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Lightlike orbits in Kerr spacetime



 
 
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  #1  
Old October 31st 12, 06:36 AM posted to sci.physics.relativity,sci.physics,sci.math,sci.astro
Koobee Wublee
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Posts: 815
Default Lightlike orbits in Kerr spacetime

On Oct 30, 9:08 am, Tom Roberts wrote:
On 10/30/12 10/30/12 - 10:18 AM, Mahipal wrote:
Koobee Wublee wrote:
On Oct 20, 4:33 pm, Tom Roberts wrote:


When people say "black hole" without qualification, they
invariably mean a SCHWARZSCHILD black hole, in which light
orbits at r=3M. Hence my remarks above.


That should be r=2M, not 3M. Typo?


No typo. The only closed orbits for light in Schw. spacetime are
at r=3M, 1.5 times the value of r at the horizon.


That depends on what Lagrangian you decide to use. Consider the
following segment of Schwarzschild spacetime. For everything to work,
all coordinate system must be of the distant observers. shrug

** ds^2 = c^2 (1 – 2 U) dt^2 – dr^2 / (1 – 2 U) – r^2 dO^2

Where

** U = G M / c^2 / r
** dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2

It should not be too difficult to rewrite the above equation into the
following.

** ds = sqrt(c^2 (1 – 2 U) – (dr/dt)^2 / (1 – 2 U) – r^2 (dO/dt)^2)
dt

Since the above equation satisfies the action in accumulated amount of
spacetime moving from a fixed point in spacetime to another fixed
point, the Lagrangian can trivially be identified as the following and
nothing else. Good luck proving that is wrong. shrug

** L = sqrt(c^2 (1 – 2 U) – (dr/dt)^2 / (1 – 2 U) – r^2 (dO/dt)^2)

Where

** Action = Integral(L dt)

The Lagrangian is in a perfect form devoid of any variables other than
observer’s own coordinate system. Why is this valid? Because all
transforms that satisfy the null results of the MMX say the AETHER
MUST EXIST, and spacetime is composed from these equations that say
the Aether must exist. The Lorentz transform is fudged mathemaGically
from Larmor’s transform. The Lorentz transform is only a fantasy that
finds no experimental support. Koobee Wublee has talked about this
many times over in the recent past. shrug

Since all the conditions leading to the Euler-Lagrange equations are
valid, thus after much grungy derivations, the Euler-Lagrange
equations can be written into the following simple equation. shrug

** (d^2r/dO^2 – 2 (dr/dO)^2 / r – r) (dO/dt)^2 = c^2 (U / r) (1 – 2 U
– (dr/dt)^2 / c^2 / (1 – 2 U))

For circular motions, the above equations drastically simplifies into
the following.

** r^2 (dO/dt)^2 / c^2 = U (1 – 2 U)

Or

** B^2 = U (1 – 2 U)

Where

** d^2r/dO^2 = 0
** (dr/dO)^2 = 0
** (dr/dt)^2 = 0
** B^2 = r^2 (dO/dt)^2 / c^2 = Normalized orbital speed

Under weak curvature of spacetime where (1 2 U), (B^2 = U) which
indeed agrees with Newtonian limit. However, there is no solution for
U when (B^2 = 1). The highest orbital speed under the Schwarzschild
metric is when (B^2 = 1/8) that yield an orbit of (U = 1/2). So,
under which Lagrangian is (r = 3 G M / c^2)? If you did not utilize
any Lagrangian, shame on you, and you deserve to be publically
spanked. shrug

the subtlety which the original poster [KW] missed is the
difference between the r coordinate and "distance" (which the original poster
incorrectly used). The r coordinate is most definitely NOT radial distance (this
is NOT Euclidean geometry).


By the way, if you insist on your mystical way of mix-and-match
coordinate systems in a simple equation describing distance, there is
no fvcking hope of deriving anything --- certainly not your (r = 3 G
M / c^2) bull****. shrug

Any attempt to measure radial distance from the horizon will
fail, because no ruler can remain in one piece at the horizon
(and at rest wrt the black hole); radar also fails because you
cannot fix a reflector at the horizon, and even if you could
no return pulse would ever reach you. Computationally, the
integral of ds is singular along any path that includes r=2M
(the horizon).


