I am trying to derive the Noether theorem from the following integral action: \begin{equation} S=\int_{\mathbb{\Omega}}d^{D}x~\mathcal{L}\left( \phi_{r},\partial_{\nu}% \phi_{r},x\right) , \tag{II.1}\label{eq1}% \end{equation} where $\phi_{r}\equiv\phi_{r}\left( x\right) $ represents the $r$-th field of set $\left\{ \phi_{r}\right\} $, while $\partial_{\nu}\phi_{r}\equiv \partial\phi_{r}/\partial x^{\nu}$ represents its fisrt partial derivatives. The functional $\mathcal{L}\left( \phi_{r},\partial_{\nu}\phi_{r},x\right) $ is the Lagrangian density of the theory and has, as usual, energy density dimension, whereas $d^{D}x$ is the volume element of $D$-dimensional spacetime. For simplicity, we have represented by the dependence of the spacetime coordinates by $x$, such that $x \equiv x^{\mu}$.

I have assumed that the total variation occurs under the following coordinate transformation \begin{equation} x^{\prime\mu}=f^{\mu}\left( \varepsilon,x^{\nu}\right) , \tag{II.2}\label{eq2}% \end{equation} where $\varepsilon$ is a paramenter responsible for variation. Naturaly, which if $\varepsilon=0$, so \begin{equation} x^{\mu}=f^{\mu}\left( 0,x^{\nu}\right) . \tag{II.3}% \end{equation}

If the parameter $\varepsilon$ is sufficiently small, it is possible to use a Taylor expansion to rewrite the expression (\ref{eq2}) as \begin{equation} x^{\prime\mu}\approx x^{\mu}+\varepsilon\xi^{\mu}\left( x\right) , \tag{II.4}\label{eq4}% \end{equation} where only the terms of the first order in $\varepsilon$ are considered. $\xi^{\mu}\left( x\right) $ It is a field vector, contravariant, which in general can be defined by \begin{equation} \xi^{\mu}\left( x\right) =\dfrac{\partial x^{\prime\mu}}{\partial \varepsilon}\rule[-0.35cm]{0.02cm}{0.9cm}_{\varepsilon=0}.\tag{II.5}% \end{equation} In the literature, it is common to denote $\varepsilon\xi^{\mu}$ by $\delta{x^{\mu}}$, i.e., $\varepsilon\xi^{\mu}\equiv\delta{x^{\mu}}$.

At this point, I will denote the integral action (\ref{eq1}) in terms of the prime coordinates, such that, \begin{equation} S^{\prime}=\int_{\mathbb{\Omega}^{\prime}}d^{D}x^{\prime}~\mathcal{L}\left( \phi_{r}^{\prime},\partial_{\nu}^{\prime}\phi_{r}^{\prime},x^{\prime}\right) \text{.} \tag{II.6}\label{eq6}% \end{equation}

As it is known, the volume element in the coordinates prime is connected to the volume element of the nonprime coordinates by means of the following expression \begin{equation} d^{D}x^{\prime}=\left\vert \dfrac{\partial x^{\prime}}{\partial x}\right\vert d^{D}x,\tag{II.7}\label{eq7}% \end{equation} where the Jacobian $\left\vert \dfrac{\partial x^{\prime}}{\partial x}\right\vert $ can be calculated by the following expression \begin{equation} \left\vert \dfrac{\partial x^{\prime}}{\partial x}\right\vert =\dfrac{\left( -1\right) ^{s}}{D!}\epsilon_{\alpha_{1}\alpha_{2}\cdots\alpha_{D-1}\alpha _{D}}\epsilon^{\beta_{1}\beta_{2}\cdots\beta_{D-1}\beta_{D}}\dfrac{\partial x^{\prime\alpha_{1}}}{\partial x^{\beta_{1}}}\dfrac{\partial x^{\prime \alpha_{2}}}{\partial x^{\beta_{2}}}\cdots\dfrac{\partial x^{\prime \alpha_{D-1}}}{\partial x^{\beta_{D-1}}}\dfrac{\partial x^{\prime\alpha_{D}}% }{\partial x^{\beta_{D}}}.\tag{II.8}\label{eq8}% \end{equation} Here, the parameter $s$ corresponds to the number of negative eigenvalues of the metric.

