New answers tagged

2

According to the fuzzball proposal in string theory, black hole are actually horizonless and regular solutions. For some systems in five dimensions made with bound states of intersecting branes this has been already proved directly in supergravity: there are solutions without horizons and singularities, with the same asymptotic charges (Mass, Angular ...


2

At best things are pretty speculative. Cumrun Vafa has proposed that black holes have condensates of tachyons. In some sense you can understand this without much complexity. The Schwarzschild metric has a physical singularity that is a spatial surface. The Penrose conformal diagram for the Schwarzschild metric illustrates this The bosonic string has two ...


0

The holes you are used to seeing most likely occur on Earth. Since Earth's surface is a 2-dimensional object, holes in it are also 2-dimensional. Given that we live in a 3-dimensional world (spatially/geometrically), these holes would then be traversable; leading to a "pit" below. Since it isn't realistic to have a pit that goes completely through Earth, you ...


0

When we talk about a (non-spinning) black hole we normally mean the spacetime geometry called the Schwarzschild metric. This is one of the simpler solutions to the Einstein equations, but it's important to understand that the solution is based on a number of assumptions and as a result it is only an approximation to a real black hole. The assumption that is ...


1

An object swallowed billions of years ago thus will retain some incredibly tiny solid angle over which a sufficiently vertical photon will eventually reemerge into the outside world... For a galactic center black hole, how exactly does "pump" mass-energy from the singularity back out to the event horizon... ? ... the apparent horizon at which the ...


4

This question has already been answered by a Randall Monroe of XKCD and Dr. Cindy Keeler of the Niels Bohr Institute. It forms a Naked Singularity. Which is an infinitely dense object, from which light can escape. Source: https://what-if.xkcd.com/140/ You Reissner–Nordström metric for this question, as opposed to the more well known Schwarzschild metric. ...


7

I think this is one of those situations where it is cleanest to think relativistically and concentrate on gravity as spacetime geometry, not gravity as a force. If the stress energy tensor's components reach a such a magnitude that an horizon forms, then the mass/energy inside the horizon is doomed to stay inside evermore. This is a question of spacetime ...


5

The strong force is responsible for neutron stars not becoming black holes. Although attractive at nuclear densities, it becomes repulsive at somewhat higher densities. This repulsion is many times greater than the electromagnetic repulsion between protons at similar densities. Yet we know that if a star is too large, it's supernova event will lead to a ...


1

I am only answering part I of your question. Please note that following explanation is highly simplified. A burning star has huge mass mostly of hydrogen and helium. This huge mass creates a gravitational pull which tries to collapse the star to a single point. However the thermal pressure produced by the burning hydrogen mass (nuclear fusion) counter the ...


8

Is this simply the ratio of the angular momentum that the blackhole is observed to have as a ratio of the maximal angular momentum as limited by some Physics (Kerr Metric?)? Yes, exactly. For a spinning black hole there are two event horizons, an inner and an outer horizon. The positions of the horizons are given by: $$ r = \tfrac{1}{2}\left(r_s \pm \sqrt{...


18

Yes, the dimensionless spin such as $0.2$ in this case is simply the ratio $$ a= 0.2 = \frac{|\vec J_{\rm measured}|}{|\vec J_{\rm max}(M)|}$$ where the denominator is the maximum angular momentum allowed for the same value of the mass (as the measured mass). For the $d=4$ Kerr black hole, the maximum (the angular momentum of the extremal Kerr black hole) is ...


2

After a binary merger the resulting black hole will experience a ringdown. The ringdown is characterized by a spectrum of quasi-normal modes (QNM) which, according to GR (specifically black hole perturbation theory), depend only on the final black hole mass, angular momentum, and charge. This is effectively the no-hair theorem. QNMs are oscillations that ...


0

Is there any change of the mass of a black hole if time passes ? The present day thinking is that largest black holes are in existence located at the center of galaxies, and may have masses equivalent to about a billion suns. According to General Theory of Relativity there is no lower limit to the size of a black hole. As black holes eats up "materials"...


12

Check out this figure from the most recent LIGO paper. This is the reconstructed gravitational wave signal determined by the analysis. Focus on the zoom of the end of the signal. This is the merger. The merger part of the GW151226 looks a lot like the merger of GW150914. The differences tell us about the sources of the gravitational waves in each case....


