From 58907b42d26910edff8cc2d1420295f7a02b1524 Mon Sep 17 00:00:00 2001 From: cyqm <2784841400@qq.com> Date: Thu, 14 Dec 2023 17:22:05 +0800 Subject: [PATCH] complete 2023/12/13 notes --- src/20231213.md | 86 +++++++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 86 insertions(+) create mode 100644 src/20231213.md diff --git a/src/20231213.md b/src/20231213.md new file mode 100644 index 0000000..a5310f2 --- /dev/null +++ b/src/20231213.md @@ -0,0 +1,86 @@ +## 2023/12/13 + +*Elon Musk (1971~ ): First-principle thinking 第一性原理思考：从某一体系的物理本质思考问题* + +*越南在明朝之前是中国领土。之后官员搜刮民脂民膏导致当地造反，官员将此事瞒了下来。这是不符合第一性原理的！* + +### 1. D'Almbert's principle (1743) 达朗贝尔原理 + +#### (1) Contents 内容 + +This principle is hailed as the first principle of theoretical mechanics (理论力学的第一性原理). + +In the system, there are $s$ particles, and for particle $i$, its position vector is $$\boldsymbol{r}_i = \boldsymbol{r}_i (q_1, q_2, \dots, q_n; t).$$ + +Particle $i$ can make real displacements (实位移): $$\mathrm{d} \boldsymbol{r}_i = \frac{\partial \boldsymbol{r}_i}{\partial t} \mathrm{d}t + \sum_{j = 1}^{n} \frac{\partial \boldsymbol{r}_i}{\partial q_j} \mathrm{d}q_j$$ and we can imitate its virtual displacements (虚位移): $$\delta \boldsymbol{r}_i = \sum_{j = 1}^{n} \frac{\partial \boldsymbol{r}_i}{\partial q_j} \delta q_j.$$ + +From real displacement $\mathrm{d} \boldsymbol{r}_i$, we can know the real velocity (实速度) of particle $i$: $$\boldsymbol{v}_i = \frac{\mathrm{d} \boldsymbol{r}_i}{\mathrm{d}t} = \frac{\partial \boldsymbol{r}_i}{\partial t} + \sum_{j = 1}^{n} \frac{\partial \boldsymbol{r}_i}{\partial q_j} \frac{\mathrm{d}q_j}{\mathrm{d}t} = \frac{\partial \boldsymbol{r}_i}{\partial t} + \sum_{j = 1}^{n} \frac{\partial \boldsymbol{r}_i}{\partial q_j} \dot{q}_j.$$ + +From this, we can know that $$\frac{\partial \boldsymbol{v}_i}{\partial \dot{q}_j} = \sum_{j = 1}^{n} \frac{\partial \boldsymbol{r}_i}{\partial q_j}.$$ + +Through this, we have reduced a dynamic problem to a static one (化动为静). + +According to the principle of virtual work, we can know $$\sum_{i = 1}^{s} \boldsymbol{F}_i \cdot \delta \boldsymbol{r}_i = \boldsymbol{0}.$$ According to the Newton's 2nd law of motion, we know that $$\boldsymbol{F}_i = \frac{\mathrm{d} \boldsymbol{p}_i}{\mathrm{d}t}.$$ From the 2 equations above, we know that $$\sum_{i = 1}^{s} \left( \boldsymbol{F}_i - \frac{\mathrm{d} \boldsymbol{p}_i}{\mathrm{d}t} \right) \cdot \delta \boldsymbol{r}_i = 0.$$ + +This is the **D'Almbert's principle (达朗贝尔原理)**. It can also be written in this form: $$\sum_{i = 1}^{s} \left( \boldsymbol{F}_i - m_i \ddot{\boldsymbol{r}}_i \right) \cdot \delta \boldsymbol{r}_i = 0.$$ + +#### (2) Using D'Almbert's principle to derive the Euler-Lagrange equations 用达朗贝尔原理推导欧拉-拉格朗日方程 + +Generalized force $Q_j$ is defined as follows: $$\sum_{i = 1}^{s} \boldsymbol{F}_i \cdot \delta \boldsymbol{r}_i = \sum_{i = 1}^{s} \boldsymbol{F}_i \cdot \left( \sum_{j = 1}^{n} \frac{\partial \boldsymbol{r}_i}{\partial q_j} \delta q_j \right) = \sum_{j = 1}^{n} Q_j \delta q_j = 0. \quad(1)$$ + +Also, we have $$\begin{align*} + \sum_{i = 1}^{s} \boldsymbol{F}_i \cdot \delta \boldsymbol{r}_i & = \sum_{i = 1}^{s} \frac{\mathrm{d} \boldsymbol{p}_i}{\mathrm{d}t} \cdot \left( \sum_{j = 1}^{n} \frac{\partial \boldsymbol{r}_i}{\partial q_j} \delta q_j \right) \\ + & = \sum_{i = 1}^{s} \sum_{j = 1}^{n} m_i \frac{\mathrm{d}^2 \boldsymbol{r}}{\mathrm{d}t^2} \cdot \frac{\partial \boldsymbol{r}_i}{\partial q_j} \delta q_j \\ + & = \sum_{i = 1}^{s} \sum_{j = 1}^{n} m_i \frac{\mathrm{d}}{\mathrm{d}t} \left( \frac{\mathrm{d} \boldsymbol{r}_i}{\mathrm{d}t} \right) \cdot \frac{\partial \boldsymbol{r}_i}{\partial q_j} \delta q_j \\ + & = \sum_{i = 1}^{s} \sum_{j = 1}^{n} \left[ \frac{\mathrm{d}}{\mathrm{d}t} \left( m_i \frac{\mathrm{d} \boldsymbol{r}_i}{\mathrm{d}t} \cdot \frac{\partial \boldsymbol{r}_i}{\partial q_j} \right) - m_i \frac{\mathrm{d} \boldsymbol{r}_i}{\mathrm{d}t} \cdot \frac{\mathrm{d}}{\mathrm{d}t} \left( \frac{\partial \boldsymbol{r}_i}{\partial q_j} \right) \right] \delta q_j \\ + & = \sum_{i = 1}^{s} \sum_{j = 1}^{n} \left[ \frac{\mathrm{d}}{\mathrm{d}t} \left( m_i \boldsymbol{v}_i \cdot \frac{\partial \boldsymbol{v}_i}{\partial \dot{q}_j} \right) - m_i \boldsymbol{v}_i \cdot \frac{\partial}{\partial q_j} \left( \frac{\mathrm{d} \boldsymbol{r}_i}{\mathrm{d}t} \right) \right] \delta q_j \\ + & = \sum_{i = 1}^{s} \sum_{j = 1}^{n} \left[ \frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial}{\partial \dot{q}_j} \left( {1 \over 2} m_i {v}_i^2 \right) - \frac{\partial}{\partial q_j} \left( {1 \over 2} m_i {v}_i^2 \right) \right] \delta q_j \\ + & = \sum_{i = 1}^{s} \sum_{j = 1}^{n} \left( \frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial T}{\partial \dot{q}_j} - \frac{\partial T}{\partial q_j} \right) \delta q_j \\ + & = 0. \quad(2) +\end{align*}$$ + +*编者：凑，就硬凑* + +$(1) - (2)$, and we get: $$\sum_{j = 1}^{n} \left( Q_j - \frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial T}{\partial \dot{q}_j} + \frac{\partial T}{\partial q_j} \right) \delta q_j = 0.$$ + +Because $\delta q_j$ ia arbitrary (随意选取的), we need $$ Q_j - \frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial T}{\partial \dot{q}_j} + \frac{\partial T}{\partial q_j} = 0, \quad(*)$$ where $Q_j$ is conservative, generalized force (保守的广义力). + +Correspondingly, we have a pair of work conjugates (功的共轭): $$\text{Work conjugate} +\left\{ +\begin{array}{ll} + \boldsymbol{F} & \mathrm{d} \boldsymbol{r} \\ + Q & \delta q +\end{array} +\right.$$ + +Also we know that $$\frac{\partial V}{\partial \dot{q}_j} = 0$$ and $$Q_j = - \frac{\partial V}{\partial q_j}.$$ Substitute these two equations in $(*)$, and we get $$- \frac{\partial V}{\partial q_j} - \frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial (T - V)}{\partial \dot{q}_j} + \frac{\partial T}{\partial q_j} = 0$$ $$\frac{\partial (T - V)}{\partial q_j} - \frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial (T - V)}{\partial \dot{q}_j} = 0$$ $$\frac{\partial L}{\partial q_j} - \frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial L}{\partial \dot{q}_j} = 0,$$ where $L$ is the Lagrangian. This is the Euler-Lagrange equations. + +### 2. Vibration of a string 弦的振动 + +*编者按：此部分涉及到欧拉-拉格朗日方程的多变量推广。