Volcano Seismic Signals, Source Quantification of

Article Outline

Glossary

Definition of the Subject

Introduction

Phenomenological Representation of Seismic Sources

Waveform Inversion

Spectral Analysis

Fluid-Solid Interactions

Future Directions

Acknowledgment

Appendix A: Green's Functions

Appendix B: Moment Tensor for a Spherical Source

Appendix C: Moment Tensor for a Cylindrical Source

Bibliography

Appendices

Appendix A: 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

$$ \rho\frac{\partial^2 G_{11}}{\partial t^2} = (\lambda + 2\mu)\frac{\partial^2 G_{11}}{\partial x_1^2} + (\lambda + \mu)\frac{\partial^2 G_{21}}{\partial x_2 \partial x_1}\\ + \mu\frac{\partial^2 G_{11}}{\partial x_2^2} + \delta(\boldsymbol{x}-\boldsymbol{\eta})\delta(t-\tau)\:, $$

(A1)

$$ \rho\frac{\partial^2 G_{21}}{\partial t^2} = (\lambda + 2\mu)\frac{\partial^2 G_{21}}{\partial x_2^2} + (\lambda + \mu)\frac{\partial^2 G_{11}}{\partial x_2 \partial x_1} + \mu\frac{\partial^2 G_{21}}{\partial x_1^2}\:, $$

(A2)

$$ \rho\frac{\partial^2 G_{12}}{\partial t^2} = (\lambda + 2\mu)\frac{\partial^2 G_{12}}{\partial x_1^2} + (\lambda + \mu)\frac{\partial^2 G_{22}}{\partial x_2 \partial x_1} + \mu\frac{\partial^2 G_{12}}{\partial x_2^2}\:, $$

(A3)

$$ \rho\frac{\partial^2 G_{22}}{\partial t^2} = (\lambda + 2\mu)\frac{\partial^2 G_{22}}{\partial x_2^2} + (\lambda + \mu)\frac{\partial^2 G_{12}}{\partial x_2 \partial x_1}\\ + \mu\frac{\partial^2 G_{22}}{\partial x_1^2} + \delta(\boldsymbol{x}-\boldsymbol{\eta})\delta(t-\tau)\:. $$

(A4)

Note that ${(G_{11},G_{21})}$ represents the x ₁ and x ₂ components of the wavefield at ${(\boldsymbol{x},t)}$, which is excited by the impulse in the x ₁ direction applied at ${\boldsymbol{x}=\boldsymbol{\eta}}$ and ${t=\tau}$. Similarly, ${(G_{12},G_{22})}$ represents the x ₁ and x ₂ components of the wavefield excited by the impulse in the x ₂ direction at ${\boldsymbol{x}=\boldsymbol{\eta}}$ and at ${t=\tau}$. 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 ${G_{ij}(\boldsymbol{x},t;\boldsymbol{\eta},\tau)}$. We obtain ${G_{ij}(\boldsymbol{x},t;\boldsymbol{\eta},\tau) = G_{ij}(\boldsymbol{x},t-\tau;\boldsymbol{\eta},0)}$, since Green's functions are independent of the time of origin.

The displacement ${u_i(\boldsymbol{x},t)}$ satisfies Eq. (3), which is explicitly written as

$$ \rho\frac{\partial^2 u_1}{\partial t^2} = (\lambda+2\mu)\frac{\partial^2 u_1}{\partial x_1^2} + (\lambda +\mu)\frac{\partial^2 u_2}{\partial x_1 \partial x_2}\\ + \mu\frac{\partial^2 u_1}{\partial x_2^2} + f_1^S(\boldsymbol{x},t)\:, $$

(A5)

$$ \rho\frac{\partial^2 u_2}{\partial t^2} = (\lambda+2\mu)\frac{\partial^2 u_2}{\partial x_2^2} + (\lambda +\mu)\frac{\partial^2 u_1}{\partial x_1 \partial x_2}\\ + \mu\frac{\partial^2 u_2}{\partial x_1^2} + f_2^S(\boldsymbol{x},t)\:. $$

(A6)

