When one first encounters Noether’s theorem in classical mechanics, it is generally presented this way: an invariance of a Lagrangian implies a corresponding conserved quantity. This is followed by development of the standard examples: translational invariance implies conservation of linear momentum, rotational invariance implies conservation of angular momentum, and temporal invariance implies conservation of energy.

In the process of learning how to solve the two-body problem for the Kepler potential, another conserved quantity is introduced, the vector that appears under the names of Runge, Laplace and Lenz. One invariably then asks: what is the Lagrangian invariance corresponding to the Runge vector? This is a far from trivial question. The bottom line is that the Runge vector does not correspond to an invariance of the Lagrangian itself, but rather an invariance of the action integral.

Surprisingly, the stark difference separating this conserved quantity from the others is not discussed in graduate-level classical mechanics textbooks in common use. There is generally a discussion of the group structure of the Poisson brackets of the components of the Runge vector with the components of the angular momentum vector as a connection to quantum mechanics. There may also be some demonstration of transformations generated by components of the Runge vector to show that they alter the eccentricity of the orbit without changing its energy. There is, however, no explicit discussion of the origin of the Runge vector as part of a variational procedure. The following presentation is intended to fill this gap.

In the mathematical equations below, repeated indices represent sums over those indices, as is commonly done in coordinate-based presentations of classical general relativity. The development follows part of a paper by Struckmeier and Riedel (copy here), but in a less baroque manner. An English translation of the paper that is the source of the theorem is available here.

First review the derivation of the Lagrange equations. Begin with a Lagrangian

$I=\int dt\phantom{\rule{.3em}{0ex}}L\left[{q}_{k}\right(t),{\stackrel{\xb7}{q}}_{k}(t\left)\right]$

where the endpoints of the integration are two arbitrary points on the Lagrangian manifold of the coordinates and their time derivatives. If the endpoints of the path are fixed, allow small variations in the coordinates along the path that vanish at the endpoints:

${q}_{i}^{\prime}={q}_{i}+\delta {q}_{i}$

The time derivative of the coordinates will likewise vary with the time derivative of the variation:

${\stackrel{\xb7}{q}}_{i}^{\prime}=\frac{d{q}_{i}^{\prime}}{dt}={\stackrel{\xb7}{q}}_{i}+\delta {\stackrel{\xb7}{q}}_{i}$

Now let the Lagrangian be a function of the coordinates with variations, and expand it to first order in terms of these small quantities in order to define the variation of the Lagrangian:

$\begin{array}{c}L({q}_{k}^{\prime},{\stackrel{\xb7}{q}}_{k}^{\prime})=L({q}_{k},{\stackrel{\xb7}{q}}_{k})+\frac{\partial L}{\partial {q}_{k}}\delta {q}_{k}+\frac{\partial L}{\partial {\stackrel{\xb7}{q}}_{k}}\delta {\stackrel{\xb7}{q}}_{k}+\cdots \\ \\ \delta L=\frac{\partial L}{\partial {q}_{k}}\delta {q}_{k}+\frac{\partial L}{\partial {\stackrel{\xb7}{q}}_{k}}\delta {\stackrel{\xb7}{q}}_{k}\end{array}$

When the coordinates and Lagrangian vary, the action integral will undergo a first-order change of

$\delta I=\int dt\phantom{\rule{.2em}{0ex}}[L({q}_{k}^{\prime},{\stackrel{\xb7}{q}}_{k}^{\prime})-L({q}_{k},{\stackrel{\xb7}{q}}_{k})]=\int dt\phantom{\rule{.3em}{0ex}}\delta L$

and setting this change equal to zero provides a constraint under which the action integral is invariant. Substituting the Lagrangian variation and performing an integration by parts on the second term,

$\begin{array}{l}\delta I=\int dt\phantom{\rule{.2em}{0ex}}[\frac{\partial L}{\partial {q}_{k}}\delta {q}_{k}+\frac{\partial L}{\partial {\stackrel{\xb7}{q}}_{k}}\delta {\stackrel{\xb7}{q}}_{k}]\\ \phantom{\delta I}=\int dt\phantom{\rule{.2em}{0ex}}[\frac{\partial L}{\partial {q}_{k}}-\frac{d}{dt}\frac{\partial L}{\partial {\stackrel{\xb7}{q}}_{k}}]\delta {q}_{k}+\frac{\partial L}{\partial {\stackrel{\xb7}{q}}_{k}}\delta {q}_{k}{|}_{\mathrm{beginning}}^{\mathrm{end}}\\ \delta I=\int dt\phantom{\rule{.2em}{0ex}}[\frac{\partial L}{\partial {q}_{k}}-\frac{d}{dt}\frac{\partial L}{\partial {\stackrel{\xb7}{q}}_{k}}]\delta {q}_{k}\equiv 0\end{array}$

where the second term in the middle step is zero because the coordinate variations vanish by definition at the endpoints. Since the variations are arbitrary, the change in the action integral can only be zero if the quantity in brackets is identically zero, giving the Lagrange equations

