理论力学总复习

26 年 6 月 25 日 星期四 (已编辑)
5606 字
29 分钟

这篇笔记我是一边复习一边写的,所以有些章节我没重点复习的就没有喵~

第二章:分析力学基础——拉格朗日方程

约束与自由度

  • 完整约束f(r1,,rn,t)=0f(\boldsymbol{r}_1,\dots,\boldsymbol{r}_n,t)=0,自由度 s=3nks = 3n - k
  • 非完整约束:含不可积分的微分约束
  • 理想约束:约束反力虚功为零,iNiδri=0\sum_i \boldsymbol{N}_i \cdot \delta\boldsymbol{r}_i = 0

达朗贝尔方程

考虑理想约束情况,从牛顿第二定律出发,引入惯性力,消去约束力:

i(Fimir¨i)δri=0\begin{align} \sum_i (\boldsymbol{F}_i - m_i \ddot{\boldsymbol{r}}_i) \cdot \delta \boldsymbol{r}_i = 0 \end{align}

这就是达朗贝尔方程,是理想约束体系的普遍方程.

静力学特例为虚功原理

iFiδri=0\begin{align} \sum_i \boldsymbol{F}_i \cdot \delta\boldsymbol{r}_i = 0 \end{align}

拉格朗日方程

广义力

Qα=iFiriqα\begin{align} Q_\alpha = \sum_i \boldsymbol{F}_i \cdot \frac{\partial \boldsymbol{r}_i}{\partial q_\alpha} \end{align}

一般形式

ddt(Tq˙α)Tqα=Qα\begin{align} \boxed{ \frac{\mathrm{d}}{\mathrm{d}t} \left( \frac{\partial T}{\partial \dot{q}_\alpha} \right) - \frac{\partial T}{\partial q_\alpha} = Q_\alpha } \end{align}

保守体系L=TVL = T - V):

ddt(Lq˙α)Lqα=0\begin{align} \boxed{ \frac{\mathrm{d}}{\mathrm{d}t} \left( \frac{\partial L}{\partial \dot{q}_\alpha} \right) - \frac{\partial L}{\partial q_\alpha} = 0 } \end{align}

其中 L=TV=L(q,q˙,t)L = T - V = L(q,\dot{q},t) 称为体系的拉格朗日函数

拉格朗日函数的不确定性LLL=L+df/dtL' = L + \mathrm{d}f/\mathrm{d}t 给出相同运动方程.

拉格朗日函数的不确定性的证明如下:

我们从充分性必要性两方面来证明:

充分性证明(若 LL=f˙\mathcal{L} - \mathcal{L}' = \dot{f},则它们给出相同运动方程)

LL=dfdt=i=1sfqiq˙i+ft\mathcal{L} - \mathcal{L}' = \dfrac{\mathrm{d}f}{\mathrm{d}t} = \sum_{i=1}^s \dfrac{\partial f}{\partial q_i} \dot{q}_i + \dfrac{\partial f}{\partial t},其中 f(q1,,qs,t)f(q_1,\dots,q_s,t) 是任意可微函数。

L=L+f˙\mathcal{L} = \mathcal{L}' + \dot{f} 代入拉格朗日方程:

ddt(Lq˙i)Lqi=ddt(Lq˙i+fqi)(Lqi+qijfqjq˙j+2fqit)\frac{\mathrm{d}}{\mathrm{d}t}\left( \frac{\partial \mathcal{L}}{\partial \dot{q}_i} \right) - \frac{\partial \mathcal{L}}{\partial q_i} = \frac{\mathrm{d}}{\mathrm{d}t}\left( \frac{\partial \mathcal{L}'}{\partial \dot{q}_i} + \frac{\partial f}{\partial q_i} \right) - \left( \frac{\partial \mathcal{L}'}{\partial q_i} + \frac{\partial}{\partial q_i}\sum_j \frac{\partial f}{\partial q_j} \dot{q}_j + \frac{\partial^2 f}{\partial q_i \partial t} \right)

由于

ddt(fqi)=j2fqjqiq˙j+2ftqi,\frac{\mathrm{d}}{\mathrm{d}t}\left( \frac{\partial f}{\partial q_i} \right) = \sum_j \frac{\partial^2 f}{\partial q_j \partial q_i} \dot{q}_j + \frac{\partial^2 f}{\partial t \partial q_i},

代入后,所有涉及 ff 的附加项相互抵消。因此:

ddt(Lq˙i)Lqi=ddt(Lq˙i)Lqi.\frac{\mathrm{d}}{\mathrm{d}t}\left( \frac{\partial \mathcal{L}}{\partial \dot{q}_i} \right) - \frac{\partial \mathcal{L}}{\partial q_i} = \frac{\mathrm{d}}{\mathrm{d}t}\left( \frac{\partial \mathcal{L}'}{\partial \dot{q}_i} \right) - \frac{\partial \mathcal{L}'}{\partial q_i}.

所以 L\mathcal{L}L\mathcal{L}' 给出完全相同的拉格朗日方程,即描述同一个力学体系。


必要性证明(若两个拉格朗日量给出相同运动方程,则它们必相差一个全导数)

如果 L\mathcal{L}L\mathcal{L}' 给出相同的运动方程,则它们的变分之差为零:

δt1t2(LL)dt=0\delta \int_{t_1}^{t_2} (\mathcal{L} - \mathcal{L}') \,\mathrm{d}t = 0

因为两个作用量泛函的极值条件相同。

由变分法基本引理,若 δt1t2F(q,q˙,t)dt=0\delta \int_{t_1}^{t_2} F(q,\dot{q},t) \,\mathrm{d}t = 0 对所有端点固定的变分路径都成立,则 FF 必须满足欧拉-拉格朗日方程的零解,即 FF 本身可以写成某个函数的全导数:

LL=dfdt.\mathcal{L} - \mathcal{L}' = \frac{\mathrm{d}f}{\mathrm{d}t}.

