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Q: Kerr singularity calculations ( Answered 5 out of 5 stars,   2 Comments )
Question  
Subject: Kerr singularity calculations
Category: Science > Physics
Asked by: simanonok-ga
List Price: $200.00
Posted: 30 May 2004 03:21 PDT
Expires: 29 Jun 2004 03:21 PDT
Question ID: 353813
BACKGROUND:

A singularity (or 'black hole') is a point in spacetime where matter
has condensed to a theoretical zero volume by the force of its own
gravity.  Surrounding it is a region with an outer boundary called the
event horizon and having a radius proportional to its mass, such that
anything entering the region is condemned to fall into the singularity
at the instant it crosses the event horizon.  And supposedly nothing
can get out, not even light, although there is a pecular phenomenon
called Hawking Radiation that singularities, especially small
singularities, can 'evaporate' through the loss of.

A singularity that spins is called a Kerr singularity,
and it is said that its event horizon takes on a toroidal shape. 
But actually a Kerr singularity has two event horizons, an inner event
horizon which can't be crossed without falling into the singularity
and an outer event horizon encompassing a larger volume of spacetime. 
The region between the inner and outer event horizon is called the
ergosphere.  Objects may enter the ergosphere and leave, but they may
not pass the inner event horizon and do anything but be sucked into
the singularity.  Spacetime however is distorted inside the
ergosphere, spacetime being dragged around by the rotating
singularity.  Supposedly it may be possible for an object to enter the
ergosphere of a singularity and emerge at a different time,
accelerated time if it enters in one direction and retarded time if it
enters another.

MY QUESTION:
The root aspect of my question is that I want to know how and be able
to calculate the dimensions of the inner and outer event horizons of a
Kerr singularity given the input variables of mass and rpm, and
possibly (if applicable) charge.

Four things are required to adequately answer this question:

1) The relevant equation(s), with all variables and constants defined
and units supplied.

2) Two step-by-step worked examples of the equation using these
greatly different masses and rotation rates: 100 Kg and ten solar
masses, you pick the rotation rates where the lighter singularity
rotates very fast and the heavy one rotates more slowly.

3) Citations for the originator(s) of the equations.

4) An Excel spreadsheet or other program which allows for the input of
a Kerr singularity's mass (Kg), rotation rate (rpm), and charge if
applicable, and outputting the radii of the inner and outer event
horizons.

Request for Question Clarification by tox-ga on 16 Jun 2004 15:04 PDT
simanonok-ga,
I have a solution of sorts for you, which is why I posted it as a
clarification first. If it is acceptable, then I will post it as an
answer. Feel free to ask for clarifications. I will do my best to
answer them.

The differential equations outlined in the sites that you have posted, namely:
http://scienceworld.wolfram.com/physics/KerrBlackHole.html
Deal with the path of particles caught in gravitational field of the
singularity. Since you are merely looking for the event horizon, such
extensive path information is not relevant to what you are looking
for.

The two equations you need are:
Outer Event Horizon (OEM) = M + sqrt( M^2 - a^2 - e^2) 
the Inner Event Horizon (IEM) = M - sqrt( M^2 - a^2 - e^2)
where a = J/M, J = angular momentum, M = mass, e = charge
These equations were developed by Kerr, generalizing on Schwarzschild
theories on stationary black holes, adding angular momentum J.  Later,
Newman, in 1965 further generalized the solution to incorporate
charge, e.