3

There are two complementary correspondences to black holes. One is a metric where far from the black hole horizon there is $AdS_3$ spacetime. The other occurs quite oppositely where the $AdS_2\times S^2$ spacetime occurs in the near horizon condition of an extremal black hole. For the asymptotic condition, usually seen as the AdS black hole one simply has ...


1

On top of the other excellent answers I'd like to point out that the accretion rate of dark matter particles is believed to be much smaller. The reason matter in accretion disk is being accreted rapidly is because they lose energy from electromagnetic radiation. For dark matter particles, in practice the only way it can be accreted is if the particle ...


2

The local escape velocity is $$v_{esc} = \sqrt{\text{2 G M}/\text{r}}$$ At infinty you observe that velocity slower by a factor of $$1-\text{r}_s/\text{r}$$ so at infinity you observe $$\text{v}_{esc} = \sqrt{\text{2 G M}/\text{r}} \cdot (1-\text{r}_s/\text{r})$$ because of gravitational length contraction radial to the mass and gravitational time ...


3

A magnifying glass will produce an upright, enlarged (virtual) image when the distance to the object is smaller than the focal length, and an inversed image when the distance is larger than the focal length (in which case the size of the image decreases rapidly with increasing distance). (image by cmglee, wikimedia commons) For a black hole, the ...


2

Fleshing out my answer in the comment above: The light from all the stars in the picture is lensed (i.e. it's path bent) by the black hole (or BH). The closer it passes to the BH, the larger the angle it's path is bent by. Stars far (in projection) from the BH, like A, are hardly lensed at all (in fact I significantly over-egged it in my diagram, should ...


2

The answer is that it has nothing to do with light, c, black holes, event horizons or relativity. It is simply escape velocity. We know that two bodies attract each other with a force $f = G\frac{M m}{d^2}$ and we know that something is in a stable orbit when its potential energy $PE = -GMm/d$ matches its kinetic energy $KE = m{v^2}/2 $. Solve these two for ...


2

The explanation I like is thus: In GR, all things, from planets to photons, travel in straight lines through curved space bent by mass. Black holes bend and distort spacetime so severely that the curvature captures the photon. Scale things down and it behaves much the same way passing asteroids can be captured by a star. For us, the speed of the asteroid(...


1

The picture I always liked is for an observer free-falling into the black hole, when they're just outside of the event horizon, it looks like the event horizon is propagating outward at nearly the speed of light. After the observer falls just inside, the event horizon now looks like it's propagating outward at greater than the speed of light.


0

Your question differs from the title, in that an "exploding" black hole happens when the the mass is very tiny and thus the Hawking effect causes an almost instantaneous release of all the remaining mass as energy. This would not have much effect on anything, although if it happened right inside your body the total ionizing radiation would be enough to ...


2

You would not want to be very close to a black hole merger. 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 ...


-3

The short answer is the speed of light is constant until its not. Its constant until it runs into a wall and its constant until effected by gravity. The extreme gravity of a black hole will deflect the path of the photons more and more until finally at the event horizon all deflection is toward the black hole.


33

Suppose you are floating in a river, and you have with you a model boat, called the SS Lightray, that can do 3 m/sec through the water. When you set the boat travelling upstream as far as you're concerned it is doing 3 m/sec. But I'm standing on the bank watching the river flowing at 1 m/sec, so when I look at your boat I see it travelling at a net speed of ...


6

To explain it in layman's terms, without using advanced concepts: Space is warped in the "inside" of the black hole (that is, under the event horizon) so much, that it behaves completely different than what we perceive hear on Earth. The "outward direction" simply does not exist. For example, here on Earth, we can go in the three spacial dimensions in both ...


-3

Isn't the saying "The speed of light is constant 'in a vacuum'"? The 'in a vacuum' bit is commonly missed off (in the same way, "'The love of' money is the root of all evil", is often commonly misquoted). If a black hole isn't a vacuum (because at some point, particles are close enough together it is no longer a vacuum), then the speed of light slows ...


66

The speed $c$ that is constant is so when measured locally relative to a freefalling frame (i.e. one for which all points follow spacefime geodesics wrt to the metric $g$). Local means that the frame's extent must be "small" enough that it can be thought of as flat: think of this as zooming in on the spacetime manifold, which is a smooth object, with enough ...


