First, I apologize for my bad English.

Second, I am an amateur whose knowledge means bits from fundamentals - so to speak.

 

The black hole has an "infinite" gravitational atraction - within its boundaries which is the event horizon.

Is there an explanation why there is a strong - although a finite one - gravitational pull outside the black hole ? - the one which "feeds" the balck holes with star matter. How a "abnormal" "body" - like the gravitational singularity within the black hole - generates an infinite gravitational field and a finite one ?

Thank you in advance.

 

You can have an infinite (closed) gravitational warpage of spacetime without having an infinite gravitational "field" ...

The theory is that when a black hole forms by the collapse of a massive star, there is enough mass within a small volume that the "edge" (the inner boundary of the event horizon) is curved so steeply that even light cannot escape from inside.

This essentially closes off the inner portion of the black hole from its surroundings.

The collapse of the star sheds at least some mass from the original star. Therefore, the overall gravity field in the region of the star is a bit less than it was originally. So, if you had other bodies in orbit around the star originally, they should still be in orbit around the black hole if they were sufficiently far away to escape blast effects or the event horizon.

Black holes can eat their cake, and have it, too ... pardon the mangled metaphor!

Thank you for answering me. 

I have a limited access to internet, so my answers might come late.

But I need additional clarifications; also, I want to be on the safe side, so I might ask questions about details which otherwise might appear obvious. So, sorry for being too insistent.

 

"You can have an infinite(closed) gravitational warpage of spacetime without having an infinite gravitational "field"".

I assume this statement is general information and not connected with my question ; interesting , though.

If this is the case, then:

which is this case "without" infinite gravity ?   this warpage is the same like in black holes' case ? it is still considered a black hole ? no amount of mass "causes" it ?

 

 About the 4th paragragh:

The collaps of the star happens simultaneously with the supernova of that star - what does not collapses explodes. The remains of this explosion are "thrown" away a huge distance from their original location; so, where this gravitational field comes from? from these remains ? or I err here somehow ? 

 

iulianm:

"You can have an infinite(closed) gravitational warpage of spacetime without having an infinite gravitational "field"".

...

This is related to your question only because you included the idea of "infinite gravity". Infinite gravity would consume the universe, if it existed. Infinite gravity requires infinite mass.

However, a black hole can curve spacetime sufficiently to trap light inside it. If nothing travels faster than light, then it would appear that nothing can escape the gravity of a black hole. But this is not the case. Gravity, itself, is felt outside the event horizon. Matter that was just outside the Schwartzschild Radius isn't pulled into the black hole unless it was moving in the direction of the black hole initially.

So, I was merely making the distinction between "infinite gravity" and "infinitely curved space" (the region inside the black hole). From a practical standpoint, it appears that spacetime is curved infinitely (it is closed off to our view) but we have no idea whether this is actually the case (since we can't see inside).

If this is the case, then:

which is this case "without" infinite gravity ?   this warpage is the same like in black holes' case ? it is still considered a black hole ? no amount of mass "causes" it ?

Yes, it's the case of a black hole. The original star's mass is at least slightly reduced. It does not grow at the time the black hole is created, only later as it encounters mass that becomes trapped inside it.

 About the 4th paragragh:

The collaps of the star happens simultaneously with the supernova of that star - what does not collapses explodes. The remains of this explosion are "thrown" away a huge distance from their original location; so, where this gravitational field comes from? from these remains ? or I err here somehow ? 

The correct term to use here is "implode". The star collapses so rapidly that it leaves some of its outer envelope behind. Only the core of the star actually collapses. During the collapse -- particularly if the collapse is not symmetrical -- some blast effects might throw out matter. As the black hole forms and begins to spin, it may cause matter near the event horizon to swirl rapidly in magnetic fields created by the collapsing matter's spin. This may form polar jets that push some matter away faster -- and ultimately, further.

The gravitational field of the black hole forms primarily from the mass of the core of the original star. The amount of mass in the star determines the gravitational force measured just outside the event horizon. This gravitational force is a bit less than before the collapse.

The gravitational field is curved. Outside the event horizon it is curved to the same extent it would have been for the original star, or perhaps a bit less -- depending on how much mass the star sheds during collapse. At the event horizon, the curvature is determined by the mass remaining inside the black hole. The black hole has a very much smaller radius than the star originally had -- therefore, the curvature of the field is very much larger.