These are outlined in several books and sources, including:
Kerr, R. P. "Gravitational Field of a Spinning Mass as an Example of
Algebraically Special Metrics." Phys. Rev. Let. 11, 237-238, 1963.
Shapiro, S. L. and Teukolsky, S. A. "Kerr Black Holes." 12.7 in Black
Holes, White Dwarfs, and Neutron Stars: The Physics of Compact
Objects. New York: Wiley, pp. 357-364, 1983.
http://scienceworld.wolfram.com/physics/KerrBlackHole.html
http://www.innerx.net/personal/tsmith/BlackHole.html
http://www.physics.ubc.ca/%7Epsih/kerr-metric/p407.html

Every solution, including the differential, requires the knowledge of
the angular momentum of the singularity. This can easily be estimated
by knowing the initial mass of the object and the moment of inertia of
a sphere, given by:
I = 2/5*MR^2
where M = mass, R = radius
Since angular momentum is conserved, and generally, singularity
formation involves only a slight change in mass, the inner and outer
event horizons, given the mass, radius and spin of the planet before
its collapse can give you the required J value.

There are three cases for the event horizon solution:
if e^2 + a^2 is less than M^2 then the above two equations are used
if e^2 + a^2 = M^2, then the outer and inner event horizons coincide
at a distance of M
e^2 + a^2 is greater than M^2 then the event horizon solution is
complex and current has no confirmed meaning. It is theorized that
such singularities can be used for time travel. Both Hawking and Ellis
propose that space-time tunnels require the use of such "negative
material" to open. (http://www.innerx.net/personal/tsmith/BlackHole.html#coinevhor)

For 100 kg object
-----------------
Assuming an initial mass of 100 kg, charge of 1 C, radius of 5 m and
initial rotation of 60 rpm.

angular velocity = w
                 = 60 rpm 
                 = 1 rps (rotation/second) 
                 = 2(pi) rad/s

moment of inertia = I
                  = 2/5 * MR^2
                  = 2/5 * (100 kg) * (5 m)^2
                  = 1000 kgm^2

angular momentum = J
                 = Iw
                 = 2000(pi) kgm^2/s
                 
a = J/M
  = (2000(pi) kgm^2/s)/(100kg)
  = 62.83185307
                 
OEM = M + sqrt( M^2 - a^2 - e^2) 
    = 177.789191 m

IEM = 22.21080898


For 10 solar masses
-------------------
Assuming an initial mass of 1.98892  10^31 kg, charge of 1000 C,
radius of 7000000000 m and initial rotation of 0.00001 rpm or
1.0472E-06 rad/s.

moment of inertia = I
                  = 2/5 * MR^2
                  = 3.89828E+50 kgm^2
                  
angular momentum = J
		 = Iw
		 = 4.08227E+44 kgm^2/s
		 
a = 2.05251E+13

OEM = 3.97784E+31 m

IEM = 0 (approximately)


As you can see, the angular momentum and the charge must be quite
signficant before there is a noticeable inner event horizon. The excel
spreadsheet is available at:
http://www.maxlin.ca/tos/ga/singularity.xls

Let me know if this is what you are looking for.

Cheers,
Tox-ga

Request for Question Clarification by tox-ga on 16 Jun 2004 20:14 PDT
simanonok-ga,
Actually, after some further research, knowing the rpm of a black hole
is very non-standard. Black holes are characterized by their mass,
charge and angular momentum
(http://www.campusprogram.com/reference/en/wikipedia/b/bl/black_hole.html,
http://www.phyast.pitt.edu/~ajc/teaching/chap24/chapter24.pdf), all of
which are measured values and are incorporated into calculating the
inner and outer event horizons.

Let me know if this and the above post constitute an acceptable answer.

Cheers,
Tox-ga

Clarification of Question by simanonok-ga on 19 Jun 2004 07:22 PDT
If only I could clone your mathematical ability and inject it into my brain...

