Trying to figure out how an approximation of a logarithmic equation works

Question

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

Answer 1

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.

Answer 2

It actually seems to me they use

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

and

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

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.