Think of it this way:

Put a basketball in the center of a rubber sheet and measure the radius of the stretched rubber fabric. Also measure the "depth" of the dent made in the rubber.

Now, put a billiard ball of the same mass as the basketball in place of the basketball. If you measure the radius of the stretched rubber fabric, it will be smaller. If you measure the depth of the dent made in the rubber it will be larger.

Finally, put a marble made of iron having the same mass as the basketball in place of the basketball. When you measure again, you find the radius reduced further still, and the depth is increased further still.

Extrapolate this to the point where a "ball" the size of an atom -- with the mass of the basketball -- is on the rubber sheet -- you would be hard-pressed to measure the radius of the stretched fabric (because it would be very tiny), but the depth would be very great.

The mass has remained the same throughout. The radius has diminished greatly. This means the density has increased dramatically and, with it, the amount of mass in a smaller volume. The rubber sheet represents spacetime, which as you can see is greatly changed by the reduction in radius of the balls representing the stars. Still, the mass hasn't increased, nor has the "gravity" -- only the gravitational field (the curvature of spacetime).

Does that help?

Wow ... I have read it briefly; it's a consistent answer. I need time to read it properly.

Thank you.

Sorry, I was "tricked" by the Server maintenance message during weekend; when, Sunday, I thought I try,  I did login, but it was late: my internet time had expired.

I spent time analyzing your rubber example. 

I assume "At the event horizon" - from your last post -  means "outside event horizon", doesn't it ?

I meant "infinite gravity" in nonstrict meaning, as replacing the no escape effect specific to a black hole. For me infinite gravity describes the Einstein model, "infinite" being important. So, I understood the case you wrote as being a second case of infinite curvature, without mass this time. But I was wrong: You meant only the case of a black hole with mass, didn't you ?

"The collapse of a star sheds at least some mass from the original star. "  :I assume the gravity outside the black hole is generated from "inside" the black hole, as you wrote many times after this statement.

 Just one more preliminary question: in rubber example, what is the meaning of rubber dent - it's a third element outside the streched rubber itself and the ball ( the mass of the core) - ?

 

Now the main question: why there is gravity outside the black hole - and finite one ? Even your example does not explain that - appearence of a black hole put aside.

I have a course of electrodynamics; I did not read it properly, but the section about Schwartz. solution appears to calculate the curvature generated by a certain mass body. And their results are infinite only - when the radius goes to zero, of course. There is no additional finite curvature there.

thank you.

 

 

iulianm:

I assume "At the event horizon" - from your last post -  means "outside event horizon", doesn't it ?

Yes, everything we can see or measure lies outside the event horizon. You could theoretically get extremely close to the actual horizon, but "at" would include the horizon -- beyond which we can't sense.

... You meant only the case of a black hole with mass, didn't you ?

Yes.

"The collapse of a star sheds at least some mass from the original star. "  :I assume the gravity outside the black hole is generated from "inside" the black hole, as you wrote many times after this statement.

Yes. Gravity is a curvature (in extreme cases, a warpage or complete closing off) of spacetime. So, the black hole doesn't "trap" gravity in the same sense that it traps light.

 Just one more preliminary question: in rubber example, what is the meaning of rubber dent - it's a third element outside the streched rubber itself and the ball ( the mass of the core) - ?

The rubber sheet analogy isn't really sufficient. When you see illustrations of this example, you're not looking at a simultaneous stretching in all dimensions, so it's incomplete. But it serves to illustrate the point. The "dent" is the funnel-like hole with the mass at the bottom of it.

Now the main question: why there is gravity outside the black hole - and finite one ? Even your example does not explain that - appearence of a black hole put aside.

For the same reason there was gravity from the star which formed the black hole: because there is mass inside the black hole. When the star collapses, it does so very rapidly but not in zero time. Therefore, as the collapse ensues the fabric of spacetime is continuously stretched. Since the mass of the star doesn't disappear (only its light disappears) the gravity is still there and can still be felt outside the event horizon.

Going back to the stretched rubber sheet: imagine that the stretching proceeds to the point where the black hole forms and the event horizon traps light. At that point, spacetime has reached a curvature due to the star's mass (now the black hole's mass) and the "fabric" of spacetime retains that shape unless the black hole gains or loses mass. It is as if the tip of the dent in the fabric is "pinched off" from all electromagnetic radiation, but not from gravity. This is one clue that gravity is the curvature of spacetime.

