諾特定理

提到變分法就一定要與諾特定理做搭配。諾特定理表明如果系統中Lagrangian的物理量\(\phi\) 存在連續變\(\phi \to \phi +\delta \alpha\)，使得\(\mathcal{L}\to \mathcal{L}+{\delta \mathcal{L}\over\delta \alpha}\)，卻保持對稱性（不變性）\(\delta S=0\)，那麼就存在一個守恆流與之對應。這可以應用到：
$$時間平移不變\leftrightarrow 能量守恆$$ $$空間平移不變\leftrightarrow 動量守恆$$ 廣義的證明如下，給定Lagrangian dendity \(\mathcal{L}[\phi (x^\alpha ),\partial _\mu \phi (x^\alpha ),x^\mu ]\)，之前的變分法是針對物理量\(\phi \) 做變分，現在我們同時針對物理量\(\phi\) 和時空\(x^\mu\) 變分 $$x^\mu \to \bar{x} ^\mu =x^\mu +\delta x^\mu $$ $$\phi \to \bar{\phi}=\phi +\delta \phi $$ 但是\(\phi =\phi (x^\alpha )\)是時空的函數，經過物理量\(\phi\) 和時空\(x^\mu \)同時變分後，包含自身的變分和受到時空變分的影響\(\phi (x^\alpha )\to \bar{\phi} ( \bar{x} ^\alpha )\)，令\(\bar{\phi} ( \bar{x} ^\alpha )\equiv \phi (x^\alpha )+\Delta\phi \)， $$\Delta\phi =\bar{\phi} ( \bar{x} ^\alpha )-\phi (x^\alpha )$$ $$=\bar{\phi} ( \bar{x} ^\alpha )-\phi ( \bar{x} ^\alpha )+\phi ( \bar{x} ^\alpha )-\phi (x^\alpha )$$ $$=\delta \phi +{\partial \phi \over \partial x^\alpha } \delta x^\alpha$$ 諾特定理探討變分後Action的差別： $$\delta S=\delta \int \mathcal{L}[\phi (x^\alpha ),\partial _\mu \phi (x^\alpha ),x^\mu ] d^4 x $$ $$=\int \mathcal{L}[\bar{\phi} ( \bar{x} ^\alpha ),\partial _\bar{\mu} \bar{\phi} ( \bar{x} ^\alpha ), \bar{x} ^\mu ] d^4 \bar{x} -\int \mathcal{L}[\phi (x^\alpha ),\partial _\mu \phi (x^\alpha ),x^\mu ] d^4 x $$ 很明顯可以發現到連積分的4-volune \(d^4 \bar{x}\)都發生變化。可以用乘法的微分來思考 $$\delta S=\delta \int \mathcal{L}d^4 x =\int \delta\mathcal{L} d ^4 x +\int \mathcal{L}\delta d^4 x $$ 第一項\(\delta \mathcal{L}\)在處理時要非常小心 $$\delta \mathcal{L}=\mathcal{L}[\bar{\phi} ( \bar{x} ^\alpha ),\partial _\bar{\mu} \bar{\phi} ( \bar{x} ^\alpha ), \bar{x} ^\mu ]-\mathcal{L}[\phi (x^\alpha ),\partial _\mu \phi (x^\alpha ),x^\mu ]$$ $$=\mathcal{L}[\bar{\phi} ( \bar{x} ^\alpha ),\partial _\bar{\mu} \bar{\phi} ( \bar{x} ^\alpha ), \bar{x} ^\mu ]-\mathcal{L}[\phi ( \bar{x} ^\alpha ),\partial _\bar{\mu} \phi ( \bar{x} ^\alpha ), \bar{x} ^\mu ]+\mathcal{L}[\phi ( \bar{x} ^\alpha ),\partial _\bar{\mu} \phi ( \bar{x} ^\alpha ), \bar{x} ^\mu ]-\mathcal{L}[\phi (x^\alpha ),\partial _\mu \phi (x^\alpha ),x^\mu ]$$ $$={\mathcal{L}[\phi +\delta \phi ,\partial _\bar{\mu} \phi +\delta (\partial _\bar{\mu} \phi ), \bar{x} ^\mu ]-L[\phi ,\partial _\bar{\mu} \phi , \bar{x} ^\mu ]}+{\mathcal{L}[\phi ,\partial _\bar{\mu} \phi , \bar{x} ^\mu ]-L[\phi ,\partial _\mu \phi ,x^\mu ]}$$ $$={{\partial \mathcal{L}\over\partial \phi} \delta \phi +{\partial \mathcal{L}\over\partial (\partial _\mu \phi )} \delta (\partial _\mu \phi )}+{{\partial \mathcal{L}\over \partial x^\mu } \delta x^\mu }$$ 利用\(\partial _\mu \)拉到前面去 $$={{\partial \mathcal{L}\over\partial \phi} \delta \phi -\partial _\mu \left({\partial \mathcal{L}\over\partial (\partial _\mu \phi )} \right)\delta \phi +\partial _\mu \left({\partial \mathcal{L}\over\partial (\partial _\mu \phi )} \delta \phi \right)}+{{\partial \mathcal{L}\over \partial x^\mu } \delta x^\mu }$$ 可以注意到\(\left[{\partial \mathcal{L}\over\partial \phi} -\partial _\mu \left({\partial \mathcal{L}\over\partial (\partial _\mu \phi )} \right)\right]\delta \phi\) 是Euler-Lagrange eq等於零。