In this third part of the series, I will continue the deduction of Noether's theorem initiated in the previous post - Does it make sense to speak in a total derivative of a functional? Part II.
Situation 1
Here, I will consider the validity of the total derivative \begin{equation} \frac{d\mathcal{L}}{dx^{μ}} =\frac{\partial\mathcal{L}}{\partialφ_{r}}\partial_{\mu}φ_{r}+\frac{\partial\mathcal{L}}{\partial\big(\partial_{ν}φ_{r}\big)}\partial_{\mu}\big(\partial_{ν}φ_{r}\big)+∂_{μ}\mathcal{L}.\tag{III.1}\label{eq1} \end{equation}
We have expressed in Eq. (\ref{eq24}) of the previous post (Does it make sense to speak in a total derivative of a functional? Part II) that \begin{multline} \dfrac{S^{\prime}-S}{\varepsilon} \approx \int_{\mathbb{\Omega}}d^{D}x~\left\{ \dfrac{\partial\mathcal{L}}{\partial\phi_{r}}\zeta_{r} + \dfrac{\partial \mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\partial_{\nu}\zeta_{r}\right. \\ \left. + \xi^{\mu }\left( \dfrac{\partial\mathcal{L}}{\partial\phi_{r}}\partial_{\mu}\phi _{r}+\dfrac{\partial\mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\partial_{\mu }\partial_{\nu}\phi_{r}+\partial_{\mu}\mathcal{L}\right) +\partial_{\mu}% \xi^{\mu}\mathcal{L}\right\}, \tag{II. 24}\label{eq24}% \end{multline} where I'd like to remember that $\zeta_r\equiv\zeta_r(x)$ and $\xi^{\mu}\equiv\xi^{\mu}(x)$.
If what we ask about Eq. (I.$9$) in the first post of this Series (Does it make sense to speak in a total derivative of a functional? Part I) has a yes as an answer, then the following identifications must be valid: \begin{equation} \frac{d\zeta_{r}}{dx^{\mu}}=\partial_{\mu}\zeta_{r} \quad\text{and}\quad \frac{d\xi^{\mu}}{dx^{\mu}}=\partial_{\mu}\xi^{\mu}.\tag{III.2} \end{equation} Thus, the Eq. (\ref{eq24}) becomes \begin{equation} \dfrac{S^{\prime}-S}{\varepsilon}\approx\int_{\mathbb{\Omega}}d^{D}x~\left\{ \dfrac{\partial\mathcal{L}}{\partial\phi_{r}}\zeta_{r}+\dfrac{\partial \mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\dfrac{d\zeta_{r}}{dx^{\nu}} +\xi^{\mu}\dfrac{d\mathcal{L}}{dx^{\mu}}+\mathcal{L}\dfrac{d\xi^{\mu}} {dx^{\mu}}\right\}.