Noether's theorem relates symmetries to conserved quantities. For a central potential $V \propto \frac{1}{r}$, the Laplace-Runge-Lenz vector is conserved. What is the symmetry associated with the conservation of this vector?

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    $\begingroup$ The wikipedia article on the Runge-Lenz vector says the symmetry is the result of an isomorphism between the Kepler problem and a free particle constrained to move on the 3D surface of a 4D sphere. $\endgroup$ – Dan Dec 21 '11 at 7:19

1) Hamiltonian Problem. The Kepler problem has Hamiltonian

$$ H~=~T+V, \qquad T~:=~ \frac{p^2}{2m}, \qquad V~:=~- \frac{k}{q}, \tag{1} $$

where $m$ is the 2-body reduced mass. The Laplace–Runge–Lenz vector is (up to an irrelevant normalization)

$$ A^j ~:=~a^j + km\frac{q^j}{q}, \qquad a^j~:=~({\bf L} \times {\bf p})^j~=~{\bf q}\cdot{\bf p}~p^j- p^2~q^j,\qquad {\bf L}~:=~ {\bf q} \times {\bf p}.\tag{2}$$

2) Action. The Hamiltonian Lagrangian is

$$ L_H~:=~ \dot{\bf q}\cdot{\bf p} - H,\tag{3} $$

and the action is

$$ S[{\bf q},{\bf p}]~=~ \int {\rm d}t~L_H .\tag{4}$$

The non-zero fundamental canonical Poisson brackets are

$$ \{ q^i , p^j\}~=~ \delta^{ij}. \tag{5}$$

3) Inverse Noether's Theorem. Quite generally in the Hamiltonian formulation, given a constant of motion $Q$, then the infinitesimal variation

$$\delta~=~ -\varepsilon \{Q,\cdot\}\tag{6}$$

is a global off-shell symmetry of the action $S$ (modulo boundary terms). Here $\varepsilon$ is an infinitesimal global parameter, and $X_Q=\{Q,\cdot\}$ is a Hamiltonian vector field with Hamiltonian generator $Q$. The full Noether charge is $Q$, see e.g. my answer to this question. (The words on-shell and off-shell refer to whether the equations of motion are satisfied or not. The minus is conventional.)

4) Variation. Let us check that the three Laplace–Runge–Lenz components $A^j$ are Hamiltonian generators of three continuous global off-shell symmetries of the action $S$. In detail, the infinitesimal variations $\delta= \varepsilon_j \{A^j,\cdot\}$ read

$$ \delta q^i ~=~ \varepsilon_j \{A^j,q^i\} , \qquad \{A^j,q^i\} ~=~ 2 p^i q^j - q^i p^j - {\bf q}\cdot{\bf p}~\delta^{ij}, $$ $$ \delta p^i ~=~ \varepsilon_j \{A^j,p^i\} , \qquad \{A^j,p^i\}~ =~ p^i p^j - p^2~\delta^{ij} +km\left(\frac{\delta^{ij}}{q}- \frac{q^i q^j}{q^3}\right), $$ $$ \delta t ~=~0,\tag{7}$$

where $\varepsilon_j$ are three infinitesimal parameters.

5) Notice for later that

$$ {\bf q}\cdot\delta {\bf q}~=~\varepsilon_j({\bf q}\cdot{\bf p}~q^j - q^2~p^j), \tag{8} $$

$$ {\bf p}\cdot\delta {\bf p} ~=~\varepsilon_j km(\frac{p^j}{q}-\frac{{\bf q}\cdot{\bf p}~q^j}{q^3})~=~- \frac{km}{q^3}{\bf q}\cdot\delta {\bf q}, \tag{9} $$

$$ {\bf q}\cdot\delta {\bf p}~=~\varepsilon_j({\bf q}\cdot{\bf p}~p^j - p^2~q^j )~=~\varepsilon_j a^j, \tag{10} $$

$$ {\bf p}\cdot\delta {\bf q}~=~2\varepsilon_j( p^2~q^j - {\bf q}\cdot{\bf p}~p^j)~=~-2\varepsilon_j a^j~. \tag{11} $$

6) The Hamiltonian is invariant

$$ \delta H ~=~ \frac{1}{m}{\bf p}\cdot\delta {\bf p} + \frac{k}{q^3}{\bf q}\cdot\delta {\bf q}~=~0, \tag{12}$$

showing that the Laplace–Runge–Lenz vector $A^j$ is classically a constant of motion

$$\frac{dA^j}{dt} ~\approx~ \{ A^j, H\}+\frac{\partial A^j}{\partial t} ~=~ 0.\tag{13}$$

(We will use the $\approx$ sign to stress that an equation is an on-shell equation.)

