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Largest Black Hole Measured
porkpickle tips us to a BBC article on the quasar OJ287, a binary object containing largest black hole yet discovered, weighing in at 18 billion times the mass of Sol. Researchers were able to estimate its mass due to the presence of a smaller black hole in orbit around it. When the smaller companion's orbit intersects OJ287's accretion disk, once every 12 years, it triggers a burst of radiation that was detected by the Spitzer Space Telescope. More detail and a diagram are available on the Turku University site.
How large can a singularity be?
I mean, if they used the word "massive" I'd get it. But large?
Proctologists across the globe swoon!
The simple truth is that interstellar distances will not fit into the human imagination
- Douglas Adams
Please keep the goatse.cx references to a minimum.
mmmmm. Donut.
Self proclaimed typo king, and inventor of the bear destroying coffee table (patent not pending).
Which one weighs 18 billion times our sun, and which ones weighs 100 million times our sun?
"Know but never fear the consequences of your actions."
That was serious, here's the link to the non-serious.More there...
mcgrew's razor: Never attribute to stupidity that which can be explained by greedy self-interest
My googling says its even more impressive (http://curious.astro.cornell.edu/question.php?number=31) 100 billion stars in the Milky Way and most are smaller than the sun, so 18 billion makes it very greedy indeed!
I pine for Sol, not a massive black hole. Otherwise, we'll have a massive cleanup job? Oh, wait...
Previously: "Linux... Toward the Sunrise..." Now: "Linux... Toward the-- No, now, part of Every Sunrise"
That's not a solar system, so what do you call it, a blackhole system?
Seems like /. is going down one of them two holes...
Seven Days with Ubuntu Unity
"largest black hole yet discovered, weighing in at 18 billion times the mass of Sol."
Yes, but how many Twinkies is that?
Vincent J. Murphy
Spandex Justice
What exactly is the mass of a hole? Besides a donut hole.
This is truly the holiest of holes!
BIG 2.7 km per solar mass.
used miles instead of km for AU :-)
Mass of an ordinary Twinkie: 36.4 g
http://www.mctague.org/carl/fun/twinkie/
Mass of the Sun: 1.99 × 10^33 g
http://www.google.com/search?hl=en&q=mass+of+the+sun&btnG=Google+Search
1 solar mass = 5.47 × 10^31 twinkies
Mental glitch - 18 billion = 1.8 × 10^9 ... not 18 × 10^18
I'm glad it's Friday. *headdesk*
Yes, there is much gravitational distortion around a black-hole,
which looks like a very light-bright sphere (maybe a little
physically distorted) to all humans, and within the absence of
light there is much levity to consider.
Tell me again, why is it a big black-hole and not a big bright-spot?
In the absence of levity there is gravity.
In the absence of gravity there is levity.
Unaccountable leaders are masters, and unrepresented people are slaves. How do US and EU fare?
Why do people say 'sol' instead of 'sun'. Is there some fundamental difference, or are they just trying to sound smart?
This story makes me want to play gridwars2 again.
And again, and again...
factor 966971: 966971
One question I have about gravity and black holes is this: If nothing can escape the event horizon, how can gravity escape it? In other words, would objects outside the event horizon ever feel the pull of gravity from that which is inside the event horizon?
Using this illustration and my trusty piece of paper straight edge, I estimate the long axis of the orbit to be 21000 AU and the minor axis to be 16000 AU. Using Ramunjan's Approximation for the circumference of the elliptical orbit and converting to light years, I guesstimate the circumference of the orbit to be ~1.99 (call it 2) light years.
For a 12 year orbital period this means that the orbiting black hole is AVERAGING 1/6c (~49965km/sec, call it 50k km/sec)... meaning at periquaserion it's really booking! Much faster than The Dash!
Tom.
I think this finally means that we have a definition for the SI unit "fuck-ton."
Gifts for Geeks - Stuff that really matters!
Hawking: Homer, your theory of a donut shaped universe intrigues me
"Our opponent is an alien starship packed with atomic bombs. We have a protractor."
Further corrected to 3.6 x 10^2 AU.
Google never lies.
Interestingly, this means the relevant density of the black hole (assuming it is spherical, which it almost certainly isn't, but that's a nigh-negligible correction) is 5.7 x10^-5 g cm^-3, or roughly comparable to the density of hydrogen at room temperature.
How fast is the smaller black hole traveling at perigee, relatively speaking?
If the numbers in the BBC article are correct, the orbit has an extent of 1.5 light years and period of 12 years. That means the orbital velocity is a sizable fraction of the speed of light, an average of at least 0.25c!
Off-topic: for anyone that has played with the Q20, what are some interesting things you got it too guess? It doesn't seem to do well with risque things (think of the children!). I'll give you a hint ... Q20 guessed 'toy'. My reply: Yes, sometimes that is how they are marketed.
Atlas Shrugged : Thematic Story
You know, I'm really surprised no one has posted a goatse joke yet...
...otherwise known as *the Sun*. I mean, can we get just a little more pretentious?
In Soviet Russia, Chuck Norris will still kick your ass.
