# Tag Info

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As I know, Field Theory, that to what appeals the topic creator cannot explain the very powerful gravitation fields . So trying to understand what happens with a photon there are inside the Black Hole in meaning of Field Theory, or Special Relativity, isn't a good idea. The Nature has no the alone space , and the alone time , you can abstractly image ...

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By definition, the horizon corresponds to the locations where the escape speed would be $c$, the speed of light, and maximum speed of causality.

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If you mean the black hole event horizon then the escape speed is $c$, the speed of light. This is in fact the definition of the event horizon radius, known as the Schwarzschild Radius and given by the following formula: $$R = \frac{2 G M}{c^2}$$ You can use this formula for computing escape velocities for larger radius values. Instead of the letter $c$ ...

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For better clarity, let's define the following: Axial direction = the direction the person & light beam are drawn into the BH. Radial direction = the direction perpendicular to the axial direction. If we, looking in the same direction as the person & light are being drawn into the BH, watch the light beam as it is drawn into the BH, we will see the ...

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Now there is a light ray moving outward at the speed of light. I'm afraid that isn't the case; within the event horizon of a Schwarzschild black hole, the radial coordinate is timelike and so, moving 'outward' toward the horizon is as impossible as moving 'backward' in time. This plain to see in the Kruskal–Szekeres coordinates: Image credit See ...

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I think a possible analogy would be to imagine that the singularity is a waterfall. By emitting light, you are trying to send a signal upstream using a tame fish. Outside the event horizon the fish is able to make headway against the current. But the river flows so fast within the event horizon as it approaches the waterfall, that your fish ends up going ...

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How does light behave within a black hole's event horizon? It doesn't behave at all. If the event horizon of a black hole is the distance from the center from within which light cannot escape, imagine a person with a flashlight falls into the black hole. I've explored this with a variety of relativists, and posed this question. The answer comes ...

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Orbiting matter will eventually fall in because viscosity in the disk extracts angular momentum allowing gas to move inwards. As it does so it will release gravitational potential energy, about half of which is radiated and half goes into kinetic energy. Here's another back of the envelope calculation. I'll go with Chris White's notation that the disk mass ...

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Matter does fall in, but it's not trivial for this to happen, and the rate varies a lot between different cases. The thing is black holes don't suck matter in, just as the Sun isn't sucking in the Earth. A test particle in vacuum placed in orbit around a black hole will orbit forever. In an accretion disk, interactions between particles allow for angular ...

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First of all, you should know that anti particle and anti matter are two different names of the same thing. When matter and anti matter pop up near an event horizon, the theory days that sometimes one of them may get absorbed by the black hole and the other one runs free. Since both the particles are not eliminated because they didn't collide with each ...

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The infalling observer can 'see' whatever events are in its past light cone. The past lightcone of the infalling observer at the point of intersection with the horizon does not enclose the entire exterior region. In fact, no point on the infalling trajectory does, even at the singularity. Therefore the infalling observer unambiguously does not see the "end ...

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You can't "scootch the material from the event horizon" because in the coordinates of anything approaching the hole, the matter does in fact fall in. However, you could study for example radiation from the matter. This is thought not to resolve the paradox for several reasons (note I gave a very similar answer to Can the event horizon save conservation laws ...

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My interpretation is that you are raising the following objection to the black hole information paradox: According to observers distant from the hole, causal lines take infinite coordinate time to cross the event horizon. To these observers, infalling information is thus never lost, but only very strongly redshifted; in essence it remains "painted" on the ...

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I wouldn't know that any of the answers above has shown that, from an outsider's perspective, anything can ever reach the horizon, which was essentially the question of the OP. From an in-falling observer's point of view, there is no problem because kinematic time dilation and gravitational time contraction of the rest of the universe ("looking" in the rear ...

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The obvious interpretation of black hole density is the mass of the black hole divided by the volume inside the event horizon. We need to be a bit cautious about taking this too literally because the volume inside the horizon is not coordinate independant so different observers will measure different densities. However we can easily calculate the density ...

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You are thinking classical physics + General Relativity. In the centre of a black hole is a gravitational singularity, a one-dimensional point which contains infinite mass in an infinitely small space, where gravity become infinite and space-time curves infinitely, and where the laws of physics as we know them cease to operate. As the eminent American ...

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The volume of your dust cloud will not change. The shape changes but the volume does not, so the density of the cloud remains constant. The Schwarzschild solution is a vacuum solution and the Ricci tensor is everywhere zero. If you look at section 5.2 of this paper it shows that the change in volume of an infinitesimal volume element is proportional to the ...

