Has the Event Horizon a Grip ?

[Spyman]

And to your first comment, GR doesnt break down beyond EH??

well, I guess I couldn't disagree with you more

In Einstein's theory of general relativity, the Schwarzschild solution (or the Schwarzschild vacuum) describes the gravitational field outside a spherical, non-rotating mass such as a (non-rotating) star, planet, or black hole. It is also a good approximation to the gravitational field of a slowly rotating body like the Earth or Sun.

 

The Schwarzschild solution appears to have singularities at r = 0 and r = rs; some of the metric components blow up at these radii. Since the Schwarzschild metric is only expected to be valid for radii larger than the radius R of the gravitating body, there is no problem as long as R > rs. For ordinary stars and planets this is always the case. For example, the radius of the Sun is approximately 700,000 km, while its Schwarzschild radius is only 3 km.

 

One might naturally wonder what happens when the radius R becomes less than or equal to the Schwarzschild radius rs. It turns out that the Schwarzschild solution still makes sense in this case, although it has some rather odd properties. The apparent singularity at r = rs is an illusion; it is an example of what is called a coordinate singularity. As the name implies, the singularity arises from a bad choice of coordinates. By choosing another set of suitable coordinates one can show that the metric is well-defined at the Schwarzschild radius.

http://en.wikipedia.org/wiki/Schwarzschild_metric

 

Well, I can be wrong and I don't know GR enough to argue with you...

(But I trust Wikipedia more than a complete stranger on a forum. )

[lakmilis]

Then why do you not just read Wiki instead of writing on a forum full of strangers

 

The Schw. solution would be long gone if one can't find reasons for it to be reasonable. I reiterate, There is nothing wrong with it mathematically. I just think with physics involved, the Kerr hole is the viable one (i.e. a black hole which is not artificially created) will most likely have angular momentum.

 

I mean hey, do you know any other celestial object with no angular momentum? It is the fallacy of , like you say most of us people, who don't really know GR to get all caught up in the Schwarzchild, but just for technicality , I prefer thinking of EH's through Kerrs way

thats all.

 

If the metric was not well defined on the radius itself is predicting, it kinda would be shooting itself in the foot, don't you think? May I also point out that the S. metric is *not* equivalent to GR, but a solution of the Einstein tensor where the boundary of GR goes. Thus within it, GR breaks down I believe. This is why I say GR does not hold within the EH and I am pretty sure it is a known fact that GR predicts its own break down in these areas

 

lak

[timo]

Some semi-random comments:

... would the internal forces keeping the bar togheter still act through the Event Horizon, ...

There's no light cone leading from r<r_s to r>r_s, meaning there is no causal connection from the inside to the outside. I'd assume that also means that none of the intermolecular forces can act from the inside to the outside.

 

In popular science it is often mentioned that you could pass the Event Horizon of a Supermassive Black Hole without noticing.

I'd really like seeing the calculations behind that statement. You can easily show that a point-sized particle reaches r=0 within finite eigentime (should be possible with slight alterations to the program at the bottom of this page). But without a more sophisticated investigation for an object with finite size that possibly even takes into account a simplified assumption of the intermolecular forces working I would not transfer the result for a point-sized object to an extended one.

 

As a note on the event horizont and the coordinate singularity in the Schwarzschild solution: It is often emphasized that this singularity is only a singularity caused by the coordinate system and had no real physical meaning.

-I agree, because the singularity is only in the coordiante system. The physical properties (the metric tensor and its second derivatives which we call curvature) do not pose a problem, there. It's a bit like the north-pole where our coordinate system latitude and longitude causes problems even though there's -topologically- nothing special about that point).

-I disagree, because r=r_s seperates two spacetime-volumes via a causal one-way street. I find that a pretty remarkable border, so I always have some gripes when I read people claiming that there was nothing special about r=r_s. Sounds too much like parroting what you learn in the GR lectures (examiners want to hear that you understood that the 1-r_s/r part does not mean a singularity in spacetime at r=r_s).

[Spyman]

Then why do you not just read Wiki instead of writing on a forum full of strangers

Hehe, I don't have to choose, so I do both of course...

 

I reiterate, There is nothing wrong with it mathematically.

Thank you.

 

I mean hey, do you know any other celestial object with no angular momentum?

