Noether's Theorem

We denote \(p\) is a point in spacetime, \(\mathbf{B}(p)\) as Lie algebra valued gauge potential 1-form of the Yang-Mills field, and the field strength is \(\mathbf{F} \equiv d\mathbf{B} + [\mathbf{B} \wedge \mathbf{B}]\). We will discuss the effect of the variation on gauge potential and the spacetime variation. We denote the variation on gauge potential, \(\mathbf{B} \rightarrow \mathbf{\tilde{B}}= \mathbf{B} + \delta \mathbf{B}\), and the spacetime variation drag by a vector field \(\delta x\) denote as:

\[ p \rightarrow \tilde{p} = f_{\delta x}(p) \]

The total variation of gauge 1-form is

\[ \Delta \mathbf{B} = \mathbf{\tilde{B}}(\tilde{p}) - \textbf{B}(p) = \delta \mathbf{B} + \hat{\mathcal{L}}_{\delta x} \mathbf{B} \]

In local coordinate \(p \rightarrow \{x^\mu\}\), the expressions are

\[ \mathbf{B}(p) \rightarrow B^a_\mu(x^\gamma) \hat{T}_a \]

where \(\hat{T}_a\) is the generator of Lie algebra. The field strength in local coordinate

\begin{align} \label{eq:Guv} \mathbf{F} \rightarrow \mathbf{F}_{\mu\nu} &= \hat{T}_a \partial_\mu B_\nu^a - \hat{T}_a \partial_\nu B_\mu^a + \lambda [B_\mu^a \hat{T}_a, B_\nu^b \hat{T}_b] \nonumber \\ &= \hat{T}_a \left( \partial_\mu B_\nu^a - \partial_\nu B_\mu^a+ \lambda f_{bc}^a B_\mu^b B_\nu^c \right) \end{align}

where

\[ [\hat{T}_a, \hat{T}_b] = f_{ab}^c \hat{T}_c \]

and

\[ F_{\mu\nu}^a = \partial_\mu B_\nu^a - \partial_\nu B_\mu^a + \lambda f^a_{bc}B^b_\mu B^c_\nu \]

. \(f^c_{ab}\) is the structure constant of Lie algebra. The local coordinate representation of variations are

\[p \rightarrow x^\nu = x^\nu + \delta x^\nu\]

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

The Lagrangian \(\mathscr{L}_{YM}\) of Yang-Mills field is

\begin{align} \label{eq:Lagrangian} \mathscr{L}_{YM} = Tr\left( -\frac{1}{16\pi c} g^{\mu\alpha} g^{\nu\beta} \mathbf{F}_{\mu\nu} \mathbf{F} _{\alpha\beta} \sqrt{-g} \right) = -\frac{1}{16\pi c} K_{ab} g^{\mu\alpha} g^{\nu\beta} F^a_{\mu\nu} F^b_{\alpha\beta} \sqrt{-g} \end{align}

Here, we denote the Killing form/metric as

\[ K_{ab} = Tr(f_{ad}^c f_{be}^d) = f^c_{ad} f^d_{bc} \]

The action \(S\)

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

We derive the equation of motion (EoM) and the Noether theorem follow the standard procedure [1]. The variation of action \(\Delta S\) divides into two terms:

\[ \Delta S = \int \Delta d^4 x * \mathscr{L}_{YM} + \int d^4 x * \Delta \mathscr{L}_{YM} \]

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

\[ \Delta d^4 x = \partial_\gamma \delta x^\gamma \cdot d^4x \]

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

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

Hence the \(\Delta S\) is

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

The EoM is

\[ \frac{\partial \mathscr{L}_{YM}}{\partial B^a_\nu} - \partial_\mu \left( \frac{\partial \mathscr{L}_{YM}}{\partial(\partial_\mu B^a_\nu)}\right) = 0 \]

Using \(\delta B_\nu^a = \Delta B_\nu^a - B_{\nu,\gamma}^a \delta x^\gamma - B_\gamma^a \delta x_{,\nu}^\gamma\):

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

Evaluate the \((*)\) term:

