Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Dec 19, 2014 #1
    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!
  2. jcsd
  3. Dec 19, 2014 #2


    Staff: Mentor

    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.
  4. Dec 20, 2014 #3
    Thanks, Mark. But I am afraid, I do not follow. For example, how does the l^2 emerge?
    Last edited: Dec 20, 2014
  5. Dec 20, 2014 #4


    Staff: Mentor

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook