According to David Tong's lectures on QFT, Noether's current is given by $$ J^\mu = \left( \sum_{\forall\text{ fields }\phi_{i}} \delta \phi_{i} \frac{\partial L}{\partial(\partial_{\mu}\phi_{i})} \right)-F^\mu \tag{1.38}$$ Given an infinitesimal transformation parametrized by $\varepsilon\in\mathbb{R}$ such that $x \to x^\prime = f(x,\varepsilon)$ and $\phi \to \phi^\prime = g(\phi,\varepsilon)$, how do I find $F^\mu$?

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    $\begingroup$ Eq. 1.35: $\delta\mathcal L=\partial_\mu F^\mu$. $\endgroup$ – AccidentalFourierTransform May 23 '18 at 23:30
  • $\begingroup$ So if $ \mathcal{L}\to\mathcal{L}^{\prime}=h(\mathcal{L},\varepsilon)\approx\mathcal{L}+\frac{\partial h}{\partial\varepsilon}\varepsilon\Rightarrow\delta\mathcal{L}=\frac{\partial h}{\partial\varepsilon}\varepsilon $ then $F^\mu$ is obtained by solving $ \frac{\partial h}{\partial\varepsilon}\varepsilon=\partial_{\mu}F^{\mu} $ ? $\endgroup$ – user171780 May 23 '18 at 23:44

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