Invariance under rotations in space and conservation of angular momentum

In a previous note I studied in detail the mathematical setup of Noether’s Theorem and its proof. I briefly illustrated the mathematical machinery by considering invariance under translations in time, giving the law of conservation of energy, and invariance under translations in space, giving the law of conservation of linear momentum. I briefly mentioned that invariance under rotations in space would also yield the law of conservation of angular momentum but I  did not work this out explicitly. I want to quickly do this in the present note.

We imagine a particle of unit mass moving freely in the absence of any potential field, and tracing out a path \gamma(t) in the (x, y)-plane of a three-dimensional Euclidean coordinate system between times t_1 and t_2, with the z-coordinate everywhere zero along this path. The angular momentum of the particle at time t with respect to the origin of the coordinate system is given by

\mathbf{L} = \mathbf{r} \times \mathbf{v}

= (\mathbf{i} x + \mathbf{j} y) \times (\mathbf{i} \dot{x} + \mathbf{j} \dot{y})

= \mathbf{k} x \dot{y} - \mathbf{k} y \dot{x}

= \mathbf{k} (x \dot{y} - y \dot{x})

where \times is the vector product operation. Alternatively, we could have obtained this as

\mathbf{L} = \mathbf{r} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \ \\x & y & 0 \\ \ \\\dot{x} & \dot{y} & 0 \end{vmatrix}

= \mathbf{k} (x \dot{y} - y \dot{x})

In terms of Lagrangian mechanics, the path \gamma(t) followed by the particle will be a stationary path of the action functional

S[\gamma(t)] = \int_{t_1}^{t_2} dt \frac{1}{2}(\dot{x}^2 + \dot{y}^2)

(in the absence of a potential field the total energy consists only of kinetic energy).

Now imagine that the entire path \gamma(t) is rotated bodily anticlockwise in the (x, y)-plane through an angle \theta. This corresponds to a one-parameter transformation

\overline{t} \equiv \Phi(t, x, y, \dot{x}, \dot{y}; \theta) = t

\overline{x} \equiv \Psi_1(t, x, y, \dot{x}, \dot{y}; \theta) = x \cos \theta - y \sin \theta

\overline{y} \equiv \Psi_2(t, x, y, \dot{x}, \dot{y}; \theta) = x \sin \theta + y \cos \theta

which reduces to the identity when \theta = 0. We have

d\overline{t} = dt

\dot{\overline{x}}^2 = \dot{x}^2 \cos^2 \theta + \dot{y}^2 \sin^2 \theta - 2 \dot{x} \dot{y} \sin \theta \cos \theta

\dot{\overline{y}}^2 = \dot{x}^2 \sin^2 \theta + \dot{y}^2 \cos^2 \theta + 2 \dot{x} \dot{y} \sin \theta \cos \theta

and therefore

\dot{x}^2 + \dot{y}^2 = \dot{\overline{x}}^2 + \dot{\overline{y}}^2

so the action functional is invariant under this rotation since

S[\overline{\gamma}(t)] = \int_{t_1}^{t_2} d\overline{t} \frac{1}{2}(d\dot{\overline{x}}^2 + d\dot{\overline{y}}^2) = \int_{t_1}^{t_2} dt \frac{1}{2}(\dot{x}^2 + \dot{y}^2) = S[\gamma(t)]

Therefore Noether’s theorem applies. Let

F(t, x, y, \dot{x}, \dot{y}) = \frac{1}{2}(\dot{x}^2 + \dot{y}^2)

Then Noether’s theorem in this case says

\frac{\partial F}{\partial \dot{x}} \psi_1 + \frac{\partial F}{\partial \dot{y}} \psi_2 + \big(F - \frac{\partial F}{\partial \dot{x}} \dot{x} - \frac{\partial F}{\partial \dot{y}} \dot{y}\big) \phi = const.

where

\phi \equiv \frac{\partial \Phi}{\partial \theta} \big|_{\theta = 0} = 0

\psi_1 \equiv \frac{\partial \Psi_1}{\partial \theta} \big|_{\theta = 0} = -y

\psi_2 \equiv \frac{\partial \Psi_2}{\partial \theta} \big|_{\theta = 0} = x

We have

\frac{\partial F}{\partial \dot{x}} = \dot{x}

\frac{\partial F}{\partial \dot{y}} = \dot{y}

Therefore Noether’s theorem gives us (remembering \phi = 0)

-\dot{x} y + \dot{y} x = const.

The expression on the left-hand side of this equation is the angular momentum of the particle (cf. the brief discussion of angular momentum at the start of this note), so this result is precisely the statement that the angular momentum is conserved. Noether’s theorem shows us that this is a direct consequence of the invariance of the action functional of the particle under rotations in space.