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bachnga
2006-Sep-03, 01:40 AM
When an clock falls into a black hole, its time rate, as viewed by an external observer, slows down to a stop as it approach the horizon.
If the black hole does not accumulate more matter, the event horizon actually shrinks due to mass loss from to Hawking radiation. So the distance between the falling clock and the horizon would increase, except the clock would fall further, but would still go infinitely slow as it nears the horizon. Eventually, in 10^66 years, a solar mass black hole explodes into pure radiation without the clock ever crossing the event horizon. My question is:
1) Does an object falling into a non-growing black hole ever pass the event horizon and actually enter the black hole interior?

Of course matter falling into the black hole expands the event horizon. Thus:
2) Does an object actually enter by the event horizon expanding to engulf the object?

THUS, IS IT IMPOSSIBLE TO FALL INTO A BLACK HOLE OR MUST A FALLING OBJECT BE ENGULFED BY GROWING EVENT HORIZON TO GET INSIDE?

Light and presumably gravity traveling parallel to the event horizon surface must travel slow, as measured from the outside. If fact the time retardation should be infinite at the surface. Thus a pebble falling into the hole (really approaching it but its nearby mass might locally expand and distort the horizon into a sphere with a bump on it.: So:
3) How long does it take the addition of matter on one side of the hole to cause the event horizon on the other side to expand and engulf our object? Less than 10^66 years?

An uneven mass distribution, due to falling pebbles or rings of gas, will cause the hole horizon to vibrate. A vibration that may seem like a millisecond to the hole might seem like a megayear to an external observer for reasons mentioned above. Thus:
4) How long, as measured from the outside, does it take an asymmetrical event horizon to become spherical again?

I would suspect these questions have been studied and answered decades ago. Are there references available? (not in German, please).

Bachnga

Ken G
2006-Sep-03, 01:47 AM
These are tough questions. The black hole FAQ part of the forum goes into them, and gives good links too. I think some of your questions are still debated, in fact. No doubt you are aware that in the frame of the pebble, it falls into the hole in short order, but your question is posed from the frame of a distant observer, and so you have to be careful to distinguish whose time you are talking about, and even more importantly, that time is a local quantity and extending one observer's sense of time to distant places can get semantically imprecise.

bachnga
2006-Sep-03, 05:30 PM
Thank you Ken,
If the black hole does not get bigger (no incoming matter), it should get smaller due to Hawking radiation. Thus the falling object suffers a terrible case of time retardation, but doesn't go into the hole. The hole evaporates first. If I can "detect" the object during the life of the hole, isn't it safe to say it never entered?

Ken G
2006-Sep-03, 05:42 PM
It's a good question, and one that most likely requires a full general relativity solution, with Hawking radiation included, to answer. Few are capable of that calculation, and those that are don't always get the same answer (witness Hawking's famous "bet"). For the rest of us, black holes are as black holes do, and what they do is act like an ultracompact source of gravity accounting for all the matter that has been accumulated at or near the event horizon. I think the problem at the core of your question is that what we want, out here at Earth, is to know how the black hole affects its surroundings, well away from the event horizon. Thus we are not interested in local concepts of time near the event horizon, we want to stick with our own concept of time. It is known how to handle that in general relativity, but I personally have never been too clear on it-- it's mighty tricky!

Jeff Root
2006-Sep-03, 07:32 PM
When an clock falls into a black hole, its time rate, as viewed
by an external observer, slows down to a stop as it approach the
horizon.
Yes, but it also being redshifted, and the redshift increases
faster and faster. The clock is actually accelerating toward
the black hole, and actually reaches the speed of light away
from you as it crosses your event horizon. Some people say
that to you it would appear to slow down and even stop, but
my understanding is that you will see the clock continue to
accelerate away from you even as its hands appear to stop.

You might see something like this, as the clock falls into a
supermassive black hole:

One second before the clock reaches your event horizon, it is
moving away from you at 10% the speed of light, and the hands
are slowed down noticeably. It is accelerating away from you
and becoming smaller with distance. Blue light from the clock
is shifted to green.

One tenth second before the clock reaches your event horizon,
it is moving away from you at 80% the speed of light, the hands
are hardly moving at all, and blue light is shifted to red.
The clock is becoming dim because the rate at which photons
from its face reach you is falling rapidly, like turning off
a light switch.

During the final hundredth of a second, blue light from the
clock face is shifted through the entire radio spectrum from
microwaves to short waves to long waves to waves too long to
detect with any existing device. It doesn't matter anyway
since the rate of photons reaching your receiver is orders
of magnitude too low to detect in any part of the spectrum.
The clock has become invisible to you. Though a few photons
may still reach you, they are too weak and too few in number
to detect.

As the clock disappeared, it was moving away from you at
virtually the speed of light, and still accelerating. The
distance between you and the clock was increasing in accord
with that speed.



If the black hole does not accumulate more matter, the event
horizon actually shrinks due to mass loss from to Hawking
radiation.
Utterly negligible for this question, by many, many orders
of magnitude.



So the distance between the falling clock and the horizon
would increase, except the clock would fall further, but
would still go infinitely slow as it nears the horizon.
The apparent speed of the clock away from you is not affected
by the relative time dilation which appears to slow the clock's
ticking. Not only does the clock not slow down, it continues
to speed up, without limit. Except that it disappears as it
reaches the speed of light.



1) Does an object falling into a non-growing black hole ever pass
the event horizon and actually enter the black hole interior?
You see the clock fade out as it approaches your event horizon.

The clock sees the event horizon recede away from it, toward the
center of the black hole, but curve around it on the sides, so
that there is only a small circle of the outside Universe still
visible in the last moment before it gets too close to the
center and is spaghettified.

Never pass up an opportunity to use the word "spaghettified".
The FSM appreciates it.



2) Does an object actually enter by the event horizon expanding
to engulf the object?
The increase in size of the event horizon is utterly trivial
in comparison to the size of the clock. The event horizon is
not really a surface, anyway. It is a deep, thick region in
which it is progressively more difficult for light to escape.
At the bottom -- what is called the "event horizon" -- light
has to be moving vertically away from the black hole to escape.
A little higher up, light can be moving at a slight angle and
still be able to escape. The farther up, the lower the angle
can be. At the "photon sphere" the light can move horizontally,
orbiting at a constant distance from the center. If the angle
is higher than horizontal, it will escape. Farther up yet, and
light can have a slight downward angle, and still escape from
the black hole. That progression continues to infinity.



Light and presumably gravity traveling parallel to the event
horizon surface must travel slow, as measured from the outside.
I can't say anything about the speed of gravity, nor can I say
for sure anything about the speed of a photon moving away from
the black hole, but a photon travelling parallel to the event
horizon would not be slowed, as measured from the outside.

-- Jeff, in Minneapolis

publius
2006-Sep-03, 10:01 PM
Jeff and I have a little disagreement about what a stationary observer sees of a freefaller into a black hole. I maintain that the velocity must reach a peak and come to zero, which means the (coordinate) acceleration must go to zero and change sign at some point. That is, if we drop an object, it will appear to accelerate as expected, but at some point must appear to start slowing down before it reaches the horizon. Jeff doesn't agree with this.

I found the following expression for the (coordinate) acceleration field in the Schwarzschild metric for a stationary observer far away in polar coordinates. This is the modified gravitational "force law" that will give you the precession of the perhelion of Mercury and all that. It doesn' include gravitomagnetic effects or anything like that, as the Schwarzchild metric is for a stationary, non-rotating source mass.

a = d^r/dt^2 =

-GM/r^2 * [ [1 + 2R/r + v^2/c^2] *(r_n) - 4*v^2/c^2 *(v_n dot r_n)*(v_n) ]

That expression is a bit cumbersome as I was too lazy to use bold and subscripts. R is the Schwarzschild radius, and r_n is a unit radial vector while v_n is a unit vector along the particle's velocity vector.

I'm going to play with that and compare to my "rest energy to kinetic energy" logic with the slowing coordinate speed of light reasoning and see how it compares.

Note one gets an inverse cube term plus a v^2/c^2 increase in the radial force. But the second term (with a factor of 4 on it which I wonder is related to where the factor of 4 on the gravitomagnetic permeability comes from) is directed backwards along the velocity vector, multiplied by the dot product of r and v. When v is perpendicular to r, as would be for a circular orbit, that term is zero. However, for a radial velocity, that term is maximum.

And it appears that this term is what will turn the acceleration around and slow the velocity. I'll have to play with it on paper to see.

-Richard

publius
2006-Sep-05, 12:14 AM
Well, well -- this agrees with that Felber/Farber/whatever-his-name's "repulsive" solution that he made such a big deal about.

For a straight line free-fall, no L, no tangential component of velocity, the above Schwarzschild coordinate acceleration formula reduces to. If you're following through, be careful about the signs of the dot product and the velocity vector. :)

a = -GM/r^2 * [1 + 2R/r - 3 (v/c)^2 ].

You can see the v/c term reduces the acceleration, and sort of looks like a velocity dependent "drag" term. And the question is at what velocity will the term in brackets become zero. This is the turning point at which the free-falling test particle will appear to start slowing down. And that is just:

(v/c)^2 = 1/3 *(1 + 2R/r).

At r = infinity, that's just v/c = Sqrt(1/3) ~ 58%. Which means a test particle launched with this velocity or greater will only appear to slow down, not accelerate, as it approaches the source. Note this velocity increases for "launching points" deeper in the well, and at r = R, v/c becomes 1. The turnaround velocity is lower the farther away you launch. If I can still cut the differential equation mustard, I'm going to look at r(t) solutions for the above and see how this plays out for the maximum and final near-horizon velocities.

This agrees with Felber, although in his frame, he was thinking about a moving gravitating mass approaching a stationary one and seeing a repulsive "force".

I'm confident in this value now. But I wonder what is the flow in my "gamma and mc^2" reasoning in this thread, where yielded
v/c = sqrt(2/3). Note that in the above, we'd reach this value for r = 2R.

http://www.bautforum.com/showthread.php?t=45553

If I had to hazard a guess, I'd say that reasoning didn't account for the "curvature". :)

-Richard

Ken G
2006-Sep-05, 12:20 AM
It's impressive to only have that much of a discrepancy, I'm sure it's just a minor detail. No doubt this new insight will serve you well in future threads as issues like this keep reappearing!

publius
2006-Sep-05, 12:38 AM
Ken,

I'm still not exactly sastified. Note the "launch velocity" at which the acceleration will be zero increases as r decreases. That would seem to indicate it would still increase speed, so long as v/c never got above the value determined by 1 + 2R/r. And that may the difference between this and my E=mc^2 thing.

The above is a second order, non-linear differential equation in r(t), and I have little confidence that I can easily find a analytic solution. :) But what I think I will try is to convert that from r(t) to v(r) and see how that compares to my E=mc^2 v(r) equation. If that's easily possible, that is. If you're half-way interested, I wouldn't mind if you (and anyone else) would play with the equation.

And something else about the above Schwarzschild acceleration. I don't know for sure if this is not *proper time* rather the stationary observer's time. The source I got the above from is not clear about that. It doesn't matter for r >> R anyway and low v/c, and so wouldn't be of concern for calculating something like Mercury's orbit. But it would matter for high v/c and near R.

You've always got to be careful about that in GR, and in many cases the proper coordinates are desired as that would be what would be locally measured (which is what you've pointed out many times in many threads.)

-Richard

neilzero
2006-Sep-05, 01:35 AM
I'm skeptical about some of the axioms and postulates of black holes, especially as they apply to black holes with one billion solar mass or thereabouts. It seems to me there are several concentric event horizons. The outermost is the limit of escape by a ballistic object. A powered object can get farther, perhaps several light years away. Another is the horizon where a distant observer observes that time has stopped. A possible third horizon is where light can not escape as observed by the distant observer and a possible 4th is where the orbital speed around the singularity reaches 0.9999 c. I suspect the radius of some of these horizons depends at least slightly on the speed, direction and distance of the distant observer.
The observer riding the clock ship suffers deadly tidal stress near the event horizon of a one solar mass black hole even if a point in his body is in free fall, but he does not experience a time perception change exceeding his distance from the free fall/gravity center of the clock ship.
For a billion solar mass black hole the clock rider experenses hardly anything until he is thousands of kilometers inside the event horizon, as he is essentially at rest with respect to falling clock ship, or so it seems to me. An exception would be quarks and other subatomic particles that resently did a sling shot manuver around the singularity.The clock rider would experience this as dangerous radiation. It seems to me that the singularity is too small to be hit and too small to eat matter. It can however convert matter to photons of electromagnetic radiation, so there may be lots of mass near the singularity, but much of the mass would be moving at high speed at a considerable distance from the singularity. The radiation from near the singularity would tend to divert matter falling in the general direction of the singularity with respect to an observer inside the event horizon. Neil

publius
2006-Sep-05, 02:01 AM
Neil,

All of those "three horizons" are the same thing, the event horizon for a Schwarzschild black hole, which is basically the GR solution for a true uncharged, non-rotating, point mass that has been there for an infinite time. The singularity is the point mass, finite mass compressed to a point.

Once inside the horizon, *all paths* lead to the singularity. There are no slingshot manuevers or orbits inside. Everything goes right to the singularity, and time and space "end" right there for the object.

Rotating and charged black holes are different. The metric is different and things behave differently, but there is still an event horizon, a point of no return to the outside universe.

And this may all be academic anyway, if the ECO (Eternally Collapsing Object) theory is correct. This basically says "in the real world" a collapse to a singularity in its own frame is not possible in finite proper time. It is trying to collapse, but is limited by how fast it can "vent" radiation, which gets slower and slower the more it collapses.

And ECO would pretty much look and behave like a black hole for all practical purposes, although radiation could still escape and there would be no horizon.

-Richard

publius
2006-Sep-05, 03:36 AM
If anyone wants to tackle some messy algebra and some derivatives to help check what I'm about to do, well here's the deal.

My E=mc^2 reasoning, using the Schwarzchild formula for the coordinate speed of light yields this:

(v/c)^2 = u*(1-u^2), where u = (1 - R/r)^2, R being the Schwarzschild radius. If you read the thread where I first posted this, you can see the details of where I got the various "max speed" factors. Now, this is v(r), giving v as a function of the r coordinate, not v(t). The above formula I posted in this thread gives us the acceleration, and involves v(t) and r(t).

That sucker is non-linear and second order, a mess to solve. I have no idea if an analytic solution is possible. But you could plug that into a numerical routine and see what it looked like. But I wondered if the two might be equivalent, and the way to see that is to convert it to v(r) form.

Now, we can write dv/dt = dv/(dr/v) = vdv/dr, and convert the above acceleration to an equation for v(r). When we do that, we get,

v dv/dr = -GM/r^2 * [ 1 + 2R/r - 3 *(v/c)^2 ]

which now gives an equation for v(r).

So the question is to take the first v/c expression and see if it agrees with this last one. That is some messy algebra. :) I have been working with the first in v(u) and du/dr terms and have gone through two pages of tiny little scribbling, then discovered I made a sign error half way through that threw everything off. :lol:

First, to get rid of GM, note that R = 2GM/c^2 or GM = c^2/2R. That will get everything in terms of R and c^2.

