$\renewcommand{\rm}{\mathrm} \newcommand{\bm}[1]{{\boldsymbol{#1}}} \newcommand{\ii}[0]{{\mathrm{i}}} \newcommand{\ee}[0]{{\mathrm{e}}} \newcommand{\upi}[0]{{\uppi}} \newcommand{\dag}[0]{{\dagger}} \renewcommand{\cal}{\mathcal} \renewcommand{\scr}{\mathscr} \newcommand{\frak}[1]{{\mathfrak{#1}}} \newcommand{\bb}[1]{{\mathbb{#1}}} \renewcommand{\sf}{\mathsf} \newcommand{\diag}[1]{{\mathrm{diag}\left(#1\right)}} \renewcommand{\d}{{\dd}} \newcommand{\const}[0]{{\mathrm{const}}} \newcommand{\eqref}[1]{{\mbox{式}\,\left(\ref{#1}\right)}}$

间隔不变性的一般证明

$c=1$ $\d s^2=\d t^2-\qty|\d \bm{r}|^2=\eta_{\mu\nu}\d x^\mu\d x^\nu$ $\d s^{\prime 2}=\d t^{\prime 2}-\qty|\d \bm{r}^\prime|^2=\eta_{\mu\nu}\d x^\mu\d x^\nu$ $x^{\prime\mu}=\phi^\mu\qty(x)$ $\pdv{\phi^\mu}{x^\nu}={U^\mu}_\nu\qty(x)$ $\d s^2$ $\d s^{\prime 2}$ $x$ $\d x$ $\d x^\mu=\epsilon^\mu\qty(x)$ $I\qty(v)=\eta_{\mu\nu}v^\mu v^\nu$ $\d s^2=I\qty(\epsilon\qty(x))$ $\d s^{\prime 2}=I\qty(U\qty(x)\epsilon\qty(x))$ $f_{\qty[\epsilon]}\qty(x)=\dfrac{I\qty(U\qty(x)\epsilon\qty(x))}{I\qty(\epsilon\qty(x))}$ $\epsilon:x\mapsto \d x$ $x$ $f_{\qty[\epsilon]}\qty(x)$ $\epsilon$ $x$ $f_{\qty[\epsilon]}\qty(x)$ $\dfrac{1}{f_{\qty[\epsilon]}\qty(x)}$ $f_{\epsilon}\qty(x)$ $f_{\qty[\epsilon]}\qty(x)$ $\mathop{\lim}\limits_{\epsilon\to0}f_{\qty[\epsilon]}\qty(x)$ $F\qty(x)$ $I\qty(U\qty(x)\epsilon\qty(x))=F\qty(x)I\qty(\epsilon\qty(x))+o\qty(\epsilon\qty(x)^2)$ $o\qty(\epsilon\qty(x)^2)=C^{\qty(3)}_{\mu\nu\rho}\epsilon^\mu\qty(x)\epsilon^\nu\qty(x)\epsilon^\rho\qty(x)+C^{\qty(4)}_{\mu\nu\rho\sigma}\epsilon^\mu\qty(x)\epsilon^\nu\qty(x)\epsilon^\rho\qty(x)\epsilon^\sigma\qty(x)+\cdots$ $C^{\qty(n)}$ $\epsilon$ $U$ $C^{\qty(n)}=C^{\qty(n)}\qty(U)$ $I\qty(\epsilon\qty(x))=0\Leftrightarrow I\qty(U\qty(x)\epsilon\qty(x))=0$ , 故有

\begin{matrix} (1) & \begin{array}{r} 0 = 0 + C_{μ ν ρ}^{(3)} ϵ^{μ} (x) ϵ^{ν} (x) ϵ^{ρ} (x) + \dots, \end{array} \end{matrix}

