# 守恆量

## 動量

${\displaystyle \mathbf {F} =0}$

${\displaystyle \mathbf {F} ={\frac {\mathrm {d} \mathbf {p} }{\mathrm {d} t}}}$

## 角動量

${\displaystyle {\boldsymbol {\tau }}=0}$

${\displaystyle {\boldsymbol {\tau }}={\frac {\mathrm {d} {\boldsymbol {\ell }}}{\mathrm {d} t}}}$

## 能量

${\displaystyle E=T+V}$

${\displaystyle T=mv^{2}/2}$

${\displaystyle \mathbf {F} =-\nabla V}$

${\displaystyle {\frac {\mathrm {d} E}{\mathrm {d} t}}=m\mathbf {v} \cdot {\frac {\mathrm {d} \mathbf {v} }{\mathrm {d} t}}+\mathbf {v} \cdot \nabla V=\mathbf {v} \cdot (m\mathbf {a} -\mathbf {F} )=0}$

## 能量函數

${\displaystyle {\mathcal {L}}=T-V}$

${\displaystyle {\frac {d}{dt}}\left({\frac {\partial {\mathcal {L}}}{\partial {\dot {q}}_{i}}}\right)-{\frac {\partial {\mathcal {L}}}{\partial q_{i}}}=0}$

${\displaystyle {\frac {d{\mathcal {L}}}{dt}}=\sum _{i}{\frac {\partial {\mathcal {L}}}{\partial q_{i}}}{\dot {q}}_{i}+\sum _{i}{\frac {\partial {\mathcal {L}}}{\partial {\dot {q}}_{i}}}{\ddot {q}}_{i}+{\frac {\partial {\mathcal {L}}}{\partial t}}}$

{\displaystyle {\begin{aligned}{\frac {d{\mathcal {L}}}{dt}}&=\sum _{i}{\frac {d}{dt}}\left({\frac {\partial {\mathcal {L}}}{\partial {\dot {q}}_{i}}}\right){\dot {q}}_{i}+\sum _{i}{\frac {\partial {\mathcal {L}}}{\partial {\dot {q}}_{i}}}{\ddot {q}}_{i}+{\frac {\partial {\mathcal {L}}}{\partial t}}\\&=\sum _{i}{\frac {d}{dt}}\left({\frac {\partial {\mathcal {L}}}{\partial {\dot {q}}_{i}}}{\dot {q}}_{i}\right)+{\frac {\partial {\mathcal {L}}}{\partial t}}\\\end{aligned}}}

${\displaystyle {\mathit {h}}\ {\stackrel {def}{=}}\ \sum _{i}{\frac {\partial {\mathcal {L}}}{\partial {\dot {q}}_{i}}}{\dot {q}}_{i}-{\mathcal {L}}}$

${\displaystyle {\frac {d{\mathit {h}}}{dt}}=-{\frac {\partial {\mathcal {L}}}{\partial t}}}$

## 參考文獻

1. ^ Morin, David. Introduction to classical mechanics: with problems and solutions. Cambridge University Press. 2008: 138. ISBN 9780521876223.
2. Goldstein, Herbert, Classical Mechanics 3rd, United States of America: Addison Wesley: pp. 2–5, 61, 312–324, 1980, ISBN 0201657023 （英语）