Why is a spinning Black Hole distinct from one that doesn't? How does a Black Hole even spin? Why is it important?

Simply “yes”. The spinning one distorts spacetime differently from the non-spinning one. This is true whether the objects are black holes or just boring planets or something. You can conduct experiments to observe the differently distorted spacetime.

Through its effects on the surrounding spacetime.

Whereas Newtonian gravity depends only on the amount of mass present, the curvature of spacetime in GR depends on the presence of energy and momentum and on the flow of that energy and momentum through space and time. So, moving sources “gravitate” differently from non-moving ones. GR’s dependence on this wider range of features relates to fundamental relativistic principles (like how physics shouldn’t depend on coordinate system choices). That’s a short answer to “how”.

Echoing @HMS_Irruncible, if there is angular momentum, there is not a point-like singularity.

Yeah, the conservation of angular momentum is one-and-the-same with the principle of rotational invariance of the laws of physics, and this is not overturned by any GR principle, so angular momentum has to be conserved during collapse.

The limit of angular momentum of a (Kerr solution) black hole is

|J| \leq M^2

where J is angular momentum and M is the mass-energy of the singularity. The derivation of this is somewhat involved (it’s all algebraic geometry but more than I want to code up in LaTeX this morning) but in essence it is where the circumference of the event horizon is not positive (i.e. doesn’t exist) resulting in a ‘naked singularity’.

It’s really not correct to think of the singularity as a physical object with ordinary mechanical properties, at least not in our universe; it is a solution of the Einstein field equations in which the volume of the mass goes to zero, and thus the equations of general relativity break down in describing spacetime at that point. Whether the mass in a collapsing black hole can ever be compressed to a zero volume is an open question (the consensus is that some other physical theory ‘takes over’ at that point) but regardless, the spacetime around the black hole behaves as if the mass has angular momentum and an effective moment of inertia.

Stranger

I get that the two black holes warp spacetime near them but how do you twist space (so to speak)? How does that work? What is a spinning black hole doing to space that a non-spinning one is not? (really asking)

Note that the answer does not rely on the mass being in black holes. The gravitational field of a black hole outside an event horizon is no different than the gravitational field of any other mostly spheroidal object.

You can look at it, for one thing. The black “shadow” will be asymmetric if the hole is rotating:

It matters to the matter orbiting the hole due to the “dragging”.

Indeed, and that’s one of the aspects of General Relativity, as I understand it: all objects with mass “curve” or “bend” space-time around them, and the more massive they are, the more they do so.

Looking into this, it all appears to be correct, but I’m intrigued by the fact that the formula you gave (which is apparently correct as I’ve seen it elsewhere as a way of expressing the angular momentum limit) doesn’t appear to pass the test of unit consistency. I’m probably missing something here, but I don’t see how it’s derived or how to interpret it.

The alternate way of expressing the limit which does pass the units test is

j = Jc / GM^2

Here J is angular momentum, c is the speed of light, G is the gravitational constant, and M is the mass-energy.

Typical units would be:

J = kg * m^2/s
c = m/s
G = m^3 * kg^{-1} * s^{-2}
M = kg

So then, unless I screwed up, we get:

Numerator = kg * m^3/s^2
Denominator = (m^3 * kg^{-1} * s^{-2} ) * kg^2
= kg * m^3/s^2

So the units cancel out, yielding the dimensionless quantity j. AIUI the rule is that to preserve the event horizon, j cannot be greater than 1.

In 3D Euclidean space, the distance (squared) between two points is the sum of the squares of coordinate separations – i.e., the Pythagorean theorem. So the distance s between (x_1,y_1,z_1) and (x_2,y_2,z_2) is given by s = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}. Hereafter, we can talk just about s^2 and not s to avoid the clumsy square root sign. We can also zoom in to an infinitesimal displacement so that we can talk about how the space behaves locally as a point. For this we can use differential symbols like “dx” to mean “an infinitesimal step in the x direction”, which is the calculus/limiting case of a more chunky step \Delta x.

Putting all that together, the “metric” – how you measure in the space – in boring 3D Euclidean space is

ds^2 = dx^2 + dy^2 +dz^2 .

Keep in mind that s is related to the concept of distance between points (and ds the infinitesimal version that encapsulates the local behavior of the geometry at hand).

In (3+1)D special relativity, distances in just the space portion aren’t fundamental. We are instead concerned with distances in spacetime. In special relativity, distances in spacetime are related to the metric

ds^2 = -(c\,dt)^2+dx^2 + dy^2 +dz^2 .

