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publius
2006-Aug-10, 05:31 AM
Spurred on by the question of gravitational lensing, and how, according to Maxwell in gravitational field as derived by Landau & Lifsh-itz, one can see the gravitational field as a "medium" of increasing permeability, I arrived at what to me is a rather strange conclusion.

For a an object that starts a free-fall far away, but at relativistic initial velocity, *it will appear to slow down*, not accelerate from a stationary frame watching it*. From that frame, the coordinate speed of light decreases with depth in the field. But nothing can appear to be going faster than the local speed of light. So if light appears to slow down, then a relativistic free-faller would have to slow down as well to keep its speed below the coordinate speed of light.

We all know that an object free-falling toward a BH will appear to stop at the event horizon. I had never thought about the consequences of this too much, but at the event horizon, the coordinate speed of light goes to zero. Hence everything has to appear to stop.

So, for a free faller starting out at zero velocity at infinity, it will have accelerate, but at some distance it will have to appear to start slowing down. But this has to depend on the initial velocity, as I just realized. If it starts out at nearly light speed, it cannot accelerate much before it would exceed the coordinate speed of light.

Above, I'm thinking about a straight line radial free fall, L = 0. If we have angular momentum for a relativistic initial velocity, gravity still appears attractive in the directional sense, pulling the object toward it, yet its (scalar)speed must decrease as it falls into the gravity well!

And that's my sanity check, here. Is this correct or am I messing up big time somewhere?

-Richard

publius
2006-Aug-10, 06:01 AM
Playing with E = mc^2, one can come up with something interesting about a free fall. In the GR view, there is no force -- no potential energy being converted to kinetic energy, just an object following its geodesics in curved space-time. It only appears to accelerate because the directions of time and space are changing relative to a stationary frame. Putting it very roughly, an object's "motion through time" becomes motion through space.

Now, consider E = mc^2 = ym0c^2, where m0 is the rest mass. Let's consider a mass of m0 which free falls into a BH from infinity. Let E = m0*c^2 be the initial energy, of course. Now equate that constant with ym0*u^2, where u^2 is the *coordinate speed of light*, which decreases as a function of r in a gravitational field. Solve for v in the 'y' (gamma).

One gets this:

E = ymc^2 -> 1/y = mc^2/E

(1 -v^2/u^2) = (mu^2)^2/E^2 = u^4/c^4

v^2/u^2 = 1 - u^4/c^4

v^2 = u^2(1 - u^4/c^4)

v = u*sqrt(1 - u^4/c^4)

v(r) = c*(1 - 2GM/rc^2) *sqrt[1 - (1 - 2GM/rc^2)^4]

:) How about that? Note that at r = infinty, the square root term is zero, as u = c. Thus v = 0. But at r = Schwarzchild radius, v(r) = 0 as well, because u itself is 0 here.

This idea of playing with E=mc^2 in this fashion just hit me, and it appears to give the correct behavior at the limits. I'd have to investigate it further to see if it made sense in between. That 4th power term may give us the desired "sharp" flip past a maximum. That is v(r) would increase as expected to some maxium at some small r, then quickly drop back down to zero. The above assumes we start out with zero initial velocity. We would have to modify this a bit to take initial velocity into account.

If anyone here knows the actual GR solution, does this agree?

-Richard

Ken G
2006-Aug-10, 02:20 PM
I don't know the true formula when coordinatized in this fashion, I only know that one can get all kinds of different pictures when using different coordinates. So when you talk about the "coordinate speed of light", note that this is not a unique concept. There are also types of global coordinates where distances become infinite as you approach the event horizon, so then c stays the same.

Squashed
2006-Aug-10, 03:09 PM
...For a an object that starts a free-fall far away, but at relativistic initial velocity, *it will appear to slow down*, not accelerate from a stationary frame watching it*. From that frame, the coordinate speed of light decreases with depth in the field. But nothing can appear to be going faster than the local speed of light. So if light appears to slow down, then a relativistic free-faller would have to slow down as well to keep its speed below the coordinate speed of light.

We all know that an object free-falling toward a BH will appear to stop at the event horizon. I had never thought about the consequences of this too much, but at the event horizon, the coordinate speed of light goes to zero. Hence everything has to appear to stop.

