# Tag Info

0

The important quantity associated with a black hole is the event horizon area. The volume contained inside is not what one would think of as $V = 4\pi r^3/3$. More on the volume later. The important quantitiy is the area of the event horizon. The reason is that from the perspective of an exterior observer this is the limit of observation. Everything that ...

1

There's nothing different about a supermassive black hole except that its mass is large. So, if things work for small black holes they work for large ones. It turns out that, by the measure of density given, the density decreases with the mass of the object. This follows immediately from the formula for the Schwarzschild radius, $r=2MG/c^2$: this goes as ...

0

Nice question. Gravitational waves are solutions of the linearized equations of motion and as such I don't expect quantum gravity effect to change in a significative amount the entropy content of the waves. Nevertheless there are quantum gravity approaches, for instance the fuzzball proposal in string theory, where the discrepancy with classical physics ...

0

The collapsing of the black hole is impossible. It is like trying to burn ashes. Ashes are already burnt. Similarly, the black hole is already collapsed. The black hole can evaporate and get bigger, but not collapse.

7

Thanks to the hint given by knzhou I figured out that if one wants to give the particle a proper initial velocity of $v_0$, the initial velocity in terms of Schwarzschild coordinates $v_i$ would then be $$\dot{\theta}(0)\cdot r(0) = \frac{{v\perp}_0}{ \color{green}{\sqrt{ 1-v_0^2/c^2}}}$$ for the transversal component, and $$\dot{r}(0) = ... 0 Anna, you are more or less right. It is pressure from matter at higher densities that can stop the gravitational collapse. It depends on the state of the matter, and in a simplistic description the equation of state. As it collapses as a hot gas after it exhausts it nuclear fuel, and maybe after a supernova explosion, it'll collapse. The first point at ... 17 There seem to be several confusions here. Massive and massless particles behave qualitatively differently, even if the massive particle is traveling very fast. The minimum radius for a stable orbit for a massive particle is 3 r_s. Circular orbits above this radius are all stable. Massless particles only have circular orbits at the photon sphere, (3/2) ... 1 A black hole in the Schwarzschild case has a radius$$ r=\frac{2GM}{c^2} $$Let us solve for the mass with a radius of about Fermi or 10^{-15}m, or the radius of a baryon such as a neutron,$$ M = \frac{rc^2}{2G}=\frac{10^{-15}m\times 9\times 10^{16}m^2/s^2}{2\times 6.67\times 10^{-11}Nm^2/kg^2}  =6.7\times 10^{11}kg. $$This is close to a billion ... 3 When black holes form from the collapse of stars, we think they usually produce bright supernovae explosions which release tremendous amounts of energy. Some models suggest that a non-negligible fraction of stars which produce black holes may not produce normal supernovae, these are often called 'failed supernovae', see for example astrobites: Gone Without ... 1 A black hole m\approx300,000M_{sun} at the center of galaxy NGC 4845 is thought to have a giant gas planet with the mass no larger than that of a brown star in orbit, though it is currently being drained into the black hole. (And by currently I mean what we are seeing right now, which happened a long time ago). The gas giant is thought to have been ... 0 The quasinormal mode frequencies are complex-valued numbers. For a given type of field (for example, scalar, vector, gravitational, fermionic, etc) there are a discrete infinity of these frequencies which are of course independent of the metric used to describe the background. If the background possess some symmetry, for example spherical symmetry, then the ... 2 The general relativity model for an electron is the Kerr-Newman solution, that is a charged rotating black holes. Unfortunately the radius of the exterior horizon in geometric units is given by:$$r_{+}=M + \sqrt{M^2-Q^2-P^2-\left(\frac{J}{M}\right)^2}$$Here P is the magnetic charge, zero in this case. If you convert this relation to standard units ( for ... 3 If you believe classical parameters for black holes down to a neutron's scale, then it turns out that a neutron has too much angular momentum to form a black hole, and would be best interpreted as a naked singularity. The reason why is that the radius of a black hole's event horizon is predicted to be:$$M \pm \sqrt{M^{2} - a^{2}}$$where M is the mass ... 1 This is indeed possible! With a risk of overstatement, this fact is extremely important for astrophysics because it turns out that by dipping in and out of the ergogregion (the region between the surface of infinite redshift and the event horizon) one may extract quite a bit of energy and angular momentum from a rotating black hole in a process known as the ... 1 Traversable - Overlapping (actually intersecting) region would not be Traversable even if the gravity at some parts of the region may be zero. For exampple, between earth and moon, gravity will be zero at some point. That does not mean something in that region can go out of earth/moon system. As soon as an observer leaves that region, it either falls towards ... 2 Torsion is not frame dragging. Torsion is having an anti-symmetric spacetime connection. As you do parallel transport in general relativity (GR) you drag frames the frames roll as they move. With torsion they would twist. The connection is GR is the Christopher symbols, symmetric in the two bottom indices. The torsion is an anti-symmetric tensor. It will ... 0 If the event horizons overlap you get one big horizon. EM forces can not counteract gravity if the curvature is too large since the force required to counteract gravity becomes infinite at the horizon. You can see this in the equation$$F=\frac{G\cdot M\cdot m}{r^2\cdot\sqrt{1-r_s/r}} $$which becomes infinite at the horizon r_s. Since from the outside ... 0 If you are using words like instantly and think it even means something in general relativity, then you need to learn more about general relativity. In special relativity you already learn that simultaneity depends on frame. Black holes form from collapsing matter. And the event horizon is a one way surface. Not because it is magic, but because it ... 