Talk:Kerr metric

Image Showing Static Limit and Event Horizon
The showing the static limit and (outer) event horizon appears twice. Moreover, in its second incarnation, the caption incorrectly refers to the outer event horizon as the static limit. I'm not bold enough to decide which copy should go. Certainly one must, and the correct caption used. Warrickball (talk) 23:09, 13 March 2008 (UTC)

Objection to introduction
Does anyone else object to this section of the text, or is it just me?

"Such black holes have two event horizons where the metric appears to have a singularity. The outer horizon encloses the ergosphere and has an oblate spheroid shape, a flattened sphere similar to a discus. The inner horizon is spherical and marks the "radius of no return"; objects passing through this radius can never again communicate with the world outside that radius."

I would say it's both wrong and misleading. Firstly, the outer surface that encloses the ergosphere is the "stationary limit surface" and not an event horizon (it even says so explicitly in Hawking and Ellis). Secondly, I would say that the metric exists independently of which coordinates you use to describe it and hence whether it appears to have a singularity in a given coordinate system or not isn't really relevant. In Boyer-Lindquist coordinates the metric is not singular at the stationary limit surface/boundary of ergosphere and it is singular at the true singularity (which isn't a horizon). Thirdly, if we are going to call something the "inner horizon" then we really ought to be talking about the inner horizon as the inner Cauchy horizon. I believe that is standard usage - it is where I come from.

--Eujin16 (talk) 05:17, 6 December 2007 (UTC)

I would add a new aspect to the introduction. "quasispherical event horizon" has really no meaning. The event horizon is spherical. Who has ever heard the word "quasispherical event horizon" should come with a source otherwise I would suggest that it should be changed. — Preceding unsigned comment added by 200.120.147.186 (talk) 18:50, 6 October 2020 (UTC)

The Boyer/Lindquist coordinate chart metric is wrong
I am certain that the metric as given in the section "Boyer/Lindquist coordinate chart" is wrong. I think that the $$r^2+a^2$$ [in $$(r^2+a^2)\sin^2\theta d\phi^2 $$] needing to be squared and divided by &rho;2 is the only issue, but I can't be 100% sure without a chance to work it all through.

I admit that I could be wrong, but I am certain enough that there is a problem here that I would rather have the red flag thrown back in my face than let this go unchallenged. --EMS | Talk 17:08, 27 July 2005 (UTC)


 * Why are you certain? I didn't write this part of the article, but whoever did just copied (5.29) from Hawking & Ellis or (21.1) in Stephani or (8.32) in Ohanian & Ruffini, or the same expression as given in some other standard textbook.  Even better, long years ago, I verified that this does give a vacuum solution.


 * I guess the confusion might have arisen if you have seen the terms in the BL line element collected in a different way (I'll give an example below), but all expressions for the BL line element should agree if you multiply everything out. Another possibility is that you saw the line element in a chart which looks like BL but isn't (for example, de Felice and Clarke introduce a "rotating" BL chart before they get to the standard BL chart).


 * This "controversy" may be moot since I've been planning to rewrite the article anyway to discuss the Kerr vacuum in much greater detail using various coframes. For example, from the line element as given in (19.27) in D'Inverno or (217) in Chandrasekhar we can read off the following coframe:


 * $$\sigma^0 = -\frac{\delta}{\rho} \, \left( dt + a \sin(\theta)^2 d\phi \right)$$
 * $$\sigma^1 = \frac{\rho}{\delta} \, dr$$
 * $$\sigma^2 = \rho \, d\theta$$
 * $$\sigma^3 = \frac{\sin(\theta)}{\rho} \, \left( -a \, dt + (r^2 + a^2) d\phi \right)$$
 * where
 * $$ \delta^2 = r^2 - 2 m r + a^2, \; \; \rho^2 = r^2 + a^2 \cos(\theta)^2$$
 * Then, the metric is
 * $$ -\sigma^0 \otimes \sigma^0 + \sigma^1 \otimes \sigma^1 + \sigma^2 \otimes \sigma^2 + \sigma^3 \otimes \sigma^3$$
 * This is one of the simpler coframes, but several others are of great interest. For example, the LNR coframe is given in the textbook by de Felice & Clarke.  The Doran coframe is a generalization of the LeMaitre coframe for Schwarzschild.  Other charts in common use include Eddington charts, Doran chart (generalization of Painleve chart), Plebanksi chart, Weyl canonical chart, the original Kerr chart, etc., and the coframes can be written in all of these charts.  So you might see many different ways of writing down the Kerr solution.---CH  (talk) 22:49, 27 July 2005 (UTC)


 * I don't think that it was copied correctly. I come up with $$g_{\phi\phi} = \left [ \left (r^2 + a^2 \right ) + 2mra^2/\rho \right] \sin^2 \theta$$ from the metric as given in the article.  That is not correct. --EMS | Talk 23:46, 27 July 2005 (UTC)
 * Ah ha! Now I see what I was missing:  The $$sin^2 \theta$$ was squaring also.  Now it works out OK.  This just is not a form of Kerr that I am used to.  The coframe version and the spelled-out versions are ones that I familiar with.  --EMS | Talk 23:54, 27 July 2005 (UTC)

Sorry for the confusion
I was exhausted yesterday, and did not realize just how tired I was. I just cannot do any intellectual "heavy lifting" when I'm like that. I knew that $$(r^2 + a^2)^2/\rho^2$$ was involved with $$g_{\phi\phi}$$. I had lost track of the fact that it was involved with the coframe version and sorts itself out into a form such as is in the article.

