3

Let's consider an arbitrary infinitesimal transformation of the fields and their coordinates :

$$x'^{\mu}= x^{\mu} + \delta x^{\mu} = x^{\mu} + \frac{\delta x^{\mu}}{\delta{\omega}^a}{\omega}^a\tag{1}$$

$${\Phi}'(x') = {\Phi}(x) + \delta\Phi= {\Phi}(x) + \frac{\delta\Phi}{\delta\omega^a}{\omega}^a\tag{2}$$

where ${\{\omega}^a\}$ is a set of infinitesimal parameters. The corresponding variation of the Action is:

$$\delta S= \delta (\int{dx \,\mathcal{L}[\Phi,\partial^{\mu}\Phi]})=\int{(dx \, \delta \mathcal{L}} +\mathcal{L}\,\delta dx)= \int{dx (\delta \mathcal{L}+\mathcal{L}\,\partial_{\mu}\delta x^{\mu})}\tag{3}$$

Now if

$$\delta=\delta_0 + \delta x^{\mu}\partial_{\mu},\tag{4}$$

where:

$\delta_0 \phi (x)=\phi'(x)-\phi(x)$ at first order, then the variation becomes:

$$\delta S=\int{dx\,(\delta\mathcal{L}+\mathcal{L}\,\partial_{\mu}\delta x^{\mu})}=\int{dx(\delta_0\mathcal{L}+\delta x^{\mu}\partial_{\mu}\mathcal{L}+\mathcal{L}\,\partial_{\mu}\delta x^{\mu})}$$ $$=\int{dx\,(\delta_0\mathcal{L}+ \partial_\mu(\mathcal{L}\delta x^{\mu}))}=\int{dx\,(\frac{\partial\mathcal{L}}{\partial{\Phi}}\delta_0\Phi}+\frac{\partial{\mathcal{L}}}{\partial[\partial_\mu\Phi]}\delta_0\partial_\mu\Phi+ \partial_\mu(\mathcal{L}\delta x^{\mu})).\tag{5}$$

Using the Euler-Lagrange equations and commuting $\delta_0$ and $\partial$, the last integral becomes:

$$\delta S=\int{dx\,\partial_\mu(\frac{\partial\mathcal{L}}{\partial[\partial_\mu\Phi]}\delta_0\Phi+\mathcal{L}\delta x^\mu)}.\tag{6}$$

Now we go back to $\delta$ from $\delta_0$:

$$\delta S =\int{dx\,\partial_\mu(\frac{\partial\mathcal{L}}{\partial[\partial_\mu\Phi]}\delta\Phi-\frac{\partial\mathcal{L}}{\partial[\partial_\mu\Phi]}\delta x^\nu\partial_\nu\Phi+\mathcal{L}\delta x^\mu)}\tag{7}$$

and finally:

$$\delta S =\int{dx\,\partial_\mu(\frac{\partial\mathcal{L}}{\partial[\partial_\mu\Phi]}\delta\Phi-\frac{\partial\mathcal{L}}{\partial[\partial_\mu\Phi]}\delta x^\nu\partial_\nu\Phi+\mathcal{L}\delta x^\mu)} \\ =\int{dx\,\partial_\mu[\frac{\partial\mathcal{L}}{\partial[\partial_\mu\Phi]}\delta\Phi+(\mathcal{L}\delta^\mu_\nu-\frac{\partial\mathcal{L}}{\partial[\partial_\mu\Phi]}\partial_\nu\Phi)\delta x^\nu]}\\ =\int{dx\,\partial_\mu\{[\frac{\partial\mathcal{L}}{\partial[\partial_\mu\Phi]}\frac{\delta\Phi}{\delta\omega^a}+(\mathcal{L}\delta^\mu_\nu-\frac{\partial\mathcal{L}}{\partial[\partial_\mu\Phi]}\partial_\nu\Phi)\frac{\delta x^\nu}{\delta\omega^a}]\delta\omega^a\}}.\tag{8}$$

Now defining:

$$j^{a\mu} = (\frac{\partial\mathcal{L}}{\partial[\partial_\mu\Phi]}\partial_\nu\Phi-\mathcal{L}\delta^\mu_\nu)\frac{\delta x^\nu}{\delta\omega^a} - \frac{\partial\mathcal{L}}{\partial[\partial_\mu\Phi]}\frac{\delta\Phi}{\delta\omega^a}\tag{9} $$

we eventually get:

$$\delta S = - \int{dx\,\partial_\mu(j^{a\mu}\delta\omega^a)}.\tag{10}$$

Let's suppose for now that $\{\omega^a\}$ are $x$-independent, such that:

$$\delta S = - \int{dx\,\partial_\mu(j^{a\mu})\delta\omega^a}.\tag{11}$$

  1. Here's my first question: do I need to ask that $S$ is invariant under my transformations in order to get $$\partial_\mu\,j^{a\mu}=0~?\tag{12}$$ Or can the transformation be arbitrary given that in every case $$\delta S = 0\tag{13}$$ for all infinitesimal variations of the fields satisfying Euler-Lagrange dynamical equations?

  2. Then it comes my second question. My book$^{(1)}$ says that, even not taking EL equations into account, the variation of the action under infinitesimal arbitrary transformation $(1),(2)$ is: $$\delta S=-\int{dx\,j^{\mu a}\partial_{\mu}\omega^a}\tag{14}.$$ Now I don't know how $(14)$ could be derived without taking EL equations into account beacause if $j^{\mu a}$ does not have $\frac{\partial\mathcal{L}}{\partial\Phi}$ terms is just thanks to those equations.

$^{(1)}$Philippe Di Francesco, Pierre Mathieu, David Sénéchal - Conformal Field Theory, page 41.

Qmechanic
  • 220,844
Alessandro
  • 105

1 Answers1

1

OP asks good technical questions about the proof of Noether's first theorem.

  1. Yes, eqs. (10) and (11) indeed hold on-shell for $x$-dependent ($x$-independent) infinitesimal parameters, respectively, without assuming that the action $S$ is a quasi-symmetry.

    But there is no free lunch. Without assuming that the action $S$ is a quasi-symmetry, one can not conclude an on-shell conservation law (12) from eq. (11).

    The point is that on-shell the infinitesimal variations $\delta S$ and infinitesimal vertical variations $\delta_0 S$ are not necessarily zero, but could contain boundary terms. This is because the Noether variations do not necessarily satisfy the boundary conditions that we usually impose when deriving the Euler-Lagrange (EL) eqs.

  2. Eq. (14) is valid off-shell (modulo boundary terms) for $x$-dependent infinitesimal parameters (with the implicit understanding that the Noether current entering eq. (14) is in general the full Noether current rather than the bare Noether current (9)). For the derivation of eq. (14) and the trick of how to derive the Noether current via $x$-dependent infinitesimal parameters, see e.g. this Phys.SE post. Specifically, the disappearance of the $\frac{\partial\mathcal{L}}{\partial\Phi}$-term is an indirect consequence of the assumption that the action $S$ is a quasi-symmetry.

Qmechanic
  • 220,844