\tag{III.3}\label{eq3} \end{equation} Now, we do use of identity \begin{equation} \dfrac{\partial\mathcal{L}}{\partial\partial_{\mu}\phi_{r}} \dfrac{d\zeta_{r}}{dx^{\mu}}=\frac{d}{dx^{\mu}}\left(\zeta_{r}\frac{\partial\mathcal{L} }{\partial\partial_{\mu}\phi_{r}}\right)-\zeta_{r}\frac{d}{dx^{\mu}}\frac{\partial\mathcal{L}}{\partial\partial_{\mu}\phi_{r}},\tag{III.4}\label{eq4} \end{equation} such that \begin{equation} \dfrac{S^{\prime}-S}{\varepsilon}\approx\int_{\mathbb{\Omega}}d^{D}x~\left\{ \left(\dfrac{\partial\mathcal{L}}{\partial\phi_{r}}-\dfrac{d}{dx^{\nu} }\dfrac{\partial\mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\right) \zeta _{r}+\dfrac{d}{dx^{\nu}}\left( \zeta_{r}\dfrac{\partial\mathcal{L}}{\partial\partial_{\nu}\phi_{r}}+\xi^{\mu}\mathcal{L}\right) \right\},\tag{III.5}\label{eq5} \end{equation} where we have used \begin{equation} \xi^{\mu}\dfrac{d\mathcal{L}}{dx^{\mu}}+\mathcal{L}\dfrac{d\xi^{\mu}}{dx^{\mu }}=\dfrac{d}{dx^{\mu}}\left( \xi^{\mu}\mathcal{L}\right). \end{equation}
We have to said in Does it make sense to speak in a total derivative of a functional? Part II, Eq.(\ref{II19}), that \begin{equation} \zeta_{r}\left( x\right) +\xi^{\mu }\left( x\right) \partial_{\mu}\phi_{r}\left( x\right) =\dfrac{\tilde{\delta}\phi_{r}}{\varepsilon}=\chi_{r}\left( x\right) ,\tag{II.19}\label{II19} \end{equation} so that (\ref{eq5}) becomes \begin{multline} \dfrac{S^{\prime}-S}{\varepsilon}\approx\int_{\mathbb{\Omega}}d^{D}x~\left( \dfrac{\partial\mathcal{L}}{\partial\phi_{r}}-\dfrac{d}{dx^{\nu}} \dfrac{\partial\mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\right) \zeta _{r}\\ +\int_{\mathbb{\Omega}}d^{D}x~\dfrac{d}{dx^{\mu}}\left[ \dfrac {\partial\mathcal{L}}{\partial\partial_{\mu}\phi_{r}}\chi_{r}-\left( \dfrac{\partial\mathcal{L}}{\partial\partial_{\mu}\phi_{r}}\partial_{\nu} \phi_{r}-\delta_{\nu}^{\mu}\mathcal{L}\right) \xi^{\nu}\right].\tag{III.6}\label{eq6} \end{multline}
And now comes the question: how can we apply the generalized divergence theorem in the second integral on the right side-hand if instead of a partial derivative we have a total derivative?
Situation 2
Before asking the question, let's see what happens if we do not use Eq. (\ref{eq1}). In this case, we can rewrite the Eq. (\ref{eq24}) as: \begin{equation} \dfrac{S^{\prime}-S}{\varepsilon}\approx\int_{\mathbb{\Omega}}d^{D}x~\left\{ \dfrac{\partial\mathcal{L}}{\partial\phi_{r}}\chi_{r}+\dfrac{\partial \mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\left( \partial_{\nu}\zeta _{r}+\xi^{\mu}\partial_{\mu}\partial_{\nu}\phi_{r}\right) +\partial_{\nu }\left( \xi^{\nu}\mathcal{L}\right) \right\},\tag{III.7}\label{eq7} \end{equation} where we have used (\ref{II19}).