So I was tired and irritated (due to other things) and was not able to respond to my own self-red-flag of "I know I could be wrong". --EMS | Talk 22:03, 28 July 2005 (UTC)

Roy P. Kerr
Hi, Taxman, don't worry, I am certain the initial is indeed "P."---CH (talk) 20:11, 5 August 2005 (UTC)

Proposed biographical subcategory
There are many professional biographies on leading figures which have appeared in the journals over the years. I am pretty sure I have seen one on Kerr, for example (can't recall where). Since I have my hands full with writing more technical articles, would someone who lives near a good physics research libraray be interested in systematically searching for such biographies, looking them up, and writing wikibiographies? The biographical articles should cite the published biography as a source. I see someone has written a brief biography of Roy P. Kerr; to avoid cluttering up a category (this one) which is already large (and I plan to enlarge it considerably), I have created the new subcategory "Contributors to general relativity".

To Merge or not to Merge?
I'm against merging rotating black hole with this article. They are parent subjects, not the same one. The content of these articles are enough different to support so. nihil 09:25, 25 November 2005 (UTC)


 * Nihil may have a point. The "no hair theorem" is or should be surprising, and Price's theorem is often summarized by saying that any inhomogeneties or anisotropies outside a black hole which can be radiated away, will be radiated away (which restores the Kerr exterior vacuum of a relaxed rotating black hole).  So, perhaps the article Rotating black hole can discuss these facts, while referring to Kerr vacuum for the relaxed state.---CH  &#91;&#91;User_talk:Hillman&#124;(talk)]] 18:26, 25 November 2005 (UTC)

The spin parameter
The explanation is incomprehensible as the number a = J/M is not dimensionless. Somebody please clarify. Bo Jacoby 08:56, 26 January 2006 (UTC)
 * units where c=1. If it troubles you, then use a=J/Mc2. -lethe talk 09:11, 26 January 2006 (UTC)

Thanks! But the SI-unit of a = J/Mc2 is (m2&times;kg/s)/(kg&times;m2/s2) = s. Still not dimensionless. Bo Jacoby 10:43, 26 January 2006 (UTC)
 * Hmm... you're totally right. I do apologize for answering so hastily and incorrectly.  It is not dimensionless.  And in fact, I guess it shouldn't be dimensionless, it should have the same dimension as r and s.  I don't know what the problem is. -lethe talk 10:54, 26 January 2006 (UTC)
 * So that stuff was added by an anonymous editor. I now think it's mistaken, but I won't remove it unless I know what it's supposed to say.  I suspect that what is meant is that a can range from 0 to m, rather from 0 to 1, and the anonymous had used a reference with a different notation convention.  But I'm not sure, so.... -lethe talk 11:12, 26 January 2006 (UTC)


 * First of all a indeed is not dimensionless at all, nor should it be. In geometrized units (where c=G=1), it is a length, just as M and r are. Secondly, the angular momentum J must be an area so that a=J/M can also be a length.  To get these geometrized lengths, you need to multiply a mass by G/c2 and an angular momentum by G/c3. --EMS | Talk 15:53, 26 January 2006 (UTC)
 * P.S. The anon mentioned above was most likely CH, who sometimes forgets to log in.  if not, it was someone else who is quite familiar with the Kerr metric. --EMS | Talk 16:01, 26 January 2006 (UTC)
 * Oops, I mean thanks! -lethe talk 18:16, 26 January 2006 (UTC)
 * FYI - Turns out that I was wrong about the anon being CH. (I should have guessed once I realized what the error was.)  My apologies to Chris.  Let's just say that this talk page seems to be jinxed for me.  :-( --EMS | Talk 04:55, 28 January 2006 (UTC)
 * I'm the anon. Sorry about the error, I usually use the Kerr metric with M normalized to be 1, and measure everything in R_G 's.  In those units a ranges from 0 to 1.  By writing the M's in explicitly in the metric, yes, the range changes from 0 to M.  What's there now is what I indended, if not what I actually wrote. Sfuerst 22:53, 17 February 2006 (UTC)

The angular momentum J defines a length J/Mc, and the mass M defines another length, the half Schwarzschild radius GM/cc. The ratio Jc/GMM is dimensionless. Is that the spin parameter? Bo Jacoby 12:49, 29 January 2006 (UTC)


 * Here is an easy way to read off the geometrized units of parameters appearing in metric coefficients (or in one-forms defining a cobasis field). Since metric coefficients are dimensionless in geometric units, a quantity such as m/r must be dimensionless.  Since r has units of length, so must m.  Angles and trig functions of angles should be dimensionless (as should the argument of any exponential).  Thus, $$a \, \sin(\theta)^2/\sqrt(r^2+a^2)$$ must be dimensionless, so since $$\sqrt(r^2+a^2)$$ has units of length, so must a.  Last but not least, Bo please note that angular momentum has geometrized units of area, just as you would expect from Newtonian theory!  See geometrized units. ---CH 23:26, 17 February 2006 (UTC)