So, for a free faller starting out at zero velocity at infinity, it will have accelerate, but at some distance it will have to appear to start slowing down. But this has to depend on the initial velocity, as I just realized. If it starts out at nearly light speed, it cannot accelerate much before it would exceed the coordinate speed of light.

Above, I'm thinking about a straight line radial free fall, L = 0. If we have angular momentum for a relativistic initial velocity, gravity still appears attractive in the directional sense, pulling the object toward it, yet its (scalar)speed must decrease as it falls into the gravity well!

And that's my sanity check, here. Is this correct or am I messing up big time somewhere?

-Richard


Not being a mathematician but knowing a little about the theoretical implications of relativity I have inquired about the statement in red in regard to increasing size of blackholes and this observational discrepancy with relativity. I conclude that relativity must be wrong at the extremes, as represented by blackholes and event horizons, but I have insufficient mathematical skills to make a valid case for my conclusion.

Jeff Root
2006-Aug-10, 05:19 PM
We all know that an object free-falling toward a BH will appear
to stop at the event horizon.
My understanding is that this statement is an incorrect
description of what actually happens, originating in poorly-
written popular descriptions of relativistic effects. More
accurate to say that an object red-shifts to invisibility.
It would be gone without a trace in a few milliseconds.

-- Jeff, in Minneapolis

publius
2006-Aug-10, 05:32 PM
Jeff,

No, it's not just an illusion. It takes an infinite amount of time in the frame of a stationary observer for a free faller to cross the horizon. It's not just an illusion of red shift.

Now, in the frame of the free-faller, he crosses that horizon in short order, and if he's watching a stationary observer, he sees that observer accelerating away and redshifting himself. There is a total breakdown of "reciprocity" between the two observers. And there's no problem with that at all. Non-inertial frames play even stranger tricks with simultaneity that SR does.

-Richard

Ken G
2006-Aug-10, 05:40 PM
I don't think publius' statements are incorrect, it is only that they come from a nonlocal reference frame. Relativity basically comes in two parts, local rules about what you can actually observe, and global rules about how to patch together locally coordinatized reference frames into a global coordinate system. publius has chosen a particular coordinatization, and although I'm not clear on the details of that system, other references talk about various ways to do this and I would tend to suspect publius is adopting those other perspectives faithfully.

publius
2006-Aug-10, 07:59 PM
Ken,

You remember the thread about the guy whose name escapes who made a PR release that he had discovered a "repulsive" solution for a gravitational source approaching a test mass at high speed?

Well, this falls right out of the above reasoning, and is actually nothing all that spectacular at all, just gravity behaving like GR says it does.

Switch to the rest frame of our relativistic free-faller, who has some initial velocity close to c at infinity. In that frame, the BH is approaching at high speed. Since the relativistic free-faller must appear to slow down, it will appear to be repelled by the approaching BH in that rest frame. The black hole will catch it, and the free-faller will freeze at the horizon and both will continue moving away at high speed.

But I'll point out the slowing down (or repulsion) is just a "cool trick" of curved space time. If you look the kinetic energy part of the E = mc^2 playing above, you will see it increases, and becomes equal to the original rest energy at infinity. The reson is appears to slow down is because light is slowing down or "space-time is becomming denser", however you like to think of it.

But the KineticEnergy(r) function always increases as r -> 0, it's just that kinetic energy becomes very low velocity in that "dense" space near the horizon.

IIRC, he pulled out a factor v/c = Sqt(3)/2 when things turned "repulsive" -- I wonder if that will fall out of the above E=mc^2 playing.

-Richard

publius
2006-Aug-10, 08:17 PM
I don't know the true formula when coordinatized in this fashion, I only know that one can get all kinds of different pictures when using different coordinates. So when you talk about the "coordinate speed of light", note that this is not a unique concept. There are also types of global coordinates where distances become infinite as you approach the event horizon, so then c stays the same.

Ken,

I can see that -- we choose a yardstick that shrinks with decreasing radius (it might be just a simple as saying our unit length "undoes" the Schwarzchild (1 - 2GM/rc^2) factor). This will undo the "compression" of space-time, and so allow our velocity to act more normally, with c being constant. Near horizon, the free faller gets close to c, but he has infinite distance to travel to make it to the horizon.

