Noether's Theorem

The Lagrangian \(\mathscr{L}_{EM}\) and the action \(S\) of electromagnetic field are

\begin{align*} \mathscr{L}_{EM} = -\frac{1}{16\pi c} g^{\mu\alpha}g^{\nu\beta}F_{\alpha\beta}F_{\mu\nu}\sqrt{-g} \end{align*}

and

\[ S = \int d^4x \, \mathscr{L}_{EM}[A_\mu(x^\gamma), A_{\nu,\mu}(x^\gamma), x^\gamma] \]

We derive the equation of motion (EoM) and Noether theorem follow standard procedure. The variation of action \(\Delta S\) divides into two terms:

\[ \delta S = \int \delta d^4x \cdot \mathscr{L}_{EM} + \int d^4x \cdot \delta \mathscr{L}_{EM} \]

The first term is the variation of volume form, which is \autocite{ryder1996quantum}

\[ \delta d^4x = {\delta{X}^\gamma }_{,\gamma} \cdot d^4x \]

The second term is the variation of \(\mathscr{L}_{EM}\)

\begin{align} \label{eq:ab_lag} \Delta \mathscr{L}_{EM} &= \mathscr{L}_{EM}[\tilde{A}_\nu(\tilde{x}^\gamma), \tilde{A}_{\nu,\mu}(\tilde{x}^\gamma), \tilde{x}^\gamma] - \mathscr{L}_{EM}[A_\nu(x^\gamma), A_{\nu,\mu}(x^\gamma), x^\gamma] \\ & = \left[ \frac{\partial \mathscr{L}_{EM}}{\partial A_\nu} \delta A_\nu + \frac{\partial \mathscr{L}_{EM}}{\partial (\partial_\mu A_\nu)} \delta (\partial_\mu A_\nu) \right] (x^\gamma) + \left[ {\mathscr{L}_{EM}}_{, \gamma} \delta x^\gamma \right] (x^\gamma) + O(\delta^2) \nonumber \end{align}

Hence, \(\Delta S\) is

\begin{align} \label{eq:Delta S_ab} \Delta S = \int \left[ \frac{\partial \mathscr{L}}{\partial A_\nu} - \partial_\mu\left(\frac{\partial\mathscr{L}_{EM}}{\partial(\partial_\mu A_\nu)}\right) \right] \delta A_\nu d^4 x + \int \left[ \partial_\mu\left(\frac{\partial \mathscr{L}_{EM}}{\partial (\partial_\mu A_\nu)}\delta A_\nu\right) + (\mathscr{L}_{EM} \delta x^\gamma)_{,\gamma}\right] d^4 x \end{align}

The EoM is

\[ \frac{\partial \mathscr{L}_{EM}}{\partial A_\nu} - \partial_\mu \left( \frac{\partial \mathscr{L}_{EM}}{\partial(\partial_\mu A_\nu)} \right) = 0 \]

Now we introduce the total variation:

\[\Delta A = \delta A + \hat{\mathcal{L}}_{\delta x} A \]

, instead of the traditional

\[\Delta A_\nu=\delta A_\nu +\delta x^\mu \partial_\mu A_\nu\]

. Using

\[\delta A_\nu = \Delta A_\nu - A_{\nu,\gamma} \delta x^\gamma - A_\gamma \delta x^\gamma_{,\nu}\]

,

\begin{align*} \Delta S &= \int \{EoM\} \delta A_\nu d^4 x + \int \partial_\mu\left[\frac{\partial \mathscr{L}_{EM}}{\partial(\partial_\mu A_\nu)} (\Delta A_\nu - A_{\nu,\gamma}\delta x^\gamma - A_\gamma \delta x_{,\nu}^\gamma) + \delta^\mu_\gamma\mathscr{L}_{EM} \delta x^\gamma\right]d^4x\\ &= \int \{EoM\} \delta A_\nu d^4 x + \int \partial_\mu \left[ \frac{\partial \mathscr{L}_{EM}}{\partial(\partial_\mu A_\nu)}\Delta A_\nu - \left( \frac{\partial \mathscr{L}_{EM}}{\partial(\partial_\mu A_\nu)}A_{\nu,\gamma} \delta x^\gamma + \underbracket[0.4pt][0pt]{\frac{\partial \mathscr{L}_{EM}}{\partial(\partial_\mu A_\nu)}A_\gamma \delta x^\gamma_{,\nu}}_{(*)} - \delta^\mu_\gamma \mathscr{L}_{EM}\delta x^\gamma\right)\right]d^4x \end{align*}

Evaluate the \((*)\) term:

\begin{align*} \underbracket[0.4pt][0pt]{\partial_\mu \left[ \frac{\partial \mathscr{L}_{EM}}{\partial(\partial_\mu A_\nu)} A_{\gamma} \delta x^\gamma_{,\nu} \right]}_{(*)} &= \underbracket[0.4pt][0pt]{\left[ \frac{\partial \mathscr{L}_{EM}}{\partial (\partial_\mu A_\nu)} A_\gamma \delta x^\gamma \right]_{,\nu \mu}}_{(*1)} -\partial_\mu \left[ \underbracket[0.4pt][0pt]{ \left( \frac{\partial \mathscr{L}_{EM}}{ \partial (\partial_\mu A_\nu)}\right)_{,\nu} A_\gamma \delta x^\gamma}_{(*2)}\right] -\partial_\mu \left[ \underbracket[0.4pt][0pt]{\frac{\partial \mathscr{L}_{EM}}{\partial(\partial_\mu A_\nu)}A_{\gamma,\nu} \delta x^\gamma}_{(*3)} \right] \end{align*}

