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Black Hole's "Point of No Return" Found

samzenpus posted about 2 years ago | from the not-coming-back dept.

Space 130

dsinc writes "Using a continent-spanning telescope, an international team of astronomers has peered to the edge of a black hole at the center of a distant galaxy. For the first time, they have measured the black hole's 'point of no return' — the closest distance that matter can approach before being irretrievably pulled into the black hole. According to Einstein's theory of general relativity, a black hole's mass and spin determine how close material can orbit before becoming unstable and falling in toward the event horizon. The team was able to measure this innermost stable orbit and found that it's only 5.5 times the size of the black hole's event horizon. This size suggests that the accretion disk is spinning in the same direction as the black hole. The observations were made by linking together radio telescopes in Hawaii, Arizona, and California to create a virtual telescope called the Event Horizon Telescope, or EHT. The EHT is capable of seeing details 2,000 times finer than the Hubble Space Telescope."

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130 comments

Editors (5, Informative)

Anonymous Coward | about 2 years ago | (#41649141)

What in the name of everything you hold holy were you thinking when posting this?

Sure, the news is interesting, but while we're getting used to spelling errors and broken links on the front page, a blatantly mis-formatted link is something new, I think.

Re:Editors (4, Funny)

K. S. Kyosuke (729550) | about 2 years ago | (#41649159)

What in the name of everything you hold holy were you thinking when posting this?

This is Slashdot. They hold plaintext holy.

Re:Editors (2)

AliasMarlowe (1042386) | about 2 years ago | (#41649183)

Sure, the news is interesting, but while we're getting used to spelling errors and broken links on the front page, a blatantly mis-formatted link is something new, I think.

Not in a Slashdot summary... The "editors" post them fairly regularly.

Re:Editors (5, Funny)

Rhinobird (151521) | about 2 years ago | (#41649207)

What in the name of everything you hold holy were you thinking when posting this?

I think you meant to say:

What in 'http://news.harvard.edu/gazette/story/2012/10/the-name-of-everything-you-hold-holy-were-you-thinking/the name of everything you hold holy' were you thinking when posting this?

Re:Editors (3, Insightful)

Blue Stone (582566) | about 2 years ago | (#41650527)

I think at this point in time, you could probably randomly assign slashdot editorship to anyone with an internet connection and get a more dedicated, professional, giving-even-the-slightest-shit approach to the task.

Re:Editors (3, Funny)

Anonymous Coward | about 2 years ago | (#41650563)

I was thinking the black hole sucked the HTML tags out of the original post.

I need a plugin (1)

paiute (550198) | about 2 years ago | (#41649157)

Great. My browser has a find option, but now I'll need an unfind option to read the comments on this story to get rid of all the goatse links.

That link cleaned up (5, Informative)

madcarrots (308916) | about 2 years ago | (#41649161)

If you're too lazy to cut and paste. :)

http://news.harvard.edu/gazette/story/2012/10/point-of-no-return-found/ [harvard.edu]

Re:That link cleaned up (2)

SternisheFan (2529412) | about 2 years ago | (#41649301)

Thanks for that. Not everyone can view /. via a home computer screen with mouse. I was going to try to meticulously "select text" myself on my 2 1/2" smartphone screen and post the link, an excercise in futility at times.

Re:That link cleaned up (-1)

Hal_Porter (817932) | about 2 years ago | (#41650925)

First world problems.

Re:That link cleaned up (4, Interesting)

PNutts (199112) | about 2 years ago | (#41651433)

Thanks for that. Not everyone can view /. via a home computer screen with mouse. I was going to try to meticulously "select text" myself on my 2 1/2" smartphone screen and post the link, an excercise in futility at times.

First world problems.

Besides not being funny any more, your statement demonstrates a lack of knowledge of mobile devices in developing countries.

Re:That link cleaned up (2, Funny)

Hal_Porter (817932) | about 2 years ago | (#41651491)

Yeah I bet non clickable links are a major problem when you're forced to work as a slave labourer underground for Kali cultists until their high priest tears your heart out in a savage ceremony.

Re:That link cleaned up (1, Informative)

siride (974284) | about 2 years ago | (#41653401)

Not everybody in the "third world" lives like that.

Re:That link cleaned up (0)

Anonymous Coward | about 2 years ago | (#41651705)

Thanks... I had the same problem on my prototype iPad Nano.

Re:That link cleaned up (1)

magic maverick (2615475) | about 2 years ago | (#41650683)

Or, you know, if you use Firefox you can select the link and right click...

Re:That link cleaned up (2)

b4dc0d3r (1268512) | about 2 years ago | (#41653227)

Did you use CTRL-X or CTRL-C? Or Edit/Copy vs. Edit/Cut?

I'm guessing it was copy. Copy and paste. Like Xerox, a copier. Not a cutter.

"Cut and paste job" refers to the older method of physically cutting apart something to make a new work. Like Thomas Jefferson's Bible. It is also pejorative, implying something that can be done by one with little brain.

You can say the piss poor editing is a "cut and paste" job, because it is. A user being too lazy to "copy and paste" is pejorative enough, going the extra mile is not just unnecessary but actually clouds your meaning.

Language evolves, and I have already lost this fight. But hopefully this helps people.

