Noether's Theorem

When discussing variational principles, Noether's theorem is an essential companion. Noether's theorem states that if a system's Lagrangian \(\phi\) possesses a continuous transformation \(\phi \to \phi +\delta \alpha\) that leaves the action invariant (\(\delta S=0\)), there exists a corresponding conserved current. This can be applied to: $$Time-translation-invariance \leftrightarrow Energy-conservation$$ $$Spatial-translation-invariance \leftrightarrow Momentum-conservation$$ The generalized proof is as follows: Given the Lagrangian density \(\mathcal{L}[\phi (x^\alpha ),\partial _\mu \phi (x^\alpha ),x^\mu ]\), we now consider variations with respect to both the field \(\phi\) and spacetime coordinates \(x^\mu\): $$x^\mu \to \bar{x} ^\mu =x^\mu +\delta x^\mu $$ $$\phi \to \bar{\phi}=\phi +\delta \phi $$ However, since \(\phi =\phi (x^\alpha )\) is a function of spacetime, after the variations of both \(\phi\) and \(x^\mu\), \(\phi (x^\alpha )\to \bar{\phi} ( \bar{x} ^\alpha )\) can be expressed as \(\bar{\phi} ( \bar{x} ^\alpha )\equiv \phi (x^\alpha )+\Delta\phi \), where: $$\Delta\phi =\bar{\phi} ( \bar{x} ^\alpha )-\phi (x^\alpha )$$ $$=\bar{\phi} ( \bar{x} ^\alpha )-\phi ( \bar{x} ^\alpha )+\phi ( \bar{x} ^\alpha )-\phi (x^\alpha )$$ $$=\delta \phi +{\partial \phi \over \partial x^\alpha } \delta x^\alpha$$ Noether's theorem examines the difference in the action after variation: $$\delta S=\delta \int \mathcal{L}[\phi (x^\alpha ),\partial _\mu \phi (x^\alpha ),x^\mu ] d^4 x $$ $$=\int \mathcal{L}[\bar{\phi} ( \bar{x} ^\alpha ),\partial _\mu \bar{\phi} ( \bar{x} ^\alpha ), \bar{x} ^\mu ] d^4 x -\int \mathcal{L}[\phi (x^\alpha ),\partial _\mu \phi (x^\alpha ),x^\mu ] d^4 x $$ It is clear that even the integral's 4-volume \(d^4 \bar{x}\) changes. Consider the differential product: \(\delta S=\delta \int \mathcal{L}d^4 x =\int \delta\mathcal{L} d ^4 x +\int \mathcal{L}\delta d^4 x \) The first term, \(\delta \mathcal{L}\), must be handled with care: $$\delta \mathcal{L}=\mathcal{L}[\bar{\phi} ( \bar{x} ^\alpha ),\partial _\mu \bar{\phi} ( \bar{x} ^\alpha ), \bar{x} ^\mu ]-\mathcal{L}[\phi (x^\alpha ),\partial _\mu \phi (x^\alpha ),x^\mu ]$$ $$=\mathcal{L}[\bar{\phi} ( \bar{x} ^\alpha ),\partial _\mu \bar{\phi} ( \bar{x} ^\alpha ), \bar{x} ^\mu ]-\mathcal{L}[\phi ( \bar{x} ^\alpha ),\partial _\mu \phi ( \bar{x} ^\alpha ), \bar{x} ^\mu ]+\mathcal{L}[\phi ( \bar{x} ^\alpha ),\partial _\mu \phi ( \bar{x} ^\alpha ), \bar{x} ^\mu ]-\mathcal{L}[\phi (x^\alpha ),\partial _\mu \phi (x^\alpha ),x^\mu ]$$ $$={\mathcal{L}[\phi +\delta \phi ,\partial _\mu \phi +\delta (\partial _\mu \phi ), \bar{x} ^\mu ]-L[\phi ,\partial _\mu \phi , \bar{x} ^\mu ]}+{\mathcal{L}[\phi ,\partial _\mu \phi , \bar{x} ^\mu ]-L[\phi ,\partial _\mu \phi ,x^\mu ]}$$ $$={{\partial \mathcal{L}\over\partial \phi} \delta \phi +{\partial \mathcal{L}\over\partial (\partial _\mu \phi )} \delta (\partial _\mu \phi )}+{{\partial \mathcal{L}\over \partial x^\mu } \delta x^\mu }$$ Using \(\partial _\mu \) to bring to the front: $$={{\partial \mathcal{L}\over\partial \phi} \delta \phi -\partial _\mu \left({\partial \mathcal{L}\over\partial (\partial _\mu \phi )} \right)\delta \phi +\partial _\mu \left({\partial \mathcal{L}\over\partial (\partial _\mu \phi )} \delta \phi \right)}+{{\partial \mathcal{L}\over \partial x^\mu } \delta x^\mu }$$ Note that \(\left[{\partial \mathcal{L}\over\partial \phi} -\partial _\mu \left({\partial \mathcal{L}\over\partial (\partial _\mu \phi )} \right)\right]\delta \phi\) is the Euler-Lagrange