Thank you for an excellent response and I apologize for the time it
has taken me to work through it.  I do have a problem with one part
however:

The part that I don't understand and which seems highly
counterintuitive is how the inner event horizon (IEM) of the
10-solar-mass Kerr singularity can be nearly zero radius while the IEM
of a 100 Kg singularity can have a 22 meter radius.  It's my
understanding that the inner event horizon of a Kerr singularity has
more or less the characteristics of the inner event horizon of a
nonspinning singularity, in that passing it is a one-way ticket into
the singularity.  In other words, anything which enters the volume
encompassed by a Schwarzchild radius gets sucked into the singularity,
not even light can escape.  And this is true for both spinning and
nonspinning singularities, correct?  The geometries of the two may
differ because the Kerr singularity has an inner event horizon that is
toroidal, but I assume that when we are discussing the radius of the
Kerr singularity's inner event horizon, we are talking about a
measurement perpendicular to the axis of rotation.  Without
belabouring the point of the toroidal distortion (unless I am
seriously missing something), those radii ought to be more or less
proportional to the masses of the singularities, such that a
singularity massing ten suns should have an inner event horizon with a
radius far greater than one massing only ten Kg, REGARDLESS of whether
we are talking about Kerr singularities or nonspinning ones (in which
case we wouldn't be referring to an 'inner' event horizon of course,
just an 'event horizon').  Or am I missing something?
Answer  
Subject: Re: Kerr singularity calculations
Answered By: tox-ga on 19 Jun 2004 14:25 PDT
Rated:5 out of 5 stars
 
simanonok-ga,
After your clarification, I went back to think about my previous
calculations and I realized I missed a few small things that made a
big difference in the resulting answer. The fact that the event
horizon for a small object is almost two hundred meters is, as you
know, slightly ridculous. I have since revised my calculation, as well
as consulted with a physics professor. Here are my results:

The calculation for angular momentum, J, remains unchanged, as that is
done with classical mechanics, nothing extremely tricky. However, to
calculation "a", the formula is a = J/Mc, where c is the speed of
light.
To calculate e, given electric charge, q, in coulombs, the formula is
e^2 = (Gq^2)/(4(Pi)(Eo)c^4). These constants are all defined in Sheet2
of the excel spread sheet downloadable below.

The outer radius is determined by:
r+ = (GM/c^2)(1+(1-a^2-e^2)^0.5)

The inner radius is determined by:
r- = (GM/c^2)(1-(1-a^2-e^2)^0.5)

For additional knowledge, the static limit (outer edge of the
ergosphere) can be determined by:
re = (GM/c^2)(1+(1-(a^2)(sin^2(theta)))^0.5)
Though charge is not incorporated into this forumla.

You may note that a significant amount of charge is required before it
affects the size of the inner and outer event horizons. However, since
any charge that black hole has is quickly neutralized by incoming
material, black holes typically exhibit zero charge
(http://archive.ncsa.uiuc.edu/Cyberia/NumRel/BlackHoleEvolution.html)

The sources for these equations are in pdf format, downloadable at:
http://www.maxlin.ca/tos/ga/blackhole1.pdf
http://www.maxlin.ca/tos/ga/blackhole2.pdf
http://www.maxlin.ca/tos/ga/blackhole3.ps
http://www.engr.mun.ca/~ggeorge/astron/blackholes.html (Based upon a
resentation to a meeting of the St. John?s (Newfoundland) centre of
the Royal Astronomical Society of Canada, 1989 March 15)

An updated excel spreadsheet can be found at:
http://www.maxlin.ca/tos/ga/singularity.xls

The first entry in the excel spreadsheet is for a star with 10 solar
masses which is spinning very slowly. The third entry is the same
star, without spin. Note the very similar outer even horizons. To
answer your question, the inner event horizon arises as a result of
charge/spin.

The even horizons (both inner and outer) are spherical in shape. The
ergosphere is elliptical and extends outward along the equator of the
black hole's spin. The Kerr-Newman metric describes a singularity
which is torroidal in shape (so the singularity itself is torroidal,
not the inner event horizon). The donut shape has given rise to many
interesting theories about time travel if one were to pass through the
torroid.

I hope this is the answer you are looking for. Let me know if it is,
or ask for clarification if you have any more questions.

