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  4. Is there a possible Killing vector?
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Is there a possible Killing vector?

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Offline jeffreyH (OP)

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Is there a possible Killing vector?
« on: 31/12/2018 21:29:27 »
Is it possible to define a killing vector at the north or south pole of the event horizon of a Kerr black hole?
« Last Edit: 31/12/2018 21:34:57 by jeffreyH »
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guest47899

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Re: Is there a possible Killing vector?
« Reply #1 on: 01/01/2019 15:49:44 »
an uncharged, rotating black hole, with diminishing angular momentum, implies no electromagnetism, implies gravity without an EM light component.

a killing vector implies, a velocity in which rotation can no longer sustain mass in continuity with gravity.

a kerr bh has skewed rotational angular momentum, implying that "poles" are in continual shift

so the function for determining a killing vector must include, poles angular momentum shift, velocity of rotation, bh mass, gravity, kinetic energy, em light

s = angular momentum shift, vr = rotational velocity, b = bh mass, g = gravity, k = kinetic energy, em = light

when does vr slow to destabilize s

when does b diminish to the point where it's g recedes

what is the reason for these conditions in a kerr bh

the kerr bh hole is a dying bh

the kerr bh is a primodial galaxy bh or a neutron star that has lost or in the process of losing its EM light

k = EM * g2

with a reduction of em light, the bh gravity field no longer produces kinetics energy

with the reduction of kinetic energy, the bh consumes it's accumulation disk

with a consumed accumulation disk, the bh dies.

a neutron star is a smaller example of a kerr bh in the making.  in that it has not expended the last of it's em potential
it spins at incredible speeds that exceed its gravitational mass capabilities. is emits gamma waves in death pulses
it's rotational spin has not deteriorated to the point of losing angular momentum

when two unstable black holes collide, you have two attractive gravitational forces

one attractive gravitational force is greater than the other

the lesser kerr bh gravitational force acts as a negative

both are lacking EM light as a component, this reduces their complexity to the point where only the attractive forces of gravity is in play

the accumulative gravitational forces without em light of the two forces produce a gravity wave

the dynamics of the gravity waves can be used as a static component in a function with em light to explain how the Sun's gravity interacts with planetary gravity.










 
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guest47899

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Re: Is there a possible Killing vector?
« Reply #2 on: 01/01/2019 16:08:09 »
does LIGO point its gravity wave telescopes in the direction of the earliest known portion of the Universe?
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Offline jeffreyH (OP)

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Re: Is there a possible Killing vector?
« Reply #3 on: 01/01/2019 22:50:59 »
@Pesq Does any of that nonsense resemble any of this.
http://mathworld.wolfram.com/KillingVectors.html
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Offline jeffreyH (OP)

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Re: Is there a possible Killing vector?
« Reply #4 on: 01/01/2019 22:53:30 »
See also.
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guest47899

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Re: Is there a possible Killing vector?
« Reply #5 on: 01/01/2019 23:45:20 »
Quote from: jeffreyH on 01/01/2019 22:50:59
@Pesq Does any of that nonsense resemble any of this.
http://mathworld.wolfram.com/KillingVectors.html

did you ask the original question because you didn't understand what the math meant or its applications?
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guest47899

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Re: Is there a possible Killing vector?
« Reply #6 on: 01/01/2019 23:49:33 »
Quote from: jeffreyH on 01/01/2019 22:53:30
See also.

listen to the video, and feel comfortable that the scenario I laid out was a practical application example for the killing vector in regards to a kerr blackhole. I am not trying to be pretentious.
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Offline jeffreyH (OP)

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Re: Is there a possible Killing vector?
« Reply #7 on: 02/01/2019 19:56:03 »
Quote from: Pesq on 01/01/2019 23:49:33
Quote from: jeffreyH on 01/01/2019 22:53:30
See also.

listen to the video, and feel comfortable that the scenario I laid out was a practical application example for the killing vector in regards to a kerr blackhole. I am not trying to be pretentious.

No, you were posting nonsense.
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Offline jeffreyH (OP)

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Re: Is there a possible Killing vector?
« Reply #8 on: 02/01/2019 19:59:02 »
You may need to better understand the symmetry implied in both the Schwarzschild and Kerr solutions. Also to understand why the symmetry relates to the killing vector. Go study it.
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Offline jeffreyH (OP)

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Re: Is there a possible Killing vector?
« Reply #9 on: 02/01/2019 20:03:27 »
Oh and just in case you think this is about dying black holes ....
https://en.m.wikipedia.org/wiki/Wilhelm_Killing
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Offline jeffreyH (OP)

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Re: Is there a possible Killing vector?
« Reply #10 on: 02/01/2019 20:10:36 »
I think the PDF at the following link mentions singularities at the poles of a Kerr black hole. Does this look like a respectable paper?
https://arxiv.org/abs/1408.6316
« Last Edit: 02/01/2019 20:15:18 by jeffreyH »
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Offline Bored chemist

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Re: Is there a possible Killing vector?
« Reply #11 on: 05/01/2019 02:10:07 »
Quote from: jeffreyH on 31/12/2018 21:29:27
Is it possible to define a killing vector
I don't know.

Can you define it?
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Re: Is there a possible Killing vector?
« Reply #12 on: 05/01/2019 03:51:50 »
You mean a vector field consisting of? A Kerr black hole is not electrically charged so I guess you're thinking of gravity?

" A gravitational field generated by any massive object is also a vector field. For example, the gravitational field vectors for a spherically symmetric body would all point towards the sphere's center with the magnitude of the vectors reducing as radial distance from the body increases. " https://en.m.wikipedia.org/wiki/Vector_field
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Offline jeffreyH (OP)

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Re: Is there a possible Killing vector?
« Reply #13 on: 05/01/2019 17:24:51 »
Quote from: Bored chemist on 05/01/2019 02:10:07
Quote from: jeffreyH on 31/12/2018 21:29:27
Is it possible to define a killing vector
I don't know.

Can you define it?

In Minkowski spacetime the tangent to a particles worldline is zero since it matches the worldline. For the Schwarzschild metric there is a spherical symmetry to the metric so the tangent does not change due to any spherical rotation. This is not the case for the Kerr metric. The symmetry is a rotation about the equator of the metric.  A rotation around the poles other than 180 degrees breaks the symmetry. It is a special case. Since relativistic jets appear at the poles of rotating black holes this must impact the definition of the killing vector at this point.

EDIT I need to correct something here. A 180 degree rotation still breaks symmetry since the object then rotates in the opposite direction and so will impact the worldline.
« Last Edit: 05/01/2019 23:38:34 by jeffreyH »
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Offline jeffreyH (OP)

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Re: Is there a possible Killing vector?
« Reply #14 on: 05/01/2019 17:27:46 »
BTW It may be simply the magnetic field that inhibits the definition.
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