equation, which equals zero. Therefore: $$\delta \mathcal{L}=\partial _\mu \left({\partial \mathcal{L}\over\partial (\partial _\mu \phi )} \delta \phi \right)+\partial _\mu \mathcal{L}*\delta x^\mu $$ The second term, \(\delta d^4 x\), involves the Jacobian: $$\int \mathcal{L}\delta d^4 x =\int \mathcal{L}d^4 \bar{x} -\int \mathcal{L}d^4 x $$ $$d^4 \bar{x}=J\left({\partial \bar{x} ^\mu \over\partial x^\nu }\right) d^4 x$$ \(J\left({\partial \bar{x} ^\mu \over\partial x^\nu }\right)\) is the Jacobian determinant: $${\partial \bar{x} ^\mu \over\partial x^\nu }={\partial (x^\mu +\delta x^\mu )\over\partial x^\nu }= \delta ^\mu _\nu +\partial _\nu \delta x^\mu $$ The Jacobian determinant is calculated as: $$J\left({\partial \bar{x} ^\mu \over\partial x^\nu }\right)=1+\partial _\mu \delta x^\mu $$ So: $$\int \mathcal{L}\delta d^4 x =\int \mathcal{L}*\partial _\mu \delta x^\mu d^4 x $$ Combining the two terms: $$\delta S=\delta \int \mathcal{L}d^4 x =\int \left[\partial _\mu \left({\partial \mathcal{L}\over\partial (\partial _\mu \phi )} \delta \phi \right)+\partial _\mu \mathcal{L}*\delta x^\mu \right] d^4 x +\int \left[\mathcal{L} *\partial _\mu \delta x^\mu \right] d^4 x $$ $$=\int \partial _\mu \left({\partial \mathcal{L}\over\partial (\partial _\mu \phi )} \delta \phi \right)+\partial _\mu (\mathcal{L}\delta x^\mu ) d^4 x $$ $$=\int \partial _\mu \left({\partial \mathcal{L}\over\partial (\partial _\mu \phi )} \delta \phi +\mathcal{L}\delta x^\mu \right) d^4 x $$ $$=\int \partial _\mu \left({\partial \mathcal{L}\over\partial (\partial _\mu \phi )} \Delta\phi -\left({\partial \mathcal{L}\over\partial (\partial _\mu \phi )} \partial _\alpha \phi - \delta ^\mu _\alpha \mathcal{L}\right)\delta x^\alpha \right) d^4 x$$ Set $$\left(\begin{matrix} \delta x^\alpha \\ \Delta\phi \end{matrix}\right)=\varepsilon\left(\begin{matrix} Χ^\alpha \\ \Psi \end{matrix}\right)$$ \(Χ^\alpha\) and \(\Psi\) are symmetry generators: $$\delta S=\varepsilon \int \partial _\mu \left({\partial \mathcal{L}\over\partial (\partial _\mu \phi )} \Psi-\left({\partial \mathcal{L}\over\partial (\partial _\mu \phi )} \partial _\alpha \phi - \delta ^\mu _\alpha \mathcal{L}\right) Χ^\alpha \right) d^4 x =0$$ Since \(\varepsilon \) is arbitrary: $$\partial _\mu \left({\partial \mathcal{L}\over\partial (\partial _\mu \phi )} \Psi-\left({\partial \mathcal{L}\over\partial (\partial _\mu \phi )} \partial _\alpha \phi - \delta ^\mu _\alpha \mathcal{L}\right) Χ^\alpha \right)\equiv \partial _\mu N^\mu =0$$ Define the Noether current: $$N^\mu \equiv {\partial \mathcal{L}\over\partial (\partial _\mu \phi )} \Psi-\left({\partial \mathcal{L}\over\partial (\partial _\mu \phi )} \partial _\alpha \phi - \delta ^\mu _\alpha \mathcal{L}\right) Χ^\alpha $$ Where \({\partial \mathcal{L}\over\partial (\partial _\mu \phi )}\) corresponds to the conserved current for the symmetry transformation \(\Psi\) of the field \(\phi\), and \(\left({\partial \mathcal{L}\over\partial (\partial _\mu \phi )} \partial _\alpha \phi - \delta ^\mu _\alpha \mathcal{L}\right)\) corresponds to the conserved current under the transformation \(Χ^\alpha\) of spacetime \(x^\mu\).

For vector fields \(\alpha^\nu\): $$N^\mu \equiv {\partial \mathcal{L}\over\partial (\partial _\mu \alpha^\nu )} \Psi^\nu-\left({\partial \mathcal{L}\over\partial (\partial _\mu \alpha^\nu )} \partial _\alpha \alpha^\nu- \delta ^\mu _\alpha \mathcal{L}\right) Χ^\alpha $$

Originally written in Chinese by the author, these articles are translated into English to invite cross-language resonance.