# Noether's Theorem in Non-Local Field Theories
## 2 Symmetries and the Conserved Currents
However, the order of the matrices \(S(a)\) and \(U(\omega)\) is interchangeable as the transformations of internal symmetries commute with those of external symmetries. Thus, the first term on the right-hand side of Equation (6) corresponds to transformations of internal symmetries, the second corresponds to translations, and the third corresponds to the Lorentz transformations. Returning to Equation (1) for the infinitesimal parameters \(\omega^{a}\), \(\varepsilon^{\mu\
u}\), and \(b^{\mu}\), one can now write the variation of the Lagrangian, \(\delta\mathcal{L}(x)=\mathcal{L}^{\prime}(x)-\mathcal{L}(x)\), as  
\[-b^{\
u}\partial_{\sigma}\mathcal{L}-\varepsilon^{\
u\
u}x_{\
u}\partial_{ \sigma}\mathcal{L}=\frac{\partial\mathcal{L}}{\partial\Psi}\delta\Psi+\sum_{n \geq 1}\frac{\partial\mathcal{L}}{\partial(\partial_{\mu_{1}}\ldots\partial_{ \mu_{n}}\Psi)}\partial_{\mu_{1}}\ldots\partial_{\mu_{n}}\delta\Psi. \tag{7}\]  
To derive the expression for the conserved current, one must use the generalized higher-order Euler-Lagrange equation  
\[\frac{\partial\mathcal{L}}{\partial\Psi}+\sum_{n\geq 1}(-)^{n}\partial_{\mu_{ n}}\ldots\partial_{\mu_{1}}\frac{\partial\mathcal{L}}{\partial(\partial_{ \mu_{1}}\ldots\partial_{\mu_{n}}\Psi)}=0. \tag{8}\]  
By replacing the first term on the right-hand side of Equation (7) with the corresponding expression from the Euler-Lagrange Equation (8), one can rewrite the right-hand side of Equation (7) as follows:  
\[\text{r.h.s.}=-\sum_{n\geq 1}(-)^{n}\partial_{\mu_{n}}\ldots\partial_{\mu_{1}} \frac{\partial\mathcal{L}}{\partial(\partial_{\mu_{1}}\ldots\partial_{\mu_{ n}}\Psi)}\delta\Psi+\sum_{n\geq 1}\frac{\partial\mathcal{L}}{\partial(\partial_{ \mu_{1}}\ldots\partial_{\mu_{n}}\Psi)}\partial_{\mu_{1}}\ldots\partial_{\mu_{ n}}\delta\Psi. \tag{9}\]  
The purpose is to represent the expression (9), using the form of the divergence of some quantity. The \(n\)-th order term under the first summation symbol in Equation (9) can be rewritten in the form  
\[ \begin{split}&-(-)^{n}\partial_{\mu_{n}}\ldots\partial_{\mu_{1}} \frac{\partial\mathcal{L}}{\partial(\partial_{\mu_{1}}\ldots\partial_{\mu_{n}} \Psi)}\delta\Psi\\ =&-(-)^{n}\partial_{\mu_{n}}\left(\partial_{\mu_{n-1 }}\ldots\partial_{\mu_{1}}\frac{\partial\mathcal{L}}{\partial(\partial_{\mu_{1 }}\ldots\partial_{\mu_{n}}\Psi)}\delta\Psi\right)\\ &-(-)^{n+1}\partial_{\mu_{n-1}}\ldots\partial_{\mu_{1}}\frac{ \partial\mathcal{L}}{\partial(\partial_{\mu_{1}}\ldots\partial_{\mu_{n}}\Psi) }\partial_{\mu_{n}}\delta\Psi.\end{split}  \tag{10}\]The first term has the form of a divergence, whereas in the second term, the derivative \(\partial_{\mu_{n}}\) is shifted to the right and acts on \(\delta\Psi\). By rewriting the second term of (10) in the same way,  
\[-(-)^{n+1}\partial_{\mu_{n-1}}\dots\partial_{\mu_{1}}\frac{ \partial\mathcal{L}}{\partial(\partial_{\mu_{1}}\dots\partial_{\mu_{n}}\Psi)} \partial_{\mu_{n}}\delta\Psi \tag{11}\] \[= -(-)^{n+1}\partial_{\mu_{n-1}}\left(\partial_{\mu_{n-2}}\dots \partial_{\mu_{1}}\frac{\partial\mathcal{L}}{\partial(\partial_{\mu_{1}}\dots \partial_{\mu_{n}}\Psi)}\partial_{\mu_{n}}\delta\Psi\right)\] \[-(-)^{n+2}\partial_{\mu_{n-2}}\dots\partial_{\mu_{1}}\frac{ \partial\mathcal{L}}{\partial(\partial_{\mu_{1}}\dots\partial_{\mu_{n}}\Psi)} \partial_{\mu_{n-1}}\partial_{\mu_{n}}\delta\Psi,\]  
we again obtain a divergence and one more derivative of \(\delta\Psi\) in the second term. This implies that through such recursion, one can shift the derivative to the right until it lies immediately before \(\delta\Psi\). With each such procedure, the second term in the rewritten part of the expression changes sign and the first term has the form of a divergence. Finally, the last term in the recursion can be obtained by shifting over \(n\) derivatives; this term will have an additional sign \((-1)^{n}\), and consequently, it will have a sign opposite to that of the second term of Equation (9) and will therefore vanish.