更直接地,将 LL\mathcal{L} - \mathcal{L}' 记为 ΔL\Delta L。如果 ΔL\Delta L 不满足拉格朗日方程,则其变分不会为零,与前提矛盾。因此 ΔL\Delta L 的欧拉-拉格朗日表达式为零,这意味着 ΔL\Delta L 是一个全导数项

证毕

平衡条件

Qα=0Vqα=0\begin{align} Q_\alpha = 0 \quad \text{或} \quad \frac{\partial V}{\partial q_\alpha} = 0 \end{align}

稳定平衡:势能取极小值(二阶偏导数矩阵正定).

释放约束法求约束力:解除约束 → 约束力视为主动力 → 虚功原理求解.


第四章:刚体

刚体运动学

运动分类:平动(质点的运动)、定轴转动、平面平行运动、定点转动(欧拉角)、一般运动.

自由度:自由刚体6个(3平动+3转动),定点转动3个(欧拉角),平面平行运动3个.

欧拉角

  • 进动角 φ\varphi :绕固定轴 z0z_0 的转动,描述刚体对称轴在空间中的方位角
  • 章动角 θ\theta :绕节线(xx' 轴、ONON)的转动,描述刚体 zz 轴与固定轴之间的夹角(0θπ0 \le \theta \le \pi
  • 自转角 ψ\psi :绕刚体自身 zz 轴的转动,描述刚体绕其对称轴的自转

欧拉运动学方程(定点转动):

随动坐标系 OxyzOxyz

{ωx=θ˙cosψ+φ˙sinθsinψωy=θ˙sinψ+φ˙sinθcosψωz=ψ˙+φ˙cosθ\begin{align} \begin{cases} \begin{aligned} \omega_x &= \dot{\theta} \cos\psi + \dot{\varphi} \sin\theta \sin\psi \\ \omega_y &= -\dot{\theta} \sin\psi + \dot{\varphi} \sin\theta \cos\psi \\ \omega_z &= \dot{\psi} + \dot{\varphi} \cos\theta \end{aligned} \end{cases} \end{align}

世界坐标系 Ox0y0z0Ox_0y_0z_0

{ω0,x=ψ˙sinθsinφ+θ˙cosφω0,y=ψ˙sinθcosφ+θ˙sinφω0,z=ψ˙cosθ+φ˙\begin{align} \begin{cases} \begin{aligned} \omega_{0,x} &= \dot{\psi} \sin\theta \sin\varphi + \dot{\theta} \cos\varphi \\ \omega_{0,y} &= -\dot{\psi} \sin\theta \cos\varphi + \dot{\theta} \sin\varphi \\ \omega_{0,z} &= \dot{\psi} \cos\theta + \dot{\varphi} \end{aligned} \end{cases} \end{align}

基点法求速度与加速度

{vP=vC+ω×rCPaP=aC+dωdt×rCP+ω×(ω×rCP)\begin{align} \begin{cases} \begin{aligned} \boldsymbol{v}_P &= \boldsymbol{v}_C + \boldsymbol{\omega} \times \boldsymbol{r}_{CP} \\ \boldsymbol{a}_P &= \boldsymbol{a}_C + \frac{\mathrm{d} \boldsymbol{\omega}}{\mathrm{d} t} \times \boldsymbol{r}_{CP} + \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \boldsymbol{r}_{CP}) \end{aligned} \end{cases} \end{align}

角速度 ω\boldsymbol{\omega} 与基点选择无关.

动力学方程

质心运动定律(3个):

mdvcdt=iFi(e)\begin{align} m \frac{\mathrm{d} v_c}{\mathrm{d} t} = \sum_i \boldsymbol{F}_i^{(e)} \end{align}

角动量定理(3个):

dLdt=iri×Fi(e)\begin{align} \frac{\mathrm{d} \boldsymbol{L}}{\mathrm{d} t} = \sum_i \boldsymbol{r}_i \times \boldsymbol{F}_i^{(e)} \end{align}

动能定理:

dT=i(Fi(e)+Fi(i))dri\begin{align} \mathrm{d} T = \sum_i (\boldsymbol{F}_i^{(e)} + \boldsymbol{F}_i^{(i)}) \cdot \mathrm{d} \boldsymbol{r}_i \end{align}

从这7个方程中选取6个即可建立刚体运动的动力学方程.

瞬时转心:平面平行运动中速度为零的点,纯滚动时接触点为瞬心.

转动惯量

Iij=(r2δijxixj)dm\begin{align} I_{ij} = \int (r^2 \delta_{ij} - x_i x_j) \,\mathrm{d}m \end{align}

Ikl=IlkI_{kl} = I_{lk}I\boldsymbol{I} 称为惯量矩阵惯量张量. 非对角元素成为惯量积.

惯量主轴:使惯量积为零的坐标轴. 对称轴必为主轴.

平行轴定理IP=IC+md2I_P = I_C + md^2.

回转半径:令 I=mρ2I = m\rho^2 ,其中 ρ\rho 称为对该转动轴的回转半径. 主回转半径 ρC\rho_C 是最小的回转半径.