 * If you want the spin parameter in dimensionless units that stays below 1 for black holes you have to define it by
 * $$\rm a=J\cdot c/G/M^2$$
 * Then you get around a=0.23 for the sun (for objects whose cartesian radius is larger than a times GM/c² you can also have a>1, the limitation only prevents already existing event horizons from blowing up and forces collapsing stars to throw out some of their excess angular momentum before becoming black holes).
 * --Yukterez (talk) 04:20, 17 July 2017 (UTC)

Will the real monopole please stand up?
I have elaborated slightly on this admittedly confusing issue. As I have noted elsewhere, WP requires articles on relativistic moments, moments in weak field gtr (including Weyl moments and the multi-index mass/momomentum/stress moments), and an improved article on multipole moments in Newtonian gravitation. Ultimately, the new section should be linked to this forthcoming articles and can perhaps then be shortened.---CH 22:24, 2 June 2006 (UTC)

trying to use Kerr solution for rotating stars is doomed
On the article page it says "Open problems

The Kerr vacuum is often used as a model of a black hole, but if we hold the solution to be valid only outside some compact region (subject to certain restrictions), in principle we should be able to use it as an exterior solution to model the gravitational field around a rotating massive object other than a black hole, such as a neutron star--- or the Earth. This works out very nicely for the non-rotating case, where we can match the Schwarschild vacuum exterior to a Schwarzschild fluid interior, and indeed to more general static spherically symmetric perfect fluid solutions. However, the problem of finding a rotating perfect-fluid interior which can be matched to a Kerr exterior, or indeed to any asymptotically flat vacuum exterior solution, has proven very difficult. "

No wonder it is difficult. The Kerr solution has only 2 independent parameters: mass and angular momentum. These determine all the gravitational multipole moments. But a rotating star has a complicated structure and its multipole moments will be controlled by the rotational (angular) velocity within it (which can vary with distance from the center and with latitude or angle from the plane of symmetry), the pressure and density variations with depth and possibly more things like internal magnetic fields. So the external field or various example rotating stars is more than a 2-parameter family. Whoever wrote that stuff should probably fix it. Carrionluggage 06:20, 17 October 2006 (UTC)

Contravariant components of metric tensor wanted
I think this article is written OK. I checked the metric "ds^2 = ..." and it looks good to me, however, I suggest to include contravariant components of $$g_{ik}$$ tensor too. greg park avenue 18:33, 4 June 2007 (UTC)

unit error in the Kerr metric
if i'm not mistaken, there is an error in the Kerr metric. I'm no expert on Relativity, but the last summand in the line element has unit length*time, where it should have length^2 like the other summands. the last summand should probably read (2*r_s*r*alpha/rho^2)*dPhi*(c*dT), since all other uses of time-variables in the line element are scaled by c, too. sorry in advance if i'm talking bullshit, but it puzzled me. i also find the mixed use of "a" and "alpha" in the line element misleading. am i right in assuming they should be the same? Could someone with more knowledge on this subject please verify this? thx Catskineater 14:32, 15 August 2007 (UTC)


 * Those were little things, but they did need fixing. It is done now. --EMS | Talk 17:30, 15 August 2007 (UTC)

For non-physicists,
the article is practically unreadable. I came here from the article on Tipler Cylinders, which is readable and easily understandable to a non-physicist. I'm not saying the equations should be taken out, and I also realise this is a fairly specialist article. However, perhaps the introduction should be expanded, in order to further explain to non-physicists like myself what a Kerr metric is, why it is important, and give a general idea of "how it works" as it were. At the moment the introduction is this:


 * In general relativity, the Kerr metric (or Kerr vacuum) describes the geometry of spacetime around a rotating massive body, such as a rotating black hole. This famous exact solution was discovered in 1963 by the New Zealand born mathematician Roy Kerr.

OK... that's all very well, but it doesn't tell me much. All I get from that is the idea that a Kerr metric is to do with describing how gravity works around a rotating black hole. That's all very well, but why is it important? Does it have application to anything else? I don't know.

WikiReaderer 00:22, 27 September 2007 (UTC)


 * You certainly have an interesting take on this, not that it is a bad one. To be honest with you, this article is so dense that I even have trouble with it, and I have studied general relativity is all of its glory.
 * In a nutshell, the Kerr solution is important because it is the simplest case of the spacetime around a rotating massive object. In astrophysics, it helps to predict phenomena in the vicinity of a gravitationally collapsed and rotating massive object.  It also has a number of interesting features, some of which are being checked for by Gravity Probe B.
 * Rewriting this article is something that I would like to do someday to make it more accessible. This cannot and will not be a topic that a non-physicist can get a good grip on, but that is no reason not to have a fairly non-techinical introduction that can give someone like you a sense of what this metric is about before diving into the more "gory" details.
 * FTI - Some features are:
 * A frame dragging effect that pulls spacetime around the rotating object. This means that a beam of light traveling with the rotation will be found by distant observer to go around the object faster than a beam of light that travels against the rotation.  it also means that an object that appears to a distant observer to be going around the obejct may locally be falling straight into it.
 * There is a region called the ergosphere close to the object (near the outer event horizon) where an object cannot be at rest with respect to a distant observer but instead must rotate with the object. (Even light must do so in the ergophere, including light that is locally traveling against the rotation.)
 * There are two event horizons for a "slowly" rotating black hole: An outer one and an inner one.  Entry into the outer horizon demands that you pas though the inner one, after which you are stuck there.
 * The inner region of a Kerr black hole includes a ring singularity and also "closed timelike curves". The latter expression means that you can come back not only to the same place but also the same time!  In essense, you can do time travel in the inner region of a Kerr black hole.
 * For a "rapidly" rotating black hole, the event horizons are not present, and the area where time travel is theoretically possible potentially becomes accessible.
 * Hopefully this is some help to you. (If it is, an idea may be to put this list into the article.) --EMS | Talk 02:43, 27 September 2007 (UTC)

