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In an exercise in general relativity, I am trying to show that in the limit as $\hat{r}\rightarrow\infty$,

\begin{equation*} \frac{1-GM/2\hat{r}}{1+GM/2\hat{r}} \approx 1 - \frac {2GM}{\hat{r}},\ \ \text{or}\ \ \frac{1-\phi}{1+\phi} \approx 1-4\phi\ \text{if}\ \phi\ll 1 \end{equation*}

At least, this is the approximation I deduce I need to show. I don't see a path to it, however. I see ways to show through Taylor expansion that to first order it's approximately $1-2\phi$, but I need $1-4\phi$.

HallaSurvivor
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Frank Harris
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    Or without Taylor $\frac{1-\phi}{1+\phi}=\frac{(1-\phi)^2}{1-\phi^2}=\frac{1-2 \phi + \phi^2}{1-\phi^2} \approx 1 - 2\phi$ to the first order. You won't be able to find $1-4\phi$ because that's not the correct approximation here. – dxiv May 06 '21 at 01:25
  • Makes sense when you put it like that. :) Oh well, back the drawing board. Thanks! – Frank Harris May 06 '21 at 01:26
  • Are you sure it's $\frac{GM}{2\hat{r}}$? This actually works if both uses are $\frac{GM}{\hat{r}}$... – Eric Towers May 06 '21 at 01:44

3 Answers3

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\begin{align} & 1+x \approx1 + x + x^2+ x^3 + \cdots = \frac 1 {1-x}. \\[8pt] & \frac 1 {1+x} \approx 1-x. \\[8pt] & \frac{1-x}{1+x} \approx(1-x)^2 = 1 - 2x+x^2 \approx 1 - 2x. \end{align}

Michael Hardy
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What dxiv said in the comment is correct. One more idea for calculating $\frac{1-\phi}{1+\phi}$ is to assume $\tan^2(x)=\phi$ and hence using the formula $\cos(2x)=\frac{1-\tan^2(x)}{1+\tan^2(x)}$ and hence we have $\frac{1-\phi}{1+\phi} = \cos(2 (\tan^{-1}(\sqrt{\phi})))$. This is exact expression and not an approximation.

Michael Hardy
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Balaji sb
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To demonstrate my comment (that this approximation works if both combinations containing $\hat{r}$ are $\frac{GM}{\hat{r}}$)...\begin{align*} \frac{1 - GM/\hat{r}}{1 + GM/\hat{r}} &= \frac{1 + GM/\hat{r} - 2GM/\hat{r}}{1 + GM/\hat{r}} \\ &= 1 - \frac{2 G M/\hat{r}}{1 +GM/\hat{r}} \cdot \frac{\hat{r}}{\hat{r}} \\ &= 1 - \frac{2 G M}{\hat{r} + GM} \\ &\approx 1 - \frac{2 G M}{\hat{r}} \text{,} \end{align*} for $\hat{r} \gg GM$.

Eric Towers
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