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A tidal force happens because different parts of the infalling object are trying to move along different geodesics. Suppose we take the 3-vector $\eta$ to be the distance between two points, then we can calculate how $\eta$ changes as the object falls inwards. If $\eta$ is constant then the distance between the points isn't changing and there is no tidal ...

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A singularity is a hole, cut, rip or tear in a spacetime manifold. Any geodesics coming to it necessarily end there because the manifold ends there. In the worst case, the curvature resulting from the Schwarzschild solution tips all light cones inside the event horizon towards the singularity. Since all and any time-like geodesics and worldlines are bounded ...

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The solution can be found on this wiki page in section "Example: Null tetrad for Schwarzschild metric in Eddington-Finkestein coordinates". Your metric is exactly in the same form (to see that just calculate the inverse of the given $g^{\mu\nu}$). The value of $F$ is slightly different and there is $c$ in several places, but because the metric is exactly in ...

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On the scale of galactic spiral arms, the central black hole is gravitationally utterly insignificant. I'll illustrate with an example, NGC 524. Of spiral galaxies (this is technically an S0, but there's still spiral structure) with measured black hole masses, NGC 524 has one of the most massive. Here's a picture of the galaxy: The visible disk has a ...

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You start from a common and stubborn misconception: that objects acquire mass at high, relativistic speeds. In reality it is the object's momentum $p$ that is relativistic: $$p=\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}},$$ where $m$ is mass, $v$ is velocity and $c$ is the speed of light. See relativistic mass. So your relativistic object could never acquire ...

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Not exactly a swap, time becomes an imaginary number a square root of a negative number. "Formula The Schwarzschild radius is proportional to the mass with a proportionality constant involving the gravitational constant and the speed of light: $r_\mathrm{s} = \frac{2 G M}{c^2}$ where: rs is the Schwarzschild radius; G is the ...

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With your description, if no radiation comes back from the surface of the black hole, the temperature should be the vacuum classical temperature, 0 Kelvin. BUT Hawking predicted a radiation coming out from quantum mechanical interactions with the vacuum at the limits of the event horizon. Hawking showed that quantum effects allow black holes to emit ...

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I think it's the same reason as for the effect of mass: angular momentum is conserved and made of the properties of what is going in the BH. See from the exterior, everything seems (radially) frozen on the event horizon, and you should simply consider that this is the stuff causing the various effects you can measure from the exterior.

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It is a misperception by (non-scientist) writers. The black hole physical model is a model based on General Relativity only. This tells something about real black holes, and have many interesting and complicated properties. Still, all scientists know that in the local conditions of very small scale and high density it is the realm of quantum mechanics, that ...

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Just a small correction, the solution is called Kerr-Newman-dS (without Schwarzschild, because it is just non-rotating Kerr). And a quick search gives a couple of papers: paper 1 paper 2 Hope this helps and answers your question.

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If we consider the total angular momentum of the whole system, i.e. the Kerr black hole plus the satellite, then the total angular momentum must be conserved. So yes, any change in the angular momentum of the satellite will be matched by an equal and opposite change in the angular momentum of the black hole. This is the principle behind the Penrose process ...

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It would not give the impression the universe was expanding. The expansion of the universe is related to the Ricci tensor and scalar. For a black hole of the type you describe the Ricci tensor and scalar are both zero. If you mark out any volume of space and watch it then its volume will not increase with time. However in the black hole geometry the Weyl ...

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So I take particle A and place it in space, then I place particle B 1,000,000 light years away from particle A. Alright. But, just to be sure: In astronomy, and (prehaps somewhat more recently) in cosmology and in physics in general, we understand this measure of "having been apart" as chronometric mutual separation. In your example this means ...

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Thinking from conservation of energy, It seems like for an object moving toward them, they should be able to "swing" into and out of a black Hole's event horizon no matter how strong the gravitational attraction is inside so long as it didn't collide with the singularity. ... @Javier mentioned that if a potential has greater than 1/r^2 dependence, it ...

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Assuming you are stationary with respect to the black hole, the light source appears dimmer due to the fact that the distance between you and it is increasing. This is described by the the equation for luminous intensity. see https://en.wikipedia.org/wiki/Intensity_(physics)

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Suppose I am a long way from a black hole watching you are hovering near the event horizon, then your time is dilated with respect to mine. I won't go into the details since lots of questions hereabouts involve this calculation. I'll just mention the result: $$\frac{d\tau}{dt} = \sqrt{1 - \frac{2GM}{c^2r}} \tag{1}$$ In this equation $d\tau$ is the number ...

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A black hole's gravity has so much force that light doesn't have enough speed to escape the gravitational pull being emitted by the black hole.

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