I ponder about geodesics in static spherically symmetric perfect fluid spheres. My first thought was that only radial geodesics, i.e. geodesics with zero angular momentum ($l=0$) are possible because of zero angular velocity. However, in radial direction there is no movement as well. Moreover, as I have learned, geodesics and particle trajectories are two different things. Does it mean that circular geodesics with zero angular momentum are "orbits" of the spacetime in static perfect fluid spheres?

  • $\begingroup$ Could you explain why you think only radial geodesics are possible? Or why you say that there is no movement in the radial direction? In general, all sorts of geodesics are possible. $\endgroup$
    – Javier
    Mar 26 at 17:49
  • $\begingroup$ @Javier, I would like to use congruence of geodesics to characterize the spacetime of interior solutions (like, for example, Schwarzschild interior solution). My idea was to use for this purpose circular geodesics. However, I have difficulty to understand what meaning would have $l^2$ if I do not treat it as particle trajectory. In static fluid its particle trajectories are not geodesics, see arxiv.org/abs/2011.05891. $\endgroup$ Mar 26 at 18:36
  • $\begingroup$ @Javier, by saying "there is no movement in radial direction" I have meant that fluid is static. $\endgroup$ Mar 27 at 10:38

1 Answer 1


Geodesics, as integration curves of the geodesics equation, depend only on metric. My confusion has arisen from imagination of perfect fluid sphere as an impermeable body which prevents some test particles from following the geodesics. However, the real matter is not continuous and can be penetrated without interaction. For example, a neutrino particle can traverse unhindered a matter clump following time-like geodesic. The same is valid for a gravitation wave which would follows null geodesic. By geodesics only the metric is important.


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