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Let $\Bbb R^3$ be our configuration space. Consider the Lagrangian $L\colon T\Bbb R^3 \cong \Bbb R^6 \to \Bbb R$ given by$$L(x,y,z,\dot{x},\dot{y},\dot{z}) = \frac{m}{2}(\dot{x}^2+\dot{y}^2+\dot{z}^3) - V(x^2+y^2+z^2),$$where $m>0$ is a fixed mass and $V\colon \Bbb R \to \Bbb R$ is smooth. Let ${\rm SO}(3)$ act on $\Bbb R^3$ via evaluation. Since we have a Lie algebra isomorphism $$\Bbb R^3 \ni v=(a,b,c) \mapsto {\sf A}_v = \begin{pmatrix} 0 & -c & b \\ c & 0 & -a \\ -b & a & 0\end{pmatrix} \in \mathfrak{so}(3),$$we can think of the infinitesimal action generated by vectors in $\Bbb R^3$ instead of matrices in $\mathfrak{so}(3)$. If $v \in \Bbb R^3$, the action field $v^\# \in \mathfrak{X}(\Bbb R^3)$ is given by $(v^\#)_p = v\times p \in T_p(\Bbb R^3)$. Since the Lagrangian $L$ is ${\rm SO}(3)$-invariant, Noether's theorem says that along a motion $\gamma(t) = (x(t),y(t),z(t))$ of the mechanical system considered, for any $v \in \Bbb R^3$ the Noether charge $$t\mapsto \det(\gamma(t),\dot{\gamma}(t), v)$$is a constant (which depends on $v$, of course). Choosing $v \in \{e_1,e_2,e_3\}$, we can write $$\begin{cases} c_1 = y(t)\dot{z}(t) - \dot{y}(t)z(t) \\ c_2 = \dot{x}(t)z(t)-x(t)\dot{z}(t) \\ c_3 = x(t)\dot{y}(t) - \dot{x}(t)y(t) \end{cases}$$

  • What is the physical meaning, if any, of these constants $c_1$, $c_2$ and $c_3$? I'm thinking that this resembles some sort of angular momentum, but I never studied physics properly and I don't know any rigorous mathematical definition of momentum.
  • I also assume that we can actually solve for $\gamma$ by using convenient quotient rules to find relations between $x(t)$, $y(t)$ and $z(t)$, but I'm guessing that this should be something standard and I'd like to avoid painful computations if there's an easier way to see what will happen next. Is there such a way?
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I figured this out a long time ago, should probably have deleted this or just added my own answer. Anyway, it is really the angular momentum and I should have never passed to coordinates. Also, I should have kept the mass $m$, for psychological reasons. The point is that $m \det(\gamma(t),\dot{\gamma}(t), v) = \langle m\gamma(t)\times \dot{\gamma}(t), v\rangle$ being constant for all $v$ implies that $m\gamma(t)\times \dot{\gamma}(t)$ is a constant (vector) itself, end of discussion.

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