To follow, we must then take the partial derivatives of Eq. (\ref{eq4}), which leads us to: \begin{equation} \dfrac{\partial x^{\prime\alpha_{i}}}{\partial x^{\beta_{i}}}\approx \delta_{\beta_{i}}^{\alpha_{i}}+\varepsilon\partial_{\beta_{i}}\xi^{\alpha _{i}}.\tag{II.9}\label{eq9}% \end{equation} Substituting (\ref{eq9}) into (\ref{eq8}), we have found, after laborious calculations, that \begin{multline} \left\vert \dfrac{\partial x^{\prime}}{\partial x}\right\vert \approx\left( -1\right) ^{s}\left[ \dfrac{1}{D!}\epsilon_{\alpha_{1}\alpha_{2}\cdots \alpha_{D-1}\alpha_{D}}\epsilon^{\alpha_{1}\alpha_{2}\cdots\alpha_{D-1}% \alpha_{D}}\right.\\ \left.+\dfrac{1}{\left( D-1\right) !}\varepsilon\epsilon_{\alpha _{1}\alpha_{2}\cdots\alpha_{D-1}\alpha_{D}}\epsilon^{\alpha_{1}\alpha _{2}\cdots\alpha_{D-1}\beta_{D}}\partial_{\beta_{D}}\xi^{\alpha_{D}}\right] \tag{II.10}\label{eq10}% \end{multline} Using relations \begin{equation} \epsilon_{\alpha_{1}\alpha_{2}\alpha_{3}\cdots\alpha_{D-1}\alpha_{D}}% \epsilon^{\beta_{1}\alpha_{2}\alpha_{3}\cdots\alpha_{D-1}\alpha_{D}}=\left( -1\right) ^{s}\left( D-1\right) !\delta_{\alpha_{1}}^{\beta_{1}}% ,\tag{II.11}\label{eq11}% \end{equation} and \begin{equation} \epsilon_{\alpha_{1}\alpha_{2}\alpha_{3}\cdots\alpha_{D-1}\alpha_{D}}% \epsilon^{\alpha_{1}\alpha_{2}\alpha_{3}\cdots\alpha_{D-1}\alpha_{D}}=\left( -1\right) ^{s}D!,\tag{II.12}\label{eq12}% \end{equation} we can show, without much difficulty, that: \begin{equation} \left\vert \dfrac{\partial x^{\prime}}{\partial x}\right\vert =\left( -1\right) ^{2s}\left( 1+\varepsilon\partial_{\alpha_{D}}\xi^{\alpha_{D}% }\right) .\tag{II.13}\label{eq13}% \end{equation} Now, whatever the value of $s$, $\left( -1\right) ^{2s}=+1$, and so that, we have that the volume elements are relationship by: \begin{equation} d^{D}x^{\prime}=\left( 1+\varepsilon\partial_{\alpha}\xi^{\alpha}\right) d^{D}x.\tag{II.14}\label{eq14}% \end{equation}

Now, returning Eq. (\ref{eq6}) and making use of Eq. (\ref{eq14}), we have:%

\begin{equation} S^{\prime}=\int_{\mathbb{\Omega}}d^{D}x~\mathcal{L}\left( \phi_{r}^{\prime },\partial_{\nu}^{\prime}\phi_{r}^{\prime},x^{\prime}\right) +\varepsilon \int_{\mathbb{\Omega}}d^{D}x~\mathcal{L}\left( \phi_{r}^{\prime}% ,\partial_{\nu}^{\prime}\phi_{r}^{\prime},x^{\prime}\right) \partial_{\alpha }\xi^{\alpha}.\tag{II.15}\label{eq15}% \end{equation}