If we add and subtract the term $\partial_{\mu} \phi_{r}\partial_{\nu}\xi^{\mu}$ in the expression in parentheses of the second term, that last equation becomes \begin{equation} \dfrac{S^{\prime}-S}{\varepsilon}\approx\int_{\mathbb{\Omega}}d^{D}x~\left\{ \dfrac{\partial\mathcal{L}}{\partial\phi_{r}}\chi_{r}+\dfrac{\partial \mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\partial_{\nu}\chi_{r} -\dfrac{\partial\mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\partial_{\mu} \phi_{r}\partial_{\nu}\xi^{\mu}+\partial_{\nu}\left( \xi^{\nu}\mathcal{L} \right) \right\}.\tag{III.8} \end{equation} Now, using the identities \begin{align} \dfrac{\partial\mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\partial_{\nu} \chi_{r}&=\partial_{\nu}\left( \chi_{r}\dfrac{\partial\mathcal{L}} {\partial\partial_{\nu}\phi_{r}}\right) -\chi_{r}\partial_{\nu} \dfrac{\partial\mathcal{L}}{\partial\partial_{\nu}\phi_{r}},\tag{III.9}\label{eq9}\\ -\dfrac{\partial\mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\partial_{\mu} \phi_{r}\partial_{\nu}\xi^{\mu}&=-\partial_{\nu}\left( \dfrac{\partial \mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\partial_{\mu}\phi_{r}\xi^{\mu }\right) +\xi^{\mu}\partial_{\nu}\left( \dfrac{\partial\mathcal{L}} {\partial\partial_{\nu}\phi_{r}}\partial_{\mu}\phi_{r}\right),\tag{III.10}\label{eq10} \end{align} we obtain \begin{multline} \dfrac{S^{\prime}-S}{\varepsilon}\approx\int_{\mathbb{\Omega} }d^{D}x~\xi^{\mu}\partial_{\nu}\left( \dfrac{\partial\mathcal{L}} {\partial\partial_{\nu}\phi_{r}}\partial_{\mu}\phi_{r}\right)+\int_{\mathbb{\Omega}}d^{D}x~\left( \dfrac{\partial\mathcal{L}}{\partial\phi_{r}}-\partial_{\nu}\dfrac {\partial\mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\right) \chi_{r} \\+ \int_{\mathbb{\Omega}}d^{D}x~\partial_{\nu}\left[ \dfrac{\partial \mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\chi_{r}-\left( \dfrac {\partial\mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\partial_{\mu}\phi _{r}-\xi^{\nu}\mathcal{L}\right) \xi^{\mu}\right].\tag{III.11}\label{eq11} \end{multline}
Here, considering the validity of Euler-Lagrange's equation \begin{equation} \dfrac{\partial\mathcal{L}}{\partial\phi_{r}}-\partial_{\nu}\dfrac {\partial\mathcal{L}}{\partial\partial_{\nu}\phi_{r}}=0, \tag{III.12}\label{eq12} \end{equation} and the applicability of divergence theorem over to third integral (Which now seems to be quite reasonable!) \begin{equation} \int_{\mathbb{\Omega}}d^{D}x~\partial_{\nu}J^{\nu}=\oint_{\partial \mathbb{\Omega}}dS_{\nu}~J^{\nu}=0,\tag{III.13}\label{eq13} \end{equation} with \begin{equation} J^{\nu}=\dfrac{\partial\mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\chi _{r}-\left( \dfrac{\partial\mathcal{L}}{\partial\partial_{\nu}\phi_{r} }\partial_{\mu}\phi_{r}-\delta_{\mu}^{\nu}\mathcal{L}\right) \xi^{\mu},\tag{III.14}\label{eq14} \end{equation} when $\varepsilon\rightarrow 0$, we have found \begin{equation} \lim_{\varepsilon\rightarrow 0} \dfrac{S^{\prime}-S}{\varepsilon}\approx\int_{\mathbb{\Omega}}d^{D}x~\xi^{\mu }\partial_{\nu}\left( \dfrac{\partial\mathcal{L}}{\partial\partial_{\nu} \phi_{r}}\partial_{\mu}\phi_{r}\right),\tag{III.15}\label{eq15} \end{equation} which at first seems to be non-zero.
As we know, it is hoped that \begin{equation} \lim_{\varepsilon\rightarrow 0} \dfrac{S^{\prime}-S}{\varepsilon}\approx 0.\tag{III.16}\label{eq16} \end{equation}
Questions
We have, therefore, two questions:
In the situation (1), when we use the total derivative (\ref{eq1}), the divergence theorem seems nonapplicable over the second integral of the Eq. (\ref{eq6}), so the question is: Is it still possible to apply the divergence theorem the second integral (Eq.(\ref{eq6}))?
In the situation (2), when we do not use the total derivative, we have a remaining term that is apparently is not null. The question is: Could this term become null? What does it really represent?
Of course, I am considering a possibility of that I have committed some mistake in all the way follow at here, but, at the point of view mathematical, all my calculations seem to be correct. I would be very grateful if anyone could see something besides what I have seen.