We need to get v dv/dr from the first equation and see if that equals the second.

-Richard

kzb
2006-Sep-05, 11:52 AM
Can a black hole even form in the first place, as seen by a distant observer?

Ken G
2006-Sep-05, 12:08 PM
I've heard it said that it cannot, but again I think this really depends on how you extrapolate time to other places from where is the actual clock you are using.

Squashed
2006-Sep-05, 12:12 PM
Can a black hole even form in the first place, as seen by a distant observer?

This is a profound mystery because according to the formulas no matter can ever enter a blackhole because it will take forever to reach the event horizon (in our frame) and so as soon as the event horizon forms it becomes a shield; but then when we observe the universe we see blackhole candidates of various sizes which means that a blackhole can definitely grow - but how?

The redshifting of light from an object falling into a blackhole is a red herring argument because the light traveling from the object to us has no bearing upon the outcome of the object - it only affects our view of the object.

The actual formula, as publius demonstrates, declares that time dilation affects the velocity of objects but I do not see this in actuality.

kzb
2006-Sep-05, 05:19 PM
<<but then when we observe the universe we see blackhole candidates of various sizes which means that a blackhole can definitely grow - but how?>>

Well let's take a step back. What we have, is observations of phenomena that have been explained by astronomers as the effects of black holes.

When they were first thought up by theorists, BH's were the favourite catch-all explanation of everything. Now that honour has fallen to exotic dark matter!

publius
2006-Sep-05, 11:57 PM
"Oh so close, yet oh so far..." That might be another way to describe trying to dip your fishing hook into the event horizon, but it's how my
E=mc^2 v(r) compares to the above dv/dt expression I found.

This gets messy, and it took me a lot of scribbling and going over several times to condense it to following, but here's how it goes, and please check my work if anyone is interested.

My E=mc^2 reasoning yielded the following expression for v(r) for a Schwarzschild free-fall starting out from infinity with zero initial velocity. And this (supposedly) would be from a frame at infinity.

v^2(r) = c^2 * u(1 - u^2), where u = (1 - R/r)^2, R being the Schwarzschild radius.

Now, differentiate the above with respect to u, and get the following:

2v dv/du = c^2 * (1 - 3u^2).

We're after dv/dr, so we need to multiply both sides by du/dr. That comes out to be du/dr = 2(1 - R/r)*(R/r^2). This cancels the factor of 2 and we have:

v dv/dr = dv/dt = Rc^2/r^2 * [1-3u^2]*(1 - R/r)

If this were correct, this would be an equation for dv/dt that didn't include v itself, just r, which would be nicer. However, we want to compare that with the above dv/dt equation I found. We want a (v/c)^2 in that sucker. We can get that by a little trick of back substituting for u in terms of v. From the u equation, u^2 = 1 - (v/c)^2/u. Substitute that in for u^2, and we get:

dv/dt = Rc^2/r^2 * [1 - 3(1 - v^2/uc^2)]*(1 - R/r)

Now, 1 - R/r = srt(u). Multiply that out in brackets and rearrange and one gets:

dv/dt = Rc^2/r^2 *[ -2 + 2R/r + 3(v^2/c^2)/(1 + R/r)^2 ]

Now, recognize that Rc^2 = 2GM and pull out a minus sign and we have

dv/dt = -GM/r^2 * [ 4 - 4R/r -6(v/c)^2/(1 + R/r)^2 ]

See what I mean about "Oh so close, yet oh so far"? :)
The GM/r^2 fell out nicely, but the mess in brackets didn't work out right. It's darn close, we got a constant, 2R/r and 3 (v/c)^2, but it's mulitplied by another factor of 2, the sign is wrong on R/r.

The difference, what we would have to add to the above in brackets to get the first dv/dt formula is this:

3 [ (v/c)^2 * (1 - R/r)/(1 + R/r) - (1 - 2R/r) ]

Doesn't look familiar. :lol: So something is awry, but it's something small I imagine, and I don't have a clue what it could be.

-Richard

Squashed
2006-Sep-06, 12:26 AM
"Oh so close, yet oh so far..."
.
.
.
-Richard

publius (to respect your privacy I won't use your real name),

I appreciate all your mathematical skills because I am not that adept at the "art" but would like to learn more about what you explore ... but don't count on me backchecking your work. A lot of what I have read about the universe has been popular literature, and pure theory, that does not delve into the analytical side and so I am glad to see it posted.

Squashed

publius
2006-Sep-06, 09:56 PM
Squashed,

Thanks for respecting my privacy. :lol: I first started with online BBS on CompuServe. That used to be pretty good years ago, but it's gone now -- AOL bought it and it went downhill fast. I quit it several years ago. Anyway, on the old Compuserve fora, the standard rule was one had to use your real name, no "handles" or screen names and that sort of stuff. The idea was real names discouraged bad behavior. But that was before everyone and his brother had internet access. I never liked Web based fora at first, but that's the way it's gone, and I found this one here, which I like pretty darn well. :)

Anyway, the math I used above is nothing more than some simple calculus plus algebra. Nothing to brag about at all, and when I do it, it makes me mad because I remember how much I've forgotten. "I used to be good", as the saying goes.

The math of General Relativity is some of the most advanced, and it's generally only at the graduate level does one tackle it. I never went that far. Several times I've said I was going to learn it, so I could sastify my curiousity about this kind of stuff, but well, this old dog doesn't learn new tricks too well.

-Richard

publius
2006-Sep-06, 10:03 PM
Back to business, I think, *think* I've found the flaw in my E = mc^2 reasoning. It has to do with the 'v', the velocity in the gamma factor. I was equating that with the coordinate speed dr/dt as seen by a distant observer.

That is apparently not kosher and doesn't work. That velocity has to be the *local* (stationary frame, not free fall, mind you) velocity, apparently to "make sense". Like I was saying to Squashed, I wish I would learn GR enough to know about stuff like this.

And there is sometimes a difference between the time coordinate and the local distance coordinate that comes into play. That is sometimes one really wants 'dx/dt', where x is local distance (not r), but 't' is still the time in the distant frame. That may actually be what the 'v' has to be, but I'm still not clear on that.

Coordinates are a big issue in GR, and you've got to be careful!

-Richard

publius
2006-Sep-07, 04:23 AM
Well, this makes me feel better :lol: :

http://www.physicsforums.com/showthread.php?t=126996

The guy wants to write an accurate GR (Schwarzchild metric) gravity simulator, and wanted the equations of motion in X-Y form (for a stationary frame far away). You will see how much fun they're having doing that. Eliminating the proper time from the geodesic equations and expressing everything in coordinate time is something they're trying to do.

I've got a feeling they're making this way too complicated........But it's complicated by any measure, of course.

Again, I'll say no wonder Einstein's hair looked like it did! :)

-Richard

Ken G
2006-Sep-07, 06:20 AM
Yeah, I try to stay clear of that stuff. Different coordinate systems, different physical pictures of what is going on. It's really tricky. I think things are only easy if you always use the time and distance measured by the local clock and ruler, co-moving with the local free falling matter. Then GR is not about what is happening in those frames, it is about how to transition from frame to frame as you attempt to connect two well separated events (like emission and absorption of some light). So it's thinking in terms of reference frames, and transformations, rather than a global coordinate system (since the latter will almost always involve the generation of "ficticious forces" that are a bear to figure out!). Note that the transformations are a bear too, only the people who do this for a living really understand it (often differently).

publius
2006-Sep-08, 05:57 PM
Well, I managed to get this figured out to my sastisfaction, and learn exactly what was wrong with my original E=mc^2 reasoning. This is the correct expression for (v/c)^2 as a function of r, where 'v' is dr/dt, the *coordinate radial speed* of a distant observer. This comes directly from solutions to the geodesic equations, and is surprisingly simple:

(v/c)^2 = (1 - R/r)^2*[ 1 - (1 - R/r)/E^2 *(1 + L(R/r)^2 ]

There is another expression for the theta coordinate (and phi if one doesn't want to restrict it to a plane), but we won't worry with it.

In the above, E is the constant total "specific" energy (E/m0c^2)of the test particle. Now this E (not specific)is given by the following expression:

E^2 = (1 - R/r)m0*c^2/(1 - (v/c)^2) = gamma^2*(1-R/r)*mc^2

v in this expression is the *local speed* of the object, although one could use the coordinate speed and use the coordinate speed of light. Or just convert coordinate speed to local speed. Either way one gets the same factor. This is the regular SR "gamma".

My reasoning about gamma was actually correct. However, my reasoning about modifying the mc^2 part in the numerator was not. The rest energy of a mass at rest in the Schwarzschild field is just

E = sqrt(1 - R/r)*m0*c^2.

I was attempting to use the square of the coordinate speed of light there, which gave a 4th power of (1 - R/r), when it reality it's actually the square root. IOW, I was using the 8th power of the actual value. :lol: The square root term is just the time dilation factor. The rest energy is only "adjusted" by the time dilation factor, and looking at how mc^2 comes about, one can see this. The 'c' here is more of a universal constant than an actual speed.

The 'c' in gamma can be seen as the coordinate speed of light, but not the c in the mc^2, IOW.

That was my error, and if one uses the correct E expression, the above v(r) drops out directly. Anyway, remembering the E above is the specific energy (E/m0c^2), E = 1 for a test particle dropped with zero initial velocity from infinity, and v(r) becomes simply:

(v/c)^2 = (1 - R/r)^2 * R/r

You will see that is 0 at r = R. A test particle appears to slow down and stop near the event horizon. Now, I won't post the derivation unless requested, but it's fairly simple to find the maximum speed occurs at r = 3/2R,
which is the *photon sphere*. That works out nicely.

This maximum coordinate speed is 1/3 *sqrt(2/3) ~ 27% c. That seems low, but light itself is only moving at 1/3c (radially, tangentally, it would be slowed only by the square root of (1/3) = 58%c).

But the local speed is sqrt(2/3)c, which is the original "max speed at infinity" my erroneous formular came up with!

And you'll note something else fascinating. The local speed is exactly the Newtonian sqrt(R/r)! The (1 - R/r) factor is divided out to get the local speed and one is left with the pure Newtonian expression for v(r).

This is why the Netwonia formula for the event horizon works. The local speed is indeed c at r = R (but it never gets there in the time of any external observer).

-Richard

publius
2006-Sep-08, 09:29 PM
And something else worth pointing out: The key insight in all this is the "geometric view" of gravity. The total energy of a free-falling particle does not change. In Newtonian mechanics, one defines the total energy as the sum of the kinetic and potential energies. However, in this geodesic view, there is no "potential". All the energy is "with the particle itself". With EM fields, the "potential" corresponds to actual field energy, so when a particle looses kinetic energy to potential energy (working "uphill" so to speak), there is a transfer of energy from mechanical to field.

There is no such thing here in Einstein's geodesic view of gravity. The rest energy gets "converted" to kinetic energy relative to stationary observer because space and time are "curving", and the free-falling particle's "motion through time" becomes motion through space relative to the stationary observer.

This is the key insight. When you lift something against gravity the work you are doing is actually going into the rest energy of the mass as you raise it to a region with "faster time". As you raise it, the direction of time is changing a little bit, and you must supply the energy difference.

In this view the gravitational field does not have any energy. However this gets very complex and very subtle. The gravitational "field" can carry energy away from a system in the form of gravitational radiation. For example, when a mass is dropped, that "excess" rest energy that is converted to kinetic energy must go somewhere when the object hits the ground. It usually goes into thermal energy ultimately; heat.

However, gravitational radiation can carry some of that away and transfer it to other distant masses that interact with the radiation. I was reading where some originally thought this wasn't possible because the field had no energy. But it has been demonstrated that a gravitational wave can pass through an arrangement of masses in initially flat space-time and change the total energy of the system after it passes. After it passes space-time is flat again, but the total energy of the system is greater than before. The proper view of how that works is not trivial. If the field doesn't have energy content, then how can it carry it away?

And related to this is a deep question: Would you "feel" the gravitational radiation reaction force? :lol: Does radiation reaction make a radiating mass deviate from geodesics?

All these questions are way above may pay grade, but they are not trivial, and get to the heart of the fundamentals of what GR's geometric interperation really means.

-Richard

Ken G
2006-Sep-08, 10:19 PM
When you lift something against gravity the work you are doing is actually going into the rest energy of the mass as you raise it to a region with "faster time".
I've debated this issue with others, as I like to think of this as being the case. But it was quite correctly pointed out to me rest energy is really a local quantity, so not only must the object be at rest but you must be in all ways in its reference frame, so you must be at the same point in the gravity field as the object. Using that definition of rest energy, it does not change as you lift something, it is what it is. But you can use a kind of nonlocal extension of rest energy in the way you are talking about, though I'm not sure if you are using standard nomenclature at that point. What is cute, though, is that in this picture, when you drop something, what stays the same is its relativistic mass. Almost all physicists would instead say it is the rest mass that stays constant. Which one is right does depend a bit on nomenclature, however.




And related to this is a deep question: Would you "feel" the gravitational radiation reaction force? :lol: Does radiation reaction make a radiating mass deviate from geodesics?

Yeah, I'd like to have a way to see the answer to those kinds of questions in GR.

publius
2006-Sep-09, 08:34 PM
Ken,

You know, a lot of this stuff is indeed pure convention, sort of like the definition of 'planet'. :) In reading up on this trying to find the correct Schwarzschild geodesic equations (and in the simplest form! :) ), I read several quotes of Einstein as he was developing all this.

Einstein was pefectly content with the view that "gravity" was a relative thing, its presence or absence entirely a coordinate issue/convention. An accelerating observer would be perfectly fine to say he was experiencing gravity.

And the rub is, that, when you merge space and time together, an acclerating coordinate system is no different from any other "curved" coordinate system where surfaces and curves of constant coordinate values are not straight lines or flat surfaces. Geodesics are straight lines in flat space-time, but those straight lines in a polar coordinate system do not correspond to linear coordinate expressions.

Einstein was perfectly fine with allowing an observer using those coordinates to say there was "gravity" present. It was completely a relative, coordinate dependent thing to him.

But now, I think most of those in the field define gravity to be "real curvature" of space-time, and would at best call that of an acclerating coordinate system "psuedo-gravity", but would not call a curved space coordinate system any sort of gravity at all.

But Einstein apparently was happy to say that if inertial objects didn't follow linear paths in your coordinate system, then you could call that gravity no matter what the source of that curvature. I don't think he would be such a stickler about "coordinate effects" vs "real effects". An observer sees what he sees, and that's fine and dandy.

-Richard

publius
2006-Sep-10, 12:23 AM
Ken,

I'm learnin' all sorts of cool stuff I didn't know 'bout Schwarzschild frames. For instance, the force on a stationary observer is given by,

F = GM/r^2 * 1/sqrt(1 - R/r), and r is the distant observer's radius, not local, as always here (and this is taken as positive since it must point up)

So this shows that the force is indeed infinite at the event horizon. And showing the "coordinate tricks" GR plays on us, this is not 'ma' in terms of our distant frame. It could be seen as an 'ma' in local coordinates defined appropriately, so the observer in the windowless room could think he was accelerating and not stationary in a g-field.

