# Tag Info

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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} ... 6 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 ... 3 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, ... 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 ... 3 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 ... 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 ... 3 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 ... 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 ... 3 This presumably stems from the fact that in the coordinate system of an external observer nothing can ever cross the event horizon of a black hole. This is perfectly true, but if you were watching an object fall onto a stellar mass black hole it would red shift to invisibility in a few microseconds and it would look to you just like it crossed the horizon. ... 2 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 ... 2 As stated in the LIGO discovery paper (pdf), the event is placed at 410^{+160}_{-180}\ \mathrm{Mpc} luminosity distance, equivalent to a redshift of z = 0.09^{+0.03}_{-0.04}. This gives a clue as to how one measures the distance for this event. If we know how intrinsically luminous an object (like a star, or a supernova) is, we can compare this to how ... 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 ... 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 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 ... 1 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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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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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 ...

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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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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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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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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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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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Electric fields don't start somewhere and move somewhere else. The field lines with arrows are just a picture telling you the direction (tangent to line and in the direction of the arrow) and the magnitude (stronger where field lines are closer together). Really, there is just two vectors (or one bivector) at each point in space at each time, for the ...

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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 ...

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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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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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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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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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I would have to see the source to answer completely but my first guess would be that the relativistic jets from the black holes are pushing against each other. This might cause some strange wobbling but I would think eventually the equators would lineup and they would begin to orbit each other.

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Most orthodox theory is that black hole has singularity at it's center. A gravitational singularity or spacetime singularity is a location where the quantities that are used to measure the gravitational field of a celestial body become infinite in a way that does not depend on the coordinate system. These quantities are the scalar invariant ...

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