“Thus it seems Einstein was doubly wrong when he said, God does not play dice. Not only does God definitely play dice, but He sometimes confuses us by throwing them where they can’t be seen.” -Stephen Hawking

You've no doubt heard that Stephen Hawking is claiming that the black hole information paradox has now been resolved, with the information encoded on the event horizon and then onto the outgoing radiation via a new mechanism that he'll detail in a paper due out next month, along with collaborators Malcom Perry and Andrew Strominger.

Only, that's not really what's happening here. While he does have a new idea and there is a paper coming out, its contents do not solve the information paradox, but merely provide a hypothesis as to how it may be solved in the future.

Come get all the available details here on this edition of Ask Ethan.

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Thank you for a great and detailed explanation.

And very nicely formed question by Denier. Should give credit when it's deserved :)

That's part of what is annoying. The ability is certainly there. Just the will insists to never apply it in certain hot-button topics.

IOW, not dumb, but DELIBERATELY *being* dumb. Ignorance by intent.

And it really buggers up any useful claim he makes, because it cannot be worked out in advance if he's just making up the dumb for ideology or actually has a valid point.

Rather like the fake "I have a question" that is really "I want to complain that science is entirely wrong here". The prevalence of the fake questions means GENUINE questions are greeted, at best, defensively, at worst with hostility (and justified by the very high frequency of fakers).

It's a disservice to those who really don't know yet to insist you don't know but have done so through deliberate choice.

Since the extreme gravitation at the "event horizon" would compress time and space, we could NEVER SEE anything fall into one. And these relativistic effects would start before an EH formed, so actually, time retardation has never allowed one to form in all the time since the big bang. Things have been piling up near so-called "event horizons" since enough mass collected to theoretically form them. The radial dimension of all infalling objects is compressed, but still exists near the assumed event horizon.

@Gary S #3: I'm going to take a wild guess here, and suggest that you're familiar only with popularizations, and not with the actual tensor mathematics of general relativity, with general coordinate transformations, or with the Schwarzchild metric solution to the Einstein equations.

Why is there a caveat for light in regards to nothing escapes a black hole?

@Ragtag Media #5: What "caveat" are referring to? Are you talking about the diagram which shows how light is bent around a black hole, how there is a minimum distance where light goes into a closed orbit, and how inside that orbit light falls into the black hole past the event horizon?

"Since the extreme gravitation at the “event horizon” would compress time and space"

There is no necessarily extreme gravitation at the *event horizon*. The gravitational law STILL applies and is proportional to the mass and the distance to that mass squared.

So for a supermassive black hole, the gravitational force at the event horizon can be very low indeed.

*To a small extent* you are correct that to an outside observer, the matter falling in slows down as it approaches the event horizon and therefore the "information" is smeared out at that location and this is the source of the "Holographic Universe" idea, but the model isn't as simple as that.

@#6 Mr.Kelsey

Michael, I was more just thinking out loud,

I will have about 20 tabs open with the same topic and start to see a theme. So after multiple readings a theme developing was "nothing escapes a black hole, not even light".

Perhaps it gives an exclamation point to it's impressiveness.

So "None", then, just confused babble.

@Ragtag #5 & #8

The use of light in that statement is because nothing can go faster. Light goes 380,000 miles per second the instant it is created. Think of the space around a black hole like a treadmill. The treadmill is running so fast that everything gets spit off the back and into the black hole. Even ray of light, the crowned speed champion of our universe traveling at 380,000 miles per second, cannot stay on the treadmill. Even light gets spit off the back of the treadmill and into the black hole, and if light can’t stay on the treadmill then nothing can.

@#9 Women, get back in the house before I backhand ya again.

@#10 Denier, I just read Black holes are spinning near the speed of light, is that true?

@Ragtag #11

They don't spin quite that fast. They have a spin speed limit (which might be an interesting @Ethan article if he hasn't already written it) of around 84% the speed of light. The fastest spinning black hole I know of is super massive center of NGC1365, and it is spinning at that .84c limit.

"The compacting process halts, according to Dr. Poplawski, because black holes spin. They spin extremely rapidly, possibly close to the speed of light. And this spin endows the compacted seed with a huge amount of torsion"

Denier, is there anymore study you know of about this torsion effect?

This is huge, all that mass spinning that fast. does it have torsion if it's so compact?

At what point does classical mechanics vs quantum pass the baton?

@Ragtag Media #13: You don't provide a source for your quote, but looking up "Popalawski", it appears that he's a somewhat fringe theorist. You probably want to take his comments with more than a few grains of salt.

Not all black holes have maximal spin; in fact, most probably do not. You may wish to lookup the "Kerr metric" for more detailed information on black hole spin, the difference between the event horizon (for the zero spin Schwarzchild metric) and ergosphere, and the spin upper limit.

You may also enjoy the possibility of so-called "naked singularities" in the case of "super-extremal" black holes, and the relationship of that case to the cosmic censorship hypothesis.

@Ragtag #13

I second everything @Michael Kelsey just wrote.

Classical Mechanics works up to the Event Horizon. Beyond that, people are guessing. Currently there is no consensus Quantum theory to hand the baton off to. The majority opinion among Astrophysicists whose work I've read doubt there is a singularity in the center of a black hole as Classical Mechanics would predict, but don't know what to replace it with. @Sabine Hossenfelder, whose focus is on curved space-time and black holes, wrote that same sentiment in her most recent article (see topic #4).

There are no shortage of people spitballing about the interior structure of a black hole, but until the math shows up the best advise is that of @Michael Kelsey. Take it with a grain of salt.

There is a predicted torsion effect on the space-time around the black hole which has an effect on the inner edge of the accretion disk. It is that effect astronomers have been attempting to measure to determine the spin rates of various black holes. The results indicate the rotational rate of black holes does vary from black hole to black hole.

Ah, we see raggie's weakness displayed. Can only rhetorically beat up women. But wants to.

Meanwhile boasting about how he goes armed.

Why?

Scared little pussy syndrome.

Pitiful.

@#16 I done told you women, watch the mouth. The Only disgusting fat pig Women that I have ever verbally abused are you and Rosie O'Donnell. So STFU and know your place. This thread is for Denier Not you ya disgusting PIG.

@#14, 15 Michael and Denier. Regarding the “Kerr metric”.

