New answers tagged

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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\...


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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 ...


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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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A $10^{12}$ kg Black Hole would be putting out a constant 356 megawatts of Hawking radiation in all directions and it would have an Event Horizon of only 1.4 femtometers (smaller than an atomic nucleus) No particles would even manage to get close to this thing, let alone absorbed by it. However, it's not going to evaporate any time soon either. By the ...


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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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So, to be brief there are going to be two things resisting the entirety of the matter falling into the black hole. The first kind of leads to the second so I'll begin with the first. Assuming the black hole is at the centre of the Earth (these arguments stand up in an orbiting situation as well), as matter near the centre of the earth falls in it will become ...


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A merger process will generate gravitational waves, which can carry angular momentum away from the system. Whether or not this means that the whole Earth is consumed may in part be a question of perspective. The gravitational waves will also carry away mass-energy, so that will cause the final black hole to be a bit less than if the waves weren't created. ...


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You are asking about two things: from the viewpoint of Earth itself As an observer passes the event horizon, the observer experiences nothing special (except for spaghettification). In reality, the observer might not even realize he passed the event horizon. If you disregard the effects of extreme curvature (on the different parts of Earth, and on its ...


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Hawking radiation is the same for every black-hole in the last second of its existence regardless of initial mass. The energy released prior to this last second is minuscule to what comes afterwards. The integration of power over a human time-scale will not be more than the binding energy of the earth and so a single black-hole is not capable of destroying ...


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I belive that, in theory, all you really need is to compress matter enough so that it shrinks below it's Schwartzschild radius, $R_S=2Gm/c^2$, where $G$ is the gravitational constant, $m$ the mass you want to compress and $c$ the speed of light. For small stuff this is practically impossible to achieve: the wiki page lists that to compress 70 kg to it's ...


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Time passes slower around massive objects due to relativity. You are a physical coalescence of matter and energy into "being". Your thoughts and awareness are simply electrochemical reactions unfolding throughout your nervous system over time. Time dilation around a black hole will effect all matter (whether that matter is a part of an egocentric observer or ...


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There are two main cases: SR time dilation, time elapses differently for two observers, because of their relative speed GR time dilation, time elapses differently for two observers, because they are in a different gravitational zone https://en.wikipedia.org/wiki/Time_dilation Now in your case, you are talking about GR time dilation. If you are near a ...


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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 ...


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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 ...


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Gravitational waves occurs due to the rotation of BH non symmetrically ... GW are created by the event horizon of the BH because inside the event horizon nothing is there at singularity.. BH are creating a tunnel into space and time simultaneously.. all the inside information is erased from classical point of view but all the information in the BH are ...


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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 ...


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It is possible to realize lower entropy states through projective measurements or simple measurements where the environment has minimal degrees of freedom. In such a scenario the blackhole micro states are not same as white hole microstes,and thermalisation due to time reversal scenrio can happen


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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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This question has some deep aspects to it, and at this time I really do not think there is a definitive answer to it. This touches on issues of quantum information of black holes and the final state quantum process, which approximately is a form of Hawking radiation explosion of a quantum black hole. So I will not pretend to answer this question in any ...


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The interaction between the BH and the particle antiparticle pair is so, that the negative energy particle will fall into the BH. Now the particle antiparticle pair is in a superposition, they are indistinguishable. It is very important to understand that the particle antiparticle pair is not the same as the negative and normal energy particle. Particles ...


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The most complete but rigorous text on black holes is "The Mathematical Theory of Black Holes" by Chandrasekhar. However, this one is terse and requires knowledge of advanced General Relativity. Other texts include "A Relativist's Toolkit" by Eric Poisson which discusses the topic and the relevant mathematics in sufficient details and "Gravitation:...


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Black holes pull anything and everything that enters past the event horizon into the singularity Ok. unknown output where all of this infinitely condensed matter goes Here's your problem. You probably read some terrible explanation about how black holes crush everything down into the singularity, this point of infinite... whatever. That's a crap ...


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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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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 ...


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There are three distinct but related solutions of Einstein field equations with positive cosmological constant: Nariai solution (N); extreme Schwarzschild–de Sitter (SdS) solution; near-extreme SdS black hole. (1) could be seen as an asymptotic geometry of the (2) or as an approximation of the metric between two horizons of (3). Nariai solution is the ...


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Ben Crowell's answer is right. I hope I can clarify a little. If one starts from the following two assumptions: $R_{ab} = 0\;$ (i.e. field equation in vacuum) The metric can be written in the form $$ {\rm d}s^2 = g_{uu}(u,v) {\rm d}u^2 + 2 g_{uv}(u,v) {\rm d}u {\rm d}v + g_{vv}(u,v) {\rm d}v^2 + f^2(u,v) \left( {\rm d}\theta^2 + \sin^2(\theta) \, {\rm d}\...


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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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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 ...


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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 ...


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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:...


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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.


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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 ...


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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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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.


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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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I'm also interested in the answer to this question, this is how far I got: The spacecraft follows a geodesic, and if it does an impulsive boost at a point it will now follow a different geodesic from that point but with a different 4-velocity. The incoming trajectory starts with velocity $v_0$ at infinity and the new one ends at velocity $v_1$, so the ...


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I'll be using the notation followed by Poisson in his book, "A Relativist's Toolkit". Let $\Sigma$ denote the bifurcation 2-surface, which is null. As you have already mentioned, it will have two normal null vectors, $\xi^a$ (a Killing vector) and $n^a$, along with two tangential spacelike vectors, $e^a_A$, ($A={1,2}$). The vectors $\xi^a$ and $n^a$ are ...


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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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I believe you are referring to the fact that, in the exterior Schwarzschild metric, the speed of light at the event horizon of a black hole is zero. The Schwarzschild solution is static, so the Schwarzschild coordinates are those used by a remote observer who is stationary relative to the black hole. This solution does not apply to a black hole that is ...


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A temperature below absolute zero: Atoms at negative absolute temperature are the hottest systems in the world On the absolute temperature scale, which is used by physicists and is also called the Kelvin scale, it is not possible to go below zero – at least not in the sense of getting colder than zero kelvin. According to the physical meaning of temperature, ...


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I also recently learned about Arago spots and had the very same thought regarding black holes. Concerning your retracting part of your comments, it's true that there isn't evidence black boles generally produce Arago spots. That said, while the variables make it unlikely most black holes would produce an Arago spot they also make it likely that a few black ...


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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 ...


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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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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.


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As you see from the comments, black holes are a solution of GR. You are asking why they keep attracting things. The answer is stress-energy (not mass contrary to popular belief). Though, this could be true for any kind of star, that keeps attracting things. Why are black holes special? There are a few reasons: black holes are black because nothing (not ...


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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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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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I think brightness is more likely related to the amount of infalling matter available from the accretion disc. All quasars are powered by very large, supermassive black holes, but the biggest variable is the amount of matter available. Our own SMBH in Sagittarius is, perhaps fortunately, rather quiet, due to lack of fuel. Many billions of years ago it was ...


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The density of a black hole is not a well-defined thing. Depending on what you mean by density, and what kind of black hole you're talking about, the "density" can be zero, infinity, or anything in between. A Schwarzschild black hole is a vacuum solution to the Einstein field equations, meaning that this type of black hole spacetime consists of nothing but ...


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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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