物体(质量为 mm坐标轴(惯量主轴)主转动惯量
长方形板(边长为 2a2a2b2bOxOxOyOy 依次平行于 2a2a \\2b2bOzOz 垂直于板面I11=mb2/3I22=ma2/3I33=m(a2+b2)/3I_{11} = mb^2/3 \\ I_{22} = ma^2/3 \\ I_{33} = m(a^2+b^2)/3
实心长方体(2a2a2b2b2c2cOxOxOyOyOzOz依次平行 \\2a2a2b2b2c2cI11=m(b2+c2)/3I22=m(c2+a2)/3I33=m(a2+b2)/3I_{11} = m(b^2+c^2)/3 \\ I_{22} = m(c^2+a^2)/3 \\ I_{33} = m(a^2+b^2)/3
圆板(半径为 aaOxOxOyOy 在板面上,\\ OzOz 垂直于版面I11=I22=ma2/4I33=ma2/2I_{11}=I_{22}=ma^2/4 \\ I_{33}=ma^2/2
椭圆板(长短轴为 2a2a2b2bOxOxoyoy依次沿半长、 \\ 短轴,OzOz 垂直于板面I11=mb2/4I22=ma2/4I33=m(a2+b2)/4I_{11}=mb^2/4 \\ I_{22}=ma^2/4 \\ I_{33}=m(a^2+b^2)/4
实心圆柱体(半径为 aa\\llOxOxOyOy垂直于对称轴,\\ OzOz 为其轴线I11=I22=m(3a2+l2)/12I33=ma2/2I_{11}=I_{22}=m(3a^2+l^2)/12 \\ I_{33}=ma^2/2
实心椭圆柱体(椭圆半轴为 \\ aabb,长 llOxOxOyOy为椭圆半长、\\ 短轴,OzOz 为其轴线I11=m(3b2+l2)/12I22=m(3a2+l2)/12I33=m(a2+b2)/4I_{11}=m(3b^2+l^2)/12 \\ I_{22}=m(3a^2+l^2)/12 \\ I_{33}=m(a^2+b^2)/4
球壳(外半径为 aa,内半径为 bb任意三条互相垂直的轴线I=25ma5b5a3b3\displaystyle I=\frac{2}{5}m\frac{a^5-b^5}{a^3-b^3}
薄球壳(半径为 aa任意三条互相垂直的轴线I=2ma2/3I=2ma^2/3
实心球体(半径为 aa任意三条互相垂直的轴线I=2ma2/5I=2ma^2/5
实椭球体(半轴为 aabbcc三条半轴I11=m(b2+c2)/5I22=m(c2+a2)/5I33=m(a2+b2)/5I_{11}=m(b^2+c^2)/5 \\ I_{22}=m(c^2+a^2)/5 \\ I_{33}=m(a^2+b^2)/5
圆锥(高 hh ,底面半径 aaOxOxOyOy轴在底面上,\\ OzOz 轴为其轴线I11=I22=m(3a2+2h2)/20I33=3ma2/10I_{11}=I_{22}=m(3a^2+2h^2)/20 \\ I_{33}=3ma^2/10

角动量与转动动能

一般系:

[LxLyLz]=[I11I12I13I21I22I23I31I32I33][ωxωyωz],T=12Lω\begin{align} \begin{bmatrix} L_x \\ L_y \\ L_z \end{bmatrix} = \begin{bmatrix} I_{11} & -I_{12} & -I_{13} \\ -I_{21} & I_{22} & -I_{23} \\ -I_{31} & -I_{32} & I_{33} \end{bmatrix} \begin{bmatrix} \omega_x \\ \omega_y \\ \omega_z \end{bmatrix}, \quad T = \frac{1}{2} \boldsymbol{L}\cdot\boldsymbol{\omega} \end{align}

惯量主轴系下:

L=I1ωxi+I2ωyj+I3ωzk,T=12(I1ωx2+I2ωy2+I3ωz2)\begin{align} \boldsymbol{L} = I_1\omega_x\boldsymbol{i} + I_2\omega_y\boldsymbol{j} + I_3\omega_z\boldsymbol{k}, \quad T = \frac{1}{2} (I_1 \omega_x^2 + I_2 \omega_y^2 + I_3 \omega_z^2) \end{align}

欧拉动力学方程:

随动坐标系中的角动量定理:

dLdtL×ω=M\begin{align} \frac{\mathrm{d}\boldsymbol{L}}{\mathrm{d}t} - \boldsymbol{L} \times \boldsymbol{\omega} = \boldsymbol{M} \end{align}

分量形式(欧拉方程):

{I1ω˙x(I2I3)ωyωz=MxI2ω˙y(I3I1)ωzωx=MyI3ω˙z(I1I2)ωxωy=Mz\begin{align} \begin{cases} \begin{aligned} I_1 \dot{\omega}_x - (I_2 - I_3) \omega_y \omega_z &= M_x \\ I_2 \dot{\omega}_y - (I_3 - I_1) \omega_z \omega_x &= M_y \\ I_3 \dot{\omega}_z - (I_1 - I_2) \omega_x \omega_y &= M_z \end{aligned} \end{cases} \end{align}

惯量椭球:

刚体绕 转轴 ee 做定轴转动时,有

Ie=I11α2+I22β2+I33γ2+2I12αβ+2I13αγ+2I23βγ\begin{align} I_e = I_{11}\alpha^2 + I_{22}\beta^2 + I_{33}\gamma^2 + 2I_{12}\alpha\beta + 2I_{13}\alpha\gamma + 2I_{23}\beta\gamma \end{align}

在转动轴 ee 上取一线段 ON=1IeON=\frac{1}{\sqrt{I_e}} ,则 NN 点的坐标为

x=αIe,y=βIe,z=γIe\begin{align} x=\frac{\alpha}{\sqrt{I_e}},\quad y=\frac{\beta}{\sqrt{I_e}},\quad z=\frac{\gamma}{\sqrt{I_e}} \end{align}

代入即可得到:

I11x2+I22y2+I33z2+2I12xy+2I13xz+2I23yz=1\begin{align} I_{11}x^2 + I_{22}y^2 + I_{33}z^2 + 2I_{12}xy + 2I_{13}xz + 2I_{23}yz = 1 \end{align}

这个椭球称为惯量椭球,它的对称轴就是惯量主轴. 任意轴线 ee 与其交点 NN ,那么 ONON 即为绕这个轴的转动惯量 IeI_e 的平方根的倒数.

