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

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?

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).

• I understand, thank you. Maybe a short follow-up question: where does then the amount of independent conserved quantities equal to $2n-1$ come from (e.g. physics.stackexchange.com/questions/8626/…) ? As I understand, this is because of the relation between angular and linear momentum, which leads to some momenta being a combination of the rest. But I do not see how one can deduct that from this reasoning. Jan 26, 2023 at 8:29
• @TheHunter Ah okay, I didn't quite answer your question about independent conserved quantities. The number of dynamical variables which any conserved quantity is written in terms of is $2N$ (positions and velocities/momenta). Therefore at most we can have $2N$ independent conserved quantities. In my answer I deduced $\frac{n(n+1)}{2}+1$ conserved quantities arising from independent symmetries, but certainly at most $2n$ can be independent. The key to determining how many are independent is to see how many conserved quantities commute (i.e. Poisson bracket equals zero) with all the others. Jan 26, 2023 at 19:40

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 $$2n-1$$.

• And how does this amount $2n-1$ then relate to the amount $n(n+1)/2 +1$, explained in in @arturodonjuan answer? Jan 26, 2023 at 17:06
• I do not know. I did not read that answer in deatail since that number might be false: For large n it is greater than the number of independent coordinates and this implies that these constants of motion, if exist, cannot be independent. Jan 26, 2023 at 18:06
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.