Calculating the Riemann-Christoffel tensor by~\Eq{(34.5)}, since the first derivatives vanish, \begin{align*} B_{\mu\nu\sigma\rho} &= \frac{1}{2}\left(\frac{\dd g_{\sigma\rho}}{\dd x_{\mu}\, \dd x_{\nu}} + \frac{\dd g_{\mu\nu}}{\dd x_{\sigma}\, \dd x_{\rho}} - \frac{\dd g_{\mu\sigma}}{\dd x_{\nu}\, \dd x_{\rho}} - \frac{\dd g_{\nu\rho}}{\dd x_{\mu}\, \dd x_{\sigma}}\right) \\ &= a_{\mu\nu} a_{\sigma\rho} - a_{\mu\sigma} a_{\nu\rho}. \Tag{(65.51)} \end{align*} \PageSep{151} Hence, remembering that the~$g^{\tau\rho}$ have Euclidean values~$-g_{\sigma}^{\rho}$, \[ G_{\mu\nu} = g^{\sigma\rho} B_{\mu\nu\sigma\rho} = -a_{\mu\nu} (a_{11} + a_{22} + a_{33} + a_{44}) + a_{\mu\sigma} a_{\nu\sigma}. \Tag{(65.52)} \] In particular \begin{align*} G_{11} &= -a_{11} (a_{11} + a_{22} + a_{33} + a_{44}) + a_{11}^{2} + a_{12}^{2} + a_{13}^{2} + a_{14}^{2} \\ &= (a_{12}^{2} - a_{11} a_{22}) + (a_{13}^{2} - a_{11} a_{33}) + (a_{14}^{2} - a_{11} a_{44}). \Tag{(65.53)} \end{align*} Also %[** TN: Re-breaking] \begin{align*} G &= g^{\mu\nu} G_{\mu\nu} = -G_{11} - G_{22} - G_{33} - G_{44} \\ &= -2 \bigl\{(a_{12}^{2} - a_{11} a_{22}) + (a_{13}^{2} - a_{11} a_{33}) + (a_{14}^{2} - a_{11} a_{44}) \\ &\qquad + (a_{23}^{2} - a_{22} a_{33}) + (a_{24}^{2} - a_{22} a_{44}) + (a_{34}^{2} - a_{33} a_{44})\bigr\}. \Tag{(65.54)} \end{align*} When the principal axes are taken as in~\Eq{(65.3)}, these results become \[ \left. \begin{aligned} G_{11} &= -k_{1}(k_{2} + k_{3} + k_{4})\Add{,} \\ G_{22} &= -k_{2}(k_{1} + k_{3} + k_{4});\ \text{etc.} \end{aligned} \right\} \Tag{(65.55)} \] and \[ G = 2(k_{1}k_{2} + k_{1}k_{3} + k_{1} k_{4} + k_{2} k_{3} + k_{2} k_{4} + k_{3} k_{4}). \Tag{(65.6)} \] The invariant~$G$ has thus a comparatively simple interpretation in terms of the principal radii of curvature. It is a generalisation of the well-known \index{Curvature!radius of spherical}% \index{Spherical curvature, radius of}% invariant for two-dimensional surfaces $1/\rho_{1}\rho_{2}$, or~$k_{1} k_{2}$. But this interpretation is only possible in the simple case of five dimensions. In general five dimensions are not sufficient to represent even the small portion of the surface near the origin; for if we set $G_{\mu\nu} = 0$ in~\Eq{(65.55)}, we obtain $k_{\mu} = 0$, and hence by~\Eq{(65.51)} $B_{\mu\nu\sigma\rho} = 0$. Thus it is not possible to represent a natural gravitational field ($G_{\mu\nu} = 0$, $B_{\mu\nu\sigma\rho} \neq 0$) in five Euclidean dimensions.