Cheers,
Tox-ga

Request for Answer Clarification by simanonok-ga on 21 Jun 2004 09:50 PDT
There seems to be some possible confusion between definitions of the
features of a Kerr black hole by different authors. In your
explanation and also at
http://www.innerx.net/personal/tsmith/BlackHole.html#ergohor12ring
where the graphic shows the ergosphere OUTSIDE the outer event
horizon, the 'static limit' refers to the outer boundary of the
ergosphere.  Somewhere inside the static limit is the outer event
horizon, inside the outer event horizon is an inner event horizon, and
at the center of the onion is the singularity (actually a torus
however for Kerr sigularities).   It is my understanding from this
description that the static limit represents the boundary where
objects (including light) will be dragged around the singularity by
the spacetime distortion on the inside of that boundary and they will
not be dragged around if on the outside of the static limit.  In this
depiction the inner event horizon (which is also apparently sometimes
referred to as the "Cauchy" boundary), has more or less the properties
of the single event horizon of a static black hole in that anything
passing the boundary gets sucked into the singularity.  But I am not
able to understand how the 'outer event horizon' means anything in
this context.  What happens at its boundary?

It has previously been my impression as described in my original
question that the outer event horizon of a Kerr black hole is the same
thing as the static limit, and that the volume between the inner event
horizon and the outer event horizon is the ergosphere (call this Case
A).  Case A I can understand and it all seems to make sense, but it
does not jibe with your description and with the depictions of some
authors such as at http://www.innerx.net/personal/tsmith/BlackHole.html#ergohor12ring
above (call that Case B).  Case B is puzzling to me and I hope you can
resolve the apparent discrepancy between it and Case A.  These two
sites seem to support my original understanding about Case A:

At http://www.astronomical.org/astbook/blkhole.html the author states
"Within the stationary limit, nothing can escape being dragged around
the black hole.  The region between the stationary limit and the event
horizon is known as the ergosphere".  At
http://www.physics.ubc.ca/%7Epsih/kerr-metric/node5.html the Figure on
that page shows the stationary limit being the outer limit of the
ergosphere.  In both of these cases there is no extra event horizon in
between the inner event horizon and the stationary limit (= outer
event horizon?) to puzzle over what might be going on at its boundary.

Consequently I am uncertain as to what is represented by the 'outer
event horizon' in your spreadsheet.  Is it the outermost radius of the
ergosphere (= the static limit), or is it something inside the static
limit?  From your answer it would appear to be the latter.  I am
however more interested in the outermost radius of the ergosphere.  I
am even more interested however in understanding what the hell an
outer event horizon might represent if it is actually inside the
ergosphere and not the ergosphere's border.

Some questions and concerns about your spreadsheet:
--------------------------------------------------
In your revised spreadsheet's Sheet 1 there is no charge entered at
all for the 100 Kg singularity, and only the inner event horizon's
radius is altered by very large values entered for charge.  Is this
how it is supposed to work?  Charge has very little effect on either
dimension?

I cannot enter a significant value for rpm for the ten-solar-mass case
(such as 60 rpm instead of 2.67094017094017E-09) without a failure
occurring for both the inner and outer event horizon dimensions (I
hope it's not some built-in floating point calculation overflow in
Excel; if it is, can anything be done about that?).

I presume that Rows 5 thru 19 of Sheet 1 are for entering new cases,
correct?  But each of those row's calculations for inner and outer
event horizons reference different rows in Column B of Sheet 2, each
of those representing some value for c, but after Row 37 on Sheet 2
there are increasing values for c up to Row 3949, and I am baffled as
to what those alternate values for the speed of light might be used
for.

If there is indeed a difference between the outer event horizon and
the static limit of a Kerr black hole (i.e. Case B is correct), I
would appreciate some explanation about what kind of boundary the
outer event horizon represents and the inclusion of the static limit
in your spreadsheet.

Thanks very much so far, you are doing very well at penetrating the impenetrable.

