Question 2 of 6: Central-Force Particle Under a Fixed Rotation — Lagrangian Invariance and Noether's Theorem

Question 2: Central-Force Particle Under a Fixed Rotation — Lagrangian Invariance and Noether's Theorem (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A particle of mass $m$ moves under a central-force potential $U(|\vec x|)$ that depends on position only through the distance from the origin $|\vec x|=r$; $R$ denotes an arbitrary fixed (constant, time-independent) rotation, $\vec y=R\vec x$; the particle's actual trajectory carries some angular velocity $\vec\omega$ about a fixed axis, denoted $L$ in the question (an axis label, not to be confused with the Lagrangian $L(\vec x,\dot{\vec x})$ used throughout).

Find. (a) that $L(\dot{\vec y},\vec y)=L(\dot{\vec x},\vec x)$ for any fixed rotation $R$; (b) that the component of angular momentum along the axis $L$ (called $L_{\text{axis}}$ below to avoid clashing with the Lagrangian symbol) is a constant of the motion; (c) the Noether-theorem reading of that result.

Approach. Use the defining property of a rotation matrix, $R^{\mathsf T}R=I$, to show it preserves both the norm $|\vec x|$ and the norm $|\dot{\vec x}|$, which is all the Lagrangian $L=\tfrac12 m|\dot{\vec x}|^2-U(|\vec x|)$ depends on. Then specialize to an infinitesimal rotation about the fixed axis $\hat n$ (the axis $L$ in the question) to extract the Noether conserved charge directly from the invariance, and confirm independently that this charge is conserved by the central-force equation of motion.

1. Part (a) — $L$ is invariant under any rotation. A rotation $R$ is by definition an orthogonal matrix with $\det R=+1$, i.e. $R^{\mathsf T}R=I$. Since $R$ is fixed (time-independent), differentiating $\vec y=R\vec x$ gives $\dot{\vec y}=R\dot{\vec x}$ — the same rotation carries both the position and the velocity. Orthogonal matrices preserve the dot product: for any vector $\vec v$, $$|R\vec v|^2=(R\vec v)^{\mathsf T}(R\vec v)=\vec v^{\mathsf T}R^{\mathsf T}R\,\vec v=\vec v^{\mathsf T}\vec v=|\vec v|^2.$$ Applying this to $\vec v=\vec x$ gives $|\vec y|=|\vec x|$, and to $\vec v=\dot{\vec x}$ gives $|\dot{\vec y}|=|\dot{\vec x}|$. The Lagrangian is built only from these two norms, $$L(\dot{\vec x},\vec x)=\tfrac12 m|\dot{\vec x}|^2-U(|\vec x|),$$ so substituting the rotated variables reproduces the identical expression: $$\boxed{L(\dot{\vec y},\vec y)=\tfrac12 m|\dot{\vec y}|^2-U(|\vec y|)=\tfrac12 m|\dot{\vec x}|^2-U(|\vec x|)=L(\dot{\vec x},\vec x).}$$ This holds for every fixed rotation $R$, not just an infinitesimal one — the central-force Lagrangian has full $SO(3)$ rotational symmetry because it only ever "looks at" the particle through its radial distance and speed, both rotation-blind quantities.
2. Part (b) — the angular-momentum component about axis $L$ is conserved. Specialize part (a) to an infinitesimal rotation by angle $\varepsilon$ about the fixed unit axis $\hat n$ (the direction of the axis called "$L$" in the question, and physically the axis the particle is said to revolve about with angular velocity $\vec\omega=\omega\hat n$): to first order in $\varepsilon$, $$\vec x\ \to\ \vec x+\varepsilon(\hat n\times\vec x),\qquad \dot{\vec x}\ \to\ \dot{\vec x}+\varepsilon(\hat n\times\dot{\vec x}).$$ By part (a) (now applied at first order), $L$ is unchanged under this substitution: $\left.\dfrac{\partial L}{\partial \varepsilon}\right|_{\varepsilon=0}=0$. Expanding that derivative by the chain rule, $$0=\frac{\partial L}{\partial\dot{\vec x}}\cdot(\hat n\times\dot{\vec x}) + \frac{\partial L}{\partial\vec x}\cdot(\hat n\times\vec x) = \frac{\partial L}{\partial\dot{\vec x}}\cdot(\hat n\times\dot{\vec x}) + \dot{\vec p}\cdot(\hat n\times\vec x),$$ using the Euler–Lagrange equation $\partial L/\partial\vec x=\dot{\vec p}$ (with $\vec p=m\dot{\vec x}$) in the second term. Since $\partial L/\partial\dot{\vec x}=\vec p$, the first term is $\vec p\cdot(\hat n\times\dot{\vec x})=m\dot{\vec x}\cdot(\hat n\times\dot{\vec x})=0$ identically (a vector dotted with something perpendicular to it, by the scalar-triple-product identity $\vec a\cdot(\vec a\times\vec b)=0$). So the identity collapses to $$\dot{\vec p}\cdot(\hat n\times\vec x)=0 \quad\Longrightarrow\quad \frac{d}{dt}\Big[\vec p\cdot(\hat n\times\vec x)\Big]=\dot{\vec p}\cdot(\hat n\times\vec x)+\vec p\cdot(\hat n\times\dot{\vec x})=0+0=0,$$ where the second inner product also vanishes by the same triple-product identity applied to $\dot{\vec x}$. But $\vec p\cdot(\hat n\times\vec x)=\hat n\cdot(\vec x\times\vec p)=\hat n\cdot\vec\ell$, the component of the particle's angular momentum $\vec\ell=\vec x\times\vec p$ along $\hat n$. Hence $$\boxed{\frac{d}{dt}\big(\hat n\cdot\vec\ell\big)=0}$$ — the angular momentum about the axis $L$ ($\hat n$) is a constant of the motion, independent of the particle's instantaneous angular velocity $\vec\omega$ about that axis (which need not itself be constant in direction for this argument — only the axis $\hat n$ is fixed).
3. Part (c) — reading (b) through Noether's theorem. Noether's theorem states: for every continuous one-parameter family of transformations that leaves the Lagrangian invariant (a continuous symmetry), there is an associated conserved quantity, obtained as $Q=\dfrac{\partial L}{\partial\dot q_i}\dfrac{\partial q_i}{\partial\varepsilon}\Big|_{\varepsilon=0}$ for the transformation's generator. Here the continuous family is rotation about the fixed axis $\hat n$ by an angle $\varepsilon$ (an $SO(2)$ subgroup of the full $SO(3)$ symmetry proven in part (a)); its generator acting on $\vec x$ is exactly $\hat n\times\vec x$, and the resulting Noether charge is $$Q=\frac{\partial L}{\partial\dot{\vec x}}\cdot(\hat n\times\vec x)=\vec p\cdot(\hat n\times\vec x)=\hat n\cdot(\vec x\times\vec p)=\hat n\cdot\vec\ell,$$ — precisely the axial angular-momentum component conserved in part (b). So part (b) is not a separate fact to be independently verified; it is the concrete instance of Noether's theorem for this particular continuous symmetry. The central potential $U(|\vec x|)$ is invariant under rotation about any axis through the origin, so by the same argument every component of $\vec\ell=\vec x\times\vec p$ is separately conserved (the full vector, not just the one component along $L$) — the axis $L$ in the question is simply one particular direction singled out for the exercise; nothing about the potential prefers it over any other.