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Biggest super massive black hole 12 billion light years away
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| Watsisname | Date: Monday, 02.03.2015, 14:42 | Message # 16 |
 Galaxy Architect
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| I'll let other moderators decide that.
Added:
To elaborate, I'm not giving you time off because I disagree with you. You'll notice I've disagreed with people plenty of times without pulling moderator powers. I'm doing it here because your manner of debating is more like common internet trolling and less like a sincere attempt at rational discussion. I'll point out I'm not the only nor the first person to make that conclusion about you on this forum, and you have been warned against it in the past.
If you actually are being sincere then you need to learn how to show it. That means if you make an argument, and someone responds to you and explains that you are wrong and why, then you need to show that you made an effort of understanding their response and addressing it before making new arguments. If you don't understand it, then say so, or ask questions. If you don't believe it, ask for sources. If you don't find it convincing, explain why. In short, demonstrate that you're interested in the possibility of learning through a debate. That is how rational people exchange ideas.
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| vanguardmook | Date: Tuesday, 03.03.2015, 08:46 | Message # 17 |
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| Quote Watsisname (  ) To the observer outside the hole, it seems the collapsed star has frozen at the event horizon (though not visibly so, as described previously). But in the frame of the star, or an observer standing on its surface, the collapse does not stop at the horizon. It continues right on down to singularity in finite time. So if you are watching a star collapse from far away and wait long enough (really not long at all), then it will appear exactly like a black hole, and if you dive in, you will not encounter the star's surface. You will instead encounter a singularity.
This apparently contradictory description is simply a result of the gravitational time dilation, and if it's confusing, don't worry. It confused physicists for a long time, too, when they first started studying these objects. It usually takes a bit of effort to grasp it.
Ah, I think I understand now. Does this have anything to do with hyperbolic motion?
I read in other places that an object falling into a black hole will not see the end of the universe coming in from outside due to the shape of the worldlines when plotted. Is something similar to that going on, in that because of the acceleration you can never "catch up" to infalling matter past a certain point because of the acceleration?
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| Watsisname | Date: Tuesday, 03.03.2015, 10:23 | Message # 18 |
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| That's an interesting comparison.  I'd say they're similar, but mostly in the sense of the asymptotic behavior. To the outside observer watching a star collapse, it asymptotically approaches blackness (in the continuum approximation). Similarly, a particle with constant acceleration only asymptotically approaches c. But the actual functions describing these are different, as are the underlying mechanisms. One is special relativistic, the other is general relativistic.
The hyperbolic motion we're describing here is a concept in special relativity and has to do with the Lorentz transformations, which themselves are a consequence of the metric signature of flat space-time (that it's Minkowskian rather than Euclidean). If the metric is written ds2 = -dt2 + dr2, and we choose and solve for ds, the result is the classic hyperbolic function. And that's essentially what your figure is: a plot of the Lorentz transformations and the different regions on the space-time diagram.
The general relativistic nature of the collapsing star and how it appears to a distant observer is a bit more difficult to describe. You have to analyze the equations of motion of the infalling frame (the star), and then map that to the distant frame, thereby accounting for the gravitational redshift. The result is an extremely rapid exponential fadeout to blackness.
Quote I read in other places that an object falling into a black hole will not see the end of the universe coming in from outside due to the shape of the worldlines when plotted.
That's exactly correct. You only get to watch the future evolution of the universe if you are able to hover just above the horizon. (At least for a non-rotating black hole. Things get really crazy when you examine what happens if you fall into a rotating black hole!) And yes, the easiest way to make sense of it is that if you are falling in, then the light falling in (sufficiently long) after you does not catch up in time. You reach the singularity first. By that same reasoning, and just as you said, if you decide to take the plunge into the black hole too long after the star collapsed to form it, then you can not go fast enough to catch up to it.
edit: Fixed a slight misstatement.
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| vanguardmook | Date: Wednesday, 04.03.2015, 06:21 | Message # 19 |
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| Fascinating, I read that rotating black holes can have all sorts of theoretical stuff like orbits inside the event horizon and wormholes. Do you know if a rotating black hole could be used to create a wormhole between two connected regions of spacetime (like, say, between two galaxies)
I always wondered what a black hole wormhole would be like if one were to enter it. Would a hypothetical observer shoot out the other side from a white hole, or would there be another black hole on the other side in a sort of pocket universe?
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| Watsisname | Date: Wednesday, 04.03.2015, 23:03 | Message # 20 |
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| Yeah, one of the surprises for rotating black holes as described by the Kerr metric is that the center is gravitationally repulsive (think of this as being due to the centrifugal force from the rotation). This makes all kinds of weird things happen deep in the hole's interior. The coordinate r=0 corresponds to a ring rather than a point, the event horizon splits into two, the inner of which being a place where inward velocities decrease and light can form stable orbits.
Even more crazy is that if you consider the trajectory of an infalling test particle, you'll find that it gets deflected away from the singularity, and if you extend the manifold you can trace its path into a new region of space-time, just as if it went through a wormhole. Treating rotating black holes as passages to other parts of the universe (or other universes) became extremely popular in science fiction because of this.
Quote I always wondered what a black hole wormhole would be like if one were to enter it. Would a hypothetical observer shoot out the other side from a white hole, or would there be another black hole on the other side in a sort of pocket universe?
It is very cool to see what this looks like. Here's a movie worked out by supercomputer using this metric. (It's actually the fully extended Reissner–Nordström metric, describing an electrically charged black hole, but it has the same behavior and is much easier to work with.)
