Source Quantification of Volcanic-Seismic Signals

Appendices

Green’s Functions

Here, I briefly explain the relationship between the displacement and Green’s functions. To simplify the explanation, I use two-dimensional equations in an infinite medium. The extension of the equations into three-dimension is straightforward. Green’s functions defined by Eq. (5) are explicitly written as

$$ {\displaystyle \begin{array}{l}\rho \frac{\partial^2{G}_{11}}{\partial {t}^2}=\left(\lambda +2\mu \right)\frac{\partial^2{G}_{11}}{\partial {x}_1^2}+\left(\lambda +\mu \right)\frac{\partial^2{G}_{21}}{\partial {x}_2\partial {x}_1}\\ {}+\mu \frac{\partial^2{G}_{11}}{\partial {x}_2^2}+\delta \left(\mathbf{x}-\boldsymbol{\eta} \right)\delta \left(t-\tau \right),\end{array}} $$

(225)

$$ \rho \frac{\partial^2{G}_{21}}{\partial {t}^2}=\left(\lambda +2\mu \right)\frac{\partial^2{G}_{21}}{\partial {x}_2^2}+\left(\lambda +\mu \right)\frac{\partial^2{G}_{11}}{\partial {x}_2\partial {x}_1}+\mu \frac{\partial^2{G}_{21}}{\partial {x}_1^2}, $$

(226)

$$ \rho \frac{\partial^2{G}_{12}}{\partial {t}^2}=\left(\lambda +2\mu \right)\frac{\partial^2{G}_{12}}{\partial {x}_1^2}+\left(\lambda +\mu \right)\frac{\partial^2{G}_{22}}{\partial {x}_2\partial {x}_1}+\mu \frac{\partial^2{G}_{12}}{\partial {x}_2^2}, $$

(227)

$$ {\displaystyle \begin{array}{l}\rho \frac{\partial^2{G}_{22}}{\partial {t}^2}=\left(\lambda +2\mu \right)\frac{\partial^2{G}_{22}}{\partial {x}_2^2}+\left(\lambda +\mu \right)\frac{\partial^2{G}_{12}}{\partial {x}_2\partial {x}_1}\\ {}+\mu \frac{\partial^2{G}_{22}}{\partial {x}_1^2}+\delta \left(\mathbf{x}-\boldsymbol{\eta} \right)\delta \left(t-\tau \right)\end{array}} $$

(228)

Note that (G₁₁, G₂₁) represents the x₁ and x₂ components of the wavefield at (x, t), which is excited by the impulse in the x₁ direction applied at x = η and t = τ. Similarly, (G₁₂, G₂₂)represents the x₁ and x₂ components of the wavefield excited by the impulse in the x₂ direction at x = η and at t = τ. Therefore, i and j in G_ij represent the component and direction of the impulse, respectively. To specify the relationship between the receiver and source, we use the notation G_ij(x, t; η, τ). We obtain G_ij(x, t; η, τ) = G_ij(x, t − τ; η, 0), since Green’s functions are independent of the time of origin.

The displacement u_i(x, t) satisfies Eq. (3), which is explicitly written as

$$ {\displaystyle \begin{array}{l}\rho \frac{\partial^2{u}_1}{\partial {t}^2}=\left(\lambda +2\mu \right)\frac{\partial^2{u}_1}{\partial {x}_1^2}+\left(\lambda +\mu \right)\frac{\partial^2{u}_2}{\partial {x}_1\partial {x}_2}\\ {}+\mu \frac{\partial^2{u}_1}{\partial {x}_2^2}+{f}_1^S\left(\mathbf{x},t\right),\end{array}} $$

(229)

$$ {\displaystyle \begin{array}{l}\rho \frac{\partial^2{u}_2}{\partial {t}^2}=\left(\lambda +2\mu \right)\frac{\partial^2{u}_2}{\partial {x}_2^2}+\left(\lambda +\mu \right)\frac{\partial^2{u}_1}{\partial {x}_1\partial {x}_2}\\ {}+\mu \frac{\partial^2{u}_2}{\partial {x}_1^2}+{f}_2^S\left(\mathbf{x},t\right)\end{array}} $$

