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Author Topic: Falling into a black hole
CardassianScot
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I was looking at the Wikipedia entry on Black Holes earlier today and I thought something didn't sound right about their description. However, my physics is a bit rusty so I thought I'd see if anyone here with less rusty skills agrees with me or the Wikipedia.

It says
quote:
Consider a hapless astronaut falling feet first radially towards the center of a simple Schwarzschild-type (non-rotating) black hole. The closer he gets to the event horizon, the longer the photons he emits take to escape from the black hole's gravitational field. A distant observer will see the astronaut's descent slowing as he approaches the event horizon, which he never appears to reach.

However, in his own frame of reference, the astronaut will cross the event horizon and reach the singularity, all in a finite amount of time. Once he has crossed the event horizon he can no longer be observed from the outside universe. As he falls, he will notice his feet, then his knees, becoming increasingly red-shifted until they appear invisible. As he nears the singularity, the gradient of the gravitational field from head to foot will become considerable, and he will feel stretched, and finally torn. This process is known as spaghettification. This gradient becomes large enough, close to the singularity, to tear atoms apart. The point at which these tidal forces become fatal depends on the size of the black hole. For a very large black hole such as those found at the center of galaxies, this point will lie well inside the event horizon, so the astronaut may cross the event horizon painlessly. Conversely, for a small black hole, those tidal effects may become fatal long before the astronaut reaches the event horizon .

My dispute is in what happens when an astronaut crosses the Schwarzschild radius. Here it says that if the black hole is big enough you can do this painlessly. However, shouldn't crossing the Schwarzschild radius kill you regardless of tidal effects? Since everything has to move towards the centre in a black hole, blood will move down towards your legs (assuming feet first) but be unable to return back up the body to the heart and brain. In fact blood will be unable to get from the brain to the heart. Add to this the electrical signals from the body to the brain, or the functioning of the brain itself, none of which can work if information, energy and matter only flow in one direction. Therefore, crossing the Schwarzschild radius should kill you.

I've never heard this before and the sci.physics FAQ article on Black holes even suggests that
quote:
But there's nothing *locally* special about the event horizon; when I get there it won't seem like a particularly unusual place, except that I will see strange optical distortions of the sky around me from all the bending of light that goes on.
So I am I right or is there some flaw in my reasoning?
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EDanaII
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I think your reasoning is correct.

The event horizon of a black hole is the point at which escape velocity equals the speed of light. At this point, the gravitational forces between your head and your feet are so huge that you would be ripped to shreds.

Consider this, if you are standing on the event horizon, the pull of gravity on your feet would equal aproximately 186,000 mps but the pull of gravity on your head might be 185,999 mps. That's still 1 mile per second. Certainly fast enough to "spaghettify" you. [Smile]

Could be wrong, but that is my understanding too.

Ed.

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A. Alzabo
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quote:
Could be wrong, but that is my understanding too.

Mine too, but I'm no expert on black holes -- I've always heard that the gravitational differential between the ends of an object falling into a black hole would shred it even before the event horizon was reached.

Edited to add -- I guess the article could be correct for a nonrotating black hole that was hugely massive. The Schwarzschild Radius
would be far enough out that the gravity gradient would be nonlethal, so that part seems OK. The wierd part is the one-way flow of information. Would you be severed layer by layer as you crossed the Schwarzschild Radius? Hmmmm. I guess it would depend on how fast you crossed the boundary (from the faller's FOR), and if the parts of you still in our universe moved while you were crossing. I'd bet crossing is pretty fast (since it's just a line), so you probably wouldn't die from a lack of blood flow -- but I wonder what would happen if you crossed the Schwarzschild Radius spinning or moving in some way that expects that part of you could be moving back out of the black hole?

