Noether's Theorem and Conservation Laws

Theorem 5.1 (Noether’s Theorem). For every continuous symmetry of the action, there is a Corresponding conserved quantity.

More precisely: if the action is invariant (up to a boundary term) under the infinitesimal transformation $q_j \to q_j + \epsilon f_j(q, \dot{q}, t)$Then

$Q = \sum_j \frac{\partial L}{\partial \dot{q}_j} f_j$

Is a constant of motion.

Theorem 5.2 (Noether’s Theorem --- Full Proof). Suppose the Lagrangian transforms under an infinitesimal transformation $q_j \to q_j + \epsilon \delta q_j$ as:

$L \to L + \epsilon \frac{dF}{dt}$

For some function $F(q, t)$. Then the quantity

$Q = \sum_j p_j\, \delta q_j - F$

Is conserved.

Proof. The variation of the action is:

$\delta S = \int_{t_1}^{t_2} \left[\sum_j \left(\frac{\partial L}{\partial q_j}\delta q_j + \frac{\partial L}{\partial \dot{q}_j}\delta\dot{q}_j\right)\right] dt = \int_{t_1}^{t_2} \frac{dF}{dt}\, dt$

Where the second equality uses the assumption that the action changes by at most a boundary term. Using the Euler-Lagrange equations $\frac{\partial L}{\partial q_j} = \frac{d}{dt}\frac{\partial L}{\partial \dot{q}_j}$:

$\delta S = \int_{t_1}^{t_2} \sum_j \left[\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_j}\right)\delta q_j + \frac{\partial L}{\partial \dot{q}_j}\frac{d}{dt}\delta q_j\right] dt$

$= \int_{t_1}^{t_2} \frac{d}{dt}\left(\sum_j p_j\, \delta q_j\right) dt$

Setting this equal to $\int_{t_1}^{t_2} \frac{dF}{dt}\, dt$:

$\frac{d}{dt}\left(\sum_j p_j\, \delta q_j - F\right) = 0$

Therefore $Q = \sum_j p_j\, \delta q_j - F$ is constant. $\blacksquare$

Problem. Show that spatial translation invariance implies conservation of linear momentum.

Problem. Show that rotational invariance implies conservation of angular momentum.

Problem. Show that time translation invariance implies conservation of energy.

Problem. A particle moves in a central potential $V(r)$. Show that angular momentum is conserved.

Solution. In spherical coordinates $(r, \theta, \phi)$ with $V = V(r)$:

$L = \frac{1}{2}m(\dot{r}^2 + r^2\dot{\theta}^2 + r^2\sin^2\theta\,\dot{\phi}^2) - V(r)$

Since $L$ does not depend on $\phi$ (rotational symmetry about the $z$-axis):

$p_\phi = \frac{\partial L}{\partial \dot{\phi}} = mr^2\sin^2\theta\,\dot{\phi} = \mathrm{const}$

This is the $z$-component of angular momentum. By Noether’s theorem, the full angular momentum vector Is conserved for any central potential. $\blacksquare$