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

$$ u_1(\boldsymbol{x},t) = \int_{-\infty}^{\infty} \iint_{-\infty}^{\infty} \big[f_1^S(\boldsymbol{\eta},\tau) G_{11}(\boldsymbol{x},t-\tau;\boldsymbol{\eta},0)\\ + f_2^S(\boldsymbol{\eta},\tau) G_{12}(\boldsymbol{x},t-\tau;\boldsymbol{\eta},0)\big] \mskip2mu\mathrm{d}\eta_1 \mskip2mu\mathrm{d}\eta_2 \mskip2mu\mathrm{d}\tau\:, $$

(A7)

$$ u_2(\boldsymbol{x},t) = \int_{-\infty}^{\infty} \iint_{-\infty}^{\infty} \big[f_1^S(\boldsymbol{\eta},\tau) G_{21}(\boldsymbol{x},t-\tau;\boldsymbol{\eta},0)\\ + f_2^S(\boldsymbol{\eta},\tau) G_{22}(\boldsymbol{x},t-\tau;\boldsymbol{\eta},0)\big] \mskip2mu\mathrm{d}\eta_1 \mskip2mu\mathrm{d}\eta_2 \mskip2mu\mathrm{d}\tau\:. $$

(A8)

We can verify that the displacement given by Eqs. (A7) and (A8) satisfies Eqs. (A5) and (A6) in the following way. We can rewrite Eqs. (A5) and (A6) as

$$ f_1^S(\boldsymbol{x},t) = \rho\frac{\partial^2 u_1}{\partial t^2} - (\lambda+2\mu)\frac{\partial^2 u_1}{\partial x_1^2}\\ - (\lambda +\mu)\frac{\partial^2 u_2}{\partial x_1 \partial x_2} - \mu\frac{\partial^2 u_1}{\partial x_2^2}\:, $$

(A9)

$$ f_2^S(\boldsymbol{x},t) = \rho\frac{\partial^2 u_2}{\partial t^2} - (\lambda+2\mu)\frac{\partial^2 u_2}{\partial x_2^2}\\ - (\lambda +\mu)\frac{\partial^2 u_1}{\partial x_1 \partial x_2} - \mu\frac{\partial^2 u_2}{\partial x_1^2}\:. $$

(A10)

We denote the right-hand side of Eq. (A9) as L ₁, which is derived from Eqs. (A7) and (A8) as

$$ \begin{aligned} L_1 &= \int_{-\infty}^{\infty} \iint_{-\infty}^{\infty}\bigg[\rho\left(f_1^S\frac{\partial^2 G_{11}}{\partial t^2} + f_2^S\frac{\partial^2 G_{12}}{\partial t^2}\right)\\ &\quad-(\lambda + 2\mu)\left(f_1^S\frac{\partial^2 G_{11}}{\partial x_1^2} + f_2^S\frac{\partial^2 G_{12}}{\partial x_1^2}\right)\\ &\quad-(\lambda + \mu)\left(f_1^S\frac{\partial^2 G_{21}}{\partial x_1 \partial x_2} + f_2^S\frac{\partial^2 G_{22}}{\partial x_1 \partial x_2}\right)\\ &\quad-\mu\left(f_1^S\frac{\partial^2 G_{11}}{\partial x_2^2} + f_2^S\frac{\partial^2 G_{12}}{\partial x_2^2}\right)\bigg]\mskip2mu\mathrm{d}\eta_1 \mskip2mu\mathrm{d}\eta_2 \mskip2mu\mathrm{d}\tau\:. \end{aligned}$$

(A11)

This equation can be modified as follows:

$$ \begin{aligned} L_1 &= \int_{-\infty}^{\infty} \iint_{-\infty}^{\infty}\bigg[f_1^S\bigg\{\rho\frac{\partial^2 G_{11}}{\partial t^2} -(\lambda + 2\mu)\frac{\partial^2 G_{11}}{\partial x_1^2}\\ &\quad-(\lambda + \mu)\frac{\partial^2 G_{21}}{\partial x_1 \partial x_2}-\mu\frac{\partial^2 G_{11}}{\partial x_2^2}\bigg\}\\ &\quad + f_2^S\bigg\{\rho\frac{\partial^2 G_{12}}{\partial t^2} -(\lambda + 2\mu)\frac{\partial^2 G_{12}}{\partial x_1^2}\\ &\quad-(\lambda + \mu)\frac{\partial^2 G_{22}}{\partial x_1 \partial x_2}-\mu\frac{\partial^2 G_{12}}{\partial x_2^2}\bigg\}\bigg]\mskip2mu\mathrm{d}\eta_1 \mskip2mu\mathrm{d}\eta_2 \mskip2mu\mathrm{d}\tau\:. \end{aligned}$$