What is the relationship with the Kerr vacuum and frame dragging in Laymen terms. There's a lot of math with regards to the Kerr Metric which I don't have time to pour over. Is there an executive summary out there, TIA?

I think what is missing from the article is why it is so important that information must be always preserved.

My thinking on the subject is what is called "Conservation of Energy" (which I think the most basic law of physics) is actually "Conservation of Information".

@Frank #18: It's not conservation of energy, but rather "conservation of probability." The equations of quantum mechanics involve a mathematical object called the "wave function", which describes all of the possible outcomes of whatever interaction is going on.

One fairly intuitive rule is that _something_ has to happen -- that is, if you add up all the different probabilities for the various possible outcomes, they must add up to one. As a quantum system evolves (that is, as it undergoes various interactions), the wavefunction will change, but it's integral (the sum of all the possibilities) must always be one. In technical language, the evolution of the wavefunction is required to be unitary.

What Hawking realized was that this rule -- which is really, really basic, and built into how quantum mechanics (and quantum field theory) can be mathematically consistent -- would be violated if black holes could really evaporate to nothing. At the end, the wavefunction which previously had been describing the black hole, and all the stuff which fell into it, etc., would become zero, violating unitarity.

Does that help clarify things?

@Ragtag Media #17: Frame dragging is a consequence of evaluating GR in a rotating coordinate system. You can sort of think about it by analogy to the Coriolis force in Newtonian mechanics. Of course, in GR the Coriolis force is still there (since GR is Newtonian in the low-energy, low-velocity, low-mass limit :-); frame dragging is an additional effect.

You're not going to find a decent non-mathematical explanation, because the maths are really at the core of GR.

Fuck off you godbothering religiotard and get back to your stone age myth worship.

Just to add to Michael's post in #20, and please keep in mind this is meant to be an analogy, not an exact answer. The math is necessary for a complete answer.

Frame dragging can be thought of as an effect of the "rubber sheet" analogy that is usually given to visualize GR. If you haven't heard of it, recall that according to the GR conception of the universe, spacetime can be thought of as a rubber sheet. Massive objects (and energy) distort the rubber sheet creating curvature of spacetime. Now picture a massive spinning object. It's easy to picture that the spinning object will cause the "rubber sheet" to rotate somewhat as well. That rotation of the "rubber sheet" of spacetime is the analogy for frame dragging.

Although the cause of it is the finite speed of gravity.

@ M K #20

Thanks Michael for your always decent, insightful and thought provoking input.

@ #21 Dirt bag

"get back to your stone age myth worship"

Good Idea, I think I will go and read what the Apollo 8 crew read as they orbited the moon. Genesis verses 1-10

Yes, we get it. You're so terrified of everything that you can only bring yourself to threaten women who you will never have to face and only then while you fantasise about having a gun.

We get it

You're a pussy.

Hey, teabag medium, you got permission from your mum to talk like that to grownups?

haha, wowzer wants to be tea-bagged...Kweer

So you're running private browsing and telling mom you're buying her a present, then.

She probably reckons you're jerking off.

You would know about "private browsing" in the London bathhouses. Seeings as you live or rather sponge off of your relative pensioner's. Does Nana and Papa wonder why your on the computer all day and out all night?

Still lying about the job hunting I guess; just tell them the truth, your a shitty teacher and that is why no school will hire you. you were not even able to hold on to the lollypop crossing guard Job the school gave you because you bitched at the kids for crossing to slow and made them cry.

Such a terrible person you are, perhaps you should repent and change your pathetic life around.

That's right, teabag! Don't let those women bully you around. When you've safely distanced yourself from them so they can't harm you...

LOL!

Hates women. Can't stop thinking about men having sex with men. Is there a Ted Haggard thing going on with you too?

Question: How are baryon and lepton numbers conserved during matter-antimatter collisions?

@Denier #31: This is my specialty (I'm an experimental particle physicist in real life). Baryon and lepton number, along with lepton family number, are simple additive quantum numbers, and that's how they are conserved.

In electron-positron collisions (what I did until 2009), the total baryon number is zero, as is the total lepton number. So anything you get out will either have zero baryons, or a baryon plus an antibaryon (or multiples thereof :-).

If you collide two protons together, that's a total baryon number of two, so whatever comes out will have some mix of baryons and antibaryons, but exactly two less of the latter than the former.

@ Denier #13

sorry.. a little late to the party with this comment.. but here goes... in regard to singularity.

This often gets skipped... so I tough it would be a useful reminder. There are actually multiple singularities when dealing with a black hole, but some are of our own making. The singularity at the center is of twofold nature. One is put in by us in the first place... remember what the Schwartzschild solution is. It's a solution of einstein field equations where we assume an idealized scenario where all the mass is located in the single point, and observer "infinitely" far away... so r=0 is a singularity made by us.. but that's not so much of a problem because frankly.. in newtonian gravity it's the same.. if you set r=0... you get the same thing. The other singularity comes from real physics.. but IMO it's an "if"-y one.. and it's that to our best knowledge, no force can withstand the gravity going to infinity once you compress matter beyond neutrons... this might be true, but just as likely (my own bias) it might be the case that at i.e. 10^-23 scales there are forces that could "push back" and make equilibrium. You would still get the black hole in all it's glory.. it's just not infinity inside but some new form of state of matter.

Other singularity that arises in swartzschild solution is at 1-Rs/R... or when that is equal to 0... or the even horizon.. but this is due to the choosing of coordinates for "very far away" observer.. that part doesn't work like that for a coordinate system which is i.e. moving toward the BH... very near to it... that solution isn't a schwartschild metric solution. So that singularity physicists call "non-physical" singularity since it only arises in certain coordinate systems. But IMO it's a bit misleading, since for those coordinate systems it IS their physical reality.. and that's what we see.. out physical reality is the seeing the event horizon when observing from earth, in other words from very way away.

So when you look at it like that.. there is just one "maybe" problem with BH.. but I'm of firm belief that there are no singularities anywhere in this physical universe. On the other hand.. if at one time (and I don't think it will be in this lifetime) we really realize that there ARE singularities in the physical universe... well, that will be even more fascinating and raising some very concerning questions :)

ups.. sorry.. should be to your post #15 not 13.. my bad

p.p.s

the statements about there might being a new form of states once the neutron cores colapse, are just my own.. it's not physics. It would be a pleasure if Michael, as an expert in that area, has any thoughts on it even being possible.