[ December 02, 2004, 06:08 PM: Message edited by: A. Alzabo ]

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nyani
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um, i've never studied physics or black holes at all, so this is really sort of over my head. But the Schwarzchild Radius is just the point where the gravitational pull of the black hole becomes strong enough to block the emission of photons, right? In that case, the gradient of gravity would stay pretty constant, and crossing the radius wouldn't necessarily be traumatic. But once you're inside, does the gravity increase exponentially, such that it would be significantly different at say your feet and your head? I guess the bloodflow thing would kill you anyhow, unless you remain a closed system which i guess you don't.

By the way, if all energy and matter flow to the center of the black hole at incredible speeds, what happens to that energy? I suppose it somehow increases the pull of the black hole, but... i mean, when it gets to the center it just stops and where does that energy go, if not outward?

haha... i don't think i'm making any sense at all.

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EDanaII
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Yer definitley not making sense, Nyani. [Smile]

Consider this. The Earth's surface gravity is about 32 fps. If the Event Horizon is where gravity is so strong that light cannot escape, then we are talking about a gravitational pull of 186000 mps * 5280 or 982,080,000 fps... That's a lot faster than 32 fps. More importantly, that means gravity is 30,690,000 times what it is on Earth and a man that weighs 200 lbs here would weigh 6 billion pounds there! You would be crushed by your own weight.

Here's a little experiment: take two balls, hold them side by side, and drop them. Under normal circumstances, those balls will hit the ground at the same time. Now, take those two balls, stack one on top of the other and let them go. As they fall, the lowest one will fall faster and faster than the one above and the distance between the two will grow larger and larger as a result. This is because, as an object falls towards a gravity well, the gravitational attraction becomes stronger and stronger, and the objects are pulled faster and faster.

Next, take those two balls again, stack them on top of each other and drop them from an event horizon. The lower one will fall faster than the upper one because of the difference in gravitational weight between the two, and the speed at which they move apart would be measured in miles per second.

Finally, just consider one ball. The difference between the weight of the top part of the ball and the bottom part of the ball would be so significant that the bottom part would fly away faster than the top part -- just as the two separate balls flew apart -- and, again, the speed at which this occurred would be measured in miles per second. The ball would be torn to shreds.

Nothing can survive that.

Ed.

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Everard
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yeah, Ed's right on this one. The wikipedia entry is a mess. THe person who wrote that must not really know very much about black holes.
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Lewkowski
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If two balls fall side by side wouldn't the bigger one land first? Gravity is measured from center to center right? So wouldn't the smaller would be slightly drawn to the center of the larger one?
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The Drake
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Significant digits, Lew [Smile] But I think also the answer is indeterminate. If they're dropped from a shelf (bottoms aligned):

The balls will have two force vectors acting on it (assuming no atmospheric resistance, magnetism, or other forces are in play). They will be pulled toward center of the earth, and toward the center of the other ball. Since the bottoms of the balls were aligned, the distance from the small ball to the surface of the earth will increase compared to the large ball.

But hold on. Since no other forces are in play, there is no damping effect on gravitational forces between the balls. The small ball will oscillate between its original relative position, and a higher position.

So it is extremely unlikely that the small ball would strike at the same time, but not impossible. This is assuming that the balls are not heavy enough and far apart to orbit one another. And that they are far apart enough that they do not collide. Note too, that if the larger ball is only slightly larger, its center is being significantly drawn to the small ball. And it could arrive first.

(ed: and implicitly, the balls are of equal density)

[ December 03, 2004, 04:57 PM: Message edited by: The Drake ]

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A. Alzabo
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quote:
If two balls fall side by side wouldn't the bigger one land first? Gravity is measured from center to center right? So wouldn't the smaller would be slightly drawn to the center of the larger one?
And the bigger one is drawn to the smaller one. The "center to center" is a mathematical construct for the sake of convenience.

In a vaccuum, both balls would hit the ground at the same time (so would a feather and a bowling ball). Depending on the height of the fall and the conditions, the smaller ball would probably land slightly sooner, since the air resistance would be less. The attraction between tiny masses close together (and huge masses far apart) is negligable.

Edited to add:

I see that you mean balls falling into a black hole, so what I said above is irrelevant.

