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13

There are only four known stable black hole geometries: Schwarzschild, Reissner-Nordstrom, Kerr and Kerr-Newman. We expect that any random assemblage of matter dense enough to form a black hole will relax into one of these four geometries by emission of gravitational waves. None of these geometries has two distinct singularities, so (as far as we know) it is ...


13

I think the closest model to what you're talking about would be two colliding black holes, during the intermediate period where their horizons had merged, but the central objects had not yet collided. These systems are very different from isolated black holes, as they give off significant gravitational radiation, and have horizons that rapidly change in ...


0

That's just strong equivalence principle. Standing still on a planet is getting accelerated upward. Light is going at the speed $c$ only for inertial observers, that is, for the ones free falling. It is a big misconception to claim that light doesn't "slow down" as it ascends, for you standing still on the planet. Where do you think that red shift comes from ...


0

Note that the property "black hole" is relative to an observer. A freely falling observer will not notice anything strange like an event horizon etc. Those observers will not have an information paradox. But similar, an observer at rest relative to the black hole will have no information paradox too, since nothing ever falls into it. For a calculation ...


0

If you consider a spaceship as v → c then someone in a stationary frame observing the spaceship will see ∆ t → 0 as well (on the spaceship). The thing to note is that the spaceship is still moving forwards at close to c. Time is only frozen for things moving on the spaceship (as observed by someone in a stationary frame) but the spaceship itself is still ...


0

The thing to consider is not $t$ the 'coordinate time', but $\tau$ the 'proper time' measured by the clock of the in-falling particle. $$ \Delta\tau = \int \sqrt{-ds^2}, $$ where $ds^2$ is the spacetime interval along a time-like path. All in-falling particles take finite proper time to cross the horizon. Further, all in-falling, massive particles will ...


0

From Wikipedia, the free encyclopedia A sonic black hole (sometimes called a dumb hole) is a phenomenon in which phonons (sound perturbations) are unable to escape from a fluid that is flowing more quickly than the local speed of sound. They are called sonic, or acoustic, black holes because these trapped phonons are analogous to light in astrophysical ...


1

The theories of tidal pull, planet's rotation etc are good and promising, and almost explain the phenomenon that causes the great tsunami waves on Miller's planet. But here is another theory: Because of the time slippage. A planet that near to a black hole, if there is time slippage occurring (7 years per hour), this cannot be uniform all over the planet. ...


1

Whether you see the incoming background radiation as blueshifted depends on your relative motion compared to the large scale Hubble flow. Even in the absence of a black hole you can accelerate in a direction and see blue shift in the forward direction. And that's fundamentally what is happening in the black hole, to stay at a fixed distance from the black ...


2

There's no minimum size. There's a minimum density. Stars turn themselves into black holes when they exhaust their fuel and collapse. For that to yield a black hole, they need to start off with around 25 solar masses. If a star starts with enough mass, natural processes cause it to eventually suffer a core collapse which greatly compresses some of its mass. ...


2

As noted, the sun isn't on this path. But, the gravity from a black hole is not radically increased. If the sun became a black hole, its mass wouldn't change, and we would experience the same gravitational force. (I am not sure how close you have to get before you start to see the GR-type effects. But it's pretty close.)


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At the distance of the Earth, the gravitational pull would be the same: its mass does not change just because it's concentrated in a small region of space, and the gravitational field of the black sun at that distance would therefore be unchanged. Our friends at NASA have answered the question essentially the same way: It would exert no more ...


3

Let $\xi^\alpha$ be a Killing vector of a metric $g_{\mu\nu}$, i.e. it satisfies $$ \nabla_\mu \xi_\nu + \nabla_\nu \xi_\mu = g_{\mu\alpha} \partial_\nu \xi^\alpha + g_{\nu\alpha} \partial_\mu \xi^\alpha + \xi^\alpha \partial_\alpha g_{\mu\nu} $$ Then the quantity $$ Q = \xi^\alpha u_\alpha $$ is conserved along any geodesic. To see this, we can compute $$ ...


2

But why is it the component of the dual vector $p_\phi$ that is constant rather than $p^\phi$? From the bottom of page 189: The geodesic equation can thus, in complete generality, be written $$m \frac{dp_\beta}{d\tau} = \frac{1}{2}g_{\nu \alpha,\beta}\;p^\nu p^\alpha$$ We therefore have the following important result: if all of the ...


2

No observer will ever observe a true horizon because it takes an infinite time to form. All we real observers will ever see is an apparent horizon, though for typical black holes the difference between the real and apparent horizons is extremely small. The Hawking radiation does depend on the motion of the observer. For example a freely falling observer ...


