# 定常系統

## 應用

${\displaystyle T={\frac {1}{2}}mv^{2}}$

${\displaystyle \mathbf {v} ={\frac {d\mathbf {r} }{dt}}=\sum _{i}\ {\frac {\partial \mathbf {r} }{dq_{i}}}{\dot {q}}_{i}+{\frac {\partial \mathbf {r} }{dt}}}$

${\displaystyle T={\frac {1}{2}}m\sum _{i}\ \left({\frac {\partial \mathbf {r} }{\partial q_{i}}}{\dot {q}}_{i}+{\frac {\partial \mathbf {r} }{\partial t}}\right)^{2}}$

${\displaystyle T=T_{0}+T_{1}+T_{2}}$

${\displaystyle T_{0}={\frac {1}{2}}m\left({\frac {\partial \mathbf {r} }{\partial t}}\right)^{2}}$
${\displaystyle T_{1}=\sum _{i}\ m{\frac {\partial \mathbf {r} }{\partial t}}\cdot {\frac {\partial \mathbf {r} }{\partial q_{i}}}{\dot {q}}_{i}}$
${\displaystyle T_{2}=\sum _{i,j}\ {\frac {1}{2}}m{\frac {\partial \mathbf {r} }{\partial q_{i}}}\cdot {\frac {\partial \mathbf {r} }{\partial q_{j}}}{\dot {q}}_{i}{\dot {q}}_{j},\!}$

${\displaystyle T_{0}}$${\displaystyle T_{1}}$${\displaystyle T_{2}}$分別為廣義速度${\displaystyle {\dot {q}}_{i}}$的0次、1次、2次齊次函數。如果這系統是定常系統，位置不顯性地含時間，${\displaystyle {\frac {\partial \mathbf {r} }{\partial t}}=0}$，則只有${\displaystyle T_{2}}$不等於零。所以，${\displaystyle T=T_{2}}$，動能是廣義速度的2次齊次函數。

## 實例1：單擺

${\displaystyle {\sqrt {x^{2}+y^{2}}}-L=0}$

## 實例2：受驅擺

${\displaystyle x_{t}=x_{0}\cos \omega t}$

${\displaystyle {\sqrt {(x-x_{0}\cos \omega t)^{2}+y^{2}}}-L=0}$

## 參考文獻

1. ^ Goldstein, Herbert. Classical Mechanics 3rd. United States of America: Addison Wesley. 1980: pp. 25. ISBN 0201657023 （英语）.