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The physics books I'm reading gives $$\triangle\tau=\frac{2}{c}\left(1-\frac{2m}{r_{1}}\right)^{1/2}\left(r_{1}-r_{2}+2m\ln\frac{r_{1}-2m}{r_{2}-2m}\right).$$ We are then told $2m/r$ is small for $r_{2}<r<r_{1}$ which gives the approximation$$\triangle\tau\approx\frac{2}{c}\left(r_{1}-r_{2}-\frac{m\left(r_{1}-r_{2}\right)}{r_{1}}+2m\ln\left(\frac{r_{1}}{r_{2}}\right)\right).$$ I can see how $$\frac{2}{c}\left(1-\frac{2m}{r_{1}}\right)^{1/2}\approx\frac{2}{c}$$ but can't see how the rest of it appears. It seems to be saying that$$2\ln\frac{r_{1}-2m}{r_{2}-2m}\approx\left(-\frac{\left(r_{1}-r_{2}\right)}{r_{1}}+2\ln\left(\frac{r_{1}}{r_{2}}\right)\right)$$

I've tried getting all the lns on one side, and also expanding $\ln\frac{r_{1}-2m}{r_{2}-2m}$ to $\ln\left(r_{1}-2m\right)-\ln\left(r_{2}-2m\right)$ and generally juggling it all about but with no luck. Any suggestions or hints from anyone?

It's to do with the gravitational time delay effect. It seems a bit more maths than physics which is why I'm asking it here.

Many thanks

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Is $m$ a small mass? I.e. is $m^2/r_1$ considered negligible? – Raskolnikov Dec 24 '11 at 17:02
@Raskolnikov - Thanks. m equals Gm/c^2. It's part of the Schwarzschild metric. – Peter4075 Dec 24 '11 at 17:17
@Raskolnikov - sorry, that's nonsense. m equals GM/c^2 – Peter4075 Dec 24 '11 at 17:28
OK, for Earth, this would mean that $GM/c^2\approx (GM/R^2)\cdot (R^2/c^2)\approx 10\cdot(R/c)^2$ where $R$ is the Earth's radius. This is indeed a small number and is therefore negligible when squared. I presume the time delay is related to a GPS application or something? – Raskolnikov Dec 24 '11 at 17:32
@Raskolnikov. It's to do with bouncing a radar signal sent from Earth off Venus or Mercury and measuring the gravitational time delay effect due to the Sun curving spacetime. All planets are on the same side of the Sun. M is the mass of the Sun. r1 and r2 are the Schwarzschild radial coordinates of Earth and Venus/Mercury. The book is A short course in general relativity by Foster and Nightingale, page 130. It's on Google Books. Thanks – Peter4075 Dec 24 '11 at 17:47
up vote 2 down vote accepted

As Raskolnikov says, the first approximation is actually

$$ \frac{2}{c}\left(1-\frac{2m}{r_1}\right)^{1/2}\approx\frac{2}{c}\left(1-\frac{m}{r_1}\right) $$

This is a valid approximation because the power series for $(1-x)^{1/2}$ is

$$1 -\frac{1}{2}x+\cdots$$

So as long as $x=\frac{2m}{r_1}$ is close to zero, the above approximation is a valid first-degree approximation. Expanding this substitution,

$$ \begin{align} \Delta\tau&\approx\frac{2}{c}\left(1-\frac{m}{r_1}\right)\left(r_{1}-r_{2}+2m\ln\frac{r_{1}-2m}{r_{2}-2m}\right)\\ & = \frac{2}{c}\left(r_{1}-r_{2}-\frac{m(r_1-r_2)}{r_1}+2m\left(1-\frac{m}{r_1}\right)\ln\frac{r_{1}-2m}{r_{2}-2m}\right) \end{align} $$

So the second approximation that has been made is

$$ \begin{align} \left(1-\frac{m}{r_1}\right)\ln\frac{r_{1}-2m}{r_{2}-2m}&\approx\ln\left(\frac{r_1}{r_2}\right)\\ \end{align} $$

This is equivalent to the following approximation using logarithm rules

$$ \begin{align} \left(1-\frac{m}{r_1}\right)\left(\ln(r_1)+\ln(1-2m/r_1)-\ln(r_2)-\ln(1-2m/r_2)\right)&\approx\ln\left(\frac{r_1}{r_2}\right)\\ \end{align} $$

Now you just drop all the terms that have $\frac{m}{r_i}$, and your approximation is another logarithm rule. It is valid to drop these terms, because presumably $\ln(r_1)$, $\ln(r_2)$, and $1$ are relatively much larger.

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Yes! I would never have seen that in a month of Sundays. Thanks. – Peter4075 Dec 24 '11 at 18:28

It actually seems to me they use



$$2m\ln\frac{r_{1}-2m}{r_{2}-2m}\approx 2m\ln\left(\frac{r_{1}}{r_{2}}\right) \; .$$

EDIT: Just realized the following:

$$2m\ln\frac{r_{1}-2m}{r_{2}-2m}\approx 2m\ln\left(\frac{r_{1}}{r_{2}}\right) + 2m\left(\frac{2m}{r_2}-\frac{2m}{r_1}\right) \; .$$

Now the last term can be rewritten as

$$2m\left(\frac{2m}{r_2}-\frac{2m}{r_1}\right) = \frac{(2m)^2}{r_1 r_2}(r_1-r_2) = \left(\frac{2m}{r_1}\right)\left(\frac{2m}{r_2}\right)(r_1-r_2) $$

which is negligible.

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