7) The variation of the Hamiltonian Lagrangian $L_H$ is a total time derivative

$$ \delta L_H~=~ \delta (\dot{\bf q}\cdot{\bf p})~=~ \dot{\bf q}\cdot\delta {\bf p} - \dot{\bf p}\cdot\delta {\bf q} + \frac{d({\bf p}\cdot\delta {\bf q})}{dt} $$ $$ =~ \varepsilon_j\left( \dot{\bf q}\cdot{\bf p}~p^j - p^2~\dot{q}^j + km\left( \frac{\dot{q}^j}{q} - \frac{{\bf q} \cdot \dot{\bf q}~q^j}{q^3}\right)\right) $$ $$- \varepsilon_j\left(2 \dot{\bf p}\cdot{\bf p}~q^j - \dot{\bf p}\cdot{\bf q}~p^j- {\bf p}\cdot{\bf q}~\dot{p}^j \right) - 2\varepsilon_j\frac{da^j}{dt}$$ $$ =~\varepsilon_j\frac{df^j}{dt}, \qquad f^j ~:=~ A^j-2a^j, \tag{14}$$

and hence the action $S$ is invariant off-shell up to boundary terms.

8) Noether charge. The bare Noether charge $Q_{(0)}^j$ is

$$Q_{(0)}^j~:=~ \frac{\partial L_H}{\partial \dot{q}^i} \{A^j,q^i\}+\frac{\partial L_H}{\partial \dot{p}^i} \{A^j,p^i\} ~=~ p^i\{A^j,q^i\}~=~ -2a^j. \tag{15}$$

The full Noether charge $Q^j$ (which takes the total time-derivative into account) becomes (minus) the Laplace–Runge–Lenz vector

$$ Q^j~:=~Q_{(0)}^j-f^j~=~ -2a^j-(A^j-2a^j)~=~ -A^j.\tag{16}$$

$Q^j$ is conserved on-shell

$$\frac{dQ^j}{dt} ~\approx~ 0,\tag{17}$$

due to Noether's first Theorem. Here $j$ is an index that labels the three symmetries.

9) Lagrangian Problem. The Kepler problem has Lagrangian

$$ L~=~T-V, \qquad T~:=~ \frac{m}{2}\dot{q}^2, \qquad V~:=~- \frac{k}{q}. \tag{18} $$

The Lagrangian momentum is

$$ {\bf p}~:=~\frac{\partial L}{\partial \dot{\bf q}}~=~m\dot{\bf q} \tag{19} . $$

Let us project the infinitesimal symmetry transformation (7) to the Lagrangian configuration space

$$ \delta q^i ~=~ \varepsilon_j m \left( 2 \dot{q}^i q^j - q^i \dot{q}^j - {\bf q}\cdot\dot{\bf q}~\delta^{ij}\right), \qquad\delta t ~=~0.\tag{20}$$

It would have been difficult to guess the infinitesimal symmetry transformation (20) without using the corresponding Hamiltonian formulation (7). But once we know it we can proceed within the Lagrangian formalism. The variation of the Lagrangian is a total time derivative

$$ \delta L~=~\varepsilon_j\frac{df^j}{dt}, \qquad f_j~:=~ m\left(m\dot{q}^2q^j- m{\bf q}\cdot\dot{\bf q}~\dot{q}^j +k \frac{q^j}{q}\right)~=~A^j-2 a^j . \tag{21}$$

The bare Noether charge $Q_{(0)}^j$ is again

$$Q_{(0)}^j~:=~2m^2\left(\dot{q}^2q^j- {\bf q}\cdot\dot{\bf q}~\dot{q}^j\right) ~=~-2a^j . \tag{22}$$

The full Noether charge $Q^j$ becomes (minus) the Laplace–Runge–Lenz vector

$$ Q^j~:=~Q_{(0)}^j-f^j~=~ -2a^j-(A^j-2a^j)~=~ -A^j,\tag{23}$$

similar to the Hamiltonian formulation (16).

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  • $\begingroup$ How to compute $\delta t $ in Hamiltonian formalism? If use your formula, possion bracket of $t$ and any function is zero. $\endgroup$ – maplemaple Dec 18 '17 at 4:57
  • $\begingroup$ Yes, $\delta t =0$ is here zero. $\endgroup$ – Qmechanic Dec 19 '17 at 20:25
  • $\begingroup$ Isn't there any geometrical interpretation of (20) using the fact that newtonian orbits are closed? $\endgroup$ – Iván Mauricio Burbano Sep 8 '18 at 19:22