If the balls approached relatavistic speeds as they fell into the hole, they would be more and more attracted to each other as they fell. But if they were side-by-side, the attraction would be orthogonal to the "falling" direction, and wouldn't alter the distance along that axis in an amount greater than other effects.

[ December 03, 2004, 05:03 PM: Message edited by: A. Alzabo ]

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The Drake
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Actually I take it back. They wouldn't oscillate. It's been a long time since I've done this stuff...
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A. Alzabo
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quote:
By the way, if all energy and matter flow to the center of the black hole at incredible speeds, what happens to that energy? I suppose it somehow increases the pull of the black hole, but... i mean, when it gets to the center it just stops and where does that energy go, if not outward?

A vast amount of energy is spewed "out" of black holes in the form of radiation. I believe Stephen Hawking recently posited that information does "return" to our universe from black holes (that is, black holes don't go anywhere/when else), only in a form so mangled that it can't be reconstructed.
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frogcat
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you know... you can edit the wikipedia entry yourself and fix it up...
http://en.wikipedia.org/w/wiki.phtml?title=Black_hole&action=edit

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Lewkowski
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I sucked at HS physics but it seems to me that a larger object would hit first in a vaccum. There is more gravitational exertion above the tiny ball then below it. Say the big ball is 10x as big, well slice off the portion that is the same size as the tiny ball and look at it that way. You now have 3 objects falling, The two tiny balls and the remainder of the Big Ball. The big ball is above the initial tiny ball and will cause it to be pulled upward.

Weather whole or thinly sliced its still going to pull on the ball.

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Slander Monkey
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Whoa, I haven't seen this much sketchy science since the last creation vs. evolution thread. Let me try to make it a little more sketchy:

1. The article may not give the right impression about what it would actually be like to fall into a black hole, but not for the reasons mentioned so far -- well actually Cardassian Scot's and A. Alzabo's explanations seemed pretty close but I'll pretend like I didn't read them. Basically the properties of 'space' inside the Schwarzshild radius are different than those outside, if we are to believe the math -- and the change in properties is, again if we believe the math, instantaneous -- so, aside from the fact that we can never really know what it's to fall into a black hole, I'd say that passing the event horizon would be an unpleasant experience under any circumstances.

2. I think the problem with an analysis of a plunge into a black hole is that, if you analyze it quasi-classically, then the you can make the statements that are made in the article, and they can make sense. But if you take the math seriously, which basically says that all things inside the event horizon travel directly toward the center (I'm just repeating what I've read here), and consequently can't escape, then you can easily come up with a silly example that shows that crossing the event horizon is bad joojoo. Imagine the situation where you orbited a black hole and came in close enough to the event horizon to dip your toe in -- your toe would immediately head toward the center of the black hole (orthogonal to your orbital motion), while the rest of you would kept on orbiting. This characteristic sounds pretty non-conducive to life preservation -- all you need is some non-zero velocity not directly toward the center of the hole to mess yourself up upon crossing the event horizon. But, what am I saying -- I have no idea what it would be like to fall into a black hole. The point is that if space inside a black hole isn't space as we know it, then we probably need to redefine what we mean by: seeing, living, moving, etc.

3. EDanall, you've got your velocities, accelerations and forces mixed up. Also, the tidal force (gravitational gradient) at the schwarzshild radius decreases as the mass of the black hole increases -- if you check the math, the magnitude of the "gradient" (at the schwarzshild radius) decreases with the inverse square of the mass. So the larger the black hole, the less likely you are to be torn apart before you cross the event horizon.

