New answers tagged

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You seem to be asking two things, could a black hole's magnetic fields cool nearby matter, and could this cooling produce cold fusion. But maybe we should first ask whether "cold fusion" is a real thing. Nuclei contain protons and neutrons held together by pions. Fusion is when two nuclei become one. The barrier to this happening, is the positive electric ...


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Yes, you are right...We see object because when light hits the object, it reflect and reaches our eye through space, of which image is made at Retina.. Black Hole's gravitational field so strong that not even a light can escape from it...Though we cannot see a black hole itself, but we may see the hole's effects on nearby matter.We can detect it by help of ...


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Today, seeing an object is pretty much done by capturing the electromagnetic waves emitted or reflected by that thing and forming an image of that object. Black holes do not reflect or emit electromagnetic waves (except Hawking radiation which may be too small amount for seeing purpose). The reason for not reflecting and not emitting is that the enormous ...


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Every object has an escape velocity, the velocity that if given to an object, said object will never fall back. On Earth that is 11.19 meters per second. For a black hole that velocity is greater than the speed of light or 299,792,458 meters per second. If the escape velocity is greater than that, then light does not have sufficient velocity to escape from ...


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The event horizon of a black hole marks the limit of a gravitational field so strong that any known object in this universe, including a photon of light, would need to attain escape velocity greater than the speed of light in order to overcome the pull of the gravitational field. As the speed of light marks the upper speed limit in this universe according ...


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By definition a black hole is an object so massive that its gravity prevents anything from escaping it including light, once inside its event horizon. Its effects can be observed, however. It would appear as a black sphere against background stars and the like.


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Well there is two answers: 1. It is not infinite. At a point when the black hole has taken in lots of matter, it will throw some out of the black hole to again be able to take in matter.It is like the biggest foodie in the world who can eat a lot but at a point when his tummy is full, he needs to throw up or wait until he can take in more. 2. It is infinite. ...


1

Black holes are in the realm of General Relativity. In GR even the law of conservation of energy is under question when approaching singularities of the GR solution. Potential energy is a concept that comes with conservation of energy. Where the singularity in the black hole solutions is dominating, one cannot talk in terms of energy conservation and ...


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Your question What is the potential energy of a black hole? doesn't make sense because energy is a somewhat tricky concept to deal with in GR. If we treat the black hole as fixed we can study the motion of a test particle falling into it, and we find that there is a quantity analogous to total energy that is constant as the particle falls in. So in ...


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I'll try to boil down several of your questions and answer what I think is most fundamental, and hopefully clarify things in the process: Gravity is completely synonymous with the shape of spacetime across all 4 dimensions (3 space, 1 of time). The reason we speak of spacetime is thus: When you (having negligent mass) stand in a "gravity field" such as ...


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You are quite correct that any external observer will never see a true event horizon form, however they will see an apparent horizon and an apparent horizon generates Hawking radiation. So yes you will observe Hawking radiation from an object that is on its way to becoming a black hole.


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Good question, Rovelli in his book Quantum Gravity writes: 1.1.3 GR is the discovery that the gravitational field and spacetime are the same entity. What we call 'spacetime' is itself a physical object, in many respects the same as the Electromagnetic Field. Hence gravitational waves, as the EM field has waves; he adds: We can say that GR is the ...


3

The picture of a black hole being like a huge vacuum is pretty misleading, not to say wrong. The gravitational field around the black hole is exactly the same as around any other object of the same mass. Like any other object, it's perfectly possible to orbit a black hole, just like we orbit the Sun and don't fall in. In fact, if you were to magically ...


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You've the answer in your statement: ...black holes are giant vacuums that will absorb/consume the area around it. The key is that the area around it is a few times the event horizon of the black hole and we are very far from the event horizon of the supermassive black hole at the center of the Milky Way. The event horizon for Sag A* (the name of the ...


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Stars orbiting black holes (I assume that's what you mean) and observed from afar will have their light doppler shifted due to (i) gravitational redshift; (ii) the relativistic doppler effect due to their orbital motion. Effect (i) becomes more important the closer a star gets to the event horizon of the black hole. The redshifted frequency is given by ...


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Yes, they absolutely would. In general, a light-ray which passes at a minimum distance $x$ to the BH will have all of the same effects as a light-ray emitted at the same distance $x$.


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The faster processor would have less mass near it. Furthermore if Photons and Time are slowed by the back hole's gravity creating a red shift then phonons would also be slowed and pulled in by gravitational pressure making matter colder after passes the photon sphere. Temperature and energy would be slowed near absolute 0 before the horizon. It would be ...