While Kepler second law is simply a statement of the conservation of angular momentum (and as such it holds for all systems described by central forces), the first and the third laws are special and are linked with the unique form of the newtonian potential $-k/r$. In particular, Bertrand theorem assures that only the newtonian potential and the harmonic potential $kr^2$ give rise to closed orbits (no precession). It is natural to think that this must be due to some kind of symmetry of the problem. In fact, the particular symmetry of the newtonian potential is described exactly by the conservation of the RL vector (it can be shown that the RL vector is conserved iff the potential is central and newtonian). This, in turn, is due to a more general symmetry: if conservation of angular momentum is linked to the group of special orthogonal transformations in 3-dimensional space $SO(3)$, conservation of the RL vector must be linked to a 6-dimensional group of symmetries, since in this case there are apparently six conserved quantities (3 components of $L$ and 3 components of $\mathcal A$). In the case of bound orbits, this group is $SO(4)$, the group of rotations in 4-dimensional space.

Just to fix the notation, the RL vector is:

\begin{equation} \mathcal{A}=\textbf{p}\times\textbf{L}-\frac{km}{r}\textbf{x} \end{equation}

Calculate its total derivative:

\begin{equation}\frac{d\mathcal{A}}{dt}=-\nabla U\times(\textbf{x}\times\textbf{p})+\textbf{p}\times\frac{d\textbf{L}}{dt}-\frac{k\textbf{p}}{r}+\frac{k(\textbf{p}\cdot \textbf{x})}{r^3}\textbf{x} \end{equation}

Make use of Levi-Civita symbol to develop the cross terms:

\begin{equation}\epsilon_{sjk}\epsilon_{sil}=\delta_{ji}\delta_{kl}-\delta_{jl}\delta_{ki} \end{equation}


\begin{equation} \frac{d\mathcal{A}}{dt}=\left(\textbf{x}\cdot\nabla U-\frac{k}{r}\right)\textbf{p}+\left[(\textbf{p}\cdot\textbf{x})\frac{k}{r^3}-2\textbf{p}\cdot\nabla U\right]\textbf{x}+(\textbf{p}\cdot\textbf{x})\nabla U \end{equation}

Now, if the potential $U=U(r)$ is central:

\begin{equation} (\nabla U)_j=\frac{\partial U}{\partial x_j}=\frac{dU}{dr}\frac{\partial r}{\partial x_j}=\frac{dU}{dr}\frac{x_j}{r} \end{equation}


\begin{equation} \nabla U=\frac{dU}{dr}\frac{\textbf{x}}{r}\end{equation}

Substituting back:

\begin{equation}\frac{d\mathcal A}{dt}=\frac{1}{r}\left(\frac{dU}{dr}-\frac{k}{r^2}\right)[r^2\textbf{p}-(\textbf{x}\cdot\textbf{p})\textbf{x}]\end{equation}

Now, you see that if $U$ has exactly the newtonian form then the first parenthesis is zero and so the RL vector is conserved.

Maybe there's some slicker way to see it (Poisson brackets?), but this works anyway.

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  • $\begingroup$ it can be shown that the RL vector is conserved iff the potential is central and newtonian. Is this done by showing that the corresponding poisson bracket involving the Hamilton function vanishes? Could You expand this a bit in Your answer, please (I am just interested in this) ;-) ? $\endgroup$ – Dilaton Dec 10 '11 at 14:01
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    $\begingroup$ Comment to v1: Good to mention the somewhat hidden $SO(4)$ symmetry. Since OP mentions Noether's theorem, chances are that OP is really asking a slightly different question (which @Christoph addresses in his answer; I hope he puts it back up), namely, what is the explicit expression for the (off-shell) symmetry of the action $S$, that via Noether's theorem generates the conservation law for the RL vector. $\endgroup$ – Qmechanic Dec 10 '11 at 15:05
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    $\begingroup$ @Qmechanic: I'll put my answer back up after I've had time to do some facts checking - in particular, Emmy Noether's paper does include the case of velocity/momentum-dependant symmetries (as you mentioned in a comment), even though the formulation of the 'classical' Noether's theorem one finds in textbooks often does not; also, I've found another source for the symmetry corresponding to the LRL vector, which - at least at first glance - does not agree with the one from Wikipedia... $\endgroup$ – Christoph Dec 10 '11 at 22:36

The symmetry is an example of an open symmetry, i.e. a symmetry group which varies from group action orbit to orbit. For bound trajectories, it's SO(4). For parabolic ones, it's SE(3). For hyperbolic ones, it's SO(3,1). Such cases are better handled by groupoids.

