The Schwarzschild metric, with the simplification $c=G=0$,

$d{s}^{2}=(1-\frac{2M}{r})d{t}^{2}-(1-\frac{2M}{r}{)}^{-1}d{r}^{2}-{r}^{2}d{\theta}^{2}-{r}^{2}{sin}^{2}\theta d{\phi}^{2}$

describes the spacetime around a spherically symmetric source outside of the actual source material. It was first generalized to an arbitrary number of spatial dimensions by Tangherlini, working in standard higher-dimensional spherical coordinates. In that reference he states that the Einstein equations arise “[a]fter a standard calculation”, and while this is true it glosses over the details of the actual evaluation.

In higher-dimensional spherical coordinates, most components of the metric tensor depend on almost all independent variables. While this occurs in a regular pattern and can be handled, it is still a nontrivial calculation and rather tedious. The use of the Cartan formalism does not make this process much simpler compared to employing a traditional coordinate basis.

What does simplify the process is switching first to a different set of coordinates that go by the name of Kerr-Schild. These are algebraically special coordinates that do not apply in general to gravitating systems, but do apply in the Schwarzschild case. A metric in these coordinates has the general form

${g}_{ik}={\eta}_{ik}-\phi \left({x}^{m}\right){v}_{i}{v}_{k}$

where the first term is the diagonal Minkowski metric of flat spacetime,

${\eta}_{ik}=[1,-1,-1,-1,\dots ]$

the coefficient *φ* is a scalar function of spacetime variables, and the vector with *n*+1 components is null with respect to both the full metric and the Minkowski metric:

${g}^{lm}{v}_{l}{v}_{m}=0={\eta}^{lm}{v}_{l}{v}_{m}$

The left-hand equality defines nullity in a metric space, while the right-hand equality is additional information. By convention a repeated index implies summation.

Due to the null nature of the vector, its index is essentially raised and lowered with the Minkowski metric:

${v}_{i}={g}_{im}{v}^{m}={\eta}_{im}{v}^{m}-\phi {v}_{i}{v}_{m}{v}^{m}={\eta}_{im}{v}^{m}$

Then it is almost trivial to verify that the inverse of the metric tensor is

${g}^{ik}={\eta}^{ik}+\phi \left({x}^{m}\right){v}^{i}{v}^{k}$

with a change of sign on the second term. Differentiation of the statement of nulllity implies

${\partial}_{k}[{v}_{m}{v}^{m}=0]\phantom{\rule{1em}{0ex}}\to \phantom{\rule{1em}{0ex}}{v}^{m}{\partial}_{k}{v}_{m}={v}_{m}{\partial}_{k}{v}^{m}=0$

since the constant Minkowski metric moves through the differentiation. The null condition thus holds over single derivatives.

This is one situation where the Cartan formalism is a hindrance compared to a coordinate basis. Diagonalizing the metric as one typically does in the Cartan formalism simply returns to spherical coordinates, thereby defeating the purpose of the exercise. It is simpler to use the traditional evaluation.

Combinations of the null vector are great aids in simplifying quantities. The Christoffel symbols in raw form are

$\begin{array}{l}{\Gamma}_{kl}^{i}=\frac{1}{2}{g}^{im}[{\partial}_{k}{g}_{lm}+{\partial}_{l}{g}_{km}-{\partial}_{m}{g}_{kl}]\\ \phantom{{\Gamma}_{kl}^{i}}=-\frac{1}{2}{g}^{im}\left[{\partial}_{k}\right(\phi {v}_{l}{v}_{m})+{\partial}_{l}(\phi {v}_{k}{v}_{m})-{\partial}_{m}(\phi {v}_{k}{v}_{l}\left)\right]\end{array}$

Contracting these symbols on the lower indices with the null vector clearly gives zero, since nullity holds over single derivatives. If one assumes the null vector defines geodesics in these coordinates, then a single contraction over the geodesic equation adds another useful combination:

${v}^{k}[v^{i}{}_{;k}=v^{i}{}_{,k}+{\Gamma}_{km}^{i}{v}^{m}=0]\phantom{\rule{1em}{0ex}}\to \phantom{\rule{1em}{0ex}}{v}^{k}{\partial}_{k}{v}^{i}=0$

That the null vector is geodesic can be shown in more generality: for this presentation it is sufficient to assume it is and check explicitly. The final index can be lowered with the constant Minkowski metric.

