Curvature in space-time is shown as a "fabric"

[Markus Hanke]

1 hour ago, geordief said:

Is the geometry of spacetime universally applicable because the old geometry of space and time as measured from any one point has proved incapable of modeling events in other places and times to a sufficient degree of accuracy  under high speeds of relative motion?  (which spacetime can do except   under extreme conditions)

Do you mean Euclidean geometry by ‘old geometry’?

If so, then yes - Euclidean geometry cannot guarantee that the spacetime interval is the same for all observers; physically, this means that all observers experience the same physics, irrespective of their states of relative motion. That’s what we see happening in the real world, so we need any good model to reflect that. Minkowski geometry (and Riemann geometry in GR) do this natively and very elegantly.

[studiot]

 

Hi Markus,

I know we are always stressing there is no fabric.

Have you heard of the cosmic Fabric model of gravity ?

 

 

https://www.google.co.uk/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&cad=rja&uact=8&ved=2ahUKEwjM-ouXq4PzAhUKxRQKHfQwDd4QFnoECAIQAQ&url=https%3A%2F%2Fwww.worldscientific.com%2Fdoi%2F10.1142%2FS0218271819500962&usg=AOvVaw3Kqo9MvdIjk3xvl4NsgsTR

Edited September 16, 2021 by studiot

[geordief]

45 minutes ago, Markus Hanke said:

Do you mean Euclidean geometry by ‘old geometry’?

If so, then yes - Euclidean geometry cannot guarantee that the spacetime interval is the same for all observers; physically, this means that all observers experience the same physics, irrespective of their states of relative motion. That’s what we see happening in the real world, so we need any good model to reflect that. Minkowski geometry (and Riemann geometry in GR) do this natively and very elegantly.

Thanks.

Yes ,I thought of writing "Euclidean geometry" but was a bit uncertain as to the correct terminology

"Natively" an auto misspelling for "naturally" I am guessing.(my bold)

[md65536]

12 hours ago, Markus Hanke said:

There is no detectable information about this at any one point on the Earth’s surface. This is because the geometry concerns relationships between points, so what you do is take measurements of path lengths, areas, or angles. For example, you’ll find that the sum of the angles in a triangle on Earth’s surface is no longer exactly 180 degrees - it’s possible to directly measure this deviation. But you can’t do it at a single point, you need to measure across some distance. That’s because the effects of a non-flat metric are accumulative - mathematically, you integrate components of the metric to obtain path lengths.

Tying this in with the rubber sheet analogy: The height on the sheet or on Earth corresponds (assuming constant g) with gravitational potential, proportional to how much energy it takes to lift an object to a specific height. The derivative of that, the slope of the sheet or ground, corresponds with gravitational force---how quickly a marble would accelerate as it rolls along that point. The derivative of that represents curvature. If we imagine the rubber sheet to be semi-rigid, and you actually have to bend it into a curved shape, the severity of the bending at a given point corresponds with curvature. You can tilt a sheet, and that corresponds with a constant gravitational force but no curvature.

(Just to complete the analogy, curvature corresponds with tidal forces. If you have a toy car where the front and back are separate and connected by a spring, and roll it down a hill with constant slope, the spring doesn't stretch or compress. If you put it on a curve, eg. on top of a ball, the front and back are on different slopes and can pull away from each other.)

The "local flatness of spacetime" implies that if you're looking only at a single point, you can't detect curvature, but you also can't detect gravitational force (is that right???). So for the Earth analogy, using it only to show what you can detect locally, and not how it affects the motion of objects, we could take the real Earth and modify it so that "up" is always normal to the surface you're standing on. Like in some cartoons where if you're standing on the side of a mountain, your body is tilted so your feet remain "flat" against the slope. Then if you can only look at the ground beneath your feet, you can't tell if you're on a slope or not.

Is that a fair analogy to spacetime, or is it only curvature that is locally undetectable, but not gravitational force?

