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4

The distance at which the tidal forces froma primary start tearing apart a satellite is known as the Roche limit. In calculating the Roche limit we assume that the yield stress of the rock making up the planet is small compared to the gravitational forces at work so it can be ignored. The question is then simply whether the gravity of the body (in this case ...


0

In this, I have taken the assumption that the singularity is created within the atmosphere, perhaps by an experiment or some such, rather than a wandering black hole, and therefore starts life with much less mass than the earth. A naturally created black hole travelling in space would generally be far more massive, having been created from a collapsing star. ...


2

There is a known example where it looks like a black hole on the outside but is flat spacetime on the inside. Imagine the funnel shaped exterior of a black hole and take the entire part outside of the event horizon, which is a spherical shell of surface area $4\pi r^2$. Then take a spherical ball of Minkowski space of radius $r$ and sew the two together ...


17

The radius of the event horizon of a black hole of mass $m$ is given by: $$ r_s = \frac{2GM}{c^2} \tag{1} $$ Let's consider your idea of taking $n$ black holes of mass $M$ and arranging them into a sphere. The total mass is $nM$, and the radius of the event horizon corresponding to this mass is: $$ R_s = n\frac{2GM}{c^2} \tag{2} $$ Now let's see how ...


5

It is in principle possible, at least for some time, to have a collection of black holes gravitating along the surface of a sphere such that one can still escape from the inside of that sphere. In other words, the inside of the sphere need not be hidden behind a gravitational horizon. However, as soon as the density of black holes exceeds a critical value, ...


0

The positive mass theorem is a theorem, so you make some assumptions and fix a deductive methodology and then get a conclusion. Common assumptions include an energy condition and asymptotic flatness. The conclusion is usually something like that the ADM mass is positive. I think the idea is intuitively explained by an example. If you look at the ...


1

This is covered by a number of existing questions, but I think it does no harm to present a fresh summary. Firstly the analogy of the black hole absorbing one member of a pair of virtual particles is just an analogy, and actually a rather poor one, but let's go with it for now. In this analogy you're quite correct that equal numbers of particles and ...


0

Ummm, not exactly. The spectrum emitted by macroscopic black holes is essentially a thermal spectrum of neutral particles, i.e. photons. If neutrinos were massless, they too, would be emitted, but with current estimates for neutrino masses, stellar size black holes are way too cold to emit even neutrinos. All known black holes are much colder than the ...


1

General Relativity allows black holes of any size (though making a small one might be hard or worse than hard). So as a thought experiment that means that you can consider a small black hole that curves spacetime exactly as much as the sun does. This black hole would be much smaller than the sun, but to us out here spacetime would look the same (except we ...


0

In the Schwarzschild spacetime, the Schwarzschild time is a Killing vector field, $\xi = \partial_t$ which generates a conserved quantity for an orbit of a particle with four-velocity $u$: $$e = -g(\xi,u) = \left(1-\frac{2M}{r}\right)\frac{\mathrm{d}t}{\mathrm{d}\tau}\text{.}$$ Additionally, the spacetime is rotationally symmetric, which gives a conserved ...


1

The equation you cite is the Newtonian potential energy, and as BMS suggests I'd guess whoever is presenting the video is just being careless with the sign. Potential energy is not a well defined concept in general relativity. If you do a naive integration of $Fdr$ along a radial line to the event horizon it goes to (minus) infinity at the event horizon. ...


1

The difference between your bungee cord analogy and the train is that the bungee cord is secured to the centre of the merry go round while the train is falling freely. So unlike the bungee, the far end of the train is not supporting the weight of the parts of the train nearer the singularity. The tidal acceleration between two points separated by a small ...


4

If you measure the large-distance strength of the gravitational acceleration $g\approx \frac{GM}{r^2}$ of a star / black hole with the assumption that your distance $r$ is much further out than the various mass parts, shock wave, and ejected material; then $g\approx \frac{GM}{r^2}$ is (within a percent or so) the same before and after the supernova. This is ...


0

You're using cartoon physics. There's no "force carrying a photon". Photons will happily move on their own, without any external force being applied. This is basic Newtonian physics: F=0 <=> a=0 <=> dv/dt=0 <=> v(t) = v(0)


0

You should not take claims about the failure of MOND to reproduce strong gravitational lensing too seriously. In this review article the authors, Benoît Famaey and Stacy S. McGaugh, state: Due to the fact that all the above models were using the Bekenstein μ-function (α = 0 in Eq. 46), and that this function has a tendency of slightly underpredicting ...


9

It actually goes the other way around: when a star collapses to form a black hole, its planets (if it has any) will become unbound and fly away to infinity. Simple reason: when the star explodes to form a compact object (neutron star or black hole), it releases most of its mass in the form of a SuperNova explosion, so that the central object around which ...


