# Tag Info

-2

Firstly, gravity affecting time is completely different from gravity affecting mass. Do a search on "gravitational time dilation". Secondly, gravity affects light because it has energy. This is governed by the Einstein Field Equation. Thirdly, the black-holes tag is irrelevant. Fourthly, the visible-light tag doesn't apply here. It applies to optics only, ...

0

I remember calculating the amount of time one would take to fall radially inward with zero energy (so as to maximize the time taken) starting at the schwarzschild radius of a black hole of n solar masses to be $$n(1.55 \times 10^{-5} ) \text{ seconds}$$ And you would never be able to tell that you had crossed it in the first place, the event horizon is ...

1

The TL;DR version: even if we could form a synthetic event horizon, it wouldn't help us learn about the black hole interior. The long version: The phenomena you describe where light essentially orbits a black hole is called the "photon sphere" and it doesn't happen at the event horizon. The radius of a black hole, $R$ is where the event horizon is and the ...

1

One can see that a black hole in thermodynamic equilibrium with its environment is always unstable by looking at the Hawking temperature of a black hole, given by $$T = \frac{\hbar c^3}{8\pi k_B M G}$$ The temperature being inversely proportional to the mass means that big black holes are cold, a black hole with the mass of the sun has for example a ...

0

It is not possible, assuming that all tools we could use are "normal" in some particular sense (that is, they agree with some energy conditions). That means, they are made of particles and fields that cannot travel faster that light, for one example. Everything physicists discovered so far is "normal" in that sense, and it is highly unbelievable that we ...

1

David Zaslavsky has given a solid, relatively model-independent explanation of the empirical bounds on the size of an electron based on particle-physics experiments that probe short distance scales by using collisions at short wavelengths. There is also another way of getting at this question, which has been studied by people who have tried to model quarks ...

2

Nice question! I'll try to put together a few ideas into what might be a valid answer, but this could be wrong. The event horizon is a lightlike surface. Therefore no proper time passes at the horizon, and any photon emitted exactly at the horizon would remain exactly at the horizon until the moment when evaporation was complete, after which it would fly ...

3

Our current understanding suggests that black holes can have electric charge, and that in addition to mass and angular momentum, these are the only ways that black holes can have distinguishable physical properties. This is a result of the famous no-hair theorem, although keep in mind that no definitive proof of this theorem yet exists. As this wikipedia ...

2

These are two separate questions. It's better if you don't try to combine two questions into one. In answer to the first question, yes, a black hole can have a measurable charge. You measure it the same way you'd measure any other charge. This is all purely theoretical, however. Many real-life black holes have been observed and characterized by their ...

1

According to Thorne (Black Holes and Time Warps) all matter that approaches the black hole singularity is reduced to a common degenerate form - matter and anti-matter alike. The way I interpret it, matter ceases to retain any resemblance to what existed outside the black hole. The attributes that distinguish matter and anti-matter are stripped away. ...

1

First of all, one has to distinguish between indirect and direct predictions theories. For example concerning theories describing physics at very high energy regimes, such as energy scales where quantum gravity effects are expected to kick in, it is clear to everybody that these high energy scales are with the current technologies not DIRECTLY accessible at ...

-5

As we know that opposite charges attract each other, we consider gravitation as negative and matter as positive.. Thus there is a attraction between the two... That's y we say that real particles are always positive as they r nthng but matter/energy.. However, inside the black hole the gravitational field is so strong that even real positive particles ...

1

I will have a go at an answer. Many good theoretical physicists attempt to create the Theory Of Everything (TOE): a mathematical theory that will model anything we observe in physical reality. Very many of them are concentrating their effort on some model using string theory. Suppose that we find a TOE that fits all known elementary particle data and ...

4

Physicists are tasked to construct models of reality that are internally consistent and that have predictive power. Working on black holes and Hawking radiation fits this bill. The growing theoretical framework on black hole thermodynamics is absolutely essential to eliminate any internal inconsistencies that seemed to affect thermodynamics when applied to ...

2

The idea that every prediction should be falsifiable is good as a first approximation. But the science is complex, and in it everything is linked to everything, almost literally. So actually we never know if some unfalsifiable prediction would be of any use. It can become falsifiable later. It can lead to some ideas which would lead to some new falsifiable ...

1

Your question is essentially asking for opinions and will therefore likely be closed, but I think what you're missing is the fact that in any discipline where theoretical questions are inextricably linked with experimental questions it's impossible to judge in advance which theoretical questions are going to have practical ramifications. Physics and science ...