 

 

Also I have a feeling I've asked a similar thing but still don't get it: Is the curvature relative, so it actually is locally zero but a different value from a distant? Or does local flatness merely mean, like you suggest, that the local value of the curvature only has measurable meaning across some distance? I think it's the latter??? Can curvature be called a scalar field, and is it invariant in a static universe?

Edited September 16, 2021 by md65536

[studiot]

2 hours ago, md65536 said:

Tying this in with the rubber sheet analogy: The height on the sheet or on Earth corresponds (assuming constant g) with gravitational potential, proportional to how much energy it takes to lift an object to a specific height. The derivative of that, the slope of the sheet or ground, corresponds with gravitational force---how quickly a marble would accelerate as it rolls along that point. The derivative of that represents curvature. If we imagine the rubber sheet to be semi-rigid, and you actually have to bend it into a curved shape, the severity of the bending at a given point corresponds with curvature. You can tilt a sheet, and that corresponds with a constant gravitational force but no curvature.

The problem with the rubber sheet is that the curvature of a rubber sheet is of the wrong sort of curvature.

There are two sorts of curvature, extrinsic and intrinsic.

The curvature of the manifold in GR is intrinsic.

The curvature of the rubber sheet (manifold) is extrinsic.

A further comment is that a single point has no curvature.

A line has no intrinsic curvature but may have zero extrinsic cirvature or some value of extrinsic curvature.

A surface can have no curvature, both extrinsic and intrinsic curvature or just extrinsic or intrinsic curvature.

These situations can all be drawn or sculpted in our 3D world.

Once we move to 3D however we cannot draw or sculpt extrinsic curvature and it is very difficult to imagine.

Edited September 16, 2021 by studiot

[Markus Hanke]

19 hours ago, studiot said:

Have you heard of the cosmic Fabric model of gravity ?

No, this one is new to me. Thanks for bringing it up. Having skimmed through the link, my first impression is that this formalism is not nearly as elegant and intuitive as the standard one (and full equivalence with GR is yet to be shown). I kind of fail to see the advantage, though the point about substructure is interesting.

12 hours ago, md65536 said:

Is that a fair analogy to spacetime, or is it only curvature that is locally undetectable, but not gravitational force?

See studiot’s comments on intrinsic vs extrinsic to begin with. Furthermore, there is not really any force involved in gravity - when you have initially parallel test particles in free fall, and attach an accelerometer to them, it will always read exactly zero, so no forces; nonetheless in the presence of gravity their geodesics will begin to deviate.

12 hours ago, md65536 said:

Is the curvature relative, so it actually is locally zero but a different value from a distant? Or does local flatness merely mean, like you suggest, that the local value of the curvature only has measurable meaning across some distance? I think it's the latter???

Good question! This point is a bit subtle, and really the answer should be “both of the above, depending on context”.

The physical manifestation of curvature is geodesic deviation - meaning that initially parallel world lines will begin to deviate as they extend into the future. It is thus necessary for world lines to have at least some extension in spacetime before “parallel” and “deviate” even make sense - you can’t speak of parallelism at a single event. Thus curvature has measurable meaning only across some distance. I’m highlighting the word ‘measurable’ because counterintuitively the mathematical object describing curvature (Riemann tensor) nonetheless is a local object, like all tensors. For clarification on this point, refer back to the example about calculus in my previous post.

However, there are also scenarios where the effects of gravity are in some sense ‘relative’. Consider a hollow shell of matter, like a planet that has somehow been hollowed out (not very physical of course, but I’m just demonstrating a principle here). Birkhoffs Theorem tells us that spacetime everywhere in the interior cavity is perfectly flat, ie locally Minkowski. There’s no geodesic deviation inside the cavity. Now let’s place a clock into the cavity, and another reference clock very far way on the outside, so both clocks are locally in flat Minkowski spacetime. What happens? Even though both clocks are locally in flat spacetime (no gravity), the one inside the cavity is still gravitationally dilated with respect to the far way one! This is because while both local patches are flat, spacetime in between them is curved - if you were to draw an embedding diagram, you’d get a gravitational well with a ‘Mesa mountain’ at the bottom; and the flat top of that mountain sits at a lower level than the far away clock, thus the time dilation. So in this particular case one could reasonably say that gravitation effects are ‘relative’ between local patches. Or you can put it like this: both regions are Minkowski, but one is more Minkowski than the other  

