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Chemist@

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Chemist@

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stargazer3

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Drakkith

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snorkack

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But what happens to the topology of the ring singularities inside the common event horizon during the merger? Can a ring singularity be, for example, snapped into a singularity with loose ends? Or what happens, geometrically, on a triple junction of two loops of singularities after the two rings have touched?

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Chemist@

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Thanks for the answer.

"But what happens to the topology of the ring singularities inside the common event horizon during the merger? Can a ring singularity be, for example, snapped into a singularity with loose ends? Or what happens, geometrically, on a triple junction of two loops of singularities after the two rings have touched? "

Nicely formulated. This would be the second question.

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Nabeshin

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But what happens to the topology of the ring singularities inside the common event horizon during the merger? Can a ring singularity be, for example, snapped into a singularity with loose ends? Or what happens, geometrically, on a triple junction of two loops of singularities after the two rings have touched?

This is a good question, but one I suspect no one is able to answer at the moment. Numerical simulations like the one listed above use a variety of techniques, but they all get rid of the singularities. One way to do this is to simply not simulate anything inside the event horizon, since the details are irrelevant to the goal of the simulation.

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anorlunda

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Does antimatter-matter annihilation destroy the mass of both? If yes the collision product should cease being a black hole and remnants would stream out,

But wait! Would the radiation produced by annihilation gravitate as much as the mass would? Can we have a radiation only black hole?

But wait wait! Because time is frozen at the event horizons maybe the mater and antimatter will never come in contact in finite time. Then it would be no different than two matter black holes merging.

It makes my head hurt to think of such things.

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... Because time is frozen at the event horizons maybe the mater and antimatter will never come in contact in finite time

Whoever told you that misinformed you. That just how it looks to an outside observer, NOT to things falling in, which don't even know the EH is there.

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anorlunda

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But wait wait! Because time is frozen at the event horizons maybe to an outside observer the mater and antimatter will never come in contact in finite time. No annihilation would occur in finite time as seen by an external observer. Then it would be no different than two matter black holes merging.

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Mordred

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Chronos

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- #12

Andrea Panza

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What I mean is that the bigger BH inglobates the smaller one, but there should remain a tiny (getting ever smaller) space of regular space between the two horizon surfaces.

Lets say that we have two BH with horizons that have radii of 1 meter each.

If the two BH are distant their horizons are spherical, because in every point of the space surrounding the BH the gravitational effect of the second BH are very small.

Now think of drawing a straight line between the two singularities as the two BHs get closer. Don't stop the line at the singularities but extend it so that exits from the other side of each horizons (the two BHs approach each other in a spiral motion, and the direction of the line changes, but as external observers we can always draw a line joining the two singularities and extending it past them).

In any given moment of time particles that are on the external portion of the line (the portion that is not between the black holes) feel the gravitational force of the nearest black hole AND the other BH, this mean that they need an higher escape speed and the final effect is that the event horizon of the nearest BH further extends in space in the direction opposite to the one by which the second black hole is approaching.

Particles in the internal portion of the line feel the opposite effect: because the gravity from the more distant black hole is partially contributing to pull them away from the nearest one: their escape velocity is reduced, so an external observer should see the horizon receding.

A particle that is exactly at the midpoint between the black holes (since the two have the same size) should always be able to escape with a trajectory perpendicular to the line, no matter which is the distance between the two BHs.

The singularities will always approach each other, but I think they will never really merge, the horizons between the BHs will recede and get flatter almost to perfect circular sections and there should always be some regular space between them.

If the two BH have sufficiently different masses the smaller BH will be swallowed by the bigger one, but still they will not merge: there will always be a small region of 'regular' space surrounding the event horizon of the smaller BH.

The small black hole will eventually evaporate into the big one, but this is an extremely long process.

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Drakkith

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Remind me which point of view the BH is?

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CSSlemaker

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1. For a massive star, if one traverses a straight path from the surface toward the center, the strength of the gravitational field steadily increases, reaches a maximum value, near the boundary of the core or perhaps a short distance within the core, then rapidly drops to zero at the center of mass.

The locus of points at which this maximum occurs will form a spherical shell, centered at the center of mass of the star.

2. When this star undergoes gravitational collapse, the general shape of the field-strength curve remains the same. Although the core collapse proceeds very rapidly and, for a while, leaves the upper layers behind, and this gap may introduce a glitch in the shape of the curve, there will still be a radius at which the field strength is maximal followed by a rapid decline to zero.

3. As the collapse proceeds, the density of the core material increases, the radius of the field-max sphere decreases, and, therefore, the maximum field strength increases. When this max field strength reaches the value which corresponds to an escape velocity of C, the field-max sphere becomes the event horizon and a black hole is born. [*note below]

4. In subsequent paragraphs,

5. My conjecture is that the radius of the event horizon and mass of the black hole are frozen and never, ever change following initial appearance of the event horizon. My understanding is that velocity of any subsequent infalling matter, as seen by the Observer, rapidly declines as it closely approaches the horizon (due to gravitational time dilation) and, in fact, becomes zero at the horizon, never actually entering the black hole.

