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

It is a very well-known fact that any conservation law associated with some given PDE has an associated invariance (by Noether's Theorem). However, it is completely mysterious for me how to compute/derive these conservation laws just by knowing the invariances of the equation. For example, the one-dimensional nonlinear wave equation $$u_{tt}-u_{xx}+f(u)=0, \qquad (t,x)\in\mathbb{R}\times\mathbb{R},$$ is invariant under space translations. On the other hand, it is "well-known" that associated to this space translation invariance is the momentum conservation of the equation, that is, $$P(u,v)(t):=\int_{\mathbb{R}} u_x(t,x)v(t,x)dx=\int_{\mathbb{R}}u_{0,x}(x)v_0(x)dx=P(u,v)(0).$$ Nevertheless, I have no idea how to derive this conservation law (generally speaking) just by knowing that the equation is invariant under space translations. What about time-translations for example, what is its associated conservation law? Please, don't misunderstand me, I do know how to explicitly derive the momentum conservation directly from the equation, what I would like to know is how to derive it from the space-translations invariance. Any hint suggested reading or answer is very welcome!

• The equation you wrote is variational. So you can use the general formalism there to make the computation starting from an appropriate Lagrangian. (Here the action is $$-u_t^2 + u_x^2 + F(u)$$ where $F$ is an antiderivative of the scalar function $f$.) Following en.wikipedia.org/wiki/Noether%27s_theorem#Field_theory_version you can get a conserved current $j$. Integrating $j^0$ along $\{t = const\}$ gives you the time-independent conserved quantity. Commented Apr 3, 2020 at 21:07
• Conceptually, Noether's theorem is saying that a 1-parameter family of diffeomorphisms of the space F of fields induces a Hamiltonian flow on T^*F, and the Hamiltonian (Noether charge) is then conserved under that flow. I wrote a blog post about this a couple of years ago: jde27.uk/blog/noether.html - in particular the conservation of field momentum associated with space translation is worked out. Admittedly, I find it easier to think about going from the conserved quantity to the symmetry (from Hamiltonian to its flow) but that's just because differentiating is easier than integrating. Commented Apr 3, 2020 at 22:05
• Time-translation invariance implies the conservation of energy. Commented Apr 5, 2020 at 23:21
• I don't know if it'll help to answer your specific question, but I hope you won't mind my taking the excuse to share Baez's exposition Noether's theorem in a nutshell, which I've always found second to none on this subject. Commented Apr 6, 2020 at 0:00
• @BenMcKay Every conserved current is equivalent to one s.t. its conservation condition takes the form $\mathrm{d} j[\phi] = \xi \cdot E[\phi]$ (no derivatives of the equations $E[\phi]=0$). Applying the Helmhotz operator to both sizes gives $H(\xi \cdot E[\phi])=0$, implying differential conditions on $\xi$. On appropriate equivalence classes, $j \leftrightarrow \xi$ is a bijection. When $E[\phi]=0$ is variational, the condition on $\xi$ is the same as being a symmetry of the Lagrangian. Commented Apr 6, 2020 at 14:36

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.

It's been a while since I've thought about this stuff so take this with a grain of salt. Also, I'm only familiar with this in the context of a finite dimensional phase space, whereas the phase space is infinite-dimensional in the example you give and I'm not sure what extra subtleties that introduces. But the finite-dimensional case might still provide some useful intuition.

Those two caveats aside, I hope the following is of some use:

We can define a canonical symplectic form on phase space in terms of the Lagrangian. The symplectic form gives us a bijective correspondence between tangent vector fields and 1-forms (it works the same as with Riemannian manifolds, the key is just that we have a perfect pairing on tangent spaces). We also have a Poisson bracket operation {A, B} between scalar fields A and B. {A, B} is the Lie derivative of B along the tangent vector field corresponding to the exterior derivative of A (obtained using the correspondence between 1-forms and tangent vector fields provided by the symplectic form).

A continuous symmetry is a flow that preserves the Hamiltonian H and the symplectic form. To a continuous symmetry, we can associate a unique tangent vector field that generates it, which corresponds to a 1-form (which happens to be closed). We can then integrate that 1-form to get a scalar field, which I will call S. We have that {S, H} = 0, this basically says that the vector field corresponding to S generates a symmetry. But the Poisson bracket is anticommutative, so {H, S} = 0, implying that S is a conserved quantity (since the vector field corresponding to H generates the time-evolution flow). Therefore, continuous symmetries correspond to conserved quantities.