Light orbiting black holes
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- Enola Straight
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Light orbiting black holes
Let's start with the commonly accepted premise that the singularity has such a spacetime warping gravitational field so strong that nothing, not even light, can escape.
Furthermore, the gravity can bend incoming light in a form of gravitational lens.
For a given singularity of a certain mass, is there not an orbital radius where incoming light rays do not simply get absorbed by the singularity, not just deflected, but actually settle into a stable orbit?
If so, wouldn't this orbit eventually become saturated with EM energy?
Furthermore, the gravity can bend incoming light in a form of gravitational lens.
For a given singularity of a certain mass, is there not an orbital radius where incoming light rays do not simply get absorbed by the singularity, not just deflected, but actually settle into a stable orbit?
If so, wouldn't this orbit eventually become saturated with EM energy?
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Actually the radius where light has a "stable" orbit is around 1.5 Schwarzschild radii. Light traveling tangentally to the event horizon orbits at this distance; when you're closer there are no stable orbits because space-time is flowing into the BH. To maintain position you need to accelerate away from it, and as you near the event horizon, the acceleration needed to maintain position approaches infinity. This is why the light from, say, a flashlight, would appear to become infinately red-shifted as the flashlight fell into the black hole.
Admiral V. is right that it doesn't build up to absurd levels; the orbit is fairly easily disturbed by passing matter. Also as energy collects, it alters the mass of the black hole system, which changes the stable light orbit radius. If you passed 1.5 Schwarzschild, though, you would see a brilliant flash as you passed through all the orbiting light, as well as many other wacky effects.
Admiral V. is right that it doesn't build up to absurd levels; the orbit is fairly easily disturbed by passing matter. Also as energy collects, it alters the mass of the black hole system, which changes the stable light orbit radius. If you passed 1.5 Schwarzschild, though, you would see a brilliant flash as you passed through all the orbiting light, as well as many other wacky effects.
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The Schwarszschild Metric is dλ² = (1-2m/r)dt² - dr²/(1-2m/r) - r²[dθ²-sin²θ dφ²], which if restricted to the equatorial plane (θ = π/2, dθ = 0) and if the orbit is circular (dr = 0), then it simplifies to dλ² = [1-2m/r]dt² - r²dφ², and by Kepler's third law r²dφ² = (m/r)dt². Substituting back gives dλ² = (1-3m/r)dt², which becomes singular at r = 3m. Hence, there are no circular orbits with r < 3m, with r = 3m reserved for light. (Note that this orbit is unstable, so that any pertubation will destroy it.)
That the Schwarzchild metric preserves Kepler's third law for circular orbits is easy to verify. The geodesics satisfy [d²x^m]/dλ² + Γ^m_{ij} [dx^i dx^j]/dλ² = 0, where Γ^m_{ij} is the Christoffel Symbol of the Second Kind, where (x^0,x^1,x^2,x^3) = (t,r,θ,φ). With this knowledge, the rest is just taking partial derivatives. If the orbit is circular and in the equatorial plane, then it should also satisfy d²r/dλ² = d²φ/dλ² = 0, in which case the geodesic equation for x^0 = r simplifies to [2m/r²][dt/dλ]² = 2r[dφ/dλ]². That is equivalent to Kepler's third law Ω²r³ = m.
The non-existence of orbits with eccentricity 0<e<1 under the Schwarzschild metric is due to symmetry.
That the Schwarzchild metric preserves Kepler's third law for circular orbits is easy to verify. The geodesics satisfy [d²x^m]/dλ² + Γ^m_{ij} [dx^i dx^j]/dλ² = 0, where Γ^m_{ij} is the Christoffel Symbol of the Second Kind, where (x^0,x^1,x^2,x^3) = (t,r,θ,φ). With this knowledge, the rest is just taking partial derivatives. If the orbit is circular and in the equatorial plane, then it should also satisfy d²r/dλ² = d²φ/dλ² = 0, in which case the geodesic equation for x^0 = r simplifies to [2m/r²][dt/dλ]² = 2r[dφ/dλ]². That is equivalent to Kepler's third law Ω²r³ = m.
The non-existence of orbits with eccentricity 0<e<1 under the Schwarzschild metric is due to symmetry.
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