非保守系还有广义能量一说吗 (什么是广义能量 )

• 原文: https://zhuanlan.zhihu.com/p/50881658
• 发布日期: 2018-11-27
• 分类: 经典力学 / 分析力学

让我们先引入能量积分和广义能量积分的概念(不需要这段可以直接跳到符号▲处):

先直接点明三点:

i.二者都是在完整的的保守力系下引入的.(故使用保守系的方程 \frac{\partial L}{\partial {q}_{\alpha }=\frac{d}{dt}\frac{\partial L}{\partial {\dot{q}_{\alpha } )

ii.在i的条件下若约束是稳定的就可以推出能量积分,若约束不稳定则可以的到形式相似的广义能量积分.(稳定即约束不显含时间t,或笛卡尔坐标不显含时间,即{\vec{r}_{i}={\vec{r}_{i}\left( {q}_{1},{q}_{2},...,{q}_{s} \right) )

iii.后面基本都会涉及到这一等式: \sum\limits_{\alpha =1}^{s}{\left( \frac{\partial T}{\partial {q}_{\alpha }{\dot{q}_{\alpha }+\frac{\partial T}{\partial {\dot{q}_{\alpha }{\ddot{q}_{\alpha } \right)}=\frac{d}{dt}T ，这实际上要求 $\frac{\partial T}{\partial t}=0$ .稳定约束自然不用说: \frac{\partial {\vec{r}_{i}{\partial t}=0 说明 $T$ 肯定不显含时间; 至于非稳定约束虽说 \frac{\partial {\vec{r}_{i}{\partial t}\ne 0 ,但还是要求$\frac{\partial T}{\partial t}=0$才能给出广义能量的表达式.

好的现在开始只有完整的保守力系这一条件:

先将坐标改为广义坐标: {\dot{\vec{r}_{i}=\sum\limits_{\alpha =1}^{s}{\frac{\partial {\vec{r}_{i}{\partial {q}_{\alpha }{\dot{q}_{a}+\frac{\partial {\vec{r}_{i}{\partial t} #s为自由度数

那么动能 $T$ 就可展开为下面三项:

\begin{align} & T=\sum\limits_{i=1}^{n}{\frac{1}{2}{m}_{i}{\dot{\vec{r}_{i}^{2}=\sum\limits_{i=1}^{n}{\frac{1}{2}{m}_{i}\left( \sum\limits_{\alpha =1}^{s}{\frac{\partial {\vec{r}_{i}{\partial {q}_{\alpha }{\dot{q}_{a}+\frac{\partial {\vec{r}_{i}{\partial t} \right)\left( \sum\limits_{\beta =1}^{s}{\frac{\partial {\vec{r}_{i}{\partial {q}_{\beta }{\dot{q}_{\beta }+\frac{\partial {\vec{r}_{i}{\partial t} \right)} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sum\limits_{i=1}^{n}{\frac{1}{2}{m}_{i}\left[ \sum\limits_{\alpha =1}^{s}{\frac{\partial {\vec{r}_{i}{\partial {q}_{\alpha }{\dot{q}_{a}\cdot \sum\limits_{\beta =1}^{s}{\frac{\partial {\vec{r}_{i}{\partial {q}_{\beta }{\dot{q}_{\beta }+2\sum\limits_{\alpha =1}^{s}{\frac{\partial {\vec{r}_{i}{\partial {q}_{\alpha }{\dot{q}_{a}\cdot \frac{\partial {\vec{r}_{i}{\partial t}+{\left( \frac{\partial {\vec{r}_{i}{\partial t} \right)}^{2} \right]} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sum\limits_{\begin{smallmatrix} \alpha =1 \\ \beta =1 \end{smallmatrix}^{s}{\frac{1}{2}\sum\limits_{i=1}^{n}{m}_{i}\frac{\partial {\vec{r}_{i}{\partial {q}_{\alpha }\cdot \frac{\partial {\vec{r}_{i}{\partial {q}_{\beta }{\dot{q}_{\beta }{\dot{q}_{a}+\sum\limits_{\alpha =1}^{s}{\sum\limits_{i=1}^{n}{m}_{i}\frac{\partial {\vec{r}_{i}{\partial {q}_{\alpha }\cdot \frac{\partial {\vec{r}_{i}{\partial t}{\dot{q}_{a}+\frac{1}{2}\sum\limits_{i=1}^{n}{m}_{i}{\left( \frac{\partial {\vec{r}_{i}{\partial t} \right)}^{2} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sum\limits_{\begin{smallmatrix} \alpha =1 \\ \beta =1 \end{smallmatrix}^{s}{\frac{1}{2}{a}_{\alpha \beta }{\dot{q}_{\beta }{\dot{q}_{a}+\sum\limits_{\alpha =1}^{s}{a}_{\alpha o}{\dot{q}_{a}+\frac{1}{2}{a}_{oo}={T}_{2}+{T}_{1}+{T}_{0} \\ \end{align}

