诺特定律, 也即对称性蕴含守恒流, 更加准确的说法是

每一个局部作用的可微的对称性, 都蕴含某种守恒流.

什么是守恒流

守恒流是满足: \[\require{physics} \partial_\mu j^\mu = 0 \] 的一个场.

场在对称变换下的描述

而对称性的考量数学上就有两种看法

  • 主动视角, 对应参照系不变, 物理点运动;
  • 被动视角, 物理点不变, 参照系运动.

接下来我们会更加自然的考虑后者, 因为后者更加容易给出明确的数学表示.

也就是时空点\(P\), 坐标\(x^\mu\), 场\(f\left(x^\mu\right)\), 变为坐标\(x'^\mu\), 场\(f'\left(x'^\mu\right)\),

考虑在一个无穷小变换 \(x' \to x + \delta x\) 下, 场的变换可以写作: \[\begin{aligned} \delta f & = f'(x'^\mu) - f(x^\mu) \\ & = f'(x^\mu + \delta x^\mu) - f(x^\mu)\\ & = f'(x^\mu) - f(x) + \delta x^\mu\partial_\mu f'(x^\mu) + \mathcal{O}(\delta x^2) \\ & = f'(x^\mu) - f(x) + \delta x^\mu\partial_\mu f(x^\mu) + \mathcal{O}(\delta x^2) \end{aligned} \]

\(\partial_\mu f'\) 被换成 \(\partial_\mu f\) 带来的差异不会大于 \(\delta x\) 的一阶. 并定义 \(f'(x^\mu) - f(x)\)\(\delta_0 f\)从而可以写出: \[ \delta f = \delta_0 f + \delta x^\mu\partial_\mu f \] 或者说 \[ \delta = \delta_0 + \delta x^\mu\partial_\mu \] 其中 \(\delta_0 f\) 表述场本身的变化, 而 \(\delta x^\mu\partial_\mu f\) 表示由于点坐标的变化带来的变化.

对于时空平移变换\(x'^\mu = x^\mu + a^\mu,\,\delta\phi = 0\)

此时有 \[ \delta_0 \phi = - a^\mu\partial_\mu\phi \] 其实这就对应于场在平移下的变换: \[\phi'(x^\mu) = \phi(x^\mu - a^\mu) \]

在此之外, 场可能有内禀变换, 此时\(\delta x = 0\), 变化的只有场.

诺特流的推导

考虑一个场, 其作用量可以写作 \[S(\phi(x)) = \int \dd[4]{x} \mathcal{L}(\phi, \partial_\mu\phi,x) \]

系统的演化路径遵循\(\delta S = 0\), 则: \[\begin{aligned} 0 = \delta S &= \int\left[\delta(\dd[4]{x})\mathcal{L}+\dd[4]{x}\delta\mathcal{L}\right]\\ & = \int\dd[4]{x}(\partial_\mu\delta x^\mu\mathcal{L}+\delta\mathcal{L}) \end{aligned} \] 而根据前文以及链式法则 \[\begin{aligned} \delta \mathcal{L} &= \delta x^\mu\partial_\mu\mathcal{L} + \delta_0\mathcal{L}\\ &= \delta x^\mu \partial_\mu\mathcal{L}+ \frac{\partial\mathcal{L}}{\partial\phi}\delta_0\phi + \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)}\delta_0(\partial_\mu\phi) \\ &= \delta x^\mu \partial_\mu\mathcal{L} + \left[\frac{\partial\mathcal{L}}{\partial \phi} - \partial_\mu \frac{\partial\mathcal{L}}{\partial (\partial_\mu \phi)} \right] \delta_0\phi + \partial_\mu\left(\frac{\partial\mathcal{L}}{\partial (\partial_\mu \phi)}\delta_0\phi \right) \end{aligned} \] 最后一步是使用了分部积分, 并且使用了\(\delta_0(\partial_\mu\phi) = \partial_\mu(\delta_0\mu\phi)\). 同时注意到有: \[\frac{\partial\mathcal{L}}{\partial \phi} - \partial_\mu \frac{\partial\mathcal{L}}{\partial (\partial_\mu \phi)} = 0 \] 这是场的欧拉 - 拉格朗日运动方程. 则 \[\begin{aligned} 0 = \delta S & = \int \dd[4]{x} \partial_\mu\left(\delta x^\mu \mathcal{L} + \frac{\partial\mathcal{L}}{\partial (\partial_\mu \phi)}\delta_0\phi\right) \end{aligned} \] 考虑到变分的任意性, 则有\(\partial_\mu\left(\delta x^\mu \mathcal{L} + \frac{\partial\mathcal{L}}{\partial (\partial_\mu \phi)}\delta_0\phi\right) = 0\), 对扩号内的部分使用\(\delta_0 = \delta - \delta x^\mu\partial_\mu\), 则:

