New answers tagged black-holes
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With Hawking radiation being thermal (carries no information), black hole evaporation is fundamentally irreversible. This is incompatible with quantum theory, as @Andrew says. So, the problem is to find out what happens with the information.
Solution Attempts Summary:
Denial: Nothing falls in/black holes don’t form. Nope. Just wrong. Note, this kind of ...
1
Entropy argument:
A black hole has the maximum possible entropy for a given volume (or, by extension, mass - see here https://en.wikipedia.org/wiki/Black_hole_thermodynamics )
This holds true for any size of a black hole, including the mass interval where the neutron stars are stable.
In order to decay into a neutron star, a black hole has to dispose off ...
5
I will add some details on what joseph h means by "once a black hole forms, it stays a black hole"
Our current understanding of black hole formation is that once a black hole is formed, the light cones flip at 90° inside and point out toward the singularity. In fact in that Schwarzschild metric
\begin{equation}
ds^2=\left(1-\frac{r_s}{r}\right)dt^2-...
10
Once a black hole forms, it stays a black hole. There are two possible final states of a black hole that emits Hawking radiation:
At some point it stops radiating Hawking radiation and becomes a permanent object called a "remnant"$^1$, with a very small mass, roughly the Planck mass $m_{pl}$ where $$m_{pl} =\sqrt{\frac{\hbar c}{G}}\approx 2.2\...
2
Because the evaporation just shrinks the black hole.
In short, while there might be a minimum size for a star to become a black hole, there's no minimum size for a black hole itself; as a result, the evaporation of a black hole through Hawking radiation will simply result in the black hole growing steadily smaller. Since the intensity of Hawking radiation ...
0
Hawking’s theorem (S. W. Hawking, CMP 25, 152-166 (1972)) states that Surfaces of the event horizon in (3+1)D
asymptotically flat stationary black hole spacetimes obeying the dominant energy condition are topologically 2-spheres.
In addition, if you consider an Euclidean integral of Gauss-Bonnet scalar of a BH ($r$ from horizon to infinity and Euclidean time ...
8
Quantum mechanical evolution via the Schrodinger equation (or whatever your favorite formalism is) is unitary. We can start with a pure state with some particles and no black holes, and throw these particles at each other to form a black hole. We then wait for the black hole to evaporate. Eventually it evaporates completely and, according to Hawking, we get ...
3
The decay of the ringdown modes is exponential, with a decay "half-time" determined by the mass and spin of the black hole (and the mode numbers of the mode). So (at least in the context of classical general relativity) this decay takes forever.
However, at some point the decaying field will be so weak that we need to start worrying about the ...
2
You can think of the general (Riemann) curvature as being made up of two parts: Ricci curvature and Weyl curvature. Ricci curvature is determined by the stress-energy tensor, and vice versa. Since Schwarzschild is a vacuum solution, meaning its stress-energy tensor vanishes, its Ricci curvature is also zero. Weyl curvature, on the other hand, is independent ...
6
There is no physical singularity. $r{=}0$ isn't part of the manifold. This isn't just nitpicking, because the behavior you get in the $r{\to}0$ limit doesn't really fit with the idea of the singularity as a point at the center.
In the Schwarzschild case, two geodesics that approach $r{=}0$ at the same Schwarzschild $t$ but different $θ,\phi$ end up out of ...
2
It's worth noting that the Reissner-Nordstrom singularity is timelike, like how the Kerr singularity is. (and unlike the Schwarzschild singularity, which is spacelike)
but the metric is spherically symmetric, so the singularity cannot be a ring, because that would choose a special plane in the spacetime.
2
The singularity in a Reissner-Nordström geometry is at $r = 0$. With a little assistance from Mathematica I find the Kretschmann scalar to be:
$$ K = \frac{56r_q^4}{r^8} - \frac{48 r_q^2 r_s}{r^7} + \frac{12 r_s^2}{r^6} $$
and this is finite everywhere except at $r = 0$.
theoretically, the existence of the Kerr singularity could be a mathematical artifact.
...
1
I think I have solved this by realising that we can define the energy by the invariant $E^{(B)} = \textbf{u}_{obs} \cdot \textbf{p}^{(B)}$, where $\textbf{u}_{obs}$ is the observer who is measuring the energy. In this case, we have implicitly placed the observer fixed at infinity, such that $[u_{obs}^{\mu}] = (1,0,0,0)$, and so clearly $E^{(B)}$ must still ...