That was a quick response. Thanks! All the above is useful, and pretty informative. I have already heard of some of these things before (mostly from Stephen Baxter sci-fi!), but hadn't really connected them up with this Kerr metric thing. I would definitely recommend that you insert this list into the article: the only question is where...

WikiReaderer 23:12, 30 September 2007 (UTC)

I feel that a quick summary of what each constant represents primarily the rates of change would make the page much more readable also it would not be too hard to create a diagram for the entire metric so that less savvy readers like myself can quickly grasp how to use the equation. I have spent the last few days trying to piece the equation together and I can not find the information anywhere. —Preceding unsigned comment added by 72.195.141.216 (talk) 02:49, 20 May 2011 (UTC)


 * I must say that I also find it hard to understand what some of the constants ("distance scales") and variables mean.
 * Although Schwarzschild metric contains an explanation of the variables it uses, the truth is that their meaning can (and must) be deduced from the metric itself, so strictly speaking that explanation is redundant. I assume that the same is true for the Kerr metric. JRSpriggs (talk) 17:18, 20 May 2011 (UTC)

1 November 2007 changes
I've clarified some of the material on this page and on the Kerr-Newman and rotating black hole pages. I believe it is now more accurate. Unfortunately, I've noticed in editing this that there is now some duplication (for instance, the figure on this page). If any frequent wikipedians want to clean these pages further, that would be great. 91.37.241.229 20:44, 1 November 2007 (UTC)

errors
So, this page has some real problems. For instance, from the introduction(!)
 * The Kerr metric is often used to describe rotating black holes, which exhibit even more exotic phenomena. Such black holes have two event horizons where the metric appears to have a singularity. The outer horizon encloses the ergosphere and has an oblate spheroid shape, a flattened sphere similar to a discus.

The two event horizons are the outer and inner event horizons, the outer of which cannot be crossed from inside out (I'm not sure about the inner one). In Boyer-Lindquist coordinates, the ergosphere is described by r = M + \sqrt{ M^2 + a^2 \cos^2 \theta }. Not, by the way, an oblate spheroid, as far as I can tell, because as a->1 it looks more like a peanut. Moreover, it is also neither of the horizons, which are located at r_\pm = M \pm \sqrt{ M^2 - a^2 }.

I'm not an expert, I just have a copy of Hartle's Gravity: An introduction to Einstein's general relativity, and have read the chapter on the Kerr metric.

Could someone put up a warning to folks that the info on this page is questionable and that the page needs help from an expert?

Or, of course, correct me if I'm wrong :/

Thanks&#8201;—&#8201;gogobera (talk) 05:05, 2 April 2008 (UTC)

PS: Actually, this has been brought up earlier on this talk page!


 * This was my fault, sorry! I got the math correct, as you can see from the article, but my description of it was incorrect.  My basic confusion was describing a "surface where the metric is singular" as "event horizon", which are not the same concept.  I also didn't mean oblate spheroid in its exact mathematical sense, but qualitatively; I was trying to describe the shape as a flattened sphere.  I'm pretty sure the the outer surface doesn't always have a dimple — consider the limit as the angular momentum gets small, that is, as the ratio of length-scales α/rs gets small.  In that limit, we can expand the square root in a binomial expansion to obtain (if I did the math correctly!)



r_{outer} = r_{s} \left[ 1 - \frac{\alpha^{2} \cos^{2}\theta}{r_{s}^{2}} + \ldots \right] $$


 * which approximates an oblate spheroid, wouldn't you agree? The dimple appears as the angular momentum gets larger.


