03_acoustic_and_elastic_waves.md

Acoustic wave equation

We derive the acoustic wave equation from the stress-strain relationship in tensorial form, we'll start by considering the fundamental equations of motion, the stress-strain relationship, and the constitutive relations for a linear elastic medium. Here's a step-by-step derivation:

1. Stress-Strain Relationship and Hooke's Law

For a linear elastic material, the stress tensor $\sigma_{ij}$ is related to the strain tensor $\epsilon_{ij}$ through Hooke's law:

$\sigma_{ij} = 𝐶_{ijkl} \ \epsilon_{kl}$

Where:

• $\sigma_{ij}$ is the stress tensor.
• $\epsilon_{ij}$ is the strain tensor.
• $𝐶_{ijkl}$ is the fourth-order stiffness tensor that relates stress and strain.

For an isotropic material, Hooke's law simplifies to:

$\sigma_{ij} = \lambda \delta_{ij} \ \epsilon_{kk} + 2\mu \ \epsilon_{ij}$

Where:

• $\lambda$ and $\mu$ are the Lamé constants.
• $\delta_{ij}$ is the Kronecker delta.
• $\epsilon_{kk}$ is the trace of the strain tensor (sum of the diagonal components), which is the volumetric strain.

2. Strain-Displacement Relationship

The strain tensor $\epsilon_{ij}$ is related to the displacement vector $u_{i}$ by the strain-displacement relationship:

$\epsilon_{ij} = \frac{1}{2}\bigg(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}\bigg)$

where,

• $u_{i}$ is the displacement component in the $i$-th direction.
• $x_{j}$ ​is the position coordinate in the $j$-th direction.

3. Equation of Motion

Newton's second law $(F=ma)$ for a continuous medium gives us the equation of motion, which relates the stress tensor to the acceleration of the displacement:

$\rho \frac{\partial^2 u_i}{\partial t^2} = \frac{\partial \sigma_{ij}}{\partial x_j}$

where,

• $\rho$ is the density of the medium.
• $\frac{\partial^2 u_i}{\partial t^2}$ is the acceleration in the i-th direction.

4. Substitute Stress-Strain Relationship into the Equation of Motion

Substituting the stress-strain relationship $\sigma_{ij} = 𝐶_{ijkl} \epsilon_{kl}$ into the equation of motion gives:

$\rho \frac{\partial^2 u_i}{\partial t^2} = \frac{\partial}{\partial x_j} (C_{ijkl} \epsilon_{kl})$

5. Substitute Strain-Displacement Relationship

Substituting the strain-displacement relationship $\epsilon_{ij} = \frac{1}{2}\bigg(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}\bigg)$ into the equation:

$\rho \frac{\partial^2 u_i}{\partial t^2} = \frac{\partial}{\partial x_j} \bigg(C_{ijkl} \quad \frac{1}{2}\bigg(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}\bigg)\bigg)$

6. Simplify the Equation

Assuming isotropic materials where the stiffness tensor simplifies (Lamé's constants $\lambda$ and $\mu$), we get:

$\sigma_{ij} = \lambda \delta_{ij} \sum_{k} \frac{\partial u_k}{\partial x_k} + 2 \mu \epsilon_{ij}$

Substituting this into the equation of motion:

$\rho \frac{\partial^2 u_i}{\partial t^2} = \frac{\partial}{\partial x_j} \bigg(\lambda \delta_{ij} \sum_{k} \frac{\partial u_k}{\partial x_k} + 2 \mu \epsilon_{ij} \bigg)$

This simplifies to:

$\rho \frac{\partial^2 u_i}{\partial t^2} = (\lambda + 2\mu) \frac{\partial}{\partial x_i} \sum_{k} \frac{\partial u_k}{\partial x_k} + \mu \nabla^2 u_{i}$

7. Wave Equation for Displacement

If we take the divergence of the displacement field $\nabla \cdot u$ and substitute it into the above equation, the acoustic wave equation in an isotropic medium is:

$\rho \frac{\partial^2 u_i}{\partial t^2} = (\lambda + 2\mu) \nabla (\nabla \cdot \textbf{u}) + \mu \nabla^2 \textbf{u}$

In the special case where the medium is homogeneous and isotropic (same properties in all directions), and assuming small perturbations, this reduces to:

$\boxed{\nabla^2 u - \frac{1}{c^2} \frac{\partial^2 u}{\partial t^2} = 0}$

where $c$ is the speed of sound in the medium, given by $c = \sqrt{\frac{(\lambda + 2\mu)}{\rho}}$

This is the acoustic wave equation for the displacement field $u$ in a homogeneous, isotropic, and linear elastic medium.

Elastic wave equation

To derive the elastic wave equation for an elastic medium, we'll follow a similar process to the acoustic wave equation derivation but consider the vector nature of displacements in an elastic solid. The elastic wave equation describes how mechanical waves propagate through a solid medium and accounts for both longitudinal (compressional) and shear (transverse) waves.