典型陀螺运动

欧拉陀螺M=0\boldsymbol{M}=0,即自由转动):L2L^2TT 守恒.

拉格朗日陀螺(对称陀螺 I1=I2I_1=I_2 在重力场中):

L=12I1(θ˙2+φ˙2sin2θ)+12I3(ψ˙+φ˙cosθ)2mglcosθ\begin{align} \boxed{ L = \frac{1}{2} I_1 (\dot{\theta}^2 + \dot{\varphi}^2 \sin^2\theta) + \frac{1}{2} I_3 (\dot{\psi} + \dot{\varphi} \cos\theta)^2 - mgl \cos\theta } \end{align}

φ\varphiψ\psi 为循环坐标,因此得到两个守恒量

Pφ=Lφ˙=Le3=(I1sin2θ+I3cos2θ)φ˙+I3cosθψ˙=常数Pψ=Lψ˙=Lk=I3(ψ˙+φ˙cosθ)=常数\begin{gather} P_\varphi = \frac{\partial L}{\partial \dot{\varphi}} = L_{e_3} = (I_1 \sin^2\theta + I_3 \cos^2\theta)\dot{\varphi} + I_3 \cos\theta\dot{\psi} = \text{常数} \\ P_\psi = \frac{\partial L}{\partial \dot{\psi}} = L_k = I_3(\dot{\psi} + \dot{\varphi}\cos\theta) = \text{常数} \end{gather}

同时还有能量守恒:

E=I12(θ˙2+φ˙2sin2θ)+Lk22I3+mglcosθ=常数\begin{align} E = \boxed{\frac{I_1}{2} (\dot{\theta}^2 + \dot{\varphi}^2\sin^2\theta) + \frac{L_{k'}^2}{2I_3} + mgl\cos\theta} = \text{常数} \end{align}

这就是拉格朗日陀螺的三个动力学方程.

Veff(θ)=(Le3Lkcosθ)22I1sin2θmgl(1cosθ)E=ELk22I3mgl\begin{gather} V_{\text{eff}}(\theta) = \frac{(L_{e_3} - L_k \cos\theta)^2}{2I_1 \sin^2\theta} - mgl(1- \cos\theta) \\ E' = E - \frac{L_k^2}{2I_3} - mgl \end{gather}

则可以化成:

I12θ˙2+Veff(θ)=E\begin{align} \frac{I_1}{2}\dot{\theta}^2 + V_{\text{eff}}(\theta) = E' \end{align}

θ\theta 的极值由条件 E=Veff(θ)E'=V_{\text{eff}}(\theta) 决定.

快速陀螺近似φ˙mgl/(I3ψ˙)\dot{\varphi} \approx mgl / (I_3 \dot{\psi}) (进动角速度和初角速度 ω\omega 近似成反比).


第五章:非惯性参考系

平动加速参考系

平移惯性力:F=ma0\boldsymbol{F}_{\text{惯}} = -m\boldsymbol{a}_0.

转动参考系

矢量导数关系:

(dAdt)S=(dAdt)S+ω×A\begin{align} \left( \frac{\mathrm{d}\boldsymbol{A}}{\mathrm{d}t} \right)_S = \left( \frac{\mathrm{d}\boldsymbol{A}}{\mathrm{d}t} \right)_{S'} + \boldsymbol{\omega} \times \boldsymbol{A} \end{align}

加速度合成:

aS=aS+2ω×vS+ω×(ω×r)+ω˙×r\begin{align} \boldsymbol{a}_S = \boldsymbol{a}_{S'} + 2\boldsymbol{\omega} \times \boldsymbol{v}_{S'} + \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \boldsymbol{r}) + \dot{\boldsymbol{\omega}} \times \boldsymbol{r} \end{align}

惯性力

名称表达式特点
科里奥利力F=2mω×vS\boldsymbol{F}_{\text{科}} = -2m\boldsymbol{\omega} \times \boldsymbol{v}_{S'}与速度垂直,不改变速率
离心力F=mω×(ω×r)\boldsymbol{F}_{\text{离}} = -m\boldsymbol{\omega} \times (\boldsymbol{\omega} \times \boldsymbol{r})方向垂直于转轴向外
欧拉力F=mω˙×r\boldsymbol{F}_{\text{欧}} = -m\dot{\boldsymbol{\omega}} \times \boldsymbol{r}角速度变化时出现

经典应用:傅科摆 Ω=ωearthsinλ\Omega = \omega_{\text{earth}}\sin\lambda,落体偏东,地转风.


