I'd like to know whether, given a system, there's a way to obtain all the conserved quantities. For instance if the system consists of electric and magnetic fields, the fields must satisfy Maxwell's equations. These equations are invariant under many transformations (Lorentz transformation, rotations, spatial and temporal translations, etc. By the way is there a way, maybe from group theory, to find all the possible transformations that leaves the equation(s) invariant?) which imply as many conserved quantities thanks to Noether's theorem. In wikipedia I can see an equation that seems to give all the conserved quantities (wiki's article) but it involves the Lagrangian and I'm not sure whether the formula is valid for all systems whose Lagrangian is possible to obtain.
[Physics] systematic way to obtain all conserved quantities of a system
conservation-lawsfield-theorygroup-theorynoethers-theoremsymmetry
Related Solutions
I will answer for the 1-D case, or particle mechanics, instead of field mechanics, but the idea is the same. The approach is similar to that of getting the Killing vector field of a metric, and this approach reduces to that when applied to purely kinetic Lagrangians. The objective is to get the so called Rund-Trautman identity
Theory and set up
Let us assume the following: the system is characterised by a configuration manifold $\mathcal Q$ of dimension $n\equiv dim \mathcal Q$ and a Lagrangian $L:T\mathcal Q \times\mathbb R\rightarrow \mathbb R$, possibly dependent on time.
Definition.- We say that a Lagrangian is (quasi-)invariant with respect to the $r-$parameter transformation $$ \bar{x} = \phi(x,t) = x + \varepsilon^s\xi_s + o(\varepsilon) \simeq x^i + \varepsilon^s\xi_s^i + o(\varepsilon) , $$ $$ \bar{t} = \psi(x,t) = t + \varepsilon^s\tau_s + o(\varepsilon) \simeq t + \varepsilon^s\tau_s, $$ with $s=1,\dots,r$ ; $\simeq$ means that we drop the higher orders on $\varepsilon$ and we use an index expression; and (quasi-) refers to if there is a divergence term $dG$ or not; if and only if
$$ S[x(t)]- S[\bar{x}(\bar t)] = \varepsilon^s G_s(x(t),t)|_a^b + o(\varepsilon). $$
Lemma.- A Lagrangian L is invariant under a transformation $\iff$ the following $k$ equations hold:
$$ \frac{\partial L}{\partial t} \tau_s + L\frac{d \tau_s}{dt} + \frac{\partial L}{\partial x^i} \xi^i_s +\frac{\partial L}{\partial \dot{x}^i}(\frac{d \xi^i_s}{dt} - \dot x^i \frac{d\tau_s}{dt} ) = \frac{d}{dt}G_s. $$
Sketch of the Proof
We can write the invariance condition as if we change the integration over $\bar t$ to $t$ in $S[\bar x]$ as
$$
L\left(\bar{x}(\bar t),\frac{d}{d\bar t}\bar{x}(\bar t), \bar{t}\right)\frac{d\bar{t}}{dt} - L(x,\dot x, t) \simeq \varepsilon^s\frac{d}{dt}G_s(x,t).
$$
where $G_s$ is a divergence term. After this differentiate this equation w.r.t. $\varepsilon^s$ at $\varepsilon^s=0$.
This lemma gives us a general relation between the transformation and the Lagrangian. It can be used in different ways:
- To check if a known transformation $(\phi,\psi)$ is a symmetry of a known Lagrangian $L$, and from here derive the Noether conserved quantities.
- If the Lagrangian is unknown, but the transformations are, you get a system of $r$ PDE's on the Lagrangian L, an can be used to impose symmetries,
- Lastly, we can find the symmetry transformations of a given Lagrangian L, considering that the $\dot x^i$ and their powers are independent, so the coefficients in the polynomial $P(x^i)$ are a system of PDE's to obtain $\xi$ and $\tau$.
The third point of view is the one that you want to find the symmetries of a Lagrangian.
