleration. Equation (9) may be evaluated for the spherically symmetric case. It is, however, more convenient to integrate the whole differential equation system (10) directly without making use of (9). This system allows for two types of solutions, toroidal and spheroidal solutions. The toroidal solutions obey ∇· u = ∇· (ρu) = 0 and are obtained from the ansatz

\begin{matrix}u = & W_{n}(r) e_{r} × ∇_{1}Y_{nm}(ϑ, χ) exp(iωt) , \\ φ = & 0 , \end{matrix}  (11)

where we have agreed to set ∇_{1} =^{def} r[∇ − e_{r}(e_{r} · ∇)] and Y_{nm}(ϑ, χ) is the spherical harmonic of degree n and order m given by

Y_{nm}(ϑ, χ) = \sqrt{\frac{2n+1}{4π} \frac{(n−m)!}{(n+m)!}} P_{n}^{m}(cos ϑ) e^{imχ} ,
Y_{n −m}(ϑ, χ) = (−1)^{m} Y_{nm}^{∗}(ϑ, χ) ,  with
P_{n}^{m}(cos ϑ) = (−1)^{m} \frac{(1−cos^{2} ϑ)^{m/2}}{2^{n} n!} \frac{d^{n+m}}{d(cos ϑ)^{n+m}} (cos^{2} ϑ − 1)^{n} .

(12)

For this ansatz we have

∇ · σ = \{\frac{d}{dr} [μ (W′_{n}(r) − \frac{1}{r}W_{n}(r))]
+\frac{μ}{r} (3W′_{n}(r) − \frac{1 + n(n + 1)}{r}W_{n}(r))\} e_{r} × ∇_{1}Y_{nm}(ϑ, χ) exp(iωt)  (13)

with W′_{n}(r) =^{def} dW_{n}(r)/dr. Introducing the “solution vector”

y_{T} =^{def} \sqrt{r}\binom{W_{n}}{μ(rW′_{n} − W_{n})/(n + 1)} ,  (14)

equation (10) for toroidal solutions reads

r \frac{dy_{T}}{dr} = A_{T}y_{T} ,  (15)

where the matrix A_{T} is defined as

A_{T} = ( \frac{(n−1)(n+2)μ−ρω^{2}r^{2}}{n+1} −\frac{3}{2}^{\frac{3}{2}  \frac{n+1}{μ}} ) .  (16)

Spheroidal solutions are obtained from the ansatz

\begin{matrix}u = & \{H_{n}(r)e_{r} + T_{n}(r)∇_{1}\}Y_{nm}(ϑ, χ) exp(iωt) , \\ φ = & R_{n}(r)Y_{nm}(ϑ, χ) exp(iωt) \end{matrix}  (17)

and yuield

∇ · σ =

\{\frac{d}{dr} [λ (H′_{n}(r) + \frac{2}{r}H_{n}(r) − \frac{n(n + 1)}{r}T_{n}(r)) + 2μH′_{n}(r)] + \frac{4μ}{r} (H′_{n}(r) − \frac{1}{r}H_{n}(r))

−\frac{n(n + 1)μ}{r} (T′_{n}(r) − \frac{3}{r}T_{n}(r) + \frac{1}{r}H_{n}(r))\} e_{r}Y_{nm}(ϑ, χ) exp(iωt) +

\{\frac{λ}{r} (H′_{n}(r) + \frac{2}{r}H_{n}(r) − \frac{n(n + 1)}{r}T_{n}(r)) + \frac{d}{dr} [μ (T′_{n}(r) − \frac{1}{r}T_{n}(r) + \frac{1}{r}H_{n}(r))]

+\frac{μ}{r} (3T′_{n}(r) − \frac{1 + 2n(n + 1)}{r}T_{n}(r) + \frac{5}{r}H_{n}(r))\} ∇_{1}Y_{nm}(ϑ, χ) exp(iωt) .  (18)

For the case n > 0 we introduce the “solution vector”

y_{S} =^{def} \sqrt{r}( \begin{matrix}H_{n} \\ [(λ + 2μ)rH′_{n} + 2λH_{n}]/(n + 1) − nλT_{n} \\ nT_{n} \\ μ(rT′_{n} − T_{n} + H_{n}) \\ R_{n} \\ rR′_{n} + (n + 1)R_{n} − 4πGρrH_{n} \end{matrix} )  (19)

and arive at an equation

r \frac{dy_{S}}{dr} = A_{S}y_{S}  (20)

with a matrix^{1}

A_{S} = ( \begin{matrix}\frac{−3λ+2μ}{2(λ+2μ)} & \frac{n+1}{λ+2μ} & \frac{(n+1)λ}{λ+2μ} & 0 & 0 & 0 \\ \frac{4μ(3λ+2μ)}{(n+1)(λ+2μ)} − \frac{4ρgr+ρω^{2}r^{2}}{n+1} & \frac{3λ−2μ}{2(λ+2μ)} & ρgr − \frac{2μ(3λ+2μ)}{λ+2μ} & n & ρr & \frac{−ρr}{n+1} \\ −n & 0 & \frac{3}{2} & \frac{n}{μ} & 0 & 0 \\ ρgr − \frac{2μ(3λ+2μ)}{λ+2μ} & \frac{−(n+1)λ}{λ+2μ} & \frac{4μ(n+1)(λ+μ)}{λ+2μ} − \frac{2μ+ρω^{2}r^{2}}{n} & − \frac{3}{2} & −ρr & 0 \\ 4πGρr & 0 & 0 & 0 & − (n + \frac{1}{2}) & 1 \\ 4πGρ(n + 1)r & 0 & −4πGρ(n + 1)r & 0 & 0 & n + \frac{1}{2} \end{matrix} ) .  (21)

These equations have first been given by Wiggins ?. For n = 0 the term T_{0}∇_{1}Y_{00} is identical zero and the incremental potential φ is equal to the primary potential φ^{P} outside the earth and ∇φ = 4πGρu + const. inside the earth since it must be singularity free there. We may therefore introduce the simplified “solution vector”

y_{0} =^{def} ( \begin{matrix}\sqrt{r}H_{0} \\ \sqrt{r}[(λ + 2μ)rH′_{0} + 2λH_{0}] \\ R_{0} \end{matrix} ) ,  (22)

which yields the equation

r \frac{dy_{0}}{dr} = A_{0}y_{0}  (23)

with

A_{0} = ( \begin{matrix}\frac{−3λ+2μ}{2(λ+2μ)} & \frac{1}{λ+2μ} & 0 \\ \frac{4μ(3λ+2μ)}{λ+2μ} − 4ρgr − ρω^{2}r^{2} & \frac{3λ−2μ}{2(λ+2μ)} & 0 \\ 4πGρ\sqrt{r} & 0 & 0 \end{matrix} ) .  (24)

The case of a liquid layer has to be considered seperately^{2}. For a liquid compressible layer the Adams-Williamson condition

\frac{dρ}{dr} + ρ^{2} \frac{g}{λ} = 0  (25)

must be satisfied in a state of thermodynamic equilibrium. Since for μ = 0 the Lamé parameter λ is identical with the compression modulus which in turn is nothing else but the adiabatic compressibility

λ = ρ (\frac{∂P}{∂ρ} )_{S} ,  (26)

where the subscript S denotes constancy of entropy, (25) may be derived from the mechanical equilibrium condition (5) for the static pressure P by the assumption that density variations in the liquid layer are isentropic.

For μ = 0 the r-component of the first equation (10) together with the ansatz (17) yields

−ρω^{2}H_{n} = ρ\frac{d}{dr} (R_{n} − gH_{n}) + ρg \{H′_{n} + \frac{2}{r}H_{n} − \frac{n(n + 1)}{r}T_{n}\} +
\frac{d}{dr} (λ \{H′_{n} + \frac{2}{r}H_{n} − \frac{n(n + 1)}{r}T_{n}\}) ,  (27)

while the ϑ- or χ-components of the first equation (10) yield

−ρω^{2}rT_{n} = ρ(R_{n} − gH_{n}) + λ \{H′_{n} + \frac{2}{r}H_{n} − \frac{n(n + 1)}{r}T_{n}\} .  (28)

By differentiating this equation with respect to r, adding ρg/λ times this equation, substituting dρ/dr from equation (25) and substracting equation (27) we obtain

T′_{n} = \frac{H_{n} − T_{n}}{r} .  (29)

With the ansatz (17) and the “solution vector”

y_{L} = \sqrt{r}( \begin{matrix}4πGρrH_{n} \\ nrT_{n} \\ nR_{n} \\ rR′_{n} + (n + 1)R_{n} − 4πGρrH_{n} \end{matrix} ) ,  (30)

1The matrices A_{S} and A_{L} given here are the (n + 1)-fold of the matrices _{A} and _{B} respectively given in Appendix A of ? corrected for a mistake made there in _{A_{11}}.

2Here, a layer is meant to be liquid if the shear stress vanishes (μ = 0) throughout the layer. This is not fully equivalent to a liquid described by hydrodynamic equations since the deformations uwere still supposed to remain small for all times.

we then get the equation

r \frac{dy_{L}}{dr} = A_{L}y_{L}  (31)

with a matrix

A_{L} = ( \begin{matrix}−\frac{1}{2} & 4πGρ (n + 1 − \frac{ρω^{2}r^{2}}{nλ} ) & \frac{−4πGρ^{2}r^{2}}{nλ} & 0 \\ \frac{n}{4πGρ} & \frac{1}{2} & 0 & 0 \\ n & 0 & − (n + \frac{1}{2}) & n \\ n + 1 & −4πGρ(n + 1) & 0 & n + \frac{1}{2} \end{matrix} ) .  (32)

However, for n = 0 this ansatz does not work. In this case we have to take the ansatz (22) and get the equation (23) for the liquid layer as well.

