# Tag Info

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Yes, it would remain stable. But these black holes are incredibly hot and there is no known matter that could feed them fast enough to balance the mass loss that they experience through evaporation. Assuming that with 'micro black hole' you mean a TeV-sized black hole, the mass loss is of the order of a TeV/fm. In addition these black holes are also small, ...

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Micro black holes have been hypothesized in some large dimension string phenomenological models and are searched for in the experiments at the CERN LHC. The first approach to the decays was thermodynamic with Hawking radiation diminishing them rapidly. Their lifetimes are very short so there is no way to gather and contain them and experiment with feeding ...

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I recently went to a colloquium with the theme "98 years of black hole physics" by string theorist Jan de Boer from the university of Amsterdam. I asked him this question and he replied that there have been lattice computations for black hole thermodynamics, yielding precisely Hawking's factor of $1/4$. Furthermore the result has been obtained using ...

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Whenever an amount of mass (or mass equivalent in energy) finds itself inside a volume smaller than the event-horizon for a Schwarzschild black hole of that mass ($R = \frac{Gm}{c^2}$), then you have the necessary and sufficient condition for a black hole. In fact, the black hole may form before the matter reaches this point, provided that it will reach it ...

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There is no 'escape velocity' inside the black hole, the singularity is in the future of all observers crossing the horizon the black hole and is thus unavoidable. Historically, application of Newtonian dynamics to the light of massive bodies goes back to works of John Michell and Pierre-Simon Laplace (note that light then was thought to be corpuscular, ...

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It's true that for a Schwarzschild black hole the radius of the event horizon is at the distance where Newton's law tells you the escape velocity is the speed of light. However you should not let this apparent success lead you into thinking that Newton's law has anything useful to say about black holes. For a start the Schwarzschild radius $r$ is not the ...

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It is generally adopted that for a 'reasonable' class of 'regular' initial data the vacuum spacetime outside a collapsed object will settle down to a stationary Kerr black hole. This means that at late time the metric is just the regularly perturbed Kerr metric, and the perturbations decay. As was formulated by John Wheeler: 'Black holes have no hair'. So ...

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Some remarks : If we consider a Schwarzschild black hole, a local measure of the curvature is the square root of the Kretschmann scalar $K = R_{abcd}R^{abcd}$. If you consider the inverse of the curvature, you have : $K(r)^{-\frac{1}{2}} \sim \dfrac{r^3}{GM} \tag{1}$ On the other way, the total entropy of the black hole is : $S \sim GM^2 \tag{2}$ ...

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The simplest answer is that the horizon doesn't hold the charged particles, but rather, the information about the charges of particles that have fallen into the horizon lives on the horizon itself. In simpler terms, the electric monopole moment of all of the matter inside of the horizon can be inferred from the geometry and field in a neighborhood of the ...

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In addition to Danu's answer, it is worth mentioning that there is another class of solutions in GR which arise as the artifacts of special symmetry: naked sigularities (that is, singularities without event horizons). The cosmic censorship hypothesis states that for a 'reasonable' matter, dynamical evolution from generic initial conditions can never produce ...

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I believe that both questions are easily answered even without a GR background. Saying that something is an artifact of spherical symmetry means, in this context, that it was expected that the singularity would not occur in cases where no perfect symmetry is assumed. Since perfect spherical symmetry only ever occurs in theory, this would solve the problem of ...

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Answer: slightly less mass. In general relativity, mass/energy of any system is conserved from the point of view of a distant observer (I forget the name of this theorem). We use this mass when we talk about how heavy a black hole is because we are (very) far away. If you start with particles that are stationary, the total mass of the system is slightly ...

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The debate that seems to be happening right now is what does it mean that '... the formation of the black hole does not create nor destroy entropy, so the black hole must contain zero or nearly zero entropy as well.' This is correct, of course, except that the material we observe with zero or nearly zero entropy is 'Bose-Einstein condensate' (BEC) and BEC ...

1

As something falls towards the event-horizon Lorentz transformation on the object becomes apparent to an outside observer. Specifically, space contracts increasingly tighter as the object nears the event-horizon; time for the object slows, slowing to the point that it either appears stopped relative to the age of the universe, or is actually stopped; all ...

