Knowing that the Runge vector exists in classical mechanics for any spherically symmetric potential, as an invariance of the action integral, one is tempted to ask in which other physical systems a similar constant vector might arise. Since it is well known that the geodesic equations of general relativity can be easily derived from a Lagrangian, it is natural to ask what other aspects of the derivation of the Runge vector carry over into general relativity.

In the mathematical equations below, repeated indices again represent sums over those indices, as is commonly done in coordinate-based presentations of classical general relativity. Only equations and features of general relativity that are germane to the restricted question of whether a Runge can arise in this context will be introduced and used.

In coordinate-based presentations of classical general relativity, the square of the invariant interval is written in terms of a metric tensor as

$d{s}^2=g_{kl}d{x}^{k}d{x}^l$

where the components g_kl of the metric tensor can be functions of any or all of the variables of the space-time being considered. Our visually realizable reality consists of one temporal variable and three spatial variables, but the structure of general relativity allows any number of either kind of variable, and need not be specified for the current presentation. A central feature of the invariant interval is that it does not change for arbitrary transformations of the space-time coordinates.

Since the development of the Runge vector for classical mechanics worked in terms of generalized coordinates, the presentation holds just as well for general relativity up to the point where a specific Lagrangian was introduced, with one proviso. In general relativity, time is itself one of the space-time variables, and must be included along with the spatial variables among the generalized coordinates of the Lagrangian. One can, however, use invariant interval in the place of time as an affine parameter. That means that in reading that development the variable t should be replaced by s, and the dots interpreted as derivatives with respect to this latter variable.

In terms of invariant interval, the Lagrange equations become

$\frac{d}{ds}\frac{\partial L}{\partial \left(\frac{d{x}^i}{ds}\right)}-\frac{\partial L}{\partial x^i}=0$

To derive the geodesic equations, one can simply choose a Lagrangian of the form

$L=\frac{1}{2}{g}_{kl}\frac{d{x}^k}{ds}\frac{d{x}^l}{ds}$

where the numerical factor provides closer correspondence to the classical mechanical development. Evaluate the Lagrange equations with respect to the space-time variables and their derivatives, noting that each individual derivative appears twice in the summation:

$\begin{array}{c}\frac{1}{2}\frac{d}{ds}[g_{il}\frac{d{x}^l}{ds}+g_{ki}\frac{d{x}^k}{ds}]-\frac{1}{2}\frac{\partial g_{kl}}{\partial x^i}\frac{d{x}^k}{ds}\frac{d{x}^l}{ds}=0\\ g_{im}\frac{d^2{x}^m}{d{s}^2}+\frac{1}{2}[\frac{\partial g_{il}}{\partial x^k}+\frac{\partial g_{ki}}{\partial x^l}-\frac{\partial g_{kl}}{\partial x^i}]\frac{d{x}^k}{ds}\frac{d{x}^l}{ds}=0\end{array}$

Remembering that the metric tensor and its inverse are both symmetric and satisfy $g^{ik}{g}_{kj}={\delta }_j^i$ , complete the rearrangement of the equations by multiplying through with the inverse of the metric tensor:

$\frac{d^2{x}^i}{d{s}^2}+{\Gamma }_{kl}^i\frac{d{x}^k}{ds}\frac{d{x}^l}{ds}=0$

These are precisely the general relativistic geodesic equations, where the Christoffel symbol (or connection coefficient) is

${\Gamma }_{kl}^i=\frac{1}{2}{g}^{im}[{\partial }_k{g}_{ml}+{\partial }_l{g}_{mk}-{\partial }_m{g}_{kl}]$

It is a bit curious to note that the geodesic equations have been derived without recourse to the notion of covariant differentiation.

