I am currently studying the geodesics of different type of spacetimes and I'm not sure if I'm doing it in the correct way for Schwarzschild de Sitter space (SdS).

The metric in SdS is given by: $$ ds^2=-(1-\frac{2M}{r}-\frac{\Lambda r^2}{3})dt^2+(1-\frac{2M}{r}-\frac{\Lambda r^2}{3})^{-1}dr^2+r^2d\Omega^2 $$ with the usual definitions.

Solving for radial null geodesics directly from the metric gives $$ \frac{dr}{(1-\frac{2M}{r}-\frac{\Lambda r^2}{3})}=\pm dt $$ Integrating both sides will give a very nasty integral and I'm not sure this is what I should be doing.

Is there another way to calculate radial null geodesics for SdS?

The only way I can analytically solve this problem is by requiring an extremal black hole with $\Lambda=\frac{1}{9M^2}$ but that is not what I'm looking for.

The problem gets even worse when you look at the Reissner-Nordström in de Sitter space, because an extra term $-\frac{Q^2}{r^2}$ gets added inside the brackets.

  • $\begingroup$ What you're doing seems reasonable. No particular reason to expect a closed-form solution in terms of elementary functions. $\endgroup$ – user4552 Oct 9 '19 at 13:51
  • $\begingroup$ Note that the left hand side integrand can be written $\frac{r}{(r-r_H)(r-r_\Lambda)(r+r_H+R_\Lambda)}$ where $r_H$ and $r_\Lambda$ are the black hole and cosmological horizons respectively. After this the integral can be solved using partial fractions. $\endgroup$ – mmeent Oct 10 '19 at 6:26

Although there are other ways of solving for radial geodesics (For example you could exploit the existence of a timelike Killing vector to first solve for $r(\tau)$ and $t(\tau)$), none seem like they will be simpler than the method you are already trying.

With the steps you took you ended up with an integral of a rational function. In general, such integrals can be solved using partial fractions. To do so one needs the roots of the denominator,

$$ \frac{\Lambda}{3}r^3 -r +2M.$$

You should already know two of these roots, they are the black hole horizon $r_H$ and the cosmological horizon $r_\Lambda$ of Schwarzschild-de Sitter. Since there is no $r^2$ term we know the sum of the roots has to vanish. Hence the third root is $-r_H-r_\Lambda$. The integrand thus becomes


Having factorized the denominator, decomposition in partial fractions is straight forward,

$$\frac{3}{\Lambda} \Big( \frac{r_H}{(r_\Lambda-r_H)(2r_H+r_\Lambda)}\frac{1}{r-r_H}+ \frac{r_\Lambda}{(r_\Lambda-r_H)(r_H+2r_\Lambda)}\frac{1}{r_\Lambda-r} +\frac{r_H+r_\Lambda}{(2r_H+r_\Lambda)(r_H+2r_\Lambda)}\frac{1}{r+r_H+r_\Lambda} \Big). $$

The integrals of the individual terms are elementary giving,

$$\Big( \frac{r_H}{(r_\Lambda-r_H)(2r_H+r_\Lambda)}\log(r-r_H) +\frac{r_\Lambda}{(r_\Lambda-r_H)(r_H+2r_\Lambda)}\log(r_\Lambda-r) +\frac{r_H+r_\Lambda}{(2r_H+r_\Lambda)(r_H+2r_\Lambda)} {\log(r+r_H+r_\Lambda)} \Big)= \pm \frac{\Lambda}{3}(t-t_0), $$

assuming $r_H < r < r_\Lambda$.

The RNdS problem can be solved in similar fashion, except that you now have to black hole horizons appearing as roots.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.