The boundary conditions (0.7) for a frictionless boundary between a solid and a liquid layer may be given explicitly as follows:

\{ \begin{matrix}y_{S}_{1} = & \frac{1}{4πGρ_{L}r} y_{L}_{1} , \\ y_{S}_{2} = & \frac{g}{4πG(n+1)}y_{L}_{1} − \frac{ρ_{L}ω^{2}r}{n(n+1)}y_{L}_{2} − \frac{ρ_{L}r}{n(n+1)}y_{L}_{3} , \\ y_{S}_{4} = & 0 , \\ y_{S}_{5} = & \frac{1}{n}y_{L}_{3} , \\ y_{S}_{6} = & y_{L}_{4} . \end{matrix} \}  (33)

0.1.3 Energy integrals

Here we shall take the potential energy of a deformation, as was given by Pekeris and Jarosch, Jobert, Backus, Backus and Gilbert and by Dahlen ?????, for the spherically symmetric model and evaluate it to a simple integral over the radial distance. The energy released directly in a load will be given also. The work done by the displacement field against pressure and gravity of the undeformed system is

P =−\frac{1}{2} ∫_{V} u^{∗} · \{∇(u · ∇P) − (∇· u)∇P − (u · ∇)∇P + ρ(u · ∇)∇Φ\} d^{3}r
=−\frac{1}{2} ∫_{V} ρu^{∗} · \{ge_{r}(∇· u) − ∇ (g(u · e_{r}))\} d^{3}r ,  (34)

where V denotes the volume of the earth and P is the pressure, the gradient of which is ∇P = ρ∇Φ = −ρge_{r} by (5). The energy stored in the incremental potential is given by

G = −\frac{1}{2} ∫_{V} ρu^{∗} · ∇φ d^{3}r − \frac{1}{2} ∫_{∂V} φτ^{∗}d^{2}r

= \{ \begin{matrix}−\frac{1}{8πG}_{∫_{IR_{3}}}∇φ^{∗} · ∇φ d^{3}r & for loading tides \\ −\frac{1}{8πG} ∫_{V} ∇φ^{∗} · ∇φ d^{3}r + \frac{1}{8πG} ∫_{∂_{+}V} φe_{r} · ∇φ^{∗}d^{2}r & for body tides \end{matrix} ,  (35)

where the upper line in the second expression is only valid for loading tides because the gradient of the incremental potential only then decreases sufficiently fast at infinity. This part of the potential energy may be further decomposed into two parts, G = G^{P}+G^{S}, according to the decomposition φ = φ^{P} + φ^{S} of the potential into a primary and secondary one. In this decomposition, the primary potential φ^{P} is due to other celestial bodies or to the masses on the surface and the secondary potential φ^{S} is due to the mass redistribution within the earth. These parts of the energy are

G^{P} = −\frac{1}{2} ∫_{V} ρu^{∗} · ∇φ^{P} d^{3}r − \frac{1}{2} ∫_{∂V} φτ^{∗}d^{2}r

= \{ \begin{matrix}−\frac{1}{8πG}_{∫_{IR_{3}}}∇φ^{∗} · ∇φ^{P} d^{3}r − \frac{1}{2} ∫_{∂V} φ^{S}τ^{∗}d^{2}r & for loading tides \\ −\frac{1}{8πG} ∫_{V} ∇φ^{∗} · ∇φ^{P} d^{3}r + \frac{1}{8πG} ∫_{∂_{+}V} φ^{P}e_{r} · ∇φ^{∗}d^{2}r & for body tides \end{matrix}  (36)

and

G^{S} = −\frac{1}{2} ∫_{V} ρu^{∗} · ∇φ^{S} d^{3}r

= \{ \begin{matrix}−\frac{1}{8πG}_{∫_{IR_{3}}}∇φ^{∗} · ∇φ^{S} d^{3}r + \frac{1}{2} ∫_{∂V} φ^{S}τ^{∗}d^{2}r & for loading tides \\ −\frac{1}{8πG} ∫_{V} ∇φ^{∗} · ∇φ^{S} d^{3}r + \frac{1}{8πG} ∫_{∂_{+}V} φ^{S}e_{r} · ∇φ^{∗}d^{2}r & for body tides \end{matrix}.  (37)

The elastic deformation energy is

E = −\frac{1}{2} ∫_{V} u^{∗} · (∇· σ) d^{3}r + \frac{1}{2} ∫_{∂V} u^{∗} · (e_{r} · σ)d^{2}r
= \frac{1}{2} ∫_{V} \{λ|∇· u|^{2} + \frac{μ}{2} ∑_{i,j=1}^{3} |∂_{i}u_{j} + ∂_{j}u_{i}|^{2}\} d^{3}r  (38)

and the kinetic energy is

K = \frac{1}{2} ∫_{V} ρu̇^{∗} · u̇ d^{3}r .  (39)

Finally, the potential energy of the load lift is

L = g(a) ∫_{∂V}(u · e_{r})τ^{∗}d^{2}r  (40)

while the kinetic energy of the load is small of third order. For spherical degree n with the combined ansatz (11) and (17) the integral (34) is, executing the ϑ- and χ-integration and using Poisson’s equation g′(r) = 4πGρ(r) − 2g(r)/r, evaluated to

P_{n} = \frac{1}{2} ∫_{0}^{a} \{ (4πGρ(r)r − 4g(r)) |H_{n}(r)|^{2}
+n(n + 1)g(r) [H_{n}(r)^{∗}T_{n}(r) + H_{n}(r)T_{n}(r)^{∗}] }ρ(r)rdr ,  (41)

whereas, with the ansatz

φ^{P} = C_{P} (\frac{r}{a})^{n} Y_{nm}(ϑ, χ) exp(iωt) .  (42)

for body tides and

φ^{P} = \{ \begin{matrix}\frac{4πGa}{2n + 1}τ_{nm} (\frac{r}{a})^{n} Y_{nm}(ϑ, χ) exp(iωt) & for r ≤ a \\ \frac{4πGa}{2n + 1}τ_{nm} (\frac{a}{r})^{n+1} Y_{nm}(ϑ, χ) exp(iωt) & for r > a \end{matrix}  (43)

for loading tides, the energy G is evaluated from (35) to

G_{n} = −\frac{1}{2} ∫_{0}^{a} \{rH_{n}(r)^{∗}R′_{n}(r) + n(n + 1)T_{n}(r)^{∗}R_{n}(r)\} ρ(r)rdr − \frac{1}{2}a^{2}R_{n}(a)τ_{nm}^{∗}
= −\frac{1}{8πG}\{ ∫_{0}^{a} [r^{2}|R′_{n}(r)|^{2} + n(n + 1)|R_{n}(r)|^{2}] dr
+aR_{n}(a)[(n + 1)R_{n}(a) − (2n + 1)C_{P}]^{∗}}  (44)

where, for loading tides, C_{P} = 0 is meant. The primary and secondary part of the this energy are

G_{n}^{P} = \{ \begin{matrix}−\frac{nC_{P}}{2} ∫_{0}^{a} \{H_{n}(r)^{∗} + (n + 1)T_{n}(r)^{∗}\} (\frac{r}{a})^{n} ρ(r)rdr \\ = −(2n + 1)\frac{aC_{P}}{8πG} [R_{n}(a) − C_{P}]^{∗}  for body tides \\ −\frac{2πGanτ_{nm}}{2n + 1} ∫_{0}^{a} \{H_{n}(r)^{∗} + (n + 1)T_{n}(r)^{∗}\} (\frac{r}{a})^{n} ρ(r)rdr − \frac{a^{2}}{2} R_{n}(a)τ_{nm}^{∗} \\ = −\frac{a^{2}}{2} [τ_{nm}R_{n}(a)^{∗} + τ_{nm}^{∗}R_{n}(a) − \frac{4πGa}{2n + 1}|τ_{nm}|^{2}]  for loading tides \end{matrix}  (45)

and

G_{n}^{S} = \{ \begin{matrix}−\frac{1}{2} ∫_{0}^{a} \{H_{n}(r)^{∗} [rR′_{n}(r) − nC_{P} (\frac{r}{a})^{n}] \\ +n(n + 1)T_{n}(r)^{∗} [R_{n}(r) − C_{P} (\frac{r}{a})^{n}]\} ρ(r)rdr \\ = −\frac{1}{8πG}\{ ∫_{0}^{a} \{rR′_{n}(r)^{∗} [rR′_{n}(r) − nC_{P} (\frac{r}{a})^{n}] \\ +n(n + 1)R_{n}(r)^{∗} [R_{n}(r) − C_{P} (\frac{r}{a})^{n}]\} dr \\ +a[(n + 1)R_{n}(a) − (2n + 1)C_{P}]^{∗}[R_{n}(a) − C_{P}]\} for body tides \\ −\frac{1}{2} ∫_{0}^{a} \{H_{n}(r)^{∗} [rR′_{n}(r) − \frac{4πGan}{2n + 1}τ_{nm} (\frac{r}{a})^{n}] \\ +n(n + 1)T_{n}(r)^{∗} [R_{n}(r) − \frac{4πGa}{2n + 1}τ_{nm} (\frac{r}{a})^{n}]\} ρ(r)rdr \\ = −\frac{1}{8πG}\{ ∫_{0}^{a} \{rR′_{n}(r)^{∗} [rR′_{n}(r) − \frac{4πGan}{2n + 1}τ_{nm} (\frac{r}{a})^{n}] \\ +n(n + 1)R_{n}(r)^{∗} [R_{n}(r) − \frac{4πGa}{2n + 1}τ_{nm} (\frac{r}{a})^{n}]\} dr \\ +\frac{a}{2n + 1} [(2n + 1)R_{n}(a) − 4πGaτ_{nm}] \\ × [(n + 1)R_{n}(a) − 4πGaτ_{nm}]^{∗} \}  for loading tides \end{matrix}  (46)

respectively. From the ansatz (11) and (17) with (13) and (18) and the boundary condition (0.8a) the elastic deformation energy is evaluated as

