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0

Per Neil deGrasse's explanation of the Hawking effect, the radiation that leaves the black hole via this effect is indeed the of the matter that previously fell in.


0

What about this derivation? Is this derivation of Black Hole entropy viable? In short, Bekenstein's entropy is integer amount of bits, which cannot be true for variable measured at Planck scale, where the fundamental unit of information is nat, that is $1/\ln 2 $ bit


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By black hole most of us mean the Schwarzschild metric, but the Schwarzschild metric is time independant i.e. it represents a black hole that has existed for ever and will continue to exist for ever. So it has no age in any useful sense of the word. The problem with real black holes is that, as discussed at some length following Hawking's recent paper, ...


0

It is possible that the assumption about a photon "it does not stop" may not be totally true. Light certainly interacts with the environment such as in an eye or solar generator. It may not stop in the traditional sense but may be captured both in an eye and in a black hole. For more information on the possible properties of light http://youtu.be/B1DCP4C4MnY ...


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If the black hole formation was accompanied by some type of explosion, say a supernova, you could estimate the age of the black hole by inspecting the properties of the supernova remnant. Here, you aren't looking at the black hole directly, but the stuff that was sent out at the time of its creation. This, method, coupled with knowledge of light-travel time ...


1

The answer is "maybe". Anomalous gravitational effects in black holes just exist beyond a region called "event horizon". After the collapse, only bodies moving very closely the "event horizont" can feel these bizzare gravitational effects. That's a good example about it: if the sun collapses to a black hole right now, gravitationaly, nothing would change for ...


1

Here are several thought experiments (and what happens in each). I'll ignore relativistic effects like time distortion - not for your sake, but mine :) The earth collapses to a black hole beneath our feet. We fall with it, and end up inside a black hole, presumably dead. The earth collapses to a black hole beneath our feet, but we stay in the same place. ...


2

The formula $F=G \frac{m_1 \cdot m_2}{r^2}$ is valid only for point masses. However, it can be applied to non-point masses if its spherically symmetric. Enter Shell Theorem: 1.A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its centre. So, when a spherically symmetric ...


4

During a supernova, a star blasts away its outer layers; this actually reduces the mass of the star significantly. Any star or planet has an escape velocity - the slowest an object must be traveling for it to escape the gravitational field of the star/planet. For Earth, this is 11.2 km/s. (Note that this value doesn't account for any atmospheric effects.) ...


0

In situations like this we generally assume the mass of the moving body is negligable compared to the mass of the black hole. In that case the spacetime curvature can be assumed to be just due to the black hole, and it's described by the Schwarzschild metric. Any moving body moves according to the spacetime curvature at its location. This is important to ...


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There exists semi-classical computations of the orbit of an electron in Schwarzschild spacetime, such as "The Gravitational Analogue to the Hydrogen Atom" by Koch, Kober and Bleicher. The electron has a non-vanishing wavefunction at the singularity, but on the other hand, the Hamiltonian isn't hermitian. This corresponds to the case of the electron hitting ...


-3

Black holes have light not escape them Beacuse. A black hole has so much mass and gravity can't escape it, not even light! The fastest thing in the universe!


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You have misunderstood the nature of a singularity. The singularity at the centre of a black hole is not a point in space. Instead it is a place where spacetime becomes infinity curved, and it isn't possible to describe what happens there. Well, it's not possible using General Relativity, but we hope some future theory of quantum gravity will explain what ...


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The problem here is that you need to think of it from the point of view of an observer. If we, on Earth, (or anywhere else for that matter) try to watch a photon (which travels along a radial null geodesic) approach a black hole, we will never see it 'enter' the back hole. It takes an infinite amount of time for the photon to reach the black hole. So indeed ...


-2

If a graviton was existant, then it would be infinitely fast, as gravity is felt instantly as an object is created. With infinite speed a graviton could escape the event horizon with no difficulty. Also, the graviton is essentially a representation of spacetime curvature in particle form, so is not affected by gravity


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Before the escape velocity reach the light speed, photons, which are produced on the star can escape from there and therefore star can shine (it does not have to be a visible light, though). As you have pointed out, while the mass of the star increases the escape velocity rires too. At some point of time, this velocity, as calculated, is higher then the ...


1

the black hole shrinks due to negative energy of the particle it absorbs when the other particle escapes but it does not matter whether or not it was the matter of the antimatter that was absorbed. For a virtual pair to become real particles, lets take an e+e- pair, energy must be provided by an energy source. In the simple pair creation by a real ...


2

Found this interesting read on this website: "In a classical point of view, this question is based on an incorrect picture of gravity. Gravity is just the manifestation of spacetime curvature, and a black hole is just a certain very steep puckering that captures anything that comes too closely. Ripples in the curvature travel along in small undulatory packs ...


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The situation you are describing is very similar to (but not exactly like) a Schwarzschild-de Sitter universe. That is a spacetime that is flat, infinite in size, expanding with a cosmological constant, and contains a massive body (such as a black hole). The metric for such a spacetime is: ...


2

A vacuum is a lack of matter. A black hole has a lot of matter.


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Yes, it's a misconception, or not - or both. What do you call "matter"? Let's call matter particles with a rest mass. So, everything that's made up of elementary particles is matter. Now here's the catch: To the best of our knowledge, elementary particles are pointlike, i.e. they really don't have any extend in space, they don't really "occupy" any space. ...