The negative sign on the time term is worth noting, but it’s actually not that interesting a feature for the purposes of this question. Also, you may find in your own research two different sign conventions, where the expression on the right is equal to -(ds)^2. This, too, is uninteresting for the purposes of this question, but it can be confusing.

Both of the metrics above represent “flat” – not curved – spaces, since the contribution of each coordinate step is independent of the coordinate values themselves and of the other step sizes. A step dy always contributes the same no matter what location (t,x,y,z) you are at and no matter how big a step you are taking in the other coordinate directions.

In general relativity, the presence of energy, momentum, and flows of those modify the spacetime metric. The Einstein field equations are the differential equations that relate the “sources” to the metric, although the metric itself is somewhat buried in the tidy form of those equations. (It’s in there, though!)

For spherical sources of gravity, it will be useful to switch from Cartesian coordinates to spherical coordinates. So, we’ll measure distances not in terms of steps in (t,x,y,z) but rather in (t,r,\theta,\phi). Here, r is the radial distance from the center of our source object (star or planet or whatever); \theta is the polar angle (like latitude, but with 0 at the north pole); and \phi is the azimuthal angle (like longitude).

The flat metric in these coordinates looks messier than in Cartesian coordinates:

ds^2 = -(c\,dt)^2+dr^2 + r^2d\theta^2 + r^2(\sin\theta)^2 d\phi^2

but it’s still just as flat as the previous metric. It’s just that a step in (say) \theta is worth more distance when you are at higher r, and a step in \phi is worth less distance when you are near the poles. The various extra bits in the metric are just these boring aspects of spherical coordinates in a flat space. So, this metric is our starting point now.

We can drop a spherically symmetric, non-rotating mass M at the center of our coordinate system. This leads to the metric

ds^2 = -\left(1-\frac{2GM}{rc^2}\right)(c\,dt)^2+\left(1-\frac{2GM}{rc^2}\right)^{-1}dr^2 + r^2d\theta^2 + r^2(\sin\theta)^2 d\phi^2 ,

where G is Newton’s gravitational constant. Now the distance contribution from steps in the t and r directions are modified in an r-dependent way relative to the flat case. But the angular pieces are untouched and there is no “cross” dependence anywhere, meaning the contribution of a step in r (say) doesn’t depend on how much you are stepping in \theta (say).

Finally, we introduce a spinning source, and you get the so-call Kerr metric. It is not worth reproducing the very complicated expression here – see the wiki link – but two key features include:

  • The coefficients now all also depend on the angular momentum present and the \theta coordinate.
  • There is now cross-dependence in the coordinate intervals!

For that last point, there is now a term that looks like

... +\, (\mathrm{coefficients})\, dt \,d\phi .

This means distances in the (+t,+\phi) direction will work differently than distances in the (+t,-\phi) direction, and how much this happens also depends on where you are in r and \theta (as embedded in the coefficients).

This connects directly to the previously noted feature that travel around the spinning object looks different one way versus the other. More generally, all these coordinate-value and step-value couplings lead to observable deviations in behavior from the non-spinning case. Lense-Thirring precession is another specific consequence, as mentioned above, wherein a rotating object like a gyroscope will exhibit anomalous precession when near a rotating massive object.

I love this website…you all are the best as exemplified by this answer.

It’ll take me some time to understand what was posted (and not sure I ever will) but love the effort and detail.

Thank you.

It is a fairly common convention to set certain dimension-ful constants in 1 in some settings. This is okay when the constants are, at their core, only unit conversions anyway and not actually setting the behavior of the theory. In GR, setting c=G=1 is common. It does mean that other quantities become expressed in “unusual” units.

Here’s a summarized version to perhaps guide you through the longer post:

  • The distance between two points in boring space is obtained by summing the squares of the step sizes in each coordinate direction. 3 meters in x and 4 meters in y results in a 5 meter total step since (5\ \rm{m})^2=(3\ \rm{m})^2+(4\ \rm{m})^2.
  • This can be abstracted to the concept of a “metric”, which encapsulates algebraically how the space (or spacetime) operates, how distances are measured, and how it feels to move through the space.
  • Around a non-spinning spherical mass, motions in the radial and time directions are modified relative to the flat case. This is to say that the “metric” gains additional complications in the distance terms related to the “r step size” dr and the “t step size” dt. So, “3 meters in the r direction” doesn’t contribute (3\ \rm{m})^2 to the total distance squared (ds^2). It instead contributes something that depends also on the mass of the central object and on how far away from it you are. That’s what the coefficients in front of the dr^2 piece are saying.
  • Around a spinning spherical mass, these coefficients on the different coordinate-step terms get more complicated. They depend now also on the angular momentum of the central object and on what latitude you are at. But more than this, steps taken in the time and longitudinal (\phi) directions are also tangled up. A step forward in time contributes differently to the overall spacetime “distance” covered depending on how much you are moving in the longitudinal direction, and on which direction.