$\frac{d}{dt}\frac{\partial L}{\partial {\stackrel{\xb7}{q}}_{i}}=\frac{\partial L}{\partial {q}_{i}}$

where there is a separate equation for each coordinate. If these equations are used to replace the first partial derivative in the variation of the Lagrangian,

$\delta L=\left[\frac{d}{dt}\frac{\partial L}{\partial {\stackrel{\xb7}{q}}_{k}}\right]\delta {q}_{k}+\frac{\partial L}{\partial {\stackrel{\xb7}{q}}_{k}}\delta {\stackrel{\xb7}{q}}_{k}=\frac{d}{dt}\left[\frac{\partial L}{\partial {\stackrel{\xb7}{q}}_{k}}\delta {q}_{k}\right]$

then setting this total derivative equal to zero provides a concise way of defining constants of the motion corresponding to invariances of the Lagrangian. For example, a Lagrangian

$\frac{\partial L}{\partial {\stackrel{\xb7}{q}}_{i}}\delta {q}_{i}=\mathrm{constant}=m{\stackrel{\xb7}{q}}_{i}\times \delta {c}_{i}$

for each coordinate, where in this equation there is no sum on the repeated index, and this constant is proportional to each component of linear momentum. In the same way, a Lagrangian that is invariant under rotations will lead to a constants proportional to angular momentum.

What one cannot define by this simple method, however, is an invariance of a Lagrangian corresponding to the components of the Runge vector. For that, one must consider a more complex variation of variables and its effect upon the entire action integral, not just the integrand that is the Lagrangian. This simple procedure is also not directly applicable to demonstrating energy conservation under temporal invariance, since time is treated differently from coordinates in a Lagrangian formulation. The additional advantage of the more complex variation is that it provides all possible constants of the motion in one procedure.

In the derivation of the Lagrange equations, there is no need to work out explicitly the form of small variations added to paths on the manifold. An explicit consideration of the constraints placed upon such variations in forming an extremum of the action integral not only determines the overall form of the variations, but determines the constants of the motion at the same time. Begin with a general point transformation that adds a small variation to both coordinates and time that is a function of the original variables:

${t}^{\prime}=t+\delta t({q}_{k},t)\phantom{\rule{5em}{0ex}}{q}_{i}^{\prime}={q}_{i}+\delta {q}_{i}({q}_{k},t)$

Derivatives of coordinates will acquire an additional term compared to the Lagrange equation derivation due to the explicit variation of the temporal variable,

${\stackrel{\xb7}{q}}_{i}^{\prime}=\frac{d{q}_{i}^{\prime}}{d{t}^{\prime}}=\frac{d[{q}_{i}+\delta {q}_{i}]}{d[t+\delta t]}=\frac{\frac{d}{dt}[{q}_{i}+\delta {q}_{i}]}{\frac{d}{dt}[t+\delta t]}=\frac{{\stackrel{\xb7}{q}}_{i}+\delta {\stackrel{\xb7}{q}}_{i}}{1+\delta \stackrel{\xb7}{t}}\approx {\stackrel{\xb7}{q}}_{i}+\delta {\stackrel{\xb7}{q}}_{i}-{\stackrel{\xb7}{q}}_{i}\delta \stackrel{\xb7}{t}$

where the denominator has been expanded using the geometric series, and only terms first-order in the variation have been kept. The variation of the Lagrangian also acquires an extra term compared to the Lagrange equation derivation:

$\delta L({q}_{k}^{\prime},{\stackrel{\xb7}{q}}_{k}^{\prime})=\frac{\partial L}{\partial {q}_{k}^{\prime}}\delta {q}_{k}^{\prime}+\frac{\partial L}{\partial {\stackrel{\xb7}{q}}_{k}^{\prime}}\delta {\stackrel{\xb7}{q}}_{k}^{\prime}=\frac{\partial L}{\partial {q}_{k}}\delta {q}_{k}+\frac{\partial L}{\partial {\stackrel{\xb7}{q}}_{k}}(\delta {\stackrel{\xb7}{q}}_{k}-{\stackrel{\xb7}{q}}_{k}\delta \stackrel{\xb7}{t})$