In other words, I don't know if there is anything in other theories or formula, that support that once you go past neutrons that it's turtles all the way down.. or is it just the case of not having a tech to probe that far but nothing in the theories that forbid there being a state of super compressed matter.. s super-solid if you will.. just not of infinite density.

@#35

I know that physicists think when an equation gives infinity as the answer it is a sign that the equation does not apply at that point and you need to find a new theory/equation.

I think space-time must be made of Planck Scale cells. The question is what is max mass/energy each cell can hold. If that is known someday it would be possible to calculate actual size of the core of a black hole.

@#35

Also if there is really a singularity with infinite density in the center of a black hole, then why not it start sucking in space-time around it in exponential speed (since space-time itself has massive amounts of vacuum energy equivalent to mass) and quickly destroy the Universe?

Since obviously that is not happening it is a good sign that black holes don't have singularity at their centers.

Marshall Barnes, R&D Eng Exposes Hawking's Phony Black Hole Information Paradox Solution

Large portions of what the media calls Stephen Hawking's "new solution" for the black hole information paradox are not only not new, but already included in Marshall Barnes' book on Stephen Hawking, Space Warps and Time Tunnels, published last June.

http://www.prlog.org/12488806-marshall-barnes-rd-eng-exposes-hawkings-p…

@Frank #37: Because even with infinite density, a black hole still has finite (and well-defined) total mass. See, for example, the "Dirac delta function" for how we deal with this mathematically.

The gravitational influence of a black hole is defined by its mass, not its density.

"Marshall Barnes, R&D Eng Exposes Hawking’s Phony Black Hole Information Paradox Solution"

Stephen Hawking overturns the basis of all engineering, proving that materials science is the new alchemy!

Just as believable, isn't it.

@Michael Kelsey #32

I suspected that information conservation during annihilation events had to be reasonably straight forward because no one was screaming about an information paradox. Thank you for lending your knowledge on the subject.

@Sinisa Lazarek #33 & #35

I'm with you on the super dense matter. The force I believe arrests the downward collapse is conservation of information. All matter gets compressed to the maximum information density of space-time, and there it sits.

Maybe it merely makes no more sense to say what is happening in a space smaller than the planck length than it makes sense to ask where the dark came from when you turned out the light.

wow#7

I'd understood that anything crossing the event horizon can never re-emerge. I don't see how that's compatible with (very) low gravitation there.

Well instead of closing your eyes and complaining, do something about it.

Whining won't work.

I don't see how it's incompatible with very low gravitational forces there.

Neither do reputable scientists whose job is to look at this sort of thing and whose reputation relies on accuracy and evidence based claims and whose job relies upon that reputation.

So I don't see how it can be thought of as unsupported or infeasible to be the case either.

Care to prove to me how I am wrong?

@Peter Dugdale #7: Wow used language which confused "gravitational force" (what we normally think of as the "pull" of gravity, holding us to the ground) with "tidal force" (which is the _gradient_ of gravity, or roughly the _difference_ between the force on your head vs. your feet).

Near the event horizon of a black hole, the "gravitational force" -- the net force holding you down -- is extremely strong, strong enough that the escape velocity is close to 'c'.

However, for a very large black hole, the TIDAL force can be very small, small enough that an observer fall in wouldn't notice anything at all (gravitationally) as they crossed the event horizon. Conversely, near a small (supernova-sized) black hole, the tidal forces would start to pull an observer apart long before they got near the event horizon.

Not really, Michael. The discrepancy is the change from the frame of reference well outside the black hole and the frame of reference of the thing doing the falling in.

Something falling in has zero force acting on it, only tidal differences, the difference in gravitational force at the outermost edge compared to the gravitational force at the innermost edge.

As a note: something in orbit at, just outside or inside the black hole event horizon will also be in a high-g area but experience no gravitational force: weightlessness.

Wow - the question was meant in good faith and not intended to be provocative. I apologize if you took it that way.

I'm certainly not capable of proving anything one way or another.

Michael Kelsey : I admire your courtesy and helpfulness to many people here, who like me, may put questions which may sound stupid to those who really know this stuff.

Your reply is really give me something to ponder over and I'm doing, with hope that I may get there! Many thanks!

please excuse sloppy editing in last sentence. ...has really given me...and I'm doing that...

"I’m certainly not capable of proving anything one way or another."

Then what point is there in trying to tell people that you don't see how...blah?

If you don't see how a bhuddist can think all living things are equally important, why is that of any interest at all? What is a bhuddist supposed to do with that?

If you don't know why it can't be so, why should you think it can't be so?

The question isn't provocative: it's entirely pointless and a waste of time. And it is that waste of time, not you, I'm getting pissed off at.

...and thanks to both of you for replying so promptly: I was afraid the thread had gone dormant.

No problem, asking isn't a problem.

Making a pointless claim of personal incredulity? Problem.

What Michael gave you either works now (and from your response to him, it didn't) or is going to be ignored (guess which). Since you are avowedly not knowledgeable enough to work with this (and being uneducated is not a sin either, refusing to think, not so much sinfree), the explanation that it is the tidal difference that rips things apart, not the depth of the gravitational field, and the depth of the gravitational field, not the tidal force, that stops things from escaping from the black hole, you have been given everything for a rhetorical understanding of the fact you "could not see how".

If you have a 10^25 solar mass black hole, the gravitational force at the event horizon is about 1 earth gravity.

@Wow #55: You're right that a freely falling object doesn't "feel a force" (that's why it's freely falling). But in common parlance (which is what any reasonable layperson will be using when they ask a question), the object is falling *because* there is a "force" pulling it down.

We "feel gravity" when we are standing on a floor, even though what some people would say is "really happening" (these are usually the same people who engage in pointless centripetal/centrifugal debates) is that the floor is exerting an upward force.

You even implicitly admitted this yourself when you talked about an object "in orbit around a black hole." You can't have an _orbit_ around anything without a central force inducing that orbit. And once you have that situation, it is a perfectly reasonable question to ask how strong that force is.

Michael, it's also the fact that it isn't the value of g at the event horizon that stops light.

If you collected 10^25 solar masses you'd have about 1 earth g at the event horizon, but it would still be the event horizon, with all the light-stopping characteristics that entails.

His problem is that he thinks it's a wall. That event horizon is a solid one-way trapdoor and it physically blocks.