$T_2,T_1,T_0$ 即分别为广义速度的二次项,一次项和零次项.#这里并不要求保守系

其中简写记号的定义: \left\{ \begin{align} & {a}_{\alpha \beta }=\sum\limits_{i=1}^{n}{m}_{i}\frac{\partial {\vec{r}_{i}{\partial {q}_{\alpha }\cdot \frac{\partial {\vec{r}_{i}{\partial {q}_{\beta } \\ & {a}_{\alpha o}=\sum\limits_{i=1}^{n}{m}_{i}\frac{\partial {\vec{r}_{i}{\partial {q}_{\alpha }\cdot \frac{\partial {\vec{r}_{i}{\partial t} \\ & {a}_{oo}=\sum\limits_{i=1}^{n}{m}_{i}\frac{\partial {\vec{r}_{i}{\partial t}\cdot \frac{\partial {\vec{r}_{i}{\partial t} \\ \end{align} \right.

接下来对保守系的方程 \frac{\partial L}{\partial {q}_{\alpha }=\frac{d}{dt}\frac{\partial L}{\partial {\dot{q}_{\alpha } 做一些处理：

其中 $L=T-V$

$T$ 如上面分析是广义坐标 $q$ 和广义速度 {\dot{q} 和时间 $t$ 的函数,然而实际上后面的推导都要求 $T$ 不显含时间 $t$ .

$V$ 则仅仅是广义坐标 $q$ 的函数,这是不难理解的,因为势能是形容系统各个部分相互作用的函数,所以没有外场的话将仅仅由质点系内质点的相对位置决定.

将 $L$ 拆开并乘上 {\dot{q}_{\alpha }可得\frac{\partial T}{\partial {q}_{\alpha }{\dot{q}_{\alpha }-\frac{\partial V}{\partial {q}_{\alpha }{\dot{q}_{\alpha }=\frac{d}{dt}\frac{\partial T}{\partial {\dot{q}_{\alpha }{\dot{q}_{\alpha }

对等式右边使用分部微分 \frac{\partial T}{\partial {q}_{\alpha }{\dot{q}_{\alpha }-\frac{\partial V}{\partial {q}_{\alpha }{\dot{q}_{\alpha }=\frac{d}{dt}\left( \frac{\partial T}{\partial {\dot{q}_{\alpha }{\dot{q}_{\alpha } \right)-\frac{\partial T}{\partial {\dot{q}_{\alpha }{\ddot{q}_{\alpha }

整理后对 $\alpha$ 求和 \sum\limits_{\alpha =1}^{s}{\left( \frac{\partial T}{\partial {q}_{\alpha }{\dot{q}_{\alpha }+\frac{\partial T}{\partial {\dot{q}_{\alpha }{\ddot{q}_{\alpha } \right)}-\sum\limits_{\alpha =1}^{s}{\frac{\partial V}{\partial {q}_{\alpha }{\dot{q}_{\alpha }=\sum\limits_{\alpha =1}^{s}{\frac{d}{dt}\left( \frac{\partial T}{\partial {\dot{q}_{\alpha }{\dot{q}_{\alpha } \right)}

从这里开始准备推导能量积分,故添加稳定约束这一条件:

稳定约束 \Rightarrow \frac{\partial {\vec{r}_{i}{\partial t}=0\Rightarrow T={T}_{2} 即动能是广义速度的二次齐次函数.