第六章:多自由度体系的微振动

线性化

平衡位置附近,动能和势能展开至二阶:

T=12i,jmijq˙iq˙jV=12i,jkijqiqj\begin{align} T &= \frac{1}{2} \sum_{i,j} m_{ij} \dot{q}_i \dot{q}_j \\ V &= \frac{1}{2} \sum_{i,j} k_{ij} q_i q_j \end{align}

运动方程与久期方程

j(mijq¨j+kijqj)=0\begin{align} \sum_j (m_{ij} \ddot{q}_j + k_{ij} q_j) = 0 \end{align}

qj=Ajeiωtq_j = A_j e^{\mathrm{i}\omega t},得广义本征值问题:

j(kijω2mij)Aj=0\begin{align} \sum_j (k_{ij} - \omega^2 m_{ij}) A_j = 0 \end{align}

久期方程

det(Kω2M)=0\begin{align} \det(\boldsymbol{K} - \omega^2 \boldsymbol{M}) = 0 \end{align}

简正坐标

存在线性变换使系统解耦为独立谐振子:

L=12α=1s(Q˙α2ωα2Qα2)\begin{align} L = \frac{1}{2} \sum_{\alpha=1}^s (\dot{Q}_\alpha^2 - \omega_\alpha^2 Q_\alpha^2) \end{align}

各简正模式独立,无能量交换.

一维晶格纵振动

色散关系:

ω2=4kmsin2(qa2)\begin{align} \omega^2 = \frac{4k}{m} \sin^2\left(\frac{qa}{2}\right) \end{align}

长波极限 qa1qa \ll 1ωvq\omega \approx v qv=ak/mv = a\sqrt{k/m}.


第七章:阻尼运动

恒力下的阻尼直线运动

mv˙=mgγv,v(t)=mgγ(1eγt/m),vT=mgγ\begin{align} m\dot{v} = mg - \gamma v, \quad v(t) = \frac{mg}{\gamma}(1 - e^{-\gamma t/m}), \quad v_T = \frac{mg}{\gamma} \end{align}

线性阻尼自由振动

x¨+2βx˙+ω02x=0,β=γ2m,ω0=km\begin{align} \ddot{x} + 2\beta \dot{x} + \omega_0^2 x = 0, \quad \beta = \frac{\gamma}{2m}, \quad \omega_0 = \sqrt{\frac{k}{m}} \end{align}
类型条件解的形式
欠阻尼β<ω0\beta < \omega_0x=Aeβtcos(ωdt+φ)x = Ae^{-\beta t}\cos(\omega_d t + \varphi)ωd=ω02β2\omega_d = \sqrt{\omega_0^2 - \beta^2}
临界阻尼β=ω0\beta = \omega_0x=(A+Bt)eβtx = (A + Bt)e^{-\beta t}
过阻尼β>ω0\beta > \omega_0x=Aeλ1t+Beλ2tx = Ae^{\lambda_1 t} + Be^{\lambda_2 t}λ1,λ2<0\lambda_1,\lambda_2<0

受迫振动与共振

A(ω)=F0/m(ω02ω2)2+4β2ω2\begin{align} A(\omega) = \frac{F_0/m}{\sqrt{(\omega_0^2 - \omega^2)^2 + 4\beta^2 \omega^2}} \end{align}

共振频率 ωres=ω022β2\omega_{\text{res}} = \sqrt{\omega_0^2 - 2\beta^2}.

品质因子 Q=ω0/(2β)Q = \omega_0/(2\beta).


第八章:哈密顿理论

变分法基础

泛函 J[y]=x1x2f(x,y,y)dxJ[y] = \int_{x_1}^{x_2} f(x, y, y') \,\mathrm{d}x 取极值的必要条件为欧拉-拉格朗日方程

fyddxfy=0\begin{align} \frac{\partial f}{\partial y} - \frac{\mathrm{d}}{\mathrm{d}x} \frac{\partial f}{\partial y'} = 0 \end{align}

证明: 对泛函取变分:

δJ[y(x)]=δx1x2f(y,y,x)dx=x1x2δf(y,y,x)dx=x1x2(fyδy+fyδy)dx=x1x2[fyδy+ddx(fyδy)ddx(fy)δy]dx=fyδyx1x2x1x2(ddxfyfy)δydx\begin{aligned} \delta J[y(x)] &= \delta \int_{x_1}^{x_2} f(y, y', x) \, \mathrm{d}x = \int_{x_1}^{x_2} \delta f(y, y', x) \, \mathrm{d}x \\ &= \int_{x_1}^{x_2} \left(\frac{\partial f}{\partial y}\delta y + \frac{\partial f}{\partial y}\delta y'\right) \, \mathrm{d}x \\ &= \int_{x_1}^{x_2} \left[\frac{\partial f}{\partial y}\delta y + \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\partial f}{\partial y'}\delta y\right) - \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\partial f}{\partial y'}\right)\delta y\right] \, \mathrm{d}x \\ &= \left.\frac{\partial f}{\partial y'}\delta y\right|_{x_1}^{x_2}- \int_{x_1}^{x_2} \left(\frac{\mathrm{d}}{\mathrm{d}x}\frac{\partial f}{\partial y'} - \frac{\partial f}{\partial y}\right) \delta y \, \mathrm{d}x \\ \end{aligned}

这里第一项为零,为了使得变分 δJ[y(x)]\delta J[y(x)] 为零,需要第二项括号内取零,于是得证.

ff 不显含 xx,存在初积分

fyfy=常数\begin{align} f - y' \frac{\partial f}{\partial y'} = \text{常数} \end{align}

证明: 有

ddx(fyfy)=(fyy+fyy)(fyy+yddxfy)=y(ddxfyfy)=0\begin{aligned} &\frac{\mathrm{d}}{\mathrm{d}x}\left(f - y'\frac{\partial f}{\partial y'}\right) \\ = &\left(\frac{\partial f}{\partial y}y' + \frac{\partial f}{\partial y'}y''\right) - \left(\frac{\partial f}{\partial y'}y'' + y' \frac{\mathrm{d}}{\mathrm{d}x}\frac{\partial f}{\partial y'}\right) \\ = &-y'\left(\frac{\mathrm{d}}{\mathrm{d}x}\frac{\partial f}{\partial y'} - \frac{\partial f}{\partial y}\right) = 0 \end{aligned}

得证.