E_{n}= \frac{1}{2} ∫_{0}^{a} \{λ(r) |rH′_{n}(r) + 2H_{n}(r) − n(n + 1)T_{n}(r)|^{2}
+μ(r)[2r^{2}|H′_{n}(r)|^{2} + 4|H_{n}(r)|^{2} − 2n(n + 1) (H_{n}(r)^{∗}T_{n}(r) + H_{n}(r)T_{n}(r)^{∗})
+n(n + 1) |rT′_{n}(r) − T_{n}(r) + H_{n}(r)|^{2} + 2n(n + 1)(n^{2} + n − 1)|T_{n}(r)|^{2}
+n(n + 1) |rW′_{n}(r) − W_{n}(r)|^{2} + n(n + 1)(n + 2)(n − 1)|W_{n}(r)|^{2}]} dr ,  (47)

where the surface integrals η_{nm}, θ_{nm} and ζ_{nm} over the external force per area acting on the surface were introduced in order to express the boundary term and are defined below (p. 9, equation (0.71a), (0.71b) and (0.71c)). The kinetic energy obtains the value

K_{n} = \frac{|ω|^{2}}{2} ∫_{0}^{a} \{|H_{n}(r)|^{2} + n(n + 1)[|T_{n}(r)|^{2} + |W_{n}(r)|^{2}]\} ρ(r)r^{2}dr  (48)

and at last, the load-lift energy is

L_{n} = a^{2}g(a)H_{n}(a)τ_{nm}^{∗} .  (49)

The total inner potential energy of the mass redistribution within the earth then is

{P + G^{S} + E} cos^{2}(ω(t − t_{∘}))  (50)

for a real time periodic deformation with circle frequency ω and phase t_{∘}.

0.1.4 A special property of the differential equation system

The differential equation systems (15), (20), (23) and (31) have an important property related to the formal self adjointness of the differential equation (10) which has been first pointed out by Molodenskiy ? for a slightly transformed equation system equivalent to (20). In the present formulation this property manifest itself by the similarity of the matrices in question with their negative transposed matrices:

−A_{T}^{T} = C_{T}A_{T}C_{T}^{−1} ,  (0.51a)
−A_{S}^{T} = C_{S}A_{S}C_{S}^{−1} ,  (0.51b)
−A_{L}^{T} = C_{L}A_{L}C_{L}^{−1} .  (0.51c)

The similarity transform is explicitly given by the matrices

C_{T} = ( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ) ,  (0.52a)

C_{S} = ( \begin{matrix}0 & −4πGn(n + 1) & 0 & 0 & 0 & 0 \\ 4πGn(n + 1) & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & −4πGn(n + 1) & 0 & 0 \\ 0 & 0 & 4πGn(n + 1) & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & −n \\ 0 & 0 & 0 & 0 & n & 0 \end{matrix} ) ,  (0.52b)

C_{L} = ( \begin{matrix}0 & ω^{2} & 1 & 0 \\ −ω^{2} & 0 & 0 & 0 \\ −1 & 0 & 0 & −1 \\ 0 & 0 & 1 & 0 \end{matrix} ) .  (0.52c)

For A_{0} there does not exist such an equation with a constant non-singular matrix C_{0}, but we have

−A_{0}^{T} C_{0} = C_{0}A_{0}  (53)

with a matrix

C_{0} = ( \begin{matrix}0 & −1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} ) .  (54)

Of course, these matrices are only determined up to a constant matrix factor commuting with the matrix A_{T}, A_{S}, A_{L}, A_{0} respectively. These factors have been choosen such that the equality

x_{S}^{T} C_{S}y_{S} = x_{L}^{T} C_{L}y_{L}  (55)

holds by (33) at the boundary between a liquid and a solid layer.

By differentiating such scalar products and applying (15), (20), (31) it may be concluded from (0.51a) to (0.51c) that these scalar products must be constant. Due to the continuity (55) across the boundary, the constant must be the same in all layers. Taking only solutions x and y which go to zero at the center of the earth, we therefore have

x_{T}^{T} C_{T}y_{T} = 0 , x_{S}^{T} C_{S}y_{S} = x_{L}^{T} C_{L}y_{L} = 0 .  (56)

From these equations there may be derived linear relations between Love and Shida numbers (see below, section 0.2.2).

0.1.5 Solutions for a homogeneous sphere

Solutions of the differential equation systems considered in section 0.1.2 in general diverge at the center of the earth. Therefore, these systems are numerically untractable there. To overcome this difficulty, a small sphere at the center of the earth may be taken to be homogeneous. For this homogeneous sphere the finite solutions of the differential equation systems shall be derived analytically. For a homogeneous sphere we have constant values of ρ, λ and μ and

g = \frac{4π}{3} Gρr .  (57)

The toroidal solutions may be given directly in terms of spherical Bessel functions j_{n} of the first kind (cf. ?, p. 437):

y_{T}(r) = C_{T}\sqrt{r} ( \begin{matrix}& j_{n} (\sqrt{\frac{ρ}{μ}}ωr) \\ \frac{μ}{n+1} & \{\sqrt{\frac{ρ}{μ}}ωrj′_{n} (\sqrt{\frac{ρ}{μ}}ωr) − j_{n} (\sqrt{\frac{ρ}{μ}}ωr)\} \end{matrix} )  (58)

or

W_{n}(r) = C_{T}j_{n} (\sqrt{\frac{ρ}{μ}}ωr) .  (59)

However, if the core-mantle boundary is supposed to be frictionless — and no other assumption is consistent with vanishing shear stress in the liquid core — the numerical integration has to be started at this boundary with the initial value

y_{T}^{∘} = \binom{C_{T}^{∘}}{0} .  (60)

In order to derive spheroidal solutions for n > 0 we decompose A_{S} with g replaced by the expression (57) into

A_{S} = A_{S}^{(0)} + rA_{S}^{(1)} + r^{2}A_{S}^{(2)} ,  (61)

where

A_{S}^{(0)} = ( \begin{matrix}\frac{−3λ+2μ}{2(λ+2μ)} & \frac{n+1}{λ+2μ} & \frac{(n+1)λ}{λ+2μ} & 0 & 0 & 0 \\ \frac{4μ(3λ+2μ)}{(n+1)(λ+2μ)} & \frac{3λ−2μ}{2(λ+2μ)} & \frac{−2μ(3λ+2μ)}{λ+2μ} & n & 0 & 0 \\ −n & 0 & \frac{3}{2} & \frac{n}{μ} & 0 & 0 \\ \frac{−2μ(3λ+2μ)}{λ+2μ} & \frac{−(n+1)λ}{λ+2μ} & \frac{4μ(n+1)(λ+μ)}{λ+2μ} − \frac{2μ}{n} & −\frac{3}{2} & 0 & 0 \\ 0 & 0 & 0 & 0 & − (n + \frac{1}{2}) & 1 \\ 0 & 0 & 0 & 0 & 0 & n + \frac{1}{2} \end{matrix} ) ,  (0.62a)

A_{S}^{(1)} = ( \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & ρ & −\frac{ρ}{n+1} \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & −ρ & 0 \\ 4πGρ & 0 & 0 & 0 & 0 & 0 \\ 4πGρ(n + 1) & 0 & −4πGρ(n + 1) & 0 & 0 & 0 \end{matrix} ) ,  (0.62b)

A_{S}^{(2)} = ( \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ −\frac{16πGρ^{2}+3ρω^{2}}{3(n+1)} & 0 & \frac{4π}{3} Gρ^{2} & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ \frac{4π}{3} Gρ^{2} & 0 & −\frac{ρω^{2}}{n} & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} )  (0.62c)

are constant with respect to r. Then six linear independent solutions of (20) may be given as series

y_{S}^{(i)}(r) = r^{α_{i}} ∑_{k=0}^{∞} r^{k}b_{i}^{(k)} ,  (63)

where α_{i} are the eigenvalues of A_{S}^{(0)} — α_{1} = n − 1/2, α_{2} = n + 1/2, α_{3} = n + 3/2, α_{4} = −(n − 1/2), α_{5} = −(n + 1/2), α_{6} = −(n + 3/2) — and the vectors b_{i}^{(k)} must satisfy