所以 $$\delta \mathcal{L}=\partial _\mu \left({\partial \mathcal{L}\over\partial (\partial _\mu \phi )} \delta \phi \right)+\partial _\mu \mathcal{L}*\delta x^\mu $$ 第二項\(\delta d^4 x\)要用到Jacobian $$\int \mathcal{L}\delta d^4 x =\int \mathcal{L}d^4 \bar{x} -\int \mathcal{L}d^4 x $$ $$d^4 \bar{x}=J\left({\partial \bar{x} ^\mu \over\partial x^\nu }\right) d^4 x$$ \(J\left({\partial \bar{x} ^\mu \over\partial x^\nu }\right)\)為Jacobian determinant $${\partial \bar{x} ^\mu \over\partial x^\nu }={\partial (x^\mu +\delta x^\mu )\over\partial x^\nu }= \delta ^\mu _\nu +\partial _\nu \delta x^\mu $$ Jacobian determinant計算出來為 $$J\left({\partial \bar{x} ^\mu \over\partial x^\nu }\right)=1+\partial _\mu \delta x^\mu $$ 所以 $$\int \mathcal{L}\delta d^4 x =\int \mathcal{L}*\partial _\mu \delta x^\mu d^4 x $$ 合併兩項 $$\delta S=\delta \int \mathcal{L}d^4 x =\int \left[\partial _\mu \left({\partial \mathcal{L}\over\partial (\partial _\mu \phi )} \delta \phi \right)+\partial _\mu L*\delta x^\mu \right] d^4 x +\int \left[\mathcal{L} *\partial _\mu \delta x^\mu \right] d^4 x $$ $$=\int \partial _\mu \left({\partial \mathcal{L}\over\partial (\partial _\mu \phi )} \delta \phi \right)+\partial _\mu \mathcal{L}*\delta x^\mu +\mathcal{L} *\partial _\mu \delta x^\mu d^4 x $$ $$=\int \partial _\mu \left({\partial \mathcal{L}\over\partial (\partial _\mu \phi )} \delta \phi \right)+\partial _\mu (\mathcal{L}\delta x^\mu ) d^4 x $$ $$=\int \partial _\mu \left({\partial \mathcal{L}\over\partial (\partial _\mu \phi )} \delta \phi +\mathcal{L}\delta x^\mu \right) d^4 x =0$$ 其中\(\delta \phi \)是\(\phi\) 自身的變分，不包含受到\(\delta x^\mu\) 的影響，我們想要考慮廣義一點，改寫成\(\Delta\phi \)可以包含受到\(\delta x^\mu \)的影響。因為\(\Delta\phi =\delta \phi +\partial _\alpha \phi *\delta x^\alpha\)，所以 $$\delta S=\int \partial _\mu \left({\partial \mathcal{L}\over\partial (\partial _\mu \phi )} (\Delta\phi -\partial _\alpha \phi *\delta x^\alpha )+\mathcal{L}\delta x^\mu \right) d^4 x $$ $$=\int \partial _\mu \left({\partial \mathcal{L}\over\partial (\partial _\mu \phi )} \Delta\phi -\left({\partial \mathcal{L}\over\partial (\partial _\mu \phi )} \partial _\alpha \phi - \delta ^\mu _\alpha \mathcal{L}\right)\delta x^\alpha \right) d^4 x$$ 令 $$\left(\begin{matrix} \delta x^\alpha \\ \Delta\phi \end{matrix}\right)=\varepsilon\left(\begin{matrix} Χ^\alpha \\ \Psi \end{matrix}\right)$$ \(Χ^\alpha\)、\(\Psi\)為Symmetry generators， $$\delta S=\varepsilon \int \partial _\mu \left({\partial \mathcal{L}\over\partial (\partial _\mu \phi )} \Psi-\left({\partial \mathcal{L}\over\partial (\partial _\mu \phi )} \partial _\alpha \phi - \delta ^\mu _\alpha L\right) Χ^\alpha \right) d^4 x =0$$ 因為\(\varepsilon \)是任意的，所以 $$\partial _\mu \left({\partial \mathcal{L}\over\partial (\partial _\mu \phi )} \Psi-\left({\partial \mathcal{L}\over\partial (\partial _\mu \phi )} \partial _\alpha \phi - \delta ^\mu _\alpha \mathcal{L}\right) Χ^\alpha \right)\equiv \partial _\mu N^\mu =0$$ 定義諾特流(Noether current) $$N^\mu \equiv {\partial \mathcal{L}\over\partial (\partial _\mu \phi )} \Psi-\left({\partial \mathcal{L}\over\partial (\partial _\mu \phi )} \partial _\alpha \phi - \delta ^\mu _\alpha \mathcal{L}\right) Χ^\alpha $$ 其中，\({\partial \mathcal{L}\over\partial (\partial _\mu \phi )}\) 是對應到物理量\(\phi \)在對稱變換\(\Psi\)的守恆流，\(\left({\partial \mathcal{L}\over\partial (\partial _\mu \phi )} \partial _\alpha \phi - \delta ^\mu _\alpha \mathcal{L}\right)\)是對應時空\(x^\mu\) 變化\(Χ^\alpha\)下的守恆流。

如果是向量場\(\alpha^\nu\)， $$N^\mu \equiv {\partial \mathcal{L}\over\partial (\partial _\mu \alpha^\nu )} \Psi^\nu-\left({\partial \mathcal{L}\over\partial (\partial _\mu \alpha^\nu )} \partial _\alpha \alpha^\nu- \delta ^\mu _\alpha \mathcal{L}\right) Χ^\alpha $$