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DonaldD
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quote:
However, shouldn't crossing the Schwarzschild radius kill you regardless of tidal effects? Since everything has to move towards the centre in a black hole, blood will move down towards your legs (assuming feet first) but be unable to return back up the body to the heart and brain. - CardassianScot
Ignoring tidal effects, then no – I think you are making the assumption that this hypothetical astronaut is stationary, or that he is at least undergoing some amount of ‘G” force. He is in fact in free-fall (and moving pretty darned quickly.) So, although the blood may travel upwards by half a metre (from heart to brain) in the astronaut’s frame of reference, in practice it has actually travelled much farther down towards the “hole’s” centre of gravity.

quote:
Consider this, if you are standing on the event horizon, the pull of gravity on your feet would equal aproximately 186,000 mps but the pull of gravity on your head might be 185,999 mps. That's still 1 mile per second. Certainly fast enough to "spaghettify" you - EDanaII
I started this before Slander Monkey got to it, but I basically agree with what he said: it sounds like you’re confusing escape velocity with the acceleration effects of the “gravitational” force. From a strictly Newtonian perspective, gravitic force at the event horizon (actually, at any distance) can be defined as follows: Fg = GMm/R^2, where G is the gravitic constant, M is the mass of the large body, m is the mass of the small one (also a constant in this context), and “R” is the radius (in this case, of the event horizon.) So, force is proportional to the mass of the large object, and inversely proportional to the square of the distance (radius) between the two objects. F α M / R^2

It’s evident that R is a function of the mass of the object: the more hugely massive the object, the greater R will be. In fact, the Schwarzschild radius is given by Rs = 2GM / c^2, so it is in fact directly proportional to the mass of the object. Rs α M

Taking these together, the force on the object at the Schwarzschild radius is:

F = GMm/Rs^2

F = GMm/(2GM / c^2)^2 = GMm * c^4 / 4G^2M^2

F = m c^4 / 2GM

Sooo… the force on an object at the Schwarzschild radius is actually inversely proportional to the large object’s mass (sorry Slander, not to the square of the mass…) F α 1 / M

What this means is that the more massive the object, the smaller the tidal forces will be at the event horizon.

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vulture
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Ooh - a physics thread. They're always fun. And DonaldD is right - an astronaut that was mystically suspended at the event horizon, half in and half out (and stationary relative to the center of the black hole as measured by an observer an infinite distance away), would be in trouble. An astronaut in free fall crossing the event horizon has no problem. There is no event horizon in his frame of reference. The event horizon appears in the Schwarzschild reference frame, which is based on an observer stationary relative to the black hole, and inifinitely far away. And relative to that observer, as DD says, nothing in the astronaut is moving away from the black hole at all - it is all falling in, although at very slightly different speeds.

In the reference frame of the falling astronaut, there is simply nothing special at event horizon for inward moving objects (for those of a technical inclination, this can be seen with Eddington-Finkelstein co-ordinates). Some interesting things crop up with the combination of Eddington-Finkelstein co-ordinates (advanced and retarded) and Kruschal diagrams, namely that nothing at all prohibits you from entering a black hole and leaving it again (aside from pesky issues such as tidal forces tearing everything apart, and passing through the central singularity which of course has inifinitely large tidal forces (assuming it actually exists). The curious thing is that although such a path has a finite proper time (i.e. someone following that path would experience a finite (and quite small) amount of time passing, from the point of view of any observer outisde the event horizon, the passage would take an infinite amount of time (there is nothing per se stopping things leaving a black hole, beyond the fact that it takes an infinite amount of time to do so as seen from the outside).

Rotating black holes produce other interesting effects; it is now possible to find paths that enter and leave the event horizon while missing the central singularity. Again, these take a finite amount of time as experienced by the traveller, but an infinite amount of time as measured from the outside.

What this actually means is unclear. It seems meaningless to talk about entering a BH and coming out an infinite time in the future - any moment when the traveller did emerge would necessarily be a finite time in the future. But the traveller experiences a finite time before leaving the black hole. Hence the speculation about entering a new 'pocket universe'; the infinite time issue is just the way the fact that there is no connection between the two sections that can be traversed without going through the BH. I don't think that anything concrete is really known about what the maths is telling us here though.

Incidentally, DD is right that tidal forces at the event horizon decrease as the mass of the BH increases. The tidal force at a given distance from the center however does increase with the mass.

