# 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.