And I see you are getting hot and heavy into tidal forces. Want to add GR effects into this? <grin, duck and run>

The Schwarzschild metric has some very interesting symmetries that make for some rather special relations. The GR tidal forces felt by stationary observers are the same as Newton, even though the total force at a point is not. And the tidal forces felt by radial free-fallers are equal to this as well. This is not true for general mass distributions, however.

But, tidal forces for orbiting observers are different from Newton. I was reading about standard "frame fields" of the Schwarzschild metric, which are the standard local coordinate systems of interest. Stationary is one, radial free-falling is another, and then circular orbit frames are another one of particular interest.

The radial tidal force is larger than above, and the other two are not isotropic as with the above. The component in the direction of motion is the same as the above, but the component orthogonal is greater.

And finally, something else interesting about the circular orbit frame. Most of the time you want it, pun intended "tidally locked" with the origin so that one direction is always radial. This frame is therefore rotating -- this doesn't effect the tidal forces calculated (it is done so only "real gravity" comes into play). However, if you "gyrostabilize" this circular orbit frame so it is "not rotating" (defined as feeling no centrifugal forces roughly, making it inertial), you will find that frame is still rotating with respect to our far away stationary frame!

That is the "geodetic precession" effect, which Gravity Probe B will measure as well as gravitomagnetic/frame dragging precession.

-Richard

Ken G
2006-Sep-10, 02:21 AM
But Einstein apparently was happy to say that if inertial objects didn't follow linear paths in your coordinate system, then you could call that gravity no matter what the source of that curvature. I don't think he would be such a stickler about "coordinate effects" vs "real effects". An observer sees what he sees, and that's fine and dandy.


Interesting point, this means that the fact that an inertially moving object in polar coordinates always picks up r momentum at the expense of theta momentum means you have a radially outward gravity in that coordinate system, even if your test particle is the only thing in the universe. That's a very observer-centric view, to say the least! It also means that Tycho Brahe's model of the solar system (where the Earth is still) was equivalent to Copernicus', to Einstein.

Ken G
2006-Sep-10, 02:24 AM
That is the "geodetic precession" effect, which Gravity Probe B will measure as well as gravitomagnetic/frame dragging precession.


Thanks for exposing me to this, though I don't begin to understand it.

publius
2006-Sep-10, 05:44 AM
Thanks for exposing me to this, though I don't begin to understand it.


I don't understand it either. Basically it's just another instance of curved space being different from our Euclidean intuition. It is the same thing as "parallel" lines actually diverging or converging on curved surfaces. A common example to illustrate this is local orthogonal directions defined on the surface of a sphere, the earth's surface being the typical example. You can define a closed path involving going so far in one direction, then so far in the other, which IIRC starts at the equator then goes to the poles, then comes back down and over to the starting point.

When you get back to where you started, you'll find you are rotated 90 degrees from where you started. Follow that path for four complete circuits and you've rotated around one revolution. Now this effect from equator to pole is stark which allows you to easily see the effect. So as you move along your geodesics, you are actually rotating relative to your starting point. For small path lengths relative to the size of the sphere, the effect is still there, but very small.

And that's what happens with a circular orbit. Each time it goes around, the local axes are rotated relative to the starting directions.

For the earth, that effect is so small that it is testament to the engineering of GPB. Well, actually it is greater than the frame dragging precession IIRC, but both are incredibly small. The precision of the experiment is simply amazing.

I'm going to bet that GR comes out with flying colors. It will place stricter limits of alternate (more than GR, so to speak) theories of gravity that we have now.

-Richard

Ken G
2006-Sep-10, 02:40 PM
Yes, that's a nice description of "parallel transport", one of those crucial concepts GR types understand and just seems weird to the rest of us.

publius
2006-Sep-11, 12:37 AM
One more little cute thing about the Schwarzschild metric. For light itself, the v(r)/c expression is just:

(v(r)/c)^2 = (1 - R/r)^2. You can get that from the original form with the specific energy by noting the specific energy of light is infinite. :) Specific energy is equivalent to the ratio of relativistic mass to rest mass, and for light, the rest mass is zero. So this eliminates the additional R/r term on this.

Now, let's take the derivative of this and get the acceleration of light. Let x = R/r, and move the c^2 over to the right: d/dx = dr/dx*d/dr, and we'll take the derivative of both sides with respect to x, then mulitply the right side by dx/dr:

2v*dv/dr = 2 dv/dt = c^2[ -2 + 2x]*dx/dr.

dx/dr is just -R/r^2. Factor that in, and we have:

dv/dt = Rc^2/2r^2[ -2 + 2R/r ] Now R = 2GM/c^2, so this is just

dv/dt = -GM/r^2 * [ -2 + 2R/r ]

The acceleration of radial light is twice the Netwonian value (and in the opposite direction -- it is slowing down as it falls, not speeding up!)

So there is the famous factor of 2 on the acceleration of light. Here, for a radial "fall", it is in the opposite direction. But if we looked at the general case, with L nonzero and considered the theta coordinate equation, we'd find the factor of 2 exactly toward the center for light at infinity at a tangential initial path.

We will find that highy relativistic particles (high specific energy) follow similiar paths to light. As that Felber/Farber/Whatever guy found out and made such a big deal about (well, it was hard for me to find it out, but everything he found is right there in the Schwarzschild metric and no big deal, IMHO), a radial free faller moving at faster than Sqrt(1/3)c is going to slow down, not speed up (relative to our distant stationary frame).

Does this really mean a "repulsive" force? Not at all. Gravity is still going to pull relativisitic particles and light toward it, as can be easily seen in the tangential initial case. It's just that is you're alreading travelling toward the mass greater than a certain velocity, it's going to "brake" you a bit. It still "grabs you" and pulls your world lines toward it, but there is a velocity dependent braking effect as well.

So I hope that makes Jeff feel better about this slowing down business.

-Richard

Ken G
2006-Sep-11, 02:56 PM
The acceleration of radial light is twice the Netwonian value (and in the opposite direction -- it is slowing down as it falls, not speeding up!)But if I understand correctly, this is more a statement about the chosen reference frame than it is about light. Remind me again, what is the special value of this frame, that makes this statement carry additional importance? (Or is the absence of additional importance the point you are making, as below: )


Does this really mean a "repulsive" force? Not at all. Gravity is still going to pull relativisitic particles and light toward it, as can be easily seen in the tangential initial case.
Excellent point, you are saying that "repulsiveness" should have some kind of absolute quality (such as not depending on the direction of motion), and should not just be an arbitrary attribute of a particular reference frame. This appears to be what Felber/Farber/Whatever is missing when he oversells his conclusions. You could define repulsive as meaning that the inward speed is smaller than it would have been in the same coordinates but in the absence of any force, i.e., if the coordinates had somehow magically been globally inertial.

Cougar
2006-Sep-11, 03:09 PM
No doubt you are aware that in the frame of the pebble, it falls into the hole in short order....
I agree with this. So the answer to the OP question, "Can a falling object actually enter a black hole?" is essentially YES. The only oddity, as I understand it, as much of this thread has focused on, is viewing such an event from the outside. Such viewing is severely affected by the extreme gravity on the light that one is observing, as Publius has quantified.

Squashed
2006-Sep-11, 04:08 PM
... Such viewing is severely affected by the extreme gravity on the light that one is observing, as Publius has quantified.

We are not viewing the pebble, we are calculating its trajectory according to GR which seems to indicate that it never enters the blackhole. Viewing involves seeing the photons that travel between the pebble and the viewer - after a photon has reflected off of the pebble it has no effect whatsoever upon the continued path/trajectory of the pebble.

No one will ever see the pebble enter the blackhole and according to the GR calculation it never enters the blackhole but according to observations it does enter the blackhole: we have different sized blackholes as measured by their calculated Schwarzschild radii.

Jeff Root
2006-Sep-11, 04:48 PM
We are not viewing the pebble, we are calculating its trajectory
according to GR which seems to indicate that it never enters the
blackhole. Viewing involves seeing the photons that travel between
the pebble and the viewer - after a photon has reflected off of
the pebble it has no effect whatsoever upon the continued
path/trajectory of the pebble.

No one will ever see the pebble enter the blackhole and according
to the GR calculation it never enters the blackhole but according
to observations it does enter the blackhole: we have different sized
blackholes as measured by their calculated Schwarzschild radii.
This short passage has some of everything: Parts I agree with,
parts I think are wrong, parts that may be true but misleading,
parts that are ambiguous.

What do you mean by "calculating its trajectory according to GR"?
If the result of the calculation seems to indicate that the pebble
never enters the black hole, then you made a mistake somewhere.

What do you mean by "according to observations it does enter the
blackhole"? There are no observations of anything entering a BH.

-- Jeff, in Minneapolis

publius
2006-Sep-11, 04:50 PM
We are not viewing the pebble, we are calculating its trajectory according to GR which seems to indicate that it never enters the blackhole. Viewing involves seeing the photons that travel between the pebble and the viewer - after a photon has reflected off of the pebble it has no effect whatsoever upon the continued path/trajectory of the pebble.

No one will ever see the pebble enter the blackhole and according to the GR calculation it never enters the blackhole but according to observations it does enter the blackhole: we have different sized blackholes as measured by their calculated Schwarzschild radii.

Squashed,

All this is for a classic, static Schwarzschild black hole/point mass, and even more importantly, we are assuming the mass of the stuff falling in is negligible.

If the falling mass is not neglible, it changes things as it falls, and makes a dynamic, time-varying space-time. It is no longer a simple Schwarzschild black hole, but is some complicated, time varying thing.

What happens roughly is the event horizon expands as the matter falls in. As we're watching from our far away frame, we would see the event horizon expanding to meet the matter falling in, and that matter would appear to stop at a greater radial distance. And as the horizon expands, it pushes all the other "frozen stuff" with it.

In these time-varying metrics, there is something they call the "apparent horizon" which is different from the "real" event horizon somehow. In a static situation the two are the same, but in a dynamic one, the two will be different and only merge together after everything settles down. And that settling down may actually take infinite time, but they will be close after a sufficient time.

-Richard

Ken G
2006-Sep-11, 05:00 PM
But I think Squashed has a more fundamental problem, which is that he is trying to make absolute statements like "the pebble crosses the event horizon in a finite time". Statements like that, although they sound absolute, are actually relative to the reference frame. The most obvious example of this concept that comes up even in special relativity is a statement like "a supernova occured a million light years away shortly before I was born". If a passing alien, a million years from now, wishes to verify this statement, it is actually going to depend on how fast they are moving. There are certainly reference frames where the supernova has not even occurred by "now", the time we call 2006, from the point of view of that alien frame. Talking about when things happen depends on the choice of reference frame.

publius
2006-Sep-11, 05:10 PM
But if I understand correctly, this is more a statement about the chosen reference frame than it is about light. Remind me again, what is the special value of this frame, that makes this statement carry additional importance? (Or is the absence of additional importance the point you are making, as below: )

Excellent point, you are saying that "repulsiveness" should have some kind of absolute quality (such as not depending on the direction of motion), and should not just be an arbitrary attribute of a particular reference frame. This appears to be what Felber/Farber/Whatever is missing when he oversells his conclusions. You could define repulsive as meaning that the inward speed is smaller than it would have been in the same coordinates but in the absence of any force, i.e., if the coordinates had somehow magically been globally inertial.

Ken,

About the factor of two. In gravitational lensing discussions, it will often be noted that the trajectory of light (around a spherical mass distribution) has an acceleration of twice the Netwonian value, calculated by the trajectory curve. I was wondering if this would come out in any way for the simple radial trajectories I was considering. And it did, although "backwards", and I thought that was cute.

If pressed, I couldn't really say there was any special about this frame compared to others. However, I do consider it somewhat special as this is what we, distant observers watching stuff falling in, would see.

-Richard

Squashed
2006-Sep-11, 05:46 PM
But I think Squashed has a more fundamental problem, which is that he is trying to make absolute statements like "the pebble crosses the event horizon in a finite time". Statements like that, although they sound absolute, are actually relative to the reference frame. ....

There has been a reference frame attached to the earth since it came into existence - this reference frame may have varied due to conditions around the earth but basically the earth-based reference frame is. According to this earth-based reference frame, that I and everyone on the planet live in, a blackhole can never accrete matter because time stops at the event horizon and so it takes forever to enter the blackhole.

Jeff Root says: "There are no observations of anything entering a BH." which, if true, then all blackholes should have the same radius for the event horizon because it is impossible from the earth-based reference frame to see a blackhole deviate from its initial size.

I state that we have witnessed blackholes accrete because we know of blackhole candidates that are of various sizes.

Jeff Root
2006-Sep-11, 07:35 PM
Jeff Root says: "There are no observations of anything entering a
BH." which, if true, then all blackholes should have the same
radius for the event horizon because it is impossible from the
earth-based reference frame to see a blackhole deviate from its
initial size.
Ackkkkk!!!

Don't do that!

I said that there are no observations of anything entering a
black hole. I most certainly did not say that nothing enters
a black hole!

I have never observed anything entering my stomache. That does
not mean that nothing enters my stomache!



I state that we have witnessed blackholes accrete because we
know of blackhole candidates that are of various sizes.
Candidates, you say. They are candidates because they have
never been directly observed. We observe stars and gas orbiting
at speeds which indicate large concentrations of mass. We also
see gas ejected at enormous speed from some of these locations.
But the ejection of gas is believed to occur at a considerable
distance from the event horizon. Nothing has ever been seen to
actually get close to an event horizon, simply because of the
limits of our observing apparatus at such great distances.

I think the idea that objects even appear to slow down as they
approach the event horizon is wrong. While the passage of time
in those objects does appear to slow to a stop (the hands of a
clock would appear to stop as it reaches the event horizon, in
our view), the speed of the object away from us would never be
reduced or appear to be reduced.

Absurdly, I think those may be two separate things: The speed
away from us keeps increasing, but we can't see it, while the
apparently-increasing speed away from us is an illusion caused
by the distorted geometry near the event horizon.

-- Jeff, in Minneapolis

publius
2006-Sep-11, 08:02 PM
Jeff,

Let's look at it this way. I assume you agree that the stationary external observer (and this actually includes *all* such stationary observers, no matter how close they are to the horizon relative to a distant frame) does not see the falling object cross the horizon in finite time?

Now that means the position of the falling object must never become smaller than R in finite time. IOW, r -->R as t -> infinity. r(t) must asymptotically approach R. That curve, r(t), must start at the point r0 we drop it from. It can do a lot of things between R and r0, but it must eventually converge to R, never equal or smaller.

Now, the velocity is the derivative of r(t), dr/dt, and is the slope of the r(t) curve. As r --> R, that curve must flatten and it's slope must go to zero. The velocity must go to zero as r --> R, if the object is not to enter the horizon in finite time.

If the velocity does not go to zero, the r would have to drop below R, meaning the object did cross the horizon in finite time in our frame.

