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?

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.