Unstable? (1)

jandar (304267) | about 2 years ago | (#41649163)

What does unstable mean? I hadn't thought an orbit can be unstable outside of sf-movies ;-). How does an atom in orbit loose energy to fall into the black hole? Gravitational or electromagnetic waves?

Re:Unstable? (2)

K. S. Kyosuke (729550) | about 2 years ago | (#41649193)

Gravitational. Come close enough to a black hole and the orbital mechanics goes haywire. Periapsis starts moving so fast that the word loses its sense altogether, semi-major axis starts contracting as you lose the energy and fall into the center... I'm not sure, however, that there is any sort of fixed boundary. Come to think of it, I'm not even sure how exactly the energy loss increases as you progressively approach the black hole. Anyone versed in GTR here to help?

Re:Unstable? (5, Funny)

Sponge Bath (413667) | about 2 years ago | (#41649601)

Anyone versed in GTR here to help?

When the heart rules the mind
One look and love is blind
When you want the dream to last
Take a chance forget the past

Seasons will change
You must move on
Follow your dream

Re:Unstable? (1)

tqk (413719) | about 2 years ago | (#41650077)

Anyone versed in GTR here to help?

When the heart rules the mind ...

Yes, a perfect example of why we came up with the "Blah blah blah (Bbb)" construct, so no-one would go tripping off to "Grand Trunk Railroad" when "General Theory of Relativity" was intended. :-P

Sigh. Maybe the "editors" should outsource editing to /. readers. Email the submission to five /. readers you trust, tell 'em to review it and write a summary, then the "editor" picks one of the submissions which will be the summary/submission.

"Editor" problem solved. You're welcome. I guess I should patent it now.

Re:Unstable? (0)

Anonymous Coward | about 2 years ago | (#41651153)

I'm sorry, but the correct band to reference for this article is Kansas.

I heard the men saying something
The captains tell they pay you well
And they say they need sailing men to
Show the way and leave today
Was it you that said, "How long?"

They say the sea turns so dark that
You know it's time, you see the sign
They say the point demons guard is
An ocean grave for all the brave
Was it you that said, "How long, how long
How long to the point of know return?"

Re:Unstable? (2)

vlm (69642) | about 2 years ago | (#41649307)

What does unstable mean?

Could be many things... gravitational tidal stresses exceed any known material tensile strength, maybe the Unruh effect if it really exists exceeds the vaporization temperature of everything, maybe hawking radiation vaporizes any known thing... Which one wins probably depends on total mass...

A journalist filter is like an event horizon, in that information cannot escape once it enters. Which is too bad.

Re:Unstable? (1)

jandar (304267) | about 2 years ago | (#41649343)

gravitational tidal stresses exceed any known material tensile strength

Tital stress for a free falling object is finite at the event horizon. The more massive a black hole is the lower the differential gravity at the horizon.

Re:Unstable? (1)

History's Coming To (1059484) | about 2 years ago | (#41649437)

Indeed, I doubt a human in free-fall would feel a thing when crossing the event horizon of a supermassive black hole, it's the little ones you have to watch out for.

Re:Unstable? (1)

maugle (1369813) | about 2 years ago | (#41650475)

Indeed, I doubt a human in free-fall would feel a thing when crossing the event horizon of a supermassive black hole, it's the little ones you have to watch out for.

Well, yes, that too. The bigger reason a human would never feel a thing when crossing the event horizon is that the neurons in his brain that were closer to the black hole would be unable to send signals to the neurons further away.

Re:Unstable? (1)

Anonymous Coward | about 2 years ago | (#41651159)

No, a human falling across the event horizon would not notice anything. The only thing that would signify the "exact moment" would be when the light from the outside universe reaches a certain angle. The signals from your neurons would still reach your brain, because your whole body is falling inward. In a twisted sense, this is like your head catching up to the signals from your feet. This is a nice aspect of GR, that on a small scale, any piece of space time looks like a flat piece of space that would act no different than sitting in the middle of nowhere. The only question is how small of a scale do you have to go. If GR effects were strong enough such that space didn't look flat between your head and your feet, the tidal forces would have long ago destroyed you. If the tidal forces are low enough that it doesn't destroy you, which is easily possible for large black holes, you would not notice anything special about crossing the event horizon from your perspective. Even if the tidal forces were high, there wouldn't be something special to a local observer for the event horizon, as the tidal forces could become a problem long before you reach the event horizon.

Re:Unstable? (1)

Kozar_The_Malignant (738483) | about 2 years ago | (#41650809)

Tidal stresses would have long since ripped you into tiny pieces.

Re:Unstable? (1)

Roachie (2180772) | about 2 years ago | (#41654469)

Depends on the size of the hole. There does not have to be a large gravitational gradient at the event horizon.

Black holes don't create gravity.

Re:Unstable? (2)

vlm (69642) | about 2 years ago | (#41649467)

I researched it some more and

http://en.wikipedia.org/wiki/Epicyclic_frequency [wikipedia.org]

seems to explain it.

Maybe a really poor /. car analogy is that some cars, when you try to skid pad test them to see what kind of cornering force you can muster, will smoothly take on load and spin faster and smaller circles, but at a certain radius they just fling out of control all over the place all random like, and it turns out fluid/vapor orbiting a black hole behaves the same way, smoothing spinning in until at a certain defined radius it goes all unstable.