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    $\begingroup$ I was asking for the symmetry transformation, not the symmetry group, but +1 anyway. $\endgroup$ – Dan Dec 22 '11 at 9:20
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    $\begingroup$ Could you give me some references about your saying? I'm very interested in this. Thanks. $\endgroup$ – maplemaple Dec 18 '17 at 19:40
  • $\begingroup$ @FTK: Can you please say more on this and, ideally, provide references? Why are such cases better handled by grupoids? $\endgroup$ – Michael_1812 Feb 9 '19 at 3:34

Conservation of the Runge-Lenz vector does not correspond to a symmetry of the Lagrangian itself. It arises from an invariance of the integral of the Lagrangian with respect to time, the classical action integral. Some time ago I wrote up a derivation of the conserved vector for any spherically symmetric potential:

http://analyticphysics.com/Runge Vector/The Symmetry Corresponding to the Runge Vector.htm

The derivation is at the level of Goldstein and is meant to fill in the gap left by its omission from graduate-level classical mechanics texts.

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  • $\begingroup$ Qmechanic's answer seems to imply the presence of a transformation that leaves the Lagrangian constant by the inverse Noether's theorem. Does your answer disagree with theirs, or am I missing something? $\endgroup$ – Dan Jan 11 '14 at 0:26
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    $\begingroup$ @Dan: No, my answer seems to agree with Paul Masson's answer: It is not a symmetry of the Lagrangian. Rather it is a symmetry of the action (up to boundary terms). Phrased equivalently, it is a so-called quasisymmetry of the Lagrangian, cf. this Phys.SE answer. $\endgroup$ – Qmechanic Jan 11 '14 at 1:24

(This post may be old, but we can add some precisions) The conservation of the RL vector is not trifling, it goes with the fact that you consider a central force, lead here by a Newtonian potential $\frac{1}{r}$ which has the property to be invariant under rotations (as $\frac{1}{r^n}$ but it works only for $n=1$ as shown by @quark1245).

Therefore, the S0(3) which has not 6 conserved quantities as said before but 3, the 3 generators of the symmetry $J_i$, i=1..3 such that the symmetry transformation under an infinitesimal change $x \rightarrow x + \epsilon$ is given in the canonical formalism by $$ \delta_i X = \{X, J_i(\epsilon) \} $$ and the algebra is $$ \{ J_i, J_j \} = \epsilon_{ij}^k J_k. $$ They are conserved because, at least for the Kepler problem, the system is invariant w.r.t a time translation, and the Hamiltonian is also conserved, and the calculations show that $$ \{H,J_i\}= 0. $$

Before their redefinition as shown on Wikipedia to see that the previous algebra is fulfilled, the generators of the rotations are : one is the angular momentum $L$ which shows that the movement is planar, therefore invariant under rotation around $L$, one is the RL vector which is in the plan, therefore perpendicular to $L$ and parallel to the major axis of the ellipse, and the third one has a name I don't remember, but is parallel to the minor axis.

We can see that their are only 3 degrees of freedom if we take place in the referential such that $\vec{J}_1 = \vec{L} = (0,0,L_z)$, then the planar generators are $A = (A_x,0,0)$ and $B = (0,B_y,0)$.

It has been shown that they can be constructed from the Killing-Yano tensors (which mean symmetry), and it works also at dimensions greater than 3. A nice review about the LRL vector derivation can be found in HeckmanVanHaalten

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Looking at https://arxiv.org/pdf/1207.5001.pdf one gets a very nice solution. If one is not very keen into mathematics, their basic idea is to use the infinitesimal transformation $$\delta x^i=\epsilon L^{ik}$$ where $L^{ik}=\dot{x}^ix^k-x^k\dot{x}^i$. Since angular momentum is conserved, kinetic energy won't change. On the other hand, the potential changes up to order $\epsilon^2$ like $$\frac{k}{r+\delta r}=\frac{k}{((x^i+\delta x^i)(x_i+\delta x_i))^{1/2}}=\frac{k}{r}\left(1-\frac{x_i\delta x^i}{r^2}\right)=\frac{k}{r}-\epsilon\frac{kx_iL^{ik}}{r^3}=\frac{k}{r}-\epsilon\frac{d}{dt}\left(\frac{kx^k}{r}\right).$$

Therefore, the change in the action is $$\epsilon\left[m\dot{x}_iL^{ik}\right]=[m\dot{x}_i\delta x^i]_{t_1}^{t_2}=\delta S=\epsilon\left[\frac{kx^k}{r}\right]_{t_1}^{t_2}.$$ This gives the conservation of the vector $$m\dot{x}_iL^{ik}-\frac{kx^k}{r},$$ which can be easily shown to be the Runge-Lenz vector.

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  • $\begingroup$ If someone has a geometrical interpretation for this it would be great. I believe there is something to do with the fact that orbits don't precess. $\endgroup$ – Iván Mauricio Burbano Sep 8 '18 at 23:51
  • $\begingroup$ Minor comment to the post (v1): In the future please link to abstract pages rather than pdf files. $\endgroup$ – Qmechanic Dec 9 '18 at 19:59

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