Note that \(\frac{\partial\mathscr{L}_{EM}}{\partial(\partial_\mu A_{\nu})}\) is

\begin{align} \label{eq:ab_partial} \frac{\partial \mathscr{L}_{EM}}{\partial(\partial_\mu A_\nu)} &= -\frac{1}{4\pi c} g^{\mu\alpha}g^{\nu\beta} \sqrt{-g} F_{\alpha\beta} \end{align}

The \((*1)\) term

\begin{align} \label{eq:anti_ab} \underbracket[0.4pt][0pt]{\left[ \frac{\partial \mathscr{L}_{EM}}{\partial (\partial_\mu A_\nu)} A_\gamma \delta x^\gamma\right]_{,\nu \mu}}_{(*1)} = \left[ -\frac{1}{4\pi c} g^{\mu\alpha}g^{\nu\beta} \sqrt{-g} F_{\alpha\beta} A_\gamma \delta x^\gamma \right]_{,\nu \mu} = 0 \end{align}

due to the antisymmetric of \(F_{\mu\nu}\) and the symmetric of second order derivative \( \{_{,\nu\mu}\}\).\\

The \((*2)\) term:

\begin{align} \label{eq:*2_ab} \underbracket[0.4pt][0pt]{ \left( \frac{\partial \mathscr{L}_{EM}}{ \partial (\partial_\mu A_\nu)}\right)_{,\nu} A_\gamma \delta x^\gamma}_{(*2)}=0 \end{align}

Hence \((*)\) only left \((*3)\) term:

\[ \underbracket[0.4pt][0pt]{\partial_\mu\left[\frac{\partial \mathscr{L}_{EM}}{\partial(\partial_\mu A_\nu)} A_\gamma \delta x^\gamma \right]_{,\nu}}_{(*)} =-\partial_\mu\left[ \underbracket[0.4pt][0pt]{\frac{\partial \mathscr{L}_{EM}}{\partial(\partial_\mu A_\nu)} A_{\gamma, \nu} \delta x^\gamma }_{(3)} \right] \]

The final result,

\begin{align*} \Delta S &= \int \{EoM\}\delta A_\nu d^4 x + \int\partial_\mu \left[ \frac{\partial \mathscr{L}_{EM}}{\partial(\partial_\mu A_\nu)}\Delta A_\nu - \left( \frac{\partial \mathscr{L}_{EM}}{\partial(\partial_\mu A_\nu)} A_{\nu,\gamma} \delta x^\gamma - \underbracket[0.4pt][0pt]{ \frac{\partial \mathscr{L}_{EM}}{\partial(\partial_\mu A_\nu)} A_{\gamma, \nu} \delta x^\gamma }_{(*)=(3)} \right) - \delta^\mu_\gamma \mathscr{L}_{EM}\delta x^\gamma \right]d^4x \\ &= \int \{EoM\}\delta A_\nu d^4 x + \int \partial_\mu \left[ \frac{\partial \mathscr{L}_{EM}}{\partial(\partial_\mu A_\nu)}\Delta A_\nu - \left( \frac{\partial \mathscr{L}_{EM}}{\partial(\partial_\mu A_\nu)} F_{\gamma\nu} - \delta^\mu_\gamma \mathscr{L}_{EM} \right) \delta x^\gamma \right]d^4x \end{align*}

Hence we have

\begin{align*} \left(t_{EM}\right)^\mu_\gamma &= \frac{\partial \mathscr{L}_{EM}}{\partial(\partial_\mu A_\nu)} F_{\gamma\nu} - \delta^\mu_\gamma \mathscr{L}_{EM} \\ &=-\frac{1}{4\pi c} g^{\mu\alpha}g^{\nu\beta} F_{\alpha\beta} F_{\gamma\nu}\sqrt{-g} + \delta^\mu_\gamma \frac{1}{16\pi c} g^{\varepsilon\alpha}g^{\nu\beta}F_{\alpha\beta}F_{\varepsilon\nu}\sqrt{-g}\\ &= -\frac{1}{4\pi c}F^{\mu\nu}F_{\gamma\nu}\sqrt{-g} + \delta^\mu_\gamma \frac{1}{16\pi c} F^{\alpha\beta}F_{\alpha\beta}\sqrt{-g}\\ &=\underbracket[0.4pt][0pt]{\left( -\frac{1}{4\pi c}F^{\mu\nu}F_{\gamma\nu} + \delta^\mu_\gamma \frac{1}{16\pi c} F^{\alpha\beta}F_{\alpha\beta}\right)}_{\left({T_{EM}}\right)^\mu_\gamma}\sqrt{-g} \\ &=-\frac{1}{4\pi c}\delta^\mu_\varepsilon g^{\varepsilon\alpha} g^{\nu\beta} F_{\alpha\beta} F_{\gamma\nu}\sqrt{-g} + \delta^\mu_\gamma \frac{1}{16\pi c} g^{\varepsilon\alpha}g^{\nu\beta}F_{\alpha\beta}F_{\varepsilon\nu}\sqrt{-g}\\ &=\frac{1}{4\pi c}\left(-\delta^\mu_\varepsilon F_{\alpha\beta} F_{\gamma\nu} + \delta^\mu_\gamma \frac{1}{4} F_{\alpha\beta}F_{\varepsilon\nu}\right)g^{\varepsilon\alpha} g^{\nu\beta} \sqrt{-g} \end{align*} . This symmetric and gauge invariant.

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