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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 |
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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
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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
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fd Lightlike orbits in Kerr spacetime
dfg
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#5
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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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