# Does it make sense to speak in a total derivative of a functional? Part III

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 $$$$\frac{d\mathcal{L}}{dx^{μ}} =\frac{\partial\mathcal{L}}{\partialφ_{r}}\partial_{\mu}φ_{r}+\frac{\partial\mathcal{L}}{\partial\big(\partial_{ν}φ_{r}\big)}\partial_{\mu}\big(\partial_{ν}φ_{r}\big)+∂_{μ}\mathcal{L}.\tag{III.1}\label{eq1}$$$$

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: $$$$\frac{d\zeta_{r}}{dx^{\mu}}=\partial_{\mu}\zeta_{r} \quad\text{and}\quad \frac{d\xi^{\mu}}{dx^{\mu}}=\partial_{\mu}\xi^{\mu}.\tag{III.2}$$$$ Thus, the Eq. (\ref{eq24}) becomes $$$$\dfrac{S^{\prime}-S}{\varepsilon}\approx\int_{\mathbb{\Omega}}d^{D}x~\left\{ \dfrac{\partial\mathcal{L}}{\partial\phi_{r}}\zeta_{r}+\dfrac{\partial \mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\dfrac{d\zeta_{r}}{dx^{\nu}} +\xi^{\mu}\dfrac{d\mathcal{L}}{dx^{\mu}}+\mathcal{L}\dfrac{d\xi^{\mu}} {dx^{\mu}}\right\}.\tag{III.3}\label{eq3}$$$$ Now, we do use of identity $$$$\dfrac{\partial\mathcal{L}}{\partial\partial_{\mu}\phi_{r}} \dfrac{d\zeta_{r}}{dx^{\mu}}=\frac{d}{dx^{\mu}}\left(\zeta_{r}\frac{\partial\mathcal{L} }{\partial\partial_{\mu}\phi_{r}}\right)-\zeta_{r}\frac{d}{dx^{\mu}}\frac{\partial\mathcal{L}}{\partial\partial_{\mu}\phi_{r}},\tag{III.4}\label{eq4}$$$$ such that $$$$\dfrac{S^{\prime}-S}{\varepsilon}\approx\int_{\mathbb{\Omega}}d^{D}x~\left\{ \left(\dfrac{\partial\mathcal{L}}{\partial\phi_{r}}-\dfrac{d}{dx^{\nu} }\dfrac{\partial\mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\right) \zeta _{r}+\dfrac{d}{dx^{\nu}}\left( \zeta_{r}\dfrac{\partial\mathcal{L}}{\partial\partial_{\nu}\phi_{r}}+\xi^{\mu}\mathcal{L}\right) \right\},\tag{III.5}\label{eq5}$$$$ where we have used $$$$\xi^{\mu}\dfrac{d\mathcal{L}}{dx^{\mu}}+\mathcal{L}\dfrac{d\xi^{\mu}}{dx^{\mu }}=\dfrac{d}{dx^{\mu}}\left( \xi^{\mu}\mathcal{L}\right).$$$$

We have to said in Does it make sense to speak in a total derivative of a functional? Part II, Eq.(\ref{II19}), that $$$$\zeta_{r}\left( x\right) +\xi^{\mu }\left( x\right) \partial_{\mu}\phi_{r}\left( x\right) =\dfrac{\tilde{\delta}\phi_{r}}{\varepsilon}=\chi_{r}\left( x\right) ,\tag{II.19}\label{II19}$$$$ so that (\ref{eq5}) becomes $$\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: $$$$\dfrac{S^{\prime}-S}{\varepsilon}\approx\int_{\mathbb{\Omega}}d^{D}x~\left\{ \dfrac{\partial\mathcal{L}}{\partial\phi_{r}}\chi_{r}+\dfrac{\partial \mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\left( \partial_{\nu}\zeta _{r}+\xi^{\mu}\partial_{\mu}\partial_{\nu}\phi_{r}\right) +\partial_{\nu }\left( \xi^{\nu}\mathcal{L}\right) \right\},\tag{III.7}\label{eq7}$$$$ 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 $$$$\dfrac{S^{\prime}-S}{\varepsilon}\approx\int_{\mathbb{\Omega}}d^{D}x~\left\{ \dfrac{\partial\mathcal{L}}{\partial\phi_{r}}\chi_{r}+\dfrac{\partial \mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\partial_{\nu}\chi_{r} -\dfrac{\partial\mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\partial_{\mu} \phi_{r}\partial_{\nu}\xi^{\mu}+\partial_{\nu}\left( \xi^{\nu}\mathcal{L} \right) \right\}.\tag{III.8}$$$$ Now, using the identities \begin{align} \dfrac{\partial\mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\partial_{\nu} \chi_{r}&=\partial_{\nu}\left( \chi_{r}\dfrac{\partial\mathcal{L}} {\partial\partial_{\nu}\phi_{r}}\right) -\chi_{r}\partial_{\nu} \dfrac{\partial\mathcal{L}}{\partial\partial_{\nu}\phi_{r}},\tag{III.9}\label{eq9}\\ -\dfrac{\partial\mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\partial_{\mu} \phi_{r}\partial_{\nu}\xi^{\mu}&=-\partial_{\nu}\left( \dfrac{\partial \mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\partial_{\mu}\phi_{r}\xi^{\mu }\right) +\xi^{\mu}\partial_{\nu}\left( \dfrac{\partial\mathcal{L}} {\partial\partial_{\nu}\phi_{r}}\partial_{\mu}\phi_{r}\right),\tag{III.10}\label{eq10} \end{align} we obtain $$\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 $$$$\dfrac{\partial\mathcal{L}}{\partial\phi_{r}}-\partial_{\nu}\dfrac {\partial\mathcal{L}}{\partial\partial_{\nu}\phi_{r}}=0, \tag{III.12}\label{eq12}$$$$ and the applicability of divergence theorem over to third integral (Which now seems to be quite reasonable!) $$$$\int_{\mathbb{\Omega}}d^{D}x~\partial_{\nu}J^{\nu}=\oint_{\partial \mathbb{\Omega}}dS_{\nu}~J^{\nu}=0,\tag{III.13}\label{eq13}$$$$ with $$$$J^{\nu}=\dfrac{\partial\mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\chi _{r}-\left( \dfrac{\partial\mathcal{L}}{\partial\partial_{\nu}\phi_{r} }\partial_{\mu}\phi_{r}-\delta_{\mu}^{\nu}\mathcal{L}\right) \xi^{\mu},\tag{III.14}\label{eq14}$$$$ when $$\varepsilon\rightarrow 0$$, we have found $$$$\lim_{\varepsilon\rightarrow 0} \dfrac{S^{\prime}-S}{\varepsilon}\approx\int_{\mathbb{\Omega}}d^{D}x~\xi^{\mu }\partial_{\nu}\left( \dfrac{\partial\mathcal{L}}{\partial\partial_{\nu} \phi_{r}}\partial_{\mu}\phi_{r}\right),\tag{III.15}\label{eq15}$$$$ which at first seems to be non-zero.

As we know, it is hoped that $$$$\lim_{\varepsilon\rightarrow 0} \dfrac{S^{\prime}-S}{\varepsilon}\approx 0.\tag{III.16}\label{eq16}$$$$

## Questions

We have, therefore, two questions:

1. 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}))?

2. 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.

## 2 Answers

Concerning situation 1, the main point seems to be that the generalized divergence theorem works with total derivatives, not partial derivatives.

According to the conclusions obtained from Part II, we have concluded that situation 1 is, in fact, the correct situation and there, in the second integral of equation \eqref{eq6}, it is valid to make use of the divergence theorem, since it is a total partial derivative.