4
The Schwarzschild radius of an electron is
\begin{equation}
r_{s,e}=\frac{2 G m_e}{c^2} = 1.4 \times 10^{-55}\ {\rm cm}
\end{equation}
This is actually almost two dozen orders of magnitude smaller than the Planck length, $1.6 \times 10^{-33}\ {\rm cm}$, which is the length scale at which quantum fluctuations in spacetime itself become so large that we need a ...
2
The description of the gravity of an electron requires quantum gravity, a theory which does not yet exist. Any statements on gravity at subatomic scale are speculation. See also https://en.wikipedia.org/wiki/Black_hole_electron.
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A common feature of almost all proposals to solve the black hole information loss problem (except horizon complementarity) is that their proponents don't seem to understand the reasoning that led Hawking to conclude that there was a problem in the first place.
Schwarzschild coordinates behave pathologically near the event horizon (and don't cover the horizon ...
0
There is a subtlety in expressing the periodicities of variables, because sometimes, there may be a "shift" or "twist" in one variable as we go along a cycle in the other. This is indeed what is happening here.
The picture above is a torus in the complex plane where $z = x + i y$. The torus is defined by two equivelence relations,
\begin{...
2
It is impossible to use Reissner-Nordström black holes in you scenario, because these black holes assume spherical symmetry of spacetime. This is a good approximation if other sources are far away and produce only small distortions around the black hole (BH) you analyze, but in your case the symmetry is highly broken. Just the fact that the two event ...
27
In addition to the other answers, here is an illustration showing how the collision might look like as seen from earth:
(Open image in new tab to enlarge)
This series of photo illustrations shows the predicted merger between our Milky Way galaxy and the neighboring Andromeda galaxy.
First Row, Left: Present day.
First Row, Right: In 2 billion years ...
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Most likely nothing special will be seen from Earth within the lifetime of the Sun. The most recent results (see discussion in Will Milky way and Andromeda collide for sure? ) suggests that the Milky Way and Andromeda will have an initial "glancing blow" with a pericentre (i.e. closest approach) of 75 kpc (220,000 light years) about 5 billion years ...
46
The merger would be indirectly noticeable due to a dramatic burst of star formation and supernovas. The gas of the two galaxies will meet at high velocity, clump, and produce new stars. Some will be very heavy and bright, resulting in supernovas and gamma ray bursts: the merged galaxy may become a bit too risky for planet-bound civilizations dependent on an ...
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The Andromeda galaxy is now approaching us at $110$ km/s. Do you notice anything? No. It takes millions of years for the position to change noticeably.
It will be the same when the collision occurs. Our galaxy is somewhere around $1000$ light years thick. Since the Andromeda galaxy is traveling at about $10^{-3}$ c, it would take about a million years to ...
3
The key flaw in your intuition is that once you are inside the event horizon, there is no amount of pressure or force that can keep you from moving further inwards. (It can be shown that the amount of force required to keep something at a constant radius goes to infinity as you approach the event horizon, and closer to the black hole than that it’s simply ...
1
No. Neutron stars (and quark stars, if they exist) are not black holes. There are many observational and theoretical differences. Some examples:
Light is emitted from neutron stars (see eg: https://en.wikipedia.org/wiki/Pulsar), but cannot escape a black hole.
There is a maximum spin of a neutron star before the star breaks up, which is much less than the ...
0
To derive the asymptotic behavior of $\phi$, you have to start with the Eq.(3.17);
First, you should solve $\phi$ from Eq.(3.17) and replace $x_\pm$ by $\sigma$, then you arrive at $\phi = 2^{-1} \ln\left[\lambda /\left(\lambda e^{2 \lambda \sigma }+M\right)\right]$;
Second, you should manipulate this formula, such that you can expand it as $\sigma\to\...
2
A couple of points not covered in answers by ChiralAnomaly and mmeent:
Since before observer crosses the horizon the light from previously fallen beacons would appear redshifted and at the horizon the redshift is zero, after the horizon crossing the light from previously fallen beacons would appear blueshifted (but coming from the same direction) to the ...
6
In additon to the tidal disruption described by ProfRob, there would be accretion. Basically, most of Earth will be trying to get into an apple-sized volume. An Earth mass falling under 14g over several thousand km acquires tremendous amount of kinetic energy (hundreds of megajoule per kg or more). That flow and compression will heat things up. A lot.
The ...
8
Very roughly, the tidal acceleration would be of order
$$a_{\rm tidal} \sim 2 \left(\frac{GM_{\rm BH}}{r^3}\right)l\ ,$$
where $l$ is the length of your arm and $r$ is how far away your elbow is from the black hole.
So say you got within touching distance, with $l \sim 0.5$ m and $r \sim 0.3$ m, and with $M_{\rm BH} \sim 2\times 10^{25}$ kg, then $a_{\rm ...