Sorry, you were right, I missed the meaning of the "dent".

So, the collapsing mass of the core has two gravitational-like effects: it warps in an "infinite" manner a region of spacetime - the black hole which traps light - ( and in addition to) maintains outside a finite "gravitational field" or a (finite) spacetime curvature.

The reason why I insist with these question(s) is that your answer appears to me to be in contradiction with Einstein's model; and I think you are right. I asked about the outside "gravity" because I saw recent articles about recent discoveries about black holes "sucking" matter from nearby stars. And this, in turn, implies a gravitational field - a spacetime curvature - outside the black hole, which atracts that matter to the black hole region - the one which you describe.  His basic assumption about mass bending light was proven, but he made and other assumptions; if whatever little I understood from the book I mentioned in my last post is right, then the mathematics he uses models only the region which will become a black hole, and does not refer to this outside curvature; this is neither calculated nor predicted by his model.   Am I right  ? 

  - I use "gravitational field" and "spacetime curvature" interchangeably

The problem with models like the example I used is that they're not easy to envision in three dimensions.

The curvature of the rubber sheet is the same in all three dimensions. For example, if you took a cross-section of the model in one dimension it would resemble a circle (a conic section), but if you took the cross-section in either of the other two dimensions it would resemble a twinned hyperbolic arc, coming to a point.

It is only the point which is "closed off". It isn't so much closed off as a sphere as it is closed off as an infinitely steep curve. The problem is trying to envision that same curved slope in all dimensions (not to mention trying to imagine it in time).

So, I don't think it's a contradiction as long as you leave aside the notion that the black hole itself is spherical. It's just an infinitely steep curve (no matter which direction you look at it from).

The reason I sometimes answer with an apology first, and with the "real" answer later, is because I saw I get answer almost immediately, so I feel I need to at least give a sign I saw it, if not answer properly. So, again my apologize for a possible incomplete answer - I might make mistakes when I see an answer at first glance. My access to net is real scarce and my lack of theoretical knowledge is delaying my answers.

I try not to assume a "usual" shape for the black hole; I am aware - at statement level at least - we don't know if the meanings of our words describe the black hole characteristics. This is why I used - like yourself - quotation marks and (). I "know" the fabric of spacetime is warped in all the directions; because all the spacetime is influenced by the collapsing mass. My observation was about Einstein's model - or whatever I get from it - that it describes only the formation of the black hole, but not this additional "detail": the curvature just outside black hole or "at the horizon level". It reffers only to one region - the future black hole - by two assumptions: mass bends space(light) (1) and the degree of this bending for every value of the radius(r) of the above-mentioned  mass distribution (2). The Schwartztschild radius - associated with a solution of E's relations - models only the first region and not the second. And the collapse is modeled with the radius r going to 0, so it predicts the black hole region, and not an additional - and "unchanged" - second region. That is my first impression, at least.

- if there is a mistake in Einstein's theory, then this is it; if I am over my capabilities - which very well coud be the case - , then my mistake. As I said, this is an answer at a first reading/glance, without to much analysis.

I forgot to ask you in my last post: the metaphor you used in your first answer referes to the existence of both the black hole and the outside "gravitational field" - as in "the star dissapears, but not it's gravitational field" ?

I don't think there is a contradiction if you treat the curvature of spacetime as a continuum ... that is, it simply curves, curves more, curves yet more, etc., until it's finally closed off at the singularity. We have no way of knowing the shape of spacetime inside the event horizon, but we know that the radius of the event horizon is not zero -- while that of the singularity is zero. So, there must be some shape to spacetime inside the blackhole, assuming it still exists.

One current theory gaining acceptance is that gravity is the shape of spacetime. That's also consistent with Einstein's ideas. If this theory is correct, then there is no contradiction between the singularity closing off everything inside it from outside view and the continued existence of gravity from the collapsed star.

The star disappears from view, but we don't know what happens to its mass. If the mass remains centered on the black hole, then we'd expect no large change to its effect outside the event horizon ... so, in other words, outside the event horizon the shape of spacetime should be very much the same as it was before the collapse -- except right at the event horizon. Remember, the collapse is near-instantaneous ... as near as makes no difference. From the standpoint of objects orbiting the original star, unless they were orbiting originally within the region now enclosed by the event horizon, they would see no difference: almost all the original mass of the star is still there -- it is just much more dense (distributed in a much smaller radius).