变分算符 δ\delta 与微分算符 d\mathrm{d} 可交换:δ(dy)=d(δy)\delta(\mathrm{d}y) = \mathrm{d}(\delta y),且 δx=0\delta x = 0.

哈密顿正则方程

广义动量pα=L/q˙αp_\alpha = \partial L / \partial \dot{q}_\alpha

哈密顿函数(体系的广义能量)

H=α=1spαq˙αL\begin{align} H = \sum_{\alpha=1}^s p_\alpha \dot{q}_\alpha - L \end{align}

哈密顿函数有一条重要的性质:

dH(q,p,t)dt=L(q,q˙,t)t\begin{align} \frac{\mathrm{d} H(q,p,t)}{\mathrm{d}t} = - \frac{\partial L(q,\dot{q},t)}{\partial t} \end{align}

证明如下:

对哈密顿函数求时间的全导数:

dHdt=α(p˙αq˙α+pαq¨α)dLdt=α(p˙αq˙α+pαq¨α)[α(Lqαq˙α+Lq˙αq¨α)+Lt]=α(p˙αq˙α+pαq¨α)[α(p˙αq˙α+pαq¨α)+Lt]\begin{aligned} \frac{\mathrm{d} H}{\mathrm{d} t} &= \sum_\alpha \left( \dot{p}_\alpha\dot{q}_\alpha + p_\alpha\ddot{q}_\alpha \right) - \frac{\mathrm{d} L}{\mathrm{d} t} \\ &= \sum_\alpha \left( \dot{p}_\alpha\dot{q}_\alpha + p_\alpha\ddot{q}_\alpha \right) - \left[ \sum_\alpha \left( \frac{\partial L}{\partial q_\alpha}\dot{q}_\alpha + \frac{\partial L}{\partial \dot{q}_\alpha}\ddot{q}_\alpha \right) + \frac{\partial L}{\partial t} \right] \\ &= \sum_\alpha \left( \dot{p}_\alpha\dot{q}_\alpha + p_\alpha\ddot{q}_\alpha \right) - \left[ \sum_\alpha \left( \dot{p}_\alpha\dot{q}_\alpha + p_\alpha\ddot{q}_\alpha \right) + \frac{\partial L}{\partial t} \right] \end{aligned}

这里用到了广义动量的定义 pα=Lq˙α\displaystyle p_\alpha = \frac{\partial L}{\partial \dot{q}_\alpha} 以及拉格朗日方程 p˙α=ddtLq˙α=Lqα\displaystyle \dot{p}_\alpha = \frac{\mathrm{d}}{\mathrm{d} t}\frac{\partial L}{\partial \dot{q}_\alpha} = \frac{\partial L}{\partial q_\alpha} ,于是有

dHdt=Lt\begin{aligned} \Rightarrow \frac{\mathrm{d} H}{\mathrm{d} t} &= - \frac{\partial L}{\partial t} \end{aligned}

得证.

哈密顿正则方程

{q˙α=Hpα,p˙α=Hqα,α=1,2,,s.\begin{align} \begin{cases} \begin{aligned} \dot{q}_\alpha &= \frac{\partial H}{\partial p_\alpha}, \\ \dot{p}_\alpha &= -\frac{\partial H}{\partial q_\alpha}, \end{aligned} \end{cases} \qquad \alpha = 1,2,\cdots,s. \end{align}

相空间(qα,pα)(q_\alpha, p_\alpha) 张成的 2s2s 维空间. 若 HH 不显含时间,相轨迹不相交,这就是能量守恒定律.

刘维尔定理:相空间体积在演化中保持不变.

哈密顿原理

通过泛函分析可以得出

S=t1t2L(q,q˙,t)dt,δS=0\begin{align} S = \int_{t_1}^{t_2} L(q,\dot{q},t) \,\mathrm{d}t, \quad \delta S = 0 \end{align}

SS 称为哈密顿作用量,由 δS=0\delta S = 0 可导出拉格朗日方程或正则方程.

哈密顿正则方程的导出:

将哈密顿量的定义 H=αpαq˙αLH = \sum_\alpha p_\alpha \dot{q}_\alpha - L 代入 δS=0\delta S = 0 即可得到

δS=t1t2δ[αpαq˙αH]  dt=t1t2[α(pαδq˙α+q˙αδpα)α(Hpαδpα+Hqαδqα)]  dt=t1t2[α(q˙αHpα)δpαα(p˙α+Hqα)δqα]  dt=0\begin{aligned} \delta S &= \int_{t_1}^{t_2} \delta \left[ \sum_\alpha p_\alpha \dot{q}_\alpha - H \right] \; \mathrm{d} t \\ &= \int_{t_1}^{t_2} \left[ \sum_\alpha \left( p_\alpha \delta \dot{q}_\alpha + \dot{q}_\alpha \delta p_\alpha \right) - \sum_\alpha \left( \frac{\partial H}{\partial p_\alpha}\delta p_\alpha + \frac{\partial H}{\partial q_\alpha}\delta q_\alpha \right) \right] \; \mathrm{d} t \\ &= \int_{t_1}^{t_2} \left[ \sum_\alpha \left( \dot{q}_\alpha - \frac{\partial H}{\partial p_\alpha} \right) \delta p_\alpha - \sum_\alpha \left( \dot{p}_\alpha + \frac{\partial H}{\partial q_\alpha} \right) \delta q_\alpha \right] \; \mathrm{d} t = 0 \end{aligned}

于是得到

q˙α=Hpα,p˙α=Hqα,α=1,2,,s\begin{aligned} \dot{q}_\alpha = \frac{\partial H}{\partial p_\alpha}, \quad \dot{p}_\alpha = -\frac{\partial H}{\partial q_\alpha} , \quad \alpha = 1,2,\cdots,s \end{aligned}

哈密顿原理比较的是真实路径和其无限邻近的各种可能路径的作用量的值.