 * Anyway, I'm sorry for my errors and grateful that you pointed them out. I'm keenly conscious that I'm no expert, although I'm trying my best.  It would be great if an expert could come here and make the subject more intelligible for lay-people. (hint, hint) ;) Willow (talk) 09:36, 2 April 2008 (UTC)

It appears to me that the initial expressions for the locations of the event horizon and the Cauchy are already in natural units, and that the equations entitled "or in natural units" are arbitrarily normalised wrt rs/2. If "correct", this needs an explanation at minimum, as well as a specific reference PhysicistQuery (talk) 08:46, 7 April 2020 (UTC)

Notational consistancy
This page needs to be edited for notational consistency. In the first part in the definition of the metric, the length scales Lambda and alpha are introduced. Then in the rest of the article, the dimensionless parameter a and the parameter Delta are used without definition. Either they need to be defined, or more probably, the Kerr metric should be written using the conventional parameters Delta and a instead of with Lambda and alpha. —Preceding unsigned comment added by 142.103.234.23 (talk) 18:40, 7 April 2008 (UTC)

Duplicate figure
I commented out the duplicate copy of the ergosphere picture. Caption was:


 * Two important surfaces around a rotating black hole. The inner sphere is the static limit (the event horizon). It is the inner boundary of a region called the ergosphere. The oblate spheroidal surface, touching the event horizon at the poles, is the outer boundary of the ergosphere. Within the ergosphere a particle is forced (dragging of space and time) to rotate and may gain energy at the cost of the rotational energy of the black hole (Penrose process)

in case someone wants to put some of it back into the other copy. Wwheaton (talk) 16:27, 29 April 2008 (UTC)

What is 'a'?
"... Schwarzschild solution, with a=0, the inner..." "When M < a, there are no..."

What is 'a'? I've gone over it carefully and can't find it.

Długosz (talk) 20:36, 12 March 2009 (UTC)


 * This has now been fixed. It's my understanding that most people in relativity use "dimensionless units" in which G=c=1, which simplifies many of the formulas. The Kerr metric is usually written in that notation. In that case, $$a=\alpha(c^2/G)$$ plays the role of $$\alpha$$ in many of the formulas and generally makes things easier to read. 86.153.33.146 (talk) 11:32, 6 April 2009 (UTC)

What is the Kerr vacuum?
The expression "Kerr vacuum" is suddenly introduced and not explained. One can try in principle to understand an expression made of 2 or more words by thinking about the meanings of the individual words in the expression - this is quite standard because most such expressions have no definition and don't really need one. But here I think it's not sufficient. Being myself a mathematician interested in physics and astronomy (and just reading an astronomy book) I came into the WP-articles about black hole metrics to help myself understand more about these things. So I could not make any useful edits myself. After reading this disc. page, I think I begin to understand what is meant, but I'm not sure. The idea is that since the Schwarzschild / Kerr / Reissner-Nordström / Kerr-Newman solutions contain analytic functions (away from apparent / true singularities) they (I guess) discribe only the (supposed vacuum) outside of a massive object with a definite (spherical) boundary (Planet / Star / black hole), which seems confirmed by the considerations about e.g. "match(ing) the Schwarschild vacuum exterior to a Schwarzschild fluid interior" - so the "Kerr vacuum" might just be another name for the Kerr solution which insists on this fact. Of course it must be an approximation to reality because the outside of real astronomic objects isn't void till infinity ..., approx. that might be fairly good in the neighborhood of the object. (Here I'm ignoring the problems with the may-be complicated inner structure of the object and with the until now not realized merging of Generel Relativity with Quantum physics.) My argument about analyticity is based upon the fact that the boundary of said objects - with the vacuum imagined outside - is a surface where the matter density is not analytical.--UKe-CH (talk) 09:40, 5 July 2009 (UTC)

Where is the physical singularity?
As far as I can tell, the article never identifies the coordinates at which the physical singularity occurs. My understanding is that it is a ring (or cylinder if you include time) so that it is extended in t and &phi;. And I believe that it needs &theta;=&pi;/2. However, the value of r is not apparent to me. It should be an increasing function of &alpha;. Does anyone know what it is? And how to justify that value? JRSpriggs (talk) 16:53, 16 November 2010 (UTC)


 * The obvious possibility is r = &alpha;, the classical radius of a ring (rotating at the speed of light) with that mass and angular momentum. However, nothing special seems to happen at that radius in the Kerr metric as far as I can see. JRSpriggs (talk) 00:29, 17 November 2010 (UTC)

The Kerr metric has an obvious singularity at r=0 and &theta;=&pi;/2 because &rho; gets zero. --B wik (talk) 08:08, 12 January 2013 (UTC)


 * The inner ergosphere in unphysical Boyer-Lindquist coordinates is at $$r = 1-\sqrt{1-a^2 \cos ^2(\theta )}$$.
 * With the transformation rule from $$r,\theta,\phi$$ to $$x,y,z$$ we get the surface of the inner ergosphere in cartesian coordinates
 * $$x = \sin (\theta ) \cos (\varphi ) \sqrt{\left(1-\sqrt{1-a^2 \cos ^2(\theta )}\right)^2+a^2}$$
 * $$y = \sin (\theta ) \sin (\varphi ) \sqrt{\left(1-\sqrt{1-a^2 \cos ^2(\theta )}\right)^2+a^2}$$
 * $$z = \cos (\theta ) \left(1-\sqrt{1-a^2 \cos ^2(\theta )}\right)$$
 * Since the ring singularity is in the plane of the equator we take the limit of $$\lim_{\theta \to \pi/2}$$
 * $$x,y,z \to a \cos (\varphi ), \ a \sin (\varphi ), \ 0$$
 * Therefore the ring singularity has a cartesian radius of $$a$$, which is the value of the spin parameter.
 * So if $$a$$ were maximal, the ring singularity would have a radius of $$1 G M/c^2$$, see comparison.
 * --Yukterez (talk) 01:54, 18 June 2017 (UTC)

gps
I have heard that the Kerr metric is used by gps. If someone with some expertise could expand on this in the article, I think this would be a great addition. — Preceding unsigned comment added by 24.85.82.38 (talk) 02:31, 19 September 2011 (UTC)