To follow, we make use of Taylor's expansion to write \begin{equation} \phi_{r}^{\prime}\left( x^{\prime}\right) =\phi_{r}^{\prime}\left( x+\varepsilon\xi\right) \approx\phi_{r}^{\prime}\left( x\right) +\varepsilon\xi^{\mu}\left( x\right) \partial_{\mu}\phi_{r}^{\prime}\left( x\right) .\tag{II.16}\label{eq16}% \end{equation} We now denote the functional variation of the $\phi_{r}$ field at the same point in space-time by \begin{equation} \phi_{r}^{\prime}\left( x\right) =\phi_{r}\left( x\right) +\varepsilon \zeta_{r}\left( x\right) .\tag{II.17}\label{eq17}% \end{equation} Here, we point out that it is usual in the literature to identify $\varepsilon\zeta_{r}\left( x\right) $ with $\delta{\phi}$, i.e., $\varepsilon\zeta_{r}\left( x\right) \equiv\delta{\phi\big(x\big)}$. Substituting (\ref{eq17}) into (\ref{eq16}), we have: \begin{equation} \phi_{r}^{\prime}\left( x^{\prime}\right) \approx\phi_{r}\left( x\right) +\varepsilon\left[ \zeta_{r}\left( x\right) +\xi^{\mu}\left( x\right) \partial_{\mu}\phi_{r}\left( x\right) \right] ,\tag{II.18}\label{eq18}% \end{equation} where we can identity the total variation of the $\phi$ by \begin{equation} \zeta_{r}\left( x\right) +\xi^{\mu}\left( x\right) \partial_{\mu}\phi _{r}\left( x\right) =\frac{\tilde{\delta}{\phi}}{\varepsilon}.\tag{II.19}% \label{eq19}% \end{equation} Similarly, knowing that \begin{equation} \partial_{\nu}^{\prime}=\left[ \delta_{\nu}^{\rho}-\varepsilon\partial_{\nu }\xi^{\rho}\left( x\right) \right] \partial_{\rho},\tag{II.20}\label{eq20}% \end{equation} we can show that \begin{equation} \partial_{\nu}^{\prime}\phi_{r}^{\prime}\left( x^{\prime}\right) \approx\partial_{\nu}\phi_{r}\left( x\right) +\varepsilon\partial_{\nu}% \zeta_{r}\left( x\right) +\varepsilon\xi^{\mu}\left( x\right) \partial_{\nu}\partial_{\mu}\phi_{r}\left( x\right) .\tag{II.21}\label{eq21}% \end{equation} Now, from of the Eq.(\ref{eq4}), (\ref{eq18}) and (\ref{eq21}), we can, by means of Taylor expansion, to write \begin{equation} \mathcal{L}\left( \phi_{r}^{\prime},\partial_{\nu}^{\prime}\phi_{r}^{\prime },x^{\prime}\right) =\mathcal{L}\left( \phi_{r}+\varepsilon\left( \zeta _{r}+\xi^{\mu}\partial_{\mu}\phi_{r}\right) ,\partial_{\nu}\phi _{r}+\varepsilon\left( \partial_{\nu}\zeta_{r}+\xi^{\mu}\partial_{\nu }\partial_{\mu}\phi_{r}\right) ,x+\varepsilon\xi\right) ,\tag{II.22}% \label{eq22}% \end{equation} \begin{multline} \mathcal{L}\left(\phi_{r}^{\prime},\partial_{\nu}^{\prime}\phi_{r}^{\prime },x^{\prime}\right) \approx \mathcal{L}\left(\phi_{r},\partial_{\nu}\phi _{r},x\right) + \varepsilon\dfrac{\partial\mathcal{L}}{\partial\phi_{r}}\left( \zeta_{r} + \xi^{\mu}\partial_{\mu}\phi_{r}\right) \\ + \varepsilon \dfrac{\partial\mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\left( \partial_{\nu}\zeta_{r}+\xi^{\mu}\partial_{\mu}\partial_{\nu}\phi_{r}\right) +\varepsilon\partial_{\mu}\mathcal{L}\xi^{\mu}.\tag{II.23}\label{eq23}% \end{multline} We now use (\ref{eq23}) in (\ref{eq15}) and after some development, we get \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} This is where my doubt lies! The first two terms lead to the Euler-Lagrange equation plus a term of total divergence. The other terms must be written in the form of a total divergence that will also "absolve" the divergence term that comes from the Euler-Lagrange equation. The term in parentheses suggests that we may write that term as a total derivative in relation to L. However, I am not sure that this is correct. Such doubt motivated the exposition and inquiries exposed in the post: Does it make sense to speak in a total derivative of a functional? Part I.


2 Answers 2

  1. The parameter $s$ below eq. (\ref{eq8}) is non-standard. Noether's theorem and its Lagrangian formalism do in general not rely on a metric. Nevertheless, we only need eq. (\ref{eq14}), which is indeed correct.

