Exercise 2.11: Conservation of Angular Momentum

Fill in the details of the argument that Noether’s theorem implies that vector angular momentum is conserved by the motion of the free rigid body.


According to Noether’s theorem, for any continuous symmetry in a system (aka a parametric family of symmetries), there is a conserved quantity.

In the case of rigid bodies, rotations about any axis is a symmetry. This can be proved as shown below:

The Lagrangian in rectangular coordinates for a rigid body with particles indexed by $\alpha$ is:

$$ L(t; x,y,z; v_x, v_y, v_z) = \frac{1}{2} \sum_\alpha m_\alpha \left( \dot{x}_\alpha^2 + \dot{y}_\alpha^2 + \dot{z}_\alpha^2\right) \tag{1} $$

Consider a parameteric rotation about the $z$-axis:

$$ \begin{pmatrix}x_\alpha \\y_\alpha \\z_\alpha \end{pmatrix} = R_z(s)\begin{pmatrix}x_\alpha' \\y_\alpha' \\z_\alpha' \end{pmatrix} = \begin{pmatrix}x_\alpha' \cos{s} - y_\alpha'\sin{s}\\x_\alpha' \sin{s} + y_\alpha'\cos{s}\\z_\alpha'\end{pmatrix}\tag{2} $$

Since a rotation is an orthogonal transformation, it does not change the magnitude of the vector,

$$ x_\alpha^2 + y_\alpha^2 + z_\alpha^2 = (x_\alpha')^2 + (y_\alpha')^2 + (z_\alpha')^2\\ $$

Similarly, differentiating Eq.2 along a path, we get:

$$ \begin{pmatrix}\dot{x}_\alpha\\ \dot{y}_\alpha\\ \dot{z}_\alpha\end{pmatrix} = R_z(s)\begin{pmatrix}\dot{x}_\alpha'\\ \dot{y}_\alpha'\\ \dot{z}_\alpha'\end{pmatrix} $$


$$ \dot{x}_\alpha^2 + \dot{y}_\alpha^2 + \dot{z}_\alpha^2 = \dot{x}_\alpha'^2 + \dot{y}_\alpha'^2 + \dot{z}_\alpha'^2 $$ Combining these, we can see that the post-transformation Lagrangian $L'$ is: {% mathjax() %} $$ L'(t; x_\alpha',y_\alpha',z_\alpha'; \dot{x}_\alpha',\dot{y}_\alpha',\dot{z}_\alpha') = \frac{1}{2} \sum_\alpha m \left(\dot{x}_\alpha'^2 + \dot{y}_\alpha'^2 + \dot{z}_\alpha'^2 \right)\tag{3} $$

Therefore $L’$ in Eq.3 is the exact same function as $L$ in Eq. 1 and hence there is a conserved value corresponding to the rotational symmetry about the z-axis. The momenta are defined as:

$$ \partial_2 L = \left[\sum_\alpha m_\alpha \dot{x}, \sum_\alpha m_\alpha \dot{y}, \sum_\alpha m_\alpha \dot{z}\right] $$


$$ D\widetilde{F}(0)(t;x,y,z)=D\widetilde{R}_z(0)(x,y,z) = [ y, -x, 0]\\ $$

Therefore the Noether integral is:

$$ \begin{align*} \mathscr{I}(t; x_\alpha,y_\alpha,z_\alpha; \dot{x}_\alpha,\dot{y}_\alpha,\dot{z}_\alpha) &= ((\partial_2 L)(D\widetilde{F}(0))) (t; x_\alpha,y_\alpha,z_\alpha; \dot{x}_\alpha,\dot{y}_\alpha,\dot{z}_\alpha) \\ &= \sum_\alpha \left(m_\alpha \dot{x}_\alpha~y_\alpha -m_\alpha \dot{y}_\alpha~x_\alpha + (m_\alpha \dot{z}_\alpha)(0) \right) \\ &= \sum_\alpha m_\alpha \left(y_\alpha \dot{x}_\alpha - x_\alpha \dot{y}_\alpha \right) \end{align*} $$

This is the $z$ component of the angular momentum vector $\sum_\alpha \vec{\xi_\alpha} \times (m_\alpha \dot{\vec{\xi}}_\alpha)$

In the following program, we compute the noether integrals for rotations around all three coordinate axes.

(defn RotX [angle]
    (fn [[x, y, z]]
      (let [ca (cos angle)
            sa (sin angle)]
          (up x
              (- (* ca y) (* sa z))
              (+ (* sa y) (* ca z))))))

(defn RotY [angle]
    (fn [[x, y, z]]
      (let [ca (cos angle)
            sa (sin angle)]
          (up (+ (* ca x) (* sa z))
              (+ (- (* sa x)) (* ca z))))))

(defn RotZ [angle]
    (fn [[x, y, z]]
      (let [ca (cos angle)
            sa (sin angle)]
          (up (- (* ca x) (* sa y))
              (+ (* sa x) (* ca y))

;; Coordinate transformation with three angular "inputs" for rotations about
;; all three axes
;; Composing with `coordinate`, extracts the second element of the tuple that is passed in as the argument
(defn F-tilde [angle-x angle-y angle-z]
  (compose (RotX angle-x) (RotY angle-y) (RotZ angle-z) coordinate))

;; Lagrangian for motion of single particle in rigid body
(defn L-rigid-body_particle [m]
    (fn [[t q v]]
        (- (* 1/2 m (square v)))))

;; Define the Noether integral
(def the-Noether-integral
  (let [L (L-rigid-body_particle 'm_alpha)]
    (* ((partial 2) L) ((D F-tilde) 0 0 0))))

  (up 't
      (up 'x_alpha 'y_alpha 'z_alpha)
      (up 'xdot_alpha 'ydot_alpha 'zdot_alpha))))
\begin{bmatrix}\displaystyle{- m_{\alpha}\,y_{\alpha}\,{\dot z}_{\alpha} + m_{\alpha}\,{\dot y}_{\alpha}\,z_{\alpha}} \cr \cr \displaystyle{m_{\alpha}\,x_{\alpha}\,{\dot z}_{\alpha} - m_{\alpha}\,{\dot x}_{\alpha}\,z_{\alpha}} \cr \cr \displaystyle{- m_{\alpha}\,x_{\alpha}\,{\dot y}_{\alpha} + m_{\alpha}\,{\dot x}_{\alpha}\,y_{\alpha}}\end{bmatrix}

The results above correspond to the components of the angular momentum vector for a single particle in a rigid body. Therefore the conserved quantities for a rigid body are:

$$ \begin{align*} \sum_\alpha m_\alpha & \left( z_\alpha{\dot{y}_\alpha} - \dot{z}_\alpha y_\alpha \right)\\ \sum_\alpha m_\alpha & \left( x_\alpha{\dot{z}_\alpha} - \dot{x}_\alpha z_\alpha \right)\\ \sum_\alpha m_\alpha & \left( y_\alpha{\dot{x}_\alpha} - \dot{y}_\alpha x_\alpha \right)\\ \end{align*} $$

This proves that all three components of the angular moementum vector are conserved for a rigid body in free-rotation.

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