Talking of "massive gravity pushing you down merely cements that this event horizon is some solid surface.

@Wow #57: Ahhh....I didn't see Peter Dugdale's #44. His follow up question does make it sound like he's envisioning the event horizon as a "physical barrier" of some kind, which of course it is not.

Of course, that is a *very* common misunderstanding among laypeople, probably at least as common as the idea of black holes "sucking things in." Neither one of these misunderstandings means the person is stupid, nor willfully ignorant. It means they've read some popular descriptions of black holes and are trying to fit that reading into their existing understanding of how gravity "works."

Well, the problem is that if you use ACCURATE, you're bamboozling them or, if you have to resort to the maths, an ivory tower un-person. If you tell them that they have to learn how to understand it, you're hiding something behind BS. And if you simplify it, they complain about how it doesn't work in one specific way dependent on how to misapply the simplified message.

And sometimes they even use the simplification to tell you you MUST be wrong because reality isn't that simple and you've left something out deliberately, you scammer!!!!

His complaint was manufactured deliberate doubt.

He may find out a way to work it out for himself and change, so trying definitely isn't wrong.

But he doesn't want to know and is looking for how it doesn't.

Not so he can believe it is wrong consciously, but subconsciously he doesn't want it to be true, because what he knows "fits", so sees no need to "risk" being wrong by changing or being found a laughing stock by his peers who believe the way he does now.

In-group homogeneity fear peer pressure. Which, if recognised, is self-ostracising, so you WANT to fool yourself.

Cliff note version: many people who believe honestly and sincerely they want to learn subconsciously do not want to learn, and will search for ways to continue the same known and well trod path.

@Wow #55 & #57

Just checking but you do mean a 1 earth g tidal differential on the scale of a person, right? You don't actually think that if there were a platform built at the Event Horizon of a 10^25 solar mass black hole that a person could walk abound on that platform as if walking across a room on Earth. Just checking.

"Just checking but you do mean a 1 earth g tidal differential on the scale of a person, right? "

No, an acceleration of approximately 10m/s^2.

One gee.

The Schwartzchild radius goes up with mass linearly, the gravitational acceleration goes down with radius squared. Therefore the acceleration at the event horizon calculated goes down with the mass of the black hole linearly.

Yes, you could walk on a dyson sphere built around the black hole and feel as though you were walking on planet earth.

A planet that had a radius of approximately 3x10^28m, but one earth gravity keeping your socks on the ground.

NOTE for the nitpicky: the actual calculation is slightly different because of the curvature of space even that far away from such a massive object.

But it only changes the figures, not the consequence.

@Wow #61-63 and @Denier #60: Yes, indeed. The calculation for gravity with such a large black hole (unphysically large, see below :-) is very simple, and approximately Newtonian, as Wow wrote in #61.

What's interesting is that even with the low "surface gravity" (so to speak), it's still a black hole, and the "surface" is still an event horizon. Because this object is so huge (3e+28 m is about 3 trillion light years), the "surface gravity" stays nearly constant out to very very large distances. You'd have to get 2 trillion ly away from the event horizon for the local gravity to drop from 1 g to 0.5 g. If you work through the math (just do it Newtonian) you'll see that the escape velocity is 'c', just as required for an event horizon.

Now, a little math to support my "unphysical" contention above. The latest estimate from Hubble is that there are a few 10^11 galaxies in the observable universe. The average (mean) galaxy has a mass of a few 10^11 solar masses (that's stars, dust, etc. plus the dark matter halo). So a rough estimate for the total mass in the observable universe is about 10^23 solar masses. Wow's hypothetical object has two orders of magnitude more mass, and an event horizon about 20 times larger than the current (comoving) size of the observerable universe (radius about 46 billion light years).

Let me see if I can phrase this in a way that doesn't sound like I'm complaining about science.....

Oh, I know.....how about.... WHAAAAAAAtttt?! How can that possibly be right? If you are at one place then escape velocity is achieved with 1G of acceleration, but move 1mm to the left and a million Gs of acceleration won't get you free. How the 'F' does that work?!?

Overnight I'd checked Wikipedia on black holes and found the linear relationship between mass and schwarzschild (event horizon) radius. And yes, using the Newtonian relationship, acceleration due to gravity will decline with the square of the radius.

And I'd also come to the conclusion that Wow's putative hole with 1g at the EH is an implausible beast in this universe.

There are other sites, apparently populated by physicists, where it's not universally accepted that gravitation can be weak at the EH:

http://physics.stackexchange.com/questions/36251/gravity-on-supermassiv…

one of a couple of similar quotes :"As @Alfred said, your notion that gravitation is weaker at the horizon for larger black holes is incorrect"

Michael Kelsey must be a top-flight physicist, but it seems he wasn't completely aware of the situation, either.

So Wow's sneering arrogance is response to a good-faith question is completely unjustified.

...in response..(no edit function!)

@Denier #65: I think you don't have a good picture in your head of Wow's hypothetical gigantic black hole. I also think you may not have the maths background to work out how escape velocity works.

For the latter, it's the speed you would have to start with, moving radially away from a gravitating object, such that you could end up all the way at infinity without falling back. To work it out, you have to use calculus: you start with speed v moving upward; after a short time, gravity has slowed your speed to v-dv, but you're at a height h+dh such that gravity is just a bit lower (goes like 1/r^2). So the acceleration of the next step slows you a bit less, but your speed was a bit less. You can work out the maximum height you reach, where your speed goes to zero; if your speed is at or above escape velocity, then your speed _never_ goes to zero.

With Wow's hypothetical gigantic black hole, as you move upward, the acceleration due to gravity (which is what is slowing you down) basically never changes. It's always ~1 g. Consequently, no matter how fast your speed, that gigantic black hole can always slow you to a stop and have you fall back.

Of course, that's true with any black hole! If you start near the event horizon, then no matter how fast you jump up to move away, you always get slowed to a stop and fall back. With most black holes, the pull at the event horizon is huge. The counterintutive thing here is that with Wow's gigantic hypothetical BH, the actual "pull" of gravity is tiny, no more than here on earth.

@Peter #66: Gravity _can_ be "weak" at the event horizon of extremely large black holes, but that sort of depends on your definition of "weak."

Wow's calculation shows that if you want to try to build a BH with a pull of just 1 g at the event horizon, you need more mass than is available in the (observable) Universe!