{(α_{i} + k)1 − A_{S}^{(0)}}b_{i}^{(k)} = A_{S}^{(1)}b_{i}^{(k−1)} + A_{S}^{(2)}b_{i}^{(k−2)}  (64)

with b_{i}^{(j)} = 0 for j < 0. Only the first three solutions are finite at the center of the earth, their coefficients explicitly read

b_{1}^{(0)} = ( \begin{matrix}n \\ \frac{2n(n−1)}{n+1} μ \\ n \\ 2(n − 1)μ \\ 0 \\ 0 \end{matrix} ) , b_{1}^{(1)} = ( \begin{matrix}0 \\ 0 \\ 0 \\ 0 \\ \frac{4π}{3} Gρn − ω^{2} \\ \frac{8π}{3} Gρn(n − 1) − (2n + 1)ω^{2} \end{matrix} ) , b_{1}^{(k)} = 0 for k ≥ 2  (0.65a)

for α_{1} = n − 1/2 and

b_{2}^{(0)} = ( \begin{matrix}0 \\ 0 \\ 0 \\ 0 \\ 1 \\ 2n + 1 \end{matrix} ) , b_{2}^{(1)} = \frac{nρ}{(n + 1)[n(λ + μ) − 2μ]}( \begin{matrix}0 \\ −λ \\ 1 \\ μ \\ 0 \\ 0 \end{matrix} ) ,

b_{2}^{(k)} = [(n + k + \frac{1}{2}) 1 − A_{S}^{(0)}]^{−1} \{A_{S}^{(1)}b_{2}^{(k−1)} + A_{S}^{(2)}b_{2}^{(k−2)}\} for k ≥ 2

(0.65b) for α_{2} = n + 1/2 and

b_{3}^{(0)} = ( \begin{matrix}(n + 1)[n(λ + μ) − 2μ] \\ 2μ[n(n − 1)(λ + μ) − (3λ + 2μ)] \\ n[(n + 1)(λ + μ) + 2(λ + 2μ)] \\ 2μ[(n + 1)^{2}(λ + μ) − (λ + 2μ)] \\ 0 \\ 0 \end{matrix} ) ,

b_{3}^{(1)} = −4πGρ(n + 1)( \begin{matrix}0 \\ 0 \\ 0 \\ 0 \\ μ \\ [n(λ + 3μ) + μ] \end{matrix} ) ,

b_{3}^{(k)} = [(n + k + \frac{3}{2}) 1 − A_{S}^{(0)}]^{−1} \{A_{S}^{(1)}b_{3}^{(k−1)} + A_{S}^{(2)}b_{3}^{(k−2)}\} for k ≥ 2

(0.65c)

for α_{3} = n + 3/2.

In the case n = 0 the finite solution may again be given explicitely in terms of the spherical Bessel functions j_{1} and j_{0} of the first kind:

y_{0} = C_{S} ( \begin{matrix}& \sqrt{r}j_{1}(κr) \\ \sqrt{r} & \{(λ + 2μ)κrj′_{1}(κr) − 2λj_{1}(κr)\} \\ & \frac{4πGρ}{κ}[1 − j_{0}(κr)] \end{matrix} )  (66)

or

\begin{matrix}H_{0}(r) = C_{S}j_{1}(κr) \\ R_{0}(r) = \frac{4πGρ}{κ}C_{S}[1 − j_{0}(κr)] \end{matrix}  (67)

with

κ = \sqrt{\frac{16πGρ^{2} + 3ω^{2}ρ}{3(λ + 2μ)}} .  (68)

In the case of a liquid the Adams-Williamson condition (25) is only consistent with homogeneity if the liquid is incompressible, i.e. if λ = ∞. Under this circumstances there are two linear independent finite solutions, which can be expressed by a power of r:

y_{L}^{(1)}(r) = r^{n+1/2} ( \begin{matrix}4πGρ \\ 1 \\ 0 \\ −4πGρ \end{matrix} ) , y_{L}^{(2)}(r) = r^{n+1/2} ( \begin{matrix}0 \\ 0 \\ n \\ 2n + 1 \end{matrix} ) .  (69)

The general finite solution then reads

H_{n}(r) = C_{L}^{(1)}nr^{n−1} ,
T_{n}(r) = C_{L}^{(1)}r^{n−1} ,  (0.70a)
R_{n}(r) = C_{L}^{(2)}r^{n} .

0.2 Body tides and surface load

0.2.1 Spherical harmonic expansion of surface loads

As the first order differential equation system has been decomposed into toroidal and speroidal harmonic parts by (11) and (17) the external surface force F and the surface mass density τ have to be expanded likewise. Since the spherical harmonics (12) constitute an orthonormal system, this is easily accomplished by surface integration. For a periodic load F = F^{∘} exp(iωt), τ = τ^{∘} exp(iωt) with circle frequency ω we set

η_{nm} =^{def}∫_{0}^{π}∫_{0}^{2π} F_{r}^{∘}(ϑ, χ)Y_{nm}^{∗}(ϑ, χ) sin(ϑ) dχdϑ ,  (0.71a)

θ_{nm} =^{def} \frac{−1}{n(n + 1)} ∫_{0}^{π}∫_{0}^{2π} \{\frac{∂}{∂ϑ}[sin(ϑ)F_{ϑ}^{∘}(ϑ, χ)] + \frac{∂}{∂χ}F_{χ}^{∘}(ϑ, χ)\} Y_{nm}^{∗}(ϑ, χ) dχdϑ ,  (0.71b)

ζ_{nm} =^{def} \frac{1}{n(n + 1)} ∫_{0}^{π}∫_{0}^{2π} \{\frac{∂}{∂χ}F_{ϑ}^{∘}(ϑ, χ) − \frac{∂}{∂ϑ}[sin(ϑ)F_{χ}^{∘}(ϑ, χ)]\} Y_{nm}^{∗}(ϑ, χ) dχdϑ  (0.71c)

and

τ_{nm} =^{def} ∫_{0}^{π}∫_{0}^{2π} τ^{∘}(ϑ, χ)Y_{nm}^{∗}(ϑ, χ) sin(ϑ) dχdϑ .  (0.71d)

The definitions (0.71b) and (0.71c) amount to decompose the horizontal component of the force F into spheroidal harmonic parts θ_{nm} and toroidal harmonic parts ζ_{nm}.

For the combined ansatz (11) and (17) the incremental pseudo-stress tensor components are related to the components of the solution vectors (14) and (19) as follows:

σ_{rr} = \frac{n+1}{r^{3/2}} Y_{nm}(ϑ, χ) exp(iωt) y_{S,2} ,
\frac{−1}{sin(ϑ)} \{\frac{∂}{∂ϑ}[sin(ϑ)σ_{rϑ}] + \frac{∂}{∂χ}σ_{rχ}\} = \frac{n(n+1)}{r^{3/2}} Y_{nm}(ϑ, χ) exp(iωt) y_{S,4} ,

\frac{1}{sin(ϑ)} \{\frac{∂}{∂χ}σ_{rϑ} − \frac{∂}{∂ϑ}[sin(ϑ)σ_{rχ}]\} = \frac{n(n+1)^{2}}{r^{3/2}} Y_{nm}(ϑ, χ) exp(iωt) y_{T,2} .

(72)

From these relations we conclude that the surface boundary condition (0.8a) becomes

y_{S,2}|_{r=a−0} = \frac{a^{3/2}}{n + 1}η_{nm} ,  (0.73a)
y_{S,4}|_{r=a−0} = a^{3/2} θ_{nm} ,  (0.73b)
y_{T,2}|_{r=a−0} = \frac{a^{3/2}}{n + 1}ζ_{nm} .  (0.73c)

In order to evaluate the surface boundary condition (0.8b), we remark that φ satisfying the Laplace equation outside the earth must have the form φ ∝ r^{−(n+1)}Y_{nm}(ϑ, χ) if it is to vanish at infinity, i.e. no primary potential from sources outside the earth is present, and that φ is continuous through the surface. Therefore [∂φ/∂r]_{+} = −(n + 1)[φ/r]_{−} and equation (0.8b) may be transformed to

y_{S,6}|_{r=a−0} = 4πGa^{3/2}τ_{nm} .  (0.73d)

If the vertical component of the external surface force is solely due to the material on the surface which must have the same vertical velocity u̇_{r} as the surface themselves, the relation

η_{nm} = −g(a)τ_{nm}  (74)

must be satisfied. This is so because the vertical acceleration ü_{r} times the surface mass density τ is small of second order and must be neglected in the linearized theory studied here.

0.2.2 Love-Shida numbers for body tides and surface load

The Love numbers h_{n}, k_{n} and the Shida number ℓ_{n} are defined to be the appropriately normalized dimensionless answer of the system to a primary excitation potential of equation (42) ?. The incremental potential then is φ = φ^{P}+φ^{S} with φ^{S} = C_{S} (\frac{a}{r})^{n+1} Y_{nm}(ϑ, χ) exp(iωt) outside the earth. We then have R_{n}(a) = C_{P} + C_{S} and, according to the boundary condition (0.8c), aR′_{n}(a) − 4πGρ(a)aH_{n}(a) = nC_{P} − (n + 1)C_{S}. Therefore, y_{S}_{6}(a) = (2n + 1)\sqrt{a} C_{P}. Now the Love-Shida numbers are defined as

h_{n}(r) = \frac{g(a)H_{n}(r)}{C_{P}} = (2n + 1)g(a)\sqrt{\frac{a}{r}} \frac{y_{S}_{1}(r)}{y_{S}_{6}(a)} ,  (0.75a)

ℓ_{n}(r) = \frac{g(a)T_{n}(r)}{C_{P}} = \frac{2n + 1}{n}g(a)\sqrt{\frac{a}{r}} \frac{y_{S}_{3}(r)}{y_{S}_{6}(a)} ,  (0.75b)

k_{n}(r) = \frac{R_{n}(r)}{C_{P}} − (\frac{r}{a})^{n} = (2n + 1)\sqrt{\frac{a}{r}} \frac{y_{S}_{5}(r)}{y_{S}_{6}(a)} − (\frac{r}{a})^{n} ,  (0.75c)

for a solution y_{S} with y_{S}_{2}(a) = y_{S}_{4}(a) = 0 because of the boundary condition (0.8a) with F = 0. At the surface these numbers have the values

h_{n} = \frac{g(a)H_{n}(a)}{C_{P}} = (2n + 1)g(a)\frac{y_{S}_{1}(a)}{y_{S}_{6}(a)} ,  (0.76a)