具体的推导可以参考这篇文章：[【变分计算2】多变量系统下的变分方程 (https://zhuanlan.zhihu.com/p/357217956)](https://zhuanlan.zhihu.com/p/357217956)* + +Suppose there is an infinitesimal movement $u(x, t)$ on a piece of string whose length is $L$. + +The kinetic energy (动能) of the string is $$E_k = \int_{0}^{L} {1 \over 2} \mu \left( \frac{\partial u}{\partial t} \right)^2 \mathrm{d}x,$$ where $\displaystyle \mu = \frac{m}{L}$ is the linear mass density (质量线密度). + +For an infinitesimal length on the string from $x$ to $x + \mathrm{d}x$, when vibrating, its length becomes $$\mathrm{d}l' = \sqrt{1 + \left( \frac{\partial u}{\partial x} \right)^2} \mathrm{d}x,$$ and the change in length is $$\mathrm{d}l' - \mathrm{d}x = \left( \sqrt{1 + \left( \frac{\partial u}{\partial x} \right)^2} - 1 \right) \mathrm{d}x \approx \left[ 1 + \frac{1}{2} \left( \frac{\partial u}{\partial x} \right)^2 - 1 \right] = \frac{1}{2} \left( \frac{\partial u}{\partial x} \right)^2 \mathrm{d}x.$$ + +The potential energy is $$U = - W = \int T (\mathrm{d}l' - \mathrm{d}x) = \int_{0}^{L} \frac{1}{2} T \left( \frac{\partial u}{\partial x} \right)^2 \mathrm{d}x.$$ + +The Lagrangian $$L = E_k - U = \int_{0}^{L} \frac{1}{2} \left[ \mu \left( \frac{\partial u}{\partial t} \right)^2 - T \left( \frac{\partial u}{\partial x} \right)^2 \right] \mathrm{d}x = \int_{0}^{L} \mathcal{L} \mathrm{d}x,$$ where $\mathcal{L}$ is the Lagrangian density (拉格朗日量密度). + +$\mathcal{L}$ can be written in this form: $$\mathcal{L} = \mathcal{L} \left( u, \frac{\partial u}{\partial x}, \frac{\partial u}{\partial t}; x, t \right).$$ + +Because the Lagrangian is a functional (泛函) of $u(x, t)$, we can expand the Euler-Lagrange equations to $$\frac{\partial}{\partial t} \frac{\partial \mathcal{L}}{\partial (\partial u / \partial t)} + \frac{\partial}{\partial x} \frac{\partial \mathcal{L}}{\partial (\partial u / \partial x)} - \frac{\partial \mathcal{L}}{\partial u} = 0. \quad(*)$$ + +Here, $\displaystyle \mathcal{L} = \frac{1}{2} \left[ \mu \left( \frac{\partial u}{\partial t} \right)^2 - T \left( \frac{\partial u}{\partial x} \right)^2 \right]$ does not depend on $u$, and thus $u$ is a cyclic coordinate (循环坐标). + +Substitute $\mathcal{L}$ into $(*)$, and we get $$\mu \frac{\partial^2 u}{\partial t^2} - T \frac{\partial^2 u}{\partial x^2} = 0.$$ $$\left( \frac{1}{c^2} \frac{\partial^2}{\partial t^2} - \frac{\partial^2}{\partial x^2} \right) u = \Box u = 0.$$ Here $\Box$ is called the **d'Alembert or quabla operator (达朗贝尔算符)**: $$\Box = \frac{1}{c^2} \frac{\partial^2}{\partial t^2}- \nabla^2.$$ + +### 3. Vibration of a membrane 膜的振动 + +Suppose the membrane has an area density (质量面密度) of $\varLambda = \rho h$ (where $h$ is the thickness of the membrane), and its surface tension coefficient (表面张力系数, surface tension per unit length) is $\sigma$. + +Similar to the vibration of strings, we have the Lagrangian density $$\mathcal{L} = \frac{1}{2} \left\{ \varLambda \left( \frac{\partial u}{\partial t} \right)^2 - T \left[ \left( \frac{\partial u}{\partial x} \right)^2 + \left( \frac{\partial u}{\partial y} \right)^2 \right] \right\}$$ and equation of movement $$\frac{\partial}{\partial t} \frac{\partial \mathcal{L}}{\partial (\partial u / \partial t)} + \frac{\partial}{\partial x} \frac{\partial \mathcal{L}}{\partial (\partial u / \partial x)} + \frac{\partial}{\partial y} \frac{\partial \mathcal{L}}{\partial (\partial u / \partial y)} - \frac{\partial \mathcal{L}}{\partial u} = 0$$ and solution $$\left( \frac{1}{c^2} \frac{\partial^2}{\partial t^2} - \frac{\partial^2}{\partial x^2} - \frac{\partial^2}{\partial y^2} \right) u = \Box u = 0.$$ + +> 作业：Arnold《经典力学的数学方法》书中，为什么说所有生物跳的高度是同一量级？