对于非保守体系,有

δS=t1t2[δL(qα,qα˙,t)+αQαδqα]dt=0=t1t2[α(q˙αHpα)δpαα(p˙α+HqαQα)δqα]  dt=0\begin{align} \delta S &= \int_{t_1}^{t_2} [\delta L(q_\alpha, \dot{q_\alpha}, t) + \sum_\alpha Q_\alpha \delta q_\alpha]\mathrm{d} t = 0 \\ &= \int_{t_1}^{t_2} \left[ \sum_\alpha \left( \dot{q}_\alpha - \frac{\partial H}{\partial p_\alpha} \right) \delta p_\alpha - \sum_\alpha \left( \dot{p}_\alpha + \frac{\partial H}{\partial q_\alpha} - Q_\alpha \right) \delta q_\alpha \right] \; \mathrm{d} t = 0 \end{align}

其中的 QαQ_\alpha 为非保守力,对应的求出非保守系的哈密顿正则方程为:

q˙α=Hpα,p˙α=Hqα+Qα,α=1,2,,s\begin{aligned} \dot{q}_\alpha = \frac{\partial H}{\partial p_\alpha}, \quad \dot{p}_\alpha = -\frac{\partial H}{\partial q_\alpha} + Q_\alpha , \quad \alpha = 1,2,\cdots,s \end{aligned}

正则变换

正则变换:使得变换后新的哈密顿函数 HH^* 仍然满足正则方程

Qα=HPα,Pα=HQα,α=1,2,,s.\begin{align} Q_\alpha = \frac{\partial H^*}{\partial P_\alpha}, \quad P_\alpha = -\frac{\partial H^*}{\partial Q_\alpha}, \quad \alpha = 1,2,\cdots,s. \end{align}

的变换.

四类母函数

母函数自变量变换公式
F1(q,Q,t)F_1(q,Q,t)(q,Q)(q,Q)p=F1/qp = \partial F_1/\partial qP=F1/QP = -\partial F_1/\partial Q
F2(q,P,t)F_2(q,P,t)(q,P)(q,P)p=F2/qp = \partial F_2/\partial qQ=F2/PQ = \partial F_2/\partial P
F3(p,Q,t)F_3(p,Q,t)(p,Q)(p,Q)q=F3/pq = -\partial F_3/\partial pP=F3/QP = -\partial F_3/\partial Q
F4(p,P,t)F_4(p,P,t)(p,P)(p,P)q=F4/pq = -\partial F_4/\partial pQ=F4/PQ = \partial F_4/\partial P

四种母函数之间互相区分的关键就是所选取的独立变量不同. 对于变换公式,有以下规律可帮助记忆:

  • 若旧变量为 qq,则偏导前取正号(p=F1/qp=\partial F_1/\partial qp=F2/qp=\partial F_2/\partial q);
  • 若旧变量为 pp,则偏导前取负号(q=F3/pq=-\partial F_3/\partial pq=F4/pq=-\partial F_4/\partial p);
  • 若新变量为 QQ,则偏导前取负号(P=F1/QP=-\partial F_1/\partial QP=F3/QP=-\partial F_3/\partial Q);
  • 若新变量为 PP,则偏导前取正号(Q=F2/PQ=\partial F_2/\partial PQ=F4/PQ=\partial F_4/\partial P)。

新旧哈密顿量之间的变换关系为:

H=H+Fit,i=1,2,3,4.H^* = H + \frac{\partial F_i}{\partial t}, \quad i=1,2,3,4.

这里可以看出,在哈密顿方程中,广义坐标和广义动量只是名称上的不同,在物理意义上已经没有任何差别,因此常把它们称为正则共轭变量

用泊松括号判别正则变换

[Qα,Qβ]q,p=0,[Pα,Pβ]q,p=0,[Qα,Pβ]q,p=δαβ\begin{align} [Q_\alpha, Q_\beta]_{q,p} = 0, \quad [P_\alpha, P_\beta]_{q,p} = 0, \quad [Q_\alpha, P_\beta]_{q,p} = \delta_{\alpha\beta} \end{align}

泊松括号

定义

[f,g]=α=1s(fpαgqαfqαgpα)\begin{align} [f, g] = \sum_{\alpha=1}^s \left( \frac{\partial f}{\partial p_\alpha} \frac{\partial g}{\partial q_\alpha} - \frac{\partial f}{\partial q_\alpha} \frac{\partial g}{\partial p_\alpha} \right) \end{align}

力学量的时间演化

dfdt=[H,f]+ft\begin{align} \frac{\mathrm{d}f}{\mathrm{d}t} = [H, f] + \frac{\partial f}{\partial t} \end{align}

运动积分判据[H,f]=0    f[H, f] = 0 \iff f 是运动积分(不显含时间时).

