# Tag Info

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The theory of loop gravity would suggest that it would only become more dense in a singularity event. So It would not go away but become very very dense. So if anything the possibility arises That after the event horizon it would either become so dense it would simply just not be detected aside from its push of gravity. So if the theory dictates these events ...

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Yes. It is possible. Why it is possible is because in quantum mechanics it has been proven that they can send signals from one to another. and even appear in different places at the same time. So if you look at the space time curve theory along side of quantum mechanics you find that a beam can be sent back with no issues to the past. for quantum physics ...

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Since the question is "can you recommend a book that talks about these topics with minimal math," the answer is no. It would be even more confusing to describe quantum information and quantum computing without math than with the math, as the concepts aren't as intuitive as say general relativity, which can fairly effectively be described with mostly words. ...

0

have a look at the paper Observing gravitational lensing effects by Sgr A* with GRAVITY for a good review of this area. Bearing in mind the opacity the region around Sagittarius A* our only hope of seeing lensing is if one of the orbiting stars passes behind the the black hole and is lensed, but the expected deflection of the light is too small to be ...

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Gravitational lensing is a consequence of a distortion of space-time by the presence of a (large) mass. Every kind of light (regardless of wavelength) thinks it is traveling in a straight line in this distorted space - therefore gravitational lensing will equally affect all wavelengths.

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A Little History You ask a very good and relevant question. In fact, back in 1958 H.E. Petschek wrote an interesting paper on "Aerodynamic Dissipation". In that paper, he hypothesized that one could, in theory, produce a shock wave in a collisionless medium (like most plasmas in space). This was highly controversial, since the very concept of a shock ...

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Space isn't empty, as I'm sure you've heard before. There's always something between different bodies, such as the interstellar medium. There are also denser regions of space, including molecular clouds and H I/H II regions. Shockwaves can form in any of these places, and propagate through them. There are several different common sources of these shock waves ...

1

Not sure about overall EM radiation from colliding singularities, but one part is almost possible to answer: the Hawking Radiation (HR). As singularities approached, the amount of radiation emitted would reduce slightly in certain directions. HR emitted by Black Hole and 2 (S1 and S2) should be absorbed by the other. Assuming S1 and S2 then have a ...

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First, the video which the question links to augments real data with artistic interpretation. I am sure that the OP and other posters know that, but I just wanted to make sure there was no misunderstanding. I was part of the research group that created the data shown in the video. We worked with a scientific visualization specialist/artist to create the ...

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No, using optical means the radiation flux will always be less that that at the source (the star). Because the flux in a star is not concentrated enough to form a black hole, the same is true from any concentrated radiation that you could get by optical means

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I'm a physics enthusiast and I'm quite interested mainly in astrophysics. Recently, I've come across this very interesting video from PBS Spacetime, where they state that: A black hole is the collection of events that -for outside observers- don't happen at all, even waiting for an infinite amount of time. There is some truth to this. The outside ...

1

The traditional statement is wrong. For instance if you have a shell of collapsing matter then the event horizon that stops everything, even light, from being able to leave forms when the shell still encloses a ball of radius $2GM.$ Which is before any infinite density happens. And you can also have infinite density without a black hole. Make that shell ...

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Both of these are fine heuristic definitions of a black hole, and they are correct enough for most applications. They are both insufficient, however, when your goal is to formulate mathematical proofs (e.g. Penrose-Hawking Singularity Theorems) involving black holes. The mathematical definition of a black hole is much more subtle and clarifies many of the ...

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Annihilation conserves everything What might possibly be unintuitive is that during matter-antimatter annihilation nothing disappears - the particles simply get converted to other particles and energy. From the point of gravity, however, energy is mass - so from the point of an outside observer, if no particles escape the system, then it doesn't matter if ...

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To add just a little to what Chris has said, when the antimatter falls into the black hole - let's say it's a positron - it annihilates with regular matter. In this case, the positron would presumably annihilate an electron, creating two gamma rays (very high energy light) of energy 2*(0.511 MeV). Just as matter cannot escape a black hole, photons (our gamma ...

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Whether the infalling material is matter or antimatter makes no difference. Fundamentally, the confusion probably comes from thinking of black holes as normal substances (and thus retaining the properties of whatever matter went into making them). Really, a black hole is a region of spacetime with certain properties, notably the one-way surface we call an ...

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Your question only really makes sense for a localized tachyon, i.e. one whose wavefunction in position space is constrained to a finite region of space (i.e. has compact support) because that is the only kind that will "fit" inside a horizon of a black hole. And the answer to your question for this kind of tachyon is that it cannot escape a black hole. The ...