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Gaoics79
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Ok, this is good. I've learned that I should never try to enter a black hole, because it would be very dangerous. One of you guys should do a tour of America's pre-schools. This lecture should be right up there with "don't talk to strangers" and "always brush your teeth". The title of the lecture would be "don't go into a black hole, unless it's a really BIG one"
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nemes_ie
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Hi all
Vulture appears to be right, from my memories of my (incomplete) physics degree.

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EDanaII
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First off, guys, my argument was meant to be _illustrative_ not accurate.

Second, no one has supplied any numbers. Those formulas are nice, but if you don't apply them, they are meaningless.

Finally, from my perspective, some simple numbers. Gravity is computed by a simple formula 1/d^2. If the gravity on the Earth's surface is 32 feet per second, and the Earth is 3,500 miles from center to surface then the gravitational pull at 3,500 miles _above_ the surface is 8 feet per second (3500/7000)^2 is (1/2)^2 or 1/4.

Now, measuring a black hole from its event horizon, let's take one that is 5 miles from the singularity to that "surface." There the gravity is 186,000 miles per second. A man "standing" on the event horizon would (theoretically) have his head six feet above it so, in feet, the formula would be (26,400/26,406)^2 or 0.9997. Now, multiply that by the speed of light and you get 185,957. And the difference between that and the speed at the event horizon is still 42 miles per second!

Let's try a larger one, how about one that is 50 miles from singularity to event horizon. Such black holes MAY exist, but the earlier example is more likely. Now the formula becomes (264,000/264,006)^2 and the resulting answer is about 4 miles per second!

You tellin' me that ain't enough to rip the man apart?

Having said that, since the speed of light is an absolute and mass increases as you approach it making it harder for you to accelerate to the absolute speed of light, I can accept that this would mitigate some of that acceleration. Except that gravity doesn't push an object closer and closer to the speed of light, it _pulls_ it

Unfortunately, my understanding of physics breaks down here, so I can't make an effective argument. All I can do is illustrate the fact that, even at a meager six feet above the event horizon, the difference in gravitational force would rip you apart.

Ed.

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Slander Monkey
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EDanall,

The main problem with your analysis is that the gravitational force at the event horizon (or I should say near the event horizon) is not 186,000 miles/s -- that's the *speed* of light. DonaldD has the classical equation for gravitational force -- and, relatedly, acceleration -- correct, and it goes down as the mass of the black hole increases (at the event horizon that is). Whether a classical analysis is acceptable, I don't know at the moment, but I suspect that in reality the force would be closer to infinite due to relativistic effects [EDIT: I mean if you were somehow able to stand near the event horizon without falling in -- WHOA, if this interpretation were correct, it would actually clarify some other things now that I think about it.] -- but, seriously, what do I know. If I get a chance to ponder some specific numbers, I'll let you know.

[ December 05, 2004, 12:55 PM: Message edited by: Slander Monkey ]

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Slander Monkey
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quote:
Originally posted by vulture:
Ooh - a physics thread. They're always fun. And DonaldD is right - an astronaut that was mystically suspended at the event horizon, half in and half out (and stationary relative to the center of the black hole as measured by an observer an infinite distance away), would be in trouble. An astronaut in free fall crossing the event horizon has no problem. There is no event horizon in his frame of reference. The event horizon appears in the Schwarzschild reference frame, which is based on an observer stationary relative to the black hole, and inifinitely far away. And relative to that observer, as DD says, nothing in the astronaut is moving away from the black hole at all - it is all falling in, although at very slightly different speeds.