Re:Unstable? (3, Informative)

mbone (558574) | about 2 years ago | (#41649797)

I regard this as basically a red herring, not to mention mixing up two different things.

The epicyclic frequency and disk stability has to do with the fluid dynamics of an accretion disk - that kind of stability does not require a black hole (look at Saturn's rings, which also have sharp edges).

The key word in Innermost Stable Circular Orbit is "stable" - the meaning is not that this orbit is not decaying (it is), but that it is stable to small perturbations. Inside the ISCO, a small perturbation will cause big changes, and the orbit will rapidly decay. So, outside the ISCO, the orbit is slowly decaying - "inspiraling" - while inside the ISCO, the orbit will decay very rapidly (i.e., "plunge" into the black hole). But, still, if you had a super-duper rocket, you could escape to infinity from inside the ISCO, as long as you hadn't crossed the event horizon.

All of this ignores tidal deformations, which convert orbital energy into heat and can also rapidly decay orbits.

Re:Unstable? (1)

Kozar_The_Malignant (738483) | about 2 years ago | (#41650871)

>look at Saturn's rings, which also have sharp edges

I believe the sharp edges of Saturn's rings are attributable to the shepherd moons. This also appears to be true for the rings of Jupiter and Uranus. Not enough is known about the Nuptunian ring system to say for certain about it.

Re:Unstable? (1)

SternisheFan (2529412) | about 2 years ago | (#41652687)

>look at Saturn's rings, which also have sharp edges

I believe the sharp edges of Saturn's rings are attributable to the shepherd moons...

Black holes, Supernovas, Saturn's rings are a giant buzzsaw... The universe is scary cool!

Re:Unstable? (1)

Sulphur (1548251) | about 2 years ago | (#41650363)

A journalist filter is like an event horizon, in that information cannot escape once it enters. Which is too bad.

At last; media explained.

Re:Unstable? (3, Informative)

mbone (558574) | about 2 years ago | (#41649685)

In Newtonian gravity, 2-body orbits are stable, unless there is drag or some other non-gravitational force.

In General Relativity, orbiting bodies emit gravitational radiation, which carries away orbital energy, and so no orbit is truly stable. However, this only really becomes important near a neutron star or (even more so) near a black hole, where the gravitational radiation energy loss can be significant, and objects can spiral into each other fairly rapidly.

Of course, in either theory, the question of the stability of 3 or more body orbits is very complicated, and still an open area of research, but suffice it to say that N >2 body orbits need not be stable, although ejection of orbiting material is more likely than capture by the central body.

Re:Unstable? (5, Interesting)

Anonymous Coward | about 2 years ago | (#41650147)

Comment posting limits (and time...) won't let me respond to many individual comments, so I will see if I can address a few things at the same time here.

For a given angular momentum of something going around a black hole, you can work out what potential energy it would have at different radii. In a normal Newtonian case, you can think of having some satellite orbiting at some speed. If you try to push that satellite further in, while still maintaining its angular speed, it will try to pop back out since it is essentially going too fast to orbit at a smaller radius. There is a minimum in the potential energy of the satellite where it would have a circular orbit for that given angular momentum, as it would just stay at that radius. The potential energy about this radius would be like a bowl, if you push the satellite inward, it would roll back down toward the radius corresponding to a circular orbit. Momentum would of course carry it beyond that point, so it would oscillate in radius between some place closer and some place further from the circular orbit. This would give you an elliptical orbit where the radius goes between two values. The potential energy for over radius for a given angular momentum would look roughly like the red curve in the image here [oregonstate.edu] .

Now, for a black hole, GR gives some differences from Newtonian gravity when you get closer. The potential energy curve now looks more like this [amazonaws.com] . There is still a stable orbit, as you can see it could oscillate around the minimum there like a marble in a bowl. In other words, small pushes on a perfectly circular orbit will turn it into a slightly elliptical orbit that is still pretty close to the circular one. However, if you push it far enough inward to get over that bump, the orbital radius would be like a marble just rolling down that hill toward the black hole. Now, the size of that bump changes depending on what angular momentum you are talking about. As you increase the angular momentum, which in Newtonian gravity would just give you a smaller radius for a circular orbit, that bump gets smaller. There is a point where the bump goes away, such that you just now have a curve that decreases with decreasing radius. Hence, a particle in such an orbit would continue to move closer to the black hole, as there is lower potential energy the closer it gets.

This is all due to the geometry of space around a black hole. Weird stuff like the circumference of a circle not being 2 pi r depending on how you measure the r from the black hole, which is why orbits no longer have the same stability they have in Newtonian gravity. This is not an effect due to gravitational waves. The orbiting particle can be something like a proton where the gravitational waves would be too small to matter. However, if you are talking about the orbit of a massive object, like a star or second black hole, then the gravitational waves become significant. In that case, the orbit at any radius would slowly decay due to emitting gravitational waves. Once the decay orbit hits the radius of the innermost stable orbit, the decay would greatly accelerate.

This is also not an effect of rotation or frame dragging, as it happens with a non-spinning black hole solution too. However, spinning black holes and frame dragging do factor into it, such that for a spinning black hole, the inner most stable orbit is smaller if you are going in the right direction around the black hole. Although there are other effects that the frame dragging causes. You get things like the ergosphere, a region where due to frame dragging, you would have to go faster than light to look stationary from an outside viewer, so all matter within that region is spinning around the black hole.