If that were the case, then the "freezing" would be a light-travel illusion. We would never see its image cross the horizon, but we could calculate that it did actually cross. That is not the case.

The Schwarzschild geodesics in the frame of the distant stationary observer are for the actual position, the "instantaneous position" of the object, not its light image. If we wanted to calculate when we would actually see it, when its photons would reach our eyes, we'd have to add the light travel time.

It is perfectly fine for us to say that objects never enter the hole. They don't in any stationary frame. We just see a bunch of mass frozen near the event horizon, gravitating like crazy, but it is not inside as far as we're concerned.

In its own frame, it does enter the hole and meet its fate at the singularity in finite time.

The problem here is our own minds see this as a contradiction. We want to *demand* that everyone agree the object falls into the hole because our common sense says that something either falls in or doesn't. But curved space-time does not require that. Simultaneity is warped beyond anything in SR, and "reciprocity" is completely out the window.

-Richard

publius
2006-Sep-11, 09:25 PM
Excellent point, you are saying that "repulsiveness" should have some kind of absolute quality (such as not depending on the direction of motion), and should not just be an arbitrary attribute of a particular reference frame. This appears to be what Felber/Farber/Whatever is missing when he oversells his conclusions. You could define repulsive as meaning that the inward speed is smaller than it would have been in the same coordinates but in the absence of any force, i.e., if the coordinates had somehow magically been globally inertial.
Ken,

Negative mass would be repulsive according to this, and would repel everything including other negative mass. I can't help but starting thinking about plugging in a negative point mass M in the Schwarzschild metric. If that is the kosher, and the minus sign doesn't change the solution and we can just plug "-M' into the metric, things are going to get very weird.

The factor will be (1 + R/r), and things are going to get very strange indeed. Things "dropped" at low initial velocity will be repelled. Light on a tangential path will be curved away. However, light will be speeding up, not slowing down, and the path of radially inward directed light is going to be really weird. Just as the attractive positive mass slows it down as it falls, the negative mass is going to *speed it up toward it*. This is just crazy of course. And that light will reach infinite speed at r = 0. I think it will act as some sort of crazy mirror, reflecting all that radial light back (and perhaps in finite time). And time will speed up reaching infinity at r = 0 as well.

So I think it would be "ultimately repulsive" for radial light, but it's going to pull it in then spit it back out in the radially outward direction. And a highly relativistic particle might well follow the same path, being pulled all the way in before it was spit back out. :lol:

This negative mass is quite different from the "white hole", I caution -- that's the time reversed black hole and quite different.

-Richard

Jeff Root
2006-Sep-11, 09:42 PM
Let's look at it this way. I assume you agree that the stationary
external observer (and this actually includes *all* such stationary
observers, no matter how close they are to the horizon relative to
a distant frame) does not see the falling object cross the horizon
in finite time?
Although the idea that the image freezes at the event horizon
is widespread, my understanding is that it will continue to appear
to accelerate away from the viewer until it disappears at the
event horizon.



Now that means the position of the falling object must never
become smaller than R in finite time.
Did you mean the apparent position as seen by a stationary
observer?



IOW, r -->R as t -> infinity. r(t) must asymptotically approach R.
That curve, r(t), must start at the point r0 we drop it from. It
can do a lot of things between R and r0, but it must eventually
converge to R, never equal or smaller.

Now, the velocity is the derivative of r(t), dr/dt, and is the
slope of the r(t) curve. As r --> R, that curve must flatten and
its slope must go to zero. The velocity must go to zero as r -->
R, if the object is not to enter the horizon in finite time.

If the velocity does not go to zero, the r would have to drop
below R, meaning the object did cross the horizon in finite time
in our frame.
All that is pretty much a tautology. You are saying that if
something approaches the event horizon but never reaches it,
then its speed must fall to zero before it reaches the event
horizon, and it never reaches the event horizon. Obviously.
I disagree with the premise.



If that were the case, then the "freezing" would be a light-travel
illusion. We would never see its image cross the horizon, but we
could calculate that it did actually cross. That is not the case.

The Schwarzschild geodesics in the frame of the distant stationary
observer are for the actual position, the "instantaneous position"
of the object, not its light image. If we wanted to calculate when
we would actually see it, when its photons would reach our eyes,
we'd have to add the light travel time.
I'm making bald assertions without backing them up, so you have
every right to complain when I ask you to support an assertion
of yours, but...

Where do you get that the Schwarschild calculations give the
"actual, instantaneous position" in the frame of the distant
stationary observer? That sounds suspicious.



It is perfectly fine for us to say that objects never enter the
hole. They don't in any stationary frame. We just see a bunch of
mass frozen near the event horizon, gravitating like crazy, but
it is not inside as far as we're concerned.
Can you spell out in words what you mean by "objects never enter
the hole [from the point of view of a stationary frame]? What
is the difference, from our point of view, between a thing being
inside and that thing being outside?



In its own frame, it does enter the hole and meet its fate at the
singularity in finite time.
I have said in this thread that a thing falling into a BH sees
the event horizon move away as it approaches the center. The
falling object will quickly reach the vicinity of the singularity
(I don't know what happens very close to the singularity),
but it will of course not cross its own event horizon.



The problem here is our own minds see this as a contradiction.
We want to *demand* that everyone agree the object falls into the
hole because our common sense says that something either falls in
or doesn't. But curved space-time does not require that.
My sense says that the object rapidly falls to the center, and
the object sees itself rapidly fall to the center; we see the
object apparently accelerate into the black hole at increasing
speed and with increasing redshift and at rapidly-increasing
distance until it disappears at our event horizon.

-- Jeff, in Minneapolis

publius
2006-Sep-11, 10:40 PM
Jeff,

Okay, so you disagree that a free-falling object never reaches the horizon in a stationary frame, and the freezing is therefore a trick of light travel time. And r(t) does indeed get less than R in finite time. Fair enough.

However, that is incorrect. :)

I do not mean "apparent position". I mean the position of the object as measured by the rulers of our distant stationary frame against the time defined by our clocks in that frame. r(t) --> R as t --> infinity,
and v(t) -->0 as r --> R.

I don't have the skill to derive the Schwarzschild metric from the Einstein Equation, nor go through solving the geodesic equations from that metric. I can only assure you that they are correct. I went to a lot of trouble digging this up and making sure I had the correct expressions.

The geodesic equations are derived in the coordinate system one is using and their solutions faithfully describe the equations of motions in terms of the rulers and clocks defined by those coordinates. These are inertial paths -- paths followed by objects with no "real" forces acting on them. The curve, in terms of velocity vs position for a particle dropped from infinity with zero initial velocity is simply:

(v/c)^2 = (1 - R/r)^2 * R/r. When r = R, that goes to zero. v(R) is zero. The local velocity takes out the (1-R/r)^2 factor, and you're amazingly left with the Newtonian result, v/c = Sqrt(R/r). That goes to c at R=r. But there is no stationary frame at R=r. Any stationary frame at r > R will see the free-falling object whiz by it at the local speed, but it will see it slow down and stop as well, no matter how close. The local rulers are getting very short at r near R, and local stationary observers think the horizon is a lot farther away than we would in our distant frame..

The maximum speed of the object dropped with zero initial velocity occurs at the photon sphere, r = 3/2R. The local free-fall speed there would be Sqrt(2/3)c. A stationary observer there would see the object whizzing by him at over 80%c, but he would see that thing start to slow down and stop. The distance it travelled would be a lot greater than 1/2R in his frame, though.

-Richard

Cougar
2006-Sep-11, 10:45 PM
Well, this seemed like a good question for mathematical physicist and GR guru John Baez, and the UCR.edu site What happens to you if you fall into a black hole? (http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/fall_in.html) seems to have our answers. One of the points made there is as follows:
So if you, watching from a safe distance, attempt to witness my fall into the hole, you'll see me fall more and more slowly as the light delay increases. You'll never see me actually get to the event horizon. My watch, to you, will tick more and more slowly, but will never reach the time that I see as I fall into the black hole. Notice that this is really an optical effect caused by the paths of the light rays.

publius
2006-Sep-11, 11:11 PM
Cougar,

I'm going to have disagree with that view expressed by this quote: "Notice that this is really an optical effect caused by the paths of the light rays." That is like saying the time dilation of SR is just a trick of the doppler shift of light. When you correct for the doppler effect, you find the frequency is still slower than the emitting observer thinks it is. And that is just by the time dilation factor.

The reason the "optical illusion" is so appealing is because common sense seems to demand that if something crosses the horizon in one frame, it must cross in all frames. And GR does not require that.

He says this in that same page:

At large distances t does approach the proper time of someone who is at rest with respect to the black hole. But there isn't any non-arbitrary sense in which you can call t at smaller r values "the proper time of a distant observer," since in general relativity there is no coordinate-independent way to say that two distant events are happening "at the same time." The proper time of any observer is only defined locally.

Now, t is the time coordinate of an observer at infinity, but the proper time of closer stationary observers is about the same where R/r is very small. The author is arguing that even though this time goes to infinity, it's not real, and only the proper time of the free-faller means anything. He is actually getting close to violating a cardinal rule of relativity, which is that there is no preferred frame. He is claiming the frame of the free-faller is the one that is "real", and not that of distant stationary observers.

That is not something I agree with at all. It does take infinite time in the frame of any stationary observer, whether close or distant, and that view is a good and correct as that of the free-falling frame itself.

-Richard

Squashed
2006-Sep-12, 12:54 PM
...The author is arguing that even though this time goes to infinity, it's not real, and only the proper time of the free-faller means anything. He is actually getting close to violating a cardinal rule of relativity, which is that there is no preferred frame. He is claiming the frame of the free-faller is the one that is "real", and not that of distant stationary observers.

That is not something I agree with at all. It does take infinite time in the frame of any stationary observer, whether close or distant, and that view is a good and correct as that of the free-falling frame itself.

-Richard

That is exactly what I have thought but was unable to say convincingly. To use the "illusion" description requires a real and not-real frame determination and that is just totally wrong. I have hated the "illusion" description but have never had the mathematical backing to prove that my dislike was justified, thank you publius.

But now back to blackhole candidates of different sizes; how can we explain them because either:

1.) The matter gets inside the event horizon or
2.) The matter stacks in shells on the outside of the event horizon or
3.) Blackholes never actually form or ...

Squashed
2006-Sep-12, 01:08 PM
A thought crossed my mind while reading another thread: If the matter is converted into electromagnetic radiation at the event horizon then reference frames no longer apply to the radiation and irregardless of which way the momentum of the radiation points it will ultimately spiral into the blackhole - because the event horizon is inside the photon sphere and so the "matter" actually enters but only after it is converted into radiation.

Ken G
2006-Sep-12, 02:23 PM
He is actually getting close to violating a cardinal rule of relativity, which is that there is no preferred frame. He is claiming the frame of the free-faller is the one that is "real", and not that of distant stationary observers.


It might be more fair to say that he is arguing that the very concept of a reference frame is a local concept. This is also what I meant in a few other threads when I pointed out the difference between a reference frame and a global coordinate system. Global coordinate systems are absolutely arbitrary, as long as you know how to transform them into local reference frames (i.e., frames that have the correct spacetime structure locally). All the global coordinate system is, then, is a means for bookkeeping the transformations from local frame to local frame as you consider different events. These transformations must obey some rules to go from a reference frame to a reference frame (i.e., frames that actual local observations will obey the laws of physics), and those rules are specified in a coordinate-independent way by general relativity, but which ones you choose are arbitrary.

I would explain the above dispute as arising from the fact that what publius is doing is choosing a particular coordinate system, and its applied allowed transformations, to make statements about what is happening. That is one way to do relativity, but it is an usual way to do it, because you are never actually in a frame where any observations can be made to test your result-- you will always need to transform to a local reference frame where actual observations get done to check your answer. Put succinctly, all clocks measure their own proper time. Baez is not even considering other coordinate systems, presumably because they are not directly usable for measurements. If you restrict to reference frames (i.e., to somebody's local proper time), and you want to make inferences about somebody else's local proper time, then you do have to distinguish between what is real and what is an optical illusion (and note that in special relativity, proper time and reference frames can be extended globally, but not so in general relativity-- reference frames are local). So Baez is simply talking about something different from what publius has done.

By the way, the more standard approach (used by Baez) is to make transformations that always transform to the particle's local rest frame. Then not only is one always in a reference frame where measurements can be made, but one is also using "proper time" (the "proper" means owned by the particle, not "correct"). When you do that, the words you get to describe what is happening are totally different from the approach of publius. This is what I call pedagogy, and one should not expect the pedagogies arising from two different global coordinate systems to sound the same-- heck, they don't even sound the same between two different local reference frames.

Ken G
2006-Sep-12, 02:27 PM
If the matter is converted into electromagnetic radiation at the event horizon then reference frames no longer apply to the radiation and irregardless of which way the momentum of the radiation points it will ultimately spiral into the blackhole - because the event horizon is inside the photon sphere and so the "matter" actually enters but only after it is converted into radiation.

This is an interesting speculation, but note the matter is generally not turned into electromagnetic radiation as it falls into a black hole (just ask someone who has...!). However, you may be on to a similar idea-- there are ways of looking at this (and I can't put this on a rigorous foundation) in which the rest mass of the particle reaches zero at the event horizon. In such a system (which is not a reference frame, see my above comment), the particle is not radiation, yet does have zero rest mass so acts like radiation. So I'm not sure, but there may be some value in saying that particles can cross an event horizon as their rest mass reaches zero. (Note it will not be zero in their own frame, of course, that's how you can tell they are not converted to radiation-- they still have a rest frame).

Squashed
2006-Sep-12, 02:57 PM
This is an interesting speculation, but note the matter is generally not turned into electromagnetic radiation as it falls into a black hole (just ask someone who has...!). However, you may be on to a similar idea-- there are ways of looking at this (and I can't put this on a rigorous foundation) in which the rest mass of the particle reaches zero at the event horizon. In such a system (which is not a reference frame, see my above comment), the particle is not radiation, yet does have zero rest mass so acts like radiation. So I'm not sure, but there may be some value in saying that particles can cross an event horizon as their rest mass reaches zero. (Note it will not be zero in their own frame, of course, that's how you can tell they are not converted to radiation-- they still have a rest frame).

One of the ways of identifying a blackhole is by the x-rays that matter "screams" when it spirals into a blackhole - this fact indicates to me that a change is taking place: from matter to radiation.

As far as rest frames goes it should be annihilated because when we destroy a small amount of matter in a nuclear explosion the rest frame of that matter is destroyed when it is turned into radiation and so the same would hold true for matter that is converted into radiation upon entering a blackhole.

publius
2006-Sep-12, 04:44 PM
I think it should be noted that a Matt McIrvin wrote the FAQ Cougar posted not Baez himself. I just noticed this.