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Why is it that black holes emit Hawking radiation? We don't actually know that they do. Hawking radiation remains a hypothesis. It's been around for so long that people rather take it for granted, but there's no actual evidence for Hawking radiation. And when you read the "given" explanation, it doesn't seem to make sense. See Wikipedia: "This ...

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The reason black holes emit radiation is because virtual particles are popping into existence and popping out of existence throughout space including at the event horizon of a black hole. When they pop into existence they pop into existence in pairs that then annihilate within a fraction of a fraction of a second. When a pair of virtual particles pop into ...

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Black holes are basically neutron stars with such a gravitational force that even light cannot escape from it. A black hole is a mathematical solution. A neutron star over the critical mass gets so dense that it forms larger and larger time dilation relative to the outside universe thus we get to see what happens on short time scales. But what ...

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You'll see the object at first accelerate towards the hole (under gravity) and then slow more and more as it approaches the event horizon. It will asympotically freeze in place at the event horizon and then gradually shift redder and redder until it disappears. This is assuming that the black hole is big enough that the acceleration is similar across the ...

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Yes. Since faster-than-light travel is required to leave the black hole, and tachyons apparently propagate faster than light, such a thing would be possible. If tachyons existed...

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As the comments and other answer note, this solution does not exist in the same form in a lower dimensional spacetime. The thing that you probably want to do instead is exploit the symmetry in this spacetime so that you can think about just the parts that have "interesting" behavior. Schwarzschild is spherically symmetric, so that would mean, in this case, ...

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General relativity in low dimensions gets progressively simpler, making it difficult to really make interesting statements on it. Here is the situation : 2+1 dimensions : In 2+1 dimensions, the Riemann tensor depends only on the Ricci tensor. So if spacetime is Ricci flat at a point, it is totally flat. This means that when there is a 0 stress energy ...

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Particle antiparticle potential/hypothetical pairs exist in vacuum as a mathematical description, necessary for calculations of interactions between elementary particles. These mathematically annihilate and reappear within the heisenberg uncertainty principle.In the Hawking radiation case the virtual pairs at the event horizon have a probability one of ...

3

In principle yes, though in practice it's far beyond our current capabilities. The problem is that a charged black hole has to have a Schwarzschild radius greater than $2r_Q$, where $r_Q$ is given by: $$r_Q = \sqrt{\frac{Q^2 G}{4\pi\varepsilon_0 c^4}}$$ If this isn't true the black hole will be superextremal and this (probably) means it's unstable. For ...

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I would have to conclude that it is a black hole most of the time. I would conjecture most of the time we observe a black hole with a very strong magnetic field in addition to its powerful gravitational field. The strong magnetic field can account for a number of the observations that were pointed out in your question. I prefer to refer to it as a ...

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I usually work in the Poincare patch for which the line element is $$ds^2 = {{d {\vec x}^2 + dz^2} \over {z^2}} \ .$$ Given translation invariance in $x$, one can then assume a solution of the form $\phi(\vec x,z) = e^{i \vec k \cdot \vec x } f(z)$. The wave equation for a massive scalar in $d+1$ dimensional AdS space then reduces to $$z^{d+1} \partial_z ... 7 Could the black hole in the center of the galaxy be a white hole? I think not. IMHO there are no white holes. IMHO white holes are a mathematical fantasy. In the center of the galaxy there is a strong radio source which we call Sagittarius A*. Based on the high speed and orbit of nearby stars we have calculated that something with the mass of more ... 1 For a static black hole described by the Schwarzschild metric the escape velocity is:$$ v_e = c\left(1 - \frac{r_s}{r}\right)\sqrt{\frac{r_s}{r}} \tag{1} $$and the orbital velocity is:$$ v_o = c\sqrt{\frac{r_s}{2(r - r_s)}} \tag{2} $$where r_s is the Schwarzschild radius:$$ r_s = \frac{2GM}{c^2} $$If we graph these we get: Note that the x ... 1 The neutrino is a weakly interacting particle, thus it has some quantum numbers (weak isospin, hypercharge) which is summed to zero in the black hole. Thus, even if we don't take in sight the numerous related problems (see Annas answer above), the answer is no. 4 Lets take this a step at a time. A black hole is an object posited by the classical theory of General relativity. It is a legitimate solution for the case of a great accretion of mass which has a singularity at the origin, and it has a definition of a radius called an event horizon which once any mass goes through it cannot come out because of the great ... 4 I'd like to address the misconceptions in your quote from the wikipedia article. ”Like black holes, white holes have properties like mass, charge, and angular momentum. They attract matter like any other mass, but objects falling towards a white hole would never actually reach the white hole's event horizon" They do "have" mass, charge, and angular ... 17 There are numerous misconceptions here, but allow me to address just a few: Black holes do not have "appetite." In order for an object to be consumed by a black hole, the object's trajectory would need to literally intersect with the event horizon (i.e. be on a collision course with it), otherwise the object will continue to orbit the black hole. Because ... 1 Contrary to the comments on the original question, I think the physical situation is theoretically possible. The comments are correct that the black holes cannot maintain a constant distance, but under the scenario described, they will fall directly toward each other and the observer who started half way in between will maintain a symmetric relationship to ... 0 I think the choice of the coordinate set is based uniquely on simplicity: in fact in those coordinate set you easily understand what is the ergosphere. In fact the definition of the Ergosphere is coordinate invariant; Taken the timelike killing vector of the kerr metric$$ \xi = k + \Omega_H m $$where k is the timelike killing vector of usual non rotating ... 0 The answer to the question is simply the Carnot efficiency:$$ Efficiency = 1-{T_U \over T_B} $$Where T_U means the temperature of the universe and T_B means the temperature of the black hole, and Efficiency means the maximum efficiency that is possible. It does not matter if you place the heat engine close to the black hole or far away from it, the ... 8 Any black body in space radiates and ends up very cold, might even crystallize. The law of increasing entropy holds for closed systems, in this case the whole system: "all the radiation that left the black body + the black body itself" microstates. In the sense that a black hole behaves as a black body the same holds true, it cannot be considered a ... 4 Let's assume we are dealing with circular orbits for simplicity. In that case the orbital velocity for a satellite at a distance r from a black hole is given by:$$ v = \sqrt{\frac{GM}{r - r_s}} \tag{1} $$where r_s is the radius of the event horizon:$$ r_s = \frac{2GM}{c^2} \tag{2} $$Now let's rearrange equation (2) to get:$$ GM = ...

