# Physical meaning of constants (momenta?) generated by Noether's theorem via an ${\rm SO}(3)$-action

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?

## 1 Answer

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.