The isn’t very intuitive, but mathematically perfectly consistent - if you look at the world lines of the clocks, you’ll find that while they appear parallel in space (they’re simply at rest wrt to one another), they deviate in spacetime. In GR it is crucially important that one fully understands local vs global, or else there’ll be no end to misunderstandings and problems. This point is where most, if not all, apparent ‘paradoxes’ in GR arise.

12 hours ago, md65536 said:

Can curvature be called a scalar field, and is it invariant in a static universe?

In general, no, it’s not a scalar - it’s a rank-4 tensor field, the Riemann tensor. However, you can choose to look at only certain aspects of curvature, such as how volumes change (rank-2 Ricci tensor), or how areas differ from Euclidean counterparts (rank-0 Ricci scalar), or the average Gaussian curvature of a small region of space (rank-2 Einstein tensor). But to capture all aspects, you need the full rank-4 tensor with 20 independent components.

Tensors are not invariant, but covariant - meaning their individual components do vary in just the right ways so that the relationships between the components remain, hence the overall object is the same for all observers. Remember a tensor is all about the relationships between its components.

Edited September 17, 2021 by Markus Hanke

[studiot]

1 hour ago, Markus Hanke said:

Tensors are not invariant, but covariant - meaning their individual components do vary in just the right ways so that the relationships between the components remain, hence the overall object is the same for all observers. Remember a tensor is all about the relationships between its components.

A very insightful comment, without any need to refer to index notation.

+1

[SergUpstart]

1 hour ago, Markus Hanke said:

See studiot’s comments on intrinsic vs extrinsic to begin with. Furthermore, there is not really any force involved in gravity - when you have initially parallel test particles in free fall, and attach an accelerometer to them, it will always read exactly zero, so no forces; nonetheless in the presence of gravity their geodesics will begin to deviate.

To do this, the size of the accelerometer must be negligible compared to the curvature of ST. The accelerometer in the form of comet Shoemaker-Levy in July 1994 perfectly "discovered" the gravity of Jupiter.

 

[Markus Hanke]

16 hours ago, SergUpstart said:

To do this, the size of the accelerometer must be negligible compared to the curvature of ST.

Yes, that’s right. As pointed out earlier in the thread, spacetime is only locally flat. If the accelerometer is too large, it will begin to measure tidal effects.

[md65536]

On 9/16/2021 at 2:52 PM, studiot said:

The curvature of the rubber sheet (manifold) is extrinsic.

I think it also has intrinsic curvature in the usual setup.

 

22 hours ago, Markus Hanke said:

However, there are also scenarios where the effects of gravity are in some sense ‘relative’. Consider a hollow shell of matter, like a planet that has somehow been hollowed out (not very physical of course, but I’m just demonstrating a principle here). Birkhoffs Theorem tells us that spacetime everywhere in the interior cavity is perfectly flat, ie locally Minkowski. There’s no geodesic deviation inside the cavity. Now let’s place a clock into the cavity, and another reference clock very far way on the outside, so both clocks are locally in flat Minkowski spacetime. What happens? Even though both clocks are locally in flat spacetime (no gravity), the one inside the cavity is still gravitationally dilated with respect to the far way one! This is because while both local patches are flat, spacetime in between them is curved - if you were to draw an embedding diagram, you’d get a gravitational well with a ‘Mesa mountain’ at the bottom; and the flat top of that mountain sits at a lower level than the far away clock, thus the time dilation. So in this particular case one could reasonably say that gravitation effects are ‘relative’ between local patches. Or you can put it like this: both regions are Minkowski, but one is more Minkowski than the other

Sure, but the gravitational time dilation should depend only on their relative gravitational potential (right?) and not how space between them curves to produce that difference in potential. For example if you have 2 locations at different potentials in a constant gravitational field, you have 2 locally Minkowskian regions and no curvature anywhere, but you still get gravitational time dilation.