Application
In the case of a natural Lagrangian: $$ L \equiv \frac{1}{2} g_{ij}(x)\dot{x}^i\dot{x}^j - V(x) $$
we compute the derivatives $$ \partial_t L =0, \;\;\;\; \partial_{k}L = \frac{1}{2}\partial_{k}g_{ij}\dot{x}^i\dot{x}^j - \partial_{k}V, \;\;\;\; \partial_{,k}L = \frac{1}{2}g_{il}\left(\dot{x}^i\delta^l_k+\dot{x}^l\delta^i_k\right) $$ Then the equations become:
$$ 0\cdot \tau_s + \left( \frac{1}{2} g_{ij}\dot{x}^i\dot{x}^j - V \right) \frac{d \tau_s}{dt} + \left( \frac{1}{2}\partial_{k}g_{ij}\dot{x}^i\dot{x}^j - \partial_{k}V \right) \xi^k_s + \frac{1}{2}g_{il}\left(\dot{x}^i\delta^l_k+\dot{x}^l\delta^i_k\right) \left( \frac{d \xi^k_s}{dt} - \dot x^k \frac{d\tau_s}{dt} \right) = \frac{d}{dt}G_s. $$ taking the powers of $\dot x$ as independent and demanding the equations to be satisfied always, we get the first order PDE's:
$$ 1 \; : \;\; V\partial_t \tau_s -\partial_k V\xi^k_s = \partial_tG_s $$
$$ x^i \; : \;\; - V \partial_i\tau_s + \frac{1}{2}\left(g_{il}\partial_t\xi^l_s+g_{li}\partial_t\xi^l_s\right) = \partial_iG_s $$
$$ \dot x^i \dot x^j \; : \;\; -\frac{1}{2} g_{ij} \partial_t\tau_s + \frac{1}{2}\partial_{k}g_{ij} \xi^k_s + \frac{1}{2}\left( g_{il}\partial_j\xi^l_s + g_{lj}\partial_i\xi^l_s \right) = 0 $$
$$ \dot{x}^i\dot{x}^j\dot x^k\; : \;\; g_{ij}\partial_k\tau_s - 2g_{ik}\partial_j\tau_s= 0 $$
These are the Rund-Trautman identities, or generalised Killing equations to obtain the symmetries of the Lagrangian L. We note that for every $s=1\dots r$ has the same system of equations, and that the number of equations is highly dependant on the form of the Lagrangian.
In the next part I will illustrate one example that particularly interests me. More examples can be found in the references, particularly in [1,4].
Example
A particle in the Hyperbolic Poincaré Disc $\mathbb D$, that has a metric $\frac{2|dz|^2}{1-|z|^2}$, has a Lagrangian $$ L = \frac{\dot x^2 + \dot y^2}{(1-(x^2+y^2))^2} = \gamma^2( \dot x^2 + \dot y^2) $$ with $\gamma \equiv \frac{1}{1-(x^2+y^2)}$. Then $g_{ij}=\gamma^2\delta_{ij}$ and $V=0$. I only want time independent transformations, this means that I fix $\tau_s=0$ and $\partial_t=0$, then the R-T identities become
$$ 1 \; : \;\; 0 - 0 = \partial_tG_s $$
$$ x^i \; : \;\; 0 = \partial_iG_s $$
$$ \dot x^i \dot x^j \; : \;\; \frac{1}{2}\partial_{k}g_{ij} \xi^k_s + \frac{1}{2}\left( g_{il}\partial_j\xi^l_s + g_{lj}\partial_i\xi^l_s \right) = 0 $$
$$ \dot{x}^i\dot{x}^j\dot x^k\; : \;\; 0 = 0 $$
we only have one set of equations, the coefficients of the square terms. Taking into account that $\partial_kg_{ij} = 2\gamma (-\gamma^2)(-2x^k)= 4\gamma^3 x^k\delta_{ij} $ we have the system $$ 4\gamma x_k\delta_{ij} \xi^k_s + \partial_j\xi^i_s + \partial_i\xi^j_s = 0 $$
And in components $(x,y)$ we get 3 equations: $$ \partial_y\xi^y_s = -2 \gamma \vec{x}\vec{\xi}_s $$ $$ \partial_x\xi^x_s = -2 \gamma \vec{x}\vec{\xi}_s $$
$$ \partial_x\xi^y_s + \partial_y\xi^x_s = 0 $$
The second equation tells us directly that one family of solutions is given by the vector field $\vec{\xi}_s=s(-y,x)$, and it checks with the others. So rotations, that could be seen directly from the Lagrangian. If we sum the first two, we get $$ \nabla\cdot \vec{\xi}_s = -4 \gamma \vec{x}\vec{\xi}_s $$
Continuation
For the other possible symmetries you will have to wait, as I haven't computed them yet. Well, one and two can be writen as $$ \frac{1}{\gamma}\partial_i(\gamma \xi^i)=0, \text{ no sum over } i $$ and unfortunatley, to fulfil both at the same time we need $$ \xi = \gamma^{-1}(f_1(y),f_2(x)) $$ and this doesn't fulfil whater the choice of $f_1,f_2$ se make. So it seams that we cant extract more variational simmetries this way. Because I know that there is a transformation that leaves the Lagrangian invariant: $$ \Phi_{\alpha\in\mathbb R, a\in \mathbb C}(z) =\exp(i\alpha) \frac{z-a}{\bar{a}z-1} $$ leaves the Lagrangian invariant, but has not appeared entirely, only the rotational parti, not the one dependent in $a$. This is a mistery.