这里要清楚一个齐次函数的欧拉定理： $T_2$ 是 {\dot{q} 的二次齐次函数 \Rightarrow \frac{\partial {T}_{2}{\partial {\dot{q}_{\alpha }{\dot{q}_{\alpha }=2{T}_{2}

#上述定理最后会给出证明

则式子 \sum\limits_{\alpha =1}^{s}{\left( \frac{\partial T}{\partial {q}_{\alpha }{\dot{q}_{\alpha }+\frac{\partial T}{\partial {\dot{q}_{\alpha }{\ddot{q}_{\alpha } \right)}-\sum\limits_{\alpha =1}^{s}{\frac{\partial V}{\partial {q}_{\alpha }{\dot{q}_{\alpha }=\sum\limits_{\alpha =1}^{s}{\frac{d}{dt}\left( \frac{\partial T}{\partial {\dot{q}_{\alpha }{\dot{q}_{\alpha } \right)}

可以写为 \frac{d}{dt}{T}_{2}-\frac{d}{dt}V=\frac{d}{dt}\sum\limits_{\alpha =1}^{s}{\left( \frac{\partial {T}_{2}{\partial {\dot{q}_{\alpha }{\dot{q}_{\alpha } \right)}=\frac{d}{dt}\left( 2{T}_{2} \right)

整理得到$\frac{d}{dt}\left(T_2+V\right)=0\Rightarrow T_2+V=E_{const}$这就是能量积分

不难看出这对应着牛顿力学的机械能守恒.

总结能量积分存在条件:稳定约束的完整保守系统(势能不显含时)

从这里开始准备推导广义能量积分,故添加非稳定约束这一条件:

非稳定约束 \Rightarrow \frac{\partial {\vec{r}_{i}{\partial t}\ne 0\Rightarrow T={T}_{2}+{T}_{1}+{T}_{0}

则式子 \sum\limits_{\alpha =1}^{s}{\left( \frac{\partial T}{\partial {q}_{\alpha }{\dot{q}_{\alpha }+\frac{\partial T}{\partial {\dot{q}_{\alpha }{\ddot{q}_{\alpha } \right)}-\sum\limits_{\alpha =1}^{s}{\frac{\partial V}{\partial {q}_{\alpha }{\dot{q}_{\alpha }=\sum\limits_{\alpha =1}^{s}{\frac{d}{dt}\left( \frac{\partial T}{\partial {\dot{q}_{\alpha }{\dot{q}_{\alpha } \right)}

可以写为\frac{d}{dt}T-\frac{d}{dt}V=\frac{d}{dt}\sum\limits_{\alpha =1}^{s}{\left( \frac{\partial T}{\partial {\dot{q}_{\alpha }{\dot{q}_{\alpha } \right)} #实际上这里要求了 $T$ 不显含 $t$

分析等号右边:\begin{align} & \frac{d}{dt}\sum\limits_{\alpha =1}^{s}{\left( \frac{\partial T}{\partial {\dot{q}_{\alpha }{\dot{q}_{\alpha } \right)}=\frac{d}{dt}\sum\limits_{\alpha =1}^{s}{\left( \frac{\partial {T}_{2}{\partial {\dot{q}_{\alpha }{\dot{q}_{\alpha } \right)}+\frac{d}{dt}\sum\limits_{\alpha =1}^{s}{\left( \frac{\partial {T}_{1}{\partial {\dot{q}_{\alpha }{\dot{q}_{\alpha } \right)}+\frac{d}{dt}\sum\limits_{\alpha =1}^{s}{\left( \frac{\partial {T}_{0}{\partial {\dot{q}_{\alpha }{\dot{q}_{\alpha } \right)} \\ & \frac{d}{dt}\sum\limits_{\alpha =1}^{s}{\left( \frac{\partial T}{\partial {\dot{q}_{\alpha }{\dot{q}_{\alpha } \right)}=\frac{d}{dt}\left( 2{T}_{2} \right)+\frac{d}{dt}{T}_{1}+0 \\ \end{align}

综上各式得到: $\frac{d}{dt}\left(T_2-T_0+V\right)=0\Rightarrow T_2-T_0+V=H_{const}$ 这就是广义能量积分

=======================

可以看出广义能量积分只是形式上和能量积分一样,而 $T_2-T_0$ 不是动量 $H_{const}$ 也不是能量.

