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

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.

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?

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!

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.

Utterly negligible for this question, by many, many orders
of magnitude.

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.

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.

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.

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

__________________
http://www.FreeMars.org/jeff/

"The other planets?
Well, they just happen to be there, but the point of rockets is to explore them!"
-- Kai Yeves

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

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.

Relativistic free fall sanity check.

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

-Richard

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!

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

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

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

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.

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

Can a black hole even form in the first place, as seen by a distant observer?

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.

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.

<<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!

"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. So something is awry, but it's something small I imagine, and I don't have a clue what it could be.

-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