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!



How to find the associated conservation law from a given symmetry
To be blunt, the answer to your question **is** [Noether's theorem](https://en.wikipedia.org/wiki/Noether%27s_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.




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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.


