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Let's take the carbon nucleus as a convenient example. Its mass is $1.99 \times 10^{-26}$ kg and its radius is about $2.7 \times 10^{-15}$ m, so the density is about $2.4 \times 10^{17}$ kg/m$^3$. Your density is ten orders of magnitude too high. The Schwarzschild radius of a black hole is given by: $$r_s = \frac{2GM}{c^2}$$ and for a mass of $1.99 \... 24 The shape of a black hole's event horizon depends on who is asking. Observers who are moving quickly towards a hole, for example, will see a different shape compared to those who are not. Per @benrg the event horizon of a static black hole is not observer dependent, similarly to how the shape of an expanding flash of light is not. In the coordinates ... 21 Suppose you have two black holes of the same mass$M$and$m = GM/c^2$. The radius of each black hole is then$r = 2m$, and the horizon area is$A = 4\pi r^2 = 16\pi m^2$. Two constraints are imposed. The first is that the type-D solutions have timelike Killing vectors, which are isometries that conserve mass-energy, and with the merger the gravitational ... 21 Energy (in any form) falling into a black hole contributes to the mass of the hole, and mass is one of the many forms that energy can take, using the usual conversion factor:$E = mc^2$. 18 Yes, they are perfect spheres. But let's understand what is the sphere and why. The spherical surfaces are the horizons. They are surfaces in space that have perfect spherical geometry. Now, that only holds true for various conditions: 1) they are static black holes, and if they arose from matter/energy collapsing they are in their equilibrium states. ... 17 Assuming that the black hole is large enough that one can cross the event horizon without being spaghettified by tidal forces, and the that when the accident happened, the both of you were hovering in place above the black hole with your jetpacks, rather than orbiting it: You can still see your friend (no matter how long you dally, the part of his worldline ... 14 The total surface area of the event horizons never decreases. We won't consider charged black holes, since in real life, black holes never have very large charge. However, they can have very large angular momentum, as LIGO showed. The horizon area of a rotating and uncharged black hole (the Kerr metric) is $$8 \pi M \left(M+ \sqrt{M^2 - a^2}\right),$$ ... 9 To expand on @dmckee's answer, if we have a speacetime that has the matter concentrated in a central area, we can definte an overall conserved energy-momentum vector called the ADM energy. It can be further shown that the ADM energy does not change when the matter falls into the black hole. 6 For a spherically symmetric object the relative rate that time flows is given by: $$\frac{d\tau}{dt} = \sqrt{1 - \frac{GM}{c^2r}}$$ where$G$is Newton's constant,$M$is the mass of the object (the Sun in this case),$r$is the distance from the object and$c$is the speed of light. The mass of the Sun is$1.989 \times 10^{30}$kg and the radius of the ... 6 I will first show a simple way to think of Hawking radiation and how we might indirectly observe it. If you have a particle on an accelerated frame it is within what is called a Rindler wedge. Below is a diagram of spacetime for an accelerated frame. We think of a particle in region I The hyperbolic lines are regions of constant radius from the$45$degree ... 6 The innermost stable orbit is a theoretical concept that applies to a "test particle" orbiting a black hole. A test particle in this case would mean that it has negligible mass and so does not alter the metric (Schwarzschild, or more realistically Kerr) it is in. If an orbiting black hole were much less massive than the black hole it orbited, then perhaps ... 6 See also a separate related answer by Patrick Gupta, second answer to Did merging Black Holes in GW150914 give up entropy and information to the gravitational waves, since they lost 3 solar masses? (the question was faulty). He calculated the final horizon area for the rotating case, with entropy 1.57 times the original entropy, so entropy did grow, and the ... 4 The Eddington limit is a theoretical maximum accretion rate onto an accreting body under certain idealised circumstances. The limit is calculated by assuming that the gravitational acceleration felt by the accreted material is exactly balanced by radiation pressure, under the assumptions that the material is completely ionised hydrogen, that the opacity is ... 4 Susskind's original argument doesn't work. Alice just needs Bob to send a message to her saying "I'm still alive!" She doesn't have to illuminate Bob. Of course, it's hard to get a message out from the near-horizon region of a black hole because of the redshift, but there's no theoretical reason that this shouldn't work. Suppose you don't have a ... 3 ANSWER WITHDRAWN I am withdrawing my answer because I am persuaded by Henning and others that I am mistaken about the impossibility of catching up with someone who has crossed the Event Horizon. I have also withdrawn my Vote-to-Close. Original Answer What do you mean that your friend has "fallen into a black hole"? If you mean that he has crossed the ... 3 You've forgotten an important player in the system: the gravitational field. Here's a pretty argument that gravitational fields are physically meaningful objects that carry energy: imagine two masses accelerating towards each other from rest, from a great distance away. The rest energy of the system is$E_\text{rest} = (m_1+m_2)c^2$; the kinetic energy is ... 3 For non quantum effects an equilibrium spherical black hole horizon is smooth, perfectly symmetrical. It is defined by only one parameter (mass, if no charge and no rotation, and thus spherical) - thus it is said it has no hair. If it is perturbed it will deform, and acquire some hair. For perturbations that are not very large, and not spherically symmetric, ... 3 Outside of its event horizon the gravity from a black hole is exactly the same as the gravity from any object that isn't a black hole. So black holes just make up part of the overall stuff in the universe. There are no effects on the universe expansion that are a special result of black holes being present. The gravity of the black holes certainly does ... 2 This horizon involves time, and to be a real horizon the time of observation must tend to infinity for observing a falling probe particle. The particle field is assumed to be much smaller than that of BH. In this case the space-time geometry is smooth. 2 Hawking radiation is a theoretical result from the analysis of quantum field theory in general relativistic, curved spacetime. This sort of semi-classical analysis, combining aspects quantum mechanics of with classical GR, is a stepping stone to a full theory of quantum gravity. The theoretical result does not prove that Hawking radiation exists. If ... 2 Let me try to explain this by making an analogy with a simpler system i.e. a hydrogen atom. If you measure the mass of a hydrogen atom you find it is less than the mass of an electron plus the mass of a proton. In fact it is 13.6eV less. This happens because if you let a separated electron and proton fall together under their mutual electrostatic attraction ... 2 On top of Rob Jeffries' answer, I'd like to add that gravitational wave radiation makes the black hole lose orbital energy$E$. But for really huge black holes this won't make much of a difference ( see the calculation). Very rough estimates for the energy: Let's take an order of magnitude estimate: The gravitational wave luminosity in the 1st Post-... 2 You can show within Newtonian gravitation that the net gravitational force at the center of a continuous mass distribution with a nontrivial point group symmetry is zero. (If the body doesn't have a point group symmetry then there's no clear notion of a "center".) Black holes fail to satisfy two of these conditions - they're not Newtonian and the mass ... 2 The minimum radius of a spherical object to be a black hole is given by : r = 2Gm/ (c^2) From this, I think we may be able to calculate the minimum density for the object to be a black hole, which is: d = (21/704)((c^6)/((G^3)(m^2)) (Assuming pi = 22/7) it is, d = (7.37 x 10^79 )/(m^... 2 There are the${\bf k}_t,~{\bf k}_\phi$Killing vectors. Another condition is that $${\bf k}_t\cdot{\bf k}_\phi~=~\frac{(2mr~-~Q^2)asin\phi}{\rho^2},$$ for$a~=~J/m$. This is zero for$a~=~0$or for$Q^2~=~2mr$or$\phi~=~0,~\pi$. There is also $${\bf k}_\phi\cdot{\bf k}_\phi~=~\frac{(r^2~+~a^2)^2sin^2\phi~-~\Delta a^2sin^4\phi}{\rho^2},$$ for$\Delta~=~...

