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5

This is a curious question. Many black holes are detected and identified due to light emitted from material falling into them. When black holes accrete matter, conservation of angular momentum would usually lead to the formation of an accretion disk. The release of gravitational potential energy as material falls means that this disk can be hot, and it is ...


6

The reason has to do with time dilation, and specifically, with the resulting red shift. A black hole forms from a collapsing star, which is of course made of brightly glowing matter. The event horizon forms in the centre and moves outwards while the star-matter falls towards it. Because of gravitational time dilation, the infalling matter never crosses the ...


1

The answer to your question is that nothing can travel faster than light, and light can't escape through the event horizon. Therefore gravitational waves can't escape either. I give an algebraic proof that light can't escape in my answer to Why is a black hole black?, and a more visual proof in my answer to Would the inside of a black hole be like a giant ...


2

First, if you accept that gravitational waves can't travel at fast than the speed of light in regular space, then you can move to the inside of a black hole and then imagine letting the light and the gravitational wave race each other as you fall freely. As you fall freely then over a short time interval and a short distance everything looks normal to ...


0

Yes, the no-hair theorems are asymptotic, and yes a frozen star or red hole approach has some accuracies associated with it. However, the issue of cosmic censorship is not well understood, and is still relevant to your situation. In particular, if a collapsing system forms a singularity that is not naked, then the event horizon could introduce long range ...


0

First, I worry that you might confuse a singularity, which is a region (possibly pointlike) where the curvature is undefined (or infinite), versus a black hole, which is a special surface with a particular global property in an asymptotically flat spacetime. There is a famous conjecture (the cosmic censorship conjecture) that every singularity is within an ...


0

According to the second postulate of special relativity, light is always observed as moving at c. - However, the proper time of a photon is zero. That means, from the hypothetical point of view of the photon, no time is passing for the photon which is infalling through the event horizon towards the singularity, and afterwards emitted as Hawking radiation. ...


0

The event horizon of static black holes without charge is the same for every observer, and it is located at the Schwarzschild radius. Even for an observer falling into the black hole the event horizon is the same, but at the difference that he will not notice the presence of any event horizon when falling through it. In the meanwhile outside observers will ...


0

To take your question literally, we may speak of it because our speech is not bound by nature or reality - one walk down the 'Fiction' aisle of a bookstore will demonstrate that. We write of and speak of and theorize about many things we cannot, or will not experience (such as other multiverses, interiors of stars, etc.). Why would you think we would be ...


2

The photons do not get "stuck" at the event horizon, from their own reference frame they are still traveling at C. The event horizon is simply the point where gravity from the singularity is strong enough that the escape velocity exceeds C. Just above the event horizon, photons can still escape the black hole, and just below it they sink into the ...


0

Consider a thin spherical dust shell. By the usual arguments, the gravitational field inside the shell is exactly zero -- this is valid both for GR and Newtonian gravity. Over time, the radius of the shell will decrease under the action of gravity until eventually the radius of the shell becomes smaller than its Schwarzschild radius. At that point, the shell ...


0

Yes, it is totally possible. You need to use classical GR with continuous matter distributions and it requires energetic continuous matter with zero rest mass, and it is an unstable setup, so you have to do it 100% perfectly with absolutely no margin for any error. Let's see how. You can take a sphere of radius $R$ in Minkowski space, and take a ...


2

I am going to answer the question 'When just considering GR without evaporation nor QM, is an empty (containing no matter) black hole possible?' and I will omit the 'anything' part, because is an ambiguous term. And the answer is yes, they are possible. As stated by another person here, Schwarzschild black holes and the rotating and charged versions are ...


6

Strictly speaking the two well known uncharged black hole metrics (Schwarzschild and Kerr) are vacuum solutions. This means the stress-energy tensor is zero everywhere except at the singularity where it is undefined. This is what Alfred means when he says "A static (Schwarzschild) black hole 'contains' no matter". However this strikes me as a bit of a cheat ...


1

No, that's not possible. Even if the two bodies could be compressed to be just larger than their Schwarzschild radius (they can't really, without collapsing further to black holes), their combined Schwarzschild radius, which grows linearly with mass, is twice their individual Schwarzschild radii. That means that even if they effectively rolled on each other ...


2

This is a good idea... Dark matter by definition doesn't interact electromagnetically (i.e. it has no charge). Therefore its cross section $\sigma$ for absorbing radiation and being pushed away from an accreting object is $0$, at least to first order. You could look at higher-order effects, like its neutrino-absorption cross section, to calculate some ...


0

No, you can't send information through an event horizon with gravitational waves. When a compact body contracts to form a black hole it leaves the vacuum spacetime outside the body in a curved manner. That's because certain kinds of curvature can persist on their own, even static ones,they are called vacuum solutions, and they are curved spacetimes that ...


