201903030450 Exercise 9.1.1

The Schwarzschild metric is given by Eq. (9.3):

$\mathrm{d}s^2=-\bigg( 1-\displaystyle{\frac{r_s}{r}} \bigg) \mathrm{d}t^2+\bigg( 1-\displaystyle{\frac{r_s}{r}} \bigg)^{-1}\,\mathrm{d}r^2+r^2\,\mathrm{d}\theta +r^2\sin^2\theta\,\mathrm{d}\phi^2$

whereas the metric for spherical coordinates in flat spacetime is given by Eq. (9.2):

$\mathrm{d}s^2=-\mathrm{d}t^2+\mathrm{d}r^2+r^2\,\mathrm{d}\theta^2+r^2\sin^2\theta\,\mathrm{d}\phi^2$

Along a purely radial worldline $\mathrm{d}t=0$ , $\mathrm{d}\theta =0$ and $\mathrm{d}\phi =0$ , of the Schwarzschild metric will become

$\mathrm{d}s^2=\bigg( 1-\displaystyle{\frac{r_s}{r}} \bigg)^{-1}\,\mathrm{d}r^2$ .

Now that the Schwarzschild radius $r_s$ is $2GM$ , there is Eq. (9.15):

$\mathrm{d}s=\displaystyle{\frac{\mathrm{d}r}{\sqrt{1-2GM/r}}}$ .

The total radial distance between two events differing only by $r$ -coordinates, i.e., $r_A$ and $r_B$ is calculated by definite integration, given by Eq. (9.16):

$\Delta s=\displaystyle{\int \mathrm{d}s=\int_{r_A}^{r_B}\frac{\mathrm{d}r}{\sqrt{1-2GM/r}}}$

Trying binomial approximation:

$(1+x)^n =1+nx+\displaystyle{\frac{n(n-1)}{2!}}x^2 +\displaystyle{\frac{n(n-1)(n-2)}{3!}}x^3+O(x^4)$

then,

$\begin{aligned} &\quad\enspace (1-2GM/r)^{-\frac{1}{2}} \\ & =1+\bigg(-\frac{1}{2}\bigg)(-2GM/r) +\displaystyle{\frac{(-\frac{1}{2})(-\frac{1}{2}-1)}{2!}}(-2GM/r)^2+O(r^2) \\ \end{aligned}$

Considering only first-order approximation:

$1+\displaystyle{\frac{GM}{r}}$ .

Upon integration, there is Eq. (9.17):

$\begin{aligned} \Delta s & \approx \displaystyle{\int_{r_A}^{r_B}}\bigg( 1+\displaystyle{\frac{GM}{r}}\bigg) \,\mathrm{d}r \\ & = \bigg[ r+GM\ln \bigg( \displaystyle{\frac{r_B}{r_A}} \bigg) \bigg]\bigg|_{r_A}^{r_B} \\ & = (r_B-r_A) + GM\ln \bigg( \frac{r_B}{r_A} \bigg) \end{aligned}$

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