(A12)

From Eq. (A3), we find that the second integral of Eq. (A12) is zero. Therefore, Eq. (A12) from Eq. (A1) becomes

$$ \begin{aligned} L_1 &= \int_{-\infty}^{\infty} \iint_{-\infty}^{\infty}f_1^S(\boldsymbol{\eta},\tau)\delta(\boldsymbol{x}-\boldsymbol{\eta})\delta(t-\tau) \mskip2mu\mathrm{d}\eta_1 \mskip2mu\mathrm{d}\eta_2 \mskip2mu\mathrm{d}\tau\\ &= f_1^S(\boldsymbol{x},t)\:. \end{aligned}$$

(A13)

Similarly, we can show that the right-hand side of Eq. (A10) is equivalent to ${f_2^S(\boldsymbol{x},t)}$, and thus the displacement in the forms of Eqs. (A7) and (A8) satisfies the equation of motion (A5) and (A6).

Appendix B: Moment Tensor for a Spherical Source

Let us consider the moment density tensors for three tensile cracks in the planes ${\xi_1=0, \xi_2=0}$, and ${\xi_3=0}$. If we sum and average these three tensors and assume that ${[u_1]=[u_2]=[u_3]=D_s}$, we obtain the following moment density tensor:

$$ \boldsymbol{m}=D_s\left(\begin{array}{ccc} \lambda + 2\mu/3 & 0 & 0 \\ 0 & \lambda + 2\mu/3 & 0 \\ 0& 0 & \lambda + 2\mu/3 \end{array} \right)\:. $$

(B1)

This represents the moment density tensor for the isotropic expansion of a cubic element. Following Müller [89], we consider a spherical crack surface of radius R where a constant radial expansion ${D_s = (d_s + \Delta_s)}$ occurs (Fig. A1). Here, the inner wall of the crack moves inward by ${d_s}$ and the outer wall moves outward by ${\Delta_s}$. The moment tensor of the spherical expansion may be obtained by integration of the moment density tensor (B1) over the surface R:

Figure A1

A spherical crack surface of radius R where a constant radial expansion ${D_{\text{s}} = (d_{\text{s}} + \Delta_{\text{s}})}$ occurs. Here, the inner wall of the crack moves inward by d _s and the outer wall moves outward by ${\Delta_{\text{s}}}$ [89]

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

(B2)

where ${\Delta V}$ is the volume given as

$$ \Delta V = 4\pi R^2 D_s=4\pi R^2 (d_s + \Delta_s)\:. $$

(B3)

The volume ${4\pi R^2 d_s}$ is caused by the inward motion, which compresses the sphere. The volume $\Delta V_s = 4\pi R^2 \Delta_s$, on the other hand, is caused by the outward motion, which excites seismic waves in the region outside the sphere. ${\Delta V_s}$ 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., [121]

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

(B4)

and

$$ \sigma_{rr} = (\lambda+2\mu) \frac{\partial u}{\partial r} + \lambda\frac{2}{r}u\:, $$

(B5)

$$ \sigma_{\theta\theta} = \sigma_{\phi\phi} = \lambda \frac{\partial u}{\partial r} + (\lambda + \mu)\frac{2}{r}u\:, $$

(B6)

where u is the radial displacement and ${\sigma_{rr}}$, ${\sigma_{\theta\theta}}$, and ${\sigma_{\phi\phi}}$ are the stress components in the spherical coordinate. Equations (B4) and (B5) apply to both regions 1 and 2. Substituting Eq. (B5) into Eq. (B4) and setting ${\rho (\partial^2 u_r/\partial t^2)=0}$, we obtain the following static equilibrium equation:

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

(B7)

This equation has two solutions: ${u=ar}$ and ${u=b/r^2}$, where a and b are constants. The former is the interior solution for region 1 (${r \leq R}$), and the latter is the exterior solution for region 2 (${r \geq R}$). 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 ${r = R}$:

$$ u_e(R) - u_i(R) = \frac{b}{R^2} - aR = D_s\:, $$

(B8)

