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18

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 ...


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). ...


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 ...


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. ...


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 ...


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 - ...


5

This is an answer to the question as qualified in a comment. The stress energy tensor is a tensor field so it is a function of position in spacetime. In the Schwarzschild coordinates the geometry is time independent so the local value of the stress-energy tensor is just a function of the position in space. Everywhere outside the spherical object it is zero ...


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 ...


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 ...


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 ...


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 ...


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 ...


3

It seems that the answer to my question is that there's a difference between the pre-inflation universe "singularity" and that of other types of singularities like a black hole. It wasn't a singularity as describes a black hole, it was a point in time where the scale of the universe was zero. All that exists didn't occupy a single point in space in the way ...


3

For a freely falling observer the local geometry of spacetime is always flat i.e. described by the Minkowski metric. So the freely falling observer can never observe themself to fall through an event horizon, because that contradicts the requirement that spacetime be locally flat. In fact the freely falling observer will observe an apparent horizon that ...


3

The event horizon is a null surface, which means that if you pick a very small region of spacetime that straddles the event horizon, small enough that the equivalence principle should apply, then any local inertial frame in this region should measure the event horizon to be moving outwards at the speed of light. Slower-than-light objects in this region can ...


2

Reissner-Nordstrom is thermodynamically unstable, and so are the Schwarzschild and Kerr solutions. The thermodynamic instability is easy to see and explain: smaller black holes are hotter than larger ones. So as a black hole radiates via the Hawking process, the black hole losses mass, shrinks, and becomes hotter. If you tried to put these black holes into ...


2

First of all, this is a very good question. Classically black holes have no hair, and so are specified by a handful of charges (mass, angular momentum, etc). Quantum mechanically, they act like thermodynamic systems. They have temperature and a large entropy, and in fact all the laws of thermodynamics have black hole analogues. To explain away the apparent ...


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 ...


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 ...


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 ...


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 ...


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 ...


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 ...


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: ...


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 ...


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): ...


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 ...


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 ...


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 ...


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 ...



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