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Both vertical lines in this diagram, i.e. one "after" the evaporation as well as the vertical line in the left lower corner (and all vertical lines without "teeth" in all Penrose diagrams in the world) describe the vicinity of the point $r=0$ in polar coordinates. One can't "cross" these lines because there are no points with the radial coordinate $r\lt 0$. ...


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The answer to this relies on knowing about the extended solutions for the Schwarzschild solution, namely the Kruskal solution. In this solution, there is not just a black hole, but a pair of a black hole and a white hole. Particles move from the white hole in the distant past, and then eventually fall into the black hole. Therefore, the time-reversed ...


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Both schools, as described by you, are wrong. There seem to be lack of agreement regarding conditions (like gravitational acceleration and time dilation) at the event horizon. Gravitational acceleration isn't a thing. The metric evolves according to the Einstein equation, test particles move along geodesics determined by their tangents, and the ...


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The answer to your first question can be answered regardless of your misunderstanding of Big Bang. The time passing before atoms begin to form is 379,000 years, which is the time it took for the temperature to drop sufficiently for atoms not to be constantly ionized. – pela, in a comment.


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https://en.wikipedia.org/wiki/Geodesic_deviation This is due to tidal forces. The equations of geodesic deviation which are valid for objects that are small compared to the (radius) of curvature are the easiest way to qualitatively see that. If you plug in the coefficients for the Schwarzschild metric in the conventional distant observer form you will get ...


3

For a star to be turned into a black hole it needs some inward force compressing the matter. In nature this force is gravity, pulling the star's material inward. An additional compression occurs when the star's outer material bounces off of the dense core and is expelled outwards. This is the same type of process that happens during a core collapse ...


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If it's inside the event horizon, can this light escape it? no , it cannot escape. Anything falling cannot see the ground, nor send message back, meaning in the sphere surface direction. can you escape it by the speed of light itself? no, it not sufficient the light is red-shifted means that an object near the BH may emit or reflect this light ...


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The space between galaxies isn't that much more empty than what we have here. There is some curvature, enough so that it causes gravitational lensing, even enough to make us suspect there's some extra matter (dark matter) we can't detect by other means. What you're asking is directly measurable: it's the gravitational blueshift from whatever light sources ...


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Gravity is acceleration. Einstein's equivalence principle says that gravity (with the vector pointing toward the center of the mass) is equivalent to actual movement with acceleration pointed "outward". That's why we observe gravitational blueshift. Now, blueshift means that the frequency of the photon received is increased as compared to its frequency at ...


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Yes and no. Remember in special relativity whenever someone asks a question, they always are told to draw a spacetime diagram. The same thing happens in general relativity. If you want to see what is possible, consider drawing a Carter-Penrose diagram. For a black hole you can draw the event of a test particle crossing the event horizon. The past light cone ...


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There are actually some nifty simulations that show what you would see: http://jila.colorado.edu/~ajsh/insidebh/intro.html (Had to post as 'answer' because I don't have enough reputation to comment)


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The answer to this question is covered in the book "Exploring black holes: Introduction to General Relativity" by Taylor & Wheeler (2000), within the framework of classical General Relativity. If we are talking about a supermassive black hole, such that a free-falling observer can survive tidal forces as they approach the event horizon and the ...


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Technically: It will be full dark. you won't able to see light in any direction because all light will be absorb after event horizon. Update: The light you will see inside the black hole it will be the light before you hit the event horizon. Does someone falling into a black hole see the end of the universe? front view(full black) side view(half black) ...


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In a comment you say (I fixed a few words in this quote) "If it was a neutron star an atom would lose its electrons and protons before becoming a part of that star"... And in the question you say, "Are photons absorbed by atoms compressed out by gravity". This reminds me of Feynman's father. If you Google "feynman father photon", you should find the story ...


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Notice the photons are reduced around the smaller one. Is that happening before the photon sphere? The photon sphere by definition does not send any photons in our direction, as it is a spherical region of space where gravity is strong enough that photons are forced to travel in orbits. So the photons seen come from the region before, ...


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Since most the universe is empty space The short answer to this surprise is to see that the density $\rho = \frac M V = \frac M {\frac 4 3 {{\frac{2 G M}{c^2}}^{3}} \pi} = \frac{positive-constant} {M^2}$ decreases quickly as the mass increases. Now let's play with the values of size and mass that one can find on Wikipedia and other publications. ...