0

The speed of light is invariant, so it is not possible to accelerate it above $c$. From Newton's law we know that the escape velocity $v$ from any given mass is $v_{esc}=\sqrt{\frac{2 G M}{r}}$ so if the mass $M$ is high enough and the distance $r$ to the mass small enough $v>c$. Since nothing can have more than $c$ even light cannot escape from there. ...


3

Far away from a black hole, spacetime is curved only a little bit, and many different things could curve it like that out there. It's like if you had a dollar in your pocket, and it's been there for a long time, and you can't remember if you got it from your boss or from your friend. But a dollar is a dollar. So you could have a massive star, or a black ...


11

Although we don't have a quantum theory of gravity, we think we have some reliable knowledge about the properties of black holes from general relativity. One thing we think we know is the so-called "No-hair conjecture", which says that black holes can be described by just three numbers: mass, charge, and angular momentum (i.e. how much they are spinning). ...


0

The singularity probably does not exist, as GR likely breaks down at those size / energy scales. When we have a full quantum description of gravity we may know what's really there. By the way, the part of the black hole we fully understand is actually the vacuum solution - the Schwarzschild metric - which includes the event horizon but not the source mass. ...


5

Scientists asked the question "How does a body of arbitrary mass affect spacetime around it?" To answer this question, they took Einstein's General Relativity and applied it to the description of a spherically symmetric spacetime (meaning you can rotate any way you like and it looks the same) centered on a body of arbitrary mass, $M$. I'll spare you the ...


0

It's almost certainly incorrect that the center of a black hole is a singularity as this would be at odds with quantum mechanics. Just how exactly it looks like would be something to ask of a theory of quantum gravity! Regardless of being a singularity or not, the mass is determined by how much mass you stuff into your black hole. Hence black holes of ...


0

The density of black holes isn't infinite. Some black holes have the billionfold density of our sun (like the black holes in center of galaxies). There are big and small black holes.


19

The accretion of matter onto a compact object cannot take place at an unlimited rate. There is a negative feedback caused by radiation pressure. If a source has a luminosity $L$, then there is a maximum luminosity - the Eddington luminosity - which is where the radiation pressure balances the inward gravitational forces. The size of the Eddington ...


0

When the escape velocity of a body of mass is faster than the speed of light, you have a black hole. This is because gravity affects light the same way it affects matter.


9

A 12 billion Solar mass black hole sounds massive, but actually it's not all that big. The radius of the event horizon is given by: $$ r_s = \frac{GM}{c^2} $$ and for a 12 billion Solar mass black hole this works out to be about $1.8 \times 10^{13}$m. This seems big, but it's only about 0.002 light years. For comparison, the radius of the Milky way is ...


1

First, I highly suggest reading up on the concept of Locality. The issue is where you're measuring the speed from... Remember that it isn't so much that light can't escape due to the escape velocity, as it is that space itself is being dragged into the black hole (and anything residing in it), which happens to be falling in at the speed of light where you ...


-3

Light will no longer be light as it gets absorbed in black hole . So there is nothing left to talk about its speed.


2

Short answer: yes. But what do the original "straight lines" mean now? they cannot be defined in a nice way in the new metric, because the natural geometric entities now are geodesics, and a quadrilateral of four right angles does not exist (apart from possible special choice of corners). You must define your volume in a correct way (see below) Long answer: ...


0

Simply because the Second Law of Thermodynamics states: Entropy > 0 (always) (think of taking a basketball and squeezing it into a golf ball) so that the density gets larger as the mass gets smaller. therefore, Entropy will always increase in a black hole, gaining Entropy [heat] as it gets smaller.The stars reaching critical mass(limiting mass) ~ ...


2

The energy of any infalling mass is absorbed by the black hole. Classically, the temperature of a black hole is absolute zero, since it is a perfect absorber. If you include quantum mechanical effects, as Stephen Hawking did, you can show that black hole horizons will emit radiation in such a way that is consistent with the horizon being a hot body with ...


4

If I recall it correctly, the information sunk into black hole can be considered encoded in the ripples on black hole surface, much like egg impact parameters which could in principle be deciphered (at least partially; even quantum theories give us certain confidence intervals) from shattered egg fragments. Falling objects will necessarily have mass, and ...


1

"They can't erase it, hence they won't heat" : If they do nothing indeed they won't heat. Maybe you don't understand when erasing would be needed for a demon to work : imagine you have a box full of gas : you could put a wall in the middle, with a door controlled by a demon : the demon would let only in the high momentum particle, so that you can have 2 ...