Steps 1 to 3 are the same, also we follow the same procedure. Let's start from step no. 4

4. Substitute Stress-Strain Relationship into the Equation of Motion

Substituting the stress-strain relationship $\sigma_{ij} = \lambda \delta_{ij} \ \epsilon_{kk} + 2\mu \ \epsilon_{ij}$ into the equation of motion gives:

$\rho \frac{\partial^2 u_i}{\partial t^2} = \frac{\partial}{\partial x_j} (\lambda \delta_{ij} \ \epsilon_{kk} + 2\mu \ \epsilon_{ij})$

5. Substitute Strain-Displacement Relationship

Substituting the strain-displacement relationship $\epsilon_{ij} = \frac{1}{2}\bigg(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}\bigg)$ into the equation:

$\rho \frac{\partial^2 u_i}{\partial t^2} = \frac{\partial}{\partial x_j} \bigg(\lambda \delta_{ij}\frac{\partial u_k}{\partial x_j} + \mu \quad \bigg(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}\bigg)\bigg)$

Expanding this equation:

$\rho \frac{\partial^2 u_i}{\partial t^2} = \lambda \frac{\partial}{\partial x_i} \bigg(\frac{\partial u_k}{\partial x_k} \bigg) + \mu \frac{\partial^2 u_i}{\partial x^2_j} + \mu \frac{\partial^2 u_{j}}{\partial x_{i} \partial x_{j}} $

6. Simplify the Equation

Rearranging and simplifying, the equation becomes:

$\rho \frac{\partial^2 u_i}{\partial t^2} = (\lambda + \mu) \frac{\partial^2 u_{k}}{\partial x_{i} \partial x_{k}} + \mu \nabla^2 u_i $

where, $\nabla^2 u_i$ is the Laplacian of the displacement field $u_i$, given by $\nabla^2 u_i = \frac{\partial^2 u_i}{\partail x_j \partial x_j}$

7. Vector Form of the Elastic Wave Equation

The above equation can be written more compactly in vector form ($u$ is a vector in bold):

$\rho \frac{\partial^2 u}{\partial t^2} = (\lambda + \mu) \nabla(\nabla \cdot u) + \mu \nabla^2 u$

This is the elastic wave equation in an isotropic medium.

8. Interpreting the Elastic Wave Equation

• The term $(\lambda + \mu) \nabla(\nabla \cdot u)$ corresponds to the propagation of compressional (longitudinal) waves, also known as P-waves.
• The term $\mu \nabla^2 u$ corresponds to the propagation of shear (transverse) waves, also known as S-waves.

The elastic wave equation describes how these two types of waves propagate through a solid medium. The P-wave speed $(c_p)$ and S-wave speed $(c_s)$ are given by:

$c_p = \sqrt{\frac{\lambda + 2\mu}{\rho}}$ and $c_s = \sqrt{\frac{\mu}{\rho}}$

These wave speeds determine how fast the respective wave types travel through the medium.

Hence, the Elastic Wave Equation in Time Domain in vector form for a linear, isotropic, and homogeneous medium is:

$\rho \frac{\partial^2 u}{\partial t^2} = (\lambda + \mu) \nabla(\nabla \cdot u) + \mu \nabla^2 u$

where,

• $u(x,t)$ is the displacement vector field.
• $\rho$ is the density of the medium.
• $\lambda$ and $\mu$ are the Lamé constants.
• $t$ is time.
• $\nabla$ is the gradient operator.
• $\nabla^2$ is the Laplacian operator.

Frequency Domain of elastic wave equation

To convert the elastic wave equation from the time domain to the frequency domain, we need to perform a Fourier transform on the equation. The Fourier transform allows us to analyze the wave equation in terms of frequency components, which can be particularly useful for solving problems where time-harmonic solutions are desired.

1. Time Domain Elastic Wave Equation

$\rho \frac{\partial^2 u}{\partial t^2} = (\lambda + \mu) \nabla(\nabla \cdot u) + \mu \nabla^2 u$

2. Fourier Transform of the Displacement Field

To convert the equation to the frequency domain, we apply the Fourier transform to the displacement vector $u(x,t)$

$u(x,t) = \int_{- \infty}^{\infty} U(X,\omega) e^{-i\omega t} d\omega$

Where,

• $U(X,\omega)$ is the Fourier transform of the displacement vector, depending on spatial coordinates $x$ and angular frequency $\omega$.
• $i$ is the imaginary unit.
• $\omega$ is the angular frequency.

The inverse Fourier transform is:

$U(X,\omega) = \frac{1}{2\pi} \int_{- \infty}^{\infty} u(x,t)e^{i\omega t} dt$

3. Applying Fourier Transform to the Elastic Wave Equation

We apply the Fourier transform to each term in the elastic wave equation.

a. Time Derivative Term

The second time derivative of the displacement field transforms as:

$\frac{\partial^2 u}{\partial t^2} \rightarraow - \omega^2 U(X,\omega)$

Thus, the left-hand side of the wave equation becomes:

$\rho \frac{\partial^2 u}{\partial t^2} \rightarraow -\rho \omega^2 U(X,\omega)$

b. Gradient and Divergence Terms

The spatial derivatives do not change when we perform the Fourier transform with respect to time, so:

$\nabla(\nabla \cdot u) \rightarrow \nabla(\nabla\cdot U(X,\omega))$

$\nabla^2 u \rightarrow \nabla^2 U(X,\omega)$

4. Elastic Wave Equation in Frequency Domain

Substituting the Fourier-transformed components into the original elastic wave equation, we obtain:

$-\rho \omega^2 U(X,\omega)$

5. Simplifying the Equation

We can further simplify the frequency-domain elastic wave equation by considering the vector wave components:

• The term
• The term