$o\qty(\epsilon\qty(x)^2)=0$ $C^{\qty(n)}=0$ $\epsilon$ $I\qty(\epsilon)=0$ $I\qty(U\qty(x)\epsilon\qty(x))$ $I\qty(\epsilon(x))$ $\epsilon$ $\epsilon$ $\eta_{\rho\sigma}{U^\rho}_\mu\qty(x){U^\sigma}_\nu\qty(x)=F\qty(x)\eta_{\mu\nu}$ $F$ $U$ 的相关性, 将其更明确地写为

\begin{matrix} (2) & η_{ρ σ} {U^{ρ}}_{μ} (x) {U^{σ}}_{ν} (x) = F (U (x), x) η_{μ ν} . \end{matrix}

$\qty{x}$ $\qty{x^\prime}$ 系而言, 有以下两式同时成立,

\begin{matrix} (3) & \begin{aligned} Δ x^{i} = u^{i} Δ x^{0}; \\ Δ x^{' i} = u^{' i} Δ x^{' 0}, \end{aligned} \end{matrix}

$u^i$ $u^{\prime i}$ $x$ $\Delta x$ $u^0=u^{\prime 0}=1$ , 则以上两式可以写为

\begin{matrix} (4) & \begin{aligned} Δ x^{μ} = u^{μ} Δ x^{0}; \\ Δ x^{' μ} = u^{' μ} Δ x^{' 0}, \end{aligned} \end{matrix}

$\Delta x^\mu=x^\mu-x^\mu_0$ $\Delta x^{\prime \mu}=x^{\prime \mu}-x^{\prime \mu}_0$ $x^\mu_0$ $x^{\prime \mu}_0$ $x^{\prime \mu}$ $x^0$ 求导得

\begin{matrix} (5) & \begin{aligned} \frac{d x^{' μ}}{d x^{0}} = \frac{\partial x^{' μ}}{\partial x^{ν}} \frac{d x^{ν}}{d x^{0}} = \frac{\partial x^{' μ}}{\partial x^{ν}} u^{ν}; \\ \frac{d x^{' μ}}{d x^{' 0}} = u^{' μ} = \frac{(\frac{d x^{' μ}}{d x^{0}})}{(\frac{d x^{' 0}}{d x^{0}})} = \frac{\frac{\partial x^{' μ}}{\partial x^{ν}} u^{ν}}{\frac{\partial x^{' 0}}{\partial x^{ν}} u^{ν}} . \end{aligned} \end{matrix}

$u^{\prime \mu}=\const$ $\pdv{x^{\prime \mu}}{x^\nu}u^\nu\propto\pdv{x^{\prime 0}}{x^\nu}u^\nu$ $\ln\qty(\pdv{x^{\prime \mu}}{x^\nu}u^\nu)=\ln\qty(\pdv{x^{\prime 0}}{x^\nu}u^\nu)+\const$ , 从而

\begin{matrix} (6) & \begin{aligned} \frac{d}{d x^{0}} \ln (\frac{\partial x^{' μ}}{\partial x^{ν}} u^{ν}) = \frac{d}{d x^{0}} \ln (\frac{\partial x^{' 0}}{\partial x^{ν}} u^{ν}), \\ \frac{\frac{\partial^{2} x^{' \underset{―}{μ}}}{\partial x^{ρ} \partial x^{σ}} u^{ρ} u^{σ}}{\frac{\partial x^{' \underset{―}{μ}}}{\partial x^{ν}} u^{ν}} = \frac{\frac{\partial^{2} x^{' 0}}{\partial x^{ρ} \partial x^{σ}} u^{ρ} u^{σ}}{\frac{\partial x^{' 0}}{\partial x^{ν}} u^{ν}} . \end{aligned} \end{matrix}