Yours is an interesting example because the regions are intuitively flat. I guess the Riemann tensor is zero at those locations? But for example a region just outside the hollow shell is also locally Minkowskian in the coordinates of a particle in free fall, and locally measurably "flat" to such a particle (but spacetime is not actually flat there and the Riemann tensor isn't zero).

I think I'm too stuck on confusion about the meaning of "locally Minkowskian," and trying to visualize it as horizontal on a diagram showing local coordinates, but then don't know how distant curved space could be shown.

Now I remember you mentioned the covariance of tensors to me before, and I looked up the components of the Riemann tensor and couldn't make sense of it (same as now!). I think I need to learn more basics before understanding curvature in different observers' coordinates.

[Markus Hanke]

2 hours ago, md65536 said:

Sure, but the gravitational time dilation should depend only on their relative gravitational potential (right?) and not how space between them curves to produce that difference in potential.

Correct - but with the caveat that the concept of ‘gravitational potential’ can only be meaningfully defined in certain highly symmetric spacetimes, such as Schwarzschild. It is not a generally applicable concept.

2 hours ago, md65536 said:

I guess the Riemann tensor is zero at those locations?

Indeed. It vanishes locally in those regions, but not globally.

2 hours ago, md65536 said:

But for example a region just outside the hollow shell is also locally Minkowskian in the coordinates of a particle in free fall, and locally measurably "flat" to such a particle (but spacetime is not actually flat there and the Riemann tensor isn't zero).

Yes, correct.

2 hours ago, md65536 said:

I think I'm too stuck on confusion about the meaning of "locally Minkowskian," and trying to visualize it as horizontal on a diagram showing local coordinates, but then don't know how distant curved space could be shown.

Think back to your math lessons in high school - remember how you drew simple graphs such as y=x^2. No question that the graph is globally curved. But now imagine you were to choose some point (eg x=2), and zoom into the graph there. What happens? The more you zoom in, the flatter it will begin to look. It’s just like that.

2 hours ago, md65536 said:

Now I remember you mentioned the covariance of tensors to me before, and I looked up the components of the Riemann tensor and couldn't make sense of it (same as now!). I think I need to learn more basics before understanding curvature in different observers' coordinates.

This takes a while to really get your head around. I’m sorry I won’t try to offer a proper answer here, as typing LaTeX code on an on-screen phone keyboard is just too cumbersome and time consuming. What I will say though is don’t focus on the components, but on what objects you pass to the tensor, and what you get out as a result. Rough outline:

Imagine you have two test particles, whose world lines are initially parallel. Now choose a point on one of these world lines - take the unit tangent vector at that point (which is physically just that particle’s 4-velocity). Then, still at that same point, take the perpendicular separation vector that connects it to the other particle’s world line.

Now imagine the Riemann tensor as a machine with four slots (the four indices). Input the tangent vector into slots 2 & 4, and the separation vector into slot 3; leave the first slot empty. The output of the machine then is a new vector (because we left one index open) - it tells you how fast the separation between the test particles begins to change, and in what direction (relative acceleration between the test particles). Writing this down in math notation immediately gives you the geodesic deviation equation, bearing in mind that we need to use a covariant, not ordinary, second derivative.

So the Riemann tensor is a machine that takes the tangent vector on one world line, and the separation vector between them as input; and produces as a result an acceleration vector that tells you how that separation between particles changes over time (geodesic deviation). This deviation can be a combination of any space and time direction, and can be complicated - you can get the world lines twisting around one another in a helix configuration, and all kinds of fancy stuff like that. 

