how do black holes merge when the singularity infinite density slows time to no time

Mordred

THe universe is around 14 billion years matter falling into the singularity appear to take infinite time. So how would two black holes have had the time to complete a merger under those conditions? According to my understanding of GR we should be able to see mergers still in progress as the time reference at the event horizon and at the singularlity with itas infinite density would give us the apearance of infinite time taken which is well beyond the age of the Universe. Would the event horizon simply shift in size and encompass both singularities essentailly entailing two seperate singularities inside one event horizon? If I recall Hawkings radiation this process also takes a long time to reduce a large Black hole to nothing. So even if all the matter was pulled away from the smaller black hole to the larger, The smaller black hole would still not have had enough time to reduce to nothing. or am I correct in that last statement?

PAllen

An outside observer never sees matter cross the the event horizon. However, the event horizon itself responds approaching matter. As two collapsed stars with event horizon approach, their horizons merge in finite time for an external observer, with all matter of both stars still outside the merged event horizon.

Interestingly, if a planet approaches an event horizon, the horizon grows well before the planet reaches the horizon. The current event horizon reflects (in part) all matter that will eventually be trapped.

Mordred

side note was looking for examples of mergers when I found this article printed yesterday.

http://www.space.com/12790-monster-b...ollisions.html

I've also watched the Nasa simulation of blackhole mergers done in 2010 I'll link a post of it.

its neat how the gravity waves are shown in this.
http://www.youtube.com/watch?v=GAwO1okR074

Thanks for your reply makes more sense now

utesfan100

Here is a simplified model.

Here is a more complete model.

One thing to note is that since the radius of an event horizon is proportional to the mass, the resulting event horizon will expand to the sum of the colliding event horizons. Thus whatever the details are at the point of collision are, they will quickly be shrouded behind the horizon.

Naty1

And yet if you are falling with that matter, all takes place in "normal" (local) time.

Among the things Einstein brought to science is the understanding that there is no single "correct" frame of reference, all frames (observers) are "relative".....There is no universally correct observational frame.....So what you see is likely different from what a distant observer sees.

Mordred

Yeah I knew the part regarding normal time if your falling into the singularity, I was more interested in what we would see with the time dilation when we look at it from Earth.

Phrak

Interesting. But what you call an event horizon is not an event horizon, but the potential event horizon should the entire matter of each star be within the horizon boundary.

PAllen

No, it really is the event horizon. The event horizon is defined against a total 4-manifold. It necessarily reflects the complete future history. The definition is the surface from which no light will escape to infinity. Imagine a light wave or photon so nearly trapped (assuming no more matter falls in) that it takes a thousand years to escape far from the event horizon. Now, assume in 800 years a baseball approaches the event horizon. This can be enough so the light never escapes. Without the baseball, this light would be just outside the event horizon. However, because a baseball will fall in 800 years from now, it is actually inside the event horizon.

I was also speaking of observers well outside the event horizon, so, as I said, no matter is ever seen to cross the growing event horizon.

Phrak

I don't know what "800 years from now" means without reference to a coordinate chart.

PAllen

I am referring, of course, to the 'egocentric coordinate system' (Fermi-Normal coordinates based on my world line).

Make a silly quibble, get a silly answer.

(Apologies to Sydney Coleman for use of egocentric frames).

Phrak

Sorry if you think it's quibbling.

I don't think it is though. I take it that Fermi normal coordinates are those of a feely falling observer.

I believe I could define some coordinate system in which there are no event horizons forming, even aysmtotically. Then again, the event horizon is an artifact of Schwarzschild metric and not the Fermi metric. So we are talking about the coordinate singularity in one metric using the coordinates of another.

PAllen

I said several times already that I was speaking of observers distant from the horizon. Given that, it does not matter what coordinate system is used for the statements I made (as long as time is interpreted to be proper time, and the observer stays distant from the horizon and does not have extreme motion, which I thought goes without saying). By Fermi-normal coordinates I do not necessarily refer to a free falling observer. I mean you use a chosen worldline as time axis, and foliate spacetime with spacelike geodesics 4-orthogonal to this world line.

The event horizon singularity is coordinate dependent. The event horizon itself is a geometric feature of the manifold. A given null geodesic either reaches spatial infinity or it does not. This is not coordinate dependent, thus the horizon is not coordinate dependent. Its shape as a 2-surface does dependent on how spacetime is foliated into spacelike hypersurfaces, but you can't get rid of it or change anything substantive about its properties.

(I really thought you were just being pedantic, rather than didn't understand or disagreed with what I was saying).