ℓ_{n} = \frac{g(a)T_{n}(a)}{C_{P}} = \frac{2n + 1}{n}g(a)\frac{y_{S}_{3}(a)}{y_{S}_{6}(a)} ,  (0.76b)

k_{n} = \frac{R_{n}(a)}{C_{P}} − 1 = (2n + 1)\frac{y_{S}_{5}(a)}{y_{S}_{6}(a)} − 1 .  (0.76c)

There are also defined three load factors ??? for a surface mass load with η_{nm} = −g(a)τ_{nm} and θ_{nm} = ζ_{nm} = 0:

h′_{n}(r) = \frac{(2n + 1)g(a)}{4πGa} \frac{H_{n}(r)}{τ_{nm}} = (2n + 1)g(a)\sqrt{\frac{a}{r}} \frac{y_{S}_{1}(r)}{y_{S}_{6}(a)} ,  (0.77a)
ℓ′_{n}(r) = \frac{(2n + 1)g(a)}{4πGa} \frac{T_{n}(r)}{τ_{nm}} = \frac{2n + 1}{n}g(a)\sqrt{\frac{a}{r}} \frac{y_{S}_{3}(r)}{y_{S}_{6}(a)} ,  (0.77b)
k′_{n}(r) = \frac{2n + 1}{4πGa} \frac{R_{n}(r)}{τ_{nm}} − (\frac{r}{a})^{n} = (2n + 1)\sqrt{\frac{a}{r}} \frac{y_{S}_{5}(r)}{y_{S}_{6}(a)} − (\frac{r}{a})^{n} .  (0.77c)

Here, the solution y_{S} must satisfy y_{S}_{2}(a) = −\frac{g(a)}{4πG(n+1)}y_{S}_{6}(a) and y_{S}_{4}(a) = 0 because of the boundary conditions (0.73a), (0.73d) and (0.73b). At the surface these factors become

h′_{n} = \frac{2n + 1}{4πa^{3}} M \frac{H_{n}(a)}{τ_{nm}} = (2n + 1)g(a)\frac{y_{S}_{1}(a)}{y_{S}_{6}(a)} ,  (0.78a)
ℓ′_{n} = \frac{2n + 1}{4πa^{3}} M \frac{T_{n}(a)}{τ_{nm}} = \frac{2n + 1}{n}g(a)\frac{y_{S}_{3}(a)}{y_{S}_{6}(a)} ,  (0.78b)
k′_{n} = \frac{2n + 1}{4πGa} \frac{R_{n}(a)}{τ_{nm}} − 1 = (2n + 1)\frac{y_{S}_{5}(a)}{y_{S}_{6}(a)} − 1 ,  (0.78c)

where M is the mass of the earth. A unit point mass located at colatitude ϑ_{∘} and longitude χ_{∘} has sherical harmonic expansion

δ(ϑ, χ|ϑ_{∘}, χ_{∘}) = ∑_{n=0}^{∞} ∑_{m=−n}^{n} Y_{nm}(ϑ, χ)Y_{nm}^{∗}(ϑ_{∘}, χ_{∘}) .  (79)

If the unit point mass is located at the north pole, which, because of spherical symmetry, might be achieved by taking the coordinate system appropriate, the double sum may be reduced to a single sum over Legendre polynomials

δ(ϑ, χ|0) = ∑_{n=0}^{∞} \frac{2n + 1}{4π}P_{n}(cos ϑ) .  (80)

The system therefore reacts on a load τ(ϑ, χ) = \frac{M_{∘}}{a^{2}} δ(ϑ, χ|0) exp(iωt), which gives τ_{nm} = \frac{M_{∘}}{a^{2}} \sqrt{\frac{2n+1}{4π}}δ_{m0}, with elongations

u_{r}(a, ϑ, χ) = \frac{M_{∘}a}{M} ∑_{n=0}^{∞} h′_{n}P_{n}(cos ϑ) exp(iωt) ,  (0.81a)

u_{ϑ}(a, ϑ, χ) = \frac{M_{∘}a}{M} ∑_{n=0}^{∞} ℓ′_{n} \frac{∂}{∂ϑ}P_{n}(cos ϑ) exp(iωt) ,  (0.81b)

u_{χ}(a, ϑ, χ) = 0 and incremental potential at the undisturbed surface

φ(a, ϑ, χ) = \frac{M_{∘}G}{a} ∑_{n=0}^{∞}(k′_{n} + 1)P_{n}(cos ϑ) exp(iωt) .  (82)

For tangential forces there might be defined four more dimensionless coefficients, three for spheroidal forcing ??:

h′′_{n}(r) = \frac{2n + 1}{n(n + 1)} \frac{g(a)^{2}}{4πGa} \frac{H_{n}(r)}{θ_{nm}} = \frac{2n + 1}{n(n + 1)} \frac{g(a)^{2}}{4πG} \sqrt{\frac{a}{r}} \frac{y_{S}_{1}(r)}{y_{S}_{4}(a)} ,  (0.83a)
ℓ′′_{n}(r) = \frac{(2n + 1)}{n(n + 1)} \frac{g(a)^{2}}{4πGa} \frac{T_{n}(r)}{θ_{nm}} = \frac{2n + 1}{n^{2}(n + 1)} \frac{g(a)^{2}}{4πG} \sqrt{\frac{a}{r}} \frac{y_{S}_{3}(r)}{y_{S}_{4}(a)} ,  (0.83b)
k′′_{n}(r) = \frac{2n + 1}{n(n + 1)} \frac{g(a)}{4πGa} \frac{R_{n}(r)}{θ_{nm}} = \frac{2n + 1}{n(n + 1)} \frac{g(a)}{4πG} \sqrt{\frac{a}{r}} \frac{y_{S}_{5}(r)}{y_{S}_{4}(a)} ,  (0.83c)

for a solution y_{S} with y_{S}_{2}(a) = y_{S}_{6}(a) = 0 because of (0.73a) and (0.73d). At the surface these coefficients read

h′′_{n} = \frac{2n + 1}{n(n + 1)} \frac{GM^{2}}{4πa^{5}} \frac{H_{n}(a)}{θ_{nm}} = \frac{2n + 1}{n(n + 1)} \frac{Mg(a)}{4πa^{2}} \frac{y_{S}_{1}(a)}{y_{S}_{4}(a)} ,  (0.84a)
ℓ′′_{n} = \frac{2n + 1}{n(n + 1)} \frac{GM^{2}}{4πa^{5}} \frac{T_{n}(a)}{θ_{nm}} = \frac{2n + 1}{n^{2}(n + 1)} \frac{Mg(a)}{4πa^{2}} \frac{y_{S}_{3}(a)}{y_{S}_{4}(a)} ,  (0.84b)
k′′_{n} = \frac{2n + 1}{n(n + 1)} \frac{M}{4πa^{3}} \frac{R_{n}(a)}{θ_{nm}} = \frac{2n + 1}{n(n + 1)} \frac{M}{4πa^{2}} \frac{y_{S}_{5}(a)}{y_{S}_{4}(a)} .  (0.84c)

Finally, there is one coefficient for toroidal forcing ?:

f_{n}(r) = \frac{g(a)^{2}}{4πGa} \frac{W_{n}(r)}{ζ_{nm}} = \frac{g(a)^{2}}{4πG(n + 1)} \sqrt{\frac{a}{r}} \frac{y_{T}_{1}(r)}{y_{T}_{2}(a)} ,  (85)

which at the surface has the value

f_{n} = \frac{GM^{2}}{4πa^{5}} \frac{W_{n}(a)}{ζ_{nm}} = \frac{Mg(a)}{4π(n + 1)a^{2}} \frac{y_{T}_{1}(a)}{y_{T}_{2}(a)} .  (86)

The nine coefficients for spheroidal forcing are not independent. Applying equation (56) to any pair of the three solutions

y = ( \begin{matrix}h_{n} \\ 0 \\ nℓ_{n} \\ 0 \\ (1 + k_{n})g \\ (2n + 1)g \end{matrix} ) , y′ = ( h′_{n} ; \frac{−(2n+1)Mg}{4π(n+1)a^{2}} ; nℓ′_{n} ; 0 ; (1 + k′_{n})g ; (2n + 1)g ) , y′′ = ( h′′_{n} ; 0 ; nℓ′′_{n} ; \frac{(2n+1)Mg}{4πn(n+1)a^{2}} ; k′′_{n}g ; 0 ) ,  (87)

given at the surface of the earth, the three linear relations

k′_{n} = k_{n} − h_{n}  (0.88a)
k′′_{n} = ℓ_{n}  (0.88b)
h′′_{n} = k′′_{n} − ℓ′_{n} = ℓ_{n} − ℓ′_{n}  (0.88c)

valid at the surface may be derived ?. Furthermore, evaluating equation (9), there may be derived the following integral relations for the Love-Shida numbers:

k_{n}(a) = \frac{4πGn}{(2n + 1)g(a)} ∫_{0}^{a}\{h_{n}(r) + (n + 1)ℓ_{n}(r)\} (\frac{r}{a})^{n+1} ρ(r)dr ,  (0.89a)
k′_{n}(a) = \frac{4πGn}{(2n + 1)g(a)} ∫_{0}^{a}\{h′_{n}(r) + (n + 1)ℓ′_{n}(r)\} (\frac{r}{a})^{n+1} ρ(r)dr ,  (0.89b)
k′′_{n}(a) = \frac{4πGn}{(2n + 1)g(a)} ∫_{0}^{a}\{h′′_{n}(r) + (n + 1)ℓ′′_{n}(r)\} (\frac{r}{a})^{n+1} ρ(r)dr .  (0.89c)