This is also quite distinct from the event horizon. This concerns a particle in free fall with a bound orbit. You can fling something at the black hole in a non-bound orbit and have it get closer and still sling shot away (although there is a limit to that too). Additionally, if you had a powered rocket (with unlimited power...), you could get all the away to just outside the event horizon then leave through powered flight.

Re:Unstable? (0)

Anonymous Coward | about 2 years ago | (#41650227)

As you increase the angular momentum...

That should be decrease, not increase... should make coffee before posting.

Re:Unstable? (1)

jandar (304267) | about 2 years ago | (#41652289)

Even without coffee this was a really insightful explanation. I would have to go back to university to study GR to grasp the influence of this geometry of space but the graphs provoke a dim understanding of the unstableness (is this a word?) of an orbit. thanks.

Re:Unstable? (1)

Anonymous Coward | about 2 years ago | (#41652501)

Another subject, which I forgot to mention, is to look into Bertrand's theorem [wikipedia.org] , It basically says that in order for there to be stable, closed orbits (e.g. stable circular orbits) when you have a source of a central force, that force can only be either a inverse square force, or a linear force (e.g. Hooke's law for springs). So if for any reason something like an inverse square law force varied, you would lose stable circular orbits under some conditions. There are some subtleties and differences between what that theorem is saying and what some of the orbit stability stuff in GR is saying. But it gives a good hint that as GR effects become stronger near a black hole or other compact object, any deviation from the inverse square force would cost you circular orbit stability. The theorem is at least understandable without GR... and possibly appreciable without much math too.

samzenpus (0)

Anonymous Coward | about 2 years ago | (#41649175)

Better do some editing and fix the summary before we start to think you're as bad as timothy.

I thought they were both the same. (3, Insightful)

lazy_nihilist (1220868) | about 2 years ago | (#41649177)

What is the difference between Event Horizon and Innermost Stable Orbit?

Re:I thought they were both the same. (2, Informative)

Anonymous Coward | about 2 years ago | (#41649261)

The Event Horizon concerns massless particles (e.g. light) , the Stable Orbit massive.

Re:I thought they were both the same. (0)

Anonymous Coward | about 2 years ago | (#41649605)

Nonsense. Light can fall into a stable orbit too! And light, because it's moving, has mass too.

Maybe you meant matter, when talking about the stable orbit.

Re:I thought they were both the same. (5, Informative)

Anonymous Coward | about 2 years ago | (#41649761)

Nonsense. Light can fall into a stable orbit too! And light, because it's moving, has mass too.

Maybe you meant matter, when talking about the stable orbit.

Ugh. No. Photons have no mass. They have momentum. Relativistic mass isn't actually mass, and in fact, physicists have been trying to get rid of the term, because of the confusion it causes.

Point of no return = distance below which no stable orbit can exist. If you have thrust, you can actually get out of the "point of no return", it's further away than the event horizon. You just can't have an unpowered orbit that won't eventually decay into the event horizon.

Re:I thought they were both the same. (0)

Anonymous Coward | about 2 years ago | (#41649955)

Imagine a 1 kg beachball made out of highly reflective material. Now add 1 kg*c^2 joules of photons bouncing inside it. Will the gravitational attraction of the object you created be that of 1 kg or 2 kg?

Re:I thought they were both the same. (1)

Anonymous Coward | about 2 years ago | (#41651281)

Imagine a 1 kg beachball made out of highly reflective material. Now add 1 kg*c^2 joules of photons bouncing inside it. Will the gravitational attraction of the object you created be that of 1 kg or 2 kg?

"Mom! My beachball just exploded and vaporized the dog!"

Re:I thought they were both the same. (1)

As_I_Please (471684) | about 2 years ago | (#41653457)

2 kg. However, it is more useful to think of this as the sum of the mass of the beachball and the energy of the photons because both of these can be measured in the rest frame of the beach ball. In the rest frame of a photon, the photon doesn't exist. This is another way of saying that photons do not have a rest frame and that they do not have mass. It is clearer to equate rest mass with mass to keep it separate from measurements different observers will disagree on. All observers will agree on the rest mass of a particle: m^2 = E^2 - p^2 (in natural units).

The concept of relativistic mass would be useful if it allowed us to keep using Newtonian equations, but it doesn't. Sure, relativistic momentum is p = gamma*mv = m_rel*v, but kinetic energy is not K = (1/2)*gamma*mv^2. It's (gamma-1)mc^2.

Re:I thought they were both the same. (0)

Anonymous Coward | about 2 years ago | (#41653359)

If photons have no mass then why do they become affected by large enough gravity wells?

Re:I thought they were both the same. (2)

As_I_Please (471684) | about 2 years ago | (#41653471)

Gravity doesn't act on mass directly. A black hole (or any other large mass) warps the space around it. Light travels in a straight line through curved space (a geodesic if you want to be technical about it). It's similar to how, if you walk in a straight line on Earth, you actually walk in a circle due to the Earth being round.

Re:I thought they were both the same. (5, Informative)

madcarrots (308916) | about 2 years ago | (#41649263)

as i understand it, the Event Horizon is the singularity limit from which light cannot escape. the Innermost Stable Orbit is the closest distance a physical object in space can orbit the black hole without being sucked into it.