1

so what if a light beam was pointed somewhere behind the event horizon to the outside, Inside the horizon, the curvature of spacetime is such that the direction "to the outside" is the past time direction. In other words, to go 'back' towards the horizon is as impossible as it is to go 'back' in time. Indeed, for the same reason that we inexorably ...


1

Note that adding boundary terms to the action does not change the equations of motion! It merely changes the boundary conditions which you have to impose in order for the variational principle to yield the equations of motion. That said, the role of the Gibbons-Hawking-York boundary term is not to cancel the all the surface integrals you obtain from varying ...


3

The most elegant way I've seen to describe this is described in the paper The river model of black holes. If we write the Schwarzschild metric in Gullstrand-Painlevé coordinates we get (in units where $c = G = 1$): $$ ds^2 = -dt_{ff}^2 + \left(dr + \beta dt_{ff} \right)^2 + r^2 d\Omega^2 $$ where: $$ \beta = \frac{2M}{r} $$ This looks like the Minkowski ...


2

General relativity describes gravity as the curvature of space-time. So to understand how black holes work, we must have a basic idea of how space-time works. In regular 3D space, the length $\Delta s$ between two points is $\Delta s^2 = \Delta x^2 + \Delta y^2 + \Delta z^2$. This is just the pythagorean theorem applied in 3D. But in space-time, where we ...


1

As you said, since nothing can go out from the event horizon, if light is sent from within the event horizon it does not escape. However, it is not true, that light will slow down and go back. In fact, inside the event horizon all accessible paths point towards the center of the Black Hole. This means, that, once inside a Black Hole, you will be pushed ...


-5

Black holes pull matter in thus creating a pulling of matter or rotation (gravity) force that matter rotates or spirals around thus creating our shape form and rotational direction of our galaxy. If our sun doesn't die first then yes one day we will be pulled to the center of our galaxy. Every galactic year our solar system speeds up and grows closer to the ...


2

One of the tricky things with general relativity is that different observers may use different coordinate systems and measure very different things. The exterior geometry of any static spherically symmetric object is described by the Schwarzschild metric: $$ ds^2 = -\left(1-\frac{r_s}{r}\right)dt^2 + \frac{dr^2}{\left(1-\frac{r_s}{r}\right)} + r^2 ...


1

I assume you don't mean the speed of light, but you are essentially asking: Will light escape that strong gravitational pull? If this is your question then first: Direct quote from wikipedia -> "An object whose radius is smaller than its Schwarzschild radius ($r_s = \frac{2GM}{c^2}$) is called a black hole. The surface at the Schwarzschild radius acts as ...


1

The speed of light is a constant regardless of where or when you measure it. The speed of the light as it leaves the star will be $c=299792458\frac{m}{s}$. The speed of the same light far from the star will also be $c$. Instead of slowing down like newtonian objects, the light will instead lose energy as it attempts to leave the star. This will correspond ...


2

In 3+1 dimensions, the Ricci tensor vanishes when the stress-energy tensor vanishes indeed. Which means that, whenever there's a vacuum, the Ricci tensor vanishes as well. But it also happens that, in that number of dimension, the metric is not entirely determined by the Ricci tensor. The full curvature tensor (the Riemann tensor) is a mixture of the Ricci ...


3

Gravitational curvature is not fully described by the Einstein or stress-energy tensors. It is only fully described by the Riemann tensor. Consider, by analogy, a simple example from electromagnetism: $\nabla \cdot E =\rho/\epsilon_0$, Gauss's law for the electric field. The function $\rho$ describes the distribution of charge density. The LHS is zero ...


1

The vacuum equation $G_{\mu\nu} = 0$ holds wherever there is vacuum. This is tautological so I don't really see what the content of the question is. If you are hesiant about point masses you can consider the Schwarschild interior solution, which describes a spherically symmetric non-rotating star with uniform density. The Schwarschild interior solution, ...


-4

There is no per se 'point'. The entire 'mass' of the universe defines itself, no matter how we might perceive it, as an infinite density with no locality. In other words, the entire universe is, always has been and always will be a pointless so-called 'singularity'. We can't perceive it or measure it as such, because we are part of it, so it appears to us ...


3

Assuming that the center of the planet just collapses to a black hole, people on the surface won't notice any difference in gravity at all until the ground starts to fall out from under them (the planet will no longer be stable, as there will be no way to support the ground under your feet, ultimately). This is due to a result known as Birchoff's theorem ...


0

To the original poster: You appear to be operating under the "hollywood" misconception that a black hole somehow "sucks harder" than the same amount of mass in a non-black-hole form. However, this false "black holes produce an enormous sucking" misconception is one of the many, many concepts of physics that "hollywood" gets totally wrong; a black hole of a ...


3

Recall that momentum $\pi^{ij}$ is a tensor density in 2-space, cf. e.g. Ref. 2. In other words, $$\frac{\pi^{ij}}{\sqrt{\det g}}$$ is a symmetric (2,0) tensor. Therefore the covariant derivative is $$(\nabla_{\ell}\pi)^{ij}~=~ \partial_{\ell} \pi^{ij} + \Gamma^i_{\ell k} \pi^{kj} + \pi^{ik}\Gamma^j_{\ell k} -\pi^{ij} \partial_{\ell}\ln\sqrt{\det g}.$$ ...


0

There is an explicitly three-dimensional version of a Hawking-Page transition in AdS space. It is given by a transition between an asymptotically $AdS_3$ spacetime and the BTZ (Bañados-Teitelboim-Zanelli) black hole. The line element of thermal (euclidean) $AdS_3$ space is given by ...



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