0.2.3 Dimensionless representation of energy integrals

The energy integrals of section 0.1.3 may also be expressed entirely in terms of Love-Shida numbers. In order to do so, we choose the dimensionless normalizations

\overline{P}_{n} =^{def} \frac{4πG}{a|C_{P}|^{2}} P_{n} ,  \overline{G} _{n}^{P} =^{def} \frac{4πG}{a|C_{P}|^{2}} G_{n}^{P} ,  \overline{G} _{n}^{S} =^{def} \frac{4πG}{a|C_{P}|^{2}} G_{n}^{S} ,
\overline{E}_{n} =^{def} \frac{4πG}{a|C_{P}|^{2}} E_{n} ,  \overline{K}_{n} =^{def} \frac{4πG}{a|C_{P}|^{2}} K_{n} ,  (90)

for body tides,

\overline{P} ′_{n} =^{def} \frac{2n + 1}{4πGa^{3}|τ_{nm}|^{2}} P_{n} , \overline{G} _{n}^{P}′ =^{def} \frac{2n + 1}{4πGa^{3}|τ_{nm}|^{2}} G_{n}^{P} , \overline{G} _{n}^{S}′ =^{def} \frac{2n + 1}{4πGa^{3}|τ_{nm}|^{2}} G_{n}^{S} ,

\overline{E} ′_{n} =^{def} \frac{2n + 1}{4πGa^{3}|τ_{nm}|^{2}} E_{n} , \overline{K} ′_{n} =^{def} \frac{2n + 1}{4πGa^{3}|τ_{nm}|^{2}} K_{n} , \overline{L} ′_{n} =^{def} \frac{2n + 1}{4πGa^{3}|τ_{nm}|^{2}} L_{n}  (91)

in the case of loading tides,

\overline{P} ′′_{n} =^{def} \frac{(2n + 1)^{2}g(a)^{2}}{4πGa^{3}n^{2}(n + 1)^{2}|θ_{nm}|^{2}} P_{n} , \overline{G} ′′_{n} =^{def} \frac{(2n + 1)^{2}g(a)^{2}}{4πGa^{3}n^{2}(n + 1)^{2}|θ_{nm}|^{2}} G_{n} ,
\overline{E} ′′_{n} =^{def} \frac{(2n + 1)^{2}g(a)^{2}}{4πGa^{3}n^{2}(n + 1)^{2}|θ_{nm}|^{2}} E_{n} , \overline{K} ′′_{n} =^{def} \frac{(2n + 1)^{2}g(a)^{2}}{4πGa^{3}n^{2}(n + 1)^{2}|θ_{nm}|^{2}} K_{n}  (92)

for spheroidal surface forces and

\overline{E} _{n}^{(t)} =^{def} \frac{g(a)^{2}}{4πGa^{3}n(n + 1)ζ_{nm}^{2}} E_{n} ,  \overline{K} _{n}^{(t)} =^{def} \frac{g(a)^{2}}{4πGa^{3}n(n + 1)ζ_{nm}^{2}} K_{n}  (93)

in the case of toroidal surface forces. Then, with the definitions

ρ̄(x) =^{def} \frac{4πGa}{g(a)}ρ(xa) , ḡ(x) =^{def} \frac{g(xa)}{g(a)} ,  (0.94a)
λ̄(x) =^{def} \frac{4πG}{g(a)^{2}} λ(xa) , μ̄(x) =^{def} \frac{4πG}{g(a)^{2}} μ(xa) ,  (0.94b)

the energy integrals (41) and (44) to (47) for body tides then, performing some partial integrations, read

\overline{P}_{n} = \frac{1}{2} ∫_{0}^{1} \{(xρ̄(x) − 4ḡ(x))|h_{n}(x)|^{2}
+n(n + 1)ḡ(x) [h_{n}(x)^{∗}ℓ_{n}(x) + h_{n}(x)ℓ_{n}(x)^{∗}] }ρ̄(x)xdx ,  (0.95a)

\overline{G} _{n}^{P} = −\frac{1}{2} ∫_{0}^{1} n \{h_{n}(x)^{∗} + (n + 1)ℓ_{n}(x)^{∗}\} ρ̄(x)x^{n+1}dx
= −\frac{2n + 1}{2}k_{n}(1)^{∗} ,  (0.95b)

\overline{G} _{n}^{S} = −\frac{1}{2} ∫_{0}^{1} \{xh_{n}(x)^{∗} \frac{dk_{n}(x)}{dx} + n(n + 1)ℓ_{n}(x)^{∗}k_{n}(x)\} ρ̄(x)xdx

= −\frac{1}{2} (∫_{0}^{1} \{|x\frac{dk_{n}(x)}{dx}|^{2} + n(n + 1)|k_{n}(x)|^{2}\} dx + (n + 1)|k_{n}(1)|^{2}) ,  (0.95c)

\overline{E}_{n} = \frac{1}{2} ∫_{0}^{1} \{λ̄(x) |x\frac{dh_{n}(x)}{dx} + 2h_{n}(x) − n(n + 1)ℓ_{n}(x)|^{2} + μ̄(x) [2 |x\frac{dh_{n}(x)}{dx}|^{2}

+4|h_{n}(x)|^{2} − 2n(n + 1)[h_{n}(x)^{∗}ℓ_{n}(x) + h_{n}(x)ℓ_{n}(x)^{∗}]

+n(n + 1)[ |x\frac{dℓ_{n}(x)}{dx} − ℓ_{n}(x) + h_{n}(x)|^{2}

+2(n^{2} + n − 1)|ℓ_{n}(x)|^{2}]]} dx ,  (0.95d)

\overline{K}_{n} = \frac{1}{2} \frac{a|ω|^{2}}{g(a)} ∫_{0}^{1} \{|h_{n}(x)|^{2} + n(n + 1)|ℓ_{n}(x)|^{2}\} ρ̄(x)x^{2}dx .  (0.95e)

For loading tides one accordingly has

\overline{P} ′_{n} = \frac{1}{2} ∫_{0}^{1} \{(xρ̄(x) − 4ḡ(x))|h′_{n}(x)|^{2}
+n(n + 1)ḡ(x) [h′_{n}(x)^{∗}ℓ′_{n} + h′_{n}(x)ℓ′_{n}(x)^{∗}] \}\frac{ρ̄(x)x}{2n + 1}dx ,  (0.96a)

\overline{G} _{n}^{P}′ = −\frac{1}{2}( ∫_{0}^{1} \{h′_{n}(x)^{∗} + (n + 1)ℓ′_{n}(x)^{∗}\} \frac{nρ̄(x)x^{n+1}}{2n + 1} dx + [k′_{n}(1) + 1] )

= −\frac{1}{2} (k′_{n}(1) + k′_{n}(1)^{∗} + 1) ,  (0.96b)

\overline{G} _{n}^{S}′ = −\frac{1}{2} ∫_{0}^{1} \{xh′_{n}(x)^{∗} \frac{dk′_{n}(x)}{dx} + n(n + 1)ℓ′_{n}(x)^{∗}k′_{n}(x)\} \frac{ρ̄(x)x}{2n + 1}dx

= −\frac{1}{2(2n + 1)}( ∫_{0}^{1} \{|x\frac{dk′_{n}(x)}{dx}|^{2} + n(n + 1)|k′_{n}(x)|^{2}\} dx

−(n + 1)|k′_{n}(1)|^{2}) ,  (0.96c)

\overline{E} ′_{n} = \frac{1}{2(2n + 1)} ∫_{0}^{1} \{λ̄(x) |x\frac{dh′_{n}(x)}{dx} + 2h′_{n}(x) − n(n + 1)ℓ′_{n}(x)|^{2}

+μ̄(x) [2 |x\frac{dh′_{n}(x)}{dx}|^{2} + 4|h′_{n}(x)|^{2}

−2n(n + 1)[h′_{n}(x)^{∗}ℓ′_{n}(x) + h′_{n}(x)ℓ′_{n}(x)^{∗}]

+n(n + 1)[ |x\frac{dℓ′_{n}(x)}{dx} − ℓ′_{n}(x) + h′_{n}(x)|^{2}

+2(n^{2} + n − 1)|ℓ′_{n}(x)|^{2}]]} dx ,  (0.96d)
\overline{K} ′_{n} = \frac{1}{2(2n + 1)} \frac{a|ω|^{2}}{g(a)} ∫_{0}^{1} \{|h′_{n}(x)|^{2} + n(n + 1)|ℓ′_{n}(x)|^{2}\} ρ̄(x)x^{2}dx  (0.96e)

and in this case there is an aditional load-lift energy (49) which may be given in the dimensionless form (91) as

\overline{L} ′_{n} = h′_{n}(1) .  (0.96f)

For spheroidal surface forces the dimensionless energy integrals may be evaluated as