One can evaluate a Hamiltonian corresponding to the chosen Lagrangian, replacing time in the classical definition with invariant interval. Again, each individual derivative appears twice in the summation, giving

$\begin{array}{l}H=\frac{\partial L}{\partial (d{x}^k/ds)}\frac{d{x}^k}{ds}-L\\ \phantom{H}=g_{kl}\frac{d{x}^k}{ds}\frac{d{x}^l}{ds}-\frac{1}{2}{g}_{kl}\frac{d{x}^k}{ds}\frac{d{x}^l}{ds}=L\end{array}$

so that the Hamiltonian is equal to the Lagrangian for this choice of Lagrangian.

Just prior to selection of a Lagrangian in the development of the Runge vector, there appeared a linear combination of constants of the motion, which in terms of invariant interval and space-time variables is

$C=\frac{\partial L}{\partial \left(\frac{d{x}^k}{ds}\right)}\delta x^k+f-H\delta s$

where f is the added gauge-type function that accounts for possible changes in the Lagrangian itself due to coordinate transformations. The next step will be to take a derivative of this combination with respect to invariant interval, but first rewrite the Lagrangian in the informal manner nonmathematicians often use,

$2L=g_{kl}\frac{d{x}^k}{ds}\frac{d{x}^l}{ds}=\frac{1}{ds}\sqrt{d{s}^2}=\frac{ds}{ds}=1$

so that both the Lagrangian and the Hamiltonian are constant with respect to invariant interval. The more proper way to exhibit this property is to apply the chain rule to the Lagrangian,

$\frac{dL}{ds}=\frac{\partial L}{\partial x^k}\frac{d{x}^k}{ds}+\frac{\partial L}{\partial \left(\frac{d{x}^k}{ds}\right)}\frac{d}{ds}\left(\frac{d{x}^k}{ds}\right)$

and after substituting definitions above, including the geodesic equations, this derivative will turn out to be zero as expected. Now evaluate the derivative of the linear combination with respect to proper interval:

$\begin{array}{l}\frac{dC}{ds}=\frac{d}{ds}[g_{kl}\frac{d{x}^l}{ds}\delta x^k+f-g_{kl}\frac{d{x}^k}{ds}\frac{d{x}^l}{ds}\delta s]\equiv 0\\ \phantom{\frac{dC}{ds}}=\frac{d}{ds}\left[g_{kl}\frac{d{x}^l}{ds}\right]\delta x^k+g_{kl}\frac{d{x}^l}{ds}[\frac{\partial \delta x^k}{\partial x^m}\frac{d{x}^m}{ds}+\frac{\partial \delta x^k}{\partial s}]\\ +\frac{\partial f}{\partial x^m}\frac{d{x}^m}{ds}+\frac{\partial f}{\partial s}-g_{kl}\frac{d{x}^k}{ds}\frac{d{x}^l}{ds}[\frac{\partial \delta s}{\partial x^m}\frac{d{x}^m}{ds}+\frac{\partial \delta s}{\partial s}]\end{array}$

For this statement to hold for arbitrary coordinates, the coefficient of the term cubic in derivatives of the coordinates with respect to invariant interval must be identically zero. The function δs is then a function of invariant interval alone, and can be taken to be constant, so that the final bracketed term is identically zero.

Now consider the term quadratic in derivatives of the coordinates with respect to invariant interval. Choose its coefficient to produce zero when multiplied with the symmetric pair of derivatives, which for a general metric means

$\begin{array}{l}{g}_{kl}\frac{\partial \delta x^k}{\partial x^m}=2A_{lm}\delta \alpha A_{lm}=-A_{ml}\\ \frac{\partial \delta x^k}{\partial x^m}=2g^{kl}{A}_{lm}\delta \alpha \end{array}$

While this simple system of equations appears soluble by a single quadrature, it is by no means clear that the resulting functions would be consistent after integration against all coordinates. First take another partial derivative and compare the two results, remembering that the antisymmetric coefficient function has no diagonal components so that the indices k, m or p are all different:

