This result obtained for five dimensions is perfectly general. From the manner in which \Eq{(65.4)}~was obtained, it will be seen that each of the six~$z$'s will make contributions to~$g_{\nu\tau}$ which are simply additive; we have merely to sum $a_{\mu\nu} a_{\sigma\tau} x_{\mu} x_{\sigma}$ for the six values of~$a_{\mu\nu} a_{\sigma\tau}$ contributed by the six terms~$dz_{r}^{2}$. All the subsequent steps involve linear equations and the work will hold for six~$z$'s just as well as for one~$z$. Hence \Eq{(65.72)}~is true in the general case when the representation requires $10$~dimensions. Now consider the invariant quadric \[ (G_{\mu\nu} - \tfrac{1}{2} g_{\mu\nu} G)\, dx_{\mu}\, dx_{\nu} = 3. \Tag{(65.81)} \] Let $\rho_{1}$~be the radius of this quadric in the $x_{1}$~direction, so that $dx_{\mu} = (\rho_{1} , 0, 0, 0)$ is a point on the quadric; the equation gives \[ (G_{11} - \tfrac{1}{2} g_{11}G) \rho_{1}^{2} = 3, \] so that by~\Eq{(65.72)} \[ G_{(1)} = \frac{6}{\rho_{1}^{2}}. \Tag{(65.82)} \] But for a hypersphere of radius~$R$ of \emph{three} dimensions ($k_{1} = k_{2} = k_{3} = 1/R$; $k_{4}$~disappears) the Gaussian curvature is~$6/R^{2}$. Hence $\rho_{1}$~is the radius of spherical curvature of the three-dimensional section of the world perpendicular to the axis~$x_{1}$. Now the quadric~\Eq{(65.81)} is invariant, so that the axis~$x_{1}$ may be taken in any arbitrary direction. Accordingly we see that--- \emph{The radius of the quadric $(G_{\mu\nu} = \tfrac{1}{2} g_{\mu\nu}G)\, dx_{\mu}\, dx_{\nu} = 3$ in any direction is equal to the radius of spherical curvature of the corresponding three-dimensional section of the world.} We call this quadric the \emph{quadric of curvature}. \index{Quadric of curvature}% \Section{66.}{Interpretation of Einstein's law of gravitation} We take the later form of Einstein's law~\Eq{(37.4)} \[ G_{\mu\nu} = \lambda g_{\mu\nu}, \Tag{(66.1)} \] in empty space, $\lambda$~being a universal constant at present unknown but so small as not to upset the agreement with observation established for the original form $G_{\mu\nu} = 0$. We at once obtain $G = 4\lambda$, and hence \[ G_{\mu\nu} - \tfrac{1}{2} g_{\mu\nu} G = -\lambda g_{\mu\nu}. \] \PageSep{153} Substituting in~\Eq{(65.81)} the quadric of curvature becomes \[ -\lambda g_{\mu\nu}\, dx_{\mu}\, dx_{\nu} = 3, \] or \[ -ds^{2} = 3/\lambda. \Tag{(66.2)} \] That is to say, the quadric of curvature is a sphere of radius~$\Chg{\surd(3/\lambda)}{\sqrt{3/\lambda}}$, and the radius of curvature in every direction\footnote {For brevity I use the phrase ``radius of curvature in a direction'' to mean the radius of spherical curvature of the three-dimensional section of the world at right angles to that direction. There is no other radius of curvature \emph{associated with a direction} likely to be confused with it.} and at every point in empty space has the constant length~$\Chg{\surd(3/\lambda)}{\sqrt{3/\lambda}}$.