However, black hole researchers do not currently think this description actually corresponds to reality, and the reason is because this wormhole only works in a pure vacuum solution, where the only mass involved is at the singularity. The moment you introduce any new mass to the system, the solution breaks down, and the way in which it does is incredibly violent. Material attempting to pass both inward and outward through the inner horizon ends up crossing and piling up with itself, but the relative velocities don't decrease. They only get faster. This rapidly leads to what we call a "mass inflation instability" and the space-time becomes unpredictable.
This instability is further interesting for what it implies for you if you choose to fall into the rotating black hole. Rather than passing out of "the other side of the black hole", you instead encounter the mass inflation, which you find contains all the light that fell into the hole after it was formed but before you. This light is extremely blueshifted and energetic, and you are vaporized by it. (Ouch.) So there's no way to use the rotating hole as a wormhole. It's downright lethal. In the current literature we describe this energetic burst of light as another type of singularity, one which contains the past history of the universe from the hole's formation. There is even a third singularity which falls in after you, containing the light from the future evolution of the universe. A crazy but inevitable consequence of the time dilation combined with the gravitational repulsion of the black hole's core.
Study of the interior geometry of real astrophysical black holes (which do rotate) is very difficult for these reasons. We are confident that the Kerr metric accurately describes their exteriors, the ergoregion, and down to the outer event horizon (and probably a good portion of the way toward the inner horizon as well). But somewhere before the inner horizon the metric must fail. The mass inflation instability signals the onset of that failure. The region within the inner horizon and near the ring singularity also allows for closed time-like curves, and this is a huge no-no in physics. So we use Kerr's metric to describe the majority of what we're interested in with real black holes, but its description of the deep interior isn't accurate. We instead have to work through the details of collapse, which becomes a stupendously difficult boundary condition problem. We don't have the same luxury as we did for non-rotating black holes, where the details of the collapse proved irrelevant to Schwarzschild's metric. They are very relevant here.
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| Watsisname | Date: Wednesday, 11.03.2015, 12:02 | Message # 21 |
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| A challenge posted to expando for when he returns:
Quote expando ( ) It is incorrectly taught that once an object reaches escape velocity, it will never return. No matter what the energy supplied, if there is a gradient and you roll a ball as fast as you can up that gradient, and the gradient reduces squared the distance out, the ball will still eventually roll back.
Justify your statement. Using Newton's Second Law, produce a differential equation in one dimension for the motion of a particle of mass m, subjected to an attractive gravitational force of F=-GMm/r2. Solve your equation in terms of its initial and final limits for velocity and position, vo, vf, ro, and rf.
Study your result for the limit that final position goes to infinity. Are there real valued solutions? If so what are the boundary conditions?
What is the behavior of the final velocity for the above solution? Produce a plot in terms of the initial velocity, taking G = M = ro= unity.
What do you conclude from your plot?
Lastly, contrast your results to the concept of gravitational potential energy U=-GMm/r and kinetic energy K=mv2/2.
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| Watsisname | Date: Saturday, 21.11.2015, 08:27 | Message # 22 |
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| I suppose I'll post the solution to the challenge. If you like physics and mathematics you'll probably find this pretty neat. :)
Begin with Newton's Second Law, relating the acceleration vector a of a particle of mass m to the net force vector F acting on it:
Let a particle be acted on by a gravitational force from a mass M located a distance r away, with the force given by
Then the second law says,
Rewriting a with the chain rule, we obtain a separable differential equation:
Which we can integrate between some initial and final conditions (vo,vf,ro,rf),
to get
and finally with a little more rearranging,
This is our fundamental equation of motion for this particle. It gives us a relationship between the change in the velocity of the particle and its change in distance from the center of the gravitationally attracting mass.
There is a really cool thing about this equation. You might know that the kinetic energy (K) of a particle is 1/2mv2. Well, doesn't that look like a couple of kinetic energy terms on the left hand side of the equation? It also turns out that the term -GMm/r is the gravitational potential energy (U) of the particle. So this equation is also telling us something about the change in energy. Specifically,
I.e. the sum of the change in kinetic energy and the gravitational potential energy is zero. This is an energy conservation statement! The total energy of the particle is conserved!
What does the equation tell us about "escape speed"? Is it possible for a particle at some initial radius and velocity to fly off toward infinity, never to return? Let's check! Consider the limit that rf approaches infinity. Then the term GMm/rf will go to zero. With this as a final condition for radius, the final condition for velocity becomes
Does this have any real solutions? Yes! The square term on the left hand side is real as long as the right hand side is greater than or equal to zero. This requires that
If the square of initial velocity is less than this value, then the particle does not go toward infinity. It is bound to the gravitating mass, and eventually it turns around and falls back. If it is instead greater than this value, then the particle does go toward infinity, and its velocity "at infinity" is positive. I.e. it forever continues to move away from the gravitating mass, its speed decreasing but never to zero.
The boundary condition, where the particle "just reaches infinity", with a final velocity "at infinity" of zero, is
This is the definition of escape speed. Notice that this is a speed, not a velocity, because it does not depend on the direction (it is not a vector). The gravitational force is a conservative force which means the work done on the particle does not depend on the path taken. It only depends on the change in distance from the gravitating mass. Therefore, it doesn't matter if the particle starts off moving straight up/away from the mass, or tangentially. The speed required to escape is the same.
Finally, it is worth pointing out that this somewhat lengthy derivation (starting from Newton's Laws and applying calculus) can be done much more efficiently if we start with the energy conservation equation. If we look at the boundary condition where the particle goes off to infinity with a final velocity of zero, then the final potential energy and the final kinetic energy are both zero. Then the initial kinetic energy is simply the negative of the initial gravitational potential energy. Or, the escape speed is given by the condition that the sum of the kinetic energy and gravitational potential energy is zero! If the particle has a total energy less than zero, then it is bound to the gravitating mass. If its total energy is greater than zero, then it is unbound, and flies off never to return.
Q.E.D.
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