 * I think that that is unlikely. The deviations from a Schwarzschild metric due to mass-concentrations in the Earth are probably much (many orders of magnitude) greater than the relativistic corrections needed due to the Earth's rotation. JRSpriggs (talk) 06:48, 19 September 2011 (UTC)

Removed "time travel"
The article said, without citation:
 * Even stranger phenomena can be observed within the innermost region of this spacetime, such as some forms of time travel. For example, the Kerr metric permits closed, time-like loops in which a band of travelers returns to the same place after moving for a finite time by their own clock; however, they return to the same place and time, as seen by an outside observer.

I strongly doubt that closed time-like loops can be "seen by an outside observer," or that a "band of travelers" or any other finite physical system could traverse such a loop, whether seen by an outside observer or not. This seems like unsubstantiated hype-flavored stuff. If someone has a really good citation, thinks that it actually advances understanding, and can write a really clear description of this essentially non-physical idea, go to it. — Preceding unsigned comment added by 86.140.51.175 (talk) 00:37, 2 March 2013 (UTC)


 * No, this is correct, the precise statement is that any pair of events inside the region $$r<0$$ (the universe "behind" the hole) can be connected by a timelike future-pointing curve. So one can return to one's past. The beef here is that in the region $$g_{\phi\phi}<0$$, moving along the $$\phi$$ coordinate curve (now timelike) "fast enough" decreases the t coordinate. What would such a traveller look like to another observer in the region is a cool question, a bit of computer ray tracing should solve this. The region $$g_{\phi\phi}<0$$ is a solid torus tangent to the ring singularity on the "inside". Its cross-section is not circular but some weird distorted oval-like quartic curve — another computer plotting assignment. Finally, this solid torus is accessible by a timelike future-pointing curve from/to any event with $$r<0$$. Jan Bielawski (talk) 22:55, 13 May 2013 (UTC)

Wrong trajectories diagram
I am fairly certain (though I'm no expert) that the trajectories shown in one of the pictures in "Trajectory equations" are wrong. I mainly refer to the left picture but I have serious doubts about the right one too. In the left, the center mass is supposed to rotate clockwise; but, except one, all of the trajectories, when they enter the ergosphere, continue to "circle" the center mass in counterclockwise direction. This violates what I know about ergospheres. From the diagram I see a dPhi that multiplied with alpha becomes negative, causing the squared line element (as given at the beginning of the page) to be negative, hence imaginary proper time, hence space-like. The one trajectory that starts at r*sin(Phi)=0 falls straight through the ergosphere into the center, showing no frame-dragging at all, which is wrong IMHO too.

My understanding would be that as the particles fall in, frame dragging consumes all the orbital angular momentum until (at latest) at the ergosphere surface the orbital angular momentum changes sign to point into the same direction as J. This means that within the ergosphere the particle co-rotates with the center mass. Straight trajectories like the one shown for r*sin(Phi)=0 should not be possible.

Maybe some expert can comment on my observation and either confirm (meaning the picture should be removed) or explain why my understanding would be wrong?WikiPidi (talk) 09:06, 15 October 2014 (UTC)


 * WikiPidi's complaint looks sound. The radial infall is telling in both pictures.   The particle should be perturbed spinward as it falls into the BH. Cloudswrest (talk) 20:19, 15 October 2014 (UTC)


 * So after four months no expert has shown up to comment. But with one supporting comment I decided to remove the pictures now.WikiPidi (talk) 16:08, 24 February 2015 (UTC)

This section should be archived. --B wik (talk) 13:40, 14 August 2016 (UTC)

"Geometry" versus "vacuum"
The original author of this article and the one on the Schwarzschild metric uses the terminology "vacuum" which is used by some experts, but a minority. More use "geometry". I am an editor of the monograph "The Kerr Spacetime: Rotating Black Holes in General Relativity", a collection of articles by 12 leading experts in the field: Kerr, Carter, Penrose, ... The relative frequency of word counts from the source files gave: "Kerr black hole" 118, "Kerr metric" 52, "Kerr geometry" 14, "Kerr vacuum" 1. I realise that this article refers to the geometry, and not just the black hole which is more frequent because of its usage in the astrophysics literature. However, "Kerr vacuum" is really used by a minority, whereas "Kerr geometry" is far more common. — Preceding unsigned comment added by DavidLWiltshire (talk • contribs) 01:51, 15 February 2016 (UTC)

Spin
The text does not much deal of the spin of a Kerr black hole. I think that this addition 'd be useful (up to some native english writer to "translate" my comment is good english...)

The "rotation speed" of a Kerr black hole cannot be calculated simply based on the spin parameter (a*) as do some reporters (Some translated simply e.g. a*=0.84 in 84% of c, what is wrong, see e.g. http://www.universetoday.com/109308/how-fast-do-black-holes-spin/ ).

Here is what we could add in the text.

Rotation speed of a Kerr black hole.