For more realistic super massive black holes (say 10^9 solar masses, like the one in M87), the pull at the event horizon is still going to be huge compared to what we're used to, about 1500 g. Compared to the pull at the event horizon from a solar mass black hole (1.5 x 10^12 g), that's pretty small.

The issue of comparing events "outside" vs. "inside" the event horizon is different. You *cannot* simply use Newtonian physics for that. The Schwarzchild metric, and the existence of black holes, is intrinsically a GR issues, and you must use proper physics. The event horizon, besides having the cool "light speed escape velocity", is also the surface where the metric changes its shape qualitatively: the radial direction becomes timelike (future pointing inward) as soon as you cross the event horizon. There is no way for you to use your "intuition" to either understand it, or to reject it (that's a fallacious argument from ignorance). You have to dive into the maths in detail.

Michael #69

thanks again for your patience and careful explanations. I was labouring under the impression that the EH was a sort of brick wall.

I can do the newtonian stuff, but not relativity maths.

To me to say that something doesn't have escape velocity doesn't seem to exclude it moving outwards from the EH, even if it must eventually return.,That's presumably where your last paragraph must be taken to heart: GR geometry looks different.

"Now, a little math to support my “unphysical” contention above."

Oh aye, never said you could actually make one. Heck try moving a star. Really hard to push. Especially when it's so hot you can hardly stand putting your hands on it.

And there's nowhere to put your feet to push against...!

"Oh, I know…..how about…. WHAAAAAAAtttt?! How can that possibly be right?"

How can that possibly be incredible?

The definition of the event horizon is that where you cannot escape to infinity. Moving out to less then infinity then coming back isn't escaping.

The calculation of the metric around the event horizon causes the event horizon to be likewise defined and changes if you change where you define escape to, so back to an observer at less than infinity, the event horizon is closer in (slightly).

The Schwartzchild radius is just the simplest calculation of where, at infinity, there is no escape.

The confusion comes from the mathematics having the calculation done with "to infinity or beyond" written in it and the object falling in hypothesised being an extended object that doesn't extend out to infinity.

Lets see if I can get a word explanation why all roads lead back in to the black hole past the "to infinity" event horizon, even if it would be able to get *beyond* ifininity. By taking a rocket.

A rocket attempting to move away from the black hole will reverse and point directly away and put on full thrust to escape. So it goes "faster" (in the black hole inertial frame), but relativity means it gains weight doing so, which increases the pull back. And because the metric is so curved, the attempt to leave moves it less away from the rocket's doom than the reaction should have. Which means it's not moving out any faster, but is REACTING as going faster (special relativity gain).

So it continues the burn. And it "goes faster" but "goes nowhere" (well less out than it needed). And so it continues. And continues. And eventually the acceleration it can manage runs out and it never got anywhere near an escape path.

Note: it COULD burn to enter into orbit stably. That's definitely not escaping, though. But attempting to leave means it comes back after it exhausts its motive power and freefalls in.

For a non-motive part, the geodesic it follows, once it intersects the event horizon can no longer contain infinity because the metric curves space too much for the object, no matter its energy, to get into an orbit that leaves.

In the same way when the sun (or a large planet) captures an extrasolar body (planet asteroid or whatever), the orbit NO LONGER CONTAINS any path that leads to exit. Not for the same reason, though. Although if the body were close enough to barely captured energy, the depth of the sun's gravitational potential could lead to the same "the distant universe has dropped beyond the visible universe" event.

Maybe a graphical way of looking at it will work.

Draw the radial depth of potential well around a black hole.

An object always (absent tidal effects and therefore across suitably small regions of space) "sees" a flat spacetime.

In that flat spacetime out in outer space the energy may be enough to get to infinity with positive kinetic and/or potential energy. Draw that line flat on the graph prepared earlier. For simplicity's sake lets just give it zero excess.

The falling object feels no force and therefore no change in what it feels its potential line is. Remember this.

When it falls deep in to the potential well (which curves space depending on the depth of that well), that flat space it thinks it is on is now objectively (to the graph which is the view of an omniescent outside observer). So the flat line the object falling in is still flat and at the same height, but tilted.

When that object potential gets to zero, NORMALLY that would be at infinity, but because this flat space is "tilted" by the depth of the field, the intersect of that location is much closer in to the centre of the potential well than before.

It will never get to infinity now. "Infinity" is now the place where its perception of flat space now intersects the zero potential of the gravity well.

Or, in other words, draw the potential well of the black hole, and the zero potential is a tangent to that curve (because space curves depending on the depth of the potential) and its infinity is where that tangent hits the zero potential.

(because potential and space curve aren't equal or even linearly proportional the disappearance of the universe doesn't result from this graphical view: it is merely a conceptual, not a mathematical, model)

@Michael Kelsey #68

I think you are wrong here. Even from a common sense standpoint, light doesn't slow down. It redshifts and keep right on going at c. If a photon is moving away from an Event Horizon, it ain't coming back.

The Event Horizon of a BH is where time-space is curved to the point that, as you say "radial direction becomes timelike (future pointing inward) ". Both you and @Wow seem to be confusing the gravity inherent to curved space with gravitational acceleration on freely falling bodies.

You cannot build a Dyson Sphere 1mm above an Event Horizon and walk around on it. It doesn't matter how ridiculously huge the black hole is. I know Wikipedia isn't the ultimate authority but they get this one right:

From Wikipedia:

After 45 minutes of effort of hunting for ASCII symbol key codes, I am admitting defeat on posting in the mathematical proof. The math is solid but I'm not talented enough to write it in the unformatted comment section. You would need an infinitely large rocket to accelerate away from an Event Horizon just as common sense would dictate, and hovering (or resisting the pull of gravity by standing on a Dyson Sphere platform 1mm above an Event Horizon) would require an unfathomably large amount of energy. It is certainly more than is required to walk across a deck on Earth.

It loses energy as it travels, just like slowing down is the result of particles losing energy. And they both lose energy according to the equation:

E=mgh.

The only difference is that for photons

E=hv

and for particles

E=0.5*mv^2

This is a problem WHY?

Did you try? Did it fail when it should have worked? Is that how you know it can't be done?

Since it can only be right when you say it is right, why do you bother with Wikipedia when your only reliable source in your opinion is your own opinion?