0

No, they do not stay in constant orbit. Anything that enters a black hole must reach the singularity, and it will do so by constantly reducing its distance from it. (Think of it this way: If you've entered a black hole, your escape speed is larger than the speed of light, so one cannot possibly expect that you would remain in orbit inside it)

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In a dual time universe, black holes are the antimatter induction terminus of the graviton cycle that powers all atomic and galactic motion. So, what we have is a reverse motion antimatter core that strips gravitons from a forward motion wall and in turn the graviton's flip polarity and become antigraviton's that are being broadcasted at frequencies above ...

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Your observation is correct! The issue is one that is regrettably rather commonly left unexplained in physics texts: A single coordinate system on a manifold does not define the spacetime. Often, the coordinates useful for computations in physics do not even cover the entire space, i.e. they are not defined everywhere on the object of study: The (spacetime) ...

2

Where does our theoretical prediction of the existence of black holes come from? It comes from the singularity theorems of Hawking and Penrose. Before then, people were aware of solutions to Einstein's Field Equations that had singularities but they required absolutely perfect symmetries, such as perfect radial symmetry for Schwarzschild, or perfect ...

1

Nothing can escape a black hole Once it has crossed the event horizon. Quoting from the wikipedia article Physical insight into the process may be gained by imagining that particle-antiparticle radiation is emitted from just beyond the event horizon. This radiation does not come directly from the black hole itself, but rather is a result of virtual ...

1

From a historical perspective black holes weren't predicted. In 1916 Karl Schwarzschild found a solution to Einstein's equations for a spherically symmetric mass. It was only subsequently realised that the Schwarzschild metric is a vacuum solution with an event horizon and a curvature singularity at its centre, and that the metric describes a static ...

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The paper pointed out by Daniel's comment gave me a starting point to find more literature on this topic and do further research. After a while, it became clear to me that my question is actually an unsettled (research) question. Therefore, a definitive answer cannot really be expected. Nonetheless, I think it's valuable to provide something of an answer. ...

2

The event horizon is not a place. If it was a place you could stay there. A black hole has an inside and an outside and the event horizon is the boundary between them. A black hole's event horizon is not the only kind of event horizon. If someone started accelerating to the right at constant acceleration and never stopped accelerating they would have an ...

1

An event horizon is a "sphere-shaped surface of influence" of inescapable gravity from/towards the black hole. It is the point where the object (victim?) cannot overcome the gravity of the black hole, and will be sucked into it. This is also the point where, theoretically, you would vanish to an outside observer, since the light crossing the event horizon ...

1

Under General Relativity, Lorentzian wormholes (the kind that are traversable) require exotic matter (a kind of unobtainium which is not known to exist). This is not true. A maximally extended Kerr black hole solution for instance has a traversable wormhole to a (different) external universe and doesn't require exotic matter. However you do have to ...

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New question. New answer. I don't know know why you didn't ask a new question (more people are likely to look at a new question, so it can only help you). A nonrotating black hole has a spacelike singularity, it is like a time rather than like a place. Imagine that some paths in spacetime run to the future and forever chase the surface $t=+\infty$ and their ...

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