# Thread: Can a falling object actually enter a black hole?

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## Can a falling object actually enter a black hole?

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

2. 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.

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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?

4. 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!

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Originally Posted by bachnga
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
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.

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

Originally Posted by bachnga
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.

Originally Posted by bachnga
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.

Originally Posted by bachnga
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.

Originally Posted by bachnga
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

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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

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## (v/c)^2 = 1/3

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.

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

-Richard

8. 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!

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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

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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
Last edited by neilzero; 2006-Sep-05 at 02:02 AM.

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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

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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

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

14. 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.

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## Interesting Question

Originally Posted by kzb
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.

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<<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!

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"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

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## Thanks

Originally Posted by publius
"Oh so close, yet oh so far..."
.
.
.
-Richard

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

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Squashed,

Thanks for respecting my privacy. 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

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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

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Well, this makes me feel better :

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

22. 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).

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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. 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

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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? 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

25. Originally Posted by publius
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.

Originally Posted by publius
And related to this is a deep question: Would you "feel" the gravitational radiation reaction force? 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.

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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

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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

28. Originally Posted by publius
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.

29. Originally Posted by publius
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.

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Originally Posted by Ken G
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

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