With the two combinations of the null vector, the Christoffel symbols become

$\begin{array}{l}{\Gamma}_{kl}^{i}=-\frac{1}{2}({\eta}^{im}+\phi {v}^{i}{v}^{m})\left[{\partial}_{k}\right(\phi {v}_{l}{v}_{m})+{\partial}_{l}(\phi {v}_{k}{v}_{m})-{\partial}_{m}(\phi {v}_{k}{v}_{l}\left)\right]\\ \phantom{{\Gamma}_{kl}^{i}}=-\frac{1}{2}\left[{\partial}_{k}\right(\phi {v}^{i}{v}_{l})+{\partial}_{l}(\phi {v}^{i}{v}_{k})-{\eta}^{im}{\partial}_{m}(\phi {v}_{k}{v}_{l})-\phi {v}^{i}{v}_{k}{v}_{l}{v}^{m}{\partial}_{m}\phi ]\end{array}$

where the geodesic condition allows the null vector to move through summed derivatives like a constant. The Ricci tensor in terms of the Christoffel symbols is

${R}_{ik}={\partial}_{m}{\Gamma}_{ik}^{m}-{\partial}_{k}{\Gamma}_{im}^{m}+{\Gamma}_{lm}^{m}{\Gamma}_{ik}^{l}-{\Gamma}_{im}^{l}{\Gamma}_{kl}^{m}$

The two middle terms contain the ‘trace’ of the Christoffel symbols, which is

${\Gamma}_{kl}^{l}=-\frac{1}{2}\left[{\partial}_{l}\right(\phi {v}^{l}{v}_{k})-{\partial}_{m}(\phi {v}_{k}{v}^{m}\left)\right]=0$

so that these terms do not contribute to the Ricci tensor. The fourth term is messy at first, but the last terms in each bracket cancel out immediately:

$\begin{array}{l}{\Gamma}_{im}^{l}{\Gamma}_{kl}^{m}=\frac{1}{4}\left[{\partial}_{i}\right(\phi {v}^{l}{v}_{m})+{\partial}_{m}(\phi {v}^{l}{v}_{i})-{\eta}^{lp}{\partial}_{p}(\phi {v}_{i}{v}_{m})-\phi {v}^{l}{v}_{i}{v}_{m}{v}^{p}{\partial}_{p}\phi ]\\ \phantom{\rule{5em}{0ex}}\times \left[{\partial}_{k}\right(\phi {v}^{m}{v}_{l})+{\partial}_{l}(\phi {v}^{m}{v}_{k})-{\eta}^{mr}{\partial}_{r}(\phi {v}_{k}{v}_{l})-\phi {v}^{m}{v}_{k}{v}_{l}{v}^{r}{\partial}_{r}\phi ]\\ \phantom{{\Gamma}_{im}^{l}{\Gamma}_{kl}^{m}}=\frac{1}{4}\left[{\partial}_{i}\right(\phi {v}^{l}{v}_{m})+{\partial}_{m}(\phi {v}^{l}{v}_{i})-{\eta}^{lp}{\partial}_{p}(\phi {v}_{i}{v}_{m}\left)\right]\\ \phantom{\rule{5em}{0ex}}\times \left[{\partial}_{k}\right(\phi {v}^{m}{v}_{l})+{\partial}_{l}(\phi {v}^{m}{v}_{k})-{\eta}^{mr}{\partial}_{r}(\phi {v}_{k}{v}_{l}\left)\right]\end{array}$

Expanding derivatives it is not difficult to see that the first terms in each bracket cancel against the remaining terms, leaving

$\begin{array}{l}{\Gamma}_{im}^{l}{\Gamma}_{kl}^{m}=\frac{1}{4}\left[{\partial}_{m}\right(\phi {v}^{l}{v}_{i})-{\eta}^{lp}{\partial}_{p}(\phi {v}_{i}{v}_{m}\left)\right]\left[{\partial}_{l}\right(\phi {v}^{m}{v}_{k})-{\eta}^{mr}{\partial}_{r}(\phi {v}_{k}{v}_{l}\left)\right]\\ \phantom{{\Gamma}_{im}^{l}{\Gamma}_{kl}^{m}}=\frac{1}{2}\left[{\partial}_{m}\right(\phi {v}^{l}{v}_{i}\left){\partial}_{l}\right(\phi {v}^{m}{v}_{k})-{\eta}^{lp}{\partial}_{p}(\phi {v}_{i}{v}_{m}\left){\partial}_{l}\right(\phi {v}^{m}{v}_{k}\left)\right]\end{array}$

where again the constant Minkowski metric moves through derivatives in the second step, allowing symmetry in the two unsummed indices. The Ricci tensor is thus

