# Tag Info

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Black holes and "anti"-black holes are the same objects. A black hole resulting from the collapse of normal matter, and a black hole resulting from the collapse of antimatter, are indistinguishable. Recall that black holes only have charge, mass, and spin and there is no way to tell that a black hole originally was matter or not (e.g., we can't measure B or ...

8

Suppose you have some collection of matter that is so dense it has an event horizon where the escape velocity is greater than the speed of light. The escape velocity is obviously due to the strong gravitational field of the matter inside the event horizon, and equally obviously that matter is also pulled by it's own gravity towards it's centre of mass. Also ...

7

As dmckee says in his comment, the answer is no, a stationary spherical shell isn't possible. This is because not even the interparticle forces in neutronium are strong enough to support it. The problem is that once inside the event horizon there is no way to travel away from the singularity, or even maintain your distance from it, without travelling faster ...

6

It is true that, from an outside perspective, nothing can ever pass the event horizon. I will attempt to describe the situation as best I can, to the best of my knowledge. First, let's imagine a classical black hole. By "classical" I mean a black-hole solution to Einstein's equations, which we imagine not to emit Hawking radiation (for now). Such an ...

6

No, there's no detectable dispersion in gravitational lensing, at least not when the wavelength is much shorter than the curvature radius. The reason is simple to see: one may approximate the light by rays propagating along geodesics. They have to be null geodesics because the photons are massless. And given the location of the source and initial ...

5

We are already living in a nearly empty de Sitter space - the cosmological constant already represents 73% of the energy density in the Universe - and the Universe won't experience any qualitative change in the future: the percentage will just approach 100%. However, once the space may be approximated as an empty de Sitter space, all moments of time are ...

5

Indeed, nothing can get under the horizon. The stuff close to the event horizon does move outwards as the BH radius increases. Even more with any BH deformations such as waves on its surface, the tidal deformations or the change of the rotation speed, all the oblects close enough to the horizon remain "sticked" to it and follow all the changes of the BH ...

4

This is a case of an unwisely chosen simile taken waaaay too far. This idea, that the entire universe could be inside the event horizon of not a supermassive, but rather a superduperultrahypermegastupendouslymassive black hole, is usually introduced in introductory classes about general relativity. The instructor in this case is trying to make clear that, ...

4

NO... Once an object has reached beyond the event horizon... We don't even know what happens to it inside. It can't even try to escape. Only Space-time struggles here (inside a black-hole). An observer would see the spacecraft getting red-shifted and just disappear at the moment it nears the horizon. An Event horizon is a theoretical gravitational surface ...

4

Yes indeed, in the circumstances you describe a horizon does form, and it's called a cosmological event horizon. Googling for this term will lots of articles on the subject, though for once Wikipedia has let me down and does not have a good article on the subject. However each galaxy wouldn't be behind it's own horizon as groups of galaxies tend to be ...

4

The Hubble length $c/H_0$ does not coincide with the radius of the observable universe. Your calculation assumes a Hubble parameter that doesn't change over time. This is not correct: the Hubble parameter $H$ changes over time, and $H_0$ (the Hubble constant) indicates the current value of $H$. To refer to $H_0$ as a 'constant' is a bit of a misnomer, it ...

3

I would like to add a fact that, perhaps, is not controversial. Namely, that all the information about any infalling object will be available for the outside observer at any time. The information cannot get lost under the horizon, otherwise we have the information loss paradox. This means that it is theoretically possible for an outside observer to restore ...

3

Assume the object falling in is a blue laser that you launched directly (radially) towards the Schwarzchild (non-rotating) black hole that is aimed directly at you and that you are far from the black hole. The massive object is the laser itself, the light that you are watching is your way to "see" the object as it approaches the event horizon. First of all ...

3

When matter and antimatter meets we see an explosion but no additional mass/energy is created. What happens is that matter is converted to energy while the total mass/energy is constant. So when the "matter" black hole and the "antimatter" black hole merge, the total mass/energy of the resultant black hole will be equal to the sum of the mass/energy of the ...

3

There is no uniquely defined time axis is relativity. Locally, observers in different states of motion have time axes that don't coincide. Even if we fix a state of motion for the observer, there is both a forward and a backward time axis. A more meaningful question would be phrased in terms of the past and future light cones: ...