Conclusion
There exists a method to find the symmetries of a Lagrangian, and it is cumbersome and involve huge PDE systems. For Lagrangian densities, check 1 and 3. Have fun and report if you find something interesting.
Update: And it might be that the equations are not integrable, as my previous case. So whe have an algorithm, but is cumbersome and might be that it still misses some. And this is strange, could it be the complex part.
Bibliography
Invariant Variational Problems, D.J. Logan, Elsevier
"Classical Noether’s theory with application to the linearly damped particle", Raphaël Leone and Thierry Gourieux (LPM) arXiv:1412.7523v2 [math-ph]
Emmy Noether's Wonderful Theorem Dwight E. Neuenschwander, John Hopkings University Press
"Variational symmetries of Lagrangians", G.F. Torres del Castillo, C. Andrade Mirón, and R.I. Bravo Rojas, Rev. Mex. Fis. E 59(2) (2013) 140.
In this answer let us for simplicity restrict to the case of a regular Legendre transformation in a point mechanical setting, cf. this related Phys.SE post. (Generalizations to field theory and gauge theory are in principle possible, with appropriate modifications of conclusions.)
On one hand, the action principle for a Hamiltonian system is given by the Hamiltonian action $$ S_H[q,p] ~:= \int \! dt ~ L_H(q,\dot{q},p,t).\tag{1} $$ Here $L_H$ is the so-called Hamiltonian Lagrangian $$ L_H(q,\dot{q},p,t) ~:=~\sum_{i=1}^n p_i \dot{q}^i - H(q,p,t). \tag{2} $$ In the Hamiltonian formulation there is a bijective correspondence between conserved quantities $Q_H$ and infinitesimal (vertical) quasi-symmetry transformations $\delta$, as showed in my Phys.SE answers here & here. It turns out that a quasi-symmetry transformation $\delta$ is a Hamiltonian vector field generated by a conserved quantity $Q_H$: $$ \delta z^I~=~ \{z^I,Q_H\}\varepsilon,\qquad I~\in~\{1, \ldots, 2n\}, \qquad \delta t~=~0,$$ $$ \delta q^i~=~\frac{\partial Q_H}{\partial p_i}\varepsilon, \qquad \delta p_i~=~ -\frac{\partial Q_H}{\partial q^i}\varepsilon, \qquad i~\in~\{1, \ldots, n\},\tag{3}$$
On the other hand, if we integrate out the momenta $p_i$, we get the corresponding Lagrangian action $$ S[q] ~= \int \! dt ~ L(q,\dot{q},t),\tag{4} $$ cf. this related Phys.SE post. The Hamiltonian eqs. $$0~\approx~\frac{\delta S_H}{\delta p_i} ~=~\dot{q}^i-\frac{\partial H}{\partial p_i} \tag{5}$$ for the momenta $p_i$ yield via the Legendre transformation the defining relation $$p_i~\approx~ \frac{\partial L}{\partial \dot{q}^i}\tag{6}$$ of Lagrangian momenta. Eqs. (5) & (6) establish a bijective correspondence between velocities and momenta.
If we take this bijective correspondence $\dot{q} \leftrightarrow p$ into account it is clear that Hamiltonian and Lagrangian conserved charges $$Q_H(q,p,t)~\approx~Q_L(q,\dot{q},t) \tag{7}$$ are in bijective correspondence. Below we will argue that the same is true for (vertical) infinitesimal quasi-symmetries on both sides.
On one hand, if we start with a (vertical) infinitesimal quasi-symmetry in (Hamiltonian) phase space $$ \varepsilon \frac{df^0_H}{dt}~=~\delta L_H ~=~\sum_{i=1}^n\frac{\delta S_H}{\delta p_i}\delta p_i + \sum_{i=1}^n\frac{\delta S_H}{\delta q^i}\delta q^i + \frac{d}{dt}\sum_{i=1}^n p_i~\delta q^i ,\tag{8}$$ it can with the help of eq. (5) be restricted to a (vertical) infinitesimal quasi-symmetry within the (Lagrangian) configuration space: $$ \varepsilon \frac{df^0_L}{dt}~=~\delta L ~=~ \sum_{i=1}^n\frac{\delta S}{\delta q^i}\delta q^i + \frac{d}{dt}\sum_{i=1}^n p_i~\delta q^i ,\tag{9}$$ In fact we may take $$f^0_L(q,\dot{q},t)~\approx~f^0_H(q,p,t) \tag{10}$$ the same. The restriction procedure also means that the bare Noether charges $$Q^0_H(q,p,t)~\approx~Q^0_L(q,\dot{q},t) \tag{11}$$ are the same, since there are no $\dot{p}_i$ appearance.