Re:I thought they were both the same. (0)

Anonymous Coward | about 2 years ago | (#41649681)

as i understand it, the Event Horizon is the singularity limit from which light cannot escape. the Innermost Stable Orbit is the closest distance a physical object in space can orbit the black hole without being sucked into it.

Replace "physical object" by "object with non-zero mass" and you're right. A photon is a physical object, but with zero mass.

Re:I thought they were both the same. (1, Insightful)

hazah (807503) | about 2 years ago | (#41649279)

Objects in an Innermost stable orbit are still visible. Objects at or beyond the event horizon are undetectable.

Re:I thought they were both the same. (0)

Anonymous Coward | about 2 years ago | (#41649521)

And what is the visiblility of objects *between* the event horizon and the innermost stable orbit?

Re:I thought they were both the same. (1)

mbone (558574) | about 2 years ago | (#41649865)

Orbits outside the ISCO are slowly inspiraling into the black hole, under the effects of gravitational radiation (and in the real world also other things, like tidal deformations and accretion disk drag.)

Inside the ISCO, the orbit becomes unstable to perturbations and there is a "plunge" into the black hole. However, these objects are visible to the outside as long as they haven't crossed the event horizon. The thing is, that won't take long, and so they won't be visible for long.

For more details, see this paper : http://adsabs.harvard.edu/abs/1993PhRvD..47.3281K [harvard.edu]

Re:I thought they were both the same. (1)

hazah (807503) | about 2 years ago | (#41650509)

Actually, time dilation would take its toll and the object will appear to be slowly fading away until undetectable, however, you can never actually observe an object crossing an event horizon. The photons will take longer and longer to reach you, but never cease.

Re:I thought they were both the same. (0)

Anonymous Coward | about 2 years ago | (#41649571)

Wouldn't the PNR depend on the momentum of the satellite in question? --i.e. the more massive and slower an object is, the further the away the PNR would be, and the less massive and faster an object is, the closer to the singularity the PNR would be? How can one speculate this as a fixed vector with so many unspecified variables?

Re:I thought they were both the same. (0)

Anonymous Coward | about 2 years ago | (#41651251)

This is specific to circular or near circular orbits, in which case you would be expected to have a certain momentum related to the radius of the orbit for it to be nearly circular. The nature of accretion disks means that it is pretty applicable to them too. If you had a whole bunch of momentum, you could easily go within the last stable circular orbit distance and be sling shotted away. Although there is a minimum closest approach for doing that too that is outside the event horizon. The only way to arbitrarily approach the event horizon and get away would be with powered flight.

Re:I thought they were both the same. (0)

Anonymous Coward | about 2 years ago | (#41649395)

From my rusty memory, Event Horizon obviously being the point that even light going straight out cannot leave the black hole.

Before getting so close, there is a distance from the black hole when light going sideways cannot leave, i.e. spacetime is so curved that the "straight line" for light to go has been curved into a full circle. At that point, you may say that the orbital velocity had become the speed of light, so anything slower, such as anything having mass, can no longer stay in orbit.

Having said that, I suspect the Innermost Stable Orbit would be a little larger than the circular light path.

Re:I thought they were both the same. (1)

mbone (558574) | about 2 years ago | (#41649711)

A lot. If you had a super-duper rocket, and were orbiting at the innermost stable orbit, you could escape the black hole. If you went through the event horizon, you could not.

Re:I thought they were both the same. (0)

Anonymous Coward | about 2 years ago | (#41649759)

The event horizon and the innermost stable orbit have a band of space between them. What happens if you go there?

Re:I thought they were both the same. (4, Interesting)

mbone (558574) | about 2 years ago | (#41649999)

The event horizon and the innermost stable orbit have a band of space between them. What happens if you go there?

First, these regions near a black hole tend to be very nasty for our kind of life. Lot's of radiation, and the tidal stresses will kill you for a solar mass black hole. So, suppose you have a multi-billion solar mass black hole to play with, lots of shielding, and a super rocket as well. The ISCO orbit will be about 2 days in that case.

Could you make a regular orbit inside the ISCO? Yes, in principle, down to the Innermost _Unstable_ circular orbit, AKA the "photon orbit," as this is (at 1.5 Schwarzchild radii) where photons would orbit. It's unstable, so you will need to maneuver frequently to not fall into the black hole.

Below the IUCO, you have to fire your rockets constantly to avoid being sucked in. Better not run out of fuel !

A movie is worth a lot of words, so here are some movies of orbiting a back hole ISCO [colorado.edu] .

Re:I thought they were both the same. (0)

Anonymous Coward | about 2 years ago | (#41650181)

So your test "can you escape the black hole with a super-duper rocket?" distinguishes between both sides of the "photon orbit", and has nothing to do with either the event horizon or the innermost stable orbit as you originally claimed.

Re:I thought they were both the same. (2, Funny)

Anonymous Coward | about 2 years ago | (#41651305)

here are some movies of orbiting a back hole.

I've been on the Internet long enough to know not to click THAT link.

Re:I thought they were both the same. (1)

Maow (620678) | about 2 years ago | (#41651493)

I'd like someone to explain to me why my pet theory is "not even wrong".

