# Tag Info

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General relativity is based on the equivalence principle, and one way of stating the equivalence principle is that in small enough regions of spacetime, the effects of gravity become undetectable for free-falling observers. So if your cloud of observers is small, or if we restrict our attention to a small part of it, then nothing special happens as far as ...

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If some or all of the mass of an object inside a black hole is converted to energy, there is no effect at all on its event horizon. The total mass/energy of the BH determines the radius of the event horizon. Also, since momentum of a rocket, including the ejected matter is conserved, any accelertion inside a BH does not change its angular momentum.

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You can both ask and answer this question by introducing a coordinate choice suited to an observer freely falling towards the horizon. A good choice is Lemaitre coordinates; see https://en.wikipedia.org/wiki/Lema%C3%AEtre_coordinates To clarify the question, by "observe" do you mean "receive light signals from" or do you mean "deduce, according to an ...

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What exactly would happen depends on a lot of factors. For instance, you cannot quite assume perfect spherical symmetry of the Earth, and how exactly this symmetry is broken changes the outcome. Also, if the black hole is as massive as the Earth (and thus 9 mm in size), half of the planets material is suddenly attracted more to the black hole than to Earth, ...

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1) Think of $K$ as the value of (assuming metric signature (+---) ) $$g_{00}=1-\frac{r_G}{r}=\exp(2v)$$ at the $r$ distance from which the particle starts falling $\left(r_G=\frac{2G_{N}M}{c^2}\right.$ is the so called Schwarzschild radius, with $M$ the mass of the black hole). Note that $c dt$ and such only makes sense in a flat space. The integration ...

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A black hole in a stationary state is believed to be specified by only three parameters: its mass, charge, and angular momentum. (This is the famous “no-hair conjecture”.) These three parameters determine everything about the black hole, including where its event horizon is. In the case of an uncharged, non-rotating hole, the event horizon is a sphere ...

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This won't be a complete answer, because I'm not familiar with Brown-York, but: The Komar mass applies to stationary spacetimes. The ADM mass is for asymptotically flat spacetimes, and includes gravitational radiation going to null infinity. Bondi mass is similar to ADM, but doesn't include the radiation. Komar and ADM were later shown to be consistent ...

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Blackholes are the solution to Einstein's equation. For the introduction, how these solutions/geometries have been obtained look at the textbook Spacetime and Geometry: An Introduction to General Relativity by Sean Carroll This book is very easy to read and understand. More focus should be on the Penrose diagram (Various coordinate systems) which are very ...

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We will use the value of $10^{12}\,\text{kg}$, proposed in OP, for black hole initial mass for estimates. This value is large enough that the evaporation time through Hawking radiation for such a black hole is longer than the current age on Universe, moreover, the accretion rates if such a black hole is placed inside the Earth would be larger than the loss ...

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Short answer: Yes, if an isolated black hole is large enough (supermassive) and has an initial charge comparable to its mass, then it would lose mass through Hawking radiation much quicker than it would lose charge and would eventually reach nearly extreme state. It would still continue to lose mass and charge though at much slower rates and would remain in ...

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The scattering of gravitons of black holes is really no different than that photons (or massless scalars for that matter). The scattering/absorbtion cross-sections are readily calculated using modern black hole perturbation theory. The only difference is the spin of the underlying Teukolsky wave equation. The absorbtion crossection depends on the wavelength ...

