# Black holes and evaporation

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2 hours ago, md65536 said:

I'd say that causal connections are invariant. I think light cones are invariant.

This is a matter of fact, and I'm baffled that we're even on this subject. All of GR is designed to be coordinate invariant, and the ordering of events is obviously an absolute feature of physics. If different results are given in different analyses it's probably because they have altered the time and space definitions for mathematical convenience.

An observer hovering just above the event horizon would witness the infinite future, but an infalling observer has his gravitational effects counter-balanced by his relative velocity, and the net result is a redshifting. If memory serves, he would see distant clocks ticking "at half speed" as he passes the event horizon.

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11 hours ago, rjbeery said:

Some of these comments are presuming the existence of the event horizon, first, and then leaning on the definition of them as a defending explanation of them.

I am not using the definition you quoted, I am looking at the geodesic structure of spacetime instead. Wald’s definition is correct of course, but it’s only one particular way to define event horizons, and in my opinion not necessarily the best one.
The existence of event horizons is trivially necessary, even in old Newtonian gravity without any GR effects. In the context of GR specifically, the way to show this is to look at the set of all possible solutions to the geodesic equation, given a spacetime. What you will find is that, when a gravitational collapse takes place, spacetime gets divided into two distinct regions (let’s stick to Schwarzschild for simplicity for the moment) - one where the geodesic equation admits solutions that extend to infinity, and one where all solutions become geodesically incomplete.The boundary between these two regions is what we call an ‘event horizon’. This is true whether or not Hawking evaporation takes place, and is in fact inevitable given the process of gravitational collapse. This is mathematical fact, and not in dispute.

12 hours ago, rjbeery said:

Please refer to the diagram to see my objection.

It’s a standard Penrose diagram for an evaporating BH, I do not see anything in there that would support a conclusion such as the one you have drawn. At the moment I am still trying to figure out how you could possibly have arrived at your conclusion, because your reasoning doesn’t make any sense to me at all.

12 hours ago, rjbeery said:

The formation of event horizons takes an infinite amount of coordinate time.

Sure, that is because coordinate time is what a distant observer records on his own clock. It is not what a clock located at the horizon itself will record. Time in GR is a purely local concept, so in order to examine the physics of this, you need to use a clock that is located at or at least near the horizon. The Schwarzschild coordinate chart does not cover the horizon nor the interior region, it covers only the exterior spacetime. If you want to cover the entire spacetime, you need to use a coordinate map such as e.g. Kruszkal-Szekeres. A hypothetical observer comoving with the expanding horizon will always record a finite time for the formation process.

12 hours ago, rjbeery said:

When we attempt to use a Penrose diagram to illustrate events after an event horizon has formed, we are now showing spacetime coordinates (events) which have already occurred, and have been accounted for. (r=0, t=100) is represented twice in the diagram, for example.

And here is where I don’t follow your argument at all. Penrose diagrams are not embedding diagrams of the metric, they reflect only causal relationships between events, but not their spacetime coordinates. How could they? It’s a conformal diagram after all. That’s why you never see any Penrose diagrams drawn with axis that have a scale on it. There is a conformal equivalence between points on the diagram and the metric (meaning it accurately reflects the causal relationships between events in spacetime), but certainly not a coordinate equivalence. So of course you can have the same (r,t) at multiple locations on the drawing - that is the entire point of Penrose diagrams, because showing those causal relationships in terms of spacetime coordinates is much more difficult.

Here’s a simpler version of the same diagram:

So exactly what is your objection here?

By the way, it needs mentioning that evaporating black holes are of course not Schwarzschild black holes, since Schwarzschild spacetime is static. They are Vaidya black holes. Both spacetimes are of Petrov type D, but Vaidya spacetime does not admit any time-like Killing vector fields, so it isn’t a Weyl spacetime. Their geometries are thus different. It is important to remember this.

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16 hours ago, rjbeery said:

Agreed, but I'm talking about a specific observer who is located at a point in space which "used to" contain a black hole.

One problem is that you seem to be using a "common sense" concept of location when you talk about where the black hole used to be. That location (in formal terms) is not the same as the the location when the black hole was there. They are described by different coordinates.

Also, (spatial) locations are not the issue. It is an event horizon. So you need to consider events. No event inside the black hole is accessible even if the black hole evaporates.