\overline{P} ′′_{n} = \frac{1}{2} ∫_{0}^{1} \{(xρ̄(x) − 4ḡ(x))|h′′_{n}(x)|^{2}
+n(n + 1)ḡ(x) [h′′_{n}(x)^{∗}ℓ′′_{n} + h′′_{n}(x)ℓ′′_{n}(x)^{∗}] }ρ̄(x)xdx ,  (0.97a)
\overline{G} ′′_{n} = −\frac{1}{2} ∫_{0}^{1} \{xh′′_{n}(x)^{∗} \frac{dk′′_{n}(x)}{dx} + nx^{n} + n(n + 1)ℓ′′_{n}(x)^{∗}k′′_{n}(x)\} ρ̄(x)xdx
= −\frac{1}{2}( ∫_{0}^{1} \{|x\frac{dk′′_{n}(x)}{dx}|^{2} + n(n + 1)|k′′_{n}(x)|^{2}\} dx + (n + 1)|k′′_{n}(1)|^{2}) ,  (0.97b)
\overline{E} ′′_{n} = \frac{1}{2} ∫_{0}^{1} \{λ̄(x) |x\frac{dh′′_{n}(x)}{dx} + 2h′′_{n}(x) − n(n + 1)ℓ′′_{n}(x)|^{2} + μ̄(x) [2 |x\frac{dh′′_{n}(x)}{dx}|^{2}

+4|h′′_{n}(x)|^{2} − 2n(n + 1)[h′′_{n}(x)^{∗}ℓ′′_{n}(x) + h′′_{n}(x)ℓ′′_{n}(x)^{∗}]

+n(n + 1)[ |x\frac{dℓ′′_{n}(x)}{dx} − ℓ′′_{n}(x) + h′′_{n}(x)|^{2}

+2(n^{2} + n − 1)|ℓ′′_{n}(x)|^{2}]]} dx ,  (0.97c)
\overline{K} ′′_{n} = \frac{1}{2} \frac{a|ω|^{2}}{g(a)} ∫_{0}^{1} \{|h′′_{n}(x)|^{2} + n(n + 1)|ℓ′′_{n}(x)|^{2}\} ρ̄(x)x^{2}dx  (0.97d)

and for toroidal surface forces one has

\overline{E} _{n}^{(t)} = \frac{1}{2} ∫_{0}^{1} μ̄(x) \{|x\frac{df_{n}(x)}{dx} − f_{n}(x)|^{2} + (n + 2)(n − 1)|f_{n}(x)|^{2}\} dx ,  (0.98a)
\overline{K} _{n}^{(t)} = \frac{1}{2} \frac{a|ω|^{2}}{g(a)} ∫_{0}^{1} |f_{n}(x)|^{2}ρ̄(x)x^{2}dx .  (0.98b)

0.2.4 Love-Shida numbers in case of a fluid surface

In the case of a fluid surface the shear stress vanishes there. We, therefore, have only body tide and load Love-Shida numbers. Coefficients for tangential forces, on the other hand, do not exist. The boundary conditions (0.8) in this case amount for

\frac{g}{4πGa}y_{L,1}|_{r=a−0} − \frac{ρω^{2}}{n} y_{L,2}|_{r=a−0} − \frac{ρ}{n}y_{L,3}|_{r=a−0} = \sqrt{a} η_{nm} ,  (0.99a)

y_{L,4}|_{r=a−0} = 4πGa^{3/2} τ_{nm} + (2n + 1)\sqrt{a} C_{P} .  (0.99b)

From this and the definitions (0.76) and (0.78) we derive the following form of the two linear independent solution vectors at the surface:

y = \sqrt{a} C_{P} ( \begin{matrix}\frac{4πGρa}{g} & h̃_{n} \\ \frac{a}{g}nℓ̃_{n} \\ n(1 + k̃_{n}) \\ 2n + 1 \end{matrix} ) ,  y′ = \frac{4πGa^{3/2}}{2n + 1} τ_{nm} ( \begin{matrix}\frac{4πGρa}{g} & h̃′_{n} \\ \frac{a}{g}nℓ̃′_{n} \\ n(1 + k̃′_{n}) \\ 2n + 1 \end{matrix} ) ,  (100)

where the tilde denotes the fluid case of Love-Shida numbers. The boundary condition (0.99a) then yields the relations

h̃_{n} − \frac{aω^{2}}{g(a)}ℓ̃_{n} − k̃_{n} = 1 ,  (0.101a)

h̃′_{n} − \frac{aω^{2}}{g(a)}ℓ̃′_{n} − k̃′_{n} = 1 − (2n + 1)\frac{g(a)}{4πGρ(a)a} .  (0.101b)

Applying equation (56) to the two solutions (100) we again arrive at equation (0.88a) now for the fluid Love numbers

k̃′_{n} = k̃_{n} − h̃_{n} .  (102)

0.2.5 The special case n = 1

In the case of harmonic degree n = 1 there is a special exact solution corrsponding to rigid translations of the earth as a whole. This translationary exact solution is given by the spheroidal solution vector

y_{S}_{trans} = \sqrt{r}( \begin{matrix}1 \\ 0 \\ 1 \\ 0 \\ g(r) − ω^{2}r \\ −3ω^{2}r \end{matrix} ) .  (103)

With aid of the Poisson equation rg′(r) + 2g(r) = 4πGρr it may be checked that (103) indeed satisfies equation (20). However, this solution is physically meaningless for ω ≠ 0 because the inertia forces for this motion were supplied by an additional potential φ = −ω^{2}rY_{1m}(ϑ, χ) exp(iωt) for which there is no physical reason in an inertial system. In a liquid layer this solution has continuation

y_{L}_{trans} = \sqrt{r}( \begin{matrix}4πGρ(r)r \\ r \\ g(r) − ω^{2}r \\ −3ω^{2}r \end{matrix} ) ,  (104)

which satisfies (31). If this solution is multiplied by the constant −\frac{g(a)}{aω^{2}} it has the form of the vector y in (87) at the surface of the earth and therefore satisfies the surface boundary conditions of body tides. From this we may conclude

h_{1}(r) = −\frac{g(a)}{aω^{2}} ,  (0.105a)
ℓ_{1}(r) = −\frac{g(a)}{aω^{2}} ,  (0.105b)
k_{1}(r) = −\frac{g(r)}{aω^{2}} .  (0.105c)

From equations (0.88a) and (0.88b) we further conclude that

k′_{1} = 0 ,  (0.106a)
k′′_{1} = −\frac{g}{aω^{2}}  (0.106b)

must hold at the surface of the earth as well.

Up to now we have done all calculations in an inertial system. However, for body tides, it is customary to give the Love-Shida numbers in the center of mass system. If we transform to a coordinate system the origin of which moves by a vector

s(t) = s_{∘} exp(iωt)(e_{r} + ∇_{1})Y_{1m}(ϑ, χ) ,  (107)

the displacements and the incremental field transform to

H_{1}^{(tr)}(r) = H_{1}(r) − s_{∘} ,  (0.108a)
T_{1}^{(tr)}(r) = T_{1}(r) − s_{∘} ,  (0.108b)
R_{1}^{(tr)}(r) = R_{1}(r) − s_{∘}(g(r) − rω^{2}) ,  (0.108c)

while the toroidal displacement W_{1} is not affected.

The solution y_{S}_{trans} satisfies the boundary conditions for body tides and, therefore, body tides of harmonic degree n = 1 have constant displacement coefficients H_{1}(r) = H_{1}(a) = T_{1}(r) = T_{1}(a). Hence the center of mass in the case of body tides is given by the coefficient s_{∘} = H_{1}. With the above transformation the body tide Love-Shida numbers in the center of mass system then become

h_{1}^{(CM)}(r) = 0 ,  (0.109a)
ℓ_{1}^{(CM)}(r) = 0 ,  (0.109b)
k_{1}^{(CM)}(r) = −\frac{r}{a} .  (0.109c)

For loading tides, the center of mass of the earth together with the loading masses does not move because of conservation of linear momentum. The same is true for tangential forcing if the forces are exerted for instance by winds or ocean currents and the air and water masses are included in the definition of the center of mass. To do so is convenient since e.g. the motion of a satelite which might be used to measure the Love-Shida numbers does not seperate between the mass of the earth and the loading masses on the earth.

0.2.6 Asymptotic solutions for large harmonic degrees

Since meteorological as well as — due to topography — oceanic and even glacial loads have considerable components with large harmonic degrees in there expansion, for which numerical solutions become unstable, there is need for asymptotic solutions of the differential equations (15) and (20) for large n.

For large n the differential equation systems (15) and (20) show two general features. Firstly the solutions which are finite at the origin decay roughly as (r/a)^{n} with increasing deepness under the surface (r = a) and secondly their eigenfrequencies behave asymptotically proportional to n. For this reason we make the transformation

r = a exp (\frac{z}{n + 1})  (110)

and define

lim_{n→∞} \frac{ω}{n + 1} = \sqrt{\frac{ξμ(a)}{a^{2}ρ(a)}} .  (111)

That is to say, we set ξ = 0 when tides generated by external forces with certain periods are studied. However, when eigenmodes of the system are studied, ξ ≠ 0 is the quantity to be determined.

