Mathematics Lecture Notes

Theorem7.1Noether's Theorem (Lagrangian Mechanics)

Let $L(q,\dot{q},t)$ be a Lagrangian for a system with $n$ degrees of freedom, and suppose the action $S = \int_{t_1}^{t_2}L\,dt$ is invariant under the infinitesimal transformation $q_i \mapsto q_i + \varepsilon\xi_i(q,t)$ , $t \mapsto t + \varepsilon\zeta(q,t)$ . Then the quantity $I = \sum_{i=1}^n \frac{\partial L}{\partial\dot{q}_i}\xi_i - H\zeta$ where $H = \sum_i p_i\dot{q}_i - L$ is the Hamiltonian, is a first integral of the motion: $dI/dt = 0$ along any solution of the Euler-Lagrange equations.

Proof

Step 1: Invariance condition. The action invariance $S[Q,\tau] = S[q,t]$ to first order in $\varepsilon$ reads: $\int_{t_1}^{t_2}\left[\frac{\partial L}{\partial q_i}\xi_i + \frac{\partial L}{\partial\dot{q}_i}\left(\dot{\xi}_i - \dot{q}_i\dot{\zeta}\right) + L\dot{\zeta} + \frac{\partial L}{\partial t}\zeta\right]dt = 0$

Here we used $\delta\dot{q}_i = \dot{\xi}_i - \dot{q}_i\dot{\zeta}$ (the variation of velocity under simultaneous coordinate and time reparametrization) and $d\tau = (1+\varepsilon\dot{\zeta})\,dt$ .

Step 2: Rewrite using total derivatives. Observe that: $\frac{d}{dt}\left[\frac{\partial L}{\partial\dot{q}_i}\xi_i\right] = \frac{d}{dt}\frac{\partial L}{\partial\dot{q}_i}\cdot\xi_i + \frac{\partial L}{\partial\dot{q}_i}\dot{\xi}_i$ and $\frac{d}{dt}[L\zeta] = \dot{L}\zeta + L\dot{\zeta} = \left[\frac{\partial L}{\partial q_i}\dot{q}_i + \frac{\partial L}{\partial\dot{q}_i}\ddot{q}_i + \frac{\partial L}{\partial t}\right]\zeta + L\dot{\zeta}$

Step 3: Apply Euler-Lagrange. On solutions, $\frac{\partial L}{\partial q_i} = \frac{d}{dt}\frac{\partial L}{\partial\dot{q}_i}$ . Using this, collect terms: $\frac{\partial L}{\partial q_i}\xi_i + \frac{\partial L}{\partial\dot{q}_i}\dot{\xi}_i = \frac{d}{dt}\left[\frac{\partial L}{\partial\dot{q}_i}\xi_i\right]$

The remaining terms combine as: $-\frac{\partial L}{\partial\dot{q}_i}\dot{q}_i\dot{\zeta} + L\dot{\zeta} + \frac{\partial L}{\partial t}\zeta = -(H\dot{\zeta}) + \frac{\partial L}{\partial t}\zeta$

where $H = p_i\dot{q}_i - L$ . Adding back $\dot{L}\zeta - \dot{L}\zeta = 0$ : $= -\frac{d}{dt}[H\zeta] + \dot{H}\zeta + \frac{\partial L}{\partial t}\zeta$

Step 4: Conclude. On solutions, $\dot{H} = -\partial L/\partial t$ (standard result from Hamiltonian mechanics). Therefore: $\frac{d}{dt}\left[\sum_i\frac{\partial L}{\partial\dot{q}_i}\xi_i - H\zeta\right] = 0$

This establishes $dI/dt = 0$ . $\square$

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ExampleNoether's Theorem for Electromagnetic Fields

The electromagnetic Lagrangian density $\mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} - A_\mu J^\mu$ is invariant under spacetime translations $x^\mu \to x^\mu + a^\mu$ . The field-theoretic Noether current gives $\partial_\mu T^{\mu\nu} = 0$ with $T^{00} = \frac{1}{2}(\mathbf{E}^2 + \mathbf{B}^2)$ (energy density) and $T^{0i} = (\mathbf{E}\times\mathbf{B})^i$ (Poynting vector), recovering the conservation of electromagnetic energy and momentum.

RemarkConverse of Noether's Theorem

The converse is also true (under mild regularity conditions): every conserved quantity arises from a variational symmetry. More precisely, if $I(q,\dot{q},t)$ is a first integral, then the flow generated by $I$ via the Poisson bracket $\xi_i = \{q_i,I\}$ is a symmetry of the Lagrangian. This establishes a complete bijection between symmetries and conservation laws in Hamiltonian mechanics.