Classical Field Theory

In classical field theory, the continuous field amplitude \(\eta\) is extended to classical field quantities, which can be scalar fields \(\phi\), vector fields \(V^\mu\), tensor fields \(T^{\mu\nu}\), and so on. Starting with a simple scalar field \(\phi =\phi (x^\mu )\), which is a function of spacetime, we find the equation of motion for \(\phi\) using the variational principle. The action is written as: $$S=\int \mathcal{L}[\phi ,\partial _\mu \phi ] d^4 x $$ Assuming the Lagrangian density \(\mathcal{L}\) is a function of \(\phi\) and \(\partial _\mu \phi\), and varying \(\phi \to \phi +\delta \phi\): $$\phi \to \phi +\delta \phi ใ€\partial _\mu \phi \to \partial _\mu \phi +\delta \partial _\mu \phi $$ Neural network diagram
The variation of the action is: $$\delta S=\delta \int \mathcal{L}[\phi ,\partial _\mu \phi ] d^4 x =\int {\partial \mathcal{L} \over \partial \phi} \delta \phi +{\partial \mathcal{L}\over \partial (\partial _\mu \phi )} \delta \partial _\mu \phi d^4 x $$ $$=\int {\partial \mathcal{L} \over \partial \phi} \delta \phi -\partial _\mu \left({\partial \mathcal{L}\over \partial (\partial _\mu \phi )} \right)\delta \phi d^4 x +\int \partial _\mu \left({\partial \mathcal{L}\over \partial (\partial _\mu \phi )} \delta \phi \right) d^4 x $$ $$=\int \left[{\partial \mathcal{L} \over \partial \phi} -\partial _\mu \left({\partial \mathcal{L}\over \partial (\partial _\mu \phi )} \right)\right]\delta \phi d^4 x + โˆฎ_{\partial V} {\partial \mathcal{L}\over \partial (\partial _\mu \phi )} \delta \phi d^3 x $$ $$=\int \left[{\partial \mathcal{L} \over \partial \phi} -\partial _\mu \left({\partial \mathcal{L}\over \partial (\partial _\mu \phi )} \right)\right]\delta \phi d^4 x +0$$ Since \(\delta \phi\) is arbitrary, for any classical physical quantity \(\phi\), it must satisfy the Euler-Lagrange equation: $${\partial \mathcal{L} \over \partial \phi} -\partial _\mu {\partial \mathcal{L}\over \partial (\partial _\mu \phi )}=0$$ Similarly, we can return to the example of the continuous field \(\mathcal{L}={1 \over 2} \rho(\partial _t \eta )^2-{1 \over 2} Y( \nabla \eta )^2\), which satisfies: $${\partial \mathcal{L} \over \partial \phi} -\partial _t {\partial \mathcal{L}\over\partial (\partial _t \phi ) }- \nabla {\partial \mathcal{L}\over \partial ( \nabla \phi ) }=0$$ This quickly yields: $$-\rho\partial _t^2 \eta +Y \nabla ^2 \eta =0\to \nabla ^2 \eta ={1\over v^2} \partial _t^2 \eta $$ If we generalize the scalar field \(\phi\) to vector fields, such as 4-displacement \(x^\nu\) or 4-potential \(A^\nu\), the above variation only needs to replace \(\phi\) with \(A^\nu\), yielding: $${\partial \mathcal{L}\over \partial x^\nu}-{d\over d\tau} {\partial \mathcal{L}\over (\partial U^\nu )}=0 \quad (Lorentz-force)$$ $${\partial \mathcal{L}\over \partial A^\nu }-\partial _\mu {\partial \mathcal{L}\over\partial (\partial _\mu A^\nu ) }=0 \quad (Field-equation)$$ Returning to the action for the electromagnetic field: $$S=S_P+S_{PF}+S_F$$ $$=-\int P_\mu U^\mu d\tau -\int {e\over c} A_\mu U^\mu d\tau -\int {1 \over 16 \pi c} F_{\mu \nu} F^{\mu \nu} d^4 x $$ $$=\int -\rho_m c-{1 \over c^2} A_\mu J^\mu -{1 \over 16 \pi c} F_{\mu \nu} F^{\mu \nu} d^4 x $$ First, we calculate the Lorentz force: $${\partial \mathcal{L}\over \partial x^\nu}-{d\over d\tau} {\partial \mathcal{L}\over (\partial U^\nu )}=0$$ The only terms related to \(x^\nu\) are in \(S_P\) and \(S_{PF}\), so we handle them: $$\mathcal{L}=-P_\mu U^\mu -{e\over c} A_\mu U^\mu $$ $${\partial \left(-{e\over c} A_\mu U^\mu \right)\over\partial x^\nu} -{d\over d\tau} {\partial \left(-P_\mu U^\mu -{e\over c} A_\mu U^\mu \right)\over\partial U^\nu }=0$$ $$\to -{e\over c} U^\mu (\partial _\nu A_\mu )-{d\over d\tau} \left(-P_\nu-{e\over c} A_\nu \right)=0$$ $$\to-{e\over c} U^\mu (\partial _\nu A_\mu )+{dP_\nu \over d\tau } +{e\over c} {dA_\nu \over d\tau } =0$$ $$\to -{e\over c} U^\mu (\partial _\nu A_\mu )+{dP_\nu \over d\tau } +{e\over c} {\partial A_\nu \over \partial x^\mu } {dx^\mu \over d\tau} =0$$ $$\to -{e\over c} U^\mu (\partial _\nu A_\mu )+{dP_\nu \over d\tau } +{e\over c} U^\mu (\partial _\mu A_\nu )=0$$ $$\to {dP_\nu \over d\tau } +{e\over c} U^\mu (\partial _\mu A_\nu -\partial _\nu A_\mu )=0$$ $$\to {dP_\mu \over d\tau } =-{e\over c} U^\nu F_{\nu\mu} $$ $$\to {dP_\mu \over d\tau } ={e\over c} F_{\mu \nu} U^\nu$$ We quickly obtain the Lorentz force. Next, we handle the field equation: $$\partial ^\mu {\partial \mathcal{L}\over \partial (\partial ^\mu A^\nu ) }-{\partial \mathcal{L} \over \partial A^\nu }=0$$ $$\partial ^\mu \left[-{1 \over 8 \pi c} F_{\alpha\beta} {\partial F^{\alpha\beta}\over \partial (\partial ^\mu A^\nu) } \right]+{1 \over c^2} J_\nu=0$$ Since: $${\partial F^{\alpha\beta} \over \partial (\partial ^\mu A^\nu )} ={\partial (\partial ^\alpha A^\beta -\partial ^\beta A^\alpha ) \over \partial (\partial ^\mu A^\nu )}= \delta ^\alpha _\mu \delta ^\beta _\nu- \delta ^\beta _\mu \delta ^\alpha _\nu$$ We have: $$\partial ^\mu \left[-{1 \over 8 \pi c} F_{\alpha\beta} ( \delta ^\alpha _\mu \delta ^\beta _nu- \delta ^\beta _\mu \delta ^\alpha _\nu )\right]+{1 \over c^2} J_\nu=0$$ $$\partial ^\mu \left[-{1 \over 8 \pi c} F_{\mu\nu}+{1 \over 8 \pi c} F_{\nu\mu} \right]+{1 \over c^2} J_\nu=0$$ Since \(-F_{\mu\nu}=F_{\nu\mu}\): $$\partial ^\mu \left[{1 \over 4 \pi c} F_{\nu\mu} \right]+{1 \over c^2} J_\nu=0$$ $$\partial _\nu F^{\mu \nu} =-{4\pi \over c} J^\mu $$

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