Schwarzild orbits?

[Growl]

Can two black holes of equal mass have stable orbits at their respective schwarzild radii?

[Genady]

No. Here is what happens:

[Growl]

29 minutes ago, Genady said:

No. Here is what happens:

What of the special circumstance of the schwarzilde radius?

[Genady]

3 minutes ago, Growl said:

What of the special circumstance of the schwarzilde radius?

When their event horizons join they quickly make one black hole.

[Growl]

Quickly indicating a velocity greater than c? Not contesting your answer, who knows what goes on inside the event horizon… but how would this manifest to an outside observer?

[Genady]

14 minutes ago, Growl said:

Quickly indicating a velocity greater than c?

No, not at all. It is just short. This diagram shows how short is proper time of a falling particle between the event horizon, r/M=2, and the singularity, r/M=0:

[Markus Hanke]

17 hours ago, Growl said:

Can two black holes of equal mass have stable orbits at their respective schwarzild radii?

Two black holes in close proximity cannot be of the Schwarzschild kind - so their event horizons would be nowhere near spherical.

Also for a single Schwarzschild BH, no stable orbits exist at the horizon.

[Growl]

9 hours ago, Markus Hanke said:

Two black holes in close proximity cannot be of the Schwarzschild kind - so their event horizons would be nowhere near spherical.

Also for a single Schwarzschild BH, no stable orbits exist at the horizon.

Actually at the proximity described the orbits would be nearly if not perfectly circular. The distortion of the event horizons does bear consideration…
 

The time/space internal to the system would present a perfectly circular orbit to the comoving observer regardless the eccentricity expressed to a distant observer.

Edited September 20, 2023 by Growl

[Markus Hanke]

14 hours ago, Growl said:

The time/space internal to the system would present a perfectly circular orbit to the comoving observer regardless the eccentricity expressed to a distant observer.

How do you know this? The metric in close proximity to a BH merger is complicated enough so that it cannot be written in closed analytical form - it can only be modelled numerically. Given that such a spacetime would admit at most one spatial Killing field (and even that only approximately), and is not stationary, I think it is highly unlikely that any observer there would measure a perfectly circular orbit for the black holes.

Maybe it is possible to find some coordinate system where this might be approximately the case over a short span of time only, but even that I’m not convinced of.

[Growl]

7 hours ago, Markus Hanke said:

How do you know this? The metric in close proximity to a BH merger is complicated enough so that it cannot be written in closed analytical form - it can only be modelled numerically. Given that such a spacetime would admit at most one spatial Killing field (and even that only approximately), and is not stationary, I think it is highly unlikely that any observer there would measure a perfectly circular orbit for the black holes.

Maybe it is possible to find some coordinate system where this might be approximately the case over a short span of time only, but even that I’m not convinced of.

You are over complicating, as the two approach each other or recede, the gravitational field, time and space is distorted accordingly, so a comoving observer would always observe the same distance separating the two.

To restate, as the distance between the two diminshes, the gravitational force vector between them increases in amplitude and time/space undergoes distortion equivalent to the increase in gravitational field strength.

[Markus Hanke]

8 hours ago, Growl said:

so a comoving observer would always observe the same distance separating the two.

No he wouldn’t. The distance from horizon to horizon would explicitly depend on where it is measured, when it is measured, and also how it is measured (since the horizons are not spherical). The spacetime in between these black holes is not stationery and admits very few spatial symmetries, so figuring out what any one observer would measure here is highly non-trivial. But he certainly wouldn’t see any perfectly circular orbits.

8 hours ago, Growl said:

gravitational force vector between them

You cannot use Newtonian forces to describe or visualise this, since the radiation field here couples to a rank-2 tensor as its source. This really does require the full machinery of GR.