Let me draw a conclusion with my own words - and tell me your opinion - .

The physical phenomena goes like this -a qualitative view  - :

The core implodes / collapses. While imploding - in a short period of time - it generates two effects: it continuously warps the spacetime region  - until reaches the "state" of gravitational singularity -which "remains" a black hole and, simultaneously, maintains a finite spacetime curvature outside the (future) event horizon. And this goes with the first assumption of E: mass influences spacetime. This description, also, would be correspondent with your statement: "That's also consistent with E's idea." 

If one wants an additional quantitative explanation, this request can be applied to the "evolution" of geometry of the (future)black hole region, with another - additional - assumptions about the degree of the influence of core's mass over spacetime. This is because, the "empirical" data shows there is gravity outside b.h. and it's "measure" is known. The current accepted additional assumptions are the ones proposed by E., which lead to E's equations. So, Einstein's equations  ( and the particular solution Schwartzschild) are related only with "the inside" of the b.h. - so to speak.

 

The first part looks okay to me.

The second part is a bit non-specific and a little confusing. Which of Einstein's many equations are you talking about (the ones that "are related only with the inside of the bh") ??

The curvature of spacetime at a black hole seems to be continuous. That is, it is a continuous curve up to the event horizon (beyond that we can only conjecture). But there is no data that suggests the mass inside the event horizon is not responsible for the gravitational effects outside the event horizon. We are observing this indirectly, of course. At this point, all we can do is estimate the mass of the black hole from the gravitational effects outside. We have not yet seen the "construction" of a black hole from a pre-existing star ... so we have no empirical data about the initial conditions (starting mass).

The equations of the gravitational field, the General Relativity equations.

"The inside of a black hole" part meant the Schwartzschild solutions - and others - of E's equations are used to predict the appearence of a black hole: they describe the evolution of the spacetime curvature in a certain case. This evolution - radius of a body decreasing towards zero - leads to appearence of a black hole. Was I right ? Maybe my previous statement was not good.

So, the second part says the quantitative information related to the region in which the black hole is formed - the one which is described by an infinite warpage - are obtained thru mathematics of GR.( or, the uses of these equations are "limited" to the b.h. itself,  the outside field is learned by empirical data) 

(my neclarity was so far about the existence of gravity outside a black hole).

 

Gravity is spacetime curvature. How this model can be applied to explain star formation ? Also, the core implosion : how does fit the "curvature" explanation  with this phenomenon ?

iulianm:

Gravity is spacetime curvature. How this model can be applied to explain star formation ?

As clumping occurs in a molecular gas and dust cloud, tiny bits accrete into larger bits, and so on. As the clumps accrete, they reach a certain threshold of mass which has warped spacetime to the point that nearby matter 'falls' down the slope of the curvature quickly enough that the pressure buiilds to the point where stellar ignition occurs. By this time, the density of the proto-star's core has become so great that it effectively "contains" the products of fusion and the star lights up in a sustainable way.

That is, as the star "grows" its gravitational field curvature keeps the reaction products in the core.

Also, the core implosion : how does fit the "curvature" explanation  with this phenomenon ?

The implosion of a stellar core has a time constraint: unless it happens quickly enough (again, this is a function of stellar mass), there will be sufficient energy in the effect called neutron rebound to explode the star rather than implode it completely. Within a certain range of mass, the implosion can cause an explosion (supernova), an implosion which creates a neutron star, or a complete implosion (which proceeds rapidly enough to "close off" spacetime's curvature to form a black hole).

So, there is a continuum of stellar masses and what happens at core collapse is determine by where on the range of masses a particular star lies. It is only the very rapid (almost instantaneous) collapse of a sufficiently massive star that creates the black hole (complete implosion).

Another way to look at this is that the stellar mass needed to create a neutron star is in between the extremes. If you implode a stellar core that is just sufficiently massive to create a stable neutron mass, the original star's mass was between that needed to completely collapse and cause a black hole, and that which would merely "bounce off" the newly created neutron mass and explode as a supernova.

It's what we call a "slippery slope" ... and it's due to the amount of curvature which -- in turn -- is due to the original mass.

Here's another way to think about it: Let's let the surface tension of water stand in for the effect of gravity as the warpage of spacetime.