-Richard

publius
2006-Sep-12, 04:56 PM
This is an interesting speculation, but note the matter is generally not turned into electromagnetic radiation as it falls into a black hole (just ask someone who has...!). However, you may be on to a similar idea-- there are ways of looking at this (and I can't put this on a rigorous foundation) in which the rest mass of the particle reaches zero at the event horizon. In such a system (which is not a reference frame, see my above comment), the particle is not radiation, yet does have zero rest mass so acts like radiation. So I'm not sure, but there may be some value in saying that particles can cross an event horizon as their rest mass reaches zero. (Note it will not be zero in their own frame, of course, that's how you can tell they are not converted to radiation-- they still have a rest frame).

Ken,

This may actually be close to what the ECO/MECO theory says happens, actually. Wiki has a pretty good article on the theory:

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

Abhas Mitra is the Indian physcist who first proposed this

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

An ECO would pretty much look like a black hole from afar, but things would be very different "inside". It becomes a radiation dominated "soup" as it collapses and indeed may "burn" much matter into radiation.

I'm beginning to like this ECO theory, and wouldn't be surprised if black holes "evaporate" as possible real objects. :)

-Richard

publius
2006-Sep-12, 05:42 PM
Ken,

I certainly agree with what you say generally. Except in this case, we can never be in the local frame past the horizon unless we wish to share its fate. :) Past the horizon, there is no return, and the two "twins" cannot get back together to compare their clocks or whatever.

The events, the space-time coordinates of the local frame once it crosses the horizon do not transform to anything but "nonsense" (infinite time or even imaginary time or other nonsense past the horizon) in the external stationary frames. We can say the internal events do not exist in the accessible space-time of any observer who is not himself going into the hole.

So for us outside, the only meaningful thing is our own reference frames, and in those frames it takes infinite time for the object to cross the horizon.

I see this question of whether an object "really falls in" has been debated endlessly on other forums as well. :lol: Someone (on my side) made an argument I like.

He posed the question: Is there any time on our (external stationary) clock at which we can be sure the free-faller has crossed the horizon and can never come back? The answer is no. We'll say the faller crosses the horizon at 12:00. Until we see his clock tick 12:00, we can never be sure he didn't fire a powerful rocket and stop his fall at the last second and turn around. The closer his clock gets to 12, the force required would increase without limit, but remain finite until exactly 12. He could save himself 1000 years after he effectively "froze" in our frame, and only a twinkling of an eye passed in his.

So, there is no time on our clock in which can conclude the faller is gone forever.

-Richard

publius
2006-Sep-12, 05:59 PM
That is exactly what I have thought but was unable to say convincingly. To use the "illusion" description requires a real and not-real frame determination and that is just totally wrong. I have hated the "illusion" description but have never had the mathematical backing to prove that my dislike was justified, thank you publius.

But now back to blackhole candidates of different sizes; how can we explain them because either:

1.) The matter gets inside the event horizon or
2.) The matter stacks in shells on the outside of the event horizon or
3.) Blackholes never actually form or ...

Squashed,

The "growing" is not really a problem, as I mentioned in another post. Remember, here we're talking about free-fallers of negligible mass that do not change space-time as they fall in. When the mass in not negligible, the Schwarzschild metric no longer applies and we have a dynamic space-time that is changing with time.

Roughly what we see in our stationary frame is the event horizon (and the all the "frozen stuff" near it) expanding to meet the mass falling it. R, the Schwarzschild radius, expands and both meet in the middle so to speak.

Externally, the gravity of a spherical shell is the same as of a point mass so it doesn't matter.

Now, according the ECO theory, classical black holes can never form in finite *proper time* in their own frame. I'm almost a convert to the ECO theory now.

-Richard

Ken G
2006-Sep-12, 06:33 PM
So for us outside, the only meaningful thing is our own reference frames, and in those frames it takes infinite time for the object to cross the horizon.
Well, I think one could argue that in fact the only thing meaningful for us is even less than that-- it is our own local reference frame, since that's the only frame we can use to make any measurements. All other coordinate system that extend to include distant objects (like those falling into black holes) are then completely arbitrary-- they only need to give the right results when the light reaches our frame. So there is really no practical difference, at the end of the day, between what is real and what is an illusion.


Is there any time on our (external stationary) clock at which we can be sure the free-faller has crossed the horizon and can never come back? The answer is no. We'll say the faller crosses the horizon at 12:00. Until we see his clock tick 12:00, we can never be sure he didn't fire a powerful rocket and stop his fall at the last second and turn around. The closer his clock gets to 12, the force required would increase without limit, but remain finite until exactly 12.

But this leads to a paradox, because the black hole is not in a steady state over an infinite amount of time. What if, when the universe gets to age 20 billion years, a giant hand swoops down and causes everything to vanish everywhere. Where will the rocket be when that happens, for the rocket crew? Where will they be for us? Can that be different? It comes down to the issue of using our time for them is just a bad idea. We can calculate exactly when we would need to release a light signal that says "good luck" and reaches the rocket just when it crosses the event horizon (you can probably do this now!). If you were the one sending out this signal, would you really think they never actually get to the event horizon, even though you know your friends will receive your wish?

Jeff Root
2006-Sep-12, 07:29 PM
We can calculate exactly when we would need to release a light
signal that says "good luck" and reaches the rocket just when it
crosses the event horizon.
I believe we had this discussion before, just a few weeks ago,
but the people aboard the rocket could still receive a signal
from us after they have crossed our horizon. However, it will
stretch out more and more as it falls in after them, so the
length of the message has to be limited if they are to receive
it all before they are spaghettified.

The opposite of what you suggested is more clear-cut. We can
calculate exactly when those aboard the rocket need to send a
signal that can reach us. The end of the message would be just
before they cross our event horizon.

"My God, it's full of starrrrrrssssssssssssssssssssssssssssss!"

-- Jeff, in Minneapolis

astromark
2006-Sep-12, 07:39 PM
Without all this posturing and scientific dribble, The answer is yes.
You can fly into a black hole.
BUT... This is a very bad idea. You would be torn apart, radiated, crushed, spaghetti fide. stretched. and very dead well before you actually got to it.
Maybe you could use its enormous gravity to sling shot you away.
I am not coming, It sounds very dangerous

publius
2006-Sep-12, 07:42 PM
Well, I think one could argue that in fact the only thing meaningful for us is even less than that-- it is our own local reference frame, since that's the only frame we can use to make any measurements. All other coordinate system that extend to include distant objects (like those falling into black holes) are then completely arbitrary-- they only need to give the right results when the light reaches our frame. So there is really no practical difference, at the end of the day, between what is real and what is an illusion.



But this leads to a paradox, because the black hole is not in a steady state over an infinite amount of time. What if, when the universe gets to age 20 billion years, a giant hand swoops down and causes everything to vanish everywhere. Where will the rocket be when that happens, for the rocket crew? Where will they be for us? Can that be different? It comes down to the issue of using our time for them is just a bad idea. We can calculate exactly when we would need to release a light signal that says "good luck" and reaches the rocket just when it crosses the event horizon (you can probably do this now!). If you were the one sending out this signal, would you really think they never actually get to the event horizon, even though you know your friends will receive your wish?

Ken,

Ah, let's take the second one first. Remember the free-faller is going to see us accelerating away from him, and our clocks slow and our light redshifted (a stationary observer far down would see fast clocks and high blue shift).

We can calculate the time by our clock to send a signal that will reach him just as he crosses the horizon. But guess what? It will take infinite time, by our clock for that signal to reach him! If he fires a rocket and saves himself, he will receive that signal in our frame "before infinity", since he stops his fall. If we plot the course of that last bon voyage message chasing him down the hole, we'll see that signal itself slow to a crawl near the horizon, and essentially see two little radially compressed ants crawling at incredibly slow speed. The rear ant is travelling such a tad faster, and they will meet at the horizon after infinite time has passed. :)

Now, a signal sent just a tad after could reach the free faller in his frame *after* he crossed but before he hit the singularity. But that would take "more than infinite" time in our frame.

Now, about the hand swooping down and wiping everything out. If the hand appears near us, roughly in our frame, then any causal influences of that hand cannot travel faster down in the hole that our light signal, and will therefore take infinite time for that hand to have any effect on stuff down near the horizon.

If we say this is "magic wand" hand of some sort, then we're talking something outside of space-time as we know it.......................

-Richard

Cougar
2006-Sep-12, 07:51 PM
I think it should be noted that a Matt McIrvin wrote the FAQ Cougar posted not Baez himself. I just noticed this.
Right. I didn't intend to mislead. Dr. McIrvin is a fellow moderator at sci.physics.research with Baez, and with a Ph.D. in particle physics from Harvard, he's no slouch.

publius
2006-Sep-12, 09:34 PM
Right. I didn't intend to mislead. Dr. McIrvin is a fellow moderator at sci.physics.research with Baez, and with a Ph.D. in particle physics from Harvard, he's no slouch.

I didn't figure he was a slouch, just wanted to make clear it was not Baez himself who wrote it.

-Richard

Ken G
2006-Sep-13, 12:09 AM
We can calculate the time by our clock to send a signal that will reach him just as he crosses the horizon. But guess what? It will take infinite time, by our clock for that signal to reach him!
But this is the point, what is meant by "by our clock?" Our clock isn't there. Since we know he'll receive the signal in short order, this device is intended to show why what our clock reads is not terribly relevant. Not wrong, just a weird way to define time for him. It will never be the result of a local measurement, in any frame.

RussT
2006-Sep-13, 12:29 AM
But this is the point, what is meant by "by our clock?" Our clock isn't there. Since we know he'll receive the signal in short order, this device is intended to show why what our clock reads is not terribly relevant. Not wrong, just a weird way to define time for him. It will never be the result of a local measurement, in any frame.

And wouldn't this be why it is actually 'irrelevent' to say that 'new' black holes cannot form and also the same problem with saying that all matter must be frozen in time at the event horizon?

Ken G
2006-Sep-13, 03:00 AM
Irrelevant is probably too strong a word, but I would say unnecessary, yes.

publius
2006-Sep-13, 03:26 AM
But this is the point, what is meant by "by our clock?" Our clock isn't there. Since we know he'll receive the signal in short order, this device is intended to show why what our clock reads is not terribly relevant. Not wrong, just a weird way to define time for him. It will never be the result of a local measurement, in any frame.

Ken,

When I first read this reply, it didn't make sense to me. Well, not that I didn't understand your meaning, but it didn't make sense why you would say that. Then I realized that you (and perhaps Jeff and others) are thinking about this in an entirely different way. Your favorite word, pedagogy, maybe, but a little different.

I'm not trying to impose my distant clock on the free faller, and trying to make him use my time. I'm using my time (and my yardstick) *for me*. What I see, what events are simultaneous with my clock ticks *for me*.

If I were the free-faller (and I would NOT be one going inside the horizon :), but I might free-fall on other trajectories where it wasn't one-way), then I would be using my local clock.

So we are apparently thinking about "falling in the black hole" (or collapsing to a black hole) in entirely different ways. You (at least I think) naturally consider that question is best answered by being there with the free faller. So the answer is, we cross the horizon and on into oblivion in short order, and we don't worry about the outside world.

However, me, I'm thinking about that question as the spectator. So to me, it seems like some are trying to impose a distant, ridiculous clock on me. :) And that's how I think of sending the message. I watch that light signal slowly chase after the free-faller, never getting there in finite time.

And there's something else I'll point out. Let's the free-faller DOES bail out at the last second (or micro-s) (on his clock, of course). We'll assume he has Star Trek inertial dampeners and some enormous source of energy so he can apply enormous thrust to stop on a dime and return without crushing himself and his ship, but otherwise is constrained by space-time as always.

If he does that, and comes back where we can compare our clocks, well, my clock is going to be ages ahead of his. I'll be long dead before he gets back. And long dead before I would even see him fire the thrusters to stop. The point of no return does not happen in finite time on my clock.

And this is why I'm so darned adamant that he doesn't cross the horizon in finite time in my frame, nor any other far away frame.

ETA: The energy budget for that bail out at the lest microsecond is going to be enormous. Roughly, when he's near the horizon, almost all of his rest energy is kinetic, which he will have shed, and then he'll need that much more to climb back out of the well. So he's going to need at least 2mc^2 worth of energy to thrust with, where m is the mass that returns. :)

So apparently, the black hole gets twice the energy if he escapes than it would if he didn't.

-Richard

Ken G
2006-Sep-13, 05:23 AM
So to me, it seems like some are trying to impose a distant, ridiculous clock on me.
To me, the core of relativity has two pieces. One is the laws of physics and the measurement conventions we use to define those laws. Note that this is entirely local-- a clock measures local time, a ruler local distances, in their own frame. So the laws of physics are measureable in proper time and proper distance and in no other way. In effect, the measuring apparatus is the observer. But observers can communicate, so it is necessary for the laws to also obey transformations between reference frames, and that's the second tier of relativity. So you have laws, and transformations, and you tack on the additional requirement that the laws look the same when you transform them, and you are done-- that's relativity. Anything else is not relativity, it is a coordinate exercise. The physics is in the local reference frames, where distance and time have physical meaning, and these can be transformed into any arbitrary coordinate system, the coordinates don't have to carry any connection with the physical concepts of "distance" and "time", they can be any arbitrary combination.

I claim that what you are doing is not exporting a physical sense of time here on Earth to elsewhere in the universe, you are simply choosing a completely arbitrary coordinate system, and using words to describe the value of the coordinates you encounter. Locally, your coordinates do indeed correspond to our sense of time and distance, but that is a local comparison, like saying north,south, east, and west is locally what we mean by distance. Globally, however, that coordinate system does not give a normal sense of distance, you need to know the radius of the Earth to make any connection to a physical distance. Yet the coordinates still take on numerical values. So the issue is, on what basis can you claim that what you are using for your global measure of time really corresponds to "our sense of time"? I think you can only say that on the basis that it locally corresponds with our sense of time, but so would an infinite number of other global coordinate systems.

I'm saying there is no pedagogically unique way to extend local time into a global version of that time, other than the obviously wrong way to do it of "it happens when I see it". We both know that's the "optical illusion" approach, but at least it's unique. What other approaches are available in general relativity? This can all be compressed into a single question-- in the sense of time you are using, how do you define the simultaneity of two events? In special relativity, it is when light signals reach the midpoint at the same time. Where is the midpoint between you and the guy at the event horizon, and what event on Earth is simultaneous with his crossing that horizon? You are taking the distance from you to where the gravitational distortion just begins, and not including any of that distortion. That's the picture that has light slow down. But what's wrong with saying light doesn't slow down but there's a huge amount of distance piled into that last little bit of gravitational distortion? That's another picture, more akin to using proper distances, and in that view the midpoint is right close to the event horizon. I claim these are both completely allowable pedagogies, coupled to their own coordinate systems, and their own meaning of "distance" and "time", and they both intersect in the same way with our local sense of these terms. So how can one be said to be the "right way" to extend our local time over there?


I watch that light signal slowly chase after the free-faller, never getting there in finite time.
Yet, you can calculate exactly when they will receive that message, in their own time. So you encounter a coordinate singularity, not a real one-- your time coordinate goes infinite, yet they receive the message. The infiniteness of your time coordinate is not saying anything physical, it is only a mathematical entity.