The theory goes something like this: as something falls below the Event Horizon, the gravity slows time down until it stops, meaning the thing never reaches the centre, so the singularity is never actually reached. Hence black holes have varying sizes & masses and aren't simply a point with an EH around it.

After all, if the singularity were true, wouldn't all black holes be merely a singularity with an EH?

Can you briefly explain how wrong I am & why?

Thanks!

Re:I thought they were both the same. (0)

Anonymous Coward | about 2 years ago | (#41652257)

Frame or reference. What you are describing is from the observer's point of reference: you see the thing fall slower and slower and get dimmer and dimmer.

From the point of view of the thing falling into the black hole, it's all over pretty quick.

Re:I thought they were both the same. (0)

Anonymous Coward | about 2 years ago | (#41652581)

Can you briefly explain how wrong I am & why?

Thanks!

Somewhat wrong.
Because the Eureka Maru will be along to pull you out before you hit the EH.

Re:I thought they were both the same. (0)

Anonymous Coward | about 2 years ago | (#41652617)

With GR you can calculate the time it would take to fall to the center from outside the black hole, as measured by someone falling, and you always get a finite (and quite short) time. From the falling observer's point of view, nothing weird is going on until they run into tidal force problems (which could be well outside or well inside the event horizon depending on the size of the black hole). There are some graphical ways to demonstrate this, but I'm not sure how informative they are anyways, since they usually involve a bunch of coordinate transformations to make nice compact diagrams.

Outside observers will see time slow down and it looks like the person took forever to cross the event horizon. Although at some point they wouldn't see anything as the falling matter would be red shifted too far. But that is due to the large difference in position between an outside observer and one falling into the black hole, and what happens to the light coming from the falling matter. To the falling matter, everything near by looks the same as it is all falling together (so light from one's feet would still get to their head as they fell). To the falling matter, the singularity is just a point in space quickly approaching, no extra time dilation that would make that point suddenly slow down its approach..

Re:I thought they were both the same. (0)

Anonymous Coward | about 2 years ago | (#41653273)

To add to this question: What would happen if there was a blackhole with an EH with a radius of one light second? Gravity alone would attempt to accelerate an object faster than the speed of light and would have enough distance to actually make it to the speed of light.

It would take an object at least one second to reach the gravitational center and with the gravitational acceleration of c per second^2, I wonder what would happen. Not to mention that gravitational acceleration increases by the square of the reduction in distance. By the time the object was 1/2 way to the center, it would be experiencing 4c per second^2 acceleration.

Lets assume we have a very large BH with something like an EH radius of 1 light minute. Even if an object doesn't quite reach c, even though experiencing gravitational acceleration that great surpasses c, the mass of the object would increase like mad. Because the mass of the object has become something much much much much greater than it once was, the total mass of the BH plus the object is now *increasing*. Can this happen? Would this violate the conservation of energy some how?

Re:I thought they were both the same. (0)

Anonymous Coward | about 2 years ago | (#41654381)

With velocities near the speed of light, you need to use relativity, not Newtonian dynamics. Hence, if you start talking about accelerations that are going to make velocities near the speed of light, you need to treat acceleration via relativity. In that case, you won't exceed the speed of light, you will just approach it. This isn't specific to black holes, and you can look at stuff like proper acceleration [wikipedia.org] for discussion of cases like something under constant acceleration, but not ever reaching the speed of light (an observer on the space ship feels constant acceleration, but an outside observer sees its change in velocity slow down as it approaches c).

And yes, the mass increases when something accelerates to high speeds falling into a black hole. This isn't specific to black holes either, as it happens to any object falling, gaining velocity and kinetic energy. This is just converting potential energy to kinetic energy, same as happening on the Earth's surface when you drop something.

Re:I thought they were both the same. (1)

rts008 (812749) | about 2 years ago | (#41651873)

The difference is, you get flushed down the drain,instead of just circling the bowl.

But at least with a black hole, you don't have to dodge the floaters, and never need a plunger or plumber! ;-)

An obvious point, but... (0)

DoofusOfDeath (636671) | about 2 years ago | (#41649221)

Black holes are neat.

Re:An obvious point, but... (2)

hazah (807503) | about 2 years ago | (#41649287)

And have hair...

Re:An obvious point, but... (1)

marcosdumay (620877) | about 2 years ago | (#41649347)

It this a known fact, or speculation?

Re:An obvious point, but... (1)

hazah (807503) | about 2 years ago | (#41650597)

I'm not sure what you're asking... there are no "known facts" about blackholes, as it's in the relm of theoretical physics. String theory suggests that hawkings radiation is produced by bits of string pinching themselves off the event horizon (this description is horrible, but it's not my domain). Susskind had argued Hawkins for a good 30 years, and this is one of the outcomes of this argument, along with the holographic principle.

Re:An obvious point, but... (0)

Anonymous Coward | about 2 years ago | (#41651239)

Black holes might as well be on par with some of the particles seen in high energy particle physics. In either case we don't observe them directly, but we have a theories that describe their impact on other things around them. We have observational evidence of compact massive objects, and in some cases can exclude neutron stars. At the moment, black holes are the only theory we have that fits all of the evidence quite well. There has been some work to try to come up with other compact massive objects to replace black holes, or even to represent intermediate steps, but none of them has explained the observations we have so far.

