Radial Acceleration of Rotating Singularities: Planck Length & Beyond

[Atticuskirk]

And if so, how much? Should the radius be thought of as zero, an infinitesimal, or as the Planck length?

v²/r = ω²r

If its zero, then you immediately run into a problem when trying to calculate it with linear velocity.

v²/r = a_r
v²/0 = undefined

OR

ω²r = a_r
ω²0 = 0

Which would mean that v²/r ≠ ω²r

If its an infinitesimal, then there's still a problem.

Let ε = 1/∞

v²/r = a_r
v²/ε = ∞

ω²r = a_r
ω² = ε

v²/r = ω²r
∞=ε

The last case of r being equal to the Planck length seems to be the only one that makes sense, since its still a real number (just really, really tiny). So does this mean a black hole's singularity is not actually a mathematical singularity, or am I just misunderstanding something here?

 

[Orodruin]

The question is ill posed as the singularity of an idealised black hole is more like a moment in time than an object in space. Either way, the mere existence of a singularity should make you suspect that your theory, in this case GR, might not be fully extendable to that limit. In particular, using classical formulas to try to describe it will be utterly doomed to failure as you are deep into the regime where classical mechanics is not applicable.

 

[PeterDonis]

the singularity of an idealised black hole is more like a moment in time than an object in space

This is true of a non-rotating Schwarzschild black hole, but it is not true of a rotating Kerr black hole; the singularity in Kerr spacetime is timelike. However, I think the OP question is not well posed even taking that into account. See below.

Should the radius be thought of as zero, an infinitesimal, or as the Planck length?

The singularity itself does not have a "radius", since it is not part of the spacetime manifold at all. So your question is not well posed.

The term "singularity" actually refers to geodesic incompleteness: there are geodesics in a spacetime with a singularity that cannot be extended to arbitrary values of their affine parameter.

 

[Atticuskirk]

Thanks for the quick reply. I figured that the normal formulas for rotational motion might break down, but I wasn't personally aware of cases in which they failed, so I wasn't sure. Is this one of those cases where a theory of quantum gravity would be important, or is there an existing framework that would better fit this? Also, would you mind elaborating on black holes being more like moments in time? I know that time dilation approaches infinity as you reach the event horizon, but why would that prevent the remnants of the previous star from remaining as a physical object with infinite density?

If I'm honest, I have a fairly novice understanding of black holes and the like.

 

[PeterDonis]

Is this one of those cases where a theory of quantum gravity would be important, or is there an existing framework that would better fit this?

It is generally believed that the presence of singularities in GR is a sign that the theory breaks down in those particular regimes, and the usual expectation is that a theory of quantum gravity is what will be needed to cover such cases. There is no other existing theory that covers these regimes.

would you mind elaborating on black holes being more like moments in time?

The singularity in a non-rotating black hole is like a moment in time--it is to the future of every event inside the hole. That's why it's impossible to avoid the singularity once you fall inside the hole's horizon: because you can't avoid moving into the future.

A rotating hole is not like this, however; as I mentioned, the singularity in a rotating hole is timelike, which means it's like a point in space (actually it's like a ring when you work out the details). But for the case of Kerr spacetime, the regime where GR is expected to break down occurs, not close to the singularity, but close to the inner horizon, which an idealized observer falling in would encounter well before the singularity. So the specific nature of the singularity in this case is really a moot point.

 

[stevebd1]

And if so, how much? Should the radius be thought of as zero, an infinitesimal, or as the Planck length?

This probably doesn't answer your question but if you're talking about the radius of a singularity in rotating black holes then you might want to take a closer look at the quantity of $a$ in Kerr metric. $a$ is the spin parameter of a black hole, the spin parameter can also be written as $a/M$ where $M$ is the gravitational radius, both are in geometric units. $a/M$ provides a number between 0 and 1, 0 being static and 1 being maximal (the spin parameter is normally established by observing where the ISCO (innermost stable circular orbit) is, this ranges from 6M for 0 (static) and 1M for 1 (maximal). $a$ is defined by $a=J/M$ in geometric units, in SI units, this would be $a=j/mc$ which could be rearranged as $j=cma$ which is similar to the equation for angular momentum for a ring $j=vmr$ where $a$ is $r$ and $c$ is $v$. In a rudimentary (and abstract) way $a$ is the coordinate radius of the ring singularity and it's tangential velocity is always c. When trying to plot this coordinate radius with spherical coordinates in Kerr metric, it doesn't work, the ring singularity ends up in the space between the inner and outer horizon but if you use elliptical coordinates, then you can use the quantity $a$ as a coordinate radius, as it has been done in the following link (see the two bottom images)-

http://jila.colorado.edu/~ajsh/insidebh/waterfall.html

Though as it's already been pointed out, the Cauchy (inner) horizon is the boundary of predictability and GR breaks down at this point so this is hypothetical.