Clarification of Answer by tox-ga on 21 Jun 2004 11:32 PDT
simanonok-ga,

First, the simple things. The different values for the speed of light,
c, after Row 37 is a product of Excel's "interesting" pattern
recognition system. Those values should remain constant at 299792458
m/s. Sorry for the confusion. The Excel spreadsheet has been updated
(http://www.maxlin.ca/tos/ga/singularity.xls
).

Now, working backwards. I must impress again that the RPM value you
are entering is the initial speed of rotation of the mass, before its
collapse into a singularity. If you think about it, a planet 10 times
the solar mass of the sun spinning at 60 rpm is very fast indeed. The
centrifugal forces, in these instances, are so powerful that the mass
will fly apart as opposed to forming a black hole. In order for a
Kerr-Newman singularity to form, J/Mc must be < 1. If it is greater
(as it would be if you input 60 rpm as the initial rotation), you
arrive at a complex solution for the inner/outer event horizon radius.
In this case, we have a naked singularity (essentially, a singularity
without the surrounding event horizon, allowing us to see the true
form of the singularity). The ramifications of a naked singularity are
not completely undertstood, though many have conjectured its use for
time travel and/or teleportation. As well, it is correct that a
significant amount of charge is required before any effects are
observed. Any force exerted by the charge is also, as I mentioned
before, very quickly neutralized by incoming material, so for
practical purposes, most black holes have zero charge.

In terms of calculating the static limit, there is no fixed radius
(since it is an ellipsoid). However, I have included in the updated
spreadsheet, the maximum radius of the ellipsoid (essentially, the
radius at the equator).  This was done by setting theta=0 in the
equation for static limit listed above.

Regarding Case A/B, admittedly, I too was confused about which case
was the "correct" one. However, after further consulting with a
physics professor, as well as reviewing multiple sites regarding
Kerr-Newman singularities, I have determined that Case B is the "more
correct" model. indeed, even in your link,
http://www.physics.ubc.ca/%7Epsih/kerr-metric/node5.html, Case B (not
Case A) is the one being presented. Notice how they state that "The
area where S+ > r > E+ is called the ergosphere". A good picture of
Case B, I'm sure you've already found, is
http://www.innerx.net/personal/tsmith/ergohor12ring.gif. The beige is
the Static Limit (outer boundary of the Ergosphere. The red is the
outer event horizon, the green is the inner event horizon, the line
down the middle is the axis of rotation and the ring in the middle is
the singularity itself.

The first thing you encounter when heading towards a Kerr Newman black
hole is two rotating proton spheres. An excellent explanation for the
two rotating proton sphers is as follows:

"Unlike static black holes, rotating black holes have two photon
spheres. In a sense, this results in a more stable orbit of photons.
The collapsing star "drags" the space around it into rotating with it,
kind of like a whirlpool drags the water around it into rotating. As
in the diagram above, there would be two different distances for
photons. The outer sphere would be composed of photons orbiting in the
opposite direction as the black hole. Photons in this sphere travel
slower than the photons in the inner sphere. In a sense, since they
are orbiting in the opposite direction, they have to deal with more
resistance, hence they are "slowed down". Similarly, photons in the
inner ring travel faster since they are not going against the flow. It
is because the photon sphere in agreement with the rotation can travel
"faster" that it is on the inside. The closer one gets to the event
horizon, the faster one has to travel to avoid falling into the
singularity - hence the "slower" moving photons travel on the outer
sphere to lessen the gravitational hold the black hole has."
(http://www.scholars.nus.edu.sg/natureslaw/students/blackhole/types3.html)

Now, the ergosphere is not the space between the inner and outer
horizon, but the space between the outer event horizon and the static
limit. It is typically ellipsoid in shape and billows out from the
equator. While you will be dragged towards the black hole while within
the ergosphere, you can still escape, as long as your are travelling
with the direction of the black hole's spin. What happens at the inner
and outer event horizon seems to be nebulously defined at best. A
brief explanation is that "The outer event horizon switches time and
space as we know it. The inner event horizon, in turn, returns it to
the way we know it." More specifically, inside the outer event
horizon, no matter can escape to the outside and no signal can be sent
outside. Inside the inner event horizon, there will be violation of
causality since one gets closed time-like curves in this region.
Unphysical behaviour takes place, but it is deemed acceptable since no
signals could be sent to the outside world from inside this region.