Not because the gravitational force is infinite, though. And escape to where? And what does common sense have to do with extreme (even if it isn't infinite as you earlier claimed) gravity?

Or being in orbit. Which is always enough to "resist" the force of gravity. We do it every day here on earth with our satellites.

It does take a lot of energy to get the buggers up there, though.

Unless the space it can traverse is spherical.

You know, like if you start walking west (and have the ability to walk on water) "you ain't never coming back" is absolutely not true. You'll get further and further away from your starting point, then at some point later, you'll now be going closer and closer to your starting point.

@Wow #76

Denier wrote:

Wow wrote:

Let me start by saying how weird this conversation is as I am always the one railing against Relativity, but you and @Michael Kelsey have confused rest frames. With an object as extreme as the WBH (Wow's Black Hole), the difference is as crazy as walking across a room on Earth versus needing all the energy in the universe combined just to stand up.

I know you're not going to take my word on it, so I am going to hope some other people chime in, but I know you've got to be feeling itchy on this one. In searching around, it seems a lot of people make this exact same error. If you want to research, the key terms to look for are stationary relative to the Event Horizon versus falling through the Event Horizon.

@Wow #77

In that case it wouldn't be moving away from the EH. Besides space is only twisted like that on the singularity side of the EH, even for a WBH.

Blank assertion not answering the question "Why?".

No, you'd have one gee. Something that we humans are quite well adapted to.

Well, there's a good reason for that "Trust me" isn't proof.

They key term you miss is "Orbit".

Look it up.

Yes it would. When you, standing on the spherical earth, walk (OK, your lardarse will probably drive. Or get driven) to the shops, are you unable to leave home because you're standing on a spherical surface????

Worse, you would not be able to get off the couch! If it's impossible to get further away from something if you're on a spherical surface, YOU ARE HELD THERE FOREVER!!!!

OMG!

Your problem denier is that you think black holes ACTIVELY SUCK THINGS IN.

IT.

DOES.

NOT.

A dyson sphere created around the sun at about 3.5 million km would experience 1g pulling it in to the sun when not in orbit. It is what the use of a dyson sphere for a future civilisation is all about: we can live on the surface of that sphere and capture 100% of the sun's output with the same thing we live on.

If we built it around that black hole as described, IT WOULD BE NO DIFFERENT.

It was wrong to say that the difference was "orbit", though.

However, a dyson sphere able to hold up to 1g of force would not have a problem since only 1g acts on it from the super duper black hole.

Someone standing on the outer surface would find it no harder to jump up than they would on planet earth since they experience 1g just like on earth.

Someone climbing a set of stairs a half mile high would be no more tired than if they did the same feat here on earth.

A lift going up a half mile high skyscraper would meet no different engineering problems going up and down, carrying passengers and or freight than it would on earth.

Because 1g is 1g. Even if it's right next to an event horizon of a really really big black hole.

@Wow #82

A Dyson Sphere around a star and a Dyson Sphere around a BH are different situations. A Dyson Sphere around a star needs the outward pressure of the solar wind to maintain stable position. If a side drifts in, it experiences an increase in outward solar wind pressure that serves to push it back out into position. Decades ago there were astrophysicists doing proofs that Larry Niven's Ringworld was not stable due to its inability to use the outward solar pressure adequately to keep it in position.

With a BH, the situation is even worse. Not only is there no outward solar pressure, but if you had a Dyson Sphere built 1mm above an Event Horizon and one side drifted in just a millimeter or so, there isn't enough energy in the universe to put it back into position. The BH would rip the shell apart and gobble the whole thing up.

That being said, at a constant distance 1mm above the EH of that size of black hole, the force required to keep mass there is asymptotically close to infinite.

"A Dyson Sphere around a star and a Dyson Sphere around a BH are different situations."

No they aren't.

"but if you had a Dyson Sphere built 1mm above an Event Horizon and one side drifted in just a millimeter or so,"

And if your dyson sphere fell into the photosphere of the sun by just 1mm, you'd have molten dyson sphere with no structural rigidity and it would all fall in and be burned up.

Here;s an alternative: the dyson sphere DOES NOT fall inside the event horizon.

What now?

"That being said, at a constant distance 1mm above the EH of that size of black hole, the force required to keep mass there is asymptotically close to infinite."

Nope. Totally fabricated fiction.

The force is BY DEFINITION 1gee force.

We build bridges every week that manage to handle that.

@Wow, @Denier. I think, from some of my own research, is that the problem here among all three of us is that there are multiple, valid definitions for the same terms, and we are using _different_ definitions for them which results in confusion and apparent disagreement.

First, the problem is that "local acceleration" due to curved spacetime is not well-defined in general relativity. By definition, it's the time derivative of local velocity. In the low-curvature limit (i.e., near "ordinary objects"), you can define a locally flat Minkowski frame, use that to define a local time coordinate, and from that take a derivative of the potential (curvature) to get a local acceleration ("g force"). The problem is that near an event horizon (EH), the spacetime curvature, and in particular the mixing between spacelike and timelike coordinates, is large enough that that procedure is just wrong. It fails, and at the "coordinate singularity" of the EH (where r becomes strictly timelike) you get a divergence which is commonly interpreted as "infinite acceleration." This exercise is exactly what you do when evaluating the _true_ situation for the case of someone trying to "hover" with a rocket near the EH. You will discover that no matter how much local thrust is used, once inside the innermost stable circular orbit (ISCO), the thrust needed to hover diverges, and the rocket falls into the BH no matter what.

But there is a different, but still valid, method which is to apply Gauss's law to the mass of the black hole, and extract an effective "Newtownian" acceleration. This value is the local "pull" an observer at rest relative to the BH (Schwarzchild) frame would experience. More concretely, it's the local value of "g" which the accelerometer inside your iPhone would actually measure if it could be at rest in the BH frame (let's say, resting on the Dyson Sphere Wow built around his gigantic BH).

Both of these are valid calculations in GR, and give results which properly transform between frames. But they have wildly different meanings! And those differences lead inevitably to different, and contradictory, interpretations.

But there is one important different difference between the two, and this also comes to Wow #82's description. For the second calculation, the work is done in the Schwarzchild rest frame. If you do that inside the ISCO, and at or "just outside" EH, it is not physically possible for an observer to actually be in that rest frame. This is what the first calculation tells us. No amount of thrust (whether from a rocket, or from a built platform, or from an arch like a Dyson Sphere) can provide enough force to maintain an observer at rest. So it's a mathematically valid place to do computations, but it is not physically realizable. In particular, Wow's comments about jumping and clibming stairs in that frame aren't correct.