This photo helps to illustrate the concept:

The bright part at the top would be "normal" spacetime and the funnel-shaped part would be the "gravity well" that represents curved spacetime due to the collapse of a massive star. If the star is only massive enough to create a neutron star, then the gravity well would form a point like this one and then stop. In other words, it would be a funnel with a pointed tip (no spherical droplet at the bottom).

But, as in this case, if the mass is great enough, and the collapse rapid enough, a singularity would occur, trapping the remaining mass of the original star inside it ... closed off from the rest of the universe. The black hole in this analogy is represented by the spherical "droplet" perched at the bottom of the column in the photo: you'd have the black hole itself closed off from view (inside the sphere), the event horizon would be a donut-shaped region of space between the sphere and the "pinched off" tip of the funnel, and normal spacetime would be curved ever more strongly from well outside the event horizon all the way down to the border of the event horizon (where it would suddenly "close off" just inside the event horizon, at the crossover point).

Does that help?

I'll give it a try.

I will need some time to "struggle" with the last paragraph - my English needs improvement.

But, until then a (possibly superfluous) question: how does the "curvature" explains the implosion itself ? At the end of the "normal life" of a star, the core consists from gas of iron nuclei - how big is the radius of that "sphere" ? 10 km ? 5 km ?. So, immediatly the fusion stops, how the spacetime curvature - generated by the core,  the iron nuclei gas - acts on the core itself ? The "attraction forces" model seems straightforward : the iron particles attracts each other - without opposing forces - to the centre of the core; how the curvature works in this case ? I ask in relation to the formation of a black hole.

iulianm:

... how does the "curvature" explains the implosion itself ?

The curvature is there all the time, during the life of the star. Before the star forms, the curvature is like lots of tiny dimples and ripples caused by all the bits of matter that are coming together to form the star. Once the star forms, its mass and radius determine the amount of curvature in the star's gravitational field.

As the star begins its contraction leading to implosion at the end of its life (as a star), the curvature gets steeper and steeper as the mass shrinks in radius (imagine the star sliding down the "funnel" and dragging spacetime with it).

It is the implosion itself (the shrinking of the star to a very tiny radius compared to its former size) that steepens the curve to the point where it closes itself off from spacetime (outside the event horizon).

 

At the end of the "normal life" of a star, the core consists from gas of iron nuclei - how big is the radius of that "sphere" ? 10 km ? 5 km ?.

This depends entirely on the mass of the star. As long as the star is "normal" matter (not degenerate), then a certain mass corresponds to a certain minimum radius -- that is, normal matter can be compressed only so far. Once the star implodes, however, the immense pressure of that implosion crushes normal matter far beyond this limit and it becomes degenerate (little free space remaining inside the atoms).

As an example, the degenerate matter of white dwarf stars (our Sun is destined to become one) is very much more compressed than the original star's matter. Our Sun is about 109 times the diameter of the Earth. When it becomes a white dwarf, it will shrink to a diameter about the size of the Earth. If it were massive enough to become a neutron star (even more degenerate), it would be about the size of Manhattan Island.

So, immediatly the fusion stops, how the spacetime curvature - generated by the core,  the iron nuclei gas - acts on the core itself ?

The core is compressed so rapidly it literally implodes. It degenerates into a new form of matter where there is very little free space remaining in the atoms.

 

I don't know if I read properly your last post - sorry if I did not -, but I am "anxious" to see the a answer to this question:

Is there a curvature of the spacetime of the "interior" of the star ? I know the curvature is there all the time; but I can picture only the curvature outside the star, and not the space between the particles which forms the star. If you say "mass and radius of the star determines the amount of curvature " this looks like a description of the external gravitational field only.

So, in this context, I am not able to see how the curvature of the space inside the sphere of the gas is generated by the all the particles from that sphere of gas - at the end of the star's "normal" life. This is why I mentioned the model of "attraction forces" being "straightforward". I wished an explanation of the implosion itself being generated by the existence of a "curvature": the outside grav.field does not explain the implosion, so only the curvature of the space between the iron nuclei can explain it. And this is why I mentioned the radius: 5-10km is a significant space, so there must be a significant curvature "within" this space. But is that gas an equivalent of a "body" ? There is mass, true, but is this a "body", analog like a planet ? If it is a body, then why there is a curvature "inside" it ? And if it is not a body, why there is a curvature at all, and how it is generated, to act over the particles as attracting them to the center: during implosion ?

Do I make sense ? - I wrote it in a hurry.