S2-为什么狭义相对论中的E的定义是这样的?

引入问题

在狭义相对论中，我们定义四维动量

\begin{align} p^{\mu} & =m\frac{dx^{\mu}}{d\tau}=m\left( \frac{dt}{d\tau},\frac {d\mathbf{x}}{d\tau}\right) \tag{S2-1}\\ & =\left( \gamma m,\gamma m\mathbf{v}\right) =\left( \gamma m,\gamma \mathbf{p}\right)\tag{S2-2} \end{align}

至此，所有的结果都是平庸的，不平庸的是

\begin{equation} E\overset{def}{=}\gamma m \tag{S2-3} \end{equation}

但是为什么这样定义？

分析与解决

相对论性的自由粒子的拉式量

\begin{equation} L=-m\sqrt{1-\mathbf{v}^{2}} \tag{S2-4} \end{equation}

对于拉式量(4)关于时间平移不变性$\overline{t}=t+\epsilon$相应的广义坐标的变化

\begin{equation} q\left( t+\epsilon\right) =q\left( t\right) +\epsilon\eta \tag{S2-5} \end{equation}

且

\begin{equation} \eta=\lim_{\epsilon\rightarrow0}\frac{q\left( t+\epsilon\right) -q\left( t\right) }{\epsilon}=\dot{q} \tag{S2-6} \end{equation}

对于作用量

\begin{align} S\left( \epsilon\right) & =\int_{t_{1}}^{t_{2}}L\left( q\left( t\right) +\epsilon\eta,\dot{q}+\epsilon\dot{η},t\right) dt \tag{S2-7}\\ S\left( 0\right) & =\int_{t_{1}}^{t_{2}}L\left( q,\dot{q},t\right) dt \tag{S2-8} \end{align}

按$\epsilon$展开，

\begin{align} S\left( \epsilon\right) -S\left( 0\right) \tag{S2-9}& =\int_{t}^{t+\epsilon }L\left( q+\epsilon\eta,\dot{q}+\epsilon\dot{η},t\right) -L\left( q,\dot{q},t\right) dt \\ &\tag{S2-10} =\int_{t}^{t+\epsilon}\epsilon\left( \frac{\partial L}{\partial q} \eta+\frac{\partial L}{\partial\dot{q}}\dot{η}\right) dt\\ & \tag{S2-11}=\epsilon\int_{t}^{t+\epsilon}\frac{\partial L}{\partial q}\eta+\frac{d} {dt}\left( \frac{\partial L}{\partial\dot{q}}\eta\right) -\eta\frac{d} {dt}\left( \frac{\partial L}{\partial\dot{q}}\right) dt\\ & \tag{S2-12}=\epsilon\int_{t}^{t+\epsilon}\frac{\partial L}{\partial q}\eta-\eta\frac {d}{dt}\left( \frac{\partial L}{\partial\dot{q}}\right) dt+\epsilon\int _{t}^{t+\epsilon}\frac{d}{dt}\left( \frac{\partial L}{\partial\dot{q}} \eta\right) dt\\ & \tag{S2-13}=\epsilon\int_{t}^{t+\epsilon}\left[ \frac{\partial L}{\partial q}-\frac {d}{dt}\left( \frac{\partial L}{\partial\dot{q}}\right) \right] \eta dt+\epsilon\left( \frac{\partial L}{\partial\dot{q}}\eta\right) |_{t_{1} }^{t_{2}} \end{align}

当$q$满足运动方程时

\begin{equation} S\left( \epsilon\right) -S\left( 0\right) =\epsilon\left( \frac{\partial L}{\partial\dot{q}}\eta\right) |_{t_{1}}^{t_{2}} \tag{S2-14} \end{equation} \begin{equation} \left( \frac{\partial L}{\partial\dot{q}}\eta\right) |_{t}^{t+\epsilon }=\frac{S\left( \epsilon\right) -S\left( 0\right) }{\epsilon} \tag{S2-15} \end{equation}

即

\begin{align} \lim_{\epsilon\rightarrow0}\frac{S\left( \epsilon\right) -S\left( 0\right) }{\epsilon} & =\frac{\partial S\left( \epsilon\right) }{\partial\epsilon }|_{\epsilon0}=\int_{t}^{t+\epsilon}\frac{\partial L\left( \epsilon\right) }{\partial\epsilon}dt\nonumber\\ \tag{S2-16} & =\int_{t}^{t+\epsilon}\frac{\partial L}{\partial t}dt=L|_{t}^{t+\epsilon }% \end{align}

联立式(15)与式(16)

\begin{equation} \left( \frac{\partial L}{\partial\dot{q}}\eta\right) |_{t}^{t+\epsilon}=L|_{t}^{t+\epsilon} \tag{S2-17} \end{equation}

所以守恒荷为

\begin{equation} Q=\frac{\partial L}{\partial\dot{q}}\dot{q}-L \tag{S2-18} \end{equation}

由式(4)

\begin{align} Q & =\frac{\partial L}{\partial\mathbf{v}}\mathbf{v}-L=H\nonumber\\ \tag{S2-19} & =\frac{m\mathbf{v}^{2}}{\sqrt{1-\mathbf{v}^{2}}}+m\sqrt{1-\mathbf{v}^{2}% }\nonumber\\ \tag{S2-20} & =\frac{m}{\sqrt{1-\mathbf{v}^{2}}}=\gamma m \end{align}

可以看出这就是哈密顿量.

Noether定理要求的时间平移不变性对应能量守恒。