The problem of determining~$ds^{2}$ at points within a sphere of fluid of uniform \PageSep{169} density has been treated by Schwarzschild, Nordström and de~Donder. Schwarzschild's solution\footnote {Schwarzschild's solution is of considerable interest; but I do not think that he solved exactly the problem which he intended to solve, viz.\ that of an incompressible fluid. For that reason I do not give the arguments which led to the solution, but content myself with discussing what distribution of matter his solution represents. A full account is given by de~Donder, \Title{La Gravifique Einsteinienne}, p.~169 (Gauthier-Villars, 1921). The original gravitational equations are used, the natural curvature of space being considered negligible compared with that superposed by the material sphere.} is \[ ds^{2} = -e^{\lambda}\, dr^{2} - r^{2}\, d\theta^{2} - r^{2}\sin^{2}\theta\, d\phi^{2} + e^{\nu}\, dt^{2}, \] where \[ \left. \begin{aligned} e^{\lambda} &= 1/(1 - \alpha r^{2})\Add{,} \\ e^{\nu} &= \tfrac{1}{4}\bigl(\Chg{3\surd(1 - \alpha a^{2})}{3\sqrt{1 - \alpha a^{2}}} - \Chg{\surd(1 - \alpha r^{2})}{\sqrt{1 - \alpha r^{2}}}\bigr)^{2}, \end{aligned} \right\} \Tag{(72.1)} \] and $a$ and~$\alpha$ are constants. The formulae~\Eq{(46.9)}, which apply to this form of~$ds$, become on raising one suffix \[ \left. \begin{aligned} -8\pi T_{1}^{1} &= e^{-\lambda} \bigl(\nu'/r - (e^{\lambda} - 1)/r^{2}\bigr)\Add{,} \\ -8\pi T_{2}^{2} &= e^{-\lambda} \bigl(\tfrac{1}{2} \nu'' - \tfrac{1}{4}\lambda'\nu' + \tfrac{1}{4}\nu'^{2} + \tfrac{1}{2}(\nu' - \lambda')/r\bigr)\Add{,} \\ -8\pi T_{3}^{3} &= -8\pi T_{2}^{2}\Add{,} \\ -8\pi T_{4}^{4} &= e^{-\lambda} \bigl(-\lambda'/r - (e^{\lambda} - 1)/r^{2}\bigr). \end{aligned} \right\} \Tag{(72.2)} \] We find from~\Eq{(72.1)} that \[ (e^{\lambda} - 1)/r^{2} = \tfrac{1}{2}\lambda'/r;\quad \tfrac{1}{2} \nu'' - \tfrac{1}{4}\lambda'\nu' + \tfrac{1}{4}\nu'^{2} = \tfrac{1}{2}\nu'/r. \] Hence \begin{gather*} T_{1}^{1} = T_{2}^{2} = T_{3}^{3} = \frac{1}{8\pi} e^{\lambda}(\tfrac{1}{2}\lambda' - \nu')/r, \Tag{(72.31)} \\ T_{4}^{4} = \frac{1}{8\pi} e^{\lambda} · \tfrac{3}{2}\lambda'/r = 3\alpha/8\pi. \Tag{(72.32)} \end{gather*} Referred to the coordinate-system $(r, \theta, \phi)$, $T_{4}^{4}$~represents the density and $T_{1}^{1}$, $T_{2}^{2}$, $T_{3}^{3}$ the stress-system. Hence Schwarzschild's solution gives uniform density and isotropic hydrostatic pressure at every point. \index{Pressure!in homogeneous sphere}% On further working out~\Eq{(72.31)}, we find that the pressure is \[ p = -T_{1}^{1} = \frac{\alpha}{8\pi}\, \frac{\bigl\{\frac{3}{2}(1 - \alpha r^{2})^{\frac{1}{2}} - \frac{3}{2}(1 - \alpha a^{2})^{\frac{1}{2}}\bigr\}} {\bigl\{\frac{3}{2}(1 - \alpha a^{2})^{\frac{1}{2}} - \frac{1}{2}(1 - \alpha r^{2})^{\frac{1}{2}}\bigr\}}. \Tag{(72.4)} \]