3

It seems to me that this question is quite trivial, so I hope I am not overlooking anything. For a horizon to form, the observer needs to be "as light-like as possible" (otherwise the slope of his world-line will be more than 45 degrees and union of his past light-cones will cover all of universe). And because there is no other way of achieving that than ...

3

Yes, P will observe a thermal radiation that is locally identical to the Hawking radiation and it is called the Unruh radiation. Its temperature is $a/2\pi$ in the $\hbar=c=k_B=1$ units. http://en.wikipedia.org/wiki/Unruh_radiation A few more words about the relationship of Unruh and Hawking radiation. Unruh found his derivation after Hawking found ...

3

No, a geodesic can freely cross a timelike hypersurface. Just consider the surface $x=0$ in minkowski space--obviously, an particle is free to cross this surface--all it needs is an x-component to its velocity. The crossing geodesic will have a normal component and a tangent component, however--it won't be completely normal.

3

Well, a black hole doesn't have a solid surface, it has an event horizon that, as you might already know, represents a mathematical boundary beyond which no matter/energy can escape, which includes photons. Any light that the "star" may have released is no longer visible because it cannot escape, but incoming light may either fall into the hole, or "orbit" ...

3

You've heard a slightly garbled account of the physics inside a black hole event horizon, but what you've heard is not so far from the truth. The physics of (stationary) black holes is described by the Schwarzschild metric, though unless you're a GR nerd you'll find this a bit opaque. I'll try to describe what is going on in everyday terms, but you need to ...

3

There is nothing wrong with your calculations. From the Wikipedia article on supermassive black holes: "the average density of a supermassive black hole (defined as the mass of the black hole divided by the volume within its Schwarzschild radius) can be less than the density of water in the case of some supermassive black holes" Given that black hole ...

2

The question seems to point to the question of whether or not one can “turn on” the Unruh effect. This of course has some experimental relevance, for if we test it with highly accelerated particles then we do have to prepare a system where the acceleration is initiated. So this question is related to other funny problems, such as the physics which ...

2

Yes, in all of those cases you will get radiation. The case of the cosmological horizon is discussed here: http://math.ucr.edu/home/baez/end.html I wouldn't refer to "Rindler space" -- don't you just mean flat spacetime described in Rindler coordinates? But anyway, an observer with constant proper acceleration in a flat spacetime does see radiation from ...

2

Whatever happens, one should always remember that for any observer gravity manifests itself only through second order effecs in the distance to the observer. In other words, in the coordinates, comoving with any observer, metric is always flat along the observer's world line and is quadratic in spatial distance to the world line (see comoving Fermi ...

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A remote observer will see the matter pile up at the event horizon so the density of matter there increases (to infinity if you wait long enough). The matter falling into a black hole question has been discussed many times on this site, so I won't go into more detail here. However it's worth emphasising that you need to separate formation of a black hole ...

2

Leaving aside for the moment the dark energy and cosmic acceleration, there isn't an event horizon 13.7 billion light years away. It's true that we can't see farther than this at the moment, but we can do if we simply wait. Every year we can can see one light year farther. The reason that dark energy complicates this is that the acceleration caused by dark ...

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This question is the black hole information paradox. If you take two entangled particles, make a black hole by colliding two highly energetic photons, throw in one of the two entangled particles, and wait for the black hole to decay, is the remaining untouched particle entangled with anything anymore? In Hawking's original view, the infalling particle would ...

2

It depends on the size of the black hole. With small ones (a few solar masses) the tidal forces are strong enough to "spaghettify" your body as you approach the event horizon. With supermassive (a million solar masses) black holes the gravity gradient is very small and the tidal forces are so low that you won't feel anything until well after you have ...

2

The horizon, for any static black hole, is the surface where the escape velocity is $c$. Thus, your notion that gravitation is weaker at the horizon for larger black holes is incorrect. EDIT: Consider that the thrust required to hover goes to infinity at the horizon regardless, i.e., the proper local acceleration for a stationary observer goes to infinity ...

2

The question really boils down to the dynamics of event horizons when black holes merge. It turns out that there are some great simulations that explore these dynamics. If one scrolls down to the bottom of this black-holes.org page one can see a video of the merging of two different sized black holes. One can review the underlying paper and see that the ...

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