0 A first problem is that there is in GR no such thing as "an observers frame", except in sloppy speech. There are various systems of coordinates. Two systems of coordinates may agree for an observer as much as one likes but differ elsewhere. And all the systems of coordinates are on equal foot, none is preferred. What could replace the "observer's frame"? ... 4 The "theory" describing the black hole interior (in the classical approximation) is the same theory that implies the existence of the black holes, namely the general theory of relativity. As the OP correctly said, the singularity at the event horizon is a coordinate singularity – one that is an artifact of a bad choice of coordinates. When a coordinate ... 0 An orbiting observer is a bit problematic because there are no stable orbits for r \le 3r_s, so let's instead consider an observer hovering at some distance r from the black hole. In that case as r \rightarrow r_s the blue shift does indeed \rightarrow\infty and the observer would indeed be roasted. But this shouldn't be surprising. The acceleration ... 0 We won't see inside the horizon, but we will see up to close to the horizon. We will be able to see with much higher sensitivity when we deploy in a few years the eLISA gravitational observatory in space, with 3 satellites separated by 1 million Kms in a triad configuration. See http://www.livingreviews.org/lrr-2013-7 The article also describes the many ... 3 Yes. You're exactly right, deviations from no-hair do occur for example after BH mergers --- and hints of the "quasi-normal" mode ("ringdown") were observed in the LIGO detection. The no-hair theorem is constructed for a static, stationary BH (i.e. fully settled). In general, deviations from no-hair (magnetic fields, asymmetry, etc) will be radiated away ... -1 The asymmetry you observe will be the asymmetry of the configuration with the infalling matter - in this case, with the additional infalling black hole. It becomes more symmetric with time - a very short time, but not instantly. 0 Actually, only a gravitational wave and not a uniform gravitational field is composed of gravitons. I think that just like en electromagnetic wave, a gravitational wave is a sum of sinusodial waves with gravitons of different energy for different wavelengths in that sum. An isolated blackhole doesn't emit gravitons but a pair of orbiting blackholes does. I ... -3 If a black hole travels from point A to point B, long distance at the speed of light you can say it's a wormhole. 2 Well, plenty of black holes show jets of highly energetic gas or plasma coming out perpendicular to their rotation plane, seemingly out of the hole. It's just accreting highly energetic gas that's spiraling into the black hole but can not make it in, much of it does but much of it is expelled out. It is not from the hole, just the energetics works out that ... 0 Some misunderstandings. Black hole can be round (we call them spherically symmetric, or spherical), or sort of ellipsoidal, or axially symmetric, if they have angular momentum (i.e., if they rotate). That is true for stationary black holes, i.e., after they achieve a stationary state. In getting to that state they can be very dynamic and have deformed ... 0 Black holes, as seen in the picture are actually spheres formed by event horizon. The matter is all concentrated on the singularities (except for the matter that is falling into either singularity at a given point in time). So, individual black hole would be spherically round. During merger of two black holes, the event horizon can become non-round ... 2 That's just a very poorly written article. Nothing was observed to exit from inside the event horizon. There was a high energy event that launched a flare from near the BH. The event is unexplained, but doesn't obviously violate any known laws of physics. 0 The information paradox has no particular connection to electromagnetism. Hawking radiation is not just photons, it's any sort of particle. And Hawking radiation in itself doesn't solve the information paradox - the problem is that Hawking radiation is supposed to be thermal, so quantum information of infalling objects has been irreversibly lost, but that ... 1 Yes. Black Holes (BH) can grow from accreting anything with energy --- including dark matter (DM). I'm not entirely clear on the second part of your question, but probably the most important thing to keep in mind is that the black hole information paradox is still unresolved. Answering how information is not lost for any type of particle, including DM, ... 0 Yes the singularity of a black hole would be incredibly bright once you get to the center because the light and energy can't escape the black so it is all compressed into and infinity dense small point and even if it is just normal matter going inside the black hole it would even make it more bright because the black hole tidal forces would rip the atoms ... 2 Hawking, I believe, is referring to a more metaphorical 'hovering'. As light, or anything, approaches the event horizon, it becomes more and more redshifted---it's motion appearing to go slower and slower and slower, approaching zero apparent velocity to an outside observer (approximately) infinitely far away. Anything falling into a BH, thus appears to ... 0 Draw two circles on a piece of paper and poke holes in both of them fold the paper to where a pencil can fit through the two holes you have created. This shows how a black hole works and it is a fun thing to do with kids who are interested in this subject, like myself. And i'm only 13 and I get it. 0 The hypothesis doesn't make much sense. The resistance of an object to being ripped apart is given by its elasticity, not by how much is massive. Anyway, to the central question there is no answer yet. This is the so called information loss paradox, one of the greatest unsolved problem in theoretical physics. In general relativity the object falls in a ... 10 Note that this is an incomplete answer. Imagine an object of mass m at a distance r from the centre of a black hole of mass M. The gravitational potential energy is$$ U(r)=-G\frac{Mm}{r} $$This has its highest value when r=\infty and its lowest value when r is at the event horizon of the black hole, i.e. the Schwarzschild radius$$ R = ...

