Tag Info

Hot answers tagged

26

does it mean that at a blackhole, an object will fall at an infinite speed because of the infinitely strong gravitational pull of the blackhole? No. Actually, defining exactly what you mean by the speed an object falls into a black hole is a tricky problem. In relativity you generally find that different observers observe different things. But we can ...


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 ...


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). ...


10

There are three kinds of orbits that encompass all cases: bounded $E < E_{crit}$, extreme $E = E_{crit}$ and unbounded orbits $E > E_{crit}$. As Akano has said, the extreme case can circulate around the `photon sphere' for ever, because the energy is tuned to exactly repel the inward pull. You can still have an unbounded orbit circulate around as ...


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 ...


8

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 ...


6

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

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 ...


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

A black hole does not have an infinitely strong 'gravitational pull'; the spacetime curvature is finite at the horizon. However, the proper acceleration required to hover above the horizon diverges at the horizon. That is to say, the weight of an observer, hovering above the horizon, goes to infinity at the horizon. Nonetheless, to an observer hovering ...


3

In the comments you mention Susskind's use of a metaphor involving water flow 7 minutes into this video, but this shouldn't be understood in terms of spacetime behaving fundamentally differently around a black hole as opposed to any other gravitating body. Rather, I suspect Susskind is just referring to the analysis of a black hole in a particular type of ...


3

I'm surprised that Jacobson's notes, which are great by the way, failed to mention this explicitly, but the reason is that Hawking radiation indeed violates one of the theorems assumptions, namely a positive energy condition. Positive energy conditions normaly state that the energy density according to all observers is non-negative. When we pass to the ...


3

You can just write down $$\Gamma^i{}_{k\ell}=\tfrac{1}{2}\,g^{im} \left(\partial_{x^\ell}g_{mk} + \partial_{x^k} g_{m\ell} - \partial_{x^m} g_{k\ell} \right)$$ in Mathematica. The example blow is for the Schwarzschild metric. Here is the code. (You might have to patch the parts lost by my excessive use of display style.)


3

They are listed in here catalogue of spacetimes


3

The event horizon is not an optical effect, so changing the refractive index of the space around it will make no difference. Adding extra mass will make a difference, but it's the mass that makes the difference not the refractive index. You couldn't make your spherical shell of glass go right up to the horizon because at the horizon no material is strong ...


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

The question of AdS (in)stability is indeed a hot topic in current research of the AdS/CFT correspondence. It is a field that ties together many interesting subjects: Gravity in AdS (i.e a confining box), thermalization in QFTs, the theory of non-linear differential equations and their perturbative treatment, turbulence etc. This explains the explosion of ...


3

The only problem with your hypothetical ultralight black hole is that it would be big. Very big. First, note that if you had a black hole with mass $M$ and corresponding volume $V$ such that $M/V < \rho_\mathrm{space}$, then a volume $V$ of a typical patch of space would have more than enough mass to be a black hole. This should worry you if you thought ...


2

Note that the value of Newton's constant, $G$, is measured in the lab at distances on order of 1 meter. This puts strict limits on your $\epsilon$ and $n$, and immediately rules out significant effects on astrophysical black holes ( whose Schwartzchild radii are much larger than that).


2

The issue here is that the Schwarzschild coordinates are divided into two disconnected patches by the coordinate singularity at $r=2m$. There is no physical connection and the two pieces can be viewed as separate solutions. The Eddington-Finkelstein coordinates take the outer solution and extend it beyond the horizon, but inside the horizon it is different ...


2

I think, Kyle's link is a really good answer. For they layman (like me), you can run some numbers here: http://xaonon.dyndns.org/hawking/ a 1 earth mass black-hole (about the size of a ping-pong ball) would radiate so little energy that it would easily destroy the earth, even from a low orbit. It might take some time to devour it, but no question, it ...


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

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 ...


2

Probably only once or twice, as a very small gravitational disturbance (a nearby electron) would send the photon into an orbit with a necessary velocity not = to C.


2

I saw a some videos about the information paradox aka Hawking paradox. My understanding of it was that as soon as something is unable to leave a black hole (the event horizon where light cannot escape) that it's information is then represented as surface area (2d) on the outside of the black hole rather than our standard idea of volume (3d) and believing ...


1

A small note on the definition. It should be $$ M(S) := \sqrt{ \frac{\text{Area}(S)}{16 \pi}} \left(1- \frac 1 {16 \pi}\int_S \theta^-\theta^+ d_{\sigma_S}\right), $$ where $\theta^\pm$ are the divergences along the two null directions. It is equal to what you have written if the 2-surface $S$ lies in a space-like 3D submanifold with vanishing extrinsic ...


1

This is not possible. Please look at the following links: http://physics.stackexchange.com/a/170798/75518 http://physics.stackexchange.com/a/154051/75518 http://physics.stackexchange.com/a/170884/75518 http://en.wikipedia.org/wiki/No-communication_theorem The first describes in a good way what entanglement is (statistical correlation). The second gives a ...



Only top voted, non community-wiki answers of a minimum length are eligible