基本性质

  1. [f,g]=[g,f][f,g] = -[g,f].
  2. [f,c]=0  (c为常量)[f,c] = 0 \; (c \text{为常量}).
  3. [f1+f2,g]=[f1,g]+[f2,g][f_1 + f_2,g] = [f_1,g] + [f_2,g].
  4. [f1f2,g]=f1[f2,g]+f2[f1,g][f_1 f_2,g] = f_1 [f_2,g] + f_2 [f_1,g].
  5. t[f,g]=[ft,g]+[f,gt]\displaystyle \frac{\partial}{\partial t}[f,g] = [\frac{\partial f}{\partial t},g] + [f, \frac{\partial g}{\partial t}].
  6. [f,qα]=fpα;[f,pα]=fqα\displaystyle [f,q_\alpha] = \frac{\partial f}{\partial p_\alpha}; \quad [f, p_\alpha] = -\frac{\partial f}{\partial q_\alpha}.
  7. [pα,pβ]=0,[qα,qβ]=0,[pα,qβ]=δαβ[p_\alpha,p_\beta] = 0, \quad [q_\alpha,q_\beta] = 0, \quad [p_\alpha,q_\beta]=\delta_{\alpha\beta}.
  8. 雅可比恒等式: [f,[g,h]]+[g,[h,f]]+[h,[f,g]]=0[f,[g,h]] + [g,[h,f]] + [h,[f,g]] = 0.
  9. 正则变换下保持不变: [f,g]p,q=[f,g]P,Q[f,g]_{p,q} = [f,g]_{P,Q}.

泊松定理:若 ffgg 均为运动积分,则 [f,g][f, g] 也是运动积分.

哈密顿-雅可比方程

哈密顿主函数 S(q,t)S(q, t) (对应的就是第二类母函数的 F(q,P,t)F(q, P, t) ):

H(q1,,qs,Sq1,,Sqs,t)+St=0\begin{align} H\left(q_1, \dots, q_s, \frac{\partial S}{\partial q_1}, \dots, \frac{\partial S}{\partial q_s}, t\right) + \frac{\partial S}{\partial t} = 0 \end{align}

这种解正则方程的方法只是将解正则方程的困难转移到了解哈密顿-雅可比方程上了,但毕竟指出了另一条途径.

SS 还可以写成

dSdt=αSqαqα˙+StS=Ldt\begin{gather} \frac{\mathrm{d} S}{\mathrm{d} t} = \sum_\alpha \frac{\partial S}{\partial q_\alpha} \dot{q_\alpha} + \frac{\partial S}{\partial t} \\ S = \int L \mathrm{d} t \end{gather}

即哈密顿主函数 SS 就是积分限不确定的哈密顿作用量,因此也被称为哈密顿作用函数.

HH 不显含时间,设 S=W(q,P)Et+AS = W(q, P) - Et + AWW哈密顿特征函数

H(q,Wq)=E\begin{align} H\left(q, \frac{\partial W}{\partial q}\right) = E \end{align}

SS 求运动

  1. 轨道方程:
Wηα=ξα=常数,α=2,3,,s\frac{\partial W}{\partial \eta_\alpha} = \xi_\alpha = \text{常数}, \quad \alpha = 2,3,\cdots,s

s1s - 1 个曲面方程的交线就是粒子运动的轨道.

  1. 运动方程:
WE=tt0=f1(q,E,η)\frac{\partial W}{\partial E} = t - t_0 = f_1 (q, E, \eta)
  1. 动量:pα=W/qαp_\alpha = \partial W / \partial q_\alpha

作用量-角变量J=pdqJ = \oint p \,\mathrm{d}q(绝热不变量),w˙=ν=H/J\dot{w} = \nu = \partial H / \partial J.


第九章:哈密顿理论的应用

连续体系的拉格朗日方程

拉格朗日密度 L(η,μη)\mathcal{L}(\eta, \partial_\mu \eta)

L=Ld3x\begin{align} L = \int \mathcal{L} \,\mathrm{d}^3 x \end{align}

连续体系的拉格朗日方程

LηaμL(μηa)=0\begin{align} \frac{\partial \mathcal{L}}{\partial \eta_a} - \partial_\mu \frac{\partial \mathcal{L}}{\partial (\partial_\mu \eta_a)} = 0 \end{align}

弹性棒纵振动L=12ρu˙212Y(u)2\mathcal{L} = \frac{1}{2}\rho \dot{u}^2 - \frac{1}{2}Y (u')^2,波动方程 u¨=v2u\ddot{u} = v^2 u''v=Y/ρv = \sqrt{Y/\rho}.

从哈-雅方程到薛定谔方程

薛定谔变换 ψ=eiS/\psi = e^{\mathrm{i}S/\hbar} 将哈-雅方程转化为薛定谔方程:

(22m2+V)ψ=iψt\begin{align} \left( -\frac{\hbar^2}{2m} \nabla^2 + V \right) \psi = \mathrm{i}\hbar \frac{\partial \psi}{\partial t} \end{align}

WKB 近似0\hbar \to 0):ψ(x)1p(x)exp(±ip(x)dx)\psi(x) \approx \frac{1}{\sqrt{p(x)}} \exp\left( \pm \frac{\mathrm{i}}{\hbar} \int p(x) \,\mathrm{d}x \right).


物理理论力学复习拉格朗日力学哈密顿力学刚体

·文章标题:理论力学总复习

·文章作者:NeoWangKing

·文章概要:本文是理论力学课程的完整复习笔记,整合了全部十章内容:牛顿力学、拉格朗日方程、两体问题与中心势场、刚体、非惯性参考系、多自由度微振动、阻尼运动、哈密顿理论、哈密顿理论的应用以及流体. 按知识模块系统梳理,包含核心公式、推导思路和关键物理图像.

·文章链接:https://www.neowangking.top/posts/physics/theoreticalmechanics/00-theory-mechanics-complete-review[点击复制]

·上次修改:


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