RussT
2006-Sep-14, 01:18 AM
So we are apparently thinking about "falling in the black hole" (or collapsing to a black hole) in entirely different ways. You (at least I think) naturally consider that question is best answered by being there with the free faller. So the answer is, we cross the horizon and on into oblivion in short order, and we don't worry about the outside world.

However, me, I'm thinking about that question as the spectator. So to me, it seems like some are trying to impose a distant, ridiculous clock on me. And that's how I think of sending the message. I watch that light signal slowly chase after the free-faller, never getting there in finite time.

Let me try this once more.

Here is the problem as I see it.

You and Ken are both talking about this as though we can see light in a straight line all the way from the event horizon to us.

I don't think this can even happen, because as far as I can see 'all' black holes rotate, and as Jeff Root said somewhere far above, there is a point, in the accretion disc or the photon-sphere, or ergo-sphere, where if the light trajectory is slightly inward at wont come back out, but even the light that is slightly outward, ceratainly wont be coming at us in a straight line, and from this scenario could never be coming from the Event Horizon.

http://filer.case.edu/~sjr16/stars_blackhole.html

I just think that we need to consider all black holes (at least massive ones) from this perspective (even though I know the math is tremendously more difficult).

publius
2006-Sep-14, 02:40 AM
Russ,

We were using the classic, Schwarzschild "massenpunkt" black hole, a non-rotating, stationary point mass, because that is the simplest one. :) And I further simplified things by only considering radial trajectories, zero angular velocity. And this was good enough to serve the purposes of this debate, which was about the meaning of coordinate systems, global vs local. That applies to rotating black holes as well, but things are far more complicated.

Radial light paths are possible with the Schwarzschild BH. The photon sphere has to do with "orbiting light". That is where the circular orbit speed (in local coordinates) is just the speed of light. Light transmitted radially from any point r > R travels in a radial straight line, straight in or out.

A rotating, or "Kerr" black hole is a much, much more complex animal. The ergosphere is not the same as the photon sphere above, but a Kerr Black hole does have a photon sphere. Two of them, in fact. One for orbits in the same direction as the black hole spin, and the other for the opposite direction.

The ergosphere, which is actually an ellipsoidal shape, is the region below which stationary observers are not possible, but moving observers are possible that do not further fall radially. This is the same effect as with a rotating coordinate system, where no stationary observer is possible outside the limit where v = rw > c for an inertial observer.

And the Kerr BH has two event horizons as well. The outer horizon is similiar to the Schw. horizon, where no radially stationary observer is possible. Between here and the ergosphere, one can be radially stationary, but one must have a minimum angular velocity, which increases as you approach the outer horizon.

The region between inner and outer horizons is about the weirdest thing you've ever heard of. It is said you could play 4-player bridge with yourself in this region, as there exist timelike curves that take you to the past and future of yourself at all sorts of different points. It would be very weird to say the least.

The inner horizon is the next boundary to the "protected" ring singularity. Unlike a Schw. singularity, this one is different and you supposedly can pass right through the center of the ring to, um, "Elsewhere". Elsewhere can include another universe, a part of the existing universe, or "negative space", whatever that would be.

That is, if such a thing exists. The interior structure is thought to very unstable to the slightest perturbation.

And like I said to Ken, the ECO guys are winning me to their side, and if they're right, none of these black holes exist anyway so we don't have to worry with any of this.

-Richard

RussT
2006-Sep-14, 08:35 AM
[And like I said to Ken, the ECO guys are winning me to their side,]

Publius, Please stay on the Dark Side!:D

I know Kerr Massive Black Holes are a bear to work with, but I have even more confidence than mainstream does that GR is right, which means that ONE of the things current thinking has correct, is that MASSIVE BLACK HOLES do exist in the cores of galaxies, and that they are in fact the whole answer to the universe as a whole. In fact, I believe that if Einstein would have stayed on the macro and knew that SMBH's and String/"M" theory existed, he would have already figured this whole thing out!:clap:




And the Kerr BH has two event horizons as well. The outer horizon is similiar to the Schw. horizon, where no radially stationary observer is possible. Between here and the ergosphere, one can be radially stationary, but one must have a minimum angular velocity, which increases as you approach the outer horizon.

Where does the accretion disc come in here? I didn't particularly care for his picture representation in the above link, because it was very difficult to put the different depths into perspective, although I am sure this is not an easy thing to pictorialize!
In my understanding it goes from outside in as ergo-sphere/accretion disc/double event horizon/ring singularity/worm hole/white hole, is that your understanding as well?



The inner horizon is the next boundary to the "protected" ring singularity. Unlike a Schw. singularity, this one is different and you supposedly can pass right through the center of the ring to, um, "Elsewhere".

This is my understanding as well.

My question to you is, what size is the ring singularity?

Once you answer the question directly above this line, then maybe we can have a conversation about this...[That is, if such a thing exists. The interior structure is thought to very unstable to the slightest perturbation.]

Thanks

bachnga
2006-Sep-14, 10:54 AM
Hi gang
I'm pleased my original question produced such a flurry of intelligent well thought out replies. Since they start from different assumptions, obviously they're not all correct, but they are carefully thought out.

My point in this note is that in the real world you can make a good argument that the falling particle does go into the hole. Suppose, using the arguments given, the particle is a millimeter above the surface of a perfect hole event horizon. Maybe it stops due to time dilation or may it plunges in but we can't see it. Suppose the hole is of near solar masses and has a ten km diameter.

Let other matter continue to fall into the black hole. Suppose a cloud of dust of mass equal to a millioneth of the hole's mass falls toward the hole. If it actually does enter the event horizon, the mass of the hole is now one one millioneth (0.0001%) greater and the the Schwarzchild radius is 10 mm larger the 10 km. Perhaps for a while there woulld be an ill defined radius as it vibrates due to the sudden change in mass, but eventually when it settles down, the radius would have increased by 10 mm. Surely this expansion would engulf the original test particle which was less than a mm from the horizon. I can't believe the expanding horizon would push the first particle outwards. Thus, in the real world, where matter falls into the hole from the outside faster than energy is lost by Hawking radiation, the radius increases with time, and even if falling particles seem to stop at the surface of a static hole, they'd eventually be engulfed as the hole expands and the particle would be inside.

Thus, I ask, even if a falling object really does slow down as it approaches, so what? If the object doesn't punch through the horizon, the expanding horizon moves through and above the particle and the particle is inside. With other matter constantly and irregularrly falling, the event horizon might be non-spherical and jiggling like Santa's belly, emmitting graavty waves to become spheriical, and the particle may have a turbulent fall once inside, but it would be inside.

Any comments?

Thanks, Bachnga

Squashed
2006-Sep-14, 01:01 PM
...My point in this note is that in the real world you can make a good argument that the falling particle does go into the hole. ...

Observations declare that the particle enters the blackhole because there are blackhole candidates of different sizes ... the real conundrum is reconciling the GR calculations with observation.

We have mostly been addressing the fact that GR calculates for time to cease at the event horizon but GR also says that time moves backwards inside the event horizon and so if time goes backward then the particle should move backward in time - to a point outside the event horizon.

If, on the other hand, the matter is converted into electromagnetic radiation prior to entering the event horizon then these time calculations become meaningless because EM radiation (or light) has no reference frame and so the reference frame-ology becomes mute at the event horizon.

Ken G
2006-Sep-14, 01:24 PM
Observations declare that the particle enters the blackhole because there are blackhole candidates of different sizes ... the real conundrum is reconciling the GR calculations with observation.


Observations never catch anything entering the hole, every hole has an age and therefore a maximum time it could take us to see anything approaching the event horizon. Thus there's an observational "closest approach" for every black hole. The light we see is from the accretion disk, from places farther out than the "innermost stable orbit".

Squashed
2006-Sep-14, 01:31 PM
Observations never catch anything entering the hole, every hole has an age and therefore a maximum time it could take us to see anything approaching the event horizon. Thus there's an observational "closest approach" for every black hole. The light we see is from the accretion disk, from places farther out than the "innermost stable orbit".

What I meant by my statement is not that we actually observe the particles enter but we definitely see the aftermath of particles that entered: greater size/mass.

Although technically if the matter accumulated into a shell at the event horizon the gravitational effects would be similar; but there would be a difference because instead of originating from a point the gravity would originate from a larger "shell" mass - which is a more diffuse source of gravity.

Jeff Root
2006-Sep-14, 07:33 PM
as Jeff Root said somewhere far above, there is a point, in the
accretion disc or the photon-sphere, or ergo-sphere, where if the
light trajectory is slightly inward it wont come back out, but
even the light that is slightly outward, certainly wont be coming
at us in a straight line, and from this scenario could never be
coming from the Event Horizon.
I prefer to discuss non-rotating black holes, since I think I
understand them fairly well. :D

Using the terms on the page you linked, the inner event horizon
is where light moving vertically away from a black hole can just
barely escape.

The outer event horizon is the same as the photon sphere, where
light traveling horizontally could theoretically orbit. Anything
on a trajectory from outside the photon sphere to inside never
escapes, including light. But light emitted between the photon
sphere and the inner horizon can escape if it is moving outward
at a high enough angle.

As the page you linked says, "This marks the boundary at which
the escape velocity is greater than the speed of light".

This contradicts what publius suggested about escaping
from just above the inner event horizon with a rocket. A rocket
could only escape if it stays outside the photon sphere.

-- Jeff, in Minneapolis

Jeff Root
2006-Sep-14, 07:34 PM
The events, the space-time coordinates of the local frame once it
crosses the horizon do not transform to anything but "nonsense"
(infinite time or even imaginary time or other nonsense past the
horizon) in the external stationary frames. We can say the internal
events do not exist in the accessible space-time of any observer
who is not himself going into the hole.
As Ken said, Someone falling into a black hole can still receive
a message from us after they have crossed our horizon. We can
calculate exactly when they will receive it, according to their
clock, and when we need to send the message in order for it to
reach them, according to our clock. We can also calculate how
red-shifted the signal will be when they receive it.

Likewise, we can calculate exactly when we will recieve a
message from them, according to our clock, sent before they
have crossed our event horizon, if we know when they sent it
according to their clock, or, inversely, we can calculate when
they sent a message from the time we get it. We can calculate
exactly when they will cross our event horizon according to
their clock, and when we will see them cross our event horizon
according to our clock.



So for us outside, the only meaningful thing is our own reference
frames, and in those frames it takes infinite time for the object
to cross the horizon.
It should take infinite time for the object to reach the
singularity (meaning that it would never do so), because the
distance to the center is always increasing as the mass at the
center continues to collapse toward infinite density. All the
matter falling into the singularity gets pulled apart in the
radial direction, so whatever particles might be able to exist
there become more and more separated from each other. I don't
know how fast the singularity deepens, but it isn't limited by
the speed of light. Everything falling into a BH is already
falling at the speed of light relative to us as it crosses the
inner event horizon, and keeps on accelerating, although local
measurements still give the usual value of c.

So there is an infinite time involved, but it is the time to
reach the center, not the horizon.

I agree with the description on the web page by Matt McIrvin,
taken from Misner, Thorne, and Wheeler:


As an example, take the eight-solar-mass black hole I mentioned
before. If you start timing from the moment you see the object
half a Schwarzschild radius away from the event horizon, the light
will dim exponentially from that point on with a characteristic
time of about 0.2 milliseconds, and the time of the last photon
is about a hundredth of a second later. The times scale
proportionally to the mass of the black hole.
That is what we would see. What would actually be happening
is that the object would be moving away from us at the speed of
light, and still accelerating.



This, translated into embedding diagrams, was the insight
that came from Finkelstein's new reference frame. With this
way of thinking about the implosion, there was no more mystery.
An imploding star really does shrink through the critical
circumference without hesitation. That it appears to freeze
as seen from far away is an illusion.



I see this question of whether an object "really falls in" has
been debated endlessly on other forums as well. Someone (on my
side) made an argument I like.

He posed the question: Is there any time on our (external
stationary) clock at which we can be sure the free-faller has
crossed the horizon and can never come back? The answer is no.
We'll say the faller crosses the horizon at 12:00. Until we see
his clock tick 12:00, we can never be sure he didn't fire a
powerful rocket and stop his fall at the last second and turn
around.
Except that that isn't possible inside the photon sphere, or
outer event horizon.



The closer his clock gets to 12, the force required would
increase without limit, but remain finite until exactly 12.
He could save himself 1000 years after he effectively "froze"
in our frame, and only a twinkling of an eye passed in his.
Sorry, that is complete garbage. If someone approaches the
outer event horizon very closely in a rocket, he will have to
be in a hyperbolic orbit AND firing his super-rocket engines in
order to get away again. That means he will dip down close to
the photon sphere at nearly the speed of light and immediately
rise away again. His trip will be over in a few minutes by our
clock, and maybe half that by his clock.

The arguement is pointless. If he is on a trajectory straight
toward the black hole and has no rocket, then we can calculate
the exact time he crosses our event horizon, both by his clock
and ours.



So, there is no time on our clock in which can conclude the
faller is gone forever.
Our knowledge of events near the black hole is delayed or made
completely impossible by the light travel time and redshift.
That doesn't imply that those events never happen in our time.

-- Jeff, in Minneapolis

publius
2006-Sep-14, 09:16 PM
Jeff,

No what I said is not garbage, and I take a bit of umbrage at that. You're speaking of outer and inner horizons which are properties of rotating black holes, not the simple static Schwarzschild hole I'm considering here.

Any observer who doesn't cross the event horizon in his own frame can get back theoretically, although the force will increase without limit as he approaches. The force on a stationary observer, Sch. metric, is simply GM/r^2 *1/sqrt(1 - R/r). It is larger than Newtonian as r gets close to R, but only infinite at the horizon, r = R. Stationary frames exist all the way down to the horizon in the Sch. metric.

It does take infinite time in a stationary frame for an object to reach r = R. Oh, and Jeff, that message sent from the free-faller just as he crosses the horizon will take infinite time in our frame to reach us.

The "last photon" has to do with quantum considerations which limit how much that last photon can be stretched out.

EDIT: No sense in my getting all bent of shape about this, as it's no big deal....


Your understanding of the photon sphere is also incorrect as that has do with inertial fallers/orbiters, those without forces applied, following geodesics, without forces applied. If such a geodesic crosses the photon sphere, it goes into the hole. But it is entirely possible to apply force and stop the fall.

THe Finkelstein coordinates are a different way of looking at things, and have nothing to do with the time coordinate of a far away stationary frame. That is, they are not measuring the proper time of a stationary observer.

End EDIT.


-Richard

RussT
2006-Sep-14, 09:26 PM
I prefer to discuss non-rotating black holes, since I think I
understand them fairly well.

Yes, Publius, I was Just going to cover exactly what you have correctly dealt with above.

Jeff made this quote above and then went 'apparently', sometimes talking about 'rotating' and then 'non-rotating' scenarios...everyone needs to be very careful to include which one they are talking about.

Also, I would certainly like for someone to show me an example of where we have actually found a NON-ROTATING black hole, either stellar or massive!

Jeff Root
2006-Sep-14, 11:26 PM
No what I said is not garbage,
That one paragraph was.



and I take a bit of umbrage at that. You're speaking of outer
and inner horizons which are properties of rotating black holes,
not the simple static Schwarzschild hole I'm considering here.