  2. Note that the so-called vertical generator $\zeta_{r}\left( \phi_{r}(x),\partial\phi_{r}(x),x\right)$ in eq. (\ref{eq17}) depends on the field and derivatives thereof in important applications, not just $x$. (For a simple example from point mechanics, see e.g. this Phys.SE post.)

  3. The partial derivative $\partial_{\nu}\zeta_{r}$ in the main eq. (\ref{eq24}) should actually be a total derivative $d_{\nu}\zeta_{r}$. Then the main eq. (\ref{eq24}) leads to Noether's theorem by standard arguments. In particular, the parenthesis $(\ldots)$ in eq. (\ref{eq24}) is indeed the total spacetime derivative $d_{\mu}{\cal L}\equiv \frac{d {\cal L}}{dx^{\mu}}$, cf. OP's question.

  • $\begingroup$ @Qmechanics, I agree when you say that the parameter $s$ does not influence the final deduction of the Noether theorem. The only reason I have introduced it was because I am generalizing the deduction to $D$-dimensional spacetimes. In fact, in most of the literature on the subject, the $s$ parameter is not mentioned. The only exception known to me is the Sean Carroll's book, Space and Geometry. $\endgroup$
    – lucenalex
    Apr 23, 2019 at 12:37
  • $\begingroup$ In most of the literature, if not in all literature, the Noether theorem is deduced for $\big(1 + 3\big)$-dimensional spacetime. However, formulations of field theory in $\big(1 + 2\big)$-dimensions, with applications to phenomena of condensed matter, are very common. $\endgroup$
    – lucenalex
    Apr 23, 2019 at 12:38
  • $\begingroup$ I am writing a monograph that deals with a system like these and it is my intention to present Noether's theorem in an appendix. In order not to have to make such a demonstration in a specific spacetime, $\big(1 + 2\big)$ or $\big(1 + 3\big)$, I thought it would be more appropriate to present it in a more general form, shown that the consequence of what is deductible there is applied in any spacetime dimension. Actually, I'm still deciding whether I present in the way or whether I'll use the standard notation with the variation symbol being $\delta$. $\endgroup$
    – lucenalex
    Apr 23, 2019 at 12:40
  • $\begingroup$ on the dependence of the vertical generator $\zeta_r$ with the field and its derivatives, could you exemplify one application, please? I would like to see how this work. $\endgroup$
    – lucenalex
    Apr 26, 2019 at 4:44
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Apr 26, 2019 at 8:26

Following the hints that have been given by @Qmechanic in his reply, and after consulting some references, such that as [1], [2] and [3], I have gone the following conclusions:

  1. The general form to equation \eqref{eq2} should be written as \begin{equation} x^{\prime\mu}=f^{\mu}\left( \varepsilon,x,\phi_r\big(x\big),\partial_\nu\phi_r\big(x\big)\right) , \tag{A}\label{A}% \end{equation} which, automatically, generalized the equation \eqref{eq4} to\begin{equation} x^{\prime\mu}\approx x^{\mu}+\varepsilon\xi^{\mu}\left( x,\phi_r\big(x\big),\partial_\nu\phi_r\big(x\big)\right). \tag{B}\label{B}% \end{equation}

  2. In \eqref{eq9}, $\dfrac{\partial {x^\prime}^\mu}{\partial {x}^\nu}$ is actually a total partial derivative because the field $\phi_r$ is dependent of $x^\mu$ and it cannot be held fixed such as it would be if we have considered function instead of functionals. The same is valid to $\xi\big(x,\phi_r\big(x\big),\partial_\nu\phi_r\big(x\big)\big)$ and also to Jacobian in \eqref{eq8} [ See [1] in page 172, footnote 18].

  3. Once that the conclusion 2 to be valid is easy to conclude that the latter two terms in Eq. \eqref{eq24} leads to total (partial) derivative term, and thus, we can to conclude that situation 1 in Part III is the correct situation.

[1] I. M. Gelfand and S. V. Fomin, Calculus of Variation, Prentice-Hall, Inc, chapter 7;

[2] D. E. Neuenschwander, Emmy Noether's Wonderful Theorem, Johns Hopkins University Press, chapters 4 and 6;

[3] Nivaldo Lemos, Analytical Mechanics, Cambridge University Press, chapter 11.


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