Noether's Theorem and Conservation Laws

5.1 Statement of Noether”s Theorem

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.

5.2 Full Proof of Noether’s Theorem

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. $\blacksquare$

5.3 Worked Example: Spatial Translation and Linear Momentum

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

Solution

Consider an infinitesimal translation $x \to x + \epsilon$ I.e., $\delta x = 1$ , $\delta y = 0$ , $\delta z = 0$ .

For a free particle, $L = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2)$ Which is invariant ( $\delta L = 0$ So $F = 0$ ).

By Noether’s theorem:

$Q = p_x \cdot 1 + p_y \cdot 0 + p_z \cdot 0 - 0 = p_x = \mathrm{const}$

This is conservation of the $x$ -component of linear momentum. Translation invariance in all three directions gives conservation of the full momentum vector $\mathbf{p}$ . $\blacksquare$

5.4 Worked Example: Rotation and Angular Momentum

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

Solution

Consider an infinitesimal rotation by angle $\epsilon$ about the $z$ -axis:

$\delta x = -\epsilon y, \quad \delta y = \epsilon x, \quad \delta z = 0$

For a free particle, $\delta L = m(\dot{x}\,\delta\dot{x} + \dot{y}\,\delta\dot{y}) = m(\dot{x}(-\epsilon\dot{y}) + \dot{y}(\epsilon\dot{x})) = 0$ So $F = 0$ .

By Noether’s theorem:

$Q = p_x(-y) + p_y(x) - 0 = x\, p_y - y\, p_x = L_z$

This is the $z$ -component of angular momentum. Full rotational invariance gives conservation of the entire angular momentum vector $\mathbf{L} = \mathbf{r} \times \mathbf{p}$ . $\blacksquare$

5.5 Worked Example: Time Translation and Energy

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

Solution

Consider an infinitesimal time translation $t \to t + \epsilon$ . The coordinates transform as $q_j(t) \to q_j(t + \epsilon) \approx q_j(t) + \epsilon \dot{q}_j(t)$ So $\delta q_j = \dot{q}_j$ .

If $L$ does not depend explicitly on time, then:

$\delta L = \sum_j \left(\frac{\partial L}{\partial q_j}\dot{q}_j + \frac{\partial L}{\partial \dot{q}_j}\ddot{q}_j\right)\epsilon = \frac{dL}{dt}\epsilon = \frac{d}{dt}\left(\epsilon L\right)$

So $F = \epsilon L$ Giving $F = L$ (per unit $\epsilon$ ).

By Noether’s theorem:

$Q = \sum_j p_j \dot{q}_j - L = h$

This is the energy function, which equals $T + V$ for natural systems. $\blacksquare$

5.6 Summary: Symmetry-Conservation Correspondence

Symmetry	Transformation	Conserved Quantity
Time translation	$t \to t + \epsilon$	Energy $H$
Spatial translation	$x \to x + \epsilon$	Linear momentum $p_x$
Rotation about $z$	$\phi \to \phi + \epsilon$	Angular momentum $L_z$
Galilean boost	$x \to x + \epsilon t$	Centre-of-mass motion

5.7 Worked Example: Central Potential

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. $\blacksquare$

Mark this page as reviewed