$\det\qty(\pdv{x^{\prime \mu}}{x^\nu})\neq 0$ $\xi^\mu=\pdv{x^{\prime \mu}}{x^\nu}u^\nu$ $u^\mu=\pdv{x^\mu}{x^{\prime \nu}}\xi^\nu$ $\eqref{惯性系之间的坐标变换导出的条件}$ 式得

\begin{matrix} (7) & \begin{array}{r} \frac{\frac{\partial^{2} x^{' \underset{―}{μ}}}{\partial x^{α} \partial x^{β}} \frac{\partial x^{α}}{\partial x^{' ρ}} \frac{\partial x^{β}}{\partial x^{' σ}} ξ^{ρ} ξ^{σ}}{ξ^{\underset{―}{μ}}} = \frac{\frac{\partial^{2} x^{' 0}}{\partial x^{α} \partial x^{β}} \frac{\partial x^{α}}{\partial x^{' ρ}} \frac{\partial x^{β}}{\partial x^{' σ}} ξ^{ρ} ξ^{σ}}{ξ^{0}} . \end{array} \end{matrix}

$4$ $\xi^\mu$ 不可能全为零, 因为如果那样, 就有

\begin{matrix} (8) & u^{0} = \frac{\partial x^{0}}{\partial x^{' μ}} ξ^{μ} = 0, \end{matrix}

$u^0$ $u^0=1$ $\eqref{4个\xi不可能全为零}$ $\eqref{惯性系之间的坐标变换再次变换后导出的条件}$ $\cal{Q}^\mu\qty(\xi)$ $\cal{Q}^0\qty(\xi)$ $4$ $\cal{Q}^\mu\qty(\xi)$ $\mu$ $4$ $\xi^\mu$ $4$ $\cal{Q}^\mu$ $\xi$ $\cal{Q}^\mu$ $\cal{Q}^\mu$ $\xi$ $\mu$ $\xi^\mu=0$ $\cal{Q}^\mu$ 的分子也必定为零, 所以

\begin{matrix} (9) & \begin{array}{r} \frac{\partial^{2} x^{' μ}}{\partial x^{α} \partial x^{β}} \frac{\partial x^{α}}{\partial x^{' ρ}} \frac{\partial x^{β}}{\partial x^{' σ}} ξ^{ρ} ξ^{σ} = \sum_{ρ, σ \neq μ} \frac{\partial^{2} x^{' μ}}{\partial x^{α} \partial x^{β}} \frac{\partial x^{α}}{\partial x^{' ρ}} \frac{\partial x^{β}}{\partial x^{' σ}} ξ^{ρ} ξ^{σ} = 0, \end{array} \end{matrix}

$x$ $\xi^\nu\qty(\nu\neq\mu)$ $u$ $x$ $\pdv{x^{\prime \mu}}{x^\alpha}{x^\beta}\pdv{x^\alpha}{x^{\prime \rho}}\pdv{x^\beta}{x^{\prime \sigma}}$ $\xi^\rho$ $\xi^\sigma$ $\xi^\rho$ $\xi^\sigma$ 的任意性可知,

\begin{matrix} (10) & \frac{\partial^{2} x^{' μ}}{\partial x^{α} \partial x^{β}} \frac{\partial x^{α}}{\partial x^{' ρ}} \frac{\partial x^{β}}{\partial x^{' σ}} = 0 (ρ, σ \neq 0), \end{matrix}

