201903030439 Exercise 14.1.3

Eq. (14.9):

$\Delta s=4GM \displaystyle{\int_{0}^{u(R)}\frac{\mathrm{d}u}{(1-u^2)^2}}$

Since the indefinite integral evaluated by Eq. (14.10):

$\displaystyle{\int \frac{\mathrm{d}u}{(1-u^2)^2}=\frac{u}{2(1-u^2)}+\frac{1}{4}\ln \bigg| \frac{1+u}{1-u} \bigg|}$

and the identity given by Eq. (14.11):

$\tanh^{-1}u=\displaystyle{\frac{1}{2}\ln \bigg| \frac{1+u}{1-u} \bigg|}$

are combined to give

$\displaystyle{\int \frac{\mathrm{d}u}{(1-u^2)^2}=\frac{u}{2(1-u^2)}+\frac{1}{2}\tanh^{-1}u}$ .

From $u(R)=\sqrt{1-\displaystyle{\frac{2GM}{R}}}$ ,

$\begin{aligned} \displaystyle{\frac{u(R)}{2(1-u(R)^2)}} & = \displaystyle{\frac{1}{2\Big( 1-\Big( 1-\frac{2GM}{R} \Big) \Big)}} \sqrt{1-\displaystyle{\frac{2GM}{R}}}\\ & = \displaystyle{\frac{R}{4GM}} \sqrt{1-\displaystyle{\frac{2GM}{R}}} \end{aligned}$ .

$\Delta s=4GM \Bigg( \displaystyle{\frac{R}{4GM}} \sqrt{1-\displaystyle{\frac{2GM}{R}}} +\displaystyle{\frac{1}{2}}\tanh^{-1}\bigg( \sqrt{1-\displaystyle{\frac{2GM}{R}}}\bigg) \Bigg)$

and it checks with Eq. (14.3):

$\Delta s=R\sqrt{1-2GM/R}+2GM\tanh^{-1}\sqrt{1-2GM/R}$ .

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