It is not a direct measurement but a translation of what the local
space is projected into the observer’s own coordinate system. Gee,
Tom! Please read Riemann’s work. It is actually common sense.
shrug

Again, the geometry must be invariant regardless who is observing it.
To observe and describe the geometry, one must choose a coordinate
system in the first place. You then needs the metric to completely
describe the geometry. shrug

IOW: distance from the horizon is not well defined; differences in values of r,
however, are well defined (outside the horizon, of course).


Pure nonsense! Yes, the interior of a black hole, which black holes
themselves can only exist at the end of time to any observers
according to the Schwarzschild spacetime, is hopeless beyond any
description by any models of physics forever. However, the size of
the event horizon is not. It can still be related using the same
observer’s choice of coordinate system, and this is all within the
proper interpretation of GR not some self-styled physicists who are
witnessing his make-believe castle in the air collapsing and attempt
to spin more mysticism to prolong the agony of demystification.
shrug

It's always a good plan to study.


Koobee Wublee agrees. shrug

Do that and you'll find that KW makes
outrageously incorrect statements all the time, and essentially everything he
says is flat-out wrong.


Wrong! There remain zero corrections from Tom. Why is that? Is Tom
really intimidated by Koobee Wublee? A better explanation is that
Koobee Wublee has so strong of arguments that cause Tom to be
impotent. Come on, Tom. Don’t behave like the little professor from
Norway and be a small man. Stand up to the criticism. If you are
wrong, you need to admit it and move on. If Koobee Wublee is wrong,
you need to come up with a strong argument to show so. No bull****.
That is what science is all about --- not some cowardly little person
ranting about something he does not understand very well himself.
shrug

I simply do not understand how he and so many other people
around here take pride in their own ignorance and stupidity.


Tossing your education around does not help. It only shows your
weakness. If Koobee Wublee is wrong, you don’t have to bring up your
educational achievement to tell Koobee Wublee is wrong. shrug

Oh, by the way, your beloved perihelion advance of Mercury when
applied with this Lagrangian that Koobee Wublee has brought up does
not yield that 43” per century of advance. You can always get lucky
once but not consistently if you are indeed wrong in the first place.
shrug

Any questions? shrug
  #2  
Old November 1st 12, 05:49 AM posted to sci.physics.relativity,sci.physics,sci.math,sci.astro
Koobee Wublee
external usenet poster
 
Posts: 815
Default Lightlike orbits in Kerr spacetime

On Oct 30, 10:36 pm, Koobee Wublee wrote:

That depends on what Lagrangian you decide to use. Consider the
following segment of Schwarzschild spacetime. For everything to work,
all coordinate system must be of the distant observers. shrug

** ds^2 = c^2 (1 – 2 U) dt^2 – dr^2 / (1 – 2 U) – r^2 dO^2

Where

** U = G M / c^2 / r
** dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2

It should not be too difficult to rewrite the above equation into the
following.

** ds = sqrt(c^2 (1 – 2 U) – (dr/dt)^2 / (1 – 2 U) – r^2 (dO/dt)^2)
dt

Since the above equation satisfies the action in accumulated amount of
spacetime moving from a fixed point in spacetime to another fixed
point, the Lagrangian can trivially be identified as the following and
nothing else. Good luck proving that is wrong. shrug

** L = sqrt(c^2 (1 – 2 U) – (dr/dt)^2 / (1 – 2 U) – r^2 (dO/dt)^2)

Where

** Action = Integral(L dt)

The Lagrangian is in a perfect form devoid of any variables other than
observer’s own coordinate system. Why is this valid? Because all
transforms that satisfy the null results of the MMX say the AETHER
MUST EXIST, and spacetime is composed from these equations that say
the Aether must exist. The Lorentz transform is fudged mathemaGically
from Larmor’s transform. The Lorentz transform is only a fantasy that
finds no experimental support. Koobee Wublee has talked about this
many times over in the recent past. shrug

Since all the conditions leading to the Euler-Lagrange equations are
valid, thus after much grungy derivations, the Euler-Lagrange
equations can be written into the following simple equation. shrug

** (d^2r/dO^2 – 2 (dr/dO)^2 / r – r) (dO/dt)^2 = c^2 (U / r) (1 – 2 U
– (dr/dt)^2 / c^2 / (1 – 2 U))

For circular motions, the above equations drastically simplifies into
the following.