Noether’s theorem in its entirety states that if an integral, in this case the action integral, is invariant under a transformation, then there will exist corresponding constants of the motion. In the most general form, that means

$I=\int dt\phantom{\rule{.3em}{0ex}}L({q}_{k},{\stackrel{\xb7}{q}}_{k})=\int d{t}^{\prime}\phantom{\rule{.2em}{0ex}}[L({q}_{k}^{\prime},{\stackrel{\xb7}{q}}_{k}^{\prime})+\frac{df}{d{t}^{\prime}}]$

where there is an added function to account for changes in the Lagrangian itself due to the transformation. The appearance of this added function is similar to how gauge functions work in field theories, but since one thinks of a gauge function as something that does not alter a physical system, is seems more appropriate to refer to it as a gauge-type function. The Runge vector does not appear at the end of the variational procedure without the inclusion of this function, and that certainly constitutes an alteration of the physical content of the system.

If the added gauge-type function is taken to be of the same order of magnitude as the variations, the overall change in the integral can then be written

$\delta I=\int dt\phantom{\rule{.2em}{0ex}}[\delta L+L\delta \stackrel{\xb7}{t}+\frac{df}{dt}]\equiv 0$

where the extra middle term arises from the transformation $d{t}^{\prime}=d[t+\delta t]=dt[1+\delta \stackrel{\xb7}{t}]$ of the variable of integration. Now with the chain rule

$\frac{dL}{dt}=\frac{\partial L}{\partial {q}_{k}}{\stackrel{\xb7}{q}}_{k}+\frac{\partial L}{\partial {\stackrel{\xb7}{q}}_{k}}{\stackrel{\xb7\xb7}{q}}_{k}$

for the total time derivative of the Lagrangian, the integrand of the overall change in the action integral can be written

$\begin{array}{l}\delta L+L\delta \stackrel{\xb7}{t}+\frac{df}{dt}=\frac{\partial L}{\partial {q}_{k}}\delta {q}_{k}+\frac{\partial L}{\partial {\stackrel{\xb7}{q}}_{k}}(\delta {\stackrel{\xb7}{q}}_{k}-{\stackrel{\xb7}{q}}_{k}\delta \stackrel{\xb7}{t})\\ \phantom{\rule{8em}{0ex}}+\frac{d}{dt}[L\delta t+f]-[\frac{\partial L}{\partial {q}_{k}}{\stackrel{\xb7}{q}}_{k}+\frac{\partial L}{\partial {\stackrel{\xb7}{q}}_{k}}{\stackrel{\xb7\xb7}{q}}_{k}]\delta t\\ \phantom{\delta L+L\delta \stackrel{\xb7}{t}+\frac{df}{dt}}=\frac{d}{dt}[\frac{\partial L}{\partial {\stackrel{\xb7}{q}}_{k}}(\delta {q}_{k}-{\stackrel{\xb7}{q}}_{k}\delta t)+L\delta t+f]\\ \phantom{\rule{8em}{0ex}}-[\frac{d}{dt}\frac{\partial L}{\partial {\stackrel{\xb7}{q}}_{k}}-\frac{\partial L}{\partial {q}_{k}}](\delta {q}_{k}-{\stackrel{\xb7}{q}}_{k}\delta t)\end{array}$

The coefficients in the second square brackets are the Lagrange equations for each independent coordinate and so vanish. The first term in square brackets, as a total time derivative equal to zero, is a linear combination of the constants of the motion accompanied by variations. Labeling this linear combination *C* and rewriting slightly,

$\begin{array}{l}\frac{dC}{dt}=\frac{d}{dt}[\frac{\partial L}{\partial {\stackrel{\xb7}{q}}_{k}}(\delta {q}_{k}-{\stackrel{\xb7}{q}}_{k}\delta t)+L\delta t+f]\\ \phantom{\frac{dC}{dt}}=\frac{d}{dt}[\frac{\partial L}{\partial {\stackrel{\xb7}{q}}_{k}}\delta {q}_{k}+f-(\frac{\partial L}{\partial {\stackrel{\xb7}{q}}_{k}}{\stackrel{\xb7}{q}}_{k}-L\left)\delta t\right]\\ \frac{dC}{dt}=\frac{d}{dt}[\frac{\partial L}{\partial {\stackrel{\xb7}{q}}_{k}}\delta {q}_{k}+f-H\delta t]\end{array}$

one of these constants is the Hamiltonian. Note that the first term here is the equivalent to the statement above concerning invariances of the Lagrangian itself. To reveal the remaining constants of the motion, use a Lagrangian of standard form and its corresponding Lagrange equations:

