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66

The speed $c$ that is constant is so when measured locally relative to a freefalling frame (i.e. one for which all points follow spacefime geodesics wrt to the metric $g$). Local means that the frame's extent must be "small" enough that it can be thought of as flat: think of this as zooming in on the spacetime manifold, which is a smooth object, with enough ...


33

Suppose you are floating in a river, and you have with you a model boat, called the SS Lightray, that can do 3 m/sec through the water. When you set the boat travelling upstream as far as you're concerned it is doing 3 m/sec. But I'm standing on the bank watching the river flowing at 1 m/sec, so when I look at your boat I see it travelling at a net speed of ...


18

Yes, the dimensionless spin such as $0.2$ in this case is simply the ratio $$ a= 0.2 = \frac{|\vec J_{\rm measured}|}{|\vec J_{\rm max}(M)|}$$ where the denominator is the maximum angular momentum allowed for the same value of the mass (as the measured mass). For the $d=4$ Kerr black hole, the maximum (the angular momentum of the extremal Kerr black hole) is ...


12

Check out this figure from the most recent LIGO paper. This is the reconstructed gravitational wave signal determined by the analysis. Focus on the zoom of the end of the signal. This is the merger. The merger part of the GW151226 looks a lot like the merger of GW150914. The differences tell us about the sources of the gravitational waves in each case....


8

Is this simply the ratio of the angular momentum that the blackhole is observed to have as a ratio of the maximal angular momentum as limited by some Physics (Kerr Metric?)? Yes, exactly. For a spinning black hole there are two event horizons, an inner and an outer horizon. The positions of the horizons are given by: $$ r = \tfrac{1}{2}\left(r_s \pm \sqrt{...


7

I think this is one of those situations where it is cleanest to think relativistically and concentrate on gravity as spacetime geometry, not gravity as a force. If the stress energy tensor's components reach a such a magnitude that an horizon forms, then the mass/energy inside the horizon is doomed to stay inside evermore. This is a question of spacetime ...


6

To explain it in layman's terms, without using advanced concepts: Space is warped in the "inside" of the black hole (that is, under the event horizon) so much, that it behaves completely different than what we perceive hear on Earth. The "outward direction" simply does not exist. For example, here on Earth, we can go in the three spacial dimensions in both ...


5

The strong force is responsible for neutron stars not becoming black holes. Although attractive at nuclear densities, it becomes repulsive at somewhat higher densities. This repulsion is many times greater than the electromagnetic repulsion between protons at similar densities. Yet we know that if a star is too large, it's supernova event will lead to a ...


4

First, the Unruh and Hawking radiation aren't quite "the same thing". They have a similar origin and the Unruh radiation may be considered a flat space (large black hole) limit of the Hawking radiation. Now, the near-horizon metric of an extremal black hole is $AdS_2\times S^2$ while for a non-extremal one, the $AdS_2$ is replaced by the Rindler space. ...


4

I don't think there has even been observational evidence of Kerr-specific effects. While gravitational lensing is well known these days I don't think any of the objects studied have been rotating fast enough for the difference between the Kerr and Schwarzschild metrics to be apparent. Well, not in lensing anyway - Gravity Probe B did measure frame dragging. ...


4

You sound as though you may have heard of Gullstrand Painlevé co-ordinates, which are a particular system of co-ordinates for labelling spacetime defined by the Schwarzschild metric around a nonspinning, noncharged black hole. The analogy is often made of a "spacetime river" with this depiction; if you stand still with respect to the co-ordinates you are ...


4

This question has already been answered by a Randall Monroe of XKCD and Dr. Cindy Keeler of the Niels Bohr Institute. It forms a Naked Singularity. Which is an infinitely dense object, from which light can escape. Source: https://what-if.xkcd.com/140/ You Reissner–Nordström metric for this question, as opposed to the more well known Schwarzschild metric. ...


3

There are two complementary correspondences to black holes. One is a metric where far from the black hole horizon there is $AdS_3$ spacetime. The other occurs quite oppositely where the $AdS_2\times S^2$ spacetime occurs in the near horizon condition of an extremal black hole. For the asymptotic condition, usually seen as the AdS black hole one simply has ...


3

A magnifying glass will produce an upright, enlarged (virtual) image when the distance to the object is smaller than the focal length, and an inversed image when the distance is larger than the focal length (in which case the size of the image decreases rapidly with increasing distance). (image by cmglee, wikimedia commons) For a black hole, the ...


3

The idea that distances have to be integer multiples of the Planck length is a common misunderstanding. The actual role of the Planck length is a bit subtler than that. In quantum mechanics, the possible observable values of a physical quantity (such as a particle's position) are the eigenvalues of a Hermitian operator associated with that quantity. But the ...