因此

\begin{matrix} (11) & \begin{aligned} Q^{μ} (ξ) & = \frac{\frac{\partial^{2} x^{' \underset{―}{μ}}}{\partial x^{α} \partial x^{β}} \frac{\partial x^{α}}{\partial x^{' ρ}} \frac{\partial x^{β}}{\partial x^{' σ}} ξ^{ρ} ξ^{σ}}{ξ^{\underset{―}{μ}}} \\ = \frac{\frac{\partial^{2} x^{' \underset{―}{μ}}}{\partial x^{α} \partial x^{β}} \frac{\partial x^{α}}{\partial x^{' \underset{―}{μ}}} \frac{\partial x^{β}}{\partial x^{' \underset{―}{μ}}} {(ξ^{\underset{―}{μ}})}^{2}}{ξ^{\underset{―}{μ}}} \\ = \frac{\partial^{2} x^{' \underset{―}{μ}}}{\partial x^{α} \partial x^{β}} \frac{\partial x^{α}}{\partial x^{' \underset{―}{μ}}} \frac{\partial x^{β}}{\partial x^{' \underset{―}{μ}}} ξ^{\underset{―}{μ}} \\ = \frac{\partial^{2} x^{' \underset{―}{μ}}}{\partial x^{α} \partial x^{β}} \frac{\partial x^{α}}{\partial x^{' \underset{―}{μ}}} \frac{\partial x^{β}}{\partial x^{' \underset{―}{μ}}} \frac{\partial x^{' \underset{―}{μ}}}{\partial x^{ν}} u^{ν} \\ = \frac{\partial^{2} x^{' 0}}{\partial x^{α} \partial x^{β}} \frac{\partial x^{α}}{\partial x^{' 0}} \frac{\partial x^{β}}{\partial x^{' 0}} \frac{\partial x^{' 0}}{\partial x^{ν}} u^{ν} \\ = G_{ν} (x) u^{ν}, \end{aligned} \end{matrix}

$G_\nu\qty(x)=\pdv{x^{\prime 0}}{x^\alpha}{x^\beta}\pdv{x^\alpha}{x^{\prime 0}}\pdv{x^\beta}{x^{\prime 0}}\pdv{x^{\prime 0}}{x^\nu}$ . 从而

\begin{matrix} (12) & \begin{array}{r} \frac{\partial^{2} x^{' μ}}{\partial x^{ρ} \partial x^{σ}} u^{ρ} u^{σ} = Q^{μ} \frac{\partial x^{' μ}}{\partial x^{ν}} u^{ν} = G_{ρ} (x) u^{ρ} \frac{\partial x^{' μ}}{\partial x^{σ}} u^{σ}, \end{array} \end{matrix}

即

\begin{matrix} (13) & \begin{array}{r} (\frac{\partial^{2} x^{' μ}}{\partial x^{ρ} \partial x^{σ}} - G_{ρ} (x) \frac{\partial x^{' μ}}{\partial x^{σ}}) u^{ρ} u^{σ} = 0, \forall u . \end{array} \end{matrix}

$u$ $\rho$ $\sigma$ $1$ $0$ , 即可得

\begin{matrix} (14) & \begin{array}{r} (\frac{\partial^{2} x^{' μ}}{\partial x^{ρ} \partial x^{σ}} - G_{ρ} (x) \frac{\partial x^{' μ}}{\partial x^{σ}}) + (\frac{\partial^{2} x^{' μ}}{\partial x^{σ} \partial x^{ρ}} - G_{σ} (x) \frac{\partial x^{' μ}}{\partial x^{ρ}}) = 0, \end{array} \end{matrix}

即

\begin{matrix} (15) & \begin{array}{r} \frac{\partial^{2} x^{' μ}}{\partial x^{ρ} \partial x^{σ}} = \frac{1}{2} (G_{ρ} (x) \frac{\partial x^{' μ}}{\partial x^{σ}} + G_{σ} (x) \frac{\partial x^{' μ}}{\partial x^{ρ}}) . \end{array} \end{matrix}

$\eqref{未考虑狭义相对性原理而仅考虑光速不变原理时坐标变换所需满足的共形对称性}$ $x^\lambda$ ${U^\mu}_\nu\qty(x)=\pdv{x^{\prime \mu}}{x^\nu}$ $\eqref{坐标变换的二阶偏导所具有的形式}$ 式可得

\begin{matrix} (16) & \begin{aligned} 2 η_{μ ν} (\frac{\partial F (U (x), x)}{\partial x^{ρ}} - F (U (x), x) G_{ρ} (x)) \\ = F (U (x), x) (η_{μ ρ} G_{ν} (x) + η_{ν ρ} G_{μ} (x)), \end{aligned} \end{matrix}

