When the origin of the $1''.94$ has been traced as in the foregoing investigation, the anti-relativist who has been arguing that the observed effect is definitely caused by rotation, must change his position and maintain that it is definitely due to a gravitational perturbation exerted by the sun on Foucault's pendulum; the relativist holds to his view that the two causes are not distinguishable. \Section{45.}{Solution for a particle in a curved world} \index{Particle!gravitational field of}% In later work Einstein has adopted the more general equations~\Eq{(37.4)} \[ G_{\mu\nu} = \alpha g_{\mu\nu}. \Tag{(45.1)} \] In this case we must modify~\Eq{(38.61)}, etc.\ by inserting $\alpha g_{\mu\nu}$ on the right. We then obtain \begin{gather*} \tfrac{1}{2}\nu'' - \tfrac{1}{4}\lambda'\nu' + \tfrac{1}{4}\nu'^{2} - \lambda'/r = -\alpha e^{\lambda}, \Tag{(45.21)}\displaybreak[0] \\ e^{-\lambda}\bigl(1 + \tfrac{1}{2}r(\nu' - \lambda')\bigr) - 1 = -\alpha r^{2}, \Tag{(45.22)}\displaybreak[0] \\ e^{\nu-\lambda}(-\tfrac{1}{2}\nu'' + \tfrac{1}{4}\lambda'\nu' - \tfrac{1}{4}\nu'^{2} - \nu'/r) = \alpha e^{\nu}. \Tag{(45.23)} \end{gather*} From \Eq{(45.21)} and \Eq{(45.23)}, $\lambda' = -\nu'$, so that we may take $\lambda = -\nu$. An additive constant would merely amount to an alteration of the unit of time. Equation~\Eq{(45.22)} then becomes \[ e^{\nu} (1 + r\nu') = 1 - \alpha r^{2}. \] Let $e^{\nu} = \gamma$; then \[ \gamma + r\gamma' = 1 - \alpha r^{2} \] which on integration gives \[ \gamma = 1 - \frac{2m}{r} - \tfrac{1}{3}\alpha r^{2}. \Tag{(45.3)} \] The only change is the substitution of this new value of~$\gamma$ in~\Eq{(38.8)}. By recalculating the few steps from \Eq{(39.44)} to \Eq{(39.61)} we obtain the equation of the orbit \[ \frac{d^{2}u}{d\phi^{2}} + u = \frac{m}{h^{2}} + 3mu^{2} - \frac{1}{3}\, \frac{\alpha}{h^{2}}\, u^{-3}. \Tag{(45.4)} \] The effect of the new term in~$\alpha$ is to give an additional motion of perihelion \index{Perihelion!in curved world}% \[ \frac{\delta\varpi}{\phi} = \frac{1}{2}\, \frac{\alpha h^{6}}{m^{4}} = \frac{1}{2}\, \frac{\alpha a^{3}}{m}(1 - e^{2})^{3}. \Tag{(45.5)} \] At a place where $\gamma$~vanishes there is an impassable barrier, since any change~$dr$ corresponds to an infinite distance~$i\, ds$ surveyed by measuring-rods. The two roots of the quadratic~\Eq{(45.3)} are approximately \[ r = 2m\quad\text{and}\quad r = \Chg{\surd(3/\alpha)}{\sqrt{3/\alpha}}. \] \PageSep{101} The first root would represent the boundary of the particle---if a genuine particle \index{Horizon of world}% could exist---and give it the appearance of impenetrability. The second barrier is at a very great distance and may be described as the horizon of the world.