本文抄选自知乎文章 地球物理笔记—地震学（上） - Eden的文章 - 知乎

弹性力学基础

应变与位移

取连续介质中相邻两点 $A$, $B$. 令之形变, 得各自位移 $\mathbf{U}(\mathbf{r}), \mathbf{U}(\mathbf{r} + \Delta\mathbf{r})$. 则有相对位移

$$ \Delta\mathbf{U} = \mathbf{U}(\mathbf{r} + \Delta\mathbf{r}) - \mathbf{U}(\mathbf{r}) = \frac{\partial\mathbf{U}}{\partial x}\Delta x + \frac{\partial\mathbf{U}}{\partial y}\Delta y + \frac{\partial\mathbf{U}}{\partial z}\Delta z + \mathcal{O}(\Delta \mathbf{r}^{2}) $$

取小形变近似有

$$ \begin{aligned} \begin{bmatrix} \Delta U_{x}\\ \Delta U_{y}\\ \Delta U_{z} \end{bmatrix} = \begin{bmatrix} \frac{\partial U_{x}}{\partial x} & \frac{\partial U_{x}}{\partial y} & \frac{\partial U_{x}}{\partial z}\\ \frac{\partial U_{y}}{\partial x} & \frac{\partial U_{y}}{\partial y} & \frac{\partial U_{y}}{\partial z}\\ \frac{\partial U_{z}}{\partial x} & \frac{\partial U_{z}}{\partial y} & \frac{\partial U_{z}}{\partial z} \end{bmatrix}\begin{bmatrix} \Delta x\\ \Delta y\\ \Delta z \end{bmatrix} \end{aligned} $$

通过对称阵 + 反对称阵的分解技巧, 得到

$$ \Delta\mathbf{U} = \mathbf{e}\Delta\mathbf{r} + \mathbf{\omega}\Delta\mathbf{r}\\ e_{ij} = e_{ji} = \frac{1}{2}\left(\frac{\partial U_{i}}{\partial x_{j}} + \frac{\partial U_{j}}{\partial x_{i}}\right)\\ \omega_{ij} = -\omega_{ji} = \frac{1}{2}\left(\frac{\partial U_{i}}{\partial x_{j}} - \frac{\partial U_{j}}{\partial x_{i}}\right) $$

使用 Einstein 求和约定, 可以写作 $\Delta U_{i} = e_{ij}\Delta x_{j} + \omega_{ij}\Delta x_{j}$. $\mathbf{e}$ 为应变张量, $\mathbf{\omega}$ 为旋转张量.

取小形变近似, 有 $\Delta \mathbf{U}\approx \Delta\mathbf{r}$, 即有 $\mathbf{e}\approx \mathbf{1}, \omega\approx \mathbf{0}$.

• 旋转张量 $\mathbf{\omega}$ 为反对称阵, 写作

$$ \mathbf{\omega} = \frac{1}{2}\begin{bmatrix} 0 & \frac{\partial U_{x}}{\partial y} - \frac{\partial U_{y}}{x} & \frac{\partial U_{x}}{\partial z} - \frac{\partial U_{z}}{\partial x}\\ \frac{\partial U_{y}}{\partial x} - \frac{\partial U_{x}}{\partial y} & 0 & \frac{\partial U_{y}}{\partial z} - \frac{\partial U_{z}}{\partial y}\\ \frac{\partial U_{z}}{\partial x} - \frac{\partial U_{x}}{\partial z} & \frac{\partial U_{z}}{\partial y} - \frac{\partial U_{y}}{\partial z} & 0 \end{bmatrix}= \frac{1}{2}\begin{bmatrix} 0 & -\omega_{3} & \omega_{2}\\ \omega_{3} & 0 & -\omega_{1}\\ -\omega_{2} & \omega_{1} & 0 \end{bmatrix} $$

所以 $\mathbf{I} + \mathbf{\omega}$ 即代表绕 $\mathbf{\omega} = (\omega_{1}, \omega_{2}, \omega_{3})^{T}$ 轴微小旋转.

• 对称阵 $\mathbf{e}$ 的对角元反映了三个方向上的体形变. 体应变:

$$ \theta = e_{xx} + e_{yy} + e_{zz} = \sum\frac{\partial U_{i}}{\partial x_{i}} = \nabla\cdot\mathbf{U} $$

应力和 Hooke

• 体力: 某点处单位体积受到的力;
• 面力: 某局域平面受到的力.

$\mathbf{T}(\hat{\mathbf{n}})$ 表示以 $\hat{\mathbf{n}}$ 为外法线方向的面上, 外对内 单位面积 的作用力. 若以该平面为四面体的斜面, 其余的三个面法向即为坐标平面的法向矢量 $\hat{x}, \hat{y}, \hat{z}$.

四面体受力平衡, 即有

$$ T_{x}(\hat{n}) = \sigma_{xx}\hat{n}_{x} + \sigma_{xy}\hat{n}_{y} + \sigma_{xz}\hat{n}_{z}\\ T_{y}(\hat{n}) = \sigma_{yx}\hat{n}_{x} + \sigma_{yy}\hat{n}_{y} + \sigma_{yz}\hat{n}_{z}\\ T_{z}(\hat{n}) = \sigma_{zx}\hat{n}_{x} + \sigma_{zy}\hat{n}_{y} + \sigma_{zz}\hat{n}_{z} $$

Einstein 求和约定下, 可以写作 $T_{i}(\hat{n}) = \sigma_{ij}\hat{n}_{j}$.