Well this has been an interesting opportunity for me to learn what the deal with black holes is -- I was originally convinced of basically the same thing that you wrote, Vulture, but then I checked a few websites to make sure I wasn't being stupid... and quite unexpectedly they turned me stupid. I didn't read quite enough it seems, and I didn't try to consider what exactly the difference was between the space inside the event horizon and the space outside. So what's the difference? Apparently there's not much difference, however space itself, it seems, travels faster than light towards the singularity *inside* of the event horizon. Hence the inability of light and everything else to escape the black hole. Now this really confused me for a while -- because it seems that there would have to be a discontinuity between motion toward the center of the BH and orbital-type motion wherein classically speaking, you can orbit a body at a lower velocity than what is required to escape the gravity of the body completely (by a factor of sqrt(2) -- I think). So is this what happens at the event horizon? Well if we do a naive classical classical orbital mechanics calculation but account for relativistic effects (a lorentz trasformation) we find that orbital speed required at the schwarzshild radius to avoid falling inside of the schwarzshild radius is precisely c. Ok that makes more sense in terms of the continuity of space -- but is it right? Actually, from what I understand -- without checking it out myself, is that this is basically the same result obtained from the singularity in the schwarzshild metric. But that means that while there may not be anything special about the schwarzshild radius for a falling observer (or is there?), the radius is indeed special -- you really can't just pop in and pop out of a black hole. Really, what would happen if you were orbiting a black hole near the schwarzshild radius and dropped an anchor in? Would it drag you in slowly, or instantly pull taut -- because you certainly can't pull it back out again -- probably a slow drag.

vulture again:
quote:
The event horizon appears in the Schwarzschild reference frame, which is based on an observer stationary relative to the black hole, and inifinitely far away.
Is this infinitely far away business really a criterion? As is my new (and perhaps final) understanding of this BH stuff, the schwarzshild radius is a hard (essentially invariant) limit with regards to stuff not being able to pass outward from that point. Basically, even if you sat immediatly on the outside of the EH, you shouldn't be able to observe anything on the inside -- or have I got it wrong?

Similarly, shouldn't someone falling into a BH also see a receding event horizon due to the more and more exagerated stretching of space as you near the singularity?

DonaldD wrote:
quote:
Sooo… the force on an object at the Schwarzschild radius is actually inversely proportional to the large object’s mass (sorry Slander, not to the square of the mass…) F α 1 / M
Geez Donald, Now I feel like I must defend my honor or something -- I was talking about the tidal forces -- hence the gradient nonsense -- take the derivative of force, substitute Rs and you get dF/dRs α 1/(ornery mod's alter ego), so no need to be sorry, I was talking about stretching and you were taling about squashing. I must admit, though, that for some reason it never occured to me that force decreases as well despite the equation sitting right in front of me -- so thanks for that. Maybe one day, someone will build a restaurant at the edge of a very big black hole just for laughs.
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vulture
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quote:
Originally posted by Slander Monkey:
Well this has been an interesting opportunity for me to learn what the deal with black holes is -- I was originally convinced of basically the same thing that you wrote, Vulture, but then I checked a few websites to make sure I wasn't being stupid... and quite unexpectedly they turned me stupid. I didn't read quite enough it seems, and I didn't try to consider what exactly the difference was between the space inside the event horizon and the space outside. So what's the difference? Apparently there's not much difference, however space itself, it seems, travels faster than light towards the singularity *inside* of the event horizon. Hence the inability of light and everything else to escape the black hole. Now this really confused me for a while -- because it seems that there would have to be a discontinuity between motion toward the center of the BH and orbital-type motion wherein classically speaking, you can orbit a body at a lower velocity than what is required to escape the gravity of the body completely (by a factor of sqrt(2) -- I think). So is this what happens at the event horizon?

Don't believe everything you read on websites, especially regarding black holes. Basically, only trust what you reliably understand and can reproduce yourself from a knowledge of GR.

The Schwarzschild solution (which is the one that most commonly describes BHs) describes the region outside the event horizon well enough. It is that solution that creates the idea that the radial direction and time swap roles within the BH (so the future is literally towards the center of the BH). But, the Schwarschild solution isn't valid at or inside the event horizon. It is a description of the BH from an observer infinitely far away, but the co-ordinate system of that observer just doesn't cover the interior of the BH. That's why Eddington-Finkelstein co-ordinates were developed; they match the Schwarzschild solution exactly in the exterior (and so describe the exact same physical system), but also extened smoothly to the interior with creating any co-ordinate singularities in the way that the Schwarzschild solution does.