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

(B9)

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

$$ u_i(r) = - \frac{4\mu D_s}{3(\lambda+2\mu)}\frac{r}{R} \quad (r \leq R)\:, $$

(B10)

$$ u_e(r) = \frac{(\lambda+2\mu/3) D_s}{\lambda + 2\mu}\frac{R^2}{r^2} \quad (r \geq R)\:. $$

(B11)

Then, we obtain

$$ d_s = -u_i(R)=\frac{4\mu D_s}{3(\lambda+2\mu)}\:, $$

(B12)

$$ \Delta_s = u_e(R) = \frac{(\lambda+2\mu/3) D_s}{\lambda + 2\mu}\:. $$

(B13)

Equation (B3) can be modified as

$$ \Delta V = 4\pi R^2 \Delta_s (D_s/\Delta_s) = \frac{\lambda + 2\mu}{\lambda + 2\mu/3}\Delta V_s\:. $$

(B14)

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

$$ \boldsymbol{M} = (\lambda + 2\mu)\Delta V_s\left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right)\:. $$

(B15)

Appendix C: Moment Tensor for a Cylindrical Source

We consider the moment density tensors for two tensile cracks in the planes ${\xi_1=0}$ and ${\xi_2=0}$, where ξ₁ and ξ₂ are two horizontal axes. If we sum and average these two tensors and assume that ${[u_1]=[u_2]=D_c}$, we obtain

$$ \boldsymbol{m}=D_c\left(\begin{array}{ccc} \lambda + \mu & 0 & 0 \\ 0 & \lambda + \mu & 0 \\ 0& 0 & \lambda \end{array} \right)\:. $$

(C1)

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 ${D_c=(d_c+\Delta_c)}$ occurs (Fig. A2). The moment tensor for the radial expansion of the cylinder may be obtained by integration of the moment density tensor (C1) over the surface R:

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

(C2)

where ${\Delta V}$ is the volume given as

$$ \Delta V = 2\pi R L D_c=2\pi R L (d_c + \Delta_c)\:. $$

(C3)

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

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

(C4)

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

Figure A2

A vertical cylinder of length L and radius R. The cylinder surface can be regarded as a cylindrical crack, where the radial expansion ${D_c = (d_c + \Delta_c)}$ occurs. Here, the inner wall of the crack moves inward by ${d_c}$ and the outer wall moves outward by ${\Delta_c}$ [89]

$$ u_e(R) - u_i(R) = \frac{b}{R} - aR = D_c\:, $$

(C5)

$$ \sigma_{rr}^i (R) - \sigma_{rr}^e(R) = -\frac{2\mu}{R^2} b - 2(\lambda + \mu) a = 0\:, $$

(C6)

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

$$ d_c = -u_i(R)=\frac{\mu D_s}{(\lambda+2\mu)}\:, $$

(C7)

$$ \Delta_c = u_e(R) = \frac{(\lambda+\mu) D_c}{\lambda + 2\mu}\:, $$

(C8)

and

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

(C9)

The moment tensor for the vertical cylinder is therefore given as

$$ \boldsymbol{M} = \frac{\lambda + 2\mu}{\lambda+\mu}\Delta V_c\left(\begin{array}{ccc} \lambda + \mu & 0 & 0 \\ 0 & \lambda + \mu & 0 \\ 0 & 0 & \lambda \end{array} \right)\:. $$

(C10)

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 and (2) rotate the ξ₁ and ξ₂ axes around the ξ₃ axis through an angle ${-\phi}$, and (3) further rotate the ξ₁ and ξ₃ axes around the ξ₂ axis through an angle ${-\theta}$. The rotation matrix ${\boldsymbol{R}}$ is given as

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

(C11)

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

(C12)

Using the matrix ${\boldsymbol{R}}$, we obtain the moment tensor for a cylinder with the axis orientation angles ${(\phi,\theta)}$ as

$$ \boldsymbol{M}^\prime = \boldsymbol{R}^{\text{T}} \boldsymbol{MR}\:, $$

(C13)

which leads to Eq. (34).