Infinitesimal point symmetry of a time-dependent Lagrangian and its conserved charge

Let $(O,\mathcal{L})$ be a time-dependent Lagrangian chart over $[a,b]\times U$ in the sense of A time-dependent Lagrangian chart, its action functional and its fixed-endpoint first variations. Let $V\subseteq {\mathbb{R}}^n$ and $A\subseteq \mathbb{R}\times V$ be open sets containing respectively the configuration projection of $O$ and its time-configuration projection. Let

$\xi :V⟶{\mathbb{R}}^n,B:A⟶\mathbb{R}$

be of class $C^1$ ($C^k$ Euclidean maps and diffeomorphisms); write $D\xi (q)\in M_n(\mathbb{R})$ for the Jacobian matrix of $\xi$ at $q$ (The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case, Directional derivatives and partial derivatives of a map $U\subseteq {\mathbb{R}}^m\to {\mathbb{R}}^n$), and for a $C^1$ function $\phi (t,q)$ write

$D_t\phi (t,q,v):={\phi }_t(t,q)+{\phi }_q(t,q)\cdot v\mathrm{.}$

The pair $(\xi ,B)$ is an infinitesimal point symmetry of $\mathcal{L}$ when the identity

${\mathcal{L}}_q(t,q,v)\cdot \xi (q)+{\mathcal{L}}_v(t,q,v)\cdot D\xi (q)v=B_t(t,q)+B_q(t,q)\cdot v$

holds for every $(t,q,v)\in O$. The charge of the pair is the function

$C(t,q,v):={\mathcal{L}}_v(t,q,v)\cdot \xi (q)-B(t,q)\mathrm{.}$

Reading of the identity. The left-hand side is the derivative of the Lagrangian under the infinitesimal family of point transformations

$q⟼q+s\xi (q),t⟼t$

with the velocity transformed accordingly by $v\mapsto v+sD\xi (q)v$, and the right-hand side is the total derivative $\frac{d}{dt}B(t,q(t))$ along a $C^1$ trajectory with $q^{\mathrm{\prime }}(t)=v$; thus the identity says that the infinitesimal variation of $\mathcal{L}$ is a total time derivative of the chart function $B$. In particular:

• Taking $B=0$ reduces the identity to ${\mathcal{L}}_q(t,q,v)\cdot \xi (q)+{\mathcal{L}}_v(t,q,v)\cdot D\xi (q)v=0$, the invariance of $\mathcal{L}$ under the infinitesimal family.
• For a constant field $\xi \equiv a$ one has $D\xi =0$, and the identity reads ${\mathcal{L}}_q\cdot a=B_t+B_q\cdot v$; when $B$ is independent of $q$ this is the statement that $\mathcal{L}$ is unchanged by the translation $q\mapsto q+sa$ up to a total time derivative.

Scope. The definition is an identity imposed on the chart data; it does not assert that a given Lagrangian possesses any symmetry, nor that a symmetry of the configuration space, a frame change or any kinematic structure automatically induces one. Whether a symmetry produces a conserved quantity along the Euler–Lagrange solutions is the theorem The charge of an infinitesimal point symmetry is constant along Euler-Lagrange solutions; the general reparametrisation version is Noether identity for point transformations with time reparametrisation: the conserved charge $p\cdot \xi -E\tau -B$, and the scope of the statement for physical models is recorded in Scope of point symmetries: symmetries belong to the Lagrangian, not to the configuration space, and time translation is the energy case.

• The notation $D_t\phi$ is the derivative of $\phi$ along the curve $t\mapsto (t,q(t))$ when $v=q^{\mathrm{\prime }}(t)$: for such a curve, $D_t\phi (t,q(t),q^{\mathrm{\prime }}(t))=\frac{d}{dt}\phi (t,q(t))$. This is the reason the velocity slot of the symmetry identity is filled by $v$ rather than by a fixed vector: the infinitesimal transformation moves $q$ with velocity $\xi (q)$, and along a motion the velocity changes by $D_t\xi -v{D}_t\tau$ in the general case treated later.

• The charge is a function on the chart $O$; a conserved quantity is the composite $t\mapsto C(t,q(t),q^{\mathrm{\prime }}(t))$ along a solution. No claim is made that the charge is the only conserved quantity, or that it is invariant under changes of chart.