2
$\begingroup$

I am trying to derive the so called Gauge Identities: \begin{equation} D_\nu\frac{\delta S}{\delta\phi} = 0 \end{equation} 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: \begin{equation} \bar{\delta}\phi(x) = \varphi_\nu\lambda^\nu(x)\simeq \varphi_\nu\lambda^\nu + \varphi^\mu_\nu\partial_\mu\lambda^\nu \end{equation}

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:

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

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?

$\endgroup$
0
$\begingroup$

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 \begin{equation} \int_N d^4x ~\lambda^\nu D_\nu\frac{\delta S}{\delta\phi}= 0 \end{equation} 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?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.