Quote

The black hole resides in the past of this observer. If you can't see the contradiction here, please refer to the diagram.

The black hole does. The interior of the black hole doesn't.

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15 hours ago, Mordred said:

Agreed but let's ask a question. Where is the null surface located ? Under one coordinate choice its r_s= 2GM. However this isn't true for the Kruskal.

Now ask yourself is it the spacelike or the time like that are invariant under the Lorentz transforms ?

This question becomes important to understand the region's of the Penrose diagrams.

Here is an examination of different causal connections in different coordinate systems the article is specifically dealing with Penrose diagrams.

[...]

The null surface is the event horizon? Isn't it located at the same place in different coordinates, just with different numerical representations in the different coordinate systems?

What are you asking is invariant? There are aspects of spacelike and timelike things that are relative, and other aspects that are invariant, right?

Mordred, can you please answer my previous question so that I know I'm not just wasting my time here? Is rapidity some kind of acceleration?

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Here

Proper acceleration (the acceleration 'felt' by the object being accelerated) is the rate of change of rapidity with respect to proper time (time as measured by the object undergoing acceleration itself).

However rapidity isn't strictly acceleration. You can have other relations that can apply rapidity. The link gives a few examples.

Edited by Mordred
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4 hours ago, md65536 said:

The null surface is the event horizon? Isn't it located at the same place in different coordinates, just with different coordinate systems.

Unfortunately this is not true in the case of a coordinate singularity such as the EH.

4 hours ago, md65536 said:

What are you asking is invariant? There are aspects of spacelike and timelike things that are relative, and other aspects that are invariant, right?

A coordinate singularity is not invariant under coordinate change. The r_s=2GM is an artifact of the Schwartzchild metric.

I'm going to add a hypothetical question. Would a near c observer see the same radius for the event horizon as the at rest observer. You would see a different Blackbody temperature and as a result a different rate of Hawking radiation. Ie Unruh effect.

(PS the answer cannot rely on the Schwartzchild metric ). One can argue the Schwartzchild metric is only suitable to a far away observer.

Edited by Mordred
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2 hours ago, Mordred said:

Unfortunately this is not true in the case of a coordinate singularity such as the EH.

A coordinate singularity is not invariant under coordinate change. The r_s=2GM is an artifact of the Schwartzchild metric.

I'm going to add a hypothetical question. Would a near c observer see the same radius for the event horizon as the at rest observer. You would see a different Blackbody temperature and as a result a different rate of Hawking radiation. Ie Unruh effect.

(PS the answer cannot rely on the Schwartzchild metric ). One can argue the Schwartzchild metric is only suitable to a far away observer.

Wouldn't the radius shrink and mass increase proportionally?

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2 hours ago, Mordred said:

Unfortunately this is not true in the case of a coordinate singularity such as the EH.

A coordinate singularity is not invariant under coordinate change. The r_s=2GM is an artifact of the Schwartzchild metric.

Doesn't the null surface correspond with the event horizon, which has the same physical significance (re. light-like intervals, causality, etc.) regardless of coordinates? Whether or not the event horizon at the Schwarzschild radius is a singularity depends on choice of coordinates, but that doesn't determine its existence or behaviour. It's still an event horizon.

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4 hours ago, Endy0816 said:

Wouldn't the radius shrink and mass increase proportionally?

4 hours ago, md65536 said:

Doesn't the null surface correspond with the event horizon, which has the same physical significance (re. light-like intervals, causality, etc.) regardless of coordinates? Whether or not the event horizon at the Schwarzschild radius is a singularity depends on choice of coordinates, but that doesn't determine its existence or behaviour. It's still an event horizon.

Good answers to you both however the correct answer gets more complex.

@md65536 the Schwartzchild metric though it is still a horizon breaks down on different observers. The coordinate system cannot accurately describe the causal nature of events.

@Endy0816 the radius of the null surface would change afiak however the mass determines the radius so it would not be an inverse relation. However the null surface isn't necessarily the event horizon..... (more on that later) ( I could very well be wrong on this)

This is where we can finally get back to the Ops misunderstanding on Hawking radiation.

There is another horizon called the Apparent horizon which represents the trapped surface. This trapped surface may or may not coincide with the event horizon described by R_s=2GM.

Here is an article covering this.

