중력장 방정식의 근사적 적분법: 두 판 사이의 차이

{{c|<math>\begin{array}{c}\text{a}) \\ \\ \text{b}) \\ \\ \text{c}) \\ \\ \text{d}) \\ \\ \text{e}) \\ \\ \text{f})\end{array}</math><math>\begin{aligned} \displaystyle \frac{1}{i}t_{41} &= \frac{f'^2}{4x}(\alpha_{11}^2 + \alpha_{14}^2 + \alpha_{44}^2) = 0 \\ \displaystyle \frac{1}{i}t_{41} &= \frac{f'^2}{2x}(\alpha_{12}^2 + \alpha_{24}^2)=0 \\ \displaystyle \frac{1}{i}t_{41} &= \frac{f'^2}{2x}(\alpha_{13}^2 + \alpha_{34}^2)=0 \\ \displaystyle \frac{1}{i}t_{41} &= \frac{f'^2}{4x}\alpha_{22}^2 = \frac{1}{4x}\left(\frac{\partial \gamma\,_{22}'}{\partial t}\right)^2 \\ \displaystyle \frac{1}{i}t_{41} &= \frac{f'^2}{4x}\alpha_{23}^2 = \frac{1}{4x4\varkappa}\left(\frac{\partial \gamma\,_{23}'}{\partial t}\right)^2 \\ \displaystyle \frac{1}{i}t_{41} &= \frac{f'^2}{4x}\alpha_{33}^2 = \frac{1}{4x}\left(\frac{\partial \gamma\,_{33}'}{\partial t}\right)^2 \end{aligned}</math>}}