$\begin{array}{l}{R}_{ik}=-\frac{1}{2}{\partial}_{m}\left[{\partial}_{i}\right(\phi {v}^{m}{v}_{k})+{\partial}_{k}(\phi {v}^{m}{v}_{i})-{\eta}^{mp}{\partial}_{p}(\phi {v}_{i}{v}_{k})-\phi {v}^{m}{v}_{i}{v}_{k}{v}^{p}{\partial}_{p}\phi ]\\ \phantom{\rule{5em}{0ex}}-\frac{1}{2}\left[{\partial}_{m}\right(\phi {v}^{l}{v}_{i}\left){\partial}_{l}\right(\phi {v}^{m}{v}_{k})-{\eta}^{lp}{\partial}_{p}(\phi {v}_{i}{v}_{m}\left){\partial}_{l}\right(\phi {v}^{m}{v}_{k}\left)\right]\end{array}$

This can be simplified greatly by raising one index. First write

$\begin{array}{l}R^{i}{}_{k}={g}^{in}{R}_{nk}\\ \phantom{R^{i}{}_{k}}=-\frac{1}{2}{\partial}_{m}\left[{\eta}^{in}{\partial}_{n}\right(\phi {v}^{m}{v}_{k})+{\partial}_{k}(\phi {v}^{m}{v}^{i})-{\eta}^{mp}{\partial}_{p}(\phi {v}^{i}{v}_{k})-\phi {v}^{i}{v}^{p}{\partial}_{p}(\phi {v}^{m}{v}_{k}\left)\right]\\ \phantom{\rule{5em}{0ex}}-\frac{1}{2}\left[{\partial}_{m}\right(\phi {v}^{l}{v}^{i}\left){\partial}_{l}\right(\phi {v}^{m}{v}_{k})-{\eta}^{lp}{\partial}_{p}(\phi {v}^{i}{v}_{m}\left){\partial}_{l}\right(\phi {v}^{m}{v}_{k}\left)\right]\\ \phantom{\rule{5em}{0ex}}-\frac{1}{2}\phi {v}^{i}{v}^{n}{\partial}_{m}[{\partial}_{n}\left(\phi {v}^{m}{v}_{k}\right)+{\partial}_{k}\left(\phi {v}^{m}{v}_{n}\right)\\ \phantom{\rule{15em}{0ex}}-{\eta}^{mp}{\partial}_{p}\left(\phi {v}_{n}{v}_{k}\right)-\phi {v}^{m}{v}_{n}{v}_{k}{v}^{p}{\partial}_{p}\phi ]\end{array}$

When the factors outside of the overall derivative of the third bracket are brought inside the derivative, they will zero out the final three terms inside the bracket, leaving a single term that cancels the final term inside the first bracket. One merely needs to add compensating contributions from the derivative:

$\begin{array}{l}R^{i}{}_{k}=-\frac{1}{2}{\partial}_{m}\left[{\eta}^{in}{\partial}_{n}\right(\phi {v}^{m}{v}_{k})+{\partial}_{k}(\phi {v}^{m}{v}^{i})-{\eta}^{mp}{\partial}_{p}(\phi {v}^{i}{v}_{k}\left)\right]\\ \phantom{\rule{5em}{0ex}}-\frac{1}{2}\left[{\partial}_{m}\right(\phi {v}^{l}{v}^{i}\left){\partial}_{l}\right(\phi {v}^{m}{v}_{k})-{\eta}^{lp}{\partial}_{p}(\phi {v}^{i}{v}_{m}\left){\partial}_{l}\right(\phi {v}^{m}{v}_{k}\left)\right]\\ \phantom{\rule{5em}{0ex}}+\frac{1}{2}{\partial}_{m}\left(\phi {v}^{i}{v}^{n}\right)[{\partial}_{n}\left(\phi {v}^{m}{v}_{k}\right)+{\partial}_{k}\left(\phi {v}^{m}{v}_{n}\right)\\ \phantom{\rule{15em}{0ex}}-{\eta}^{mp}{\partial}_{p}\left(\phi {v}_{n}{v}_{k}\right)-\phi {v}^{m}{v}_{n}{v}_{k}{v}^{p}{\partial}_{p}\phi ]\end{array}$

The second and fourth terms inside the third bracket are zero, and the first and third terms inside the same bracket cancel the second bracket. This just leaves

$R^{i}{}_{k}=-\frac{1}{2}{\partial}_{m}\left[{\eta}^{in}{\partial}_{n}\right(\phi {v}^{m}{v}_{k})+{\partial}_{k}(\phi {v}^{m}{v}^{i})-{\eta}^{mp}{\partial}_{p}(\phi {v}^{i}{v}_{k}\left)\right]$

as the simplified form of the Ricci tensor in mixed indices.