This is really most of what there is to it in a GR context - Riemann has other uses as well (eg one can calculate tidal forces from it), but I won’t get into this here. It does in fact reflect all possible degrees of freedom of gravity. The individual components of the tensor represent tidal effects between various combinations of directions, but I really don’t think it’s helpful to try to look at it this way - it won’t help you understand. Better to think of it as a machine with slots that take an input, and produces an output; with the indices of the tensor being those slots. Any tensor can be conceptualised in that way - eg the Ricci tensor takes a future-pointing time-like unit vector into both slots, and produces a real number that is the rate at which a small volume changes when in free fall. In vacuum R(u,v)=0, so in vacuum a small volume in free fall is conserved (but its shape will get distorted, which is described by a different tensor). Inside a matter distribution, neither volume nor shape would be preserved. (Note carefully that this geometric interpretation only holds so long as there is no expansion, shear, or vorticity, which is true for most simple spacetimes)

Hopefully this makes any kind of sense to you.

Edited September 18, 2021 by Markus Hanke

[AbstractDreamer]

On 9/18/2021 at 9:23 AM, Markus Hanke said:

Think back to your math lessons in high school - remember how you drew simple graphs such as y=x^2. No question that the graph is globally curved. But now imagine you were to choose some point (eg x=2), and zoom into the graph there. What happens? The more you zoom in, the flatter it will begin to look. It’s just like that.

If you zoom OUT far enough, and assuming you zoom out at the same rate for the X and Y axes, then y=x^2 approaches a straight line.  Does this mean anything?  Is it possible that eventually that no global curvature can be measured from the equation?  How is it a global curvature if its not truly global and only local?

What does it mean if you zoom IN at different rates?  What if this rate itself was not a constant but another function of something.  Is it possible it may not look flatter?

Is there a loose analogy between the zoom-ratio of axes and tensors?

[studiot]

11 hours ago, AbstractDreamer said:

If you zoom OUT far enough, and assuming you zoom out at the same rate for the X and Y axes, then y=x^2 approaches a straight line.  Does this mean anything?  Is it possible that eventually that no global curvature can be measured from the equation?  How is it a global curvature if its not truly global and only local?

 

 

This is not actually true and demonstrates the difficulty with analogies.

 

The radius of curvature at any point is defined as the radius of a circle which exactly matches the curve at the point concerned.

The curvature is defined as the reciprocal of the radius of curvature.

For flat geometry or straight lines the curvature is zero.

 

If the curve is a circle then the radius of curvature is the same at every point.

If we now let that circle expand so that the radius tends to infinity, the reciprocal of that radius tends to zero ie straight or flat.

But for your parabola, the radius of curvature is different at every point and remains finite at all points between the vertex and infinity.

So its curvature is never zero, except at infinity and the vertex.

 

These two examples do nicely show the difference between local and global however.

For a circle, changing the radius affects every point on the circle equally.

That is global.

For a parabola the fact that the parabola radius of curvature approaches infinity as the parabola approaches infinity does not affect (the curvature of)  any of the points on the parabola which already have a finite curvature.

Curvature for a parabola is local to each and every point on it.

 

It's not zooming out that makes things approximately flat, it's zooming in.

 

 

 

 

 

Edited September 25, 2021 by studiot

[md65536]

On 9/24/2021 at 6:09 PM, AbstractDreamer said:

If you zoom OUT far enough, and assuming you zoom out at the same rate for the X and Y axes, then y=x^2 approaches a straight line.  Does this mean anything?

Well, the curvature of the parabola decreases with increasing positive x. There's an analogy for what you're talking about. "Local" effectively means at small enough distances that the effects of distance don't matter, and that's not a fixed size. At small x, you have to zoom in to a smaller area to make the parabola appear flat, and at large x, it can appear flat over a larger range. Analogously, near to a gravitational mass "local" might be very small, but very big much farther away. If you're free-falling near a black hole and being spaghettified, spacetime is still locally Minkowskian but at distances much smaller than a human.

