For any transformation of the fields, $$\varphi\to\varphi'=\varphi+\delta\varphi$$ the change in the Lagrangian can be written as $$\delta\mathcal L = \text{EoM} + \partial_\mu j^\mu\tag{1}$$where "EoM" represents the equations of motion (Euler-Lagrange equations) and all other terms can be written as a total derivative of some function $j^\mu$, which is a known function in terms of the Lagrangian.

I would like to distinguish the different realizations of transformations. Let's assume that the transformation (1) leaves the action invariant, $\delta S=0$.

  1. $\delta\mathcal L=0$

    • EoM $=0$, "on-shell": Noether current is conserved, $\partial_\mu j^\mu=0$.

    • EoM $=\partial_\mu b^\mu\neq0$, "off-shell": modified Noether current $J^\mu = j^\mu+b^\mu$ is conserved, $\partial_\mu J^\mu=0$.

  2. $\delta\mathcal L =\partial_\mu a^\mu \neq 0$, "quasi-symmetry"

    • EoM $=0$, "on-shell": modified Noether current $J^\mu = j^\mu-a^\mu$ is conserved, $\partial_\mu J^\mu=0$.

    • EoM $=\partial_\mu b^\mu\neq0$, "off-shell": modified Noether current $J^\mu = j^\mu-a^\mu+b^\mu$ is conserved, $\partial_\mu J^\mu=0$.

Is this listing correct?

What roles do the terms "on/off-shell" and "(quasi) symmetry" play in Noether's theorem?

Related: one, two, three, four, five.


1 Answer 1

  1. The assumption in Noether's (first) theorem is an off-shell$^1$ quasisymmetry of the action $S$. It leads to an off-shell Noether identity off-shell Noether identity $$d_{\mu} J^{\mu} ~\equiv~ - \frac{\delta S}{\delta\phi^{\alpha}} \tag{A}Y_0^{\alpha}.$$ Here $J^{\mu}$ is the full Noether current, which is necessarily non-trivial; and $Y_0^{\alpha}$ is a (vertical) symmetry generator. The off-shell identity (A) in turn implies an on-shell continuum equation/conservation law.

  2. An on-shell quasisymmetry of the action $S$ is a tautology. It has not an associated continuum equation/conservation law. Even a strict symmetry of the action $S$ (or the Lagrangian density ${\cal L}$) on-shell has not an associated continuum equation/conservation law.$^2$

  3. OP is only considering so-called vertical transformations $\delta\phi$, i.e. $\delta x^{\mu}=0$, which carries certain simplifications in the form of the Noether current.


$^1$The words on-shell and off-shell refer to whether the Euler-Lagrange (EL) equations (=EOM) are satisfied or not.

$^2$ Here is another heuristic argument: Ignoring various technical assumptions & details, there is morally speaking a bijective correspondence between off-shell quasisymmetries and on-shell conservation laws, cf. e.g. this Phys.SE post. In particular, all on-shell conservation laws are already explained by off-shell quasisymmetries alone. In other words, there is no room for on-shell quasisymmetries to play an independent role in this correspondence.

  • $\begingroup$ If the conservation law is considered on-shell, why start with an off-shell transformation? -- If for a general transformation, $\delta\mathcal L=\text{EoM}+\partial_\mu j^\mu$ and now for a particular vertical transformation, $\delta\mathcal L=\partial_\mu\Lambda^\mu$, we subtract both to get $\text{EoM}+\partial_\mu(j^\mu-\Lambda^\mu)=0$. If we assume on-shell, this becomes $\partial_\mu(j^\mu-\Lambda^\mu)=0$ and we have our conservation law. $\endgroup$
    – ersbygre1
    Nov 2, 2018 at 9:18
  • $\begingroup$ I understand that for an "on-shell" quasi symmetry, $\Lambda$ in my previous comment should become $j$, therefore the conservation law is a trivial equation. However, using this logic leads to the conclusion that also afterwards in my last step, $\Lambda\to j$ and the equation becomes trivial. What am I missing? $\endgroup$
    – ersbygre1
    Nov 2, 2018 at 9:20
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Nov 2, 2018 at 11:30
  • $\begingroup$ Thank you for your update. I think this together with another answer (physics.stackexchange.com/a/438369/127780) and this paper (arXiv:1510.07038) has solved my confusion. $\endgroup$
    – ersbygre1
    Nov 3, 2018 at 3:05

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