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17

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 ...


14

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 ...


10

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 ...


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 ...


5

Hawking radiation is such a miniscule effect we can be sure we'll never detect it for a real astrophysical black hole. The Wikipedia article gives some numbers: For a black hole of the mass of the Sun, the power emitted in Hawking radiation amounts to $9\times10^{-29}\ \mathrm{W}$. Even if all this energy were converted to visible-light photons ...


3

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 = ...


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 $$ ...


3

A black hole is a 4D object, but that's because all objects are 4D as they live in a four dimensional spacetime - three spatial dimensions and one time dimension. However what I suspect you're asking is whether there has to be an extra spatial dimension for space to bend in, making five dimensions in all. If so, then the answer is that no there is no extra ...


3

To expand a bit on the others' answers, a couple of ways to visualize/understand the difference between intrinsic and extrinsic curvature: Intrinsic curvature, as the name suggests, only deals with stuff that lies inside a surface/space/manifold etc. (I will use the term manifold) If your lines, triangles, etc. don't work the same way as they do in ...


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

Using the Schwarzschild radius for this purpose makes sense, because this is the radius of a sphere which becomes a black hole, if it has the given density. For example a sphere made of air at Earth density does not become a black hole if its radius is 1 meter. But if the radius is big enough, it will actually become a black hole. Even though the density is ...


2

A black hole is a region of spacetime enclosed by an event horizon. Thus, the singularity, while a fact about black holes as far as we understand them, is not an defining feature of what black hole is. Therefore, it makes more sense to try to calculate volume (and hence indirectly, density) of a black hole according to the extent of the event horizon rather ...


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.)


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

"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 ...


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

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 ...


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

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 ...


2

The light due to Hawking radiation will only ever be detected from very tiny black holes. The Hawking radiation scales as the inverse square of the black hole mass, but the radiation causes the black hole mass to decrease. This causes accelerated emission, such that all tiny black holes will go through a phase of emitting all their rest mass as $10^{22}$ J ...


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- ...


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.


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

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

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.


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

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 ...


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

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 ...



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