Conservation of Angular Momentum (Lagrangian Mechanics)

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Theorem

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

Let $P$ have the action $S$:

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

where:

$L$ is the standard Lagrangian

$t$ is time.

Let $L$ be invariant with respect to rotation around the $z$-axis.

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

Proof

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

$\ds T$

$=$

$\ds t$

$\ds X_i$

$=$

$\ds x_i \cos \epsilon + y_i \sin \epsilon$

$\ds Y_i$

$=$

$\ds -x_i \sin \epsilon + y_i \cos \epsilon$

$\ds Z_i$

$=$

$\ds z_i$

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:

$\ds \phi$

$=$

$\ds 0$

$\ds \psi_{i x}$

$=$

$\ds \valueat {\dfrac {\partial X_i} {\partial \epsilon} } {\epsilon \mathop = 0}$

$\ds = y_i$

$\ds \psi_{i y}$

$=$

$\ds \valueat {\dfrac {\partial Y_i} {\partial \epsilon} } {\epsilon \mathop = 0}$

$\ds = -x_i$

$\ds \psi_{i z}$

$=$

$\ds 0$

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 right hand side is the $z$ component of angular momentum of $P$ along the $z$-axis, we conclude that it is conserved.

$\blacksquare$

Sources

1963: I.M. Gelfand and S.V. Fomin: Calculus of Variations ... (previous) ... (next): $\S 4.22$: Conservation Laws