$\mu=\nu=\rho$ , 得

\begin{matrix} (17) & \frac{\partial F (U (x), x)}{\partial x^{μ}} = 2 F (U (x), x) G_{μ} (x), \end{matrix}

$\mu=\mu\neq \rho$ , 得

\begin{matrix} (18) & \frac{\partial F (U (x), x)}{\partial x^{μ}} = F (U (x), x) G_{μ} (x), \end{matrix}

$\eqref{矛盾方程1}$ $\eqref{矛盾方程2}$ 可知,

\begin{matrix} (19) & G_{μ} (x) = \frac{\partial F (U (x), x)}{\partial x^{μ}} = 0, \end{matrix}

因此

\begin{matrix} (20) & \begin{aligned} \frac{\partial^{2} x^{' μ}}{\partial x^{ρ} \partial x^{σ}} = 0, \\ \Rightarrow x^{' μ} = {C^{μ}}_{ν} x^{ν} + a^{μ}, \end{aligned} \end{matrix}

${C^\mu}_\nu$ $a^\mu$ $x$ $F\qty(U\qty(x),x)$ 就可以进一步明确地写成

\begin{matrix} (21) & F (U (x), x) = F (C, a), \end{matrix}

$\eqref{未考虑狭义相对性原理而仅考虑光速不变原理时坐标变换所需满足的共形对称性}$ 变成

\begin{matrix} (22) & η_{ρ σ} {C^{ρ}}_{μ} {C^{σ}}_{ν} = F (C, a) η_{μ ν}, \end{matrix}

$\eqref{更明确的共形对称性}$ $a^\mu$ $\mu=\nu$ , 可得

\begin{matrix} (23) & \frac{\partial F (C, a)}{\partial a^{μ}} = 0, \end{matrix}

$F$ $a$ $F\qty(C,a)=F\qty(C)$ $\eqref{更明确的共形对称性}$ 变成

\begin{matrix} (24) & η_{ρ σ} {C^{ρ}}_{μ} {C^{σ}}_{ν} = F (C) η_{μ ν}, \end{matrix}

写成矩阵形式即为

\begin{matrix} (25) & C^{T} η C = F (C) η . \end{matrix}

因此

\begin{matrix} (26) & d s^{' 2} = F (C) d s^{2} . \end{matrix}

$x^i$ $x^i+\d x^i$ $x^i$ $x^i+\d x^i$ $x^i+\d x^i$ 这一事件之前发生, 由因果性, 在另一坐标系中同样必须有上述结论成立, 因此该亚光速信号在另一个坐标系中仍然亚光速; 同样的论证也可以用于超光速信号. 从而对于由非光信号相联系的两个事件而言, 只能有以下两种情况,

\begin{matrix} (32) & \begin{aligned} {\begin{cases} d s^{2} = d t^{2} - {| d r |}^{2} = d t^{2} (1 - {| v |}^{2}) > 0 \\ d s^{' 2} = d t^{' 2} - {| d r^{'} |}^{2} = d t^{' 2} (1 - {| v^{'} |}^{2}) > 0 \end{cases} 同时成立, \\ 或 \\ {\begin{cases} d s^{2} = d t^{2} - {| d r |}^{2} = d t^{2} (1 - {| v |}^{2}) < 0 \\ d s^{' 2} = d t^{' 2} - {| d r^{'} |}^{2} = d t^{' 2} (1 - {| v^{'} |}^{2}) < 0 \end{cases} 同时成立 . \end{aligned} \end{matrix}

$F\qty(C)$ 的要求是

\begin{matrix} (33) & F (C) > 0. \end{matrix}

$\eqref{矩阵形式的共形变换}$ $F\qty(C)$ ,

\begin{matrix} (34) & F (C) = \frac{1}{d} tr (C^{T} η C η^{- 1}), \end{matrix}

所以

\begin{matrix} (35) & C^{T} η C η^{- 1} = \frac{1}{d} tr (C^{T} η C η^{- 1}) . \end{matrix}