1

This isn't something that can be answered within our current understanding of physics, since GR and QFT fail at predicting what exists beyond the event horizon of a black hole. Philosophically, the question you ask is fascinating, because it represents one of the deepest mysteries in physics. In practice, though, we are only really concerned with things that ...


2

Yes, we expect all astrophysical black holes to have nonzero rotation. If nothing else, a rotating black hole that absorbs even a single particle with net angular momentum will then have nonzero angular momentum.


2

Firstly we should note that the universe as a whole is not described by the Schwarzschild metric, so the Schwarzschild radius of the universe is a meaningless concept. However if you take the mass of the observable universe you could ask what the Schwarzschild radius of a black hole of this mass is. For a mass $M$ the Schwarzschild radius is: $$ r_s = ...


1

The point of view of a photon is not really a valid reference frame. This can be seen by asking yourself this question: if the speed of light is constant in all reference frames (a basic postulate of special relativity), then does a photon see itself traveling at the speed of light? As can be seen, this is utterly nonsensical.


1

We can't tell how matter behaves inside a black hole. I can think of at least several solutions, but there is no way to either confirm or deny them. I'd say its most likely matter forms a sphere inside the event horizon equal to the radii of the black hole. Considering physics (as we know it) don't break down inside the black hole, matter can't travel ...


2

You are correct that the status of a black hole is determined by its mass, but also by its radius. The gravitational field becomes stronger the bigger the mass and the closer you can get to that mass. A black hole forms once a mass $M$ is compressed inside the Schwarzschild radius $r_s = 2GM/c^2$. i.e. once its density achieves $$ \rho > \frac{3M}{4\pi ...


2

A stellar mass black hole usually forms during a core collapse supernova of a very massive star. Our understanding of star formation is that most stars will have some angular momentum, and some of this angular momentum will be passed on to the black hole which is produced when the core collapses. This means that, as you mention, most black holes will be ...


9

My favorite reference for these sorts of things that straddle physics and geometry is Frankel's "The geometry of physics". In the chapter on harmonic forms, you will find what he refers to simply as "Hodge's Theorem". It's a little more general than you need, because it applies to general $p$-forms, and you only need functions (0-forms). So I'll ...


0

Yeah, great question. Firstly, one of your underlying assumptions is spot on. You can send messages with gravity. If you have a brick in your hand and wave it back and forth a pendulum on the moon will have its period slightly changed because the direction and strength of the gravitational field around the pendulum will change slightly as the brick moves. ...


-1

No, a star with 3 solar masses or more, will eventually become a black hole. After exhausting all the elements inside and expanding into the next phase to then explode in a hyper or supernova, it will become a black hole. So no, even if the Sun had that mass, it will take billions of years for that to happen.


0

I think you should be more specific in the type of black hole, because the spinning (and spinning-charged) black holes can make light reach the photonsphere and not be swallowed by the black hole, but enter an orbit around it. Eventually debris that are also in orbit will be heated up and started to glow, and it's the first step for a quasar to be born.


-1

No, there's not a map of all the black holes in the galaxy, remember they're quite hard to detect, but there's one of stars in our solar interstellar neighborhood.


2

"Unfortunately, due to time dilation the time for mass to fall into the event horizon becomes infinite, so that nothing can cross the horizon within in the life time of the universe." This can not be understood as any objective statement about coordinate-independent physical realities, since it's only true in certain coordinates like Schwarzschild ...


-1

I can only offer an opinion rather than a peer-reviewed authoritative response. But I hope it's of some use. We can read various descriptions about falling into a black hole, such as this on Baez. There's also Andrew Hamilton's website. I'm afraid to say I think they're wrong. Take a look at The Formation and Growth of Black Holes on mathspages, and you can ...


0

No, relativity is about reference frames. The person falling into a black hole will pass through the event horizon, but due to time dilation, no one else can observe it.


1

The problem with this is that $$E = mc^2$$. Density, however, is given by $$density = \dfrac{mass}{volume}$$ Thus, if volume = 0, then density is infinite. Black holes have a finite mass. It is there density which is finite because all the mass is at a single point (singularity, volume = 0).


1

No, the energy of a black hole is not infinite. It depends on its mass, angular momentum and charge. Infinite density at a point does not translate to infinite energy in the $E=mc^2$ sense. It is in fact possible to extract energy from black holes by exploiting certain properties of accretion disks or ergospheres, but this is a finite process.