$\begin{array}{l}\hfill \frac{1}{2\delta \alpha }\frac{\partial \delta x^k}{\partial x^m\partial x^p}={\partial }_p\left(g^{kl}{A}_{lm}\right)={\partial }_m\left(g^{kl}{A}_{lp}\right)\hfill \\ \\ {\partial }_p\left(g^{kp}{A}_{pm}\right)+{\partial }_p\left(g^{kk}{A}_{km}\right)+\sum _{l\ne k,m,p}{\partial }_p\left(g^{kl}{A}_{lm}\right)\\ ={\partial }_m\left(g^{km}{A}_{mp}\right)+{\partial }_m\left(g^{kk}{A}_{kp}\right)+\sum _{l\ne k,m,p}{\partial }_m\left(g^{kl}{A}_{lp}\right)\end{array}$

Since the Runge vector only arises in classical mechanics for systems with particular symmetry, one cannot automatically expect to extract anything meaningful from this statement for a general metric. In the spirit of the classical mechanical development, restrict the comparison above to diagonal metric tensors, for which only the middle term on each side remains. Setting p = k then gives

${\partial }_k\left(g^{kk}{A}_{km}\right)=0\to A_{km}\sim g_{kk}$

and the antisymmetric coefficient must be proportional to a metric tensor component. The antisymmetry of the coefficient function requires

$g_{mm}\sim A_{mk}=-A_{km}\sim -g_{kk}$

and the outer proportionalities can only hold for a nonconstant metric if all of the metric tensor components are equal for the portion of the space in which the constants are constructed. This means that for there to be extensions of the classical mechanical constants to general relativity the metric must contain an isotropic portion,

$d{s}^2=\sum _{\mathrm{nonisotropic}}{g}_{mm}(d{x}^m{)}^2+g_{\mathrm{iso}}\sum _{\mathrm{isotropic}}(d{x}^k{)}^2$

and since this isotropic portion is flat, it can always be described in terms of what are called isotropic Cartesian coordinates. In this isotropic portion of the space

$A_{km}={\alpha }_{km}{g}_{\mathrm{iso}}{\alpha }_{km}=-{\alpha }_{mk}$

where the quantities α_km are analogous to the classical mechanical constant coefficients, and their indices have no tensor meaning. The single metric function in the isotropic portion of the space is potentially a function of all coordinates. The changes in isotropic coordinates are trivially integrated for

$\delta x^k=2{\alpha }_{kl}{x}^l\delta \alpha +{\beta }_k\left(s\right)\delta \beta$

The gauge-type function is then determined by zeroing out in a nontrivial manner the terms linear in derivatives of the coordinates with respect to invariant interval,

$\frac{\partial f}{\partial x^k}=-\frac{\partial \delta x^k}{\partial s}{g}_{\mathrm{iso}}=-\frac{d{\beta }_k}{ds}\delta \beta g_{\mathrm{iso}}$

and the quantities β_k would be determined by the remaining terms:

$\frac{\partial f}{\partial s}+(2{\alpha }_{kl}{x}^l\delta \alpha +{\beta }_k\delta \beta )\frac{d}{ds}\left[g_{\mathrm{iso}}\frac{d{x}^k}{ds}\right]=0$

Before considering possible solutions to this system of equations, write down the linear combination of constants of the motion as determined to this point,

$C={\alpha }_{kl}{g}_{\mathrm{iso}}(\frac{d{x}^k}{ds}{x}^l-\frac{d{x}^l}{ds}{x}^k)\delta \alpha +{\beta }_k\delta \beta g_{\mathrm{iso}}\frac{d{x}^k}{ds}+f-H\delta s$

and consider restrictions to ensure that the first ‘constant’ of this combination is truly so. For k and l in the isotropic portion of the space but no initial restrictions on summed indices, differentiate with respect to invariant interval,