(230)

Equation (6) indicates that u_i is described by the following relationships:

$$ {\displaystyle \begin{array}{l}{u}_1\left(\mathbf{x},t\right)={\int}_{-\infty}^{\infty }{\iint}_{-\infty}^{\infty}\Big[{f}_1^S\left(\boldsymbol{\eta}, \tau \right){G}_{11}\left(\mathbf{x},t-\tau; \boldsymbol{\eta}, 0\right)\\ {}+{f}_2^S\left(\boldsymbol{\eta}, \tau \right){G}_{12}\left(\mathbf{x},t-\tau; \boldsymbol{\eta}, 0\right)\Big]\;\mathrm{d}{\boldsymbol{\eta}}_1\ \mathrm{d}{\boldsymbol{\eta}}_2\mathrm{d}\tau, \end{array}} $$

(231)

$$ {u}_2\left(\mathbf{x},t\right)={\int}_{-\infty}^{\infty }{\iint}_{-\infty}^{\infty}\left[{f}_1^S\left(\boldsymbol{\eta}, \tau \right){G}_{21}\left(\mathbf{x},t-\tau; \boldsymbol{\eta}, 0\right)+{f}_2^S\left(\boldsymbol{\eta}, \tau \right){G}_{22}\left(\mathbf{x},t-\tau; \boldsymbol{\eta}, 0\right)\right]\mathrm{d}{\boldsymbol{\eta}}_1\mathrm{d}{\boldsymbol{\eta}}_2\mathrm{d}\tau . $$

(232)

We can verify that the displacement given by Eqs. (231) and (232) satisfies Eqs. (229) and (230) in the following way. We can rewrite Eqs. (229) and (230) as

$$ {\displaystyle \begin{array}{l}{f}_1^S\left(\mathbf{x},t\right)=\rho \frac{\partial^2{u}_1}{\partial {t}^2}-\left(\lambda +2\mu \right)\frac{\partial^2{u}_1}{\partial {x}_1^2}\\ {}-\left(\lambda +\mu \right)\frac{\partial^2{u}_2}{\partial {x}_1\partial {x}_2}-\mu \frac{\partial^2{u}_1}{\partial {x}_2^2},\end{array}} $$

(233)

We denote the right-hand side of Eq. (233) as L₁, which is derived from Eqs. (231) and (232) as

$$ {\displaystyle \begin{array}{l}{L}_1={\int}_{-\infty}^{\infty }{\iint}_{-\infty}^{\infty}\Big[\rho \left({f}_1^S\frac{\partial^2{G}_{11}}{\partial {t}^2}+{f}_2^S\frac{\partial^2{G}_{12}}{\partial {t}^2}\right)-\left(\lambda +2\mu \right)\left({f}_1^S\frac{\partial^2{G}_{11}}{\partial {x}_1^2}+{f}_2^S\frac{\partial^2{G}_{12}}{\partial {x}_1^2}\right)\\ {}-\left(\lambda +\mu \right)\left({f}_1^S\frac{\partial^2{G}_{21}}{\partial {x}_1\partial {x}_2}+{f}_2^S\frac{\partial^2{G}_{22}}{\partial {x}_1\partial {x}_2}\right)-\mu \left({f}_1^S\frac{\partial^2{G}_{11}}{\partial {x}_2^2}+{f}_2^S\frac{\partial^2{G}_{12}}{\partial {x}_2^2}\right)\Big]\mathrm{d}{\eta}_1\;\mathrm{d}{\eta}_2\mathrm{d}\tau .\end{array}} $$

(234)

