2
$\begingroup$

Let's say that a particle starts a radial free fall towards a Kerr black hole with zero initial energy at $r\rightarrow\infty$. The initial angular momentum of the particle is zero ($p_\phi = 0)$. From the Kerr metric, $p_\phi$ and $E$ are the constants of motion for the particle geodesic. However, we know that inside the ergosphere, we have a frame dragging effect, which means that the particle will start rotating after it enters into the ergosphere. Does it mean that inside the ergosphere, the particle angular momentum $p_\phi$ is nonzero and $p_\phi$ is not a constant of motion anymore?

$\endgroup$
  • 1
    $\begingroup$ relative to the local ZAMOs the tangential velocity of the infalling particle remains 0 all the way. If the axial angular momentum Lz=0, the particle always corotates with the local frame dragging velocity, and therefore vφ=0 in the local ZAMO's frame. It is like in this animation, but time reversed: yukterez.net/org/kerr.orbits/kerresc3.html $\endgroup$ – Gendergaga Dec 13 '19 at 0:58
4
$\begingroup$

No the angular momentum of the particle would remain zero (it is still a constant of motion). The angular velocity of the particle, on the other hand, would steadily increase as the particle approaches the Kerr black hole (i.e. the Newtonian relationship between angular momentum and angular velocity does not hold in GR!)

Note that the frame dragging does not start at the ergosphere. Frame dragging will start effecting the particle right from the beginning. The ergosphere is simply the region in which it becomes impossible to have a timelike curve with negative angular velocity.

| cite | improve this answer | |
$\endgroup$

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.