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Consider the two-dimensional surface with $r=$ constant and $t=$ constant, which has the line element $$dl^2=r^2d\theta^2+r^2\sin^2\theta d\phi^2$$ where the metric coefficients are $$g_{ij}=\begin{pmatrix} r^2 & 0 \\ 0 & r^2\sin^2\theta \end{pmatrix}; \quad g=\det(g_{ij})$$ The proper area of this 2-sphere is $$\mathcal{A}=\iint\sqrt{g}\,d\... 2 Yes, gravitons must necessarily respond to each other via gravity. This means that in addition to communicating forces between objects that bend spacetime, the force carrier particles bend spacetime themselves. But for very weak gravitational fields, the math gets easier to deal with because in this simplified case, the effect of the gravitons themselves ... 1 GR doesn't have observers who can observe things at a distance, so this type of reasoning about "from the point of view of a distant observer" is a bad conceptual trap to fall into. Observers can only receive signals (such as light rays) from distant objects. Logically connected to this is the fact that GR doesn't have a notion of simultaneity for distant ... 1 I assume the charge accumulates exactly on the inner horizon The standard black hole solutions are vacuum solutions, so there is zero charge density and zero mass density at all points in the spacetime. Real-world black hole solutions form by gravitational collapse, so they're not vacuum solutions, but the mass and charge density at a particular point in ... 1 It's just differentiation! To minimize the amount of calculation, use (1) the fact that the metric is diagonal; (2) the fact that three of the metric components depend only on one coordinate, and the other depends only on two; (3) the symmetry of the Christoffel symbols; and (4) the multiple symmetries of the Riemann tensor. You have to compute at most 40 ... 1 If the "time around" things (and people) was slowing down while the things themselves still age in the same way, then what would the idea that "time slows down" even mean? Actually the only way it makes sense to say that time slows down is precisely when two objects depart then meet again and it appears that both objects did not experience the same lapse ... 1 This will extend the answer of Brick a little. It is possible to disturb the region of spacetime beyond the horizon of a black hole, either by sending gravitational waves into it or simply by dropping matter into it. One will then have gravitational waves propagating around in the region of spacetime beyond the horizon. However, that is where they stay. All ... 1 The answer is no. Gravitational waves travel at the speed of light and are therefore also unable to escape the black hole for basically the same reasons that the EM wave cannot. It has nothing to do with the degree to which they do or do not interact with matter. It has nothing to do with scattering. In order to "escape" the wave would have to propagate ... 1 The jet is produced near the black hole, not inside it. The black hole has a surface around it called the event horizon. Nothing can move outwards through this surface except via an interesting quantum effect called Hawking effect or Hawking radiation. However the Hawking radiation is utterly negligible for the type of black hole you are asking about, so ... 1 The relativistic jets do not originate from inside the event horizon of the black hole. Instead, they originate from material of the accretion disc. While most this material finally crosses the event horizon and falls into the black hole, some part of it makes a turn before reaching the event horizon and gets thrown off in the relativistic jets. From NuSTAR:... 1 Yes, black holes are four-dimensional. Spacetime is four-dimensional (three dimensions of space and one dimension of time), and black holes are just a particular type of curved spacetime. No, black holes are not four-dimensional spheres. For example, far from a black hole, spacetime is effectively flat rather than curved. 1 You don't say anything about what kind of AGN you're looking at, in particular whether these are all in very nearby galaxies, high-redshift galaxies, or some mix of both, so it's rather hard to give a definitive answer. Basically, published stellar mass estimates (either photometric or spectroscopic) are going to be associated with particular studies, and ... 1 In addition to Sean Carroll’s Spacetime and Geometry, I will also point out this course on Coursera by Emil Akhmedov. This is a more problem and assignment oriented course. 1 Let us denote the Kruskal coordinates as u and v. Hence in the Kruskal frame, the light cone coordinates can be denoted by,$$ \begin{align} U &= u+v\\ V &= u-v \end{align} $$Now, you are right that the Kruskal frame metric has an explicit time dependence. So we can't use \frac{\partial}{\partial t} as a timelike Killing vector field to ... 1 Well the fact that you are solving R_{\mu\nu}-\frac12 R g_{\mu\nu} = T_{\mu\nu} instead of R_{\mu\nu}=0 is already a step up in difficulty. Another issue is that even T_{\mu\nu} depends on the metric tensor so the only thing that you usually hope on knowing is written in terms of your unknown. In terms of known solutions I don't know if it has ever ... 1 I think is about mass loss due to radiation. It is loosing mass only, not mechanical energy (or indirectly due to the mass lose). That's mean your K is not constant (because you have your mass in) but the angular momentum is still because radiation is assume to be isotropic. The gravitational potential drop. From there the solution is not very far. 1 Black hole is not an empty thing - it is lot of mass in the (relatively) tiny sphere. The radius of this sphere is different for different black holes - say from atomic diameter to hundreds of kilometers. You may calculate it from the mass of the black hole by the formula$$ r = {GM \over c^2} where G is the gravitational constant, M is mass, and c is ...

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Interactions between photons and mass are complex. For this reason I prefer at first to answer your question with respect to an infalling mass particle. First, you have to avoid some possible confusion: You must distinguish between your coordinate system and what you see. Both are different concepts: One simple example is a Minkowski space: If a Minkowski ...

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If the true shape and texture of the accretion disk (viewed from 45°) at the time when the photon which is closest to the observer is emitted is this: and the rings of which the disk is composed of are rotating with the prograde orbital velocity, taking the light travel time and the counterclockwise rotation of the disk components into account the observed ...

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