\begin{align} \label{eq:(*)} \underbracket[0.4pt][0pt]{\partial_\mu \left[ \frac{\partial \mathscr{L}_{YM}}{\partial(\partial_\mu B^a_\nu)} B^a_{\gamma} \delta x^\gamma_{,\nu} \right]}_{(*)} &= \underbracket[0.4pt][0pt]{\left[ \frac{\partial \mathscr{L}_{YM}}{\partial (\partial_\mu B^a_\nu)} B^a_\gamma \delta x^\gamma \right]_{,\nu \mu}}_{(*1)} -\partial_\mu \left[\underbracket[0.4pt][0pt]{ \left( \frac{\partial \mathscr{L}_{YM}}{ \partial (\partial_\mu B^a_\nu)}\right)_{,\nu} B^a_\gamma \delta x^\gamma }_{(*2)} \right] - \partial_\mu \left[\underbracket[0.4pt][0pt]{\frac{\partial \mathscr{L}_{YM}}{\partial(\partial_\mu B^a_\nu)}B^a_{\gamma,\nu} \delta x^\gamma }_{(*3)} \right] \\ &=\underbracket[0.4pt][0pt]{ 0}_{(*1)} \textcolor{red}{+}\partial_\mu \left[ \underbracket[0.4pt][0pt]{ \left( \frac{\partial \mathscr{L}_{YM}}{ \partial \left( \partial_{\textcolor{red}{\nu}} B^a_{\textcolor{red}{\mu}}\right) } \right)_{,\nu} B^a_\gamma \delta x^\gamma }_{(*2)} \right] \textcolor{blue}{ -\partial_\mu\left[\left( \frac{\partial \mathscr{L}_{YM}}{ \partial B^a_\mu} \right) B^a_\gamma \delta x^\gamma \right]} \textcolor{blue}{ +\partial_\mu\left[\left( \frac{\partial \mathscr{L}_{YM}}{ \partial B^a_\mu} \right) B^a_\gamma \delta x^\gamma \right]} -\partial_\mu \left[\underbracket[0.4pt][0pt]{\frac{\partial \mathscr{L}_{YM}}{\partial(\partial_\mu B^a_\nu)}B^a_{\gamma,\nu} \delta x^\gamma }_{(*3)} \right] \\ &=\underbracket[0.4pt][0pt]{ 0}_{(*1)} +\partial_\mu \left[ \underbracket[0.4pt][0pt]{ \left[ \left( \frac{\partial \mathscr{L}_{YM}}{ \partial \left( \partial_{\nu} B^a_{\mu}\right) } \right)_{,\nu} -\frac{\partial \mathscr{L}_{YM}}{ \partial B^a_\mu} \right] B^a_\gamma \delta x^\gamma }_{(*2)} \right] +\partial_\mu \left[\frac{\partial\mathscr{L}_{YM}}{\partial(\partial_{\mu}B_{\nu}^{a})} \lambda f_{bn}^{a} B^{b}_\gamma B_{\nu}^{n}\delta x^\gamma \right] -\partial_\mu \left[\underbracket[0.4pt][0pt]{\frac{\partial \mathscr{L}_{YM}}{\partial(\partial_\mu B^a_\nu)}B^a_{\gamma,\nu} \delta x^\gamma }_{(*3)} \right] \end{align}

We first calculate \(\frac{\partial \mathscr{L}_{YM}}{\partial (\partial_\mu B_\nu^a)}\) for later use:

\begin{align} \label{eq:Non-ab_partial} \frac{\partial \mathscr{L}_{YM}}{\partial (\partial_\mu B_\nu^a)} = -\frac{1}{4\pi c} K_{ab} g^{\alpha\mu} g^{\beta\nu} F_{\alpha\beta}^b \sqrt{-g} \end{align}

The \((*1)\) term term:

\begin{align} \label{eq:(*1)} \left( \frac{\partial \mathscr{L}_{YM}}{\partial (\partial_\mu B_\nu^a)} B_\gamma^a \delta x^\gamma \right)_{,\nu\mu} =\left( -\frac{1}{4\pi c} K_{ab} g^{\alpha\mu} g^{\beta\nu} F_{\alpha\beta}^b \sqrt{-g} B_\gamma^a \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 is:

\begin{align} \label{eq:(*2)} \underbracket[0.4pt][0pt]{ \left( \frac{\partial \mathscr{L}_{YM}}{ \partial (\partial_\mu B^a_\nu)}\right)_{,\nu} B^a_\gamma \delta x^\gamma}_{(*2)} = \frac{\partial\mathscr{L}_{YM}}{\partial(\partial_{\mu}B_{\nu}^{{a}})}\left( \lambda f_{{b}n}^{{a}}B_{\nu}^{n} B^{{b}}_\gamma \delta x^\gamma \right) \end{align}

We now have: Hence \((*)\) becomes:

