2026.9.10 动能求导等于动量的思考。
1. 动能的变分与动量
\[T(v)=\frac12m\langle v^b,v\rangle\] \[T(v+\delta v)=\frac12m\langle (v+\delta v)^b,v+\delta v\rangle\] \[=\frac12m\left(\left(\langle v,v\rangle+2\langle v,\delta v\rangle\right)^b+\langle\delta v,\delta v\rangle\right) \qquad =\text{高阶小量}\] \[\delta T=m\langle v^b,\delta v\rangle\]固定
\[v=\begin{pmatrix}v_x\\v_y\end{pmatrix},\qquad \delta v=\begin{pmatrix}\delta v_x\\\delta v_y\end{pmatrix}\] \[\delta T=mv_x\delta v_x+mv_y\delta v_y\qquad \boxed{A}\]☆
\[=p_x\delta v_x+p_y\delta v_y\] \[=(p_x\quad p_y) \begin{pmatrix}\delta v_x\\\delta v_y\end{pmatrix} =\delta T:=\langle p,\delta v\rangle\]1.1 用偏导数表示动量
\[\delta T=\frac{\partial T}{\partial v_x}\delta v_x+\frac{\partial T}{\partial v_y}\delta v_y\] \[\therefore\quad \frac{\partial T}{\partial v_x}=mv_x\qquad \frac{\partial T}{\partial v_y}=mv_y\qquad \boxed{A}\] \[v_x\to v_x+\delta v_x\] \[v_y\to v_y+\delta v_y\]2. 从直角坐标到广义坐标
\[\delta T=p_x\delta v_x+p_y\delta v_y \qquad x=x(q_1,q_2,\cdots,q_n)\] \[=p_1\delta\dot q_1+p_2\delta\dot q_2\]广义速度：
\[\dot q_1,\ \dot q_2\ \cdots\] \[\delta T=m(\dot x\quad \dot y) \begin{pmatrix} \dfrac{\partial x}{\partial q_1}\delta\dot q_1+\dfrac{\partial x}{\partial q_1}\delta\dot q_2\\[6pt] \dfrac{\partial y}{\partial q_1}\delta\dot q_1+\dfrac{\partial y}{\partial q_1}\delta\dot q_2 \end{pmatrix}\] \[=m(\dot x\quad \dot y) \begin{pmatrix} \dfrac{\partial x}{\partial q_1} & \dfrac{\partial x}{\partial q_2}\\[6pt] \dfrac{\partial y}{\partial q_1} & \dfrac{\partial y}{\partial q_2} \end{pmatrix} \begin{pmatrix}\delta\dot q_1\\\delta\dot q_2\end{pmatrix}\] \[=\left( m\dot x\frac{\partial x}{\partial q_1}+m\dot y\frac{\partial y}{\partial q_1}\quad m\dot x\frac{\partial x}{\partial q_2}+m\dot y\frac{\partial y}{\partial q_2} \right) \begin{pmatrix}\delta\dot q_1\\\delta\dot q_2\end{pmatrix}\]最初：
\[\delta T=(p_x\quad p_y) \begin{pmatrix}\delta\dot x\\\delta\dot y\end{pmatrix}\] \[\therefore\quad \delta T=(p_1\quad p_2) \begin{pmatrix}\delta\dot q_1\\\delta\dot q_2\end{pmatrix} \qquad p_i\text{为广义动量}\]2.1 广义速度与坐标变换
\[\dot x=\frac{\partial x}{\partial q_1}\dot q_1+\frac{\partial x}{\partial q_2}\dot q_2\]2.2 广义动量与坐标度量
\[m\dot x\frac{\partial x}{\partial q_1}+m\dot y\frac{\partial y}{\partial q_1}=p_1\]代入得
\[p_1=m\left(\frac{\partial x}{\partial q_1}\dot q_1+\frac{\partial x}{\partial q_2}\dot q_2\right)\frac{\partial x}{\partial q_1} +m\left(\frac{\partial y}{\partial q_1}\dot q_1+\frac{\partial y}{\partial q_2}\dot q_2\right)\frac{\partial y}{\partial q_1}\] \[=m\left[\left(\frac{\partial x}{\partial q_1}\right)^2+\left(\frac{\partial y}{\partial q_1}\right)^2\right]\dot q_1 +m\left(\frac{\partial x}{\partial q_1}\frac{\partial x}{\partial q_2} +\frac{\partial y}{\partial q_1}\frac{\partial y}{\partial q_2}\right)\dot q_2\] \[(p_1\quad p_2)=m(\dot q_1\quad\dot q_2) \begin{pmatrix} \left(\dfrac{\partial x}{\partial q_1}\right)^2+\left(\dfrac{\partial y}{\partial q_1}\right)^2 & \dfrac{\partial x}{\partial q_1}\dfrac{\partial x}{\partial q_2}+\dfrac{\partial y}{\partial q_1}\dfrac{\partial y}{\partial q_2}\\[8pt] \dfrac{\partial x}{\partial q_1}\dfrac{\partial x}{\partial q_2}+\dfrac{\partial y}{\partial q_1}\dfrac{\partial y}{\partial q_2} & \left(\dfrac{\partial x}{\partial q_2}\right)^2+\left(\dfrac{\partial y}{\partial q_2}\right)^2 \end{pmatrix}\] \[(p_1\quad p_2)=m(\dot q_1\quad\dot q_2) \begin{pmatrix}g_{11}&g_{12}\\g_{21}&g_{22}\end{pmatrix}\] \[T=\frac12m(\dot q_1\quad\dot q_2) \begin{pmatrix} g_{11}(q_1,q_2)&g_{12}(q_1,q_2)\\ g_{21}(q_1,q_2)&g_{22}(q_1,q_2) \end{pmatrix} \begin{pmatrix}\dot q_1\\\dot q_2\end{pmatrix}\] \[T=\frac12m\left(g_{11}(\dot q_1)^2+2g_{12}\dot q_1\dot q_2+g_{22}(\dot q_2)^2\right)\] \[\frac{\partial T}{\partial\dot q_1} =\frac12m\left(g_{11}\dot q_1+2g_{12}\dot q_2\right)\] \[p_1=m\left(g_{11}\dot q_1+g_{12}\dot q_2\right)\] \[\therefore\quad p_1=\frac{\partial T}{\partial\dot q_1} \qquad \star\ \text{广义动量定义}\]3. 加速度与动能微分的关系
\[a^b dx=v^b dv=d\left(\frac12v^2\right)\qquad \boxed{A}\]4. 速度与动量的 Legendre 对偶
\[p=\frac{\partial T}{\partial v}\] \[v=\frac{\partial T^*}{\partial p}\] \[\boxed{ F^b\frac{d(mx)}{dt}=p^b\frac{dp}{dt} =\frac{d}{dt}\left(\frac12p^2\right) =\frac12(mv)^2 } \qquad \text{因为v空间被拉伸m倍}\] \[\therefore\quad /m:=\frac12mv^2\]$\boxed{A}$