\[\begin{aligned} 0 = \delta S&= \int \dd[4]{x}\partial_\mu \left[\mathcal{L}\delta x^\mu + \frac{\partial\mathcal{L}}{\partial (\partial_\mu \phi)} (\delta - \delta x^\nu\partial_\nu)\phi\right] \\ & = \int \dd[4]{x}\partial_\mu\left[\left(\mathcal{L}\delta^\nu_\mu - \frac{\partial\mathcal{L}}{\partial (\partial_\mu \phi)}\partial_\nu \phi\right)\delta x^\nu +\frac{\partial\mathcal{L}}{\partial (\partial_\mu \phi)} \delta\phi \right] \end{aligned} \] 这样, 就得到了一个守恒流, 也就是本文的核心: \[j^\mu = \left(\mathcal{L}\delta^\nu_\mu - \frac{\partial\mathcal{L}}{\partial (\partial_\mu \phi)}\partial_\nu \phi\right)\delta x^\nu +\frac{\partial\mathcal{L}}{\partial (\partial_\mu \phi)} \delta\phi \]

诺特荷

对于一个在有限空间内分布的流(无穷远处, 场应该趋近于零, 这是其物理意义要求的), 考虑场的等时变分 \[\begin{aligned} 0 & = \int \dd[4]{x}\partial_\mu j^\mu \\ & = \int_{t_1}^{t_2} \dd{x}^0 \int \dd{x}^3 \left(\partial_0 j^0 + \nabla \cdot \vec{j}\right) \\ & = \int_{t_1}^{t_2} \dd{x}^0 \partial_0 \int \dd[3]{x}j^0 \\ & = Q(t_2) - Q(t_1) \end{aligned} \]

\(Q = \int \dd[3]{x}j^0\) 就是诺特荷, 是在这个对称性给出的守恒量.

举例

对于时空平移变换的特殊情况:

平移变换具有各向同性 (说人话就是朝着时空四个轴有四个 生成元) 那么, 诺特流就会升级成为能量动量张量 原来的守恒流长这样 \[j^\mu = \left(\mathcal{L}\delta^\nu_\mu - \frac{\partial\mathcal{L}}{\partial (\partial_\mu \phi)}\partial_\nu \phi\right)\delta x^\nu +\frac{\partial\mathcal{L}}{\partial (\partial_\mu \phi)} \delta\phi \] 去掉场的本身的变换:\(\delta\phi = 0\), 考虑到 \(\delta x^\nu = a^\nu\) 的任意性 \[\partial_\mu\left(\mathcal{L}\delta^\nu_\mu - \frac{\partial\mathcal{L}}{\partial (\partial_\mu \phi)}\partial_\nu \phi\right) = 0 \] 这就是场论的能量动量张量: \[\begin{aligned} T^{\mu}_{\nu} = \left(- \mathcal{L}\delta^\nu_\mu + \frac{\partial\mathcal{L}}{\partial (\partial_\mu \phi)}\partial_\nu \phi\right) \\ T^{\mu\nu} = \left(- \mathcal{L}\eta^{\mu\nu} + \frac{\partial\mathcal{L}}{\partial (\partial_\mu \phi)}\partial^\nu \phi\right) \end{aligned} \]

对于一个场的内禀变换而言, 比如复 Klein-Gordon 场的 \(U(1)\) 对称性

\[\begin{aligned} \phi &\to \mathrm{e}^{i\alpha} \phi \\ \phi^* &\to \mathrm{e}^{-i\alpha} \phi^* \end{aligned} \] 借助生成元\(\delta\phi = i\phi,\, \delta\phi^* = i\phi^*\), 可以写出 \[j^\mu = i\left[ (\partial^\mu \phi^*)\phi - \phi^*(\partial^\mu \phi)\right] \] 它对应的守恒荷就是: \[Q = \int \dd[3]{x}j^0 = i\int \dd[3]{x}\left(\dot{\phi}^*\phi - \phi^*\dot{\phi}\right) \] 可以通过正则量子化计算发现这个守恒就对应电荷守恒.


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