5
$\begingroup$

I was reading about congruences of timelike geodesics from "A Relativist's Toolkit - Eric Poisson". There is a solved example of a timelike geodesic congruence for Schwarzschild spacetime.

The kind of congruence which is considered in the example is radial, marginally bound and timelike. What are marginally bound geodesics? How are they even relevant in this case? Do we always have to consider marginally bound geodesics to observe geodesic congruence?

$\endgroup$

1 Answer 1

3
$\begingroup$

To understand what "marginally bound geodesics" are, you can gain some intuition by thinking about Newtonian gravity. In the newtonian two-body problem, the energy parameter of the orbits dictate whether they're bound or not. If $E<0$, then the orbit is an ellipse (therefore bound) while if $E>0$ the orbit is an hyperbola (unbound). The case separating the two ($E=0$, parabola) would be the "newtonian marginally bound orbit".

In many GR problems we can define a similar "energy parameter". In order to call an orbit bound, we also require the orbit to remain bound under small perturbations (therefore a circular orbit can be unbound, because it might escape to infinity when perturbed). In this sense, consider circular orbits in the Kerr metric (notation as here p. 21-22). Circular orbits can be shown to be bound if the energy parameter satisfies $E<1$, and unbound if $E>1$. The orbits with $E=1$ are therefore called "marginally bound". The reason why in GR you have $E=1$ rather than $E=0$ is that $E$ can be thought of as "energy per unit mass" in units with $c=1$, so that you always have energy due to rest mass. In fact in this case $E\to 1$ at infinity, so the "binding energy" would be $E-1$, which behaves like in the newtonian case. In any case, I don't think the concept is particularly useful, but that's just my opinion.

As to the other question, there's no need for the geodesics to be marginally bound to have a congruence. In fact, there's no requirement that the curves even be geodesics. Generally speaking, a congruence of curves is the set of integral curves of a vector field.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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