广义能量积分的存在条件:完整保守系统下$T$是广义速度的二次非齐次函数且不显含$t$.(排除能量积分这一情况,势能不显含时)

那么达到上述条件的系统大概有如下几种情况:

i.是非稳定约束,但匀速(或匀角速度).就是说可以将坐标表达为: {\vec{r}_{i}={\vec{v}_{0}t+{\vec{r}'\left( {q}_{1},{q}_{2},...,{q}_{s} \right) 这么一来虽然 \frac{\partial {\vec{r}_{i}{\partial t}=const\ne 0 但却能同时保证 $\frac{\partial T}{\partial t}=0$ 且 $T$ 是广义速度的二次非齐次函数.

ii.虽然 {\vec{r}_{i},\frac{\partial {\vec{r}_{i}{\partial t},\frac{\partial {\vec{r}_{i}{\partial {q}_{\alpha } 都显含时间, 但是有可能在 \frac{\partial {\vec{r}_{i}{\partial {q}_{\alpha } 或 \frac{\partial {\vec{r}_{i}{\partial t} 在内积求和的过程中抵消了 $t$ 的这种情况也能满足上述条件.

#下面括号内的内容是个人迷思, 请批判性的阅读.

[对应着牛顿力学的什么呢? 感觉上应该是取约束参考系的机械能守恒, 因为 $T_2$从表达式来看就知道是相对约束静止的动参考系的动能表达式,而 $V$ 则是地面静止参考系下的势能,所以可以考虑 $-T_0$ 作为附加项看作势能项的一部分,这两项的和 $V^{\prime }=-T_0+V$ 或可看作动参考系下的势能项(按照这门学科的命名风格,或该称呼它广义势能?).动参考系本身具有惯性力这一附加项,所以 $-T_0$ 或看作是惯性力对应的势能.这样一来的话自然会有一个类似于机械能守恒的式子.也就是说广义能量反映的是动参考系下的机械能守恒.这还算是蛮符合直觉的,因为在动参考系下,约束就是稳定约束了.你可能会说非稳定约束是匀速情况不具有惯性力,但这其实没有产生矛盾,从定义式 {T}_{0}=\frac{1}{2}\sum\limits_{i=1}^{n}{m}_{i}{\left( \frac{\partial {\vec{r}_{i}{\partial t} \right)}^{2} 就可以看出来,其实这一项此时是个常数,算不算上他最后都会得到一个守恒量.]

▲下面将尝试讨论非保守系下的广义能量的意义:

那自然就要用回基本形式的方程了,即 \frac{d}{dt}\frac{\partial T}{\partial {\dot{q}_{\alpha }-\frac{\partial T}{\partial {q}_{\alpha }={Q}_{\alpha }

$Q_{\alpha }$ 是广义主动力,现在将其分为保守部分和非保守部分 $Q_{\alpha }=-{\frac{\partial V}{\partial q_{\alpha }+Q^{\prime }}}_a$

得到方程为 \frac{d}{dt}\frac{\partial T}{\partial {\dot{q}_{\alpha }-\frac{\partial T}{\partial {q}_{\alpha }=-\frac{\partial V}{\partial {q}_{\alpha }+{Q}'}_{a} ,和前面比仅多出一项 {Q}'}_{a}

按照前面整理可以得到:\sum\limits_{\alpha =1}^{s}{\frac{d}{dt}\left( \frac{\partial T}{\partial {\dot{q}_{\alpha }{\dot{q}_{\alpha } \right)}-\sum\limits_{\alpha =1}^{s}{\left( \frac{\partial T}{\partial {q}_{\alpha }{\dot{q}_{\alpha }+\frac{\partial T}{\partial {\dot{q}_{\alpha }{\ddot{q}_{\alpha } \right)}+\frac{d}{dt}V=\sum\limits_{\alpha =1}^{s}{Q}'}_{a}{\dot{q}_{\alpha }

从这里开始准备推导稳定约束这一条件下的能量概念:

稳定约束 \Rightarrow \frac{\partial {\vec{r}_{i}{\partial t}=0\Rightarrow T={T}_{2} 即动能是广义速度的二次齐次函数.

则式子 \sum\limits_{\alpha =1}^{s}{\frac{d}{dt}\left( \frac{\partial {T}_{2}{\partial {\dot{q}_{\alpha }{\dot{q}_{\alpha } \right)}-\sum\limits_{\alpha =1}^{s}{\left( \frac{\partial {T}_{2}{\partial {q}_{\alpha }{\dot{q}_{\alpha }+\frac{\partial {T}_{2}{\partial {\dot{q}_{\alpha }{\ddot{q}_{\alpha } \right)}+\frac{d}{dt}V=\sum\limits_{\alpha =1}^{s}{Q}'}_{a}{\dot{q}_{\alpha }

可以写为 \frac{d}{dt}\left( 2{T}_{2} \right)-\frac{d}{dt}{T}_{2}+\frac{d}{dt}V=\sum\limits_{\alpha =1}^{s}{Q}'}_{a}{\dot{q}_{\alpha }\Rightarrow \frac{d}{dt}\left( {T}_{2}+V \right)=\sum\limits_{\alpha =1}^{s}{Q}'}_{a}{\dot{q}_{\alpha }

$still$, 我们还是默认了 $T$ 不显含 $t$ (实在是觉得重要,所以每次啰嗦了一下.)