With aid of the definitions (110) and (111) the linear differential eqaution system (20) for spheroidal modes for large n asymptotically becomes

\frac{dy_{S}}{dz} = A_{S,∞}y_{S} + O (\frac{1}{n + 1})  (112)

with a constant Matrix

A_{S,∞} = ( \begin{matrix}0 & \frac{1}{λ+2μ} & \frac{λ}{λ+2μ} & 0 & 0 & 0 \\ −ξμ & 0 & 0 & 1 & 0 & 0 \\ −1 & 0 & 0 & \frac{1}{μ} & 0 & 0 \\ 0 & \frac{−λ}{λ+2μ} & \frac{4μ(λ+μ)}{λ+2μ} − ξμ & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & −1 & 0 \\ γ & 0 & −γ & 0 & 0 & 1 \end{matrix} ) ,  (113)

where λ and μ are the surface values λ = λ(a), μ = μ(a) and we have introduced the abbreviation

γ =^{def} 4πGaρ(a) .  (114)

The matrix A_{S,∞} has for ξ ≠ 0 eigenvalues α_{1} = 1, α_{2} = \sqrt{1 − \frac{ξμ}{λ+2μ}}, α_{3} = \sqrt{1 − ξ}, α_{4} = −\sqrt{1 − ξ}, α_{5} = −\sqrt{1 − \frac{ξμ}{λ+2μ}}, α_{6} = −1 and eigenvectors

\begin{matrix}c_{1} = ( \begin{matrix}0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{matrix} ) , & c_{2} = ( \begin{matrix}√ & \overline{1 − \frac{ξμ}{λ+2μ}} \\ (2 − ξ)μ \\ 1 \\ 2μ\sqrt{1 − \frac{ξμ}{λ+2μ}} \\ 0 \\ γ \end{matrix} ) , & c_{3} = ( \begin{matrix}1 \\ 2μ\sqrt{1 − ξ} \\ √ & \overline{1 − ξ} \\ (2 − ξ)μ \\ 0 \\ −γ \end{matrix} ) , \\ c_{4} = ( \begin{matrix}1 \\ −2μ\sqrt{1 − ξ} \\ −\sqrt{1 − ξ} \\ (2 − ξ)μ \\ 0 \\ −γ \end{matrix} ) , & c_{5} = ( \begin{matrix}√ & \overline{1 − \frac{ξμ}{λ+2μ}} \\ (ξ − 2)μ \\ −1 \\ 2μ\sqrt{1 − \frac{ξμ}{λ+2μ}} \\ 0 \\ −γ \end{matrix} ) , & c_{6} = ( \begin{matrix}0 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \end{matrix} ) . \end{matrix}  (115)

The general solution which is finite at the origin with them reads

y_{S}(z) = ∑_{i=1}^{3} C_{i} exp(α_{i}z) c_{i} + O (\frac{1}{n + 1})  (116)

and from the boundary conditions (0.73a), (0.73b) and (0.73d) with η_{nm} = 0, θ_{nm} = 0, τ_{nm} = 0 we conclude that the eigenfrequency of the spheroidal mode is asymptotically given by the value of ξ satisfying

det ( \begin{matrix}0 & (2 − ξ)μ & 2μ\sqrt{1 − ξ} \\ 0 & 2μ\sqrt{1 − \frac{ξμ}{λ+2μ}} & (2 − ξ)μ \\ 1 & γ & −γ \end{matrix} ) = 0 .  (117)

This is the zero of the cubic equation

ξ^{3} − 8ξ^{2} + 8(3 − 2ν)ξ − 16(1 − ν) = 0  (118)

with ν = \frac{μ}{λ+2μ} and 0 < ξ < 1. For a proportion ν ranging from zero to 1/2 the value ξ for the eigenfrequency varies between 0.912622 and 0.763932.

For ξ = 0 the eigenvalues of the matrix A_{S,∞} are ±1 but A_{S,∞} is no longer diagonalizable. Corresponding to the eigenvalue +1 there are two eigenvectors c_{1}^{∘}, c_{2}^{∘} and a characteristic vector c_{3}^{∘} satisfying A_{S,∞}c_{3}^{∘} = c_{2}^{∘} + c_{3}^{∘} which explicitly read

c_{1}^{∘} = ( \begin{matrix}0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{matrix} ) , c_{2}^{∘} = ( \begin{matrix}1 \\ 2μ \\ 1 \\ 2μ \\ 0 \\ −γ \frac{λ+3μ}{λ+μ} \end{matrix} ) , c_{3}^{∘} = ( \frac{−μ}{λ+μ} ; 0 ; \frac{λ+2μ}{λ+μ} ; 2μ ; 0 ; 0 ) .  (119)

With them the general solution finite at the origin is

y_{S}(z) = exp(z) \{C_{1}c_{1}^{∘} + [C_{2} + (n + 1)C_{3}z]c_{2}^{∘} + C_{3}c_{3}^{∘}\} + O (\frac{1}{n + 1}) .  (120)

From equation (87) one concludes that for body tides one must have C_{1} = (2n + 1)g, C_{2} = C_{3} = 0 and, therefore, the asymptotical values of the body tide Love-Shida numbers for large n are given by

h_{n} = O(1) ,  ℓ_{n} = O (\frac{1}{n}) ,  k_{n} = O(1) .  (0.121a)

Likewise, for loading tides one has C_{1} = (2n + 1)g\{1 − \frac{gρa(λ+3μ)}{2(n+1)μ(λ+μ)}\}, C_{2} = −C_{3} = \frac{−(2n+1)g^{2}}{8πG(n+1)μ} and

h′_{n} = O(1) ,  ℓ′_{n} = O (\frac{1}{n}) ,  k′_{n} = O(1) ,  (0.121b)

and for tangential surface forces one has C_{1} = C_{2} = 0, C_{3} = \frac{(2n+1)g^{2}}{8πGn(n+1)μ}

h′′_{n} = \frac{−g^{2}}{4πG(λ + μ)n} + O (\frac{1}{n^{2}} ) , ℓ′′_{n} = \frac{−g^{2}(λ + 2μ)}{4πGμ(λ + μ)n^{2}} + O (\frac{1}{n^{3}} ) , k′′_{n} = O (\frac{1}{n^{2}} ) ,  (0.121c)

For toroidal modes one could use the explicit solution (58) to gain the asymptotes directly, but it is easier to handle it in the same way as the case of spheroidal modes using the matrix

A_{T,∞} = lim_{n→∞} \frac{1}{n + 1}A_{T} = ( \begin{matrix}0 & \frac{1}{μ} \\ μ(1 − ξ) & 0 \end{matrix} ) ,  (122)

where μ = μ(a). The equation

\frac{dy_{T}}{dz} = A_{T,∞}y_{T} + O (\frac{1}{n + 1})  (123)

then yields the asymptotic solution

y_{T}(r) = C exp(\sqrt{1 − ξ} z) ( \begin{matrix}1 \\ μ\sqrt{1 − ξ} \end{matrix} ) + O (\frac{1}{n + 1}) .  (124)

The boundary condition (0.73c) with ζ_{nm} = 0 then amounts to \sqrt{1 − ξ} = 0 and hence the eigenfrequency for toroidal modes is up to leading order asymptotically given by the value ξ = 1.

The solution (124) with ξ = 0 together with the definition (86) then yuields the asymptotical value

f_{n} = \frac{g^{2}}{4πGμn} + O (\frac{1}{n^{2}} )  (125)

for the toroidal forcing coefficient.

The case of a fluid surface is treated likewise. Equation (31) becomes

\frac{dy_{L}}{dz} = A_{L,∞}y_{L} + O (\frac{1}{n + 1})  (126)

with

A_{L,∞} = ( \begin{matrix}0 & 4πGρ & 0 & 0 \\ \frac{1}{4πGρ} & 0 & 0 & 0 \\ 1 & 0 & −1 & 1 \\ 1 & −4πGρ & 0 & 1 \end{matrix} ) ,  (127)

where now we have defined

lim_{n→∞} \frac{ω^{2}}{n} = \frac{g(a)}{a}ξ̃ .  (128)

The matrix A_{L,∞} has eigenvalues α̃_{1} = α̃_{2} = 1, α̃_{3} = α̃_{4} = −1 and corresponding eigenvectors

c̃_{1} = ( \begin{matrix}0 \\ 0 \\ 1 \\ 2 \end{matrix} ) , c̃_{2} = ( 1 ; \frac{1}{4πGρ} ; 1 ; 1 ) , c̃_{3} = ( \begin{matrix}0 \\ 0 \\ 1 \\ 0 \end{matrix} ) , c̃_{4} = ( 1 ; \frac{−1}{4πGρ} ; −1 ; −1 ) .  (129)

The general solution finite at the origin in this case is

y_{L}(z) = ∑_{i=1}^{2} C̃_{i} exp(α̃_{i}z) c̃_{i} + O (\frac{1}{n + 1})  (130)

and equation (0.99a) in conjunction with (128) yields the value ξ̃ = 1 for the asymptotic behaviour of the eigenfrequency. Comparison with (100) and (0.99a) shows that we must have C̃_{1} = \sqrt{a} C_{P}(n + 1/2 − 2πGρa/g), C̃_{2} = 4πGρa^{3/2}C_{P}/g for body and C̃_{1} = 4πGa^{3/2}τ_{nm}, C̃_{2} = −4πGa^{3/2}τ_{nm} for loading tides respectively and therefore yields the asymptotic values

h̃_{n} = O(1) ,  ℓ̃_{n} = O (\frac{1}{n}) ,  k̃_{n} = O (\frac{1}{n}) ,  (0.131a)

h̃′_{n} = −\frac{(2n + 1)g}{4πGρa} + O(1) , ℓ̃′_{n} = −\frac{g}{2πGρa} + O (\frac{1}{n}) , k̃′_{n} = −1 + O (\frac{1}{n}) .  (0.131b)

Bibliography

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