111

Nice discovery! The formula for time dilation outside a spherical body is $$\tau = t\sqrt{1-\frac{2GM}{c^2r}}$$ where $\tau$ is the proper time as measured by your object at coordinate radius $r$, $t$ is the time as measured by an observer at infinity, $M$ the mass of the spherical body, and $G$ and $c$ the gravitational constant and the speed of light. ...

0

You are asking about micro black holes. Some hypotheses involving additional space dimensions predict that micro black holes could be formed at energies as low as the TeV range, which are available in particle accelerators such as the LHC (Large Hadron Collider). Popular concerns have then been raised over end-of-the-world scenarios (see Safety of ...

0

First of all, the Universe isn't expanding according to "current theories". It is an observational fact. Second, there is no center of the Universe. Space was created, and started expanding. This expansion pulls everything away from each other. Galaxies lie approximately still in space, but space is expanding. This means that no matter where you are located ...

0

First of all, there is NO centre in the universe. I know it's not a good analogy, but think of the universe as the surface of a balloon. Forget the interior, we're only looking at 2 dimensions, whereas the real universe has 3 of them. Put some ink dots on the balloon, which represent galaxies (note: NOT planets). Now inflate that balloon. You'll see that ...

0

Yes, it is perfectly possible to move a black hole, in any reasonable sense. In particular something that is important to understand is that, far from the event horizon, black holes are not particularly special. The gravitational field of any spherically-symmetric distribution of mass is the Schwarzschild solution in the region where there is no matter for ...

0

Komar mass is associated to one asymptotically flat end. So a wormhole has two Komar mass, one for each side, and in principle they can even be different! Indeed the simplest wormhole you can imagine is simply two copies schwarzschild spacetimes glued together at the would-be horizon. As an aside, even time can run differently on the two sides (often ...

-1

Your question doesn't make much of a sense to me. Your frame of reference is everything. I can view it in such a way that it would look as if it was accelerating, and it also may seem to be stationery in some other frame. Don't mess with motion in space unless your are able to find a reference ponit.

3

The simplest answer is that you integrate the emission from annuli radiating at various temperatures. Explicitly, the luminosity per wavelength in this approximation is $$L_\lambda = 2 \int_{r_{\rm in}}^{r_{\rm out}} 2 \pi r [\pi B_\lambda(r)] dr$$ where the overall factor of 2 is for the two sides of the disk, $2 \pi r dr$ is the area of each annulus, ...

3

In addition to what @Jack Mazy said about force due to gravitational attraction, there are also charged (Reissner–Nordström) black holes. I would imagine in theory you could take another charged object and place it near the black hole which would cause Coulomb force on black hole putting it in motion.

0

One way to establish that $r$ in Schwarzschild coordinates is equivalent to the spherical radial coordinate is its asymptotic behaviour, namely for $r\to\infty$ the metric tends to Minkowski, and the weak field approximation yields Newton's gravitational potential. One way to establish that the parameter $M$ is indeed the mass of the spacetime is to ...

1

You state that: there is literally no way to squeeze more information (entropy) into a given volume than that in a black hole occupying that volume But you must keep in mind that the volume occupied by the radiation+BH system is larger than the volume of the black hole by itself. When the black hole initially forms the horizon has a radius $r$ which ...

1

There is a very general theorem that there is no stable configuration of matter that can exist above a threshhold that fundamentally just depends on the mass of the constituent particles that make up the matter (and the interatomic forces in the matter). There is some debate about the exact limit for neutron stars, but it is something like 2.5 - 3 solar ...

0

An electron being a ball of uniform mass and charge is not consistent with its observed gyromagnetic ratio. The charge must be pushed out and the mass must be pushed comparatively inwards to satisfy the existing ratio of about 2. See Classical proof of the gyromagnetic ratio $g=2$

Top 50 recent answers are included