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

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