All continuous symmetries and total number of independent conserved quantities for general classical free particle

Question

Consider the following Lagrangian $$L=\frac{1}{2}G_{ij}\dot{q}^i\dot{q}^j,$$ where $G_{ij}$ is symmetric and positive semi-definite and $i,j=1,\dots,n$. I want to determine all continuous symmetries and the total number of independent conserved quantities. I assume the conclusion will be that there are $2n-1$ independent conserved quantities (combinations of linear and angular momenta) as the configuration space Is $n$-dimensional.

My approach was to use Noether's Theorem which states that for a one-parameter group of symmetries $h_s: M\rightarrow M$, for configuration space $M$ and $s\in\mathbb{R}$ (i.e. the Lagrangian is invariant under every transformation $h_s$), there is a conserved quantity $N:TM\rightarrow \mathbb{R}$ given by coordinates

$$N(q,\dot{q})=\frac{\partial L}{\dot{q}^a}\frac{dh_s^a(q)}{ds}|_{s=0}.$$

By inspection, one can immediately see that $q$ is cyclic and so the Lagrangian is invariant under translations. We can take $h_s$ to be the infinitesimal translation $$\delta q^i = R^i_{\ j}(\alpha_s)q^j.$$ By expanding this in first order of $\alpha$, we can write

$$\delta q^i=q^i + \alpha_s\partial_s R^i_{\ j}(\alpha)q^j = q^i + \alpha_s(T_s)^i_{\ j}q^j,$$

where we defined $(T_s)=\partial_s R(\alpha)$. By plugging this into the Lagrangian, one finds

$$L\rightarrow \frac{1}{2}G_{ij}\dot{q}^i\dot{q}^j + \alpha_s(T_s)_{ij}\dot{q}^i\dot{q}^j.$$

Since we require the Lagrangian to be invariant, the last term must vanish. This happens if $T_s$ is an anti-symmetric $n\times n$ matrix. The space of $n\times n$ anti-symmetric matrices has dimensionality $\frac{n(n-1)}{2}$ and so there are that many continuous symmetries to the system.

But how can I conclude from this the total amount of independent conserved quantities is $2n-1$? Or is the total amount not $2n-1$ and is there something wrong with my reasoning?

Answer 1

Just a few comments, this felt a bit too long to merely put in the comments.

1. A finite translation in an orthogonal coordinate system would be given by $q^i \rightarrow q^i + \alpha^i$, giving $\delta q^i = \alpha^i$. Thus the change-of-basis matrix is zero, as $\delta q^i$ does not depend on $q^j$ for any $i$ or $j$.

2. For the case of transformation with nontrivial Jacobian matrix (e.g. rotation) I don't think what you wrote for $\delta q^i$ is correct. It should be $\delta q^i = \alpha T^i_j q^j$. Notice both sides should be infinitesimal.

3. For the transformed Lagrangian I get $\delta L =\alpha G_{ij} T^i_k \dot{q}^j \dot{q}^k $. This would imply the transformation would only be a symmetry if $G^T T$ was antisymmetric.

Anyway, imagine $G_{ij}$ was diagonal. Then we would have $n$ translational symmetries, $\frac{n(n-1)}{2}$ rotational symmetries, and one time-translation symmetry, giving us a total of $\frac{n(n+1)}{2} + 1$ conserved quantities. For example when $n=3$, we have 7 conserved quantities (3 spatial momenta, 3 angular momenta, and energy).

Answer 2

It does not depend on the form of the Lagrangian: Away from the equilibrium configurations, the number of functionally independent conserved quantities for an autonomous Lagrangian or Hamiltonian system is $2n-1$ and the proof is the following one.

Consider the phase space $F$ (if you prefer it is $TQ$ in Lagrangian mechanics) $$\gamma: I\ni t \mapsto x(t)\in F\:.$$ If the existence & uniqueness theorem is valid (as is the case for Lagrange/Hamilton equations) the curves above define the family of integral lines of a vector field $Z$ on $F$ $$\dot{\gamma}(t) = Z(\gamma(t))\:.$$ As is wll known from geometry, giving such an integral curve with non-vanishing tangent vector, in a neighborhood of it we can define a coordinate system of $2n$ variable where one of the coordinate is the integral parameter of the integral curves and the other $2n-1$ coordinates $y^1,\ldots, y^{2n-1}$ are transverse. The integral curves, in that coordinate system are written down $$\gamma : I \ni t \mapsto (t, y^1,y^2,\ldots, y^{2n-1})$$ where $ y^1,y^2,\ldots, y^{2n-1}$ are constants. Coming back to the initial coordinates, we have $2n-1$ functions $y^k=y^k(x^1,\ldots, x^{2n})$ which remain constant allong the motion. These functions are functionally independent because they are independent coordinates: the Jacobian matrix of the functions $y^k$ must have rank $2n-1$.

Answer 3

1. A conserved quantity/constant of motion (COM) is usually assumed$^1$ to be globally defined, cf. e.g. this Phys.SE post. (Locally, for a non-degenerate system, there trivially exist $2n$ COMs: Just express the $2n$ initial conditions via the $2n$ dynamical variables, cf. e.g. my Phys.SE answer here).

2. In this answer, we assume for simplicity that $G_{ij}$ is positive definite. Then we may rescale the $q^i$ coordinates, so that $G_{ij}=\delta_{ij}$. The COMs are $n$ momenta $p^i=\dot{q}^i$ and $n$ initial positions $q^i_{(0)}=q^i-\dot{q}^it$. The $\frac{n(n-1)}{2}$ angular momenta $L^{ij}=q^i\dot{q}^j-q^j\dot{q}^i$ can be expressed in terms of the $2n$ previous COMs.

3. An integral of motion/first integral (IOM) is a COM that doesn't depend explicitly on time. The IOMs are $n$ momenta $p^i$ and $\frac{n(n-1)}{2}$ angular momenta $L^{ij}$. However, given that we already have $n$ momenta, only $n-1$ of the $L^{ij}$ are functionally independent, so in total $2n-1$ IOMs.

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$^1$ Compare e.g. with the notion of integrability.