A one-parameter family of transformations $q_i \mapsto Q_i(q,\dot{q},t;\varepsilon)$, $t \mapsto \tau(q,\dot{q},t;\varepsilon)$ (with $\varepsilon = 0$ being the identity) is a symmetry of the Lagrangian system if the action $S = \int L\,dt$ is invariant: $\int_{t_1}^{t_2}L(q,\dot{q},t)\,dt = \int_{\tau_1}^{\tau_2}L(Q,\dot{Q},\tau)\,d\tau$ for all paths and time intervals. Infinitesimally, write $Q_i = q_i + \varepsilon\xi_i(q,t) + O(\varepsilon^2)$ and $\tau = t + \varepsilon\zeta(q,t) + O(\varepsilon^2)$ where $\xi_i$ and $\zeta$ are the generators of the symmetry.

Noether's theorem (1918): If the action is invariant under a continuous one-parameter symmetry with generators $(\xi_i,\zeta)$, then the quantity $I = \sum_i \frac{\partial L}{\partial\dot{q}_i}\xi_i + \left(L - \sum_i\frac{\partial L}{\partial\dot{q}_i}\dot{q}_i\right)\zeta$ is conserved along solutions of the Euler-Lagrange equations: $dI/dt = 0$. The term in parentheses is $-H$ (the negative Hamiltonian). Each independent symmetry yields an independent conserved quantity.

For a field theory with action $S[\phi] = \int\mathcal{L}(\phi,\partial_\mu\phi,x^\mu)\,d^4x$, a symmetry $\phi \mapsto \phi + \varepsilon\delta\phi$, $x^\mu \mapsto x^\mu + \varepsilon\delta x^\mu$ yields the Noether current: $j^\mu = \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\delta\phi + \left[\mathcal{L}\delta^\mu_\nu - \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\partial_\nu\phi\right]\delta x^\nu$. The conservation law is $\partial_\mu j^\mu = 0$, and the conserved charge is $Q = \int j^0\,d^3x$ with $dQ/dt = 0$.