Now the Apparent horizon definitely changes according to the observer. I'm unclear if the Schwartzchild event horizon does due to its limitations of valid observers and being a coordinate singularity that doesn't truly describe causality for all observers. (Future past lightcones  etc)

On 6/18/2020 at 9:53 PM, rjbeery said:

I'm not bothered by the notion of "no causal connection" but you cannot say that an event within the black hole "doesn't exist in the coordinate space". That's nonsensical. The entire point of this coordinate system is to give unique, well-ordered labels to events in spacetime.

Let's readdress this question now that we are looking at apparent horizons being involved in causal relations.

@rjbeery Don't get discouraged the Penrose diagrams are extremely confusing with regards to causal connections even with lightcones. Hawking radiation is described in more detail in the last link but note which region it occurs.

@everyone involved I am far more familiar with cosmology and particle physics than black hole dynamics. Lol unlike many I studied them to a certain extent but it's never been a focal point in my studies.

🤔

Anyways here is another article covering horizons in regards to BH's

Black hole Boundaries.

Edited by Mordred
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10 hours ago, Mordred said:

A coordinate singularity is not invariant under coordinate change. The r_s=2GM is an artifact of the Schwartzchild metric.

True, which is why it is better to define the event horizon in terms of the geodesic structure of the spacetime in question.

10 hours ago, Mordred said:

I'm going to add a hypothetical question. Would a near c observer see the same radius for the event horizon as the at rest observer. You would see a different Blackbody temperature and as a result a different rate of Hawking radiation. Ie Unruh effect.

If there is relative motion at relativistic speeds between observer and black hole, then we need to label events in that spacetime in a different way. The relevant metric for this case is called the Aichlburg-Sexl Ultraboost. In these coordinates the black hole is not spherically symmetric, so the horizon area is different, and hence also the Hawking temperature (which is consistent with the Unruh picture of this scenario). Generally speaking, temperature is an observer-dependent quantity.

It should be noted though that we are still in the same physical spacetime - Schwarzschild and Aichlburg-Sexl are related by a (more or less) simple coordinate transformation, i.e. the observer just labels events in that same spacetime differently, from his own point of view. And of course it needs to be that way, because obviously relative motion cannot affect the geometry of spacetime.

6 hours ago, Mordred said:

I'm unclear if the Schwartzchild event horizon does due to its limitations of valid observers and being a coordinate singularity that doesn't truly describe causality for all observers.

The event horizon itself is not merely an artefact of coordinates, it originates in the geodesic structure of the spacetime in question, which is of course something all observers agree on. What is an artefact of the coordinates is the (coordinate) singularity, which, in the Schwarzschild metric, happens to coincide with the physical location of the event horizon. There is no physical reason that this must be so, other than the boundary conditions of this particular metric. Once could simply choose a different coordinate chart, which eliminates the singularity, but leaves the event horizon. Spacetime itself is smooth, continuous, and geodesically complete at the horizon, which is easily shown by looking at the curvature invariants of the Riemann tensor.

Below the event horizon, there are no stationary rest frames - all geodesics are incomplete, i.e. must terminate at the singularity. This is true regardless of what coordinate chart you use to cover this spacetime, and the horizon is always at the same physical location (which of course will have different coordinate labels in different charts).

10 hours ago, Mordred said:

One can argue the Schwartzchild metric is only suitable to a far away observer.

Yes, it is the “point of view” (so to speak) of an observer at rest at asymptotically flat infinity. Schwarzschild coordinates also cover only parts of this spacetime, namely the exterior region above the horizon, excluding the horizon itself and everything below it. If one needs to examine spacetime at or below the horizon (in the classical picture of course), a different coordinate chart is needed, usually Kruszkal-Szekeres, or Gullstrand-Painleve, depending on the problem at hand.

Edited by Markus Hanke
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15 minutes ago, Markus Hanke said:

If there is relative motion at relativistic speeds between observer and black hole, then we need to label events in that spacetime in a different way. The relevant metric for this case is called the Aichlburg-Sexl Ultraboost. In these coordinates the black hole is not spherically symmetric, so the horizon area is different, and hence also the Hawking temperature (which is consistent with the Unruh picture of this scenario). Generally speaking, temperature is an observer-dependent quantity.