Something interesting to note is that it is also conjectured that the
ring formation of the singularity actually produces repulsive forces.
The only way to fall into the singularity would be to approach on a
trajectory which follows the black hole's axis. Any other angle of
approach would experience the ring's anti-gravity effects.

I hope you find this useful.

Cheers,
Tox-ga

References:
http://scienceworld.wolfram.com/physics/NakedSingularity.html
http://www.innerx.net/personal/tsmith/BlackHole.html#KerrNewman
http://home.cwru.edu/~sjr16/advanced/stars_blackhole.html
http://www.physics.ubc.ca/%7Epsih/kerr-metric/node5.html

Request for Answer Clarification by simanonok-ga on 28 Jun 2004 05:37 PDT
At this time I want to put my question in a fuller perspective for
you, and see what you think.  The reason for the question was to
attempt to clarify the feasibility of a design for a time machine that
has been presented by a purported time traveler named John Titor.  So
far, nobody's been able to conclusively prove him to be either a
hoaxer or genuine (although there are definitely hoaxers now sometimes
claiming to be him).  A fairly detailed description of Titor's time
machine can be seen at
http://www.johntitor.strategicbrains.com/TimeMachine.cfm; Titor
described it as using twin microsingularities of about 100 Kg mass
each (the 100 Kg value is a ballpark guess based on Titor's claim that
the total weight of his device was about 500 pounds).

In order for his time machine to function, it seems to me that it
would have to use effects from overlapping gravity/time distortion
fields which can extend out over several meters (see Figure 4 at
http://www.johntitor.strategicbrains.com/TimeMachine.cfm).  From your
calculations it appears that the radius of the ergosphere, the static
limit, is far too small for a 100 Kg singularity to be able to
interact with another 100 Kg singularity two or three inches away (see
Figures 6, 7, 8, 9).

On the other hand, my assumptions could be faulty.  I am assuming that
the static limits must overlap in Titor's design for them to produce
something of a linear field of gravity/time distortion that could
contain a vehicle and traveler.  I'm also assuming that the spin rates
of Titor's microsingularities are not so great as to cause the static
limit to approach the speed of light and produce a 'naked'
singularitiy, which might have very different characteristics.

If you wouldn't mind commenting on the feasibility of Titor's machine
relative to your calculations that would be most welcome; presently
however it seems that you have demonstrated that his machine probably
could not work as described.  I wanted your research to not be colored
by any bias resulting from connection to Titor's story, which is why
I'm only mentioning it now, my apologies for somewhat keeping you in
the dark.  If there are any loopholes you see that might still enable
Titor's machine to be a workable design, I'd especially like to hear
about that.  I've had an interest in Titor's story since I first heard
about it almost two years ago, and while I'd like to believe it's
true, I'm much more interested in the truth as to whether it is or not
true, especially because Titor predicted such drastic consequences
(police state and civil war starting up at the end of 2004, nuclear
war in 2015, please see the homepage summary at
http://www.johntitor.strategicbrains.com/).  That is my web site and I
would like to present the answer to this question that you have
provided as evidence that Titor's story is probably false (unless you
have any reason to believe it may yet be true).  Do you mind if I
reproduce most of your words on that site verbatim and reference this
question (I don't know how long Google will let it remain online
however) and your handle tox-ga, or would you prefer that I keep you
entirely out of it?  If you wish, you can email me from
http://www.johntitor.strategicbrains.com/EmailUs.cfm and if you would
like to continue some discussion off this Google venue, I would like
that too.  I greatly appreciate your efforts so far in helping me
understand this complex topic.