0

The proportionality between angular momentum and mass of the Kerr black hole can be shown directly by performing the Komar integral for angular momentum. In fact, you find that $J=Ma$. The parameter $a$ in the Kerr metric is thus the angular momentum per unit mass. I am not familiar with the relation between $M$ and $R_S$


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For a black hole, the temperature at infinity, in natural units $$ T = \frac{1}{8\pi M}$$ and the entropy $$S = \pi R^2 = 4\pi M^2$$ The change in heat content $$dQ = T\ dS = 8\pi M T\ dM = dM$$ So $$C_v =\frac{\partial Q}{\partial T} = \frac{\partial M}{\partial T}$$


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Lets take this one at a time: When matter enters a black hole with a singularity What is matter? Ultimately matter is a bound complex of elementary particles and energy. There exist free, not in a composite structure, elementary particles with energy too, the electrons and the neutrinos and the photons. The photons in aggregate build up the ...


2

Your case is not quite watertight - it hinges in your assertion that the optical light that is seen comes from some way out from the black hole (SMBH). The thing is that gravitational redshift can be larger than 0.2 and it is also aided by the relativistic transverse Doppler effect in the orbiting material. Some details: Gravitational redshift around a ...


2

"Space is big. Really big. You just won't believe how vastly, hugely, mind-bogglingly big it is." It seems like the night sky isn't so very big, so it should be easy to observe objects, right? Wrong. Once you use telescopes, the night sky becomes huge and if you don't know what you are searching for, you'll only find out about it by coincidence. The ...


3

All we can tell (assuming of course that the conclusions of the CalTech team are correct) is that there is a large mass in a distant orbit around the Sun. The mass could in principle be anything, but some things are more likely than others. It seems very plausible that the mass could be a planet that got ejected from an orbit nearer the Sun because: we ...


1

In principle, any body with the mass of the proposed planet would have the same gravitational effect as the planet. Therefore, it would explain the orbit of those other bodies equally well. We know of a lot of planets in orbit around stars (and we have theories about how they form). However, I don't think we've ever seen or theorized black holes in orbit ...


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Here is the abstract of the original paper by Hawking: In the classical theory black holes can only absorb and not emit particles. However it is shown that quantum mechanical effects cause black holes to create and emit particles as if they were hot bodies with temperature hκ2/πk≈10^−6(M⊙/M)∘K where κ is the surface gravity of the black hole. This ...


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As a the comments say you are setting up a rather complicated scenario. There are a few factors that make the calculation tricky, and one of the biggest is the rigidity of the earth itself. Imagine for instance that you would create the black hole at the centre of the earth; yes, matter would fall into the black hole, bu the speed at which this happened ...


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For an asymptotically flat metric, the proper time measured by a "stationary" observer (defined here as one whose path through spacetime only has changing $t$, and no changing spatial coordinates) is $$ d \tau = \sqrt{ - g_{tt}} dt, $$ where $g_{tt}$ is the time-time component of the metric. For a "weak" gravitational field, this works out to be $$ g_{tt} ...


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From any direction away from the singularity that you look, you would see a circle of light that contains light that had approached the event horizon from all directions such that the incident light could have orbited the black hole at least once.


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The region outside a star of mass $M$ has a certain type of curvature. That type of curvature gets stronger the deeper in it goes. If you had a star of mass $M$ and then had a shell of mass $m$ around it, then the curvature would be of type $M+m$ outside the shell. And it would be of type $M$ inside the shell but outside the star of mass $M.$ So a big ball ...


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In fact, it's easier than it seems. When gravity affects an object, his space-time is being curved by the mass of the one who provocates gravity. So, if time and space are being curved, the distance the object has to travel is longer, so it's his time. In conclusion, mass curve space-time, and, by curving space time, time appears, and last longer. I ...


2

As has been pointed out in the comments, it's not entirely clear how you intend to specify a metric without the use of some set of coordinates. That said, a couple of common GR texts have non-standard approaches to the Schwarzchild metric that you might find interesting. Misner, Thorne, and Wheeler's Gravitation has a fairly detailed sidebar (Box 23.3, ...


1

Yes. When you look at a (Schwarzschild) black hole with a sufficiently good telescope (according to GR), you see arbitrarily (infinitely) many "Einstein rings". Between each concentric pair of rings, there is an image of the entire surrounding sky. These images correspond to the light paths (null geodesics) that are unbound but very close to the photon ...


1

A singularity is defined as a point of infinite density and they are believed to reside at the center of a black hole . The "when does the singularity form" question, is answered concisely here.


2

Event horizon for Static non rotating charge less black hole (Schwarzschild black hole) is defined as spherical surface of radius $$r_s=\frac{2GM}{c^2}\sim 2.95\frac{M}{M_{\bigodot}} Km$$ Where $G$ is gravitational constant and $M$ is mass of the black hole and $M_{\bigodot}$ is the solar mass.So as the mass contained inside increases the event horizon ...



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