Is it just me or does it seem like AC-2 is probably AC-1's ex-husband?
Now that I think about it, I'm pretty sure everything I just said is completely wrong.
Yep..tried to think something funny...
Guns are for wimps... Use a crossbow.. this way you can pin them to their chair when you go postal.
Actually, my understanding is that the most common stars in the galaxy are Red Dwarfs, and thus smaller than our sun. (Yup, NASA confirms: http://planetquest.jpl.nasa.gov/glossary/red_dwarf.html)
Oh f***, pi*sed myself with laughter! /YAAC (Yet another AC)
So I suppose the source code isn't available... As usual.
I always thought they'd get stuck because the black hole would suck in the tape before they even got half way around it. Then it dawned on me that they probably used stiff measuring tape.
:-)
Those Turku University guys are pretty smart.
"Enjoy what you're doing! If it becomes drudgery, you're doing it wrong!" - Jim Butterfield
The Spitzer Space Telescope? For finding governor Eliot's ethics?
Huh?!?!
You use telescopes (especially "space" telescopes) to find things which are very far away.
To find things which are extremely small (if they exist at all), you use a microscope.
For those of you who don't know, the term "Sol" means "A whale's vagina."
Similes are like metaphors
I have been under the impression that existence of black holes has not yet been proven - at least, not to the degree
to which, say, existence of atoms has. So, I was wondering if anyone could clarify whether black holes' existence is still more of a hypothesis (on par with dark matter, for example) or they are real beyond any doubt.
Thanks
Move on citizens.
One man's pink plane is another man's blue plane.
Now, I don't get it. Couldn't someone past the event horizon create gravitational in space time to communicate to outsiders?
...sounds like a girl I dated in high school.
Ugh, some memories SHOULD be repressed.
If a pion (n-) collides with a proton in the woods & noone is there to hear it, does lamdba decay into the source pa
I once did some calculations about black hole gravity gradients, and calculated that you could end up with a black hole with a gravity at it's "surface" (Schwarzschild radius) of less than earth's gravity (and that the diameter would be approximately one light-month).
As I remember it, I figured that the mass of that black hole would be just a couple of GigaSols (it was a LONG time ago, so I no longer have the exact numbers). If my calculations (and memory) were correct, then this extra-super-massive blackhole should have a surface gravity that we could stand in (presuming, of course, that you could get a stationary platform at the Schwarzschild radius).
Free Software: Like love, it grows best when given away.
so "we" are sure now, that black holes actually exist, right? someone should go there and throw a spoon into it.
The surface gravity at the event horizon of a black hole is always infinite: it requires infinite thrust to hover at a horizon. (Technically: proper acceleration for a stationary observer diverges at the Schwarzschild radius.) I think you were doing a Newtonian calculation, not a relativistic calculation.
(The tidal force, however, is finite at the horizon, and of course you experience no weight if you're freely falling through the horizon.)
they haven't see your momma's!
Now, if the Black Hole were reduced to a neutron Star density, then the gravity at the theoretical physical surface of any non-primordial black hole would be enormously high, but as a black hole gets bigger, the average density, and the 'surface' gravity drop until you reach something approximating the density (and gravitational gradient) of our universe.
Remember: The Universe may be one meta-supermasive black hole. If your premise were accurate, we should all have been violently sucked into the center of The Universe a long time ago.
For a blackhole this size, the event horizon is going to have something in the range of a one light-year radius. Try calculating the Gravity for 18billion solar masses at a distance of one Light year, and see what you get. (I get .002G)
Ah, nevermind... I answered my question myself -- It looks like nominal surface gravity would be ~90G.Free Software: Like love, it grows best when given away.
I could rederive it for you if you really wanted, but in general relativity the answer works out to be:
g = GM/r^2 / sqrt(1-2GM/(rc^2))
where g is the "surface gravity" experienced by a stationary observer, r is the Schwarzschild distance, G is Newton's constant, and c is the speed of light.
This expression diverges as r->2GM/c^2, which is the Schwarzschild radius, so the "surface gravity" is infinite there. (I put "surface gravity" in quotes since of course there is no physical surface.) For r much larger than the Schwarzschild radius, it reduces to the Newtonian expression g = GM/r^2. The Universe may be one meta-supermasive black hole. No. The observable universe may be contained within a supermassive black hole, but the entire universe itself cannot be a black hole; you need to have an inside and an outside to define an event horizon, and there is no "outside the universe". If your premise were accurate, we should all have been violently sucked into the center of The Universe a long time ago. No. I didn't say that the gravitational pull was infinite everywhere outside the black hole; I said that the proper acceleration of an observer hovering at the horizon itself is infinite. For a blackhole this size, the event horizon is going to have something in the range of a one light-year radius. Try calculating the Gravity for 18billion solar masses at a distance of one Light year, and see what you get. Your calculation is purely Newtonian, and therefore inapplicable to real black holes.
By the way, an 18 billion solar mass black hole has a Schwarzschild radius of 0.00561932487 lightyears, or 355.36426 AU (calculation).
Free Software: Like love, it grows best when given away.