As the lifetime of the black hole is related to temperature, this means that the lifetime is observer dependent. Which shouldn't be surprising.

But, is that change in lifetime (and temperature) consistent with a simple calculation just using SR for time dilation? (It feels implausible that it would be but essential that it should be!)

Also, an observer in free fall does not see Hawking radiation, and so does not see the black hole evaporate. Does this mean that, from the point of view of an external observer, someone in free fall towards a black hole must always reach the event horizon before it disappears, however fast the black hole is evaporating?

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1 hour ago, Strange said:

But, is that change in lifetime (and temperature) consistent with a simple calculation just using SR for time dilation?

My earlier comments related specifically to Schwarzschild black holes - which are stationary, hence they don’t evaporate. Generalising this to the case of an evaporating black hole is not straightforward or trivial. Such black holes are called Vaidya black holes, and this represents an entirely different solution to the field equations, and thus a spacetime with a different geometry. While Schwarzschild is a vacuum solution, Vaidya is not, because now all of spacetime is filled with radiation, so we are no longer in a vacuum.

Adding relative motion between observer and a Vaidya black hole does of course not yield the aforementioned Aichlburg-Sexl ultraboost, but some different metric, which I have not encountered in my studies. Like in the Schwarzschild case, that new metric would be related to the Vaidya metric by some coordinate transformation, but I suspect the transformation will be a lot more complex than in Schwarzschild spacetime, mostly due to the presence of off-diagonal terms in the metric tensor. I’m sure it can be done though, and probably has been, though a quick search does not immediately turn up anything.

But the answer to your question will be “no” regardless, because in both cases we are in a curved spacetime, so you can’t naively use the transformation rules of SR.

1 hour ago, Strange said:

(It feels implausible that it would be but essential that it should be!)

The total time dilation here has a kinetic component from relative motion, and a gravitational component from spacetime curvature. So the lifetime a specific observer calculates for the black hole will be subject not only to his relative speed, but also to the particulars of the geodesic he traces out (i.e. to the initial and boundary conditions of his motion), and where on this geodesic he is when he performs the calculation; so it will be some (probably quite complicated) function of his trajectory and the surface area of the horizon (which will itself be some complicated function, since the horizon is no longer spherical for such an observer).

1 hour ago, Strange said:

Also, an observer in free fall does not see Hawking radiation, and so does not see the black hole evaporate.

This is a common misunderstanding based on the fundamental error of using the wrong solution to the field equations for the scenario at hand. I know that countless papers have been published showing this calculation, but ultimately these results are not physically meaningful.

If you naively go and consider free-fall observers in Schwarzschild spacetime, then yes, the maths will show you that they can’t detect Hawking radiation. However, evaporating black holes are by definition not Schwarzschild, so this point is moot. When done correctly using Vaidya spacetime, the free fall observer will definitely detect radiation (it is a non-vacuum spacetime after all!), just at a different temperature relative to other observers. The type of physical spacetime one is in is characterised by curvature invariants, which is something all observers always agree on, even if they use different coordinate charts to map that spacetime.

1 hour ago, Strange said:

Does this mean that, from the point of view of an external observer, someone in free fall towards a black hole must always reach the event horizon before it disappears, however fast the black hole is evaporating?

Like I said, one must use the correct spacetime to model this scenario. How in-fall time is related to evaporation time of a black hole in Vaidya spacetime is a question I can’t answer off hand. It would likely depend on the initial mass of the black hole, its age (as calculated by the in-falling observer), and the in-fall geodesic. For solar-mass (at the time of in-fall) black holes though the lifetime of the black hole will be much longer than the length of most in-fall geodesics, by many orders of magnitude.

Edited by Markus Hanke
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54 minutes ago, Markus Hanke said:

My earlier comments related specifically to Schwarzschild black holes ...

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Thank you for the clarifications Markus I had looked at the Vaidya spacetime before. Though it has been awhile however I had never heard of the Aichlburg-Sexl Ultraboost. Rather if I have I can't recall it. As I previously stated my studies on BH metrics are lacking compared to my primary expertise in Cosmology studies.

So many of these other coordinate systems I rarely look into. So I am glad for the clarifications +1

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5 hours ago, Markus Hanke said:

My earlier comments related specifically to Schwarzschild black holes - which are stationary, hence they don’t evaporate.[...]