6. Therefore, although an accretion disk can form and can grow in mass, the Observer can never detect, in principle, actual ingestion of any matter from that disk. In particular, this implies that two black holes can never actually coalesce.

7. Finally, unless a different mechanism for black-hole creation is proposed, all "giant black holes" must actually consist of a dense cluster of separate, but tightly bound, holes.

8. My conjecture, however, seems to contradict what I read in many places about "dynamical" black holes which are undergoing, or have recently undergone, ingestion of infalling material or merger with another hole.

My Questions:

1. Is my description of the collapse and the invariance of the initial event horizon correct, and if not, where is my error?

2. If strength of the gravitational field is not the cause of the event horizon (per the note below), then what is?

3. What is the final result, as seen by the Observer, of an "encounter" (I won't say "merger") of (a) a black hole and a 10-Solar-mass star, and (b) of two black holes?

*Note: In a 1994 book by Kip Thorne, I have read about a distinction between the "apparent event horizon" (the one I describe above) and an "absolute event horizon". The latter is said to appear initially as a point at the center of the collapsing star (in "anticipation" of its subsequent phase as a black hole) and then expand as the collapse continues until it reaches the stage that I describe above. Assuming that my description of the field-strength curve is correct (zero at the star's center), what then is the condition that precipitates initial formation of the absolute horizon at a point of zero gravity?

I hope that someone can supply explanations which do not require a PhD in math to comprehend.

Thanks.

- #15

jimgraber

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The loophole in your reasoning understandings 5-8 or so is your assumption that the horizon stays fixed. in fact, it "jumps out" and gobbles up the infalling material. So you don't ever see the infalling material make it through the old event horizon, but it makes it through the new expanded event horizon very quickly.

The 'Point" (pun intended) of your note about the black hole starting at a point of zero gravity, is simply a matter of a mathematical definition. The black hole (event horizon) is defined to be the trapped surface, ie the set of points from which a light ray or null geodesic never makes it to future infinity. This set starts with the point right in the center of the black hole and it depends on what happens in the future, and not on the zero force at the point itself at that instant.

This is a little bit sloppy, but hopefully you get the idea.

Best,

Jim Graber

- #16

CSSlemaker

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With respect to the absolute horizon's beginning as a "point", let's alter that a slight amount and consider two epsilons: a tiny elapsed time after the "point beginning" and a consequent tiny increase in the horizon radius. We now are no longer dealing with mathematical abstractions (a dimensionless point, and the instant of time zero), but the problem remains. If the epsilons are small enough, the region will still have very nearly zero net gravity. It would seem that the only escape from this problem is if the horizon depended, not upon the strength of the gravitational field, but somehow directly upon the density of the material at the center.

Best regards,

Carroll Slemaker

- #17

chill_factor

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let me make an analogy with electromagnetics. I say that magnetic fields are relative and some observers see a magnetic field, and others don't. Ridiculous? No - if you have a point charge and you are stationary with respect to it, it has a uniform radial electrical field and no magnetic field. However, if you change to a moving frame, the charge is actually a CURRENT which creates a magnetic field and simultaneously makes the electrical field nonuniform.

A rough analogy can be made for spacetime; what time actually is, depends on where you are and how fast you are moving.

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CSSlemaker

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Regards,

Carroll Slemaker

- #19

chill_factor

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Regards,

Carroll Slemaker

Basically it means that things that seem to be invariant in all frames when viewed in everyday life may not actually invariant in all frame.

It is possible to station an observer at a position at respect to the black hole if they were very far away; this observer will *see things slow down as they fall towards the black hole*. Of course, the falling things will also *exponentially redshift and dim away* which makes them visually "look" like they are falling in. Since light is actually discrete, the last photon that leaves will reach your eyes in a finite and quite short time, and then there's no more images; that's why you can "see" things fall through black holes. This is because the coordinate called "t", which corresponds to proper time in your far-away frame, does not necessarily correspond to proper time in another frame.

If YOU were the one falling in, for a black hole with a large mass, you won't even *know* when you fell through the event horizon, because your proper time τ does not correspond to the time outside t.

- #20

jimgraber

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You can read more in the second section of this reference.

But to reply directly, if the infalling object is not a mathematical point, but actually has a size, it will not reach infinite speed before it falls in, and so time won't "stop". Even a point only reaches light speed at the exact same instant it "falls in".

Also even if time "stops" for the infalling object, it does not stop for the outgoing event horizon.

Second point; A black hole space time (with one dimension suppressed) resembles a pencil. I think you should ignore the point of the pencil and concentrate on the cylindrical barrel. That is where the conventional description applies.

Further, if you include quantum effects, the object can "tunnel in".

Your original math is correct as far as it goes, but it doesn't capture the whole picture.

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