I was curious to know how fast actual black holes are really spinning, since apparently quite a few have been measured. Doing a quick Google, it appears – unsurprisingly – that they tend to spin very, very fast, which is a consequence of conserved angular momentum as big stars collapse. Some appear to be spinning at or close to the maximum theoretical limit, for example:

“We have measured the spin frequency of a famous black hole which goes by the name of GRS 1915+105,” said Ramesh Narayan from the Harvard-Smithsonian Centre for Astrophysics in the U.S. “According to theory, it can have a maximum spin rate of 1,150 times per second. Our measurements indicate that the hole is spinning between 950 and 1,150 times per second - that is, it is spinning quite close to the maximum.”
https://web.archive.org/web/20120507004507/http://www.cosmosmagazine.com/node/873

Thank you. I see that if one sets c and G to unity, then the rule that the dimensionless quantity j can never be greater than 1 makes the formula I posted exactly the same as the one posted by @Stranger_On_A_Train.

I also found mysef wondering what, intuitively, is the mechanism that prevents a black hole from exceeding the maximum possible spin rate. I think the answer is that as the maximum spin rate for any given mass approaches (mass is equivalent to diameter of event horizon) that the angular speed of the event horizon will be approaching the speed of light.

The one you feed.

An infalling particle will add both energy (mass) and angular momentum to the black hole. A particle with low energy and high angular momentum does not enter the black hole (it’s effectively in orbit), and a particle with high energy and low angular momentum moves the goal line by more than it is able to violate it. The spin limit lies at the boundary between these two cases. So, if you have a black hole that’s right near the limit and you try to push it over the edge with an infalling particle, you either can’t get your particle in at all or you can but with an increase in angular momentum that is less than the increase in squared mass. This argument can be generalized to other cases, like the collision of two spinning black holes to form one bigger one.

More precisely, spin is a kind of angular momentum. If you add up all of the angular momentum in a closed system, including spins of particles, that’s what’s conserved. If you add up just the non-spin part of angular momentum, it’s not.

No, because relativity. The angular speed does increase considerably, but it doesn’t need to go infinite.

He’s using units where c and G are both 1, which is the usual practice in GR. Both of those constants show up all over the place, and it’s inconvenient to keep writing them.

This is all true from a practical perspective: There’s no process by which such a hole could form, nor for an existing hole to develop to that state. But there’s also the issue that, if the angular momentum exceeded the limit, you’d get a naked singularity, which is, so far as we can tell, impossible in our Universe (IIRC, this was the topic of one of the infamous Thorne Wagers).

It should also be mentioned, by the way, that not only does frame dragging occur around any rotating object, but it’s actually been measured from the Earth’s rotation (and the measurements agreed with theory). That was the main mission of Gravity Probe B.

Well, the original bet was conceded by Hawking (to Preskill and Thorne) in favor of naked singularities being possible due to technicalities and contrived circumstances. But to me, it’s not really the “reason” why a spin limit exists rather an aspect of the consistent theory under practical dynamics.

Slight hijack: does there exist a black hole with zero angular momentum?

My gut says no. A topological transition from a toroidal singularity to a point-like one just feels wrong to me.

The answer to that question is…complicated. It is possible for particles (or objects) outside the event horizon but within the ergosphere of a black hole to appear to be moving faster than light (c) with respect to an external reference frame, even though in the local frame no particle can move faster than c. It isn’t really possible to explain this completely without understanding general relativity, but if you imagine a powered toy boat in a bathtub which is then floating in a fast-moving river, while the toy boat can only move that the slow speed that its propeller can push it relative to the bathtub, it is moving very fast with respect to the fixed shoreline because of how fast the bathtub is moving with the flow of the river. The river, in this analogy, is the movement of spacetime being warped by the angular momentum of the black hole which isn’t limited by general relativity because it isn’t made of mass or energy.

Stranger