2
As an addendum to Chiral Anomaly's excellent answer, I want to consider the second part of the question:
Up until they reach the event horizon, the light received by observer 2 from all the beacons would come from the direction of decreasing r. So does that mean that as observer 2 crosses the event horizon they essentially "pass through" the ...
2
An astrophysical jet forms when matter is accreted onto a black hole from an accretion disk.
The only thing that matters about the jet direction is the accretion disk: The jet is perpendicular to the accretion disk.
This is because of the magnetic field of the disk, whose field lines converge perpendicular to the accretion disk. These field lines determine ...
3
The angle at which a lightlike worldline crosses observer 2's worldline in the diagram should not be interpreted as the direction in which obsever 2 must look to see the light.
Here's one way to understand it. Consider two infalling beacons, one just ahead of observer 2 and one just behind observer 2. At any given point along observer 2's worldline, the ...
0
7 years and no satisfactory answer yet, as far as I can tell. You could only reconstruct the book from the remnants and the heat and light, even in principle, if you could know things about the remnants like position and momentum, in principle, to an arbitrarily high precision. But doesn't the uncertainty principle say that we can't know these things, even ...
12
if there is no matter, no mass, nothing with stress-energy inside the black hole, and the singularity is not part of our spacetime, then neither can cause curvature
On top of the great Dale answer - the Einstein field equations (EFEs) are local. Imagine a lone planet in the universe. Then everywhere outside the planet there is a vacuum and because EFEs are ...
1
We find ourselves in a cylindrical space, with the nose of our rocket pointing along the cylinder's axis, that is, along the coordinate 𝑡. The radius of the cylinder is gradually reducing.
I would hesitate to call it a cylindrical space because cylinders have a different metric (they are intrinsically flat). But there isn’t really a better word in English, ...
33
GR describes curvature as being caused by stress-energy.
This statement is slightly wrong and is the cause of your confusion here.
Technically, in GR the stress energy tensor is the source of curvature. That is not quite the same as being the cause.
An easy analogy is with Maxwell’s equations. In Maxwell’s equations charge and current density are the ...
0
Step 3a fails. An observer a constant distance from the black hole will have to wait an infinite amount of time to see an object fall past the horizon. Putting it another way, say the camera is snapping pictures and beaming them out via radio waves every second. Then an observer outside the black hole, will only ever receive pictures taken outside the event ...
0
Conservation of energy means that the quantity
$$ E = -g_{\mu\nu}\xi^\mu \dot x^\nu$$
along any geodesic with tangent vector $\dot x^\mu$.
4
As a warm-up, consider the non-rotating case $\Omega_H=0$. From the perspective of a distant observer, any pointlike test object that falls into a non-rotating black hole will appear to freeze at the point where its worldline intersects the horizon. The important concept here is that this is true for any timelike worldline, whether or not it represents free-...
1
There is a very strong theorem, called the Area increase theorem, which says that in any spacetime where all matter satisfies one of the energy conditions that guarantees that the matter is "normal" (there are several different technical versions of this condition, but think of it as "no negative mass matter"), then it is the case that ...
1
In practice, all black holes observed have veiled any singularities behind a trapped surface. But theoretically, the question has no definitive answer.
If a singularity is to conserve the net angular momentum of all the stuff that has fallen in, then it must not only rotate but have extension in space. Ring and linear singularities are among those which have ...
0
Search for "blackhole merger" models.
They get pretty much distorted in the process and nothing bad (from the black holes' viewpoint) happens. No singularity is exposed.
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Yes, you should be able to see out of a black hole, assuming you're still alive.
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It should be nearly impossible to do. You have to be roughly the speed of light to escape. But If your position is far from the black hole, it looks like you can still escape.
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In Schwarzschild coordinates, using the geodesic equation, the velocity of an object free-falling radially from infinity is
$$\frac{\text{d}t}{\text{d}r} = \frac{\sqrt{r/r_s}}{1-r/r_s} = \frac{\sqrt{r_s/r}}{1-r_s/r}$$
where $r_s = 2GM$. This means by defining the Gullstrand–Painlevé time coordinate as
$$\text{d}t_{GP} = \text{d}t + \frac{\text{d}t}{\text{d}...
2
There is a fundamental problem with this scheme: constructing a warp drive spacetime. The papers exploring such spacetimes all work by (1) assuming some negative energy density exotic matter to hold together the bubble, and (2) that the bubble already exists. (1) is problematic since we have never seen such exotic matter, but (2) is more fundamental: there ...
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