+1. Thank you. That's a great answer.

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Hrrm the Aichlburg-Sexl Ultraboost is rather tricky to find good articles on it. Though those I found do show the derivatives via the transformation rules. The log function I would like to get more clarity on. ( Though it makes sense to a degree)

Do you have a good source ?

Edited by Mordred
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5 minutes ago, Mordred said:

Hrrm the Aichlburg-Sexl Ultraboost is rather tricky to find good articles on it.

Aren't they a Bachman-Turner Overdrive tribute act?

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6 hours ago, Markus Hanke said:

But the answer to your question will be “no” regardless, because in both cases we are in a curved spacetime, so you can’t naively use the transformation rules of SR.

You're speaking of a general case, but if you compared two observers in flat spacetime (eg. at infinity, or in the location of a single event), wouldn't it have to be the case? I didn't see it specified what observers are being compared (eg. infalling vs. one at infinity), but if you're given the choice of which observers to compare, it should be possible?

8 hours ago, Strange said:

Also, an observer in free fall does not see Hawking radiation, and so does not see the black hole evaporate. Does this mean that, from the point of view of an external observer, someone in free fall towards a black hole must always reach the event horizon before it disappears, however fast the black hole is evaporating?

I think this is related to https://en.wikipedia.org/wiki/Black_hole_complementarity

I don't think the issue is settled in accepted science. However, seeing the infalling object reach the event horizon doesn't make sense (unless the BH has zero size). Maybe the object evaporates just like the BH does. Maybe the object simply fades out of existence as it's infinitely red-shifted to nothing.

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15 hours ago, Mordred said:

Hrrm the Aichlburg-Sexl Ultraboost is rather tricky to find good articles on it. Though those I found do show the derivatives via the transformation rules. The log function I would like to get more clarity on. ( Though it makes sense to a degree)

Do you have a good source ?

Unfortunately I don’t, sorry. This is not a metric that is often discussed in textbooks - actually, come to think of it, I have never seen it mentioned in any textbook used for teaching. I can’t remember where I came across it first, I think it might have been in Exact Solutions to Einstein’s Field Equations by Stephanie/Kramer.
And I just noticed that I misspelled it, this should be Aichelburg-Sexl Ultraboost.

15 hours ago, Strange said:

Aren't they a Bachman-Turner Overdrive tribute act?

No, it’s the latest 17-in-1 vaccine being pushed on the Austrian populace, for the purpose of easier mind control via 5G towers, by their Nazi overlords (who are really Lizard People) from the hollow interior of flat earth.

14 hours ago, md65536 said:

You're speaking of a general case, but if you compared two observers in flat spacetime (eg. at infinity, or in the location of a single event), wouldn't it have to be the case?

Yes indeed, in Minkowski spacetime they would need to be subject to the SR transformation rules.

14 hours ago, md65536 said:

I didn't see it specified what observers are being compared (eg. infalling vs. one at infinity), but if you're given the choice of which observers to compare, it should be possible?

The comparison was between an observer who is at rest relative to the black hole, and one who is in relative motion at relativistic speeds. The comparison via SR rules is very tricky here, because Vaidya spacetime is not a vacuum - it is uniformly filled with null dust (i.e. radiation), so the spacetime is not asymptotically flat. However, the discrepancy may (!) be small enough to neglect if both observers are very far from the black hole, depending on what degree of accuracy is demanded. I haven’t run the numbers though, hence I may well be wrong on this. Technically speaking though, in Vaidya spacetime there are no regions that can be considered Minkowski.

14 hours ago, md65536 said:

However, seeing the infalling object reach the event horizon doesn't make sense (unless the BH has zero size).

An external observer would not see (as in - visually witness) the infalling object reach the horizon, I think it would fade out and slow down just like in Schwarzschild spacetime, albeit at a different rate. What I am not sure about though is what the external observer would calculate in terms of coordinate infall-time on his own local clock, since Vaidya coordinate time is not the same as Schwarzschild coordinate time. I just tried a quick calculation for this, but it turns out that due to the presence of an off-diagonal term in the metric, this is actually very difficult. I will need to come back to this.

Edited by Markus Hanke
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On 6/16/2020 at 12:12 AM, rjbeery said:

Quantum mechanics is anticipated to resolve any mathematical singularity issues at the center of black holes

I'm not aware of this.