0

You don't need to get confused with black holes. Just stand outside a train and shine a laser, and have your brother inside a train shime another laser. Your light would be traveling at speed c with respect to you. If the train was going at a speed c/2, youd conclude also that his light was going again speed 3c/2, faster than you. As @WillO implied, if you ...


-1

We never observe the event horizon begin to form. The entire event horizon is a region of space time outside our past light cone. I don't see how a firewall could form. For a firewall to form, radiation at the event horizon would have to cause the firewall at another part of the event horizon in its own past light cone. Even if we do observe the black hole ...


0

No, an observer never observes the event horizon even begin to form. However, they observe regions of space closer and closer to the event horizon keep forming. When an object falls into a blackhole, we see it getting closer and closer to the event horizon. That doesn't contradict the fact that we never observe the event horizon begin to form because we ...


4

I don't think there has even been observational evidence of Kerr-specific effects. While gravitational lensing is well known these days I don't think any of the objects studied have been rotating fast enough for the difference between the Kerr and Schwarzschild metrics to be apparent. Well, not in lensing anyway - Gravity Probe B did measure frame dragging. ...


1

If I understand you correctly you are concerned that a black hole somehow manages to become less dense than the matter that made it, as if it somehow expands against its own gravity to increase its volume. However a black hole event horizon is not an object - it is just a place in spacetime. Although we can calculate a density by calculating the volume ...


2

If you look at the time scale for evaporation (which basically asks you to wait for the black hole temperature to be higher than the cosmic background radiation), you will find that binary black holes will have merged (due to emission of gravitational waves) a long, long time before there is any evaporation happening.


1

There is nothing wrong in having something more dense than a black hole, large black holes can have densities less than water. If you put a lot of iron together it might or not become a black hole. An object of any density can be large enough to fall within its own Schwarzschild radius. The larger the black hole the lower the density, so you iron ball will ...


4

You sound as though you may have heard of Gullstrand Painlevé co-ordinates, which are a particular system of co-ordinates for labelling spacetime defined by the Schwarzschild metric around a nonspinning, noncharged black hole. The analogy is often made of a "spacetime river" with this depiction; if you stand still with respect to the co-ordinates you are ...


3

The idea that distances have to be integer multiples of the Planck length is a common misunderstanding. The actual role of the Planck length is a bit subtler than that. In quantum mechanics, the possible observable values of a physical quantity (such as a particle's position) are the eigenvalues of a Hermitian operator associated with that quantity. But the ...


4

First, the Unruh and Hawking radiation aren't quite "the same thing". They have a similar origin and the Unruh radiation may be considered a flat space (large black hole) limit of the Hawking radiation. Now, the near-horizon metric of an extremal black hole is $AdS_2\times S^2$ while for a non-extremal one, the $AdS_2$ is replaced by the Rindler space. ...


2

The relationship they're claiming is a little more subtle than what comes across in the press article: The black hole masses are not directly measured. They are inferred from the $M-\sigma$ relation for black holes, which says that the larger the central velocity dispersion of stars in a galaxy, the larger the black hole. The velocity dispersion is the (...


1

In this case I would rather forget about possible no go theorems on topology changes. They are more related to how we mathematically model the problem, than to the physics. The physical fact is that matter is collapsing in a initially approximately flat space. General relativity is unable to explain in full generality what happen next, indeed the prediction ...


0

The singularity for a Schwarzschild metric is a spatial surface. It is a surface where the Weyl curvature diverges. So technically there is no topology change. For the Kerr blackhole the singularity is a ring and things are a bit different. However, the inner horizon is a Cauchy horizon where photons of arbitrary wavelength pile up. So this might in fact be ...


2

First, regardless of what physicists thought many decades ago, it seems basically clear today that traversable black holes are forbidden in Nature. They either require some negative energy density which induces instabilities, or equivalently, they allow superluminal signaling which violates the relativistic causality, and so on. Non-traversable black holes ...


-1

This is based on a more explainable or commonly understandable explanation of the theory of relativity. Imagine the space time fabric to be an infinite sized piece of cloth now if a a ball of iron is my sun then it will bend the cloth and attract the objects on the cloth that come in the range of its bending. ...


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 $...



Top 50 recent answers are included