$L=\frac{m}{2}{\stackrel{\xb7}{q}}_{k}{\stackrel{\xb7}{q}}_{k}-V\left({q}_{k}\right)\phantom{\rule{4em}{0ex}}m{\stackrel{\xb7\xb7}{q}}_{i}=\frac{d}{dt}\frac{\partial L}{\partial {\stackrel{\xb7}{q}}_{i}}=\frac{\partial L}{\partial {q}_{i}}=-\frac{\partial V}{\partial {q}_{i}}$

The undetermined functions δ*q _{k}* , δ

$\begin{array}{l}\frac{dC}{dt}=\frac{d}{dt}[m{\stackrel{\xb7}{q}}_{k}\delta {q}_{k}+f-H\delta t]\equiv 0\\ \phantom{\frac{dC}{dt}}=-\frac{\partial V}{\partial {q}_{k}}\delta {q}_{k}+m{\stackrel{\xb7}{q}}_{k}[\frac{\partial \delta {q}_{k}}{\partial {q}_{l}}{\stackrel{\xb7}{q}}_{l}+\frac{\partial \delta {q}_{k}}{\partial t}]\\ \phantom{\rule{4em}{0ex}}+\frac{\partial f}{\partial {q}_{l}}{\stackrel{\xb7}{q}}_{l}+\frac{\partial f}{\partial t}-(\frac{m}{2}{\stackrel{\xb7}{q}}_{k}{\stackrel{\xb7}{q}}_{k}+V)[\frac{\partial \delta t}{\partial {q}_{l}}{\stackrel{\xb7}{q}}_{l}+\frac{\partial \delta t}{\partial t}]\\ \frac{dC}{dt}=-\frac{m}{2}{\stackrel{\xb7}{q}}_{k}{\stackrel{\xb7}{q}}_{k}{\stackrel{\xb7}{q}}_{l}\frac{\partial \delta t}{\partial {q}_{l}}+m{\stackrel{\xb7}{q}}_{k}{\stackrel{\xb7}{q}}_{l}[\frac{\partial \delta {q}_{k}}{\partial {q}_{l}}-\frac{1}{2}\frac{\partial \delta t}{\partial t}{\delta}_{kl}]\\ \phantom{\rule{4em}{0ex}}+{\stackrel{\xb7}{q}}_{k}[m\frac{\partial \delta {q}_{k}}{\partial t}+\frac{\partial f}{\partial {q}_{k}}-V\frac{\partial \delta t}{\partial {q}_{k}}]\\ \phantom{\rule{4em}{0ex}}-\frac{\partial V}{\partial {q}_{k}}\delta {q}_{k}+\frac{\partial f}{\partial t}-V\frac{\partial \delta t}{\partial t}\end{array}$

For this statement to hold for arbitrary coordinates, the coefficient of the term cubic in time derivatives of the coordinates must be identically zero. The function δ*t* is then a function of time alone, and can be taken to be constant. The total time derivative simplifies to

$\frac{dC}{dt}=m{\stackrel{\xb7}{q}}_{k}{\stackrel{\xb7}{q}}_{l}\frac{\partial \delta {q}_{k}}{\partial {q}_{l}}+{\stackrel{\xb7}{q}}_{k}[m\frac{\partial \delta {q}_{k}}{\partial t}+\frac{\partial f}{\partial {q}_{k}}]-\frac{\partial V}{\partial {q}_{k}}\delta {q}_{k}+\frac{\partial f}{\partial t}\equiv 0$

Since the term quadratic in time derivatives of the coordinates is symmetric in its indices, its coefficient matrix must be antisymmetric to produce zero. This matrix can be taken to be constant, and will be written with a numerical factor for later convenience and an explicit variation. The result is