[Markus Hanke]

2 hours ago, md65536 said:

If you're free-falling near a black hole and being spaghettified, spacetime is still locally Minkowskian but at distances much smaller than a human.

A pretty horrible way to kick the bucket. One can only hope that the BH isn’t very massive, so that it’s over quickly.

[beecee]

20 minutes ago, Markus Hanke said:

A pretty horrible way to kick the bucket. One can only hope that the BH isn’t very massive, so that it’s over quickly.

Yeah, but realitically, it would be a Kerr metric BH, and if one could judge one's trajectory to pass at the polar regions, and the exact centre, one should pass through the ring singularity  unharmed, to where though is anyone's guess! 😉

[Markus Hanke]

23 hours ago, beecee said:

Yeah, but realitically, it would be a Kerr metric BH, and if one could judge one's trajectory to pass at the polar regions, and the exact centre, one should pass through the ring singularity  unharmed, to where though is anyone's guess! 😉

Well, depending on the total mass of the BH you’d still get ripped to shreds long before you get anywhere near the singularity. It’s worse here even, since there’s also an angular component to the tidal forces - you’d be elongated lengthwise and pulled sideways 😳 This is a cosmic meat grinder!

[beecee]

10 hours ago, Markus Hanke said:

Well, depending on the total mass of the BH you’d still get ripped to shreds long before you get anywhere near the singularity. It’s worse here even, since there’s also an angular component to the tidal forces - you’d be elongated lengthwise and pulled sideways 😳 This is a cosmic meat grinder!

There goes my dreams! 😉

How big a BH would we need? The typical SMBH at the core of galaxies?

 

[Markus Hanke]

13 hours ago, beecee said:

There goes my dreams! 😉

How big a BH would we need? The typical SMBH at the core of galaxies?

 

You’d need a situation where there’s enough angular momentum so that the event horizon sort of wraps around the ring singularity, like a donut, leaving a hole in the middle. This is theoretically possible, but I don’t know if any real-world BH ever gets to that point (I’d say not). I also don’t know how tidal forces would behave in such a scenario. My feeling is that you’d have to cross precisely through the Center of the ring, or else you’d get ripped apart and pulled into the horizon. This is a very unstable situation, and probably not realistically doable.

[beecee]

17 minutes ago, Markus Hanke said:

 My feeling is that you’d have to cross precisely through the Center of the ring, or else you’d get ripped apart and pulled into the horizon. This is a very unstable situation, and probably not realistically doable.

I won't argue with that!

[MigL]

1 hour ago, Markus Hanke said:

You’d need a situation where there’s enough angular momentum so that the event horizon sort of wraps around the ring singularity, like a donut, leaving a hole in the middle.

IIRC, when there is enough angular momentum, the event horizon disappears, leaving a 'naked', ring singularity.

" When {\displaystyle r_{s}/2<a} (i.e. {\displaystyle GM^{2}<Jc}), there are no (real valued) solutions to this equation, and there is no event horizon. With no event horizons to hide it from the rest of the universe, the black hole ceases to be a black hole and will instead be a naked singularity."

From       Kerr metric - Wikipedia

AS for the OP, the 'rubber sheet with a bowling ball on it' is a visual model, and just like all models, even mathematical ones, it is not applicable in all situations, and some predictions are non-sensical.
Just like GR, our best model, makes some non-sensical predictions, the visual rubber sheet model makes many non-sensical predictions. 
Such as the 'force' pulling the rubber sheet down.
Or that the mass is not intrinsic to the spacetime, but sits above it.
Or that there is an 'exterior' to space-time.
Or that the warping of spatial dimensions is responsible/equivalent to gravity.
Or many others where it ceases to be applicable.

[Markus Hanke]

32 minutes ago, MigL said:

IIRC, when there is enough angular momentum, the event horizon disappears, leaving a 'naked', ring singularity.