\[ \underbracket[0.4pt][0pt]{\partial_\mu \left[ \frac{\partial \mathscr{L}_{YM}}{\partial(\partial_\mu B^a_\nu)} B^a_{\gamma} \delta x^\gamma_{,\nu} \right]}_{(*)} =-\partial_\mu \left[\underbracket[0.4pt][0pt]{-\frac{\partial\mathscr{L}_{YM}}{\partial(\partial_{\mu}B_{\nu}^{a})} \lambda f_{bn}^{a} B^{b}_\gamma B_{\nu}^{n}\delta x^\gamma }_{(*2)} \right]-\partial_\mu \left[\underbracket[0.4pt][0pt]{\frac{\partial \mathscr{L}_{YM}}{\partial(\partial_\mu B^a_\nu)}B^a_{\gamma,\nu} \delta x^\gamma }_{(*3)}\right] \] \begin{align*} \Delta S &= \int \{EoM\} \delta B_\nu^a \, d^4x + \int \partial_\mu \left[ \frac{\partial \mathscr{L}_{YM}}{\partial(\partial_\mu B_\nu^a)} \Delta B_\nu^a - \frac{\partial \mathscr{L}_{YM}}{\partial(\partial_\mu B_\nu^a)} \left( B^a_{\nu, \gamma} - \underbracket[0.4pt][0pt]{B^a_{\gamma,\nu} }_{(*3)} +\underbracket[0.4pt][0pt]{ \lambda f_{bn}^{a} B^{b}_\gamma B_{\nu}^{n} }_{(*2)}\right) \delta x^\gamma + \delta^\mu_\gamma \mathscr{L}_{YM} \delta x^\gamma \right] \, d^4x \\ &=\int \{EoM\} \delta B_\nu^a \, d^4x + \int \partial_\mu \left[ \frac{\partial \mathscr{L}_{YM}}{\partial(\partial_\mu B_\nu^a)} \Delta B_\nu^a - \left( \frac{\partial \mathscr{L}_{YM}}{\partial(\partial_\mu B_\nu^a)} F^a_{\gamma\nu} - \delta^\mu_\gamma \mathscr{L}_{YM}\right) \delta x^\gamma \right] \, d^4x \end{align*}

Hence we have

\begin{align*} \left(t_{YM}\right)^\mu_\gamma &= \frac{\partial \mathscr{L}_{YM}}{\partial(\partial_\mu B_\nu^a)} F^a_{\gamma\nu} - \delta^\mu_\gamma \mathscr{L}_{YM}\\ &= \left( -\frac{1}{4\pi c} K_{ab} g^{\alpha\mu} g^{\beta\nu} F_{\alpha\beta}^b \sqrt{-g} \right)F^a_{\gamma\nu} - \delta^\mu_\gamma \left( -\frac{1}{16\pi c} K_{ab} g^{\mu\alpha} g^{\nu\beta} F^a_{\mu\nu} F^b_{\alpha\beta} \sqrt{-g} \right)\\ &= -\frac{1}{4\pi c} F^{\mu\nu}_a F^a_{\gamma\nu} \sqrt{-g} + \delta^\mu_\gamma \frac{1}{16\pi c} F^{\alpha\beta}_a F^a_{\alpha\beta} \sqrt{-g} \end{align*} \begin{align*} g_{\psi\mu}g^{\phi\gamma}\left(t_{YM}\right)^\mu_\gamma &= g_{\psi\mu}g^{\phi\gamma}\left( -\frac{1}{4\pi c} F^{\mu\nu}_a F^a_{\gamma\nu} \sqrt{-g} + \delta^\mu_\gamma \frac{1}{16\pi c} F^{\alpha\beta}_a F^a_{\alpha\beta} \sqrt{-g} \right)\\ &= -\frac{1}{4\pi c} g_{\psi\mu}g^{\phi\gamma} F^{\mu\nu}_a F^a_{\gamma\nu} \sqrt{-g} +g_{\psi\mu}g^{\psi\gamma} \delta^\mu_\gamma \frac{1}{16\pi c} F^{\alpha\beta}_a F^a_{\alpha\beta} \sqrt{-g} \\ &= -\frac{1}{4\pi c} g_{\psi\mu}g^{\phi\gamma} g^{\mu\zeta}g^{\nu\xi}F^a_{\zeta\xi} F^a_{\gamma\nu} \sqrt{-g} +\delta^\phi_\psi \frac{1}{16\pi c} F^{\alpha\beta}_a F^a_{\alpha\beta} \sqrt{-g} \\ &= -\frac{1}{4\pi c} g^{\phi\gamma} g^{\nu\xi}F^a_{\psi\xi} F^a_{\gamma\nu} \sqrt{-g} +\delta^\phi_\psi \frac{1}{16\pi c} F^{\alpha\beta}_a F^a_{\alpha\beta} \sqrt{-g} \\ &= -\frac{1}{4\pi c} g_{\psi\mu}g^{\phi\gamma} g^{\mu\zeta}g^{\nu\xi}F^a_{\zeta\xi} F^a_{\gamma\nu} \sqrt{-g} +\delta^\phi_\psi \frac{1}{16\pi c} F^{\alpha\beta}_a F^a_{\alpha\beta} \sqrt{-g} \\ &= -\frac{1}{4\pi c} F^a_{\psi\xi} F_a^{\phi\xi} \sqrt{-g} +\delta^\phi_\psi \frac{1}{16\pi c} F^{\alpha\beta}_a F^a_{\alpha\beta} \sqrt{-g} \\ &=\left(t_{YM}\right)^\phi_\psi \end{align*}

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