0

If we refer to the evaporating black hole's Penrose diagram we would see some important differences between the ideal and the evaporating black holes. For the evaporating black hole, there is a moment in spacetime, where the horizon (and the singularity) disappears, mass left at the center of coordinates is zero, and no time dilatation is left. Hense, ...

1

This is an extremely common misconception so I will try to make my answer as simple as possible. A black hole is black and light can't escape because of mass for a given volume (density), not because of mass alone. Take our solar system for example. Our sun is very massive and it pulls on all of the planets hard enough that they orbit the sun. As long as ...

4

You're assuming that the Kruskal–Szekeres (U,V) coordinates have to be defined in terms of the Schwarzschild (r,t) coordinates, but there is nothing special or fundamental about the Schwarzschild coordinates. General covariance says that we can use any coordinates we like. If the K-S coordinates had been the ones originally chosen by Schwarzschild, then ...

2

Why would you assume they do not ? Of course they do. But as you probably know, the gravitational pull decreases with distance (inverse square law). From a safe enough distance any other object (star, galaxy) would feel the normal gravitational pull of an object of the black-hole's mass at that distance; it makes no difference to the stars if the source ...

-3

When scientists say the escape speed of a black hole is greater than the speed of light, they mean that if you leave the surface of black hole at a speed which is near the speed of light, and during the trip you do not get extra boost of speed, then eventually you will fall back. That is absolutely true. But if you get a speed boost during the trip, you can ...

2

This is a great question, although unfortunately it turns out to be very difficult to interpret it in a way that allows a definite answer. The question is ambiguous because of the way mass is defined in relativity. From the way the question is posed, I assume the OP doesn't have a lot of technical background in relativity. However, there is no way to resolve ...

0

I recall a presentation (many years ago at DAMPT) where the presenter claimed that focussing gravity waves could produce a curvature singularity that bore some similarities to a black hole. I've done a quick Google and found this paper, that references two papers by Alekseev: Alekseev, G. A. and Griffiths, J. B. “Gravitational waves with spherical ...

2

You can take a Reissner-Nordström solution for the charged non-rotating black hole, and put its mass $m=0$. Then it would become a so-called naked singularity. More precisely, singularity is a point where some value ends at infinity, while density of mass being just one option. For more thorough consideration of Reissner-Nordstrom and Kerr-Newman solutions, ...

2

It's easy to forget that, in the context of relativity, there is no universal time. You write: For an outside observer the time seems to stop at the event horizon. My intuition suggests, that if it stops there, then it must go backwards inside. Is this the case? But your intuition doesn't seem to take into account that, for an observer falling ...

2

Your intuition is misguided, time does not run backwards inside a black hole. For an observer inside a black hole, time passes in a perfectly "normal" way, such as it does at the horizon. The stopping of time of time at the horizon is, as you mentioned, a phenomenon that only an outside observer experiences. It can for example be measured by noticing change ...

2

People have known for a long time about the existence of cosmological models that include a Big Crunch singularity. In these models, the density of matter in the universe is great enough to cause it to recontract. These models are no longer of interest as descriptions of the actual universe, since we now know that the expansion of the universe is ...

6

A sudden singularity is a singularity that forms in the universe in a finite time. This may see like a strange definition. After all, don't the singularities in black holes form in a finite time, so shouldn't they, and indeed all singularities be sudden? Actually, no! For observers like us, floating around the universe, in our frame of reference it takes an ...

1

This answer is kind of parallel to Brandon's, because I want to emphasise the point underlying these types of observations. We will never be able to observe a black hole, because for external observers the formation of an event horizon takes an infinite time. This may seem a bit pedantic, but it's an important point because our aim is not to directly ...

4

There are so many ways we can either directly or indirectly detect black holes that this answer will be necessarily incomplete. Although no light can escape a black hole, the effects of black holes on the space, matter, and activity around them is often very dramatic. One of the most common ways is via an accretion disk of matter spiraling around the ...

1

You question is as same as asking what happens after a spaceship accelerates to faster than the speed of light. First I would say you have to reach c and the same goes for Rs. It may only take your one second to reach Rs in your own time but in that second the whole universe would change infinitely. Why? Because according to the scharzschild metrics ...

1

They see nothing special. From the point of view of an infalling observer, the event horizon is just an invisible "point of no return". If Alice fails to start accelerating before her half of the ship crosses the line then she will be doomed along with Bob, but there is no physical measurement that either of them can make that would tell them where this ...

1

In order to escape Alice's section of the space ship has to accelerate. The required acceleration increases rapidly as you approach the horizon (to infinity right at the horizon, meaning escape from there is impossible). So what they both see initially is Alice's section of spaceship accelerating, placing the spacecraft under increasing tension until is ...

