Action Principle (Relativistic)

Since spacetime is equivalent under relativity, physicists use the Principle of Relativity and the Principle of Least Action to write the Action \(S\) in scalar form, ensuring that \(\delta S = 0\) in any coordinate system. The Actions we listed earlier are: $$S_P=-mc^2 \int _a^b d\tau $$ $$S_{PF}=-{e\over c} \int _a^b A_\mu dx^\mu $$ $$S=S_P+S_{PF}=-mc^2 \int _a^b d\tau -{e\over c} \int _a^b A_\mu dx^\mu $$ $$=\int _a^b -\gamma mc^2-\gamma e \phi+{e\over c} \overrightarrow{A} \cdot \gamma \overrightarrow{v} dt =\int _a^b Ldt$$ Although these Actions satisfy the scalar requirement, the Lagrangian \(L\) itself is not a scalar, as it varies under different coordinate systems. Physicists, therefore, aim to rewrite the Lagrangian \(L\) as a scalar. We define the 4-Volume \(d^4 x=dc\tau dV=dc\tau dxdydz\) and rewrite the original Action as: $$S_P=-mc^2 \int _a^b d\tau =-\int \rho_m dV c^2 \int _a^b d\tau $$ $$=-∬ \rho_m cdc\tau dV =\int -\rho_m cd^4 x $$ $$S_{PF}=-{e\over c} \int _a^b A_\mu dx^\mu =-{\int \rho_e dV \over c} \int _a^b A_\mu {dx^\mu \over d\tau} d\tau $$ $$=-{1 \over c} ∬ \rho_e A_\mu u^\mu d\tau dV =-{1 \over c^2} ∬ A_\mu J^\mu dc\tau d^4 x $$ $$=-{1 \over c^2} \int A_\mu J^\mu d^4 x $$ Here, \(J = J^\mu \hat{e} _\mu = \rho_e u^\mu \hat{e} _\mu\) is the 4-current density. Notably, \(d^4 x\) is an invariant, making it a scalar. Additionally, when we add the EM Field Action \(S_F\): $$S_F=-{1 \over 16\pi c} \int F_{\mu \nu} F^{\mu \nu} d^4 x $$ $$S=S_P+S_{PF}+S_F$$ $$=\int -\rho_m c-{1\over c^2} A_\mu J^\mu -{1 \over 16\pi c} F_{\mu \nu} F^{\mu \nu} d^4 x$$ $$ \equiv \int \mathcal{L} d^4 x $$ Since the 4-Volume \(d^4 x\) is a scalar and the Action is also a scalar, \(\mathcal{L}\) must be a scalar as well. We call \(\mathcal{L}\) the Lagrangian density. The Lagrangian density \(\mathcal{L}\) remains a scalar and is invariant in any coordinate system: $$\mathcal{L}=-\rho_m c-{1\over c^2} A_\mu J^\mu -{1 \over 16\pi c} F_{\mu \nu} F^{\mu \nu} $$

Advanced: \(d^4 x\) is an invariant Physical Proof Due to Time Dilation and Length Contraction being opposite effects, if \(\tau\) and \(\bar{x}\) are the proper time and proper length: $$t=\gamma \tau $$ $$x={\bar{x} \over \gamma} $$ Therefore: $$dctdx=dc\left(\gamma \tau \right)d{\bar{x} \over \gamma }=dc\tau d\bar{x}$$ Mathematical Proof Recall the Jacobian \(J\): $$dxdy=rdrdθ=J\left(r,\theta\right)drd\theta$$ Where: $$J\left(r,\theta\right)=\left| \begin{matrix} {\partial x \over \partial r} & {\partial x \over \partial \theta} \\ {\partial y \over \partial r} & {\partial y \over \partial \theta} \end{matrix} \right| =\left| \begin{matrix} {\partial rcos\theta \over \partial r} & {\partial rcos\theta \over \partial \theta} \\ {\partial rsin\theta \over \partial r} & {\partial rsin\theta \over \partial \theta} \end{matrix} \right| $$ $$=\left| \begin{matrix} { cos\theta} & { -rsin\theta} \\ { sin\theta} & { rcos\theta} \end{matrix} \right|=r $$ Similarly: $$dc\bar{t} d\bar{x}=J\left(ct,x\right)dctdx=\left| \begin{matrix} {\partial c\bar{t} \over \partial ct} & {\partial c\bar{t} \over \partial x} \\ {\partial \bar{x} \over \partial ct} & {\partial \bar{x} \over \partial x} \end{matrix} \right|dctdx$$ Recall the Lorentz transformation: $$c\bar{t} =\gamma (ct-\beta x)$$ $$\bar{x}=\gamma (x-\beta ct)$$ Substituting these back: $$dc\bar{t} d \bar{x}=\left| \begin{matrix} \gamma & -\beta \gamma \\ -\beta \gamma & \gamma \end{matrix} \right|dctdx $$ $$=\gamma ^2 (1-\beta ^2 )dctdx=dctdx$$ Thus, \(d^4 x\) is an invariant under Lorentz transformation.

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