The spin is dimensionless and it is not an angular speed and it is not linear in relation with the speed of light. So we cannot write e.g. that a*=0.84 then the rotation speed is 0.84c. It is wrong.

So, to estimate the "rotation speed" of a Kerr black hole, we have to define the angular speed ΩH at the external horizon, which is the first surface spinning like a rigid body. The formula is next :

ΩH = (a/M)(c/M)/(1+√(1-a^2/M^2)) with a=J/M.

This value expresses in units of c/M = (0.2 rad/sec)(10^6 M¤/M). When a tends to M, then ΩH tends to c/M. NB. √ = square root and M¤ = solar mass. --luxorion — Preceding unsigned comment added by 2001:7E8:D486:EC01:9F1:C67A:2A6C:F251 (talk) 12:22, 29 February 2016‎ (UTC)

I do NOT thanks members of wikipedia for having deleted my paragraph about the spin of the Kerr black hole !!!! I wonder under what authority my original text has been deleted. It is useful. But I retain the lesson. --luxorion

Nonrelativistic Limit, Cartesian Coordinates

 * In the section "Mathematical form" it says:
 * "In the non-relativistic limit where M (or, equivalently, rs) goes to zero, the Kerr metric becomes the orthogonal metric for the oblate spheroidal coordinates"
 * followed by a function depending on a.
 * But if M=0, the whole solution reduces to flat Minkowski space, even if a>0.
 * Then it continues with
 * "which are equivalent to the cartesian coordinates"
 * followed by the standard transformation Boyer-Lindquist to cartesian background coordinates(which is also valid in the fully relativistic field).
 * Therefore I will move the transformation up, below the passage
 * "where the coordinates r,θ,ϕ are standard spherical coordinate system".
 * The history of the entry is here.
 * --Yukterez (talk) 19:41, 8 July 2017 (UTC)

Trajectory equations
Have you ever used the trajectory equations you restored in the Kerr-metric article? If so, I would like to know how you dealt with the ± signs all over the place. Your equations might look more compact, but are they really usable? If not, please consider reverting your edits yourself. If you really wish to revert to the Plusminus-equations you should delete my edits completely, since it makes no sense to introduduce the r- and φ-components of the local velocity, but not the θ-component, and also the distinction between massless photons and testparticles via μ doesn't make any sense anymore since you deleted all terms where μ was taken mention of. Now we have a mish mash which I fear will not help anyone actually trying to figure out a trajectory. The equations without the ± were perfectly backed up by arxiv.org/abs/1601.02063 and arxiv.org/abs/0802.0459, by the way. The equations you seem to prefer are also mentioned on page 31, but unfortunately in a not so positive way: "We note that while these equations are concise and appealing in some ways, during numerical integration they tend to accumulate error at the turning points due to the explicit square roots in the r and θ equations, not to mention the nuisance of having to change the signs of the r and θ velocities by hand at every turning point"--Yukterez (talk) 11:52, 18 June 2017 (UTC)


 * I have used these equations, in fact, they can be solved analytically! See arXiv:0906.1420 and arXiv:1009.6117. The other form may be nicer numerical, but this has much clearer analytical properties.TR 19:46, 18 June 2017 (UTC)


 * Ok, if they are better for analytical solving you seem to have a point. I only solved numerically, and then the ± signs were really annoying, but since the numerically more stable code is in the description of the animation anyway we can keep the version you prefer. --Yukterez (talk) 20:11, 18 June 2017 (UTC)

Note that the λ in the trajectory equations is ``not'' the Mino(-Carter) time. (If it were there would be no Σ's on the LHS.) Instead (confusingly) it is (I think) τ/μ, to facilitate compatibility with the massless case. I look into clarifying this later.TR 14:58, 19 June 2017 (UTC)


 * If it is τ/μ then it is practically the proper time, and I should restore this edit saying
 * "For massive test-particles $$\mu=-1$$, while for massless particles like photons $$\mu=0$$. The affine parameter $$\lambda$$ can be interpreted as the proper time $$\tau$$ of a massive test-particle, while with massless particles (which do not have a proper time) the coordinate time $$t$$ is the only relevant time coordinate."
 * PS: it IS the proper time, at least according to the reference A Periodic Table for Black Hole Orbits, page 31 where the same equations are shown, with the only difference that τ is used instead of λ. So I revert to the previous version. So we were both wrong to undo our own statements that λ is indeed τ if μ=-1
 * --Yukterez (talk) 19:58, 19 June 2017 (UTC)
 * No, λ is not the proper time. Note that in the Levin/Perez-Giz paper they set the particle mass to 1, and hence λ = τ. However, if you want to keep the equations to be valid for massive and massless particles then you need to keep μ and λ.TR 00:40, 20 June 2017 (UTC)


