古典場論
在古典場論中，物理學將連續場的振幅\(\eta\) 推廣到古典場的物理量，可以是純量場\(\phi\) 、向量場\(V^\mu\) 、張量場\(T^{\mu\nu}\)等等。先可慮簡單的純量場\(\phi =\phi (x^\mu )\)是時空函數，通過Variational principle找到\(\phi\) 的EoM。先寫下Action，假設Lagrangian dendity \(\mathcal{L}\)是\(\phi\) ,\(\partial _\mu \phi\) 的函數，並通過\(\phi \to \phi +\delta \phi \)的變分
$$S=\int \mathcal{L}[\phi ,\partial _\mu \phi ] d^4 x $$
$$\phi \to \phi +\delta \phi 、\partial _\mu \phi \to \partial _\mu \phi +\delta \partial _\mu \phi $$
$$\delta S=\delta \int \mathcal{L}[\phi ,\partial _\mu \phi ] d^4 x =\int {\partial \mathcal{L} \over \partial \phi} \delta \phi +{\partial \mathcal{L}\over \partial (\partial _\mu \phi )} \delta \partial _\mu \phi d^4 x $$
$$=\int {\partial \mathcal{L} \over \partial \phi} \delta \phi -\partial _\mu \left({\partial \mathcal{L}\over \partial (\partial _\mu \phi )} \right)\delta \phi d^4 x +\int \partial _\mu \left({\partial \mathcal{L}\over \partial (\partial _\mu \phi )} \delta \phi \right) d^4 x $$
$$=\int \left[{\partial \mathcal{L} \over \partial \phi} -\partial _\mu \left({\partial \mathcal{L}\over \partial (\partial _\mu \phi )} \right)\right]\delta \phi d^4 x + ∮_{\partial V} {\partial \mathcal{L}\over \partial (\partial _\mu \phi )} \delta \phi d^3 x $$
$$=\int \left[{\partial \mathcal{L} \over \partial \phi} -\partial _\mu \left({\partial \mathcal{L}\over \partial (\partial _\mu \phi )} \right)\right]\delta \phi d^4 x +0$$
因為\(\delta \phi\) 是任意，所以對任何古典物理量\(\phi\) ，必須滿足Euler-Lagrange eq
$${\partial \mathcal{L} \over \partial \phi} -\partial _\mu {\partial \mathcal{L}\over \partial (\partial _\mu \phi )}=0$$
同樣的，我們可以回到連續場的例子，\(\mathcal{L}={1 \over 2} \rho(\partial _t \eta )^2-{1 \over 2} Y( \nabla \eta )^2\)，滿足
$${\partial \mathcal{L} \over \partial \phi} -\partial _t {\partial \mathcal{L}\over\partial (\partial _t \phi ) }- \nabla {\partial \mathcal{L}\over \partial ( \nabla \phi ) }=0$$
可以迅速得到
$$-\rho\partial _t^2 \eta +Y \nabla ^2 \eta =0\to \nabla ^2 \eta ={1\over v^2} \partial _t^2 \eta $$
如果我們將純量場\phi 推廣成向量場，例如4-displacement \(x^\nu\) 、4-potential \(A^\nu\) ，剛剛的變分只需將\(\phi\) 改寫成\(A^\nu\) ，即可得
$${\partial \mathcal{L}\over \partial x^\nu}-{d\over d\tau} {\partial \mathcal{L}\over (\partial U^\nu )}=0:Lorentz-force$$
$${\partial \mathcal{L}\over \partial A^\nu }-\partial _\mu {\partial \mathcal{L}\over\partial (\partial _\mu A^\nu ) }=0:Field-equation$$
回到電磁場的Action
$$S=S_P+S_{PF}+S_F$$
$$=-\int P_\mu U^\mu d\tau -\int {e\over c} A_\mu U^\mu d\tau -\int {1 \over 16 \pi c} F_{\mu \nu} F^{\mu \nu} d^4 x $$
$$=\int -\rho_m c-{1 \over c^2} A_\mu J^\mu -{1 \over 16 \pi c} F_{\mu \nu} F^{\mu \nu} d^4 x $$
先計算Lorentz force
$${\partial \mathcal{L}\over \partial x^\nu}-{d\over d\tau} {\partial \mathcal{L}\over (\partial U^\nu )}=0$$
跟\(x^\nu\)有關的只有\(S_P\) 、\(S_{PF}\)，所以處理
$$\mathcal{L}=-P_\mu U^\mu -{e\over c} A_\mu U^\mu $$
$${\partial \left(-{e\over c} A_\mu U^\mu \right)\over \partial x^\nu }-{d\over d\tau} {\partial \left(-P_\mu U^\mu -{e\over c} A_\mu U^\mu \right) \over \partial U^\nu }=0$$
$$\to -{e\over c} U^\mu (\partial _\nu A_\mu )-{d\over d\tau} \left(-P_\nu-{e\over c} A_\nu \right)=0$$
$$\to-{e\over c} U^\mu (\partial _\nu A_\mu )+{dP_\nu \over d\tau } +{e\over c} {dA_\nu \over d\tau } =0$$
$$\to -{e\over c} U^\mu (\partial _\nu A_\mu )+{dP_\nu \over d\tau } +{e\over c} {\partial A_\nu \over \partial x^\mu } {dx^\mu \over d\tau} =0$$
$$\to -{e\over c} U^\mu (\partial _\nu A_\mu )+{dP_\nu \over d\tau } +{e\over c} U^\mu (\partial _\mu A_\nu )=0$$
$$\to {dP_\nu \over d\tau } +{e\over c} U^\mu (\partial _\mu A_\nu -\partial _\nu A_\mu )=0$$
$$\to {dP_\nu \over d\tau } +{e\over c} U^\mu F_{\mu\nu}=0$$
$$\to {dP_\mu \over d\tau } =-{e\over c} U^\nu F_{\nu\mu} $$
$$\to {dP_\mu \over d\tau } ={e\over c} F_{\mu \nu} U^\nu$$
迅速得到Lorentz force。
再來處理Field equation，
$$\partial ^\mu {\partial \mathcal{L}\over \partial (\partial ^\mu A^\nu ) }-{\partial \mathcal{L} \over \partial A^\nu }=0$$
$$\partial ^\mu \left[-{1 \over 8 \pi c} F_{\alpha\beta} {\partial F^{\alpha\beta}\over \partial (\partial ^\mu A^\nu) } \right]+{1 \over c^2} J_\nu=0$$
因為
$${\partial F^{\alpha\beta} \over \partial (\partial ^\mu A^\nu )} ={\partial (\partial ^\alpha A^\beta -\partial ^\beta A^\alpha ) \over \partial (\partial ^\mu A^\nu )}= \delta ^\alpha _\mu \delta ^\beta _\nu- \delta ^\beta _\mu \delta ^\alpha _\nu$$
$$\partial ^\mu \left[-{1 \over 8 \pi c} F_{\alpha\beta} ( \delta ^\alpha _\mu \delta ^\beta _\nu- \delta ^\beta _\mu \delta ^\alpha _\nu )\right]+{1 \over c^2} J_\nu=0$$
$$\partial ^\mu \left[-{1 \over 8 \pi c} F_{\mu\nu}+{1 \over 8 \pi c} F_{\nu\mu} \right]+{1 \over c^2} J_\nu=0$$
因為\(-F_{\mu\nu}=F_{\nu\mu} \)
$$\partial ^\mu \left[{1 \over 4 \pi c} F_{\nu\mu} \right]+{1 \over c^2} J_\nu=0$$
$$\partial _\nu F^{\mu \nu} =-{4\pi \over c} J^\mu $$
王培儒