$\begin{array}{l}\frac{d}{ds}\left[g_{\mathrm{iso}}\right(x^k\frac{d{x}^l}{ds}-x^l\frac{d{x}^k}{ds}\left)\right]\\ =\frac{d{g}_{\mathrm{iso}}}{ds}(x^k\frac{d{x}^l}{ds}-x^l\frac{d{x}^k}{ds})+g_{\mathrm{iso}}(x^k\frac{d^2{x}^l}{d{s}^2}-x^l\frac{d^2{x}^k}{d{s}^2})\\ =\frac{d{g}_{\mathrm{iso}}}{ds}(x^k\frac{d{x}^l}{ds}-x^l\frac{d{x}^k}{ds})+g_{\mathrm{iso}}\frac{d{x}^p}{ds}\frac{d{x}^m}{ds}[x^l{\Gamma }_{pm}^k-x^k{\Gamma }_{pm}^l]\end{array}$

where the second derivatives have been replaced using the geodesic equations. If the components of the Christoffel symbol are now written out explicitly for a diagonal metric tensor, the right-hand side of this equation becomes

$\begin{array}{l}\\ \frac{d{g}_{\mathrm{iso}}}{ds}(x^k\frac{d{x}^l}{ds}-x^l\frac{d{x}^k}{ds})\\ +\frac{1}{2}\frac{d{x}^p}{ds}\frac{d{x}^m}{ds}[x^l({\delta }_{km}{\partial }_p{g}_{\mathrm{iso}}+{\delta }_{kp}{\partial }_m{g}_{\mathrm{iso}}-{\delta }_{pm}{\partial }_k{g}_{mm})\\ -x^k({\delta }_{lm}{\partial }_p{g}_{\mathrm{iso}}+{\delta }_{lp}{\partial }_m{g}_{\mathrm{iso}}-{\delta }_{pm}{\partial }_l{g}_{mm})]\\ =\frac{1}{2}(\frac{d{x}^m}{ds}{)}^2(x^k{\partial }_l-x^l{\partial }_k)g_{mm}\end{array}$

where the summed index runs over all space-time variables, while the partial derivatives are with respect to variables in the isotropic portion of the space. The second quantity in parentheses is the operator for spatial angular momentum, so that the ‘constant’ under consideration is only constant for a metric that respects angular momentum invariance, i.e. a spherically symmetric metric.

For the classical mechanical case, the Runge vector is only separately constant when the angular momentum is a constant of the motion. This restriction is expected to hold for the general relativistic case, so that the gauge-type function determining the Runge vector must be spherically symmetric. That means

$-\frac{d{\beta }_l}{ds}\frac{x^k}{r}\frac{\partial g_{\mathrm{iso}}}{\partial r}=\frac{1}{\delta \beta }\frac{{\partial }^{2}f}{\partial x^k\partial x^l}=-\frac{d{\beta }_k}{ds}\frac{x^l}{r}\frac{\partial g_{\mathrm{iso}}}{\partial r}$

but the appearance of different variables on either side of the equation precludes the possibility of a solution for nonzero β_k. If all of the β_k are zero then so is the gauge-type function, and there can be no general relativistic Runge vector for metrics corresponding to classical, spherically symmetric Lagrangians.

In summary, the quantities $\frac{d{x}^k}{ds}{x}^l-\frac{d{x}^l}{ds}{x}^k$ are constant for the spatial coordinates of spherically symmetric metrics in isotropic Cartesian coordinates, and form a local generalization of classical angular momentum in the isotropic portion of the space, but there is no general relativistic version of the Runge vector under these same conditions. This result makes perfect sense, for two reasons. First, in the isotropic portion of the space the invariant interval functions precisely the same way as time, reproducing the classical definition of angular momentum. Secondly, the Runge vector arises from changes in the Lagrangian itself due to variations in coordinates, and the Lagrangian here is proportional to an interval that is invariant under all coordinate transformations.

Uploaded 2012.08.23 — Updated 2014.02.11 analyticphysics.com