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Black Hole Seen Eating Star

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• The Arts
Last response: in Hobbies & Leisure
August 25, 2011 3:16:22 PM

This is the first time that a black hole has been filmed eating a star.

Pretty cool stuff, boggles the mind to think about the forces involved.

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August 25, 2011 7:52:39 PM

Now, you're thinking with portals!
August 25, 2011 11:56:50 PM

Black hole got noms.
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August 26, 2011 1:23:20 AM

Nyan cat came out during the video...explains everything!
August 26, 2011 2:28:26 AM

that... was just awesome.

Anyone informed enough about this subject, that could tell me if a Black Hole once formed - does it last forever, or is there some mechanism in which it can be un-black-holed??

Interesting how they said there is a Black Hole in every large galaxy. So much we don't yet understand about the Universe.
August 26, 2011 2:34:16 AM

knarl said:
that... was just awesome.

Anyone informed enough about this subject, that could tell me if a Black Hole once formed - does it last forever, or is there some mechanism in which it can be un-black-holed??

Interesting how they said there is a Black Hole in every large galaxy. So much we don't yet understand about the Universe.

August 26, 2011 12:03:55 PM

Yes a black hole can shrink and disappear. Ive read that not everything that gets sucked into the event horizon will be stuck there. Black holes release Hawking Radiation AKA some form of energy. So after long enough the energy its realeasing will cause it to shrink and disappear, but again its all relative some black holes will eat hundreds of stars.
August 26, 2011 5:56:55 PM

the way I see it is a BH is a vacuum in space where a gravitational system is being formed,( for some odd reason. We still do not understand what makes up gravity).

when the BH is satisfied with all the mass it needs,( light, hadrons, fermions, atoms,ect...) it will either collapse and begin fusion or release the radiation and give off a nice light show.
August 26, 2011 6:11:11 PM

A black hole doesn't need mass per se. Its a hole in the fabric of space and time. But it does release energy in the form of Hawking Radiation. Which (According to people much smarter than I) is a form of thermal radiation, it loses mass until it does BURST in a gamma ray explosion. or maybe it just goes POOF and gone.

(WARNING you may need a doctorates in physics)

ds^2 = -\left(1-{2M\over r}\right)dt^2 + {1\over 1-{2M\over r}} dr^2 + r^2 d\Omega^2 \;

The black hole is the background spacetime for a quantum field theory.

The field theory is defined by a local path integral, so if the boundary conditions at the horizon are determined, the state of the field outside will be specified. To find the appropriate boundary conditions, consider a stationary observer just outside the horizon at position r = 2M + u2 / 2M. The local metric to lowest order is:

ds^2 = - {u^2\over 4M^2} dt^2 + 4 du^2 + dX_\perp^2 = - \rho^2 d\tau^2 + d\rho^2 + dX_\perp^2 \;

which is Rindler in terms of τ = t / 4M and ρ = 2u. The metric describes a frame that is accelerating to keep from falling into the black hole. The local acceleration diverges as u\rightarrow 0.

The horizon is not a special boundary, and objects can fall in. So the local observer should feel accelerated in ordinary Minkowski space by the principle of equivalence. The near-horizon observer must see the field excited at a local inverse temperature

\beta(u)=2\pi \rho = (4\pi) u = 4\pi \sqrt{2M(r-2M)} \;,

the Unruh effect.

The gravitational redshift is by the square root of the time component of the metric. So for the field theory state to consistently extend, there must be a thermal background everywhere with the local temperature redshift-matched to the near horizon temperature:

\beta(r') = 4\pi \sqrt{2M(r-2M)} \sqrt{1-{2M\over r'} \over 1-{2M\over r}} \;

The inverse temperature redshifted to r' at infinity is

\beta(\infty) = (4\pi)\sqrt{2Mr} \;

and r is the near-horizon position, near 2M, so this is really:

\beta = 8 \pi M \;

So a field theory defined on a black hole background is in a thermal state whose temperature at infinity is:

T_H = {1 \over 8 \pi M} \;

which can be expressed more cleanly in terms of the surface gravity of the black hole, the parameter that determines the acceleration of a near-horizon observer.

T_H = \frac{\kappa}{2 \pi} \;

in natural units with G, c, \hbar and k equal to 1, and where κ is the surface gravity of the horizon. So a black hole can only be in equilibrium with a gas of radiation at a finite temperature. Since radiation incident on the black hole is absorbed, the black hole must emit an equal amount to maintain detailed balance. The black hole acts as a perfect blackbody radiating at this temperature.

In engineering units, the radiation from a Schwarzschild black hole is black-body radiation with temperature:

T = {\hbar \, c^3 \over 8 \pi G M k_b} \;\quad(\approx {1.227 \times 10^{23}\; kg \over M}\; K)\;

where \hbar is the reduced Planck constant, c is the speed of light, kb is the Boltzmann constant, G is the gravitational constant, and M is the mass of the black hole.

From the black hole temperature, it is straightforward to calculate the black hole entropy. The change in entropy when a quantity of heat dQ is added is:

dS = {dQ\over T} = 8\pi M dQ \;

the heat energy that enters serves increases the total mass:

dS = 8 \pi M dM = d(4 \pi M^2) \;.

The radius of a black hole is twice its mass in natural units, so the entropy of a black hole is proportional to its surface area:

S = \pi R^2 = {A \over 4} \;.
August 26, 2011 7:11:14 PM

I thought that would help clear up some of the confusion.
August 26, 2011 7:34:57 PM

I punched those numbers in my calculator...and made a happy face.
August 27, 2011 2:50:10 AM

It appears a black holes gravity is limited (to size?), and thermal radiation is kept within the balance of that gravity, and when more is added, more is emitted to keep it equalized
August 29, 2011 7:52:13 PM