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The Eddington luminosity is given by: $$L_{Eddington}=\frac{4\pi c GMm}{\sigma_T}.$$ This represents the maximum luminosity of a body, when there is a balance between the two primary forces - that of the radiation, acting out of the body, and that of the gravitational force of the star, acting inwards. The terms sub-Eddington and super-Eddington refer to ...

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I'm sure that an exactly spinless final black hole is theoretically possible, but it would be a very, very precise fine-tuning, akin to balancing a pencil on it's point. Also, your configuration is unlikely to produce a spinless black hole, because when the plunge phase of the merger happens, the stars will suddenly fall from the innermost stable orbit, and ...

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So the answer is no. An electric field has energy and energy generates a gravitational field, just like any mass. See the charged black hole solution is the Wikipedia article https://en.m.wikipedia.org/wiki/Charged_black_hole The charge of a black hole, if nonzero, changes the metric and solution to account for the charge and electric field. That ...

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No. When they merge their horizons will change shape, and eventually become the static or stationary shape of a BH horizon. Nothing inside either horizon while this is happening can escape. At all times the timelike curves stay inside, and the deformed horizons are where the lightlike curves end up. In each, and after they merge. The area of each horizon ...

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It loses organization, e.g. matter changes into pure energy or some such thing. It's not entirely clear what form there is (some suggest there is no form at all, but it's obvious that it doesn't follow Pauli's exclusion principle). This is nothing special, it happens all the time - when you burn carbon, for example, you get a bit of disorganised energy (heat)...

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