Tag Info

Hot answers tagged

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 ...


18

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 ...


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 ...


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.)


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 ...


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, ...


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 ...


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 ...


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 ...


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 ...


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 ...


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 ...


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 ...


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 ...


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 $r$.


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 ...


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 ...


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 ...


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 ...


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 ...


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

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 ...


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 ...


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 ...


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 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

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

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 ...


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" ...



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