$\eqref{因果性对共形变换常数因子的要求是其必须大于零}$ $\qty(\dfrac{1}{d}\tr\qty(C^T\eta C\eta^{-1}))\ge 0$ $\Omega$ $\dfrac{1}{d}\tr\qty(C^T\eta C\eta^{-1})=\ee^\Omega$ $\Omega=\ln\qty(\dfrac{1}{d}\tr\qty(C^T\eta C\eta^{-1}))$ $\eqref{更明确的矩阵形式的共形变换}$ 可写为

\begin{matrix} (36) & C^{T} η C = e^{Ω} η . \end{matrix}

$\Lambda=\ee^{-\frac{1}{2}\Omega}C$ $\eqref{更明确的矩阵形式的共形变换}$ 可得

\begin{matrix} (37) & Λ^{T} η Λ = η . \end{matrix}

$\eqref{用共性变换参数表示的矩阵形式的共形变换}$ $C$ $\eqref{洛伦兹群的条件}$ $\Lambda$ 所组成的集合也构成群, 它是整体 Weyl 变换群的一个子群, 称为 Lorentz 群.

$\d s^{\prime 2}=F\qty(C)\d s^2$ $C$ $a$ $\d s^2$ $\d x^\mu$ $\d x^{\prime \mu}$ $C$ $\d x^\mu$ $\d s^{\prime 2}=F\qty(C)\d s^2$ $\d s^{\prime 2}=\d t^{\prime 2}-\abs{\d \bm{r}^\prime}^2=\d t^{\prime 2}\qty(1-\abs{\bm{v}^\prime}^2)=F\qty(C)\qty(\d t^2-\abs{\d \bm{r}}^2)=F\qty(C)\d t^2\qty(1-\abs{\bm{v}}^2)$ $F\qty(C)$ $\d x^\mu$ $\d x^i=0$ $\bm{v}=0$ $\abs{\bm{v}^\prime}=u$ , 则有

\begin{matrix} (38) & d t^{' 2} (1 - u^{2}) = F (C) d t^{2} . \end{matrix}

$F\qty(C)$ $\d x^{\prime \mu}$ $\d x^\mu$ $C^{-1}$ $\d x^\mu$ $^\prime$ $^\prime$ $\d x^{\prime i}=0$ $\d t^{\prime 2}$ $\d t^2$ $\d t^2$ $\d t^{\prime 2}$ $\abs{\bm{v}}$ $\abs{\bm{v}^\prime}$ 相等, 从而有

\begin{matrix} (39) & d t^{2} = F (C) d t^{'} (1 - u^{2}) . \end{matrix}

$\eqref{第一个坐标系相对于第二个坐标系运动}$ $\eqref{第二个坐标系相对于第一个坐标系运动}$ 即可得

\begin{matrix} (40) & {(F (C))}^{2} = 1, \forall C, a, \end{matrix}

$\eqref{因果性对共形变换常数因子的要求是其必须大于零}$ , 可以得到

\begin{matrix} (41) & F (C) = 1, \forall C, a . \end{matrix}

$\Lambda=C$ $\d s^{\prime 2}=\d s^2$ 为惯性坐标系变换下的不变量, 而惯性坐标系之间的坐标变换为

\begin{matrix} (42) & x^{' μ} = {Λ^{μ}}_{ν} x^{ν} + a^{μ}, \end{matrix}

$\Lambda$ $\eqref{庞加莱变换}$ 所示的坐标变换称为 Poincaré 变换, 容易证明由所有 Poincaré 变换所组成的集合构成群, 称为 Poincaré 群, Lorentz 群是它的一个子群. 由此我们最终证明了惯性坐标系之间的变换为 Poincaré 变换.

$a^\mu$ $\Lambda$ $\eqref{洛伦兹群的条件}$ .