Integrating to get r^2 = l^2 + a^2

[Lapidus]

I'm reading a physics book( "Einstein's gravity in a nutshell" Zee) and at page 126 the author says:

(dr/dl)^2 + a^2 / r^2 = 1

gives r^2 = l^2 + a^2

where we absorbed an integration constant into l by setting l=0 when r=a.

Can someone explain what's going on here? How do I have to "massage" the first equation to make an integration that gives the second equation? What does Zee mean with absorbing an integration constant?

Thanks for any respons!

 

[Mark44]

I'm reading a physics book( "Einstein's gravity in a nutshell" Zee) and at page 126 the author says:

(dr/dl)^2 + a^2 / r^2 = 1

gives r^2 = l^2 + a^2

where we absorbed an integration constant into l by setting l=0 when r=a.

Can someone explain what's going on here? How do I have to "massage" the first equation to make an integration that gives the second equation? What does Zee mean with absorbing an integration constant?

Thanks for any respons!

Rewrite the DE as dr/dl = $\pm \sqrt{1 - \frac{a^2}{r^2}} = \pm \frac{\sqrt{r^2 - a^2}}{|r|}$
This equation is separable, with a fairly easy integration. The work is simpler if you can assume that r > 0 and dr/dl > 0.

 

[Lapidus]

Rewrite the DE as dr/dl = $\pm \sqrt{1 - \frac{a^2}{r^2}} = \pm \frac{\sqrt{r^2 - a^2}}{|r|}$
This equation is separable, with a fairly easy integration. The work is simpler if you can assume that r > 0 and dr/dl > 0.

Thanks, Mark. But I am afraid, I do not follow. For example, how does the l^2 emerge?

 

[Mark44]

If it's reasonable to assume that dr/dl > 0 and r > 0, we can write the DE as $dr/dl = \frac{\sqrt{r^2 - a^2}}{r}$.

Separating this, we get $\frac{r dr}{\sqrt{r^2 - a^2}} = dl$.

Now integrate both sides.