体元合力与合力矩为 0:

$$ \frac{\partial T_{i}(\hat{x}_{j})}{\partial x_{j}} = 0\\ T_{i}(\hat{x}_{j}) = T_{j}(\hat{x}_{i}) $$

代入 $T_{i}(\hat{n}) = \sigma_{ij}\hat{n}_{j}$ 即有

$$ \frac{\partial \sigma_{ij}}{\partial x_{i}} = 0\\ \sigma_{ij} = \sigma_{ji} $$

对于线性弹性体, 力变的本构关系由广义 Hooke 定律给出:

$$ \sigma_{ij} = c_{ijkl}e_{kl} $$

若介质各向同性, 可进一步精简使用 Lamé 常数 $\lambda, \mu$ 表示:

$$ \sigma_{ij} = \lambda\theta\delta_{ij} + 2\mu e_{ij} $$

波动方程

设体元的体力为 $\mathbf{f}$, 应力张量为 $\mathbf{\sigma}$, 则体元的运动方程为

$$ \rho\frac{\partial^{2}u_{i}}{\partial t^{2}} = \frac{\partial\sigma_{ji}}{\partial x_{j}} + f_{i}\\ \left(\rho\frac{\partial^{2}\mathbf{u}}{\partial t^{2}} = \nabla\cdot\mathbf{\sigma} + \mathbf{f}\right) $$

采用 各向同性 线性弹性 条件:

$$ \rho\frac{\partial^{2}u_{i}}{\partial t^{2}} = \lambda\frac{\partial\theta}{\partial x_{i}} + 2\mu\frac{\partial e_{ij}}{\partial x_{j}} + f_{i} $$

展开 $\theta, e_{ij}$:

$$ \rho\frac{\partial^{2}u_{i}}{\partial t^{2}} = \lambda\frac{\partial^{2}u_{j}}{\partial x_{i}\partial x_{j}} + \mu\left(\frac{\partial^{2}u_{i}}{\partial x_{j}\partial x_{j}} + \frac{\partial^{2}u_{j}}{\partial x_{i}\partial x_{j}}\right) + f_{i} = \left(\lambda + \mu\right)\frac{\partial^{2}u_{j}}{\partial x_{i}\partial x_{j}} + \mu\frac{\partial^{2}u_{i}}{\partial x_{j}\partial x_{j}} + f_{i} $$

写为矢量分析形式:

$$ \begin{aligned} \rho\frac{\partial^{2}\mathbf{u}}{\partial t^{2}} &= \left(\lambda + \mu\right)\nabla\left(\nabla\cdot\mathbf{u}\right) + \mu\nabla^{2}\mathbf{u} + \mathbf{f}\\ &= (\lambda + 2\mu)\nabla(\nabla\cdot\mathbf{u}) -\mu\nabla\times(\nabla\times\mathbf{u}) + \mathbf{f} \end{aligned} $$

将应变场 $\mathbf{u}$ 分解为无旋场 $\mathbf{u}_{1}$ 和无散场 $\mathbf{u}_{2}$:

$$ \mathbf{u} = \mathbf{u}_{1} + \mathbf{u}_{2}, \quad\nabla\times\mathbf{u}_{1} = \mathbf{0}, \quad\nabla\cdot\mathbf{u}_{2} = 0 $$

若忽略体力 $\mathbf{f}$, 则分解为两个波动方程

$$ \frac{\partial^{2}\mathbf{u}_{1}}{\partial t^{2}} = \frac{\lambda + 2\mu}{\rho}\nabla^{2}\mathbf{u}_{1}\\ \frac{\partial^{2}\mathbf{u}_{2}}{\partial t^{2}} = \frac{\mu}{\rho}\nabla^{2}\mathbf{u}_{2} $$

无旋场 $\mathbf{u}_{1}$ 为纵波(P), 速度为 $\alpha = \sqrt{(\lambda + 2\mu)/\rho}$; 无散场 $\mathbf{u}_{2}$ 为横波(S), 速度为 $\beta = \sqrt{\mu/\rho}$.

• 同求散度($\nabla\cdot$)

$$ \frac{\partial^{2}\theta}{\partial t^{2}} = \alpha^{2}\nabla^{2}\theta $$

• 同求旋度($\nabla\times$)

$$ \frac{\partial^{2}\mathbf{\omega}}{\partial t^{2}} = \beta^{2}\nabla^{2}\mathbf{\omega} $$

因为

$$ \nabla\times\mathbf{u}_{1} = \mathbf{0}, \quad\nabla\cdot\mathbf{u}_{2} = 0 $$

所以写作 Helmholtz 势:

$$ \mathbf{u} = \mathbf{u}_{1} + \mathbf{u}_{2} = \nabla\phi + \nabla\times\mathbf{\Psi} $$

其同样满足波动方程:

$$ \frac{\partial^{2}\phi}{\partial t^{2}} = \alpha^{2}\nabla^{2}\phi\\ \frac{\partial^{2}\mathbf{\Psi}}{\partial t^{2}} = \beta^{2}\nabla^{2}\mathbf{\Psi} $$

波动方程的解

使用分离变量法对波方程进行求解. 以无旋波势 $\phi$ 为例, 设

$$ \phi(\mathbf{r},t) = Ae^{\pm i(\omega t - \mathbf{k}_{\alpha}\cdot\mathbf{r})} $$

代表角频率为 $\omega$, 沿 $\mathbf{k}_{\alpha}$ 方向传播, $\mathbf{k}_{\alpha}$ 是 $P$ 波的 波数矢量, 且 $|\mathbf{k}_{\alpha}| = \omega/\alpha$.

类似的, 矢量势 $\mathbf{\Psi}$ 解形式为

$$ \mathbf{\Psi}(\mathbf{r},t) = \mathbf{B}e^{\pm i(\omega t - \mathbf{k}_{\beta}\cdot\mathbf{r})} $$

代表角频率为 $\omega$, 且有 $|\mathbf{k}_{\beta}| = \omega/\beta$.