201903030435 Exercise 14.1.4

Check that the physical distance from $r=3GM$ to $r=2GM$ is indeed $3.05GM$ . (If your calculator cannot handle inverse hyperbolic functions, use equation 14.11 to eliminate $\tanh^{-1}u$ .)

Eq. (14.3):

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

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 the following

$\begin{aligned} \Delta s(r) & =r\sqrt{1-2GM/r}+GM\ln \bigg| \displaystyle{\frac{1+u}{1-u}} \bigg| \\ & =r\sqrt{1-2GM/r}+GM\ln \Bigg(\Bigg| \displaystyle{\frac{1+\sqrt{1-2GM/R}}{1-\sqrt{1-2GM/R}}}\Bigg| \Bigg) \\ \end{aligned}$

which measures the physical distance $\Delta s$ from the event horizon $r=2GM$ .

Now that $r=3GM$ , the physical distance will be

$\begin{aligned} \Delta s(3GM) & =(3GM)\sqrt{\displaystyle{1-\frac{2GM}{(3GM)}}}+GM\ln \Bigg| \frac{1+\sqrt{1-\frac{2GM}{(3GM)}}}{1-\sqrt{1-\frac{2GM}{(3GM)}}} \Bigg| \\ & = \sqrt{3} GM + 1.3169 GM \\ & \approx 3.05GM\qquad (\textrm{3 s.f.}) \end{aligned}$

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