This equation can be modified as follows:

$$ {\displaystyle \begin{array}{l}{L}_1={\int}_{-\infty}^{\infty }{\iint}_{-\infty}^{\infty}\Big[{f}_1^S\left\{\rho \frac{\partial^2{G}_{11}}{\partial {t}^2}-\left(\lambda +2\mu \right)\frac{\partial^2{G}_{11}}{\partial {x}_1^2}-\left(\lambda +\mu \right)\frac{\partial^2{G}_{21}}{\partial {x}_1\partial {x}_2}-\mu \frac{\partial^2{G}_{11}}{\partial {x}_2^2}\right\}\, \\ {}+{f}_2^S\left\{\rho \frac{\partial^2{G}_{12}}{\partial {t}^2}-\left(\lambda +2\mu \right)\frac{\partial^2{G}_{12}}{\partial {x}_1^2}-\left(\lambda +\mu \right)\frac{\partial^2{G}_{22}}{\partial {x}_1\partial {x}_2}-\mu \frac{\partial^2{G}_{12}}{\partial {x}_2^2}\right\}\Big]\mathrm{d}{\eta}_1\mathrm{d}{\eta}_2\;\mathrm{d}\tau .\end{array}} $$

(235)

From Eq. (227), we find that the second integral of Eq. (235) is zero. Therefore, Eq. (235) from Eq. (225) becomes

$$ {\displaystyle \begin{array}{ll}{L}_1& ={\int}_{-\infty}^{\infty }{\iint}_{-\infty}^{\infty }{f}_1^S\left(\boldsymbol{\eta}, \tau \right)\delta \left(\mathbf{x}-\boldsymbol{\eta} \right)\delta \left(t-\tau \right)\ \mathrm{d}{\boldsymbol{\eta}}_1\ \mathrm{d}{\boldsymbol{\eta}}_2\ \mathrm{d}\tau \\ {}={f}_1^S\left(\mathbf{x},t\right).\end{array}} $$

(236)

Similarly, we can show that the right-hand side of Eq. (233) is equivalent to ${f}_2^S\left(\mathbf{x},t\right)$, and thus the displacement in the forms of Eqs. (231) and (232) satisfies the equation of motion (229) and (230).

Moment Tensor for a Spherical Source

Let us consider the moment density tensors for three tensile cracks in the planes ξ₁ = 0, ξ₂ = 0, and ξ₃ = 0. If we sum and average these three tensors and assume that [u₁] = [u₂] = [u₃] = D_s, we obtain the following moment density tensor:

$$ \mathbf{m}={D}_s\left(\begin{array}{ccc}\lambda +2\mu /3& 0& 0\\ {}0& \lambda +2\mu /3& 0\\ {}0& 0& \lambda +2\mu /3\end{array}\right). $$

(237)

This represents the moment density tensor for the isotropic expansion of a cubic element. Following Müller (2001), we consider a spherical crack surface of radius R where a constant radial expansion D_s = (d_s + Δ_s) occurs (Fig. 20). Here, the inner wall of the crack moves inward by d_s and the outer wall moves outward by Δ_s. The moment tensor of the spherical expansion may be obtained by integration of the moment density tensor (237) over the surface R:

Fig. 20

A spherical crack surface of radius R where a constant radial expansion D_s = (d_s + Δ_s) occurs. Here, the inner wall of the crack moves inward by d_s and the outer wall moves outward by Δ_s (Müller 2001)

$$ \mathbf{M}=\Delta V\left(\begin{array}{ccc}\lambda +2\mu /3& 0& 0\\ {}0& \lambda +2\mu /3& 0\\ {}0& 0& \lambda +2\mu /3\end{array}\right), $$

(238)

where ΔV is the volume given as

$$ \Delta V=4\pi {R}^2{D}_s=4\pi {R}^2\left({d}_s+{\Delta}_s\right) $$

(239)

The volume 4πR²d_s is caused by the inward motion, which compresses the sphere. The volume ΔV_s = 4πR²Δ_s, on the other hand, is caused by the outward motion, which excites seismic waves in the region outside the sphere. ΔV_s can be determined by solving an elastostatic boundary-value problem in the following way. We assume an isotropic medium and denote the regions inside and outside the sphere as regions 1 and 2, respectively. Since the motion is radial only, the equation of motion is given as (e.g., Takeuchi and Saito 1972)

$$ \rho \frac{\partial^2u}{\partial {t}^2}=\frac{1}{r^2}\frac{\partial }{\partial r}\left({r}^2{\sigma}_{rr}\right)-\frac{1}{r}\left({\sigma}_{\theta \theta}+{\sigma}_{\phi \phi}\right) $$

