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Lightlike geodesic of the Schwarzschild metric
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Lightlike geodesic of the Schwarzschild metric
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jeffreyH
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Lightlike geodesic of the Schwarzschild metric
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26/12/2015 20:06:33 »
In the thread Lambert's Cosine Law (badly named) I had determined that the lightlike circular orbit around a Schwarzschild black hole would be at 1.5 rs. I then rejected this as wrong. I am reading through Differential Forms and the Geometry of General relativity by Tevian Dray and on page 33 the lightlike orbit of the Shwarzschild solution is calculated to be at 3m. Since the event horizon is at 2m this means the lightlike orbit is in fact at 1.5 rs. Now I know I'm on the right track with some other hypotheses. Including black hole entropy.
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jeffreyH
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Re: Lightlike geodesic of the Schwarzschild metric
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26/12/2015 20:43:12 »
Around the horizon of a Planck mass black hole there exists a region the depth of 1 Planck length where only light can orbit. All massive objects, that is particles with mass, will be unable to sustain such an orbit and fall into the black hole through this entropy interface. As the mass of a black hole increases the depth of the entropy interface will also increase. With the radius of the lightlike circular orbit being the boundary. With one proviso. This is the Schwarzschild solution and so non-rotating. Finding the dimensions of this entropy interface for the Kerr solution should be the aim of research.
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Re: Lightlike geodesic of the Schwarzschild metric
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31/12/2015 14:08:24 »
The relationship between motion perpendicular to a gravitational field and a radial outward pointing geodesic swaps around at the energy convergence point, the lightlike circular orbit at 1.5rs. Usually it is far easier to move perpendicular to the field than away from it. At 1.5rs a circular orbit = c whereas escape velocity < c. So there is already a component to the geometry outside of the event horizon of a stationary black hole that acts like a directional horizon. This has to be considered important if rotation is introduced to the metric.
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