201903030446 Exercise 14.1.1 & 14.1.2

Eq. (14.7):

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

Let Eq.(14.8):

$u\equiv \sqrt{1-\displaystyle{\frac{2GM}{r}}}\Longrightarrow r=\displaystyle{\frac{2GM}{1-u^2}}$

From the event horizon $r=r_s=2GM$ to $r=R$ , it corresponds to $u=0$ and $u(R)=\sqrt{1-\displaystyle{\frac{2GM}{R}}}$ .

From $r=\displaystyle{\frac{2GM}{1-u^2}}$ , by quotient rule:

$\displaystyle{\frac{\mathrm{d}r}{\mathrm{d}u}}=\displaystyle{\frac{(1-u^2)(0)-(2GM)(-2u)}{(1-u^2)^2}}$ ,

and

$\mathrm{d}r=\displaystyle{\frac{4GMu}{(1-u^2)^2}}\,\mathrm{d}u$ .

Substituting these into Eq. (14.7), I get Eq. (14.9):

$\Delta s=\displaystyle{\int_0^{u(R)}}\displaystyle{\frac{4GM}{(1-u^2)^2}}\,\mathrm{d}u$