** r^2 (dO/dt)^2 / c^2 = U (1 – 2 U)

Or

** B^2 = U (1 – 2 U)

Where

** d^2r/dO^2 = 0
** (dr/dO)^2 = 0
** (dr/dt)^2 = 0
** B^2 = r^2 (dO/dt)^2 / c^2 = Normalized orbital speed

Under weak curvature of spacetime where (1 2 U), (B^2 = U) which
indeed agrees with Newtonian limit. However, there is no solution for
U when (B^2 = 1). The highest orbital speed under the Schwarzschild
metric is when (B^2 = 1/8) that yield an orbit of (U = 1/2). So,
under which Lagrangian is (r = 3 G M / c^2)? If you did not utilize
any Lagrangian, shame on you, and you deserve to be publically
spanked. shrug


Well, there is a typo. It should read when (B^2 = 1/8), (U = 1/4),
which means (r = 4 G M / c^4). The highest circular orbital speed
under the Schwarzschild metric is sqrt(1/8) c which occurs at (r = 4 G
M / c^2). shrug

Interestingly, the Newtonian condition where (B^2 = U) allows a
circular orbital speed of c at (r = G M /c^2). This means the
Newtonian law of gravity allows more energetic accretion disks seen
around supposedly black holes. shrug

This episode of study concludes the following about self-styled
physicists.

** FAITH IS LOGIC
** LYING IS TEACHING
** DECEIT IS VALIDATION
** NITWIT IS GENIUS
** OCCULT IS SCIENCE
** FICTION IS THEORY
** FUDGING IS DERIVATION
** PARADOX IS KOSHER
** WORSHIP IS STUDY
** BULL**** IS TRUTH
** ARROGANCE IS SAGE
** BELIEVING IS LEARNING
** IGNORANCE IS KNOWLEDGE
** MYSTICISM IS WISDOM
** SCRIPTURE IS AXIOM
** CONJECTURE IS REALITY
** HANDWAVING IS REASONING
** PLAGIARISM IS CREATIVITY
** PRIESTHOOD IS TENURE
** FRAUDULENCE IS FACT
** MATHEMAGICS IS MATHEMATICS
** INCONSISTENCY IS CONSISTENCY
** INTERPRETATION IS VERIFICATION

shrug


  #3  
Old November 2nd 12, 04:42 AM posted to sci.physics.relativity,sci.physics,sci.math,sci.astro
1treePetrifiedForestLane
external usenet poster
 
Posts: 974
Default Lightlike orbits in Kerr spacetime

atoms have angular momentum,
even if one refuses to "be classical" about it;
what does this mean for atoms,
"going at the speed of light?"

shrug


thus:
the very whiteness of the bear de pole;
could it be, that it has more to do
with the contrast with the part of the iceberg
that is not o'er the water?...
as far as evolution goes,
the "aquatic ape" theory is also good.

so, what about the "angle of total reflection off of water
-- Snell's law of refraction between two media --
at high lattitudes?"

thus:
well, of course; you do not dispute the fact that
AnIS and GrIS are both "not lowering in heighth?"

changing wind patterns is an explanation for both,
the ice loss, and the increased snow fall
at the higher elevations.
... "they are not lowering in level"

  #4  
Old November 2nd 12, 07:16 AM posted to sci.astro
Syamu Mamilla M
external usenet poster
 
Posts: 3
Default fd Lightlike orbits in Kerr spacetime

dfg

  #5  
Old November 7th 12, 07:27 PM posted to sci.physics.relativity,sci.physics,sci.math,sci.astro
1treePetrifiedForestLane
external usenet poster
 
Posts: 974
Default revenge of lightconeheads, er Nerds

thou hath mightily shrugged; well,
a whole bunch of minishruggings. anyway,
" " is the mean proportional between " " and " ."

The r coordinate is most definitely NOT radial distance (
is NOT yo momma's Euclidean geometry).

--- certainly not your (r = 3GM/cc) bull****.

shrug
* * * * (the horizon).

It is not a direct measurement but a translation of what the local
space is projected into the observer’s own coordinate system.


shrug
geometry must be invariant regardless who is observing it.
shrug
Pure nonsense! *Yes, the interior of a black hole, which black


shrugshrugshrug


thus:
maybe, it is just Wolframitism. This is totally unreadable.

thus:
virtual coffee cup;
superposition of beingfullness and
empty-being, to briefly employ yodaizm. it also
's invisible -- a superfractional haiku?

--behave just like tiny spinning bowling balls.
thanks; arivaderci pull-0-nium!
 




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