This also relates back to the old canard about "black holes sucking you in," as Wow properly rejected in his #81. Far away from the EH, in particular, outside the ISCO, black holes absolutely do NOT "suck you in." They are just ordinary "point masses" like in first year physics, and you can compute orbits, hyperbolic trajectories, even "gravity assists" if the BH is orbiting something else. But in the special, small region between the ISCO and the EH, there the BH really does "suck you in," in the sense that no amount of radial thrust can produce an outward trajectory in that curved spacetime.

Michael, I had to do some quick cramming here, so it's a little hasty.

The escape velocity of the event horizon using the Schwartschild metric varies with radius from a black hole, and the event horizon is still defined as "Where the escape velocity to reach infinity with zero remaining kinetic energy is c".

The change that happens AT this event horizon preclude falling in because the light cone becomes so narrow it never extends inside the event horizon therefore there is no non-tachyon path to enter it.

Because under this model, there is infinite time to ARRIVE at the event horizon via coordinate time (as opposed to "proper" time of the object falling in") in a Schwartschild metric.

Would this lead, as it appears to, a translightspeed shift in the frame of the infalling system (iow imaginary mass). This would swap time and space under what I can remember of SR rigourously applied.

And BECAUSE of this shift, the origin point shift of time comparisons are the cause of all these contrary discourses?

If so, WHICH coordinate system or GR model ought I look into to avoid the dichotomy of an external system NEVER falling inside the black hole in finite time, therefore precluding INSIDE the event horizon and making a mockery of any attempt to describe it being in there?

Oh, by the way, I DO believe you're incorrect on the ICSO wrt a dyson sphere because it isn't in orbit, but maybe I am misunderstanding what you mean by "escape", since all you need to do is carry away enough energy to take you over the peak of the gravitational potential metric of an orbital dropping in. ICSO only leads inevitably inward for a particle in an accretion disk which can lose energy itself and fall closer in, but the outward motion is limited by a large positive potential wall, hence a one-way seive trap. A deliberate attempt to leave would accrue and store energy until it had enough to pass over the potential well that an object falling in would have to have in excess of zero net potential to pile straight in to the event horizon.

(apologies if the language was tortured: this really needs a graph to point to)

In this specific case, being outside the event horizon in the Schwartschild metric, the dyson sphere will take infinite time to fall in the remaining meters/miles/whatever. Whether inevitable or not, it doesn't matter, since you can get a lot done in subjective infinite time...

Until that apocalypse, however, you will find yourself able to gambol and skip your way around the 100-universe-girding dyson sphere and feel no more problems in leaping or climbing than you would on earth.

No infinite gravity. Not infinite energy requirement. Though you may be glad that everything, especially light, is all safely on the other side of that sphere, because the energy gain as it falls down to your level would be punishing for anything, even very soft radio waves.

Time dilation between your brethren waiting for you in the hotel lobby and you in your 500th floor penthouse may cause some social difficulties, however.

@Michael Kelsey #88

Now you and I are on the same page.

@Wow #90

Wow wrote:

The effect you are referring to is gravitational time dilation. It only works that way because gravity is infinite there. If it were only 1G that wouldn't be the case. Falling 1mm would be like falling 1mm on Earth. Also, information storage and holography on the EH wouldn't work. Particle production via Hawking Radiation wouldn't work. All of those things are foundationally dependent on the extreme curvature of time-space.

@Wow #89: Thank you very much! Sorry these posts are getting so long (yeah, mine always are...). I'm going to try quoting yours and responding in pieces, just to reduce the number of posts. Hopefully the HTML markup works for me...

Correct! The Schwarzchild radius is the point (in r-t space, ignoring the other coordinates by symmetry) where the coordiate singularity happens. That's equivalent to the r and t axes becoming "squished together", which is equivalent to having all trajectories become null trajectories, which is equivalent to light-speed escape velocity.

Yes. To a distant observer, infalling objects would never cross the event horizon. But as you say, in the proper time of the object itself, reaching and crossing the event horizon, and even running into the "singularity" (or whatever) at the center all happens in finite time.

No. The "swapping" is an issue of matching boundary conditions at the event horizon. Inside the event horizon, the spatial coordinates all point toward the singularity; outside the event horizon, they point outward away from the singularity, so at the EH, the matching condition is that all four coordinates point "along" the EH.

I don't believe there is such a solution. The closest you might get is to look at the case of a Kerr metric with supraextremal spin, in which case you get a naked singularity. But this is where you're really going to have to dive into the maths in detail; such a system is presumed (cosmic censorship hypothesis) to be non-physical.

No, you're thinking Newtonianly here, which is not correct. Inside the ISCO, the curvature is such that the "gravitational potential" is "infinite" -- you cannot ever supply sufficient energy to get back out. The thought experiment of using rocket thrust to hover is just the simplest to analyze. It applies to _any_ kind of force being used to keep from falling in.

In particular, as you say, the Dyson sphere is not in orbit, it is a stable structure at rest around the central mass. (On a side note, that means that, just like the Ringworld, it is unstable to small perturbations in position; not a good engineering choice!) The structural dynamics of a Dyson sphere are equivalent to a geodesic dome with no endpoints (periodic boundary condition), just as the Ringworld is equivalent to a catenary arch with no endpoints. Consider a small area element of the sphere, dM. That element is being pulled downward by gravity, and that downward pull is opposed by the radial component of compressive forces from the neighboring volume elements around it (picture the keystone of an arch, which is held up by the contact with the two stones touching it on each side). If you integrate over the whole sphere, there is a net force trying to crush it inward, which is opposed by the structural strength of the material.

For a Dyson sphere built around a regular star or planet, that's no problem: the downward force is g dM (where 'g' is the local acceleration), and you just need to find a strong enough, incompressible material to take the load. But inside the ISCO, that's impossible: the downward pull is such that _no_ finite upward force (whether rocket thrust or mechanical loading) is sufficient to keep the shell element dM in place.

Ah! I think this is just what you're missing. The ISCO is not an artifact of accretion disks. It exists even for a black hole in total isolation, and is strictly a feature of the curved space time. If you want to think about it semiclassically, the ISCO is the place where the orbital velocity (the speed you have to have in order to stay in orbit against gravity) is 'c'. Inside the ISCO, an object would have to travel faster than 'c' in order to stay in orbit at all, which is impossible.