Also, Hawking radiation doesn't require string theory. It is explainable with in the well established and tested quantum field theory framework.

Re:An obvious point, but... (1)

marcosdumay (620877) | about 2 years ago | (#41651989)

We know that they exist, and now it seems that we know that they rotate. (Ok, "know" may be too strong a word, we have empiric confirmation of both. With some certainty that is smaller than 1, of corse.) My question was if we have any empiric confirmation that black holes have hair.

I asked it because last time I saw anything about it, there were only speculation either way.

How could this influence the elections? (-1)

Anonymous Coward | about 2 years ago | (#41649267)

...because if this piece of news has no connection to the presidential race, well I'm not interested and I think it shouldn't be reported at all.

LIBERA TUTEMET EX INFERNIS (0)

Anonymous Coward | about 2 years ago | (#41649333)

Where it is peering, it won't need no eyes.

"Event Horizon" (1)

SternisheFan (2529412) | about 2 years ago | (#41649341)

FTA; “Once objects fall through the event horizon, they’re lost forever,” says lead author Shep Doeleman, assistant director at the MIT Haystack Observatory and research associate at the Harvard-Smithsonian Center for Astrophysics (CfA). “It’s an exit door from our universe. You walk through that door, you’re not coming back.”

*** Yeah, I almost got married once, too. ***

Re:"Event Horizon" (2)

nomad-9 (1423689) | about 2 years ago | (#41649469)

FTA; “Once objects fall through the event horizon, they’re lost forever,”.. You walk through that door, you’re not coming back.”

*** Yeah, I almost got married once, too. ***

That would actually be a wormhole. You can come back again "on the other side", in another universe, i.e. less the house and plus the alimony payments...if you were the husband of course.

Black Hole's... (1)

noobermin (1950642) | about 2 years ago | (#41649387)

A black hole's 'http://news.harvard.edu/gazette/story/2012/10/point-of-no-return-found/point of no return'? I know a little bit about GR but I don't think I've every heard of that physical quantity before. Is this some new "mystery" technology like quantum computers, some sort of Swarzschild Uniform Resource Locators?

Starship fate (2)

epSos-de (2741969) | about 2 years ago | (#41649399)

Imagine you are on a starship and have to pass near a black hole. You read up the facts from the books and set your course. 5.5 times the size of the black hole's event horizon seems rather risky. I would take 3 times the suggested distance to pass safely.

Re:Starship fate (4, Funny)

mooingyak (720677) | about 2 years ago | (#41649451)

Imagine you are on a starship and have to pass near a black hole.
You read up the facts from the books and set your course.

5.5 times the size of the black hole's event horizon seems rather risky.

I would take 3 times the suggested distance to pass safely.

I'll keep that in mind next time I pilot my starship past one.

Re:Starship fate (0)

Anonymous Coward | about 2 years ago | (#41649539)

Now imagine you're in a starship race that passes near a black hole. Does it make sense to brag about how little distance you covered to finish the race?

Re:Starship fate (0)

Anonymous Coward | about 2 years ago | (#41649897)

Now imagine you're in a starship race that passes near a black hole. Does it make sense to brag about how little distance you covered to finish the race?

That depends on the race. For the Kessel runs, yes.

Re:Starship fate (0)

Anonymous Coward | about 2 years ago | (#41652611)

Hey, if I had the Kessel runs, I'd try to find the shortest trajectory to the loo, too!

Re:Starship fate (1)

SternisheFan (2529412) | about 2 years ago | (#41651401)

Now imagine you're in a starship race that passes near a black hole. Does it make sense to brag about how little distance you covered to finish the race?

That kinda' depends. Is one of the pilots vice presidential candidate Paul Ryan? (*ducks*)

Re:Starship fate (1)

Anonymous Coward | about 2 years ago | (#41654393)

Have you noticed how your peers cringe whenever you try to tell a joke? Trust us, you don't need to *duck* because we feel you are more to be pitied than censured.

this useless knowledge brought to you by pepsi (-1)

Anonymous Coward | about 2 years ago | (#41649523)

like what good is this knowledge , it's not like we is trippin round the center of the galaxy or somethin....

Accretion disk spinning the same way? (1)

Antique Geekmeister (740220) | about 2 years ago | (#41649909)

How is it possibly surprising that the accretion disk is spinning the same way as the black hole itself? The same stellar evolutionary forces that created the galaxy and imparted spin to it would impart mostly consistent spins to every star and to the core structure of the black hole, built out of the same intragalactic nebulae. Is there any process that could impart a different spin to the accretion disk than to the core black hole itself?

Re:Accretion disk spinning the same way? (1)

MollyB (162595) | about 2 years ago | (#41652109)

I don't know the mechanism that forms them but black holes that spin opposite to their accretion disks are called retrograde [wikipedia.org] .

inaccurate summary (4, Informative)

bcrowell (177657) | about 2 years ago | (#41650485)

The harvard.edu news article, quoted in the slashdot summary is inaccurate. It says:

For the first time, they have measured the black hole's "point of no return"-- the closest distance that matter can approach before being irretrievably pulled into the black hole.

This reads as a claim that they've resolved the event horizon. That's not true, although there are good prospects for resolving the event horizon of a black hole in the near future.

As is made clear in the rest of the article, and in the abstract [sciencemag.org] of the published paper, what they've really resolved is structure inside the innermost stable circular orbit (ISCO) [lsu.edu] .