0

The reason why a massive star does not immediately collapse to a black hole is radiation pressure. When a star is in that phase of its life called Main Sequence (MS), its luminosity depends approximately on its mass roughly as $M^4$. This means a star 10 times as massive as the Sun would be 10,000 times more luminous. This enormous luminosity is mostly ...


3

It wasn't a black hole because the density wasn't sufficiently high. The density was lower than what is needed for a black hole because the volume was larger. The volume was larger because the atoms (mostly hydrogen) were kept away from each other by the pressure produced by the fusion processes. Once the fusion processes stop, this source of repulsion ...


1

It is basically the same as two ordinary objects colliding and sticking together. The combined object's rest mass is the sum of the total energy of the original objects in the center-of-mass frame, which is their rest mass/energy plus their kinetic energy in that frame. Some of that would be carried away as gravitational radiation, but typically only a small ...


35

Actually, it doesn't have the same mass, it has significantly less mass than its precursor star. Something like 90% of the star is blown off in the supernova event (Type II) that causes the black holes. The Schwarzschild radius is the radius at which, if an object's mass where compressed to a sphere of that size, the escape velocity at the surface would be ...


30

When you watch a pop-sci TV show, you need to take everything you see with a very healthy grain of salt. This is particularly the case if the show's host isn't a scientist, but even when a scientist is the host, you need to be suspicious. Stellar black holes do not turn into monsters that reach out and pluck objects from the heavens. From far away, a black ...


0

Going out on a limb here, if string theory is right, and one can make microscopic black holes in an accelerator, there should be gravitational analogs of Stern-Gerlach (and more general spin polarization) experiments for these microscopic black holes. They could also have measurable excited states leading to decay spectra. For macroscopic black holes, of ...


1

I'll attempt an answer from a different perspective from the rest. The Schwarzschild solution is the vacuum solution for a static, spherically symmetric spacetime. The Schwarzschild coordinates are 'nice' for the solution for at least two reasons: (1) the line element far away from the spatial origin, in these coordinates, approaches the line element of ...


2

No, it is not possible. This is the whole essence of the so-called no-hair theorem: that all previous details concerning the object from which the black hole formed has no bearing whatsoever on the properties of the final object, except for mass, electric charge and angular momentum. And also, that these quantities enter only as global values, not with ...


4

Remember that the definition of an event horizon is "the past boundary of future null infinity". What this means, in common language is -- take all of the light rays that escape to infinity. Then, find the one that just barely doesn't make it back out to infinity. The surface formed by these light rays is the event horizon. Now, look at a the Kruskal ...


1

This what Penrose and Hawking proved with the Penrose-Hawking singularity theorems. Specifically, and I quote from the linked article: Penrose concluded that whenever there is a sphere where all the outgoing (and ingoing) light rays are initially converging, the boundary of the future of that region will end after a finite extension, because all the null ...


2

In Schwarzschild coordinates, if you look at the $g_{tt}$ and $g_{rr}$ parts fo the metric, they flip signs at $r=2M$. Therefore for $r<2M$ the $r$ direction is timelike and the $t$ direction is spacelike. The future-timelike light cone of any event inside the horizon points toward smaller values of $r$.


7

If I take the (corrected) formula : $$R= 2M + \sqrt{r(r-2M)}+2M\ln \left[ \sqrt{\frac{r}{2M} -1} + \sqrt{\frac{r}{2M}} \right] \tag{1}$$ You have (if no error) : $dR = \dfrac{dr}{\sqrt{ 1 - \dfrac{2M}{r}}}$, so $dR^2 = \dfrac{dr^2}{f(r)}$, and this simplifies your metrics. However, you cannot invert the formula $(1)$ to get $r$ as an explicit function of ...


-1

Imagine this sheet of paper is the known universe: The paper is the known universe. The two dots on the sheet are atoms or particles. The empty space on the paper is space/time. Space/time is continuously expanding from every point on the paper (indicated by the arrows), except for those points where a particle or atom exists. Because space/time is ...


1

There is a completely smooth exact solution to Einstein's equation known as Oppenheimer-Snyder collapse that describes this exact scenario (aside from the star being initially stable). There is no sudden local change in the metric when the event horizon forms. The only way local observers could know is if they try to send a signal to infinity and wait long ...


-1

Concrete scenarios of “local metric changes during observation of the horizon-forming events” depend on details of the latter (how rapidly, how symmetrically, etc.), but generally your material structures that provide a “non-lethal environment” would fail before you will be able to notice that an event horizon forms. Failure of these shielding structures ...


1

Just remember that a Black Hole doesn't have infinite gravity - it just has however much mass created it in the first place. Yes, anything that gets within the event horizon is trapped forever, but that event horizon will actually be smaller than the size of the equivalent amount of mass composed of ordinary matter. This is also why "microscopic black holes" ...


1

A material object may hit the (spacelike) singularity (like one inside the Schwarzschild black hole) at any speed smaller than $c$, if measured from the frame in which the singularity itself is described by $t={\rm const}$. In other words, the angle in the Penrose causal diagram between the incoming trajectory of the doomed massive object and the horizontal ...