The apparent distinction between the inner and outer regions is essentially an artefact of the co-ordinate system (in one sense; there are real physical effects assosciated with the event horizon however, but they don't preclude things moving outwards through the event horizon).

quote:

Well if we do a naive classical classical orbital mechanics calculation but account for relativistic effects (a lorentz trasformation) we find that orbital speed required at the schwarzshild radius to avoid falling inside of the schwarzshild radius is precisely c. Ok that makes more sense in terms of the continuity of space -- but is it right? Actually, from what I understand -- without checking it out myself, is that this is basically the same result obtained from the singularity in the schwarzshild metric. But that means that while there may not be anything special about the schwarzshild radius for a falling observer (or is there?), the radius is indeed special -- you really can't just pop in and pop out of a black hole. Really, what would happen if you were orbiting a black hole near the schwarzshild radius and dropped an anchor in? Would it drag you in slowly, or instantly pull taut -- because you certainly can't pull it back out again -- probably a slow drag.

I suspect the chain would snap in pretty short order...

Of course, someone orbiting the black hole, lowering an anchor down, will find that it never crosses the event horizon. It takes an infinite amount of time to do so, at least according to one of those inifintely far away observers. You can't really analyse GR problems by analogy and hand waving, unfortunately. You have to do it by solving the equations, and in GR particularly, you have to do it in a co-ordinate system that is valid over the whole range of the experiment. The Schwarzschild solution is not that co-ordinate system, so it may well give you bizarre results if you are doing anything relataing to things crossing the event horizon.


quote:
Is this infinitely far away business really a criterion? As is my new (and perhaps final) understanding of this BH stuff, the schwarzshild radius is a hard (essentially invariant) limit with regards to stuff not being able to pass outward from that point. Basically, even if you sat immediatly on the outside of the EH, you shouldn't be able to observe anything on the inside -- or have I got it wrong?

Similarly, shouldn't someone falling into a BH also see a receding event horizon due to the more and more exagerated stretching of space as you near the singularity?

The infinitely far away business is just a consequence of the Schwarzschild solution. Analysing anything with that co-ordinate system is implicity working with an observer at rest in that co-ordinate system. I.e. the results you get are only those that would be seen by such an observer, which is someone at rest relative to the BH, and not accelerating (someone, in other words, infinitely far away).

Now some consequences derived from the Schwarzschild solution are valid for other observers, but you have to be a bit careful about assuming that such things are true. It is true that any observer outside the event horizon will observe things taking an inifinite time to cross the event horizon in either direction. And, as you say, even someone immediately outside the event horizon, travelling in any manner (free fall, stationary, or whatever) will be unable to see what is happening within the event horizon.

The Schwarzschild solution is also perfectly good from describing BHs from our point of view on earth (after factoring in observational effects due to GR in an expanding universe, that is), which is a bit ad hoc, but works.

But when describing things up close and personal with a BH, you gave to be extremely careful. It is depressingly easy to misunderstand what a solution is saying because you carry over some classical assumption about the way the world is when trying to translate the equations back into physical reality.

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DonaldD
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Ed, the BHs you illustrate are pretty small – the first (with a 5 mile radius) is slightly too small to exist naturally in the universe as we know it today.

The smallest BHs that could exist (using physics and math of today) would measure about 3 solar masses. If we assume that the Schwarzschild radius is only 50 miles (80 km, your 2nd example), that would give a mass of:
M = Rs * c^2 / 2G
M = 80,000m * (3*10^8 m/s)^2 / (2 * 6.7*10^-11 m^3 kg^-1 s^-2)
M = 5.4e31Kg, or about 27 solar masses.