3

For a spherically symmetric object the relative rate that time flows is given by: $$ \frac{d\tau}{dt} = \sqrt{1 - \frac{GM}{c^2r}} $$ where $G$ is Newton's constant, $M$ is the mass of the object (the Sun in this case), $r$ is the distance from the object and $c$ is the speed of light. The mass of the Sun is $1.989 \times 10^{30}$ kg and the radius of the ...


2

After a binary merger the resulting black hole will experience a ringdown. The ringdown is characterized by a spectrum of quasi-normal modes (QNM) which, according to GR (specifically black hole perturbation theory), depend only on the final black hole mass, angular momentum, and charge. This is effectively the no-hair theorem. QNMs are oscillations that ...


2

The local escape velocity is $$v_{esc} = \sqrt{\text{2 G M}/\text{r}}$$ At infinty you observe that velocity slower by a factor of $$1-\text{r}_s/\text{r}$$ so at infinity you observe $$\text{v}_{esc} = \sqrt{\text{2 G M}/\text{r}} \cdot (1-\text{r}_s/\text{r})$$ because of gravitational length contraction radial to the mass and gravitational time ...


2

The relationship they're claiming is a little more subtle than what comes across in the press article: The black hole masses are not directly measured. They are inferred from the $M-\sigma$ relation for black holes, which says that the larger the central velocity dispersion of stars in a galaxy, the larger the black hole. The velocity dispersion is the (...


2

An object swallowed billions of years ago thus will retain some incredibly tiny solid angle over which a sufficiently vertical photon will eventually reemerge into the outside world... For a galactic center black hole, how exactly does "pump" mass-energy from the singularity back out to the event horizon... ? ... the apparent horizon at which the ...


2

First, regardless of what physicists thought many decades ago, it seems basically clear today that traversable black holes are forbidden in Nature. They either require some negative energy density which induces instabilities, or equivalently, they allow superluminal signaling which violates the relativistic causality, and so on. Non-traversable black holes ...


2

You would not want to be very close to a black hole merger. Suppose you have two black holes of the same mass $M$ and $m = GM/c^2$. The radius of each black hole is then $r = 2m$, and the horizon area is $A = 4\pi r^2$ $ = 16\pi m^2$. Two constraints are imposed. The first is that the type-D solutions have timelike Killing vectors, which are isometries that ...


2

If you look at the time scale for evaporation (which basically asks you to wait for the black hole temperature to be higher than the cosmic background radiation), you will find that binary black holes will have merged (due to emission of gravitational waves) a long, long time before there is any evaporation happening.


2

Fleshing out my answer in the comment above: The light from all the stars in the picture is lensed (i.e. it's path bent) by the black hole (or BH). The closer it passes to the BH, the larger the angle it's path is bent by. Stars far (in projection) from the BH, like A, are hardly lensed at all (in fact I significantly over-egged it in my diagram, should ...


2

The answer is that it has nothing to do with light, c, black holes, event horizons or relativity. It is simply escape velocity. We know that two bodies attract each other with a force $f = G\frac{M m}{d^2}$ and we know that something is in a stable orbit when its potential energy $PE = -GMm/d$ matches its kinetic energy $KE = m{v^2}/2 $. Solve these two for ...


2

The explanation I like is thus: In GR, all things, from planets to photons, travel in straight lines through curved space bent by mass. Black holes bend and distort spacetime so severely that the curvature captures the photon. Scale things down and it behaves much the same way passing asteroids can be captured by a star. For us, the speed of the asteroid(...


2

At best things are pretty speculative. Cumrun Vafa has proposed that black holes have condensates of tachyons. In some sense you can understand this without much complexity. The Schwarzschild metric has a physical singularity that is a spatial surface. The Penrose conformal diagram for the Schwarzschild metric illustrates this The bosonic string has two ...


2

According to the fuzzball proposal in string theory, black hole are actually horizonless and regular solutions. For some systems in five dimensions made with bound states of intersecting branes this has been already proved directly in supergravity: there are solutions without horizons and singularities, with the same asymptotic charges (Mass, Angular ...


1

I am only answering part I of your question. Please note that following explanation is highly simplified. A burning star has huge mass mostly of hydrogen and helium. This huge mass creates a gravitational pull which tries to collapse the star to a single point. However the thermal pressure produced by the burning hydrogen mass (nuclear fusion) counter the ...


1

The picture I always liked is for an observer free-falling into the black hole, when they're just outside of the event horizon, it looks like the event horizon is propagating outward at nearly the speed of light. After the observer falls just inside, the event horizon now looks like it's propagating outward at greater than the speed of light.



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