-2

I can only answer with opinion rather than facts, but for what it's worth: Is travelling through a worm hole really possible (if we can find one stable worm hole)? No. Wormholes are science fiction. So is time travel. See A World without Time: The Forgotten Legacy of Godel and Einstein. A clock doesn't literally clock up the flow of time like some cosmic ...


1

We know that light is massless so why does a black hole's gravity attract light? Because gravity doesn't just attract objects with mass. It alters the path of light too. Because gravity is caused by a concentration of energy which "conditions" the surrounding space, altering its metrical properties, whereupon we talk about spacetime curvature. But note that ...


1

A photon has a rest mass of nought (where the rest mass $m$ is the Lorentz-invariant quantity in the four-momentum's Minkowski norm squared $E^2/c^2 - p^2 = m^2 c^2$). However, a lightfield of energy $E$ gravitates and itself has a gravitational source equivalent to a mass $E/c^2$. Also, a system of photons has a nonzero rest mass (see reference), as does ...


1

light is supposed to possess relativistic moving mass even though it does not possess any rest mass. m2c2=M2c2-M2v2 where m is rest mass and M is relativistic mass and v=c. this gives m = 0 , but M is not zero the value of M can be calculated from the experimental data on radiation pressure.


-2

NO, at least by 3 reasons : recently,as 2003, it was found that 40% of the matter in the vicinity (accretion disk) of the BH will be radiated away. I'm convinced that all the matter, in excess above the limit, will be radiated away before a BH can be formed. quoting from WP-Black-Hole In the case of compact objects such as white dwarfs, ...


1

The radius of a black hole, the Schwarzschild radius, is proportional to the mass of the black hole: $$r=\frac{2GM}{c^2} \to r \propto M$$ Therefore, a black hole with a larger radius must have a greater mass. So black hole #2 is ten times as massive as black hole #1. The formula for gravitational time dilation is $$\frac{t_0}{t_f} = \sqrt{1- ...


2

The answer is no. Once a black hole is formed you no longer have access to the information stored inside it. The information is not lost, but slowly radiated away as the black hole evaporates. So, the ideal system would be one that is almost near the density for a black hole collapse, but never reach it. Otherwise you loose "easy" access to the information ...


1

With current technologies the Kerr parameter could not be precisely estimated yet. The best result today is a Kerr factor $a=0.52$. Genzel, R., Schoedel, R., Ott, T., Eckart, A., Alexander, T., Lacombe, F., Rouan & Aschenbach, B., Near-infrared flares from accreting gas around the supermassive black hole at the Galactic Centre (2003), ...


0

The short answer is: in ultra-deep gravity wells one has to bring relativistic considerations into the picture. The result of this is that an 'event horizon' forms in ultra-deep gravity wells such that an observer looking from a distance into the gravity well can only look till a finite depth. In loose terms, that specific finite depth corresponds to the ...


0

Without invoking relativity, let's look at energy. The force of gravity between two objects goes as $$F= \frac{GMm}{r^2}$$ Which implies that the force is weaker when you are far away and stronger when you get closer. potential energy is the integral of force, and we know the sum of potential and kinetic energy is constant. Putting potential energy =0 at ...


2

You make two wrong assumptions in your question, namely that if an object is accelerating the velocity would keep increasing ad infinitum without limit, and that the acceleration due to gravity on earth is always $9.8 m/s^2$ these are both not the case. First of. The theory of relativity doesn't allow for objects that have mass to go faster than the speed ...


-1

There is a concept known as entelechia, which basically means an idea so complicated and so full of border cases that no useful ideas can be obtained from it. One entelechia is the concepto of god, a súper being who created everything in just 6 days and who needs your money. And the fossil record was put there by good just to test your faith. Eventually it ...


0

You can't just create event horizons of finite area out of thin air like that. Let me explain with a very idealized thought experiment: imagine you have a thin spherically symmetric shell of dust particles of radius $R$ and total mass $M$. Say that initially $R > r_s$ (where $r_s = 2M$ in natural units). Say that the particles are initially at rest so ...


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 problem with you argument is that you are standing on the planet! When you make it denser and denser and eventually turn it into a black hole, you would still be standing on it, ie be outside the event horizon, so as you are saying correctly the light will escape. If you would like to think about being inside the event horizon then it's wrong to imagine ...


0

Well, an heuristic solution would be that it is actually impossible for anybody to cross the event horizon. As you noticed, Schwarzschild black hole is dual (at least locally, that is, if you don't make full turn) to Rindler's observer (the uniformly accelerated one). The stationnary observer in the Schwarzschild case is the uniformly accelerated one in ...


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 ...



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