If the surface gravity of a black hole is always infinite, then how does hawkings radiation word? It depends on the fact that the gravity gradient at the event horizon is much larger for small black holes than it is for large ones. I think that you're confusing the fact that equations diverges (which rather implies a problem with the chosen coordinate system) with the actual gravity that would be experienced.
Free Software: Like love, it grows best when given away.
Think about it: if you could hover at the horizon with a finite acceleration g, then with any acceleration a>g, you could escape the horizon. But you cannot escape an event horizon.
Here's a sketch of the calculation leading to the answer I gave:
The acceleration 4-vector 'a' is given by,
a = D_u u
where 'u' is the 4-velocity vector and D_u is the covariant derivative contracted in the 'u' direction. In a coordinate system, this expands to:
a^mu = d(x^mu)^2/dtau^2 + G^mu_nu_rho u^nu u^rho
where x^mu is spacetime position, tau is proper time, G^mu_nu_rho is the Christoffel symbol, and there is an implicit summation over repeated indices.
You can work out the position as a function of time easily enough (spatial position is constant for a static observer) as well as the 4-velocity (only a time component, normalized with the Schwarzschild metric). You need the Chrisftoffel symbols for the metric, which can be calculated in a straightforward if laborious way, or you can look them up.
This will give you the components of the 4-acceleration vector, and then you compute its norm with respect to the Schwarzschild metric to get the coordinate invariant proper acceleration, resulting in the expression I gave.
I've been looking for a calculation online which goes through in detail the calculation I sketched, so I don't have to rederive it myself. I found this page. The author derives it just with the line elements of the metric, without using the Christoffel symbols as I did.
The formula I gave appears below the text which reads, "Substituting this expression for dr into the above formula gives the proper local acceleration of a stationary observer", except he's using geometric units in which G=c=1. The preceding text is the derivation of that formula.
He also discusses at some length how the gravitational field can be finite at the horizon, yet the proper acceleration of a stationary observer can be infinite, as I stated. A freely falling observer will experience zero proper force at the horizon (or anywhere else where spacetime is not singular). But it requires an infinite boost to go from a freely falling frame to a stationary frame, when you're at the horizon, because the horizon is lightlike, not timelike: only light can be stationary there.
So, what I hear you saying is that the limit of the value of the acceleration as you approach the horizon is finite, but the the value at the horizon is infinite (and the coordinate system fundamentally changes once you enter)(?)
Free Software: Like love, it grows best when given away.
Schwarzschild coordinates fundamentally change once you enter the horizon, but not all coordinate systems do (see, e.g., Kruskal-Szekeres coordinates), and the statement I made about the proper acceleration is independent of any coordinate system, by definition.
As you approach the horizon, the proper acceleration diverges — increases without bound — has no finite limit.
(This is true for static observers, i.e., ones who are hovering at a fixed location. Freely falling observers will experience zero proper acceleration, i.e., they are weightless. When I say "as you approach the horizon", I don't mean physically "fall into it", but rather a sequence of static observers at locations successively closer to the horizon. It is that limit that diverges: a static observer at r=2.5 M will feel an acceleration greater than another static observer further out at r=3M; a r=2.3 M observer will feel a still greater acceleration, and so on: an observer at the horizon, r=2M, will experience an infinite proper acceleration. Just graph the function I listed earlier and see its behavior as r->2M.)
However, the gravitational field is finite at the horizon. By "gravitational field" I do not mean a Newtonian force field: I mean the spacetime curvature. (Curvature is coordinate dependent, but you can construct curvature invariants which are not, and they are finite.) The curvature may be interpreted physically as tidal forces, or loosely as a "gravitational gradient".
So, the gravitational curve that one would normally associate with orbital mechanics and acceleration is (roughly) as I calculated, but other relativistic effects mean that the acceleration needed to stay stationary diverges?
Free Software: Like love, it grows best when given away.
Fundamentally, what is going on is that spacetime inside a black hole becomes so curved that it is no longer stationary. Stationary observers cannot exist within a black hole. The horizon is the limiting case: light can remain stationary there, but no (timelike, massive) observer can. An observer near the horizon will have to boost harder and harder to remain stationary, because he's fighting the spacetime curvature. At the horizon, he loses, because only light can remain there, and no matter how hard he accelerates, he can't reach light speed.
So, although the spacetime curvature remains finite both near and at the horizon, I wouldn't say that the gravitational effects are similar to what you calculated. The acceleration required to stay stationary is much larger near a black hole than your Newtonian calculation suggests. Stable orbits don't even exist near a black hole; orbital mechanics close to a black hole are quite different from Newtonian theory. The spacetime curvature ("tidal forces", "gravitational gradient"), though finite, is very non-Newtonian near an event horizon.
I'm wondering if it's appropriate to simply say that spacetime, inside of a black hole is so bent, that it's better to describe it as 'just warped'?
Free Software: Like love, it grows best when given away.
That makes no sense. The escape velocity at the event horizon has to be greater than the speed of light, which is not the case for earth gravity.
( note that my calculations are based on newtonian physics -- but, as Ambitwistor pointed out relativity adds an additio0nal twist to the calculation).
Free Software: Like love, it grows best when given away.