Now consider transforming the Schwarzschild metric above in spherical coordinates to Kerr-Schild coordinates. In keeping with the foregoing, first write it more generally as

$d{s}^{2}=(1-\phi )d{t}^{2}-\frac{d{r}^{2}}{1-\phi}-{r}^{2}d{\theta}^{2}-{r}^{2}{sin}^{2}\theta d{\phi}^{2}$

If the temporal variable is replaced using

$dt=du+\frac{dr}{1-\phi}$

the metric becomes

$\begin{array}{l}d{s}^{2}=(1-\phi )d{u}^{2}+2dudr-{r}^{2}d{\theta}^{2}-{r}^{2}{sin}^{2}\theta d{\phi}^{2}\\ \phantom{d{s}^{2}}=(du+dr{)}^{2}-d{r}^{2}-{r}^{2}d{\theta}^{2}-{r}^{2}{sin}^{2}\theta d{\phi}^{2}-\phi d{u}^{2}\\ d{s}^{2}=(du+dr{)}^{2}-d{x}^{2}-d{y}^{2}-d{z}^{2}-\phi d{u}^{2}\end{array}$

where the completely spatial part of the metric has been rewritten in Cartesian coordinates. Now define a new temporal variable with

$d\tau =du+dr$

so that the metric becomes

$\begin{array}{l}d{s}^{2}=d{\tau}^{2}-d{x}^{2}-d{y}^{2}-d{z}^{2}-\phi \phantom{\rule{.3em}{0ex}}(d\tau -dr{)}^{2}\\ \phantom{d{s}^{2}}=d{\tau}^{2}-d{x}^{2}-d{y}^{2}-d{z}^{2}\\ \hfill -\phi \phantom{\rule{.3em}{0ex}}(d\tau -\frac{xdx+ydy+zdz}{r}{)}^{2}\end{array}$

The first part of the metric is flat Minkowski spacetime. From the second term, the vector

${v}_{i}=[1,-\frac{x}{r},-\frac{y}{r},-\frac{z}{r}]$

is null with respect to the Minkowski metric. For the nonconstant components one can easily verify that is is geodesic (here indices run over spatial variables only):

${v}^{m}{\partial}_{m}{v}_{i}=\frac{{x}_{m}}{r}{\partial}_{m}\frac{{x}_{i}}{r}=\frac{{x}_{m}}{r}[\frac{{\delta}_{im}}{r}-\frac{{x}_{i}{x}_{m}}{{r}^{3}}]=0$

The metric tensor in Kerr-Schild coordinates is thus

${g}_{ik}={\eta}_{ik}-\frac{2M}{r}{v}_{i}{v}_{k}={\eta}_{ik}-\phi \phantom{\rule{.3em}{0ex}}{v}_{i}{v}_{k}$

with the given null vector. The metric is clearly in a form easily extensible to an arbitrary number of dimensions: the problem is now to determine the dependence of the one metric function on the dimension.

The Ricci tensor as given above already holds in an arbitrary number of dimensions. One need merely pick a component that is particularly suitable for determining the single metric function and set it equal to zero. The simplest component is

$R^{0}{}_{0}=-\frac{1}{2}{\partial}_{m}{\partial}_{m}\phi =-\frac{1}{2}{\nabla}^{2}\phi $

and the function is determined from the ordinary flat-space Laplacian! First evaluate the Laplacian (here indices run over spatial variables only),

${\partial}_{m}{\partial}_{m}\phi \left(r\right)={\partial}_{m}\left(\frac{{x}_{m}}{r}{\phi}^{\prime}\right)=\frac{{x}_{m}^{2}}{{r}^{2}}{\phi}^{\u2033}+(\frac{1}{r}-\frac{{x}_{m}^{2}}{{r}^{2}}){\phi}^{\prime}$

then sum over spatial variables and set equal to zero,

$\begin{array}{c}{\phi}^{\u2033}+(n-1)r{\phi}^{\prime}=0\\ ln{\phi}^{\prime}+(n-1)lnr=lnc\\ \phi =\frac{2M}{{r}^{n-2}}\end{array}$

where the constant of integration has been set to match the three-dimensional metric. Then one need merely reverse the transformation process for the final metric

$d{s}^{2}=(1-\frac{2M}{{r}^{n-2}})d{t}^{2}-(1-\frac{2M}{{r}^{n-2}}{)}^{-1}d{r}^{2}-{r}^{2}d{\Omega}_{\left(n\right)}^{2}$

with an *n*-dimensional differential of solid angle.

This evaluation via Kerr-Schild coordinates is almost trivial compared to a direct evaluation in spherical coordinates, once one has the Ricci tensor in the same coordinates. There is admittedly a bit of work involved in obtaining that general result, but thanks the to the two combinations of the null vector it is basically a matter of watching terms go away.

*Uploaded 2021.06.07*
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