First, in the OP it's assumed, and it's not impossible, whether nature provides them or not.

Second, the rotation could be very low, so the much simpler model suffices as an excellent approximation.

 

I prefer thinking of EH's through Kerrs way

The Kerr metric of a rotating black hole is also a solution of Einstein's field equations.

 

This is why I say GR does not hold within the EH and I am pretty sure it is a known fact that GR predicts its own break down in these areas

If you really have such an urge to prove a breakdown of GR inside the EH, I suggest you start a new thread, specifically for that...

[Spyman]

There's no light cone leading from r<r_s to r>r_s, meaning there is no causal connection from the inside to the outside. I'd assume that also means that none of the intermolecular forces can act from the inside to the outside.

Thats more or less what I said in the OP...

 

because r=r_s seperates two spacetime-volumes via a causal one-way street. I find that a pretty remarkable border

Well, for a person like me, not educated in Relativity and with very little knowledge acquired from various sources, simple logic assumes the bar to be softly cut and if continued to be inserted sliced to very thin parts.

 

I'd really like seeing the calculations behind that statement.

I don't know if there is any math behind that, except for the mass to radius formula for tidal forces. But there is a lot of "creditable" sources which seems to support it, and even "papers" on how to extend your lifetime if you accidently have crossed the EH with your spaceship.

http://arxiv.org/PS_cache/arxiv/pdf/0705/0705.1029v1.pdf

(Highly speculative and quite useless if your body and/or the spaceship is separated to small individual parts.)

 

And Wikipedia claims the bar would either not be able to touch the EH or break above it:

For the case of the horizon around a black hole, observers stationary with respect to a distant object will all agree on where the horizon is. While this seems to allow an observer lowered towards the hole on a rope to contact the horizon, in practice this cannot be done. If the observer is lowered very slowly, then, in the observer's frame of reference, the horizon appears to be very far away, and ever more rope needs to be paid out to reach the horizon. If the observer is lowered quickly, then indeed the observer, and some of the rope can touch and even cross the (distant lowerer's) event horizon. If the rope is pulled taut to fish the observer back out, then the forces along the rope increase without bound as they approach the event horizon, and at some point the rope must break. Furthermore, the break must occur not at the event horizon, but at a point where the lowerer can observe it.

Attempting to stick a rigid rod through the hole's horizon cannot be done: if the rod is lowered extremely slowly, then it is always too short to touch the event horizon, as the coordinate frames near the tip of the rod are extremely compressed. From the point of view of an observer at the end of the rod, the event horizon remains hopelessly out of reach. If the rod is lowered quickly, then the same problems as with the rope are encountered: the rod must break and the broken off pieces inevitably fall in.

http://en.wikipedia.org/wiki/Event_horizon

(Can spacetime itself become physical and break the rod ???)

 

I don't have any trouble with accepting the "oneway" surface of the EH, but the question of conservation laws still bugs me, (post#20). So where does that leave me, hmm, I would say: "still very confused".

[Farsight]

Can spacetime itself become physical and break the rod ?

 

Yep. What do you think the rod is?

[Spyman]

OK, so I found this site: http://www.mathpages.com

I don't know if it's to be trusted but it seems reasonable.

(And it has math for those who like numbers. )

 

Under "Reflections on Relativity" you can find this chapter, (7.3):

"Falling Into and Hovering Near A Black Hole" http://www.mathpages.com/rr/s7-03/7-03.htm

 

Of course, for a small black hole you will have to contend with tidal forces that may induce more spatial separation between your head and feet than you'd like, but for a sufficiently large black hole you should be able to maintain reasonable point-to-point co-moving distances between the various parts of your body as you cross the horizon.

 

So in freefall towards the center you could feel and see your legs on the other side of our EH because from your frame the EH is further in, but for a distant observer, half of you have already crossed the EH.

 

At the EH the geometry of spacetime would be so warped that to move in any other direction than towards the center would aquire speeds higher than c. For the spaceship which is already moving very fast, deep down in the gravity well, touching the EH with a metal bar would be like trying to accelerate its end to c. The bar would in one end try to drag the spacetime anchored by the BH and in the other end try to slow down the spaceship. Depending on the speed and direction of the spaceship, the bar end would need to pass the speed of c already a distance above the EH, since it can't, it would deform and break, caught between the inertia of the spaceship and the BH.