2

The conceptual hang-up you have with Hawking radiation falling back into a black hole is a misunderstanding. You're right that classically "nothing can escape a black hole". The trick to Hawking radiation though is that it forms on the horizon of the black hole and saps energy from it. The radiation is not escaping from the black hole, it is stealing ...

0

You wrote: Basis an intuition around, all matter/space/time is expanding outward in a similar fashion from the start point (not a fixed point in space I realize) of the universe. And then: So does that mean that all matter inherently spins around a central axis from which the universe expanded from? But these are contradictory statements. ...

1

The metric is spherically symmetric. This means that angular momentum of the system is conserved (you can show this directly using the metric by computing the three killing vectors associated with spatial rotation and their corresponding conserved quantities) and therefore that the motion is contained to lie in a plane. If the motion is in a given plane, ...

4

Entropy isn't a force that causes things to happen. But anyway, the answer is no. Not all matter in the universe is expected to eventually collapse into black holes. See Adams and Laughlin, http://arxiv.org/abs/astro-ph/9701131 , section VD. Note also that black holes eventually evaporate, so when matter collapses into black holes, the result is that it ...

0

To quote a comment Scott Aaronson made on his blog: Can you perform an arbitrarily long computation with minimal effort, by leaving your computer on Earth, boarding a spaceship that accelerates to close to the speed of light, then turning around and returning to Earth, where you find civilization collapsed, your friends long dead, and the Sun going cold, ...

3

I remember attending a seminar by Unruh a few months ago and the same question arised. As far as I remember, he enfasized that in these hydrodynamic analogs of black holes, the flow is not quantized, it is a classical fluid, and everything is classical and that the dumb hole behaves like a quantum amplifier emitting quantum noise from the Horizon. ...

0

So far only one thing is sure about the black holes: they are exist. There is something in the galactic center that does not radiate energy but stars nearby orbit it at high speeds. It must be a black hole. But that's all we know for sure. We haven't even seen a real event horizon yet (the black disk of a black hole). But in this year we will see a larger ...

2

where does it go and what happens when it clogs up? In the context of an ideal, static black hole in general relativity, world lines end at the singularity. Consider the diagram below: This is the Schwarzschild geometry in Kruskal–Szekeres coordinates. In these coordinates, it is clear that there is no place or time that the entities falling into ...

1

Your question touches on the quantum behaviour of black holes, a subject with fundamental, unsolved problems. Classically, black holes cannot emit light, because light cannot escape their strong gravitational fields. The gravity is so strong that any matter that fell into a black hole would be torn into fundamental particles, and become part of the black ...

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

In your diagram where the black hole is just behind the observer on the line between the observer and the object being observed the answer is "yes", the apparent size of the object will shrink. The color of the object will be blue-shifted and the whole universe will appear to contract. There is a great demo of the effect at Journey into a Schwarzschild ...

1

There is no correct answer for all cases. Of course the upper-bound is 100% but in practice, even if nearly all of the matter would fall in, a huge amount of energy will be radiated as x-rays and $\gamma$-rays due to heating in the accretion disk. The dynamics of in-falling matter are quite complicated and very sensitive to initial conditions. Currently ...

1

A black hole won't form. The reason why is that the boosted particle is equivalent by a boost to a reference frame where there is no black hole, and the presence/abscence of a black hole is coordinate-independent. While the energy of, say, an object with Earth's density profile can be made arbitrarily large through a boost, the boosted Earth will still ...

2

Like the horizon of distant land or sea on Earth, it's a location beyond which you'll not receive any sight of. For the Earth, it's just the bulk of the planet being in the way. This obviously observer-dependent. For a black hole, or the cosmological boundary, it's more a matter of light can't get out, or does but is red shifted to zero frequency before ...

0

To make this question more precise, we need to define terms a little better. The standard definition of a black hole is the following. Suppose there are points in spacetime from which it's impossible to escape to a large distance. (Technically, we want these to be points from which we can't escape to future null infinity.) If we have set of such points, then ...

3

I want to start by saying that the original answer to the question does a great job, but I feel a question like this deserves a more in depth (and perhaps fun) answer. 1. The photon sphere is the radius at which the velocity of a circular orbit is the speed of light. We can no more detect it than we can detect the Moon's radius around Earth; we can measure ...

0

This is an answer to the part of the question about why smaller scales are inaccessible. Particle physicists are in the business of measuring things at very small distances. To do this, they have to use particles with wavelengths comparable to the distance scale they're trying to probe, and they have to collide those particles with the thing they're trying ...

Top 50 recent answers are included