The argument refers to \emph{similar} atoms and the question remains whether, for example, a hydrogen atom on the sun is truly similar to a hydrogen atom on the earth. Strictly speaking it cannot be exactly similar because it is in a different kind of space-time, in which it would be impossible to make a finite structure exactly similar to one existing in the space-time near the earth. But if the interval of vibration of the hydrogen atom is modified by the kind of space-time in which it lies, the difference must be dependent on some invariant of the space-time. The simplest invariant which differs at the sun and the earth is the square of the length of the Riemann-Christoffel tensor, viz.\ \[ B_{\mu\nu\sigma}^{\epsilon} B_{\epsilon}^{\mu\nu\sigma}. \] The value of this can be calculated from~\Eq{(38.8)} by the method used in that section for calculating the~$G_{\mu\nu}$. The result is \[ 48\, \frac{m^{2}}{r^{6}}. \] \PageSep{93} By consideration of dimensions it seems clear that the proportionate change of~$ds$ would be of the order \[ \frac{\sigma^{4} m^{2}}{r^{6}}, \] where $\sigma$~is the radius of the atom; there does not seem to be any other length concerned. For a comparison of solar and terrestrial atoms this would be about $10^{-100}$. In any case it seems impossible to construct from the invariants of space-time a term which would compensate the predicted shift of the spectral lines, which is proportional to~$m/r$. \Section{43.}{Isotropic coordinates} \index{Coordinate-systems!isotropic}% \index{Isotropic coordinates}% We can transform the expression for the interval~\Eq{(38.8)} by making the substitution \[ r = \left(1 + \frac{m}{2r_{1}}\right)^{2} r_{1}, \Tag{(43.1)} \] so that \begin{align*} dr &= \left(1 - \frac{m^{2}}{4r_{1}^{2}}\right) dr_{1}, \\ \gamma &= \left(1 - \frac{m}{2r_{1}}\right)^{2}\bigg/\left(1 + \frac{m}{2r_{1}}\right)^{2}. \end{align*} Then \Eq{(38.8)}~becomes \[ ds^{2} = -(1 + m/2r_{1})^{4} (dr_{1} + r_{1}^{2}\, d\theta^{2} + r_{1}^{2}\sin^{2}\theta\, d\phi^{2}) + \frac{(1 - m/2r_{1})^{2}}{(1 + m/2r_{1})^{2}}\, dt^{2}. \Tag{(43.2)} \] The coordinates $(r_{1},\theta, \phi)$ are called \emph{isotropic} polar coordinates. The corresponding isotropic rectangular coordinates are obtained by putting \[ x = r_{1} \sin\theta \cos\phi,\quad y = r_{1} \sin\theta \sin\phi,\quad z = r_{1} \cos\theta, \] giving \[ ds^{2} = -(1 + m/2r_{1})^{4} (dx^{2} + dy^{2} + dz^{2}) + \frac{(1 - m/2r_{1})^{2}}{(1 + m/2r_{1})^{2}}\, dt^{2}, \Tag{(43.3)} \] with \[ r_{1} = \Chg{\surd(x^{2} + y^{2} + z^{2})}{\sqrt{x^{2} + y^{2} + z^{2}}}. \]