"No. The “swapping” is an issue of matching boundary conditions at the event horizon. "

I think you don't comprehend.

The swapping is the problem of what happens to a particle as it passes the speed of light.

if v>c, relativistic mass is now complex and imaginary.

There's no physical swap here in this frame of reference because to get there requires transinfinite time.

Or the metric and model is never applicable and only an approximation that does not hold at the event horizon at all, and is just being extended beyond the domain it can be applied to.

"The ISCO is not an artifact of accretion disks. It exists even for a black hole in total isolation"

Not according to a solution of the Schartschild metric. Unless I'm missing something that makes it, but the treatment of a directly infalling particle gives a potential well somewhat like that of an atomic nucleus and its bound electron.

There's a positive potential resulting that ensures that an object with less than this positive energy will be unable to enter into the interior (captured) part of the system, but be scattered and progress on a hyperbolic exit orbit.

If a particle has MORE than this energy, it can pass in and continue straight into the event horizon.

A particle caught in orbit cannot get closer with positive net potential energy and MUST lose it in an accretion disk: orbital mechanics ensure it cannot enter an orbit that is captured inside the peak. It is stuck in a precessing orbit.

When at that stage, it has to lose energy to be able to fall any further in. And it has to fall further in because it still is never near the event horizon.

Once it has lowered energy enough to drop into an orbit closer in than the ICSO, it can never come out because it has to pass that positive potential again to manage to get an energetic orbit that can track outside that peak and thereby get into the precessing orbit again.

But it still can't get in it still has to lose energy.

Nothing in the model and metric indicates it will lose energy against nothing other than the warp of space. Unless it's charged or gravity waves are powerful enough to be worth including. Neither of which pertain to a stationary dyson sphere (or girdle ring) since it is electrically neutral and not moving, therefore cannot emit gravity waves.

THAT is where I believe you're wrong.

The dyson sphere is not orbiting and holds up against 1g pressure squeezing this monstrously big sphere (which is a materials problem, not theoretically insurmountable physical law breaking) by the tensile strength of the material being compressed by this force into a smaller sphere, meaning compression forces will counter (again, if strong enough, which is a materials problem, postulate adamantium if you like).

If not, then where is the bit that indicates that mere existence in this region drains energy.

"the ISCO is the place where the orbital velocity (the speed you have to have in order to stay in orbit against gravity) is ‘c’. "

Short version of the above: a dyson sphere isn't in orbit.

"On a side note, that means that, just like the Ringworld, it is unstable to small perturbations in position; "

Much more stable than a ringworld, since it has the same structure to inhibit lateral distortion, since in spherical coordinates, there is no lateral. Unlike a ring.

It's easier to twist a plastic hoop than scrunch a plastic ball.

Anyway, the point of the dyson sphere is so that you can be standing right above the event horizon and not be in orbit and have something to push against so you can experience the gravitational attraction of 1g and relate it to how it feels on earth which has a solid crust that stops you falling through to the center of the earth, during which time, absent fluid friction, you are in freefall and feel no earth gravity whatsoever.

It's to illustrate that it isn't the strength of gravity at the event horizon that makes light trapped, it's the depth of the gravity well (or alternatively the orientation of the metric of space).

I believe one problem I had with conception was the illustration of a trumpet for the space around a black hole. It had a non-vertical gradient at that point. It ought to be vertical at that point. Our view of flat space would be orthogonal to distant rest frame of an entity sitting just on the event horizon. Any movement they make there moves nowhere from our view of flat space, in the same way as moving in the y plane makes no difference to the coordinate in the x plane.

Sans dyson sphere:

You can hover over the planet earth if you can generate 1g of thrust and direct that straight downward.

You can do the same trick at the event horizon of that huge black hole, hover over, but not in orbit, around that black hole's event horizon, with an engine able to develop 1 g of acceleration for as long as your engine has power to maintain it.

Fascinating, and the search for solution in the degeneracy of the vacuum bears comparison to some of my own thoughts here:

http://moflomojo.blogspot.com/2015/09/how-real-is-real-part-1.html

Well, I'm not going to click the link. There's nothing in your post that explains what I'm going to find. It could be a site that installs malware, or a creationist website.

@Wow #99 (I was gone last weekend, and only just saw this): You wrote, "You can do the same trick at the event horizon of that huge black hole, hover over, but not in orbit, around that black hole’s event horizon, with an engine able to develop 1 g of acceleration for as long as your engine has power to maintain it."

No, you can't. That's the point I was making. Once you are inside the innermost stable circular orbit (ISCO), then the Newtonian approximation you're making is not valid. The curvature, even for your gigantic black hole, is too much. If you try to hover above the EH, in won't work. You'll continue to "slide" closer and closer to the EH, even as you crank up more and more thrust.

To prove this, you really need to go through the maths, including all of the post-Newtonian terms which are relevant.

"No, you can’t. That’s the point I was making. Once you are inside the innermost stable circular orbit (ISCO), then the Newtonian approximation you’re making is not valid. "

Yeah, right. You've claimed this, but you knew that the dyson sphere IS NOT IN ORBIT. Neither is the rocket.

Orbital velocity=0.

0<<c

"The curvature, even for your gigantic black hole, is too much"

Yeah, right. Too much for what? Dyson sphere, remember. Hovercraft.

Not moving radially. Curvature radially therefore not a problem. Since this is an ideally spherical BH, all other spherical coordinate components are invariant.

"The curvature is too high" has no more content than "The dilithium crystals are tired!".

You may be *meaning* that the 10^28 solar mass isn't enough to generate a 1g value at the event horizon because the calculation is newtonian mass times 1/r^2.

In which case the answer is "Then use a different mass until the correct equation gets 1g there".

You may be saying that "our view out here, where we can see things newtonianly" doesn't apply that close to the event horizon.

However, the answer to that is that we're not sitting out here, we're standing on a dyson sphere right in close. If our physics gives us a different answer then you've just helped denier and others prove that there's a way to tell the difference between gravitational acceleration and translational acceleration, AND that there really are special absolute frames of reference (around each black hole). You've just proved you CAN tell when you're about to pass an event horizon. Physics is nonuniversal, even outside a singularity.