(240)

and

$$ {\sigma}_{rr}=\left(\lambda +2\mu \right)\frac{\partial u}{\partial r}+\lambda \frac{2}{r}u, $$

(241)

$$ {\sigma}_{\theta \theta}={\sigma}_{\phi \phi}=\lambda \frac{\partial u}{\partial r}+\left(\lambda +\mu \right)\frac{2}{r}u, $$

(242)

where u is the radial displacement and σ_rr, σ_θθ, and σ_ϕϕ are the stress components in the spherical coordinate. Equations (240) and (241) apply to both regions 1 and 2. Substituting Eq. (241) into Eq. (240) and setting ρ(∂²u_r/∂t²) = 0, we obtain the following static equilibrium equation:

$$ \frac{\partial^2u}{\partial {r}^2}+\frac{2}{r}\frac{\partial u}{\partial r}-\frac{2}{r^2}u=0. $$

(243)

This equation has two solutions: u = ar and u = b/r², where a and b are constants. The former is the interior solution for region 1 (r ≤ R), and the latter is the exterior solution for region 2 (r ≥ R). I denote the interior and exterior solutions as u_i and u_e, respectively. The constants a and b are determined by the boundary conditions for the radial displacement and the continuity of the radial stress at r = R:

$$ {u}_e(R)-{u}_i(R)=\frac{b}{R^2}- aR={D}_s, $$

(244)

$$ {\sigma}_{rr}^e(R)-{\sigma}_{rr}^i(R)=-\frac{4\mu }{R^3}b-\left(3\lambda +2\mu \right)a=0, $$

(245)

where ${\sigma}_{rr}^i$ and ${\sigma}_{rr}^e$ are the radial stresses in regions 1 and 2, respectively. Accordingly, we obtain

$$ {u}_i(r)=-\frac{4\mu {D}_s}{3\left(\lambda +2\mu \right)}\frac{r}{R}\quad \left(r\le R\right), $$

(246)

$$ {u}_e(r)=\frac{\left(\lambda +2\mu /3\right){D}_s}{\lambda +2\mu}\frac{R^2}{r^2}\quad \left(r\ge R\right) $$

(247)

Then, we obtain

$$ {d}_s=-{u}_i(R)=\frac{4\mu {D}_s}{3\left(\lambda +2\mu \right)}, $$

(248)

$$ {\Delta}_s={u}_e(R)=\frac{\left(\lambda +2\mu /3\right){D}_s}{\lambda +2\mu }. $$

(249)

Equation (239) can be modified as

$$ \Delta V=4\pi {R}^2{\Delta}_s\left({D}_s/{\Delta}_s\right)=\frac{\lambda +2\mu }{\lambda +2\mu /3}\Delta {V}_s. $$

(250)

Finally, we obtain the moment tensor for the spherical expansion as

$$ \mathbf{M}=\left(\lambda +2\mu \right)\Delta {V}_s\left(\begin{array}{ccc}1& 0& 0\\ {}0& 1& 0\\ {}0& 0& 1\end{array}\right). $$

(251)

Moment Tensor for a Cylindrical Source

We consider the moment density tensors for two tensile cracks in the planes ξ₁ = 0 and ξ₂ = 0, where ξ₁ and ξ₂ are two horizontal axes. If we sum and average these two tensors and assume that [u₁] = [u₂] = D_c, we obtain

$$ \mathbf{m}={D}_c\left(\begin{array}{ccc}\lambda +\mu & 0& 0\\ {}0& \lambda +\mu & 0\\ {}0& 0& \lambda \end{array}\right). $$

(252)