Conversely, if we start with an infinitesimal quasi-symmetry in (Lagrangian) configuration space, we can use Noether's theorem to generate a conserved quantity $Q_L$, and in this way close the circle.
Example: Consider $n$ harmonic oscillators with Lagrangian $$ L~=~\frac{1}{2}\sum_{k,\ell=1}^n \left(\dot{q}^k g_{k\ell}\dot{q}^{\ell} - q^k g_{k\ell} q^{\ell}\right),\tag{12}$$ where $g_{k\ell}$ is a metric, i.e. a non-degenerate real symmetric matrix. The Hamiltonian reads $$H~=~\frac{1}{2}\sum_{k,\ell=1}^n \left( p_k g^{k\ell} p_{\ell} + q^k g_{k\ell} q^{\ell}\right) ~=~\sum_{k,\ell=1}^n z^{k \ast} g_{k\ell} z^{\ell},\tag{13}$$ with complex coordinates $$ z^k~:=~\frac{1}{\sqrt{2}}(q^k+ip^k), \qquad p^k~:=~\sum_{\ell=1}^ng^{k\ell}p_{\ell}, \qquad \{z^{k \ast},z^{\ell}\}~=~ig^{k\ell}. \tag{14}$$ The Hamiltonian Lagrangian (2) reads $$ L_H~=~\sum_{k=1}^n p_k \dot{q}^k - H ~=~\frac{i}{2}\sum_{k,\ell=1}^n \left( z^{k \ast} g_{k\ell} \dot{z}^{\ell} - z^{k} g_{k\ell} \dot{z}^{\ell\ast} \right) - H, \tag{15}$$ Hamilton's eqs. are $$ \dot{z}^k~\approx~-iz^k, \qquad \dot{q}^k~\approx~p^k, \qquad \dot{p}^k~\approx~-q^k. \tag{16}$$ Some conserved charges are $$ Q_H ~=~ \sum_{k,\ell=1}^n z^{k \ast} H_{k\ell} z^{\ell} ~=~\sum_{k,\ell=1}^n \left( \frac{1}{2}q^k S_{k\ell} q^{\ell} +\frac{1}{2}p^k S_{k\ell} p^{\ell}+ p^k A_{k\ell} q^{\ell}\right), \tag{17}$$ where $$ H_{k\ell}~:=~S_{k\ell}+i A_{k\ell}~=~H_{\ell k}^{\ast} \tag{18}$$ is an Hermitian $n\times n$ matrix, which consists of a symmetric and an antisymmetric real matrix, $S_{k\ell}$ and $A_{k\ell}$, respectively. The conserved charges (17) generate an infinitesimal $u(n)$ quasi-symmetry of the Hamiltonian action $$\delta z_k~=~ \varepsilon\{z_k , Q_H\} ~=~-i \varepsilon\sum_{\ell=1}^n H_{k\ell} z^{\ell},$$ $$\delta q_k ~=~ \varepsilon\sum_{\ell=1}^n \left( A_{k\ell} q^{\ell} +S_{k\ell} p^{\ell} \right), \qquad \delta p_k ~=~ \varepsilon\sum_{\ell=1}^n \left( -S_{k\ell} q^{\ell} +A_{k\ell} p^{\ell} \right). \tag{19}$$ The bare Noether charges are $$ Q^0_H ~=~\sum_{k,\ell=1}^n p^k \left( A_{k\ell} q^{\ell} +S_{k\ell} p^{\ell} \right). \tag{20}$$ Also $$ f^0_H~=~\frac{1}{2}\sum_{k,\ell=1}^n \left( \frac{1}{2}p^k S_{k\ell} p^{\ell}- q^k S_{k\ell} q^{\ell}\right). \tag{21}$$ The corresponding infinitesimal $u(n)$ quasi-symmetry of the Lagrangian action (1) is $$\delta q_k ~=~ \varepsilon\sum_{\ell=1}^n \left( A_{k\ell} q^{\ell} +S_{k\ell} \dot{q}^{\ell} \right), \tag{22}$$ as one may easily verify.
Best Answer
There is no general algorithm for doing so, and even figuring out how many conserved quantities a system has can be difficult.
A famous example is the Toda lattice. When originally proposed by Toda in 1967, this model was believed to be chaotic. It was in fact proven to be integrable (to have too many conserved quantities to be chaotic) in 1974 by Henon. See Section 3.6 of Gutzwiller's Chaos in Classical and Quantum Mechanics for more details on this story.