3

They are just saying that in our universe of 3 spatial dimensions the event horizon is a 2-sphere. Ignoring time, our universe is a 3 dimensional manifold because it takes 3 numbers to specify a point within it. Likewise, an event horison is a 2 dimensional manifold because it takes 2 numbers to specify a point within it. Judging by the comments there is ...


1

It is difficult to keep track on these things, specially since quasars are also highly variable. Trying to answer the question of the post title, I found this example that sound pretty impressive: $7\times 10^{14}\,{\rm L_\odot}$, or $1.4\times 10^{41}$ W. The ESO press release refers to the kinetic luminosity, or the kinetic energy of the outflow per unit ...


1

We do not know if the universe is closed or open, so space could very well be infinite. However, that does not mean that there is an infinite amount of space in anything. Such a conclusion does not quite make much sense in terms of a logical,mathematical (or even philosophical) argument. Take Zeno's paradox for instance: The paradox states(in summary) that ...


1

If you construct your metric in such a way that $\phi$ is a killing vector generating an axisymmetry of the spacetime, and there is a value of the other three coordinates where $g_{\phi\phi} < 0$, then you have a timelike killing vector generating an axisymmetry. The curve traced out by this vector will then be a closed timelike curve by construction. ...


1

If you did extend your arm out, you would have indeed changed your angular momentum, and for angular momentum to be conserved your orbit would change slightly in the opposite direction. In addition (I am going by your theory of "partial Spaghettification,"), if your orbit was not thrown off by the gravitational pull of the black hole(which it would be, I ...


3

People expect the singularity to not actually form, and for there to be some different structure to the object at distances where the curvature approaches $\frac{1}{\ell_{P}^{2}}$, where quantum gravitational effects should become important. There are some other researchers that expect the spacetime to be cut off at the horizon of the black hole and there ...


1

There is a logical flaw in the assumption: gravitational radiation is not the cause of gravitational attraction. The latter is due to curvature of spacetime outside the black hole. For a black hole to attract/exert redshift on something, nothing has to propagate from inside the event horizon. Gravitational waves (radiation) merely represent moving ...


1

I'll try to answer your questions, 1. Bob does in fact 'see' that Alice has died, however not directly. He sees her death through Hawking radiation. This can happen two ways, the first is that if Bib wants to see Alice at the event horizon he has to use some means of 'measuring' her whereabouts. He could do this by shining a flash light on her but, due to ...


2

Why do we care about the physics inside black holes? By Karl Popper's reckoning and the rules of general relativity, we probably shouldn't. From the outside, it's not being scientific to theorize about the inside. Whatever happens there is not falsifiable; not just difficult to falsify (like physics at the Planck scale or theories about the inside of the ...


0

So my question is how can infinite dense and infinite small object create different size black holes. The singularity within a black hole is not the crucial consideration here. If an amount of (non-spinning, non-charged) mass $M$ is contained within a volume of radius of $r_s = \frac{2GM}{c^2}$, the mass $M$ is 'hidden' within a spherical event ...


0

Different black holes have different masses. For non-rotating black holes, the distance $$r=\frac{2Gm}{c^2}$$ describes the event horizon. As you can see, the mass $m$ of the black hole enters here. This WP page on the Schwarzschild radius describes has more information.


3

A black hole has two main features: a singularity an event horizon The event horizon is a sphere with a certain radius. Most people visualize the singularity as a point at the center of the sphere, and although that's not quite rigorously right, it's good enough for the purposes of the present discussion. Using a rough Newtonian analogy, the event ...


7

That formula is an approximation derived in linearized GR; it's not correct near a black hole. (The test particle is non-Newtonian, since it's traveling at c, and that's why the answer can differ so much from the Newtonian prediction even though the gravitational field is assumed to be weak.)


1

A black hole has two main features: a singularity an event horizon The event horizon is a sphere with a certain radius. Most people visualize the singularity as a point at the center of the sphere, and although that's not quite rigorously right, it's good enough for the purposes of the present discussion. Using a rough Newtonian analogy, the event ...


1

The first point is easy - your formula is wrong. The radius of a black hole is known as the Schwarzchild Radius, and is equal to R = 2Gm / c^2 While this is very small, it is not zero. As a result, the event horizon has a non-zero area. See http://en.wikipedia.org/wiki/Black_hole for more information.


1

Well, first off, the assumption that black holes have 0 radius is wrong. It is super dense. The singularity (the center of the black hole) has infinite density. Source: http://en.wikipedia.org/wiki/Black_hole


-1

The plasma around Sgr A$^*$ is emitting in X-rays which indicates 10$^7$ K temperatures. Densities vary between 10$^6$ atoms/cm$^3$ and 10$^8$ atoms/cm$^3$, depending on which point along the accretion disk (pdf link)--possibly also depends on who you ask. I don't know that we can answer this question for many wavelengths as there can be a lot of stuff in ...



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