 * Maybe we should restore the relation between energy/angular-momentum and the local 3-velocity as it can be found here, since the conversion from one to another is not really trivial.
 * I'd say we need an explanation/definition of the conserved total energy, which would be :
 * $${\rm E = \sqrt{\frac{(\Sigma - 2 \ r) \left(\dot{\theta}^2 \ \Delta \ \Sigma +\dot{r}^2 \ \Sigma -\Delta \ \mu \right)}{\Delta \ \Sigma }+\dot{\phi}^2 \ \Delta \ \sin ^2 \theta } = \sqrt{\frac{\Delta \ \Sigma}{(1+\mu \ v^2) \ \Xi}} + \Omega \ L_z}$$
 * the conserved angular momentum on the Ф-axis :
 * $$\rm L_z = \frac{\sin ^2 \theta \ (\dot{\phi} \ \Delta \ \Sigma - 2 \ a \ E \ r)}{\Sigma -2 \ r} = \frac{v_{\phi} \ \bar R}{\sqrt{1+\mu \ v^2}}$$
 * with the radius of gyration :
 * $$\rm \bar R = \sqrt{\frac{\Xi}{\Sigma}}$$
 * and the conditions in terms of the local velocity on the r-axis :
 * $$\rm \frac{v_{r}}{\sqrt{1+\mu \ v^2}} = \dot r \ \sqrt{\frac{\Sigma}{\Delta}}$$
 * the θ-axis:
 * $$\rm \frac{v_{\theta} \ \sqrt{\Sigma}}{\sqrt{1+\mu \ v^2}} = \dot \theta \ \Sigma$$
 * and the Ф-axis:
 * $$\rm \frac{v_{\phi} \ \bar{R}}{\sqrt{1+\mu \ v^2}} = \dot \phi \ \Sigma$$
 * where
 * $$\rm \Xi= \left(a ^2+r^2\right)^2-a ^2 \ \sin ^2 \theta \ \Delta$$
 * Like Einstein himself said, "Make things as simple as possible, but not simpler."
 * --Yukterez (talk) 19:45, 8 July 2017 (UTC)


 * What is wrong with
 * $$ E/\mu = -u_t$$
 * $$ L_z/\mu = u_\phi$$
 * (with u the four-velocity). Lets not make things more complicated than necessary.TR 12:47, 9 July 2017 (UTC)


 * There's nothing wrong with it, but 1) this are only the t- and φ-components of u, not v, while the r- and θ-components are still missing and 2) the relation you quoted doesn't appear in the article at all. Using only the information which is given in the article, I'm not sure if one would find the initial conditions for a particle starting with an initial velocity of say v=0.9c (where v is the local, not shapirodelayed velocity where light is exact c) with a given inclination angle and a given vertical launch angle from a given r0, θ0 and φ0 position, would he? At least I think I wouldn't! --Yukterez (talk) 18:51, 9 July 2017 (UTC)

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Roy Kerr
I wanted to add the fact that this was discovered by Roy Kerr in the introduction, but I don't want to mess things up with my low experience. PopoDameron (talk) 12:56, 19 April 2019 (UTC)

Unused definition?: $${\displaystyle {\vec {a}}}$$ and comment on Kerr–Schild coordinates section
After Eq: 13-- the definition of $${\displaystyle {\vec {a}}}$$ ; I have to presume something like $$a^{2}=g_{uv}a^{u}a^{v}$$ and $$\vec{a}=a^{u}$$ Actually I think the article http://www.scholarpedia.org/article/Kerr-Newman_metric is much better for "Kerr–Schild coordinates". While a little longer is more intelligible. Rrogers314 (talk) 15:03, 28 May 2020 (UTC)

Kerr-NUT Paper Citation?
I found an article that refers to the Kerr-NUT metric w.r.t. axial symmetry. I think a citation should probably be added or else just have the parenthetic that refers to it removed entirely, but I'm not knowledgeable enough in this field to make the contribution. I actually thought this was a vandalism change from Newman to NUT until I looked it up, as there is no context in this Wikipedia article for it and, well... "NUT".

Article link: https://link.springer.com/article/10.1007/BF00765705 TricksterWolf (talk) 00:01, 30 July 2021 (UTC)
 * I think the clause in brackets should be removed until the Kerr-NUT article is created. Bellowhead678 (talk) 14:23, 2 August 2021 (UTC)

Important Surfaces
There is an equation for rH+ which I interpret as the radius of the outer event horizon. That simple equation seems to be for a sphere, assuming the radius is the same in all directions (unlike the ergosphere equation, there is no parameter related to latitude angle in the rH+ equation). Yet the imagine clearly shows the outer event horizon flattening, and I have seen in other places that the outer event horizon is described as oblong (https://www.space.com/black-holes-event-horizon-explained.html) as the parameter "a" increases.

Is the formula for rH+ wrong, is it just applicable to the equatorial radius, or is the shape a sphere within the identified coordinate system? If the shape is squished top to bottom and rH+ is the equator's radius, is there another equally simple formula for the pole to pole distance? 2601:648:8700:3750:E533:3A24:7957:B9EA (talk) 18:54, 29 January 2023 (UTC)
 * It would help if you specified which equation you find incorrect; the equation I find which seems to follow your description is the one specific to the Schwarzschild event horizon, which means non-rotating. The Kerr solution (which includes the latitude) is for a rotating black hole. Tarl N. ( discuss ) 21:35, 29 January 2023 (UTC)


 * If the Boyer Lindquist r is constant then the cartesian R ain't, in the equator R=√(r²+a²) while over the poles R=r, see the article on BL coordinates where everything is explained. --Yukterez (talk) 17:57, 19 August 2023 (UTC)