这是一份bath巴斯大学PH40112作业代写的成功案

It is actually easier to solve $G_{\alpha \beta}=0$ than $R_{\alpha \beta}=0$. From (4.20) we find:
$$
r\left(1-\mathrm{e}^{-2 \lambda}\right)=b \quad \rightarrow \quad \mathrm{e}^{2 \lambda}=\frac{1}{1-b / r}
$$
with $b$ constant. Substitute that in $G_{11}=0$ :
$$
2 \nu^{\prime}=\frac{b / r^{2}}{1-b / r} \quad \rightarrow \quad \mathrm{e}^{2 \nu}=A(1-b / r)
$$
since the expression on the left can be integrated to $2 \nu=\log (1-b / r)+$ const. The constant $A$ must be 1 because must be the Lorentz metric for $r \rightarrow \infty$. We insert these results :
$$
\mathrm{d} s^{2}=(1-b / r) c^{2} \mathrm{~d} t^{2}-\frac{\mathrm{d} r^{2}}{1-b / r}-r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{d} \varphi^{2}\right)
$$

PH40112 COURSE NOTES ：

$$
\frac{e^{2}}{1-r_{s} / r}-\frac{\dot{r}^{2}}{1-r_{s} / r}-\frac{h^{2}}{r^{2}}=\kappa
$$
which we may write as
$$
\dot{r}^{2}=e^{2}-V(r) ; \quad \text { with } \quad V(r)=\left(1-\frac{r_{s}}{r}\right)\left(\frac{h^{2}}{r^{2}}+\kappa\right) .
$$
For $\kappa=0$ (massless particles), $V$ has a maximum at $1.5 r_{\mathrm{s}}$. For massive particles it is necessary to distinguish between a high and a low angular momentum $h$, measured by the parameter $a \equiv r_{\mathrm{s}}^{2} / 2 h^{2}$. For $a>\frac{1}{6}$ (low angular momentum) $V(r)$ increases monotonously, and for $a<\frac{1}{6}$ (high angular momentum) $V(r)$ has two extrema at