In units where G=1 and c=1, the radius of the event horizon is 2M, where M is the mass of the black hole. The radius of the ISCO, for a nonrotating black hole, is 6M, i.e., three times the radius of the event horizon. What they've resolved is structure at 5.5M.

The first author of the paper, Doeleman, seems to post all his papers on arxiv.org, but unfortunately this one doesn't seem to be there yet, and Science has their copy paywalled.

If U gaze long into an abyss,it gazes back upon U! (1)

D4C5CE (578304) | about 2 years ago | (#41650653)

Careful what you look for: ;-)

[I]f you gaze for long into an abyss, the abyss also gazes back upon you.

(Friedrich Nietzsche)

Re:If U gaze long into an abyss,it gazes back upon (3, Funny)

wonkey_monkey (2592601) | about 2 years ago | (#41650967)

In Soviet Russia, abyss gazes upon... no, wait, you gaze upon... ah, screw it.

What does it take to be a Slashdot editor? (2)

wonkey_monkey (2592601) | about 2 years ago | (#41650707)

I know what a hyperlink is and how to use one, so I'm probably over-qualified.

The EHT sees details 2,000 times finer than HST (0)

Anonymous Coward | about 2 years ago | (#41651559)

The EHT is capable of seeing details 2,000 times finer than the Hubble Space Telescope. How much does the EHT cost ?? If ground based telescopes can see so much better that Hubble, does it really make sense to spend Billions on James Webb Telescope ??

Anon Indian Techie

Re:The EHT sees details 2,000 times finer than HST (0)

Anonymous Coward | about 2 years ago | (#41652665)

Stuff like the EHT is done in the radio to millimeter wave spectrum, so it is not something that replaces the Hubble telescope. It only works because we can record the waveform in full detail and compare the phases of the waveform to locations very far away. Interferometry can be done at optical frequencies, although requires close proximity of the telescopes, not across the world. Additionally, resolving detail is not the only thing you need, but you also need light gathering ability. Having two telescopes spread out can increase the resolving ability, but without increasing the area of the scopes, you would have the same light gathering ability.

It also doesn't help that the measurements made by such long baseline telescopes are not like what most people would imagine coming out of a telescopes. They would only be measuring small bits and slices of the Fourier transform of the image, which can limit what the observations can be used for, or make some observations much more difficult.

There are other kinds of telescopes encroaching or surpassing what the Hubble can do. However, space telescopes still win in the bands that are blocked by the atmosphere, and in some cases where the ability to observe for more than half the day increase the amount of stuff it can observe.

Mass and spin you say?! (1)

gtcodave (2581251) | about 2 years ago | (#41652937)

Is it worth mentioning that sprin does not effect gravity? Only mass does. Spin only effects the acceleration of gravity by centrifugal force.

Thought you aught to know.

Re:Mass and spin you say?! (0)

Anonymous Coward | about 2 years ago | (#41652973)

That is true under Newtonian gravity. With general relativity, the movement of mass matters too, leading to situations such as gravitoelectromagnetism, where the movement of mass has effects on mass in a way analogous to movement of charged particles, i.e. currents, have an effect on charged particles (it doesn't mean that gravity makes E&M, just that you can get analogous equations in some situations). A particular example of this is rotational frame dragging, where there is an effect being near a massive spinning object due to the spinning. This effect has been measured around Earth. Near a black hole where GR effects are even stronger, there are much stronger effects due to rotation on the space around the black hole. None of this is due to centrifugal force, and can affect a stationary object that is not spinning next to a spinning object.

Re:Mass and spin you say?! (2)

anubi (640541) | about 2 years ago | (#41653059)

You hit on one of my curiousities. The article mentions the accretion disk rotating in the same direction as the spin of a black hole.

The incoming mass has inertia, which will spin up the black hole as its pulled in.

There appear to be two huge forces at work... gravitational, electric forces, and centrifugal force. One is trying to hold everything together, the other is trying to blow it apart.

Its a pet theory of mine that what we know as the "big bang" is actually the result of a massive black hole, overspun as it consumed the universe before it. A severe case of indigestion, if you will.

My evidence is that the observed universe appears full of rotational inertia, and if everything blew out from a point, where did the rotational inertia come from?

Comments invited!

Re:Mass and spin you say?! (0)

Anonymous Coward | about 2 years ago | (#41654451)

You can find various work discussing the maximum achievable spin of black holes. Early rough work by Kip Thorn gave the value of 99.8% the maximum speed allowed by GR. This happens because the orbits around spinning black holes are different for going with rotation as opposed to against rotation. As it goes faster, the black hole will make it easier to capture mater and light that is going counter to the black hole's rotation, while making it easier to for stuff going with the rotation to escape. This is especially important with the photons, of which there will be a lot when hot matter starts falls in. His limit had some simplifications that were unrealistic, but when you add in more realistic conditions, the maximum speed achieved is even less.

Even if you consider an artificial case where you try to force the black hole to spin up as fast as possible, then you will run into a limit, where you can not throw anything fast enough at it to add to the angular momentum. It is similar to trying to make a maximally charged black hole. The maximum allowed by GR happens to correspond to the limit where you would need to shoot charged particles faster than the speed of light to get them to cross the event horizon.

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