This represents the moment density tensor for the expansion of a cubic element in the two horizontal directions. Let us consider a vertical cylinder of length L and radius R. The cylinder surface at radius R can be regarded as a cylindrical crack, where the radial expansion D_c = (d_c + Δ_c) occurs (Fig. 21). The moment tensor for the radial expansion of the cylinder may be obtained by integration of the moment density tensor (252) over the surface R:

$$ \mathbf{M}=\Delta V\left(\begin{array}{ccc}\lambda +\mu & 0& 0\\ {}0& \lambda +\mu & 0\\ {}0& 0& \lambda \end{array}\right), $$

(253)

where ΔV is the volume given as

$$ \Delta V=2\pi {RLD}_c=2\pi RL\left({d}_c+{\Delta}_c\right) $$

(254)

Fig. 21

A vertical cylinder of length L and radius R. The cylinder surface can be regarded as a cylindrical crack, where the radial expansion D_c = (d_c + Δ_c) occurs. Here, the inner wall of the crack moves inward by d_c and the outer wall moves outward by Δ_c (Müller 2001)

We obtain the static equilibrium equation for the radial expansion of the cylinder u as

$$ \frac{\partial^2u}{\partial {r}^2}+\frac{1}{r}\frac{\partial u}{\partial r}-\frac{1}{r^2}u=0. $$

(255)

This equation has internal and external solutions, which are given as u = ar and u = b/r, respectively. The constants a and b are determined by the boundary conditions

$$ {u}_e(R)-{u}_i(R)=\frac{b}{R}- aR={D}_c, $$

(256)

$$ {\sigma}_{rr}^e(R)-{\sigma}_{rr}^i(R)=-\frac{2\mu }{R^2}b-2\left(\lambda +\mu \right)a=0, $$

(257)

where superscripts i and e denote the interior and exterior solutions. We then obtain

$$ {d}_c=-{u}_i(R)=\frac{\mu {D}_s}{\left(\lambda +2\mu \right)}, $$

(258)

$$ {\Delta}_c={u}_e(R)=\frac{\left(\lambda +\mu \right){D}_c}{\lambda +2\mu }, $$

(259)

and

$$ \Delta V=\frac{\lambda +2\mu }{\lambda +\mu}\Delta {V}_c. $$

(260)

The moment tensor for the vertical cylinder is therefore given as

$$ \mathbf{M}=\frac{\lambda +2\mu }{\lambda +\mu}\Delta {V}_c\left(\begin{array}{ccc}\lambda +\mu & 0& 0\\ {}0& \lambda +\mu & 0\\ {}0& 0& \lambda \end{array}\right). $$

(261)

Let us rotate the vertical cylinder with axis orientation angles ϕ and θ (Fig. 8b). This can be done in the following steps: (1) fix the cylinder, (2) rotate the ξ₁ and ξ₂ axes around the ξ₃ axis through an angle −ϕ, and (3) further rotate the ξ₁ and ξ₃ axes around the ξ₂ axis through an angle −θ. The rotation matrix R is given as

$$ \mathbf{R}=\left(\begin{array}{ccc}\cos \theta & 0& -\sin \theta \\ {}0& 1& 0\\ {}\sin \theta & 0& \cos \theta \end{array}\right)\left(\begin{array}{ccc}\cos \phi & -\sin \phi & 0\\ {}\sin \phi & \cos \phi & 0\\ {}0& 0& 1\end{array}\right) $$

(262)

$$ \mathbf{R}=\left(\begin{array}{ccc}\cos \theta \cos \phi & -\cos \theta \sin \phi & -\sin \theta \\ {}\sin \phi & \cos \phi & 0\\ {}\sin \theta & -\sin \theta \sin \phi & \cos \theta \end{array}\right). $$

(263)

Using the matrix R, we obtain the moment tensor for a cylinder with the axis orientation angles (ϕ, θ) as

$$ {\mathbf{M}}^{\prime }={\mathbf{R}}^{\mathrm{T}}\mathbf{MR}, $$

(264)

which leads to Eq. (34).