Yes, that’s right. Just before that happens, it will wrap itself around the ring, like a donut.

Note that in Kerr spacetime there are a number of additional important surfaces, other than the event horizon. Also, the region below the EH is highly unstable, so any deviation from true vacuum will make this decay into a different type of spacetime. It’s an idealised solution.

2 hours ago, beecee said:

I won't argue with that!

But there’s a consolation: if recent results in quantum gravity (ref Netta Engelhardt etc) prove correct, then BHs do have hair after all; in fact they would sport some thick, luxurious fur! So it might (proof outstanding) be possible to reconstruct your identity from information encoded in Hawking radiation, after your in-fall (the concept is called ‘Python’s lunch’). Not only would that allow the world to know your fate, but it also preserves unitarity that way, and thus resolves the information paradox. The caveat is that the decoding algorithm requires full knowledge of all quantum gravity DoF, and may even then not be computable with any conceivable real-world computer.

All of this is work in progress at the moment.

[beecee]

1 hour ago, Markus Hanke said:

Yes, that’s right. Just before that happens, it will wrap itself around the ring, like a donut.

Note that in Kerr spacetime there are a number of additional important surfaces, other than the event horizon. Also, the region below the EH is highly unstable, so any deviation from true vacuum will make this decay into a different type of spacetime. It’s an idealised solution.

I recall the argument/bet between Hawking and Thorne on naked singularities, so what is it in simple language that stops the angular momentum of a BH reaching a stage where the singularity can be revealed? 

And which EH are we talking about? that EH enshrining the ergosphere? which theoretically at least, we would be able to extract energy from. 

Obviously they are incredibly complicated beasts when compared to the Schwarzchild metric, yet obviously are also the only real stable form of a BH...I think! charge is always negated and angular momentum obviously is the relic from the star from which it formed, which also can be negated but also added on to...are those recollections correct? Other aspects adding to their complication are the cauchy horizon [of which I know little about] and the two photon spheres operating in reverse situations to each other...complicated indeed!!

Edited September 29, 2021 by beecee

[Markus Hanke]

46 minutes ago, beecee said:

so what is it in simple language that stops the angular momentum of a BH reaching a stage where the singularity can be revealed? 

Mass, angular momentum, and charge of a BH are not entirely free parameters, but they are linked via the concept of ‘irreducible mass’ - there’s an equation linking these, which I won’t typeset here. The point is that for a given mass, there’s an upper limit to angular momentum; rendering the singularity naked would need to exceed this limit. 

57 minutes ago, beecee said:

And which EH are we talking about? that EH enshrining the ergosphere? which theoretically at least, we would be able to extract energy from. 

The EH (herein aka ‘outer horizon’) is the next horizon inside the ergosphere, and these two coincide at a point on the rotation axis, above the pole.

1 hour ago, beecee said:

yet obviously are also the only real stable form of a BH...I think!

Spacetime is stable most of the way to the Cauchy horizon (‘inner horizon’), but then becomes highly unstable.

 

1 hour ago, beecee said:

charge is always negated and angular momentum obviously is the relic from the star from which it formed, which also can be negated but also added on to...are those recollections correct?

In Kerr spacetime there’s only mass and angular momentum. If you add charge as well, then you are in Kerr-Newman spacetime.

1 hour ago, beecee said:

cauchy horizon [of which I know little about]

It’s a causality boundary - above the Cauchy horizon ordinary causality always holds, but below it causal structure becomes non-trivial; this is the region where you find closed timelike curves.

[beecee]

49 minutes ago, Markus Hanke said:

In Kerr spacetime there’s only mass and angular momentum. If you add charge as well, then you are in Kerr-Newman spacetime.

Yep, understood, what I was trying to say is that any Kerr Newman BH, would rather quickly become just a Kerr BH, as charge would be quickly negated...is that correct?  and therefor the Kerr metric would be in reality the most common?

Thanks for tidying up my understanding btw.