即 d\left( T+V \right)=\sum\limits_{\alpha =1}^{s}{Q}'}_{a}{\dot{q}_{\alpha }dt 不难看出这实际上对应着牛顿力学的功能原理.

$Lastly!$ 讨论到整篇文章的标题了,非保守系的广义能量:

先加入条件非稳定约束 \Rightarrow \frac{\partial {\vec{r}_{i}{\partial t}\ne 0\Rightarrow T={T}_{2}+{T}_{1}+{T}_{0}

则式子 \sum\limits_{\alpha =1}^{s}{\frac{d}{dt}\left( \frac{\partial T}{\partial {\dot{q}_{\alpha }{\dot{q}_{\alpha } \right)}-\sum\limits_{\alpha =1}^{s}{\left( \frac{\partial T}{\partial {q}_{\alpha }{\dot{q}_{\alpha }+\frac{\partial T}{\partial {\dot{q}_{\alpha }{\ddot{q}_{\alpha } \right)}+\frac{d}{dt}V=\sum\limits_{\alpha =1}^{s}{Q}'}_{a}{\dot{q}_{\alpha }

可以写为 \frac{d}{dt}\left( 2{T}_{2} \right)+\frac{d}{dt}{T}_{1}-\frac{d}{d}\left( {T}_{2}+{T}_{\text{1}\text{+}{T}_{0} \right)+\frac{d}{dt}V=\sum\limits_{\alpha =1}^{s}{Q}'}_{a}{\dot{q}_{\alpha }

即 d\left( {T}_{2}-{T}_{0}+V \right)=\sum\limits_{\alpha =1}^{s}{Q}'}_{a}{\dot{q}_{\alpha }dt

$well,itstillworks$, 通过前面的长篇大论不难看出这是对应着动参考系下的功能原理.

关于前面说好的齐次函数的欧拉定理的证明:

若满足式子 $f\left(a{x}_1,a{x}_2,...,a{x}_s\right)=a^{n}f\left(x_1,x_2,...,x_s\right)$

则称 $f$ 是 $x_{\alpha }$ 的n次齐次函数.

对于 $g\left(x\right)=a{x}^n$

不难想象式子 $\frac{\partial g}{\partial x}=n\frac{g}{x}$ 是成立的,当然了, $\frac{\partial g}{\partial x}x=ng$ 也成立

那么很自然的推广到 $f\left(x_1,x_2,...,x_s\right)$

就有 \sum\limits_{\alpha =1}^{s}{\frac{\partial f}{\partial {x}_{\alpha }{x}_{\alpha }=nf 好的这就是著名的齐次函数的欧拉定理. $donehere$.

#2019.02.01更新一个证明方法(总觉得上面那个到底有些敷衍):

齐次函数满足条件:$f\left(a{x}_1,a{x}_2,...,a{x}_s\right)=a^{n}f\left(x_1,x_2,...,x_s\right)$

式子两边对 $a$ 求导得到: \sum\limits_{\alpha =1}^{s}{\frac{\partial f\left( a{x}_{1},a{x}_{2},...,a{x}_{s} \right)}{\partial a{x}_{\alpha }\cdot {x}_{\alpha }=n{a}^{n-1}f\left( {x}_{1},{x}_{2},...,{x}_{s} \right)

接下来令 $a=1$ 得到: \sum\limits_{\alpha =1}^{s}{\frac{\partial f}{\partial {x}_{\alpha }\cdot {x}_{\alpha }=nf

desserts：你能证明 \sum\limits_{\beta \text{=1}^{s}{\frac{\partial }{\partial {x}_{\beta }\frac{\partial f}{\partial {x}_{\alpha }\cdot {x}_{\beta }=\left( n-1 \right)}\frac{\partial f}{\partial {x}_{\alpha } 吗?

正樹：齐次函数的欧拉定理与其一推论