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At the event horizon, the person getting sucked in will see light twice as fast. Then as he falls in he will eventually see light 3 times as fast, then 4 times as fast, then 5 times as fast, until light seems to be infinitely fast and time starts to grow infinite to him, and probably he will seem to take a split-second to fall in, then he will be destroyed ...

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Gamma rays are affected just like light rays, so they will be subject to a gravitational red shift and they will be bent by gravitational fields just as visible light is. It's important to be clear that in a gamma ray burst the gamma rays are not generated by the black hole. The process of forming the black hole heats the interior of the star to incredible ...

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The super massive black hole is creating a force acting on the material in the galaxy, but that material still has angular momentum which needs to be conserved. In a similar way, the earth is in orbit around the sun and it is bound in its orbit by the gravitational potential from the sun's mass. If you were to naively calculate the force on the earth as $F= ... 3 I think you're asking if there is some special cutoff density after which spacetime "collapses" and forms a black hole. If this is your question then the answer is no, there is no specific cutoff. Density unites are$\frac{\mathrm{mass}}{\mathrm{volume}}$but the size of black holes is dependent on the mass and the size is not proportional to the volume ... 0 At the event horizon, a test particle will appear to be moving at the speed of light, however due to time dilation to an external observer that particle will appear to have its time slowed to the point it's time is not advancing at all. At this point it will neither appear to travel faster than the speed of light, nor continue to advance in time. ... 1 Well, this is by no means settled, but that's ok (meaning any answer is likely going to be seen as controversial until further observation provide more data). John Rennie's answer doesn't appear to address whether or not singularities actually form in the first place; he does suggests that few skeptics would argue this, but in fact a few have (David ... 0 I found the answer: The apparent horizon$\mathcal{H}$is defined as the outer boundary of the region of$\Sigma$(a hypersurface of spacetime with induced metric$h_{ab}$and extrinsic curvature$K_{ab}$) which contains trapped or marginally trapped surfaces.$\mathcal{H}$itself must be a marginally trapped surface, and thus it satisfies ... 2 Assume the rocket is hovering above the hole. If the rocket loses thrust, an accelerometer on the rocket will measure less acceleration. Thus, an astronaut will feel less heavy. However, if the gravity of the hole increases, the accelerometer reading will not change; the astronaut will not feel any change in weight. But, in both cases, the rocket will ... 2 The maximum mass for a neutron star is the Tolman–Oppenheimer–Volkoff limit and is thought to be between 1.5 and 3 solar masses, the range being due to uncertainties of the equation of state of matter at these extreme densities. If the mass of a neutron star exceeds this limit then implosion to a black hole is assumed to be inevitable, there is no force that ... 2 One estimate is less than 10%. But you should see follow-up papers (click on "cited by"...) for other opinions. 1 The fine-structure constant is not determined by magnetic and electric field strengths, but by other universal constants: $$\alpha=\frac{e^2}{\hbar c}$$ (using cgs here because I'm an astrophysicist). This has a value1 of$\sim1/137$. 1 - There is some debate as to whether$\alpha$is indeed a constant; some of claimed variations over large epochs (Gyr ... 6 I decided to do the calculation and see what happens. According to Wikipedia, the radius of a Planck particle is $$r = \sqrt{\frac{2Gh}{c^3}}.$$ Apparently, the surface area of a Schwarzchild black hole is given by$4\pi r^2$, the same as a Euclidean sphere. I found this surprising until user10001 confirmed that it's because$r$is defined as ... 1 Well, the answer given by John Rennie is very good but I think there's still a problem. The r in the equality is actually the distance between two particles that is$r= |\overline{r_1}-\overline{r_2}|$. If it weren't its gradient wouldn't give the gravitational force. We can only use r itself, i.e$r= |\overline{r_1}|\$ when one body is assumed to be not ...

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I suspect you're not asking the question you're really interested in, because the answer to your question is really boring. If you jump into a black hole you'll see the event horizon retreating before you, and you'll never cross it. The distance you've travelled is an ambiguous quantity since of course in your frame you're stationary and have travelled no ...

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