The term "outer event horizon" was one that I was not familiar
with. From the description on the web page Russ linked:
http://filer.case.edu/~sjr16/stars_blackhole.html
I guessed that it is synonomous with the photon sphere.
What I was talking about when I referred to the "outer event
horizon" is the photon sphere of a nonrotating Schwarzschild
black hole.



Any observer who doesn't cross the event horizon in his own frame
can get back theoretically, although the force will increase
without limit as he approaches.
It is not possible for anything to cross the photon sphere and
escape, despite what it says on Wikipedia. The escape speed at
the photon sphere is the speed of light. Nevertheless, light
itself can escape from below the photon sphere if it initially
is on an upward trajectory.



It does take infinite time in a stationary frame for an object to
reach r = R. Oh, and Jeff, that message sent from the free-faller
just as he crosses the horizon will take infinite time in our
frame to reach us.

The "last photon" has to do with quantum considerations which
limit how much that last photon can be stretched out.
Which tell us that the last photon from the Schwarzschild
radius will reach us about a hundredth of a second after
photons from 1.5 Schwarzschild radii reach us.

You are calculating a purely theoretical time which has nothing
to do with either what can be observed or what is actually
happening.

Isn't it interesting enough to you that there really is an
infinite time involved? The time for an object to reach the
singularity is infinite. I think that's pretty cool.



I know what the devil I'm talking about, and it is not garbage.
I asked you before to say where you get that the Schwarschild
calculations give the "actual, instantaneous position" in the
frame of the distant stationary observer, and you gave no answer.
I ask again.

I have already quoted Thorne saying "An imploding star really
does shrink through the critical circumference without hesitation.
That it appears to freeze as seen from far away is an illusion."

That blows away everything you are saying, doesn't it?



Your understanding of the photon sphere is also flawed, "garbage"
I might say if I took your tact, as that has do with inertial
fallers/orbiters, those without forces applied.
Matter is held together by the electric force. The electric
force is transmitted from particle to particle by light, at the
speed of light. Applying enough force to an object below the
photon sphere to keep it from falling in would cause the object
to fall apart. Even the electrons would fall off the protons,
because the light attempting to travel between the electrons
and protons would not be able to keep up.

The best you can do inside the photon sphere is to launch
individual photons away from the black hole at a high enough
angle to escape.



The Finkelstein coordinates are a different way of looking at
things, and have nothing to do with the time coordinate of a far
away stationary frame.
Finkelstein coordinates are probably the best way to calculate
what you claim to be calculating. You are not calculating what
you claim to be calculating, though.

-- Jeff, in Minneapolis

publius
2006-Sep-14, 11:51 PM
Jeff,

I'm sorry, but you greatly misunderstand what you're talking about. I'm certainly not going to challenge Kip Thorne, but he and I are looking at this differently. He is thinking about what happens in the frame of the collapsing material. I am not. And I do disagree with his interpretation that the infinite time is an illusion.

And I am indeed calculating exactly what the heck I say I am calculating. There's no other to do this I suppose that go through the darned geodesic derivation. I can do that fairly easy for light, null geodesics, and I will later.

And once and for all, the radial escape speed from the photon sphere is not c. Locally it is sqrt(2/3)c. In our distant coordinates it is 1/3 that. The local escape velocity is only c right at the horizon. The local g force at the photon sphere for a stationary object, is GM/r^2 * 1/sqrt(1 - R/r). For
r = 3/2R, that's sqrt(3) * 4/9 * GM/R^2. That's not infinite, and no, that force does not have to be transmitted faster than light. Besides, we're neglecting tidal forces anyway, and consider very small observers, anyway.

You're having a big problem and a lot of misunderstanding about local speeds vs the speeds we measure in a far away frame as well.

Now, Jeff, you demand I go through the geodesic derivations to prove I know what I'm talking about. And I'm going to shortly. But I expect *you*, if you disagree, to show me how the local radial escape velocity exceeds that of light at the photon sphere. I say that local esacpe velocity is sqrt(2/3)c. If you say it is c, show the calculation, and we'll compare.

Furthermore, I want to see a calculation showing me the time, in our distant frame, at which a free faller crosses the horizon. I think you claimed the time he would receive a message we sent after the horizon crossing but before the singularity would be finite in our frame as well. I want to see that calculation.

My position is those times do not exist in our frame, they are infinite, and "more than infinite" in the latter case.

-Richard

RussT
2006-Sep-15, 12:33 AM
Using the terms on the page you linked, the inner event horizon
is where light moving vertically away from a black hole can just
barely escape.

Sorry Jeff, not quite.

Quote from my link above;
Next out is the outer event horizon. This marks the boundary at which the escape velocity is greater than the speed of light, and all known objects are drawn into the hole. This also marks the "outer edge" of the black hole; we cannot see into it, for no form of known radiation can escape the gravitational pull from this point inward.end

Also, it then talks about 'particle' escapability;
[The next part of the black hole is only present in a spinning black hole. The ergosphere is a region of space where all particles are drawn in a circular path that match the hole's rotation. However, within the ergosphere, matter and energy can still escape the hole's grasp. The outer edge of the ergosphere is called the static limit. This is the distance that matter must maintain in order to keep a stable orbit and not be trapped by the hole's rotation.]end

And then says that accretion discs are only in 'active black holes'??? But, this sound to me like ergo-spheres and accretion discs, are basically the same thing.
[Other parts of a black hole are present only in "active" black holes. The accretion disk is matter that has been trapped in orbit around the black hole. It will gradually be pulled into the hole. As it gets closer, its speed increases, and it also gains energy and begins to emit light. This is the radiation that astronomers can use to determine how much the black hole "weighs." By using the doppler effect, astronomers can determine how fast the material is revolving around the black hole, and thus can infer its mass. Most black holes that have been found usually weigh several million solar masses.]end

publius
2006-Sep-15, 12:49 AM
Russ,

And it should be noted a Kerr hole has two photon spheres, one for "prograde" rotation, and other, larger one for "retrograde" rotation.

The things like event horizons, "ergosphere" (static limit), and all that are part of the space-time of the hole. An accretion disk is not part of the geometry of course, but is a structure that results from matter moving in that space-time.

-Richard

publius
2006-Sep-15, 01:07 AM
Jeff,

I'm going to assume you're familiar with a metric, which is roughly the "norm" of a given space. In "curved" spaces, this metric is not a simple Pythagorean expression, but varies with the coordinates. Metrics are solutions to Einstein's field equation, which has the stree-energy tensor of mass-energy as its source.

The Schwarzschild metric is a solution to that equation for Schwarzschild's "massenpunkt", a point mass. The space coordinates are r, theta, phi of a spherical coordinate system, and t being the time coordinate. This is are the coordinates of an observer at infinity, in the asymptotic flat limit of the space-time. But these coordinates will be very close to stationary observers closer in where the curvature is neglible. Hence, roughly, these coordinates are the ruler and clock of a "distant" observer. THe metric is thus:

ds^2 = u*(c dt)^2 - 1/u *(dr)^2 - (r d(theta)^2 - (r sin(theta) d(phi))^2

I'm considering only radial trajectories (zero L), where theta and phi remain constant, and we choose the origin so they are zero. Put "Schwarzschild metric" in Google and you will see the above many, many times the world over.

Now, u = (1 - R/r), and I'm using a positive time-like convention.
R = 2GM/c^2 is the Schwarzchild radius.

Now, light-like trajectories are easy. They are the null geodesics were the "norm" is just zero. Light-like geodesics are those for which the above is identically zero. For radial geodesics, where theta and phi are constant, we have

u*(c dt)^2 = 1/u *(dr)^2.

This is the equation of motion for light in our coordinates. That is simply

(dr/dt)^2 = c^2u^2 = c^2(1 - R/r)^2, or

dr/dt = +/- c*(1 - R/r)

THis is the coordinate speed of light along radial paths. + for light coming out, - for light going in.

Note that goes to zero at r = R. Nothing can go faster than the local speed of light, and so hence any real time-like velocity must go to zero at the horizon. At the photon sphere, this is 1/3c.

These are the rulers and clocks of a distant observer, and you'll note this result was one of the first things Einstein discovered.

Now, smoke that over, and in the next message, we'll consider the local (stationary) frames. If you disagree with the above, then show your counter-math.

-Richard

publius
2006-Sep-15, 01:54 AM
Continuing:

Consider a stationary observer at some point r in the metric. dr is now zero, and the interval("norm") is just

ds^2 = u*(c dt)^2.

Just as in SR, the proper time of that observer is just the interval (over c), and we'll call that proper time T. Thus,

(c dT)^2 = u*(c dt)^2, or

dT/dt = sqrt(u) = sqrt(1 - R/r).

This is the gravitational time dilation factor (for stationary observers). Note that dT --> dt as r grow large, which shows that the clocks of (stationary) observers sufficiently far away tick about the same rate. And dT -- > 0
as r --> R. Time "stops" at the event horizon relative to distant observers.

Now, (radial) space-like paths, at constant t.

(ds)^2 = -1/u *dr^2. ds here is then the local (radial) distance, which I'll call 'X', so -(dX)^2 = -1/u*dr^2, or

dX = dr/sqrt(1 - R/r).

That is the radial distance "compression" or stretching, depending on how you like to think about it. One far distance unit is 1/sqrt(1 - R/r) local units. At the photon sphere, that would be a factor of sqrt(3). A local meterstick appears to be shrunk down to 1/1.73 meters. Distances in the tangential directions are not affected.

So, if we want local distances and local times, we have to scale by the above factors. For a local (radial) velocity, we thus want dX/dT. And by the above, that is just,

dX/dT = dr/dt * 1/(1 - R/r).

Applying that to the radial coordinate speed of light above,

dX/dT = dr/dt * 1/u = c*u/u = c.

It nicely cancels and the local speed of light remains c (as always).

One other thing we'll need. The local (radial) acceleration d^2X/dt^2, so we get another factor of srqt(u) in the numerator

d^2X/dT^2 = 1/(u^(3/2)) * d^2r/dt^2

And remember the above are all local stationary frames.

If you consider any of the above to be "garbage", I want the math, not the words to show it does not hold.

-Richard

Jeff Root
2006-Sep-15, 02:23 AM
Using the terms on the page you linked, the inner event horizon
is where light moving vertically away from a black hole can just
barely escape.
Sorry Jeff, not quite.

Quote from my link above;
Next out is the outer event horizon. This marks the boundary at
which the escape velocity is greater than the speed of light, and
all known objects are drawn into the hole. This also marks the
"outer edge" of the black hole; we cannot see into it, for no
form of known radiation can escape the gravitational pull from
this point inward.end
That does not conflict in any way with what I said above.

Editing to add:

Ah, well, obviously that was wrong. I think the statement
quoted from the web page is poorly worded. Either some
light can escape from inward of the "outer event horizon",
or it is referring to the "inner event horizon". Address my
question below about what you think the inner event horizon is.

End of edit.

The event horizon, or "inner event horizon" in the terms used
here, is the last surface where any light at all can escape to
the outside. Below that surface, nothing can escape.

The photon sphere, which I am guessing is the same as the
"outer event horizon", is the last surface where light moving
horizontally can avoid being drawn in. Below that surface,
anything headed down can never escape.

If you think I was describing the outer event horizon in the
quote above, then what do you think the inner event horizon is?



Also, it then talks about 'particle' escapability;
[The next part of the black hole is only present in a spinning
black hole. The ergosphere is a region of space where all
particles are drawn in a circular path that match the hole's
rotation. However, within the ergosphere, matter and energy can
still escape the hole's grasp. The outer edge of the ergosphere
is called the static limit. This is the distance that matter must
maintain in order to keep a stable orbit and not be trapped by
the hole's rotation.]end
Right. No disagreement with anything there.



And then says that accretion discs are only in 'active black
holes'??? But, this sound to me like ergo-spheres and accretion
discs, are basically the same thing.
No, completely unrelated.

The ergosphere is the region of space affected by the black
hole's rotation. As it says in the next passage you quote,
an accretion disk is matter (gas and possibly some dust) which
has collected around the black hole, and is orbiting it, outside
the ergosphere. When any of the matter falls into the ergosphere
it is accelerated to enormous speed and temperature, and in some
cases (apparently where there is a lot of matter falling in) a
small fraction of the matter will be accelerated away from the
disk at nearly the speed of light. We see this as a quasar or
active galactic nucleus (AGN). That is what is meant by an
"active black hole": One which puts out a lot of energy that we
detect, mostly with radiotelescopes.



[Other parts of a black hole are present only in "active" black
holes. The accretion disk is matter that has been trapped in
orbit around the black hole. It will gradually be pulled into
the hole. As it gets closer, its speed increases, and it also
gains energy and begins to emit light. This is the radiation
that astronomers can use to determine how much the black hole
"weighs." By using the doppler effect, astronomers can determine
how fast the material is revolving around the black hole, and
thus can infer its mass. Most black holes that have been found
usually weigh several million solar masses.]end
I see no conflicts between anything I've said and anything in
these quotes. Edit: Except for that bit you pointed out at
the start!

-- Jeff, in Minneapolis

publius
2006-Sep-15, 02:56 AM
No, Jeff, the two event horizons of a Kerr hole have nothing to do with the photon spheres. The outer horizon is pretty much the same as that for a Schwarzschild hole, the point of no return. The inner horizon is another boundary below the outer one. The area between the two horizons has closed time-like curves, and is very weird, so weird that many refuse to believe it is possible. The ergosphere is the stationary limit, similiar to such a limit for a Coriolis frame.

Put "Kerr metric" in Google, and start reading and learning, rather than making statements based on misunderstandings and WAGs. If one is going to use the term "garbage", one had better be well informed and ready to back that up.

I want to see your calculation showing the escape velocity from a Schwarzchild hole exceeds c below the photon sphere. Shortly, you will see mine showing it is locally sqrt(2/3)c at the photon sphere.

-Richard

Ken G
2006-Sep-15, 03:17 AM
Likewise, we can calculate exactly when we will recieve a
message from them, according to our clock, sent before they
have crossed our event horizon, if we know when they sent it
according to their clock, or, inversely, we can calculate when
they sent a message from the time we get it. We can calculate
exactly when they will cross our event horizon according to
their clock, and when we will see them cross our event horizon
according to our clock.


