Conservation of Angular Momentum (Lagrangian Mechanics)

Theorem
Let $P$ be a physical system composed of finite number of particles.

Let it have the action $S$:


 * $\ds S = \int_{t_0}^{t_1} L \rd t$

where $L$ is the standard Lagrangian, and $t$ is time.

Suppose $L$ is invariant rotation around $z$-axis.

Then the total angular momentum of $P$ along $z$-axis is conserved.

Proof
By assumption, $S$ is invariant under the following family of transformations:


 * $T = t$


 * $X_i = x_i \cos \epsilon + y_i \sin \epsilon$


 * $Y_i = - x_i \sin \epsilon + y_i \cos \epsilon$


 * $Z_i = z_i$

where $\epsilon \in \R$.

By Noether's Theorem:


 * $\nabla_{\dot {\mathbf x} } L \cdot \boldsymbol \psi + \paren {L - \dot {\mathbf x} \cdot \nabla_{\dot {\mathbf x} } L } \phi = C$

where:


 * $\phi = 0$


 * $\psi_{ix} = \intlimits {\dfrac {\partial X_i} {\partial \epsilon} } {\epsilon \mathop = 0} {} = y_i$


 * $\psi_{iy} = \intlimits {\dfrac {\partial Y_i} {\partial \epsilon} } {\epsilon \mathop = 0} {} = -x_i$


 * $\psi_{iz} = 0$

and $C$ is an arbitrary constant.

Then it follows that:


 * $\ds \sum_i \paren {\dfrac {\partial L} {\partial {\dot x}_i} y_i - \dfrac {\partial L} {\partial {\dot y}_i} x_i} = C$

Since the last term is the $z$-th component of angular momentum of $P$ along $z$-axis, we conclude that it is conserved.