# 动能

## 古典力学

$E_k = \frac{1}{2}mv^2$

$E_k = \frac{p^2}{2m}$

### 推导与定义

$W = \int \vec{F} \cdot d\vec{s}$

$\vec{F} = \frac{d\vec{p}}{dt}$

$\vec{p}=m\vec{v}$

$W = \int \frac{d\vec{p}}{dt} \cdot d\vec{s} = \int m \frac{d\vec{v}}{dt} \cdot d\vec{s} = \int m \vec{v} \cdot d\vec{v} =\frac{1}{2} \int m d (\vec{v} \cdot \vec{v}) = \frac{1}{2}mv^2 + C_0$

$E_k = \frac{1}{2}mv^2$

### 自转的物体

$E_r = \frac{1}{2} \int v^2 dm = \frac{1}{2} \int r^2 \omega^2 dm = \frac{1}{2} \omega^2 \int r^2 dm = \frac{1}{2} I \omega^2$

## 相对论

$E_\text{k} = \int \mathbf{v} \cdot d \mathbf{p}= \int \mathbf{v} \cdot d (m \gamma \mathbf{v}) = m \gamma \mathbf{v} \cdot \mathbf{v} - \int m \gamma \mathbf{v} \cdot d \mathbf{v} = m \gamma v^2 - \frac{m}{2} \int \gamma d (v^2)$

\begin{align} E_\text{k} &= m \gamma v^2 - \frac{- m c^2}{2} \int \gamma d (1 - v^2/c^2) \\ &= m \gamma v^2 + m c^2 (1 - v^2/c^2)^{1/2} - E_0 \end{align}

\begin{align} E_\text{k} &= m \gamma (v^2 + c^2 (1 - v^2/c^2)) - E_0 \\ &= m \gamma (v^2 + c^2 - v^2) - E_0 \\ &= m \gamma c^2 - E_0 \end{align}

$E_0 = m c^2 \,$

$E_\text{k} = m \gamma c^2 - m c^2 = \frac{m c^2}{\sqrt{1 - v^2/c^2}} - m c^2$

### 極限

$\lim_{v\rightarrow c}E_\text{k}=\infty$

\begin{align} E_\text{k} &= \frac{m c^2}{\sqrt{1 - (v/c)^2}} - m c^2\\ &= mc^2 (1 + \frac{1}{2} v^2/c^2 + \frac{3}{8} v^4/c^4 +\cdots) - m c^2\\ &\approx mc^2 (1 + \frac{1}{2} v^2/c^2) - m c^2\\ &= \frac{1}{2} m v^2 \end{align}

## 参考文献

1. ^ 1.0 1.1 赵志敏. 高中物理竞赛教程.基础篇. 复旦大学出版社. 2011年10月: P139. ISBN 978-7-309-08251-7.