Regards,

Karl S.

Clarification of Answer by tox-ga on 28 Jun 2004 16:50 PDT
simanonok-ga,

If you do post my answer on your site, if you could keep me anonymous,
that would be excellent. The last thing I need are staunch believers
of Titor flooding me with their side of the story and denouncing my
reasoning.

You have said so yourself. Looking at it from the perspective of pure
calculations, the event horizons and ergosphere generated by a 100 kg
object would be far too small for them to overlap as Titor has
described. Indeed, the very existence of miniature black holes
themselves are currently theoretical, nothing like it has ever been
observed. There still remains, however, the possibility of spinning
the black holes fast enough to create dual naked singularities whose
behavior would be utterly unpredictable. There is little to no
literature on the behavior of naked singularities, especially not two
mini naked singularities in close proximity. Thus, through this
avenue, Titor's claims are neither refutable, nor verifiable.

It has been conjectured that a single naked ring singularity could
facilitate time travel by passing through the torus' hole. This
appears to be what Titor's machine is attempting to emulate.

I personally do not believe that Titor's claims are feasible. While
you say he has no material gain, he has become almost immortal on the
web through the many websites and discussion groups his story has
spawned. He also has a book and a short production named after him. If
it is a hoax, perhaps all he wanted was attention.

However, I can pass you along to far more knowledgeable person who may
be able to further answer your questions on this subject. He will be
contacting you shortly.

I hope I've helped.

Cheers,
Tox-ga
simanonok-ga rated this answer:5 out of 5 stars
It may not have been the answer I had hoped to hear, but it is a good
answer nonetheless, professional and with attention to detail.  Thank
you for the several clarifications.

Comments  
Subject: Re: Kerr singularity calculations
From: me_-ga on 05 Jun 2004 08:33 PDT
 
:|

I dont think anyone would be able to give a usefull awnser.
The problem is that (all) the effects of a singularity are still
beyond the reach of most parts of physics. One example: The standard
theory states that gravity is passed allong by a force bearing
particle called a graviton. The graviton itself would be effected by
the singularity and therefor the singularity would suck in gravity...
which is very strange.
Science hasnt reached the point at which singularities can be
explained. Some physisists state that the laws of physics dont apply
on a singularity. I believe that the laws of physics we know arent
able to explain singularities yet.
Subject: Re: Kerr singularity calculations
From: simanonok-ga on 05 Jun 2004 15:55 PDT
 
Thank you for apparently trying to be helpful, &quot;me_&quot;, but
you don't really know what you're talking about.

In the first place nobody has every observed a graviton and if
Einstein was correct they don't exist, gravity being explained by him
as an epiphenomenon resulting from mass curving spacetime, not an
effect mediated directly by particles.  Yes there are people who think
there might be such things as gravitons but there is no tangible
evidence for them, and there is evidence that mass bends spacetime as
Einstein predicted.

Secondly, while you might be partly correct about all aspects of
singularities probably not being completely understood at the present
time, in terms of my question you are dead wrong.  The relevant
equations certainly exist to do the calculations I want to understand
and use, but the obstacle is that physicists only rarely define all
the variables and constants they use in their equations, writing in
kind of a secret code that only initiates in their highly specialized
fields can be expected to know.  There are plenty of good starting
points for someone who knows how to surmount this mathematical barrier
(I don't or I wouldn't be asking the question), such as:

http://www.leonllo.freeservers.com/blackworm.html

http://www.astro.ku.dk/~cramer/RelViz/text/exhib4/exhib4.html

http://perso.wanadoo.fr/lempel/trou_noir_de_kerr_uk.htm

http://scienceworld.wolfram.com/physics/KerrBlackHole.html (almost but
not quite defines all variables and constants used)

http://scienceworld.wolfram.com/physics/Kerr-NewmanBlackHole.html
(includes effects of charge, Q)

http://scienceworld.wolfram.com/physics/BlackHole.html

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