This is the misleading statement that is the cause of the problem. We on Earth can send a message that says "good luck", such that the rocket receives it just as it crosses the event horizon. But if the rocket instantly radios back "thanks", we'll never get it, ever. It is on this basis that publius is correctly saying that the event horizon is never crossed in our time, but I have pointed out that this is only true using that particular convention of what is meant by "our time"-- it is a coordinate dependent statement. Extrapolating our own sense of time to other places where clock simultaneity is an arbitrary convention is not a unique thing to do-- it is based on the choice of coordinates. A coordinate is a global thing, and we can certainly restrict our thinking to coordinates that locally map onto a sense of "distance" and "time" to some local observer, but this still begs the question-- which observer? It is arbitrary. Often, the choice is to map onto the time and distance concepts of the local people who have been in free-fall from a great distance away, and in that coordinate system, the event horizon is crossed in short order, and black holes do exist. In publius's system of coordinates, black holes never form. Neither of these are absolute statements of anyone's reality, this is the crucial message I'm putting forward. So when it is stated that:



THe Finkelstein coordinates are a different way of looking at things, and have nothing to do with the time coordinate of a far away stationary frame. That is, they are not measuring the proper time of a stationary observer.
it is important to recognize that all coordinates that locally conform to a sense of distance and time, i.e., all coordinates worth their salt, measure the proper time of a local stationary observer, because all observers are by definition stationary, and so are all proper times. But they are only local-- never global. This is because in general relativity, there is simply no way to define a unique sense of "globally stationary". For a familiar example of this principle, turn to cosmology-- in what unique sense are the distant galaxies stationary, despite their redshifts? This is only true in comoving-frame coordinates, by definition.

publius
2006-Sep-15, 03:26 AM
And now we come to time-like (radial) geodesics. Finding those gets a bit harder. Again, the interval, ds, is given by the Sch. metric (for radial paths):

ds^2 = u*(c dt)^2 - dr^2/u

From SR, one will remember that a non-accelerated geodesic, a straight line, has the *maximum* interval, which means the proper time between two events is the longest for an inertial path between them. And so it is in curved space-time. To find the time-like geodesics, one *maximizes* the interval given above. That is one finds the family of r(t) curves that maximize the integral of ds over the path.

I have gone well beyond the call of duty here, IMO, as all of the above can be found on texts on the subject. The technique used is that of the Calculus of Variations, which is intimately associated with the "principle of Least Action". Although here, we are maximizing, or technically "extremizing" the path integral. Any advanced text on mechanics will go into the techiques. Basically one gets differential equations to solve. When one does this for the above, one find the radial geodesics are given by:

(dr/dt)^2 = c^2*u^2*(1 - u/E^2), where E^2 is a constant of the motion that turns out to be the specific energy, the ratio of the total relativistic energy to the rest energy. This specific energy at any local point is given by:

E^2 = u/(1 - v^2/c^2), where v is the *local velocity*, dX/dT as defined above. For a particle dropped from infinity with zero initial velocity, E is simply 1. Substituting that in and remembering u = (1 - R/r), we have simply

(dr/dt)^2 = c^2(u^2)*(R/r).

This is the coordinate velocity, in our distant frame, of a radially free-falling particle. It also the escape velocity, as that is the velocity required to reach infinity when "thrown" upward from a point. It goes to zero at r = R, just like that of light. We will find this has a maximum at the photon sphere,
r = 3/2R

Recalling the local speed conversion just cancels out the u factor, we have the *local escape velocity*, dX/dT is just

dX/dt_escape = c*sqrt(R/r). That is exactly the Newtonian result. The Photon sphere is at r = 3/2R. Plugging that in gives simply

sqrt(2/3)c. The local escape velocity at the photon sphere, which is less than c.

While the local (escape) free-fall velocity is exactly the same as Newton, the local acceleration is not. I will get to that in a later post.

And again, if any of the above is "garbage", then show it with math, not words.

-Richard

publius
2006-Sep-15, 03:44 AM
it is important to recognize that all coordinates that locally conform to a sense of distance and time, i.e., all coordinates worth their salt, measure the proper time of a local stationary observer, because all observers are by definition stationary, and so are all proper times. But they are only local-- never global. This is because in general relativity, there is simply no way to define a unique sense of "globally stationary". For a familiar example of this principle, turn to cosmology-- in what unique sense are the distant galaxies stationary, despite their redshifts? This is only true in comoving-frame coordinates, by definition.

Ken,

I should clarify that my usage of "stationary observer" all through this mess means Schwarzschild stationary. That is stationary with respect to the center of the black hole, dr/dt = dphi/dt = dtheta/dt = 0. (These are therefore accelerated observers, of course, feeling a force). A free-faller is not stationary as I'm using it, but he is certainly stationary in his own frame, which is the sense you are using it above. Some other, far way moving observer would not call us stationary, of course, and he would have a different clock rate and yardstick.

And the proper time of a radial free-faller to the horizon will just be the integral of ds^2 from r = start to r = R, which will be finite. However, our t coordinate will not be finite in that integral.

-Richard

publius
2006-Sep-15, 04:43 AM
And continuing yet more with this mess:

We have the following expression for the radial free fall coordinate velocity with E = 1, corresponding to a particle dropped from infinity at zero initial velocity:

(dr/dt)^2 = v(r) = c^2*u^2*R/r, where u = (1 - R/r) as always.

This has a local maximum where dv/dr = 0. By taking the derivative of this with respect to r and setting it equal to zero, one finds that maximum occurs right at the photon sphere, r = 3/2R. The derivative of v^2 with respect to r also has another use, as we note dv/dt = dv/dr*dr/dt = v*dv/dr.

The derivative of v^2 is just 2v dv/dr, so the derivative of the right hand side of the above is just twice the (coordinate) acceleration of that radial free-faller. And that is just:

dv/dt = -R/2r^2 * c^2* [1- 4R/r + 3(R/r)^2].

And noting that R = 2GM/c^2 that is just:

dv/dt = -GM/r^2 * [1 - 4R/r + 3(R/r)^2], which is just Newton for
large r, much greater than R, which is a nice check to confirm this has a Newtonian limit.

-Richard

Ken G
2006-Sep-15, 06:20 AM
I should clarify that my usage of "stationary observer" all through this mess means Schwarzschild stationary. That is stationary with respect to the center of the black hole, dr/dt = dphi/dt = dtheta/dt = 0. (These are therefore accelerated observers, of course, feeling a force).
That is certainly one valid coordinate system that locally conforms to reference frames all along the way. As you have specified your coordinate system, the result is valid, and can be transformed into any other frame. It sounds like the coordinates you are using are the proper time for a chain of observers who all line up and each maintains its constant distance from the event horizon. All I'm saying is, that's not uniquely "our" time, it's an arbitrary way to cobble together local proper frames. Our time must be measured on our clock, and requires a convention for synchronizing to distant clocks. The arbitrariness appears in that convention, and you have chosen yours. What I mean by this is that one could give two different reasons for why you get an infinite time in your calculation. One is that time is indeed ticking by very slowly for the stationary observers near the edge of the horizon, and that's the picture you are using. But there is an equivalent picture, which says that time is ticking by just as quickly there as here, but there is a continously changing definition of simultaneity, such that our clocks are continuously getting ahead of theirs. Like we are continuously going more and more into "daylight savings time". The way to define simultaneity that gives you that result is if the "midpoint" between us and the event horizon is asymptotically falling into the event horizon, the view of gravity that it "sucks in" space itself. It's an equally valid interpretation of what gravity "does"-- it's just a different coordinate system.

Squashed
2006-Sep-15, 08:07 PM
According to the calculations the speed of time comes to a stop at the event horizon but along with this effect is length contraction and so at the event horizon the radial dimension of the infalling object becomes zero, right?

If the object has a zero radial thickness then at the event horizon it is coincident with that position and so the event horizon and the object become "one" - there is no incidental expansion of the event horizon needed to include the infalling object.

At that point, because of the addition of the infalling object's mass the event horizon enlarges, perhaps inperceptably but still by a value greater than zero, so that its size overtakes the zero radial distance of the object.

RussT
2006-Sep-15, 10:08 PM
Here is paper by Kip Thorne where he is examining the orbit of a compact 'massive' body on its infall to the event horizon of a Massive Black Hole (So, obviously he doen't think that tidal forces will reduce it to EM radiation by this point of its fall)! No where in here does he use the terms, ergo-sphere, photon sphere, accretion disc!

publius
2006-Sep-16, 12:55 AM
All,

To finish this mess, I was going to wrap up with the local acceleration. That, is proving not to be a PITA. :lol: In one of the above messages I sloppily said we'd get just another factor u^(1/2) on the local acceleration -- that is wrong, as I forgot the time dependence of u itself.

ie:

dX/dT = 1/u dr/dt, so d^2X/dT^2 is given by

d/dT(1/u dr/dt) = 1/sqrt(u) * d/dt(1/u dr/dt). If my derivatives are correct, that becomes

d^2X/dT^2 = 1/u^(3/2) * [ d^2r/dt^2 + R/ur^2 * (dr/dt)^2 ]

We've another term there, and this, even for the relatively simple Schwarzschild metric shows how messy GR can quickly become when you actually get down to some specific business in a specific coordinate system.

I tried a couple minutes ago to do the simplifying method of
dv/dt = vdv/dr, but in local coordinates (ie dV/dT = VdV/dX, where big 'V' is the local velocity), and that didn't work, which means I was going awry somewhere.

If I get it figured out and find the simplest way to get it, I'll post it and wrap this up. This, the local acceleration (the local 'g' ) will allow us to determine the (local) force on required to remain stationary at a given r. If anyone else wishes to follow through this, that local acceleration should be

"g_local" = c^2R/2r^2* 1/sqrt(u) = GM/r^2 * 1/sqrt(1 -R/r)

Which approached Netwon for large r, but gets larger than Newton close in, and goes to infinity at r = R.

That will complete the derivation of just about every (mathmatical and otherwise) statement I've made in this thread and the others about this subject.

-Richard

publius
2006-Sep-16, 01:29 AM
Russ and all,

Before I forget, here is Stanley Robertson's "MECO" paper, which what is tugging me away from the dark side:

http://arxiv.org/PS_cache/astro-ph/pdf/0602/0602453.pdf

And Russ, to be clear, this is not against GR -- this is using GR. Mitra was the first, and Robertson above agrees with him. Whether they are correct or not, well, as you can see how much "fun" we've been having with the simple Schwarzschild metric here, you can imagine how much "fun" the GR experts will have with this. As far as I know, none of the heavyweights have formally responded to this MECO/ECO theory yet.

In a nutshell, their argument goes the "cold dust" collapse solution cannot apply to any real matter, because of radiation trapping. "Null surfaces" can't be formed, as that would require the local velocity of matter to exceed c at some point during the collapse.

I don't know enough to make any informed comment on this, save to say they are using GR to show a collapse cannot happen.

Externally, MECO/ECOs would look and act just like we think BHs would, but inside they are very different. An ECO would have no event horizon, and would be radiating a small amount.

-Richard

RussT
2006-Sep-16, 02:47 AM
Whether they are correct or not, well, as you can see how much "fun" we've been having with the simple Schwarzschild metric here, you can imagine how much "fun" the GR experts will have with this.

No Kidd'in, Huh!

At first I thought this was a Neuker Team/Group concept that stemmed from their basic problems of being able to show that a dust cloud collapse could result in a SMBH!

But in a quick partial read, I could find nothing that would indicate that, although I can't help thinking that their work has had something to do with this, I could find nothing to indicate that either in the credits.

I did already knew they were using GR for this, and since NOBODY has been able to show how SMBH's are created (there is a very good and identifiable reason they haven't), I suppose this sort of thing was inevitable, but even after a quick partial read (I will definitely read it much more thoroughly), this already seems like the recently promoted GravStar!

More later, and thanks.

publius
2006-Sep-16, 04:52 AM
Alack and alas, there's no simple way that I've yet stumbled on to get the local acceleration. We just have to slog throught the following expression:

g_local = 1/u^(3/2) * [ d^2r/dt^2 - R/ur^2 * (dr/dt)^2 ] Note that's a minus sign on the second term, not a plus sign as I posted above.

Somewhere above (or in another thread), I posted the coordinate acceleration, (d^2r/dt^2) as a function of r, thus:

dv/dt = -c^2R/2r^2 * (1 - 4R/r + 3(R/r)^2)

We already have v^2 = c^2*u^2*(R/r). The second term in brackets is then
- R*c^2/r^2 * u * R/r
pull out -R*c^2/2r^2 and we have:

(g_local) = -c^2R/2r^2 * 1/u^(3/2) * [mess]

where [mess] is

[1 - 4R/r + 3(R/r)^2 + 2(1 - R/r)(R/r) ] =

[1 - 4R/r + 3(R/r)^2 + 2R/r - 2(R/r)^2 ] =

[1 - 2R/r + (R/r)^2] = (1 - R/r)^2 = u^2

See how all that mess of 4s and 3s coefficients reduced down to a nice u^2. When that happens, you know you're on the right track. :lol: So we're left with
(g_local) = -c^2*R/2r^2 * 1/u^(3/2) * u^2.

Recognizing Rc^2/2 is just GM, and noting the u-factor is reduced to
1/sqrt(u), we have it:

(g_local) = -GM/r^2 * 1/sqrt(1 - R/r)

That is the local acceleration field measured in local coordinates. A stationary observer must exert a local force equal to m*(g_local) to remain stationary.

-Richard

publius
2006-Sep-16, 05:15 AM
Now, after all that mess, I've derived from the scratch of the Schwarzschild metric everything thing I have said about this in this thread and the others. The one thing I did not slog through here, presenting as a "it can be shown" :) was the derivation of the time-like geodesics, which involve maximizing the path integral. There may be a trick or two to make that simple for radial paths, but I wasn't sure. In general, it involves the variational calculus techniques, the details of which can be found in advanced mechanics and math texts. To put into words, this finds the family of curves that "straight lines are bent into" by the curvature of space-time. If one maximized ds in flat space-time, one would get exactly staight lines. But I did explicitly get the light-like geodesics or null geodesics, which are fairly easy.

Also, I was considering only radial geodesics, no initial L, with the angular coordinates therefore not coming into to play. If one wishes to investigate orbits, one has to throw those in. Precession drops right out of that, as well as the bending of light-like geodesics. One finds the smallest circular orbit exists at r = 3/2R, and one finds the local speed for that circular orbit is just c. This is not the escape velocity/radial free fall velocity, but the speed of a circular orbit. This orbit is however unstable, as it is at a local maximum of the effective potential, not a minumum, and so perturbations send it one way or the other. This is the so-called "photon sphere". Stationary observers are possible between there and R, and escape is certainly possible, although the force (see g_local above) is a might bit large.

And so the relatively simple Schwarzschild metric contains just about all the "additional features" of GR gravity over Newton, save for frame-dragging/gravitomagnetism. But it does contain the "geodetic precession" effect, which one would find considering local orbiting frames.

-Richard

RussT
2006-Sep-18, 10:55 AM
I did already knew they were using GR for this, and since NOBODY has been able to show how SMBH's are created (there is a very good and identifiable reason they haven't), I suppose this sort of thing was inevitable, but even after a quick partial read (I will definitely read it much more thoroughly), this already seems like the recently promoted GravStar!

More later, and thanks.

I almost forgot about this.

This is Q&A so I can't go into some reasons, but I do have some very basic problems even with the intitial concept, but the paper is very thorough and detailed, but IMHO, has some real inconsistencies.

But I think the real hard one for me, is the concept that 'light pressure' could even come close to stopping the gravitational collapse of a 3 miilion sol mass Kerr black hole!

Publius;
Do you think it is 'light pressure' that stops a neutrons stars (that they are so willing to compare a SMBH's properties to) final collapse?