# Noether's 2nd Theorem and Local Gauge Identities

I am trying to derive the so called Gauge Identities: $$D_\nu\frac{\delta S}{\delta\phi} = 0$$ Where $D_\nu$ is an operator involving derivatives and $\frac{\delta S}{\delta\phi}$ are the usual Euler-Lagrange equations.

So far I have taken the following local field transformation: $$\bar{\delta}\phi(x) = \varphi_\nu\lambda^\nu(x)\simeq \varphi_\nu\lambda^\nu + \varphi^\mu_\nu\partial_\mu\lambda^\nu$$

And varied the action: \begin{align} \bar{\delta}S &= \int d^4x~\left(\frac{\partial\mathcal{L}}{\partial\phi}\bar{\delta}\phi + \frac{\partial\mathcal{L}}{\partial\partial_\mu\phi}\partial_\mu\bar{\delta}\phi\right)\\ &= \int d^4x~\left(\frac{\partial\mathcal{L}}{\partial\phi}\left(\varphi_\nu^0\lambda^\nu + \varphi_\nu^\rho\partial_\rho\lambda^\nu\right) + \frac{\partial\mathcal{L}}{\partial\partial_\mu\phi}\partial_\mu\left(\varphi_\nu^0\lambda^\nu + \varphi_\nu^\rho\partial_\rho\lambda^\nu\right)\right)\\ &\text{Integrate the second term by parts to get}\\ &= \int d^4x~\left(\frac{\partial\mathcal{L}}{\partial\phi}\left(\varphi_\nu^0\lambda^\nu + \varphi_\nu^\rho\partial_\rho\lambda^\nu\right) - \partial_\mu\frac{\partial\mathcal{L}}{\partial\partial_\mu\phi}\left(\varphi_\nu^0\lambda^\nu + \varphi_\nu^\rho\partial_\rho\lambda^\nu\right)\right)\\ &= \int d^4x~\left[\frac{\partial\mathcal{L}}{\partial\phi} - \partial_\mu\left(\frac{\partial\mathcal{L}}{\partial\partial_\mu\phi}\right)\right]\varphi_\nu^0\lambda^\nu + \left[\frac{\partial\mathcal{L}}{\partial\phi} - \partial_\mu\left(\frac{\partial\mathcal{L}}{\partial\partial_\mu\phi}\right)\right]\varphi_\nu^\rho\partial_\rho\lambda^\nu\\ &\text{Again integrate the second term by parts to get}\\ &= \int d^4x~\left[\frac{\partial\mathcal{L}}{\partial\phi} - \partial_\mu\left(\frac{\partial\mathcal{L}}{\partial\partial_\mu\phi}\right)\right]\varphi_\nu^0\lambda^\nu - \partial_\rho\left(\left[\frac{\partial\mathcal{L}}{\partial\phi} - \partial_\mu\left(\frac{\partial\mathcal{L}}{\partial\partial_\mu\phi}\right)\right]\varphi_\nu^\rho\right)\lambda^\nu\\ &=\int d^4x~\partial_\mu \mathcal{J}^\mu(\lambda) \end{align}

Recognising this stuff in the integral as the operator $D_\nu$, I get the following:

$$\int d^4x ~\left(\partial_\mu \mathcal{J}^\mu(\lambda) - \lambda^\nu D_\nu\frac{\delta S}{\delta\phi}\right) = 0$$

What I don't understand is how to now see that $D_\nu\frac{\delta S}{\delta\phi} = 0$ for arbitrary $\lambda$.

What if I choose a parameter that doesn't make $\mathcal{J}$ vanish on the surface for example?

• – Qmechanic Aug 14 '15 at 10:26

Lets denote the spacetime manifold by $M$. For any localized function $\lambda$ without support at the boundaries of spacetime, the first term in your last equation is vanishing, since it can be recast into a boundary integral, i.e. $$\int_M \partial_\mu \mathcal{J}^\mu(\lambda)=\int_{\partial M}d\Sigma_\mu \mathcal{J}^\mu(\lambda)=0$$ Now take a closed subregion $N\subset M$ and consider functions $\lambda$ on $M$ such that they are arbitrary on $N$ while vanishing outside. Using the above argument in your last equation implies that $$\int_N d^4x ~\lambda^\nu D_\nu\frac{\delta S}{\delta\phi}= 0$$ Since $\lambda$ is arbitrary on $N$, we find that $D_\nu\frac{\delta S}{\delta\phi}= 0$ on $N$. Repeating this argument for a sequence of patches $N_i$, which altogether cover $M$, we conclude that $D_\nu\frac{\delta S}{\delta\phi}= 0$ on the whole manifold $M$. Would you agree?