# How to find the associated conservation law from a given symmetry

To be blunt, the answer to your question **is** Noether's theorem (often precised as Noether's *first* theorem). So, essentially you already knew the answer to your own question.

However, the other answers are missing a degree of pragmatism. The calculation of the conserved current, once you know the Lagrangian and the symmetry is straightforward and mechanical. Namely, suppose you have a Lagrangian density $L[\phi] = L(x,\phi(x),\partial \phi(x), \partial^2\phi(x), \ldots)$, which depends your dynamical field $\phi(x)$. The variational principle will be $S(\phi) = \int L[\phi] \, \mathrm{d}x$, where $\mathrm{d}x$ is the coordinate volume form.^{1} An infinitesimal local field transformation $\phi^a \mapsto \phi^a + \delta_{\xi}\phi^a$ is allowed to be coordinate and field dependent, $\delta_\xi \phi^a = \xi^a[\phi] = \xi^a(x,\phi(x), \partial \phi(x), \partial^2 \phi(x), \ldots)$, and commutes with coordinate derivatives, namely $\delta_\xi \partial^n \phi^a = \partial^n (\delta_\xi \phi^a) = \partial^n \xi^a[\phi]$ for any $n\ge 0$. The example of time translation $\xi^a[\phi] = \frac{\partial}{\partial t} \phi^a$ is illustrative.

Such a local field transformation is a symmetry of the Lagrangian when its variation vanishes modulo a total divergence, $\delta_\xi L[\phi] = \partial_i J_0^i[\phi]$. The next step is a bit unintuitive, but it makes the calculation of the conserved current mechanical. Consider now the variation $\delta_{\varepsilon \xi}$, where $\varepsilon = \varepsilon(x)$ is an arbitrary function of the coordinates $x^i$. Using integration by parts, we can put the variation of the Lagrangian into the form $$ \tag{$*$} \delta_{\varepsilon \xi} L[\phi] = \varepsilon\partial_i J^i_0[\phi] + (\partial_i\varepsilon) J^i_1[\phi] + \partial_i(-)^i . $$ The leading term has to agree with $\delta_\xi L[\phi]$ when we set $\varepsilon \equiv 1$. The desired conserved current corresponding to $\xi$ is $$ J_\xi^i[\phi] = J_0^i[\phi] - J_1^i[\phi] . $$ You can get the current in one step if you use integration by parts to directly put the variation of the Lagrangian into the form $\delta_{\varepsilon \xi} L[\phi] = -J_\xi^i[\phi] (\partial_i \varepsilon) + \partial_i(-)^i$, which is a formula that can be found in some physics textbooks on QFT.

The proof of Noether's theorem in this form is also straightforward (and a reshuffling of the standard proof). It only relies on the usual lemma that any density $N[\varepsilon, \ldots]$ that linearly depends on an arbitrary function $\varepsilon = \varepsilon(x)$ (and possibly any other fields) has a unique representative modulo total divergence terms, namely $N[\varepsilon, \ldots] = \varepsilon N_0 + \partial_i(-)^i$, with $N_0$ unique. The Euler-Lagrange equations $E_a[\phi]=0$ are defined by the identity $\delta_\xi = \xi^a E_a[\phi] + \partial_i(-)^i$ for arbitrary $\xi$. So, when $\xi$ is a symmetry, using $(*)$ and one more integration by parts, we find the identity $$ \delta_{\varepsilon \xi} L[\phi] = \varepsilon \xi^a E_a[\phi] + \partial_i(-)^i = \varepsilon \partial_i J^i_\xi[\phi] + \partial_i(-)^i , $$ which implies that $\partial_i J^i_\xi[\phi] = \xi^a E_a[\phi]$, which vanishes when $E_a[\phi] = 0$. In other words, $J^i_\xi[\phi]$ is a conserved current.

^{1 If you change the independent coordinates $x^i$, the Lagrangian will change by the appropriate Jacobian. Working with differential forms allows you to keep everything more manifestly invariant.}

You can find an overview of methods to obtain conservation laws from a wave equation in On the structure of conservation laws of (3+1)-dimensional wave equation. Noether's method requires that the PDE follows from a variational principle for a Lagrangian (as pointed out by Willie Wong). A direct algorithmic method to obtain conservation laws from a PDE without variational structure is described in the cited paper.