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While Markus' instincts advocate 'following the math', one must also realize that sometimes the math leads to unphysical solutions, because of limitations with the model.
One such is obviously the singularity, which our current theory cannot handle.
Solutions such as the Schwarzschild, are unphysical because no BH is expected to be non-rotating; even the slightest rotation will be greatly increased as it collapses. The Reissner-Nordstrum solution is also nonphysical; since the accretion disc consists of ionized plasma, any charge the BH has will quickly be neutralized by infalling opposing charges, and repulsion of like charges.
Another non-physical aspects of both the Kerr ( rotating ) and Reissner-Nordstrum solutions are that for certain rotation speeds, and charge, their dual Event Horizons can disappear leaving 'naked singularity' ( which is as un-physical as you can get ).

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On 6/16/2020 at 12:12 AM, rjbeery said:

There are a variety of theories, most notably Hawking Radiation

I don't want to be a nitpicker, but Hawking's radiation is not a theory. It's an extremely clever mix-and-match semiclassical calculation, using QFT in curved backgrounds, that predicts a radiation spectrum for a black hole.

The thing about GR is that it's extremely solid. As Markus has told us (I wasn't aware of it) there is an exact solution of GR that includes the possibility of Hawking radiation (which historically was obtained using a living-dangerously kind of mixed reasoning!).

I don't think GR is completely free of problems. The most obvious one is the existence of singularities. If you want to improve GR, you must think in a perpendicular direction, so to speak. IOW, you must look for proper generalizations, rather than denial.

Another source of confusion: The Penrose diagram that you're showing patches up space time to have a BH that has a birth, as the Schwarzschild solution is eternal. So Penrose's patching up plugs in the history of the collapsing star in the past, which is not present in the Schwarzschild solution.

I'm also very interested in this:

On 6/16/2020 at 12:12 AM, rjbeery said:

Quantum mechanics is anticipated to resolve any mathematical singularity issues at the center of black holes

How?

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He might mean 'Quantum Gravity' … "is anticipated to resolve any mathematical singularity issues at the center of black holes"

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14 hours ago, MigL said:

While Markus' instincts advocate 'following the math', one must also realize that sometimes the math leads to unphysical solutions, because of limitations with the model.
One such is obviously the singularity, which our current theory cannot handle.
Solutions such as the Schwarzschild, are unphysical because no BH is expected to be non-rotating; even the slightest rotation will be greatly increased as it collapses. The Reissner-Nordstrum solution is also nonphysical; since the accretion disc consists of ionized plasma, any charge the BH has will quickly be neutralized by infalling opposing charges, and repulsion of like charges.
Another non-physical aspects of both the Kerr ( rotating ) and Reissner-Nordstrum solutions are that for certain rotation speeds, and charge, their dual Event Horizons can disappear leaving 'naked singularity' ( which is as un-physical as you can get ).

This is all true. In addition, and perhaps more critically, all these solutions rely on the assumption of asymptotic flatness, i.e. that the respective black hole exists in an otherwise completely empty universe. This is obviously not very physical.
Nonetheless, these are all useful approximations to simplify the math for special cases.

7 hours ago, MigL said:

He might mean 'Quantum Gravity' … "is anticipated to resolve any mathematical singularity issues at the center of black holes"

The issue of singularities is easily resolved by a simple modification of GR, without the need to involve any quantum physics at all. One must simply remove the requirement that the connection be torsion-free, which immediately yields Einstein-Cartan gravity. This model makes the same predictions as conventional GR for vacuum spacetimes, but it now does include a coupling between torsion and intrinsic angular momentum for the interior of mass-energy distributions. It can be shown that gravitational collapse under this model no longer leads to regions of geodesic incompleteness (i.e. there are no singularities); also, the cosmological singularity at the BB no longer occurs. Note that just like GR, this is a purely classical model.

Edited by Markus Hanke
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I fully agree with Markus on one of the better torsion included models

The Einstein Cartan method is one the better ways to understand torsion in a field. Works well with the Godel universe (with the additive of commoving coordinates.

Once you introduce rotation of a BH. You will get multiple horizons. Understanding causal zones can get quite illusive. Methods like the Penrose diagrams are a good aid.

Edited by Mordred
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