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.

  • $\begingroup$ See also this answer for getting oriented. $\endgroup$ Commented Jan 26, 2020 at 23:04

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.


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