However, the gravitic force at this point would be:
Fg = m a = G M m / R^2
And the resulting acceleration due to gravity would be

a = G M / R^2
a = 6.7e-11(m^3 kg^-1 s^-2) * 5.4e31Kg / (80,000m)^2
a = 5.6e11 (m/s^2)

This is not even close to the speed of light (3e8 m/s) - the number itself is much bigger, but it's not possible to really compare acceleration and velocity... You should always check the units of measurement when solving a problem, just to make sure the quantities being used are correct.

Supermassive black holes (measuring possibly billions of solar masses, and likely existing at the center of galaxies) would have significantly smaller accelerations at the event horizon (though still not insubstantial). A hole measuring 10 billion solar masses would still have an acceleration of about 1500 m/s^2 at the event horizon. One massing 100 billion solar masses would have an acceleration of only 150 m/s^2... and so on.

You can see that living becomes possible (if still brief) in such an area.

jasonr [Smile]

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ATW
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I used to love this stuff but its been so long that now I have trouble even following the conversation.

Eddington-Finkelstein co-ordinates: creates the mental picture of Commander Eddington from DS9 along with Mr. Finkelstein the burned-out hippie dad from Dharma and Greg.

======

Maybe I knew these at one time but


1) how would you figure out if a super-massive black hole was spinning or not?

By definition as I understand it, anything getting inside the Schwarzschild radius isn't getting back out including any information about the black hole.

2) is there any use to talking about a non-spinning super-massive black hole in space?

Spin seems to be the one constant in space. The only things I know of in space that aren't spinning would be tidal-locked planets or moons near a much more massive planet or sun.

Trying to envision a super-massive black hole tidal-locked to an ultra-super-massive black hole but not having much success.

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Hannibal
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this is very interesting stuff i allways loved phisics, problem is i can contribute any more then what being said, except that the original first article does hold some truth, about the "spagetti" effects and other things mentioned there.

in startrek to fix those problems, and the problems of breaking the spead of light virtually instantly, they have invented the "dampening field" which negates all effects of outside acceleration on the ship, thus everyone inside lives under a normal 1g gravity field

by the way.... as an ancdote, all this discussion reminds me of the extrememly obnoxious movie of "event orizon" and now i wont be ablt to sleep at night!

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Adam Masterman
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I'll leave the relativity physics to the experts, but I can answer Lew's querie about the balls falling in a vacuum. More gravitational force IS acting on the more massive ball, BUT, more is needed to overcome its greater inertia. These equal out, and the two disssimilar objects fall at exactly the same rate, 9.8 m/s2 on earth. A helpful way to imagine this is to imagine two objects, equally weighted, falling side by side at the same rate. Now, imagine a little stick is added with superglue, joining the two objects into one. Now we have one object with a much larger mass than the two previous objects. Yet it obviously continues to fall at the same rate. You can also imagine all of the atoms in an object falling at the same rate, regardless of whether they are joined together or seperate.
Adam

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EDanaII
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Unfortunately, Donald, I can't argue with you any further on the subject, since I'm no expert in either physics or math. However, what you are suggesting still flies in the face of my understanding of black holes.

Here's where we do have a fundamental difference, of course. An average sized black hole has always been described to me as being "city sized," naturally, this is up to interpretation, but I've always thought of that to mean somewhere under 20 miles in diameter. And one of 3 solar masses, again as I understand it, can only be smaller than that. And, of course, I'm refering to the event horizon, since that's the only thing you can measure, the singularity being "infinitely small."

Oh well, if I had a little more time to study the subject... But that's certainly not today. [Smile]

Ed.

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vulture
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Incidentally DonaldD, while you are right that the acceleration at the event horizon is proportional to 1/M, Slander Monkey was talking about tidal forces, and (s)he is right that they are proportional to 1/M^2. In fact, the gradient of the acceleration, da/dr = -10^10/M^s when M is measured in solar masses (and lengths in meters, times in seconds) using a Newtonian approximation (which is wrong, but probably cloe to the right order of magnitude). For 'ordinary' BHs, tidal forces close to the event horizon are massive and not survivable. For supermassive BHs they are tiny and not relevant.
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