$\begin{array}{l}\frac{\partial \delta {q}_{k}}{\partial {q}_{l}}=2{\alpha}_{kl}\delta \alpha \phantom{\rule{5em}{0ex}}{\alpha}_{kl}=-{\alpha}_{lk}\\ \\ \phantom{\rule{3em}{0ex}}\delta {q}_{i}=2{\alpha}_{ik}{q}_{k}\delta \alpha +{\beta}_{i}\left(t\right)\delta \beta \end{array}$

where the β_{i} are independent constants of integration, one for each coordinate, and an explicit variation is included for consistency. The crucial need for the gauge-type function appears now as the offset to the β_{i}, ensuring that the term linear in time derivatives of the coordinates is independently zero in a nontrivial manner:

$\frac{\partial f}{\partial {q}_{i}}=-m\frac{\partial \delta {q}_{i}}{\partial t}=-m{\stackrel{\xb7}{\beta}}_{i}\delta \beta \phantom{\rule{1em}{0ex}}\to \phantom{\rule{1em}{0ex}}f=-m{\stackrel{\xb7}{\beta}}_{k}{q}_{k}\delta \beta $

The linear combination of constants of the motion for a Lagrangian of standard form can now be written

$\begin{array}{l}C=m{\stackrel{\xb7}{q}}_{k}\delta {q}_{k}+f-H\delta t\\ \phantom{C}=m{\alpha}_{kl}({\stackrel{\xb7}{q}}_{k}{q}_{l}-{\stackrel{\xb7}{q}}_{l}{q}_{k})\delta \alpha +m({\stackrel{\xb7}{q}}_{k}{\beta}_{k}-{\stackrel{\xb7}{\beta}}_{k}{q}_{k})\delta \beta -H\delta t\end{array}$

The first set of terms are the components of the angular momentum tensor in an arbitrary number of dimensions multiplied by angles. The components of the angular momentum tensor will not be separately constant until the potential in the Lagrangian is specified as having rotational symmetry, but this structure appears immediately as part of the procedure. The last term is again the energy multiplied by a temporal shift.

The middle set of terms will determine the remaining possible constants of the motion, which will be the components of the Runge vector. This middle set does not appear without inclusion of the gauge-type function in the variation of the action integral.

Setting the final two terms of the total time derivative of the linear combination *C* equal to zero provides a differential equation determining the β_{i} :

$\frac{\partial f}{\partial t}-\frac{\partial V}{\partial {q}_{k}}\delta {q}_{k}=0=-m{\stackrel{\xb7\xb7}{\beta}}_{k}{q}_{k}\delta \beta -\frac{\partial V}{\partial {q}_{k}}(2{\alpha}_{kl}{q}_{l}\delta \alpha +{\beta}_{k}\delta \beta )$

For a potential with spherical symmetry,

$\frac{\partial V\left(q\right)}{\partial {q}_{i}}=\frac{dV}{dq}\frac{{q}_{i}}{q}\phantom{\rule{5em}{0ex}}q\equiv \sqrt{{q}_{k}{q}_{k}}$

the terms in this differential equation with antisymmetric coefficients will vanish. Since the β_{i} are independent constants of integration, this single equation can further be split apart into individual equations for each of these constants, which are linear in each constant:

${\stackrel{\xb7\xb7}{\beta}}_{i}+\frac{1}{mq}\frac{dV}{dq}{\beta}_{i}=0$

The β_{i} are determined by this equation as functions of time via the temporal dependence of the coordinates. With this equation in hand, one can in principle find a conserved vector with individual components

For the Kepler potential
_{i} has a solution

$m({\stackrel{\xb7}{q}}_{i}{\beta}_{i}-{\stackrel{\xb7}{\beta}}_{i}{q}_{i})=m{\stackrel{\xb7}{q}}_{i}({\stackrel{\xb7}{q}}_{k}{q}_{k})+{q}_{i}\phantom{\rule{.2em}{0ex}}[\frac{k}{q}-m({\stackrel{\xb7}{q}}_{k}{\stackrel{\xb7}{q}}_{k})]$

which are proportional to the components of the Runge vector in an arbitrary number of dimensions. This particular form for the β_{i} only holds for the Kepler potential.

To recap, transformations generated by the components of the Runge vector do not correspond to an invariance of the Lagrangian, but rather an invariance of the action integral. This is emphasized by the appearance of the gauge-type function representing a change of the Lagrangian itself conjugate to these transformations. The Runge vector is not restricted to the Kepler potential, but exists in general for any spherically symmetric potential, and appears naturally in a variational procedure of the same nature as that which produces the Lagrange equations. As such, it is hardly a “hidden” or “accidental” symmetry, as it is often labeled.

*Uploaded 2012.01.29 — Updated 2014.02.25*
analyticphysics.com