5

I'm not too qualified to speak about the Black Hole Information Paradox, but I think I can say a thing or two about Maxwell Daemons. I think the essential "flaw" in your description is the assumption that Maxwell Daemons destroy information. The do not, and I describe what they actually do in my answer to the Physics SE Question "How can the microstates be ...


7

If you drop an egg then it looks as though the process is irreversible and that the information about the original state of the egg is lost. However this isn't the case. The equations describing how the egg shatters are all time reversible so in principle, if not in practice, we could take the shattered egg and evolve time backwards to reconstruct it. ...


1

Singularities exist in theoretical 'perfect' solutions to General Relativity, but when you look at actual natural Kerr-like objects spinning in a noise filled background of GR waves and other incoming radiation and matter, its likely that no physical real singularities exist. Brandon Carter, referring to spinning black holes (all real black holes spin): ...


3

By the term black hole we normally mean one of four spacetime geometries, the Schwarzschild, Reissner–Nordström, Kerr or Kerr-Newman metrics. The universe is (we believe) approximately described by the Friedmann–Lemaître–Robertson–Walker metric, and it is not a black hole. The Big Bang is not the same as the singularity at the centre of a black hole. For ...


0

Since there are no extra rules for black holes they should follow the same laws. In general relativity this effects are called the de Sitter- and the Lense Thirring-effect which have been verified with big trumpets call by the famous Gravity Probe B.


1

Assuming earth mass, procession would be reduced practically to zero because tidal effects would be practically zero. Procession, at least according to this site, has to do with tidal effects and a planet being mailable. A black hole - 2/3rds of an inch in diameter would experience essentially zero tidal effects and it would be far less prone to bulging ...


0

First, let's see the density matrix of thermal mixture (in some cases is the Planck's law): $$ \rho_{Thermal}=\frac{1}{Z}\sum_{n=0}^{N}e^{-\beta E_n}|n\rangle \langle n| $$ Now we note that the expectation value $\langle \Omega|O|\Omega \rangle$ is equal to $$ \frac{1}{Z}\sum_{n=0}^{N}e^{-\beta E_n}\langle n|O|n \rangle , $$ with $Z=(1-e^{-2 \pi ...


0

The charge of the black hole isn't actually related to whatever is inside the black hole so much as a statement about the fields surrounding the black hole. This is a static field, so there's no problem or contradiction with the fact that we can't communicate with the "inside" of the black hole.


0

Well, according to the wild ER=EPR conjecture by Maldecena and Susskind, two entangled electrons are connected by a quantum wormhole. The mechanism and details of this quantum wormhole are left unspecified by these authors, though.


2

There are two ways to answer this. The first is to argue that to become a black hole an object must become sufficiently dense that it disappers inside it's own even horizon. For a star of a given mass. This means it's radius must become smaller than the Schwarzchild radius $R < 2GM/c^2$, where $M$ is the mass of the object, $G$ is the gravitational ...


0

Electrons are very close to the energy of self-capacitance of a quantum of charge. The size of the electron is very close to $r_e$, the energy supposed if one tries to charge a sphere of that radius with a single electronic charge, ie $mc^2 = e^2/4\pi\epsilon r_e$.


1

As in all extrapolations, one reaches a point where they break down. In the case of a black hole singularity and quark triplets ( neutrons and protons, pairs are quark antiquark i.e. mesons) falling into a black hole , the neutrons and protons falling in acquire energy from the gravitational energy of the black hole and at some point in energy will ...


6

There is no universally accepted quantum theory of gravity. Quantumly, the "shape" of a fundamental particle is a very fuzzy notion - we know that states are often not localized, so it is wholly unclear what it means to say "the electron is pointlike". The proper formal interpretation of a "pointlike particle" is simply a particle that is not composite - ...


2

The table you are referring to shows the Schwarzschild radius for the whole Milky Way galaxy, not it's SMBH, Sagittarius A$^*$. The total mass content of the galaxy is about $10^{12}\,M_\odot\sim10^{42}\,\rm kg$, leading to $$ r=\frac{2GM}{c^2}\approx\frac{10^{-11}\times10^{42}}{\left(10^8\right)^2}=10^{15}\,\rm m $$ For Sag A$^*$, we can estimate it to be ...


0

As stated, the charged black hole is a solution in general relativity known as a Reissner-Nordström black hole. Since you're looking for more of a conceptual overview, I'll skip the complicated maths and go straight to the end interpretations. Let's give our black hole the electric charge, $Q$, the magnetic charge, $P$, and a mass, $M$. With this, there are ...


0

Small point to add to this, but all matter has a Schwarzschild radius - the earth does (it's about 1/3rd of an inch in diameter), black holes do - it's their event horizon, and Neutron Stars do - but it's smaller than the star. Any non-black hole object has a Schwarzschild radius that is smaller than it is. Neutron Stars are dense enough that their ...



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