But I would argue the most "natural" global coordinate system to use is one stationary a large distance away, as that should give us an idea of what we would see watching stuff like this.

-Richard

Jeff Root
2006-Aug-10, 08:24 PM
No, it's not just an illusion. It takes an infinite amount of
time in the frame of a stationary observer for a free faller
to cross the horizon. It's not just an illusion of red shift.
I'm not saying that it is an illusion. I'm saying that it is
an incorrect description of what is actually seen, from popular
explanations of relativity which got the science wrong.

The time dilation effect is real, but an infalling object does
not appear to slow as it approaches the horizon!

However, I don't have a source or a link to an explanation,
so I can't support my assertion as yet. Maybe someone else?

-- Jeff, in Minneapolis

Squashed
2006-Aug-10, 08:45 PM
The time dilation effect is real, but an infalling object does
not appear to slow as it approaches the horizon!

However, I don't have a source or a link to an explanation,
so I can't support my assertion as yet. Maybe someone else?

-- Jeff, in Minneapolis

I'd be interested in reading the explanation, if ya ever find it. It seems bizarre that time dilation can affect atomic clocks and yet not affect an inertial motion: that of a freefalling body into a blackhole.

publius
2006-Aug-10, 08:46 PM
Jeff,

OK. We are in a stationary frame a large distance away from the black hole, watching something free-fall in a straight line radial trajectory, no angular momentum to worry about.

Now, if we never see the free faller cross the horizon in our frame, then its velocity as measured by our local yardstick and local clock cannot appear to continue to increase as r -> 0. That velocity must reach a maximum, then decrease and go to zero. Otherwise it would cross the horizon. As I was saying to Ken, this is an effect of space-time becomming more "dense". The coordinate speed of light (Schwarzchild metric) is decreasing as (1 -2GM/rc^2).

Now consinder the gamma factor, 1/(sqrt(1 - v^2/c^2). As the c in there decreases, the velocity at which that goes to infinity decreases as well. That behavior is what is slowing down our free faller in our frame. His kinetic energy continues to increase, but as space becomes very dense, that makes the velocity (in our frame) decrease. Early on in free fall, this effect is small, and the "conversion" of rest energy to kinetic energy (again, as seen in our frame) dominates and makes the velocity appear to increase, accelerating as we expect. But there is a turning point, where the "density" overwhelms the other, and it appears to slow down. This is very weird, and counterintuitive, which is why I posted it, of course.

There is nothing wrong with this -- it just illustrates that "strong gravitational fields" act very differently than our Newtonian intution for weak fields. And highly relativistic trajectories act differently in gravitational fields as well.

And I would say the actual misconception of all this comes from trying to reconcile the POV of such a stationary frame with that of the free-falling frame. Our notion of absolute time and absolute simultaneity is hard to shake, and it crops up in subtle ways. This applies here. One switches to the free-falling frame and says "aha, it really crosses the horizon", and then one tries to force that "really crossing" back into the stationary frame.

-Richard

publius
2006-Aug-11, 01:19 AM
Let's see if we can easily find where the above E = mc^2 equation says the turning point should occur. We start out with our velocity equation:

v(r) = c*(1 - 2GM/rc^2) *sqrt[1 - (1 - 2GM/rc^2)^4]

Now, let's divide by c, so our velocity becomes just the fraction of light speed (flat spacetime light speed, 'c0', of course). And note that the factor
'2GM/c^2' is just the Schwarzchild radius, call it 'R', so we have


v(r) = (1 - R/r) * sqrt[ 1 - (1 - R/r)^4 ]

Let's switch to 'x' = R/r, and I think we can make this simple enough to start taking derivatives without getting too messy:

v(x) = (1 - x) *sqrt[1 - (1 - x)^4]

Our free fall from infinity to R corresponds to x going from 0 to 1 here, and our v is now the fraction of light speed as measured by our far away yardstick and clock.

Since v(x) is positive over that range, we can work with the square -- the maximum of the square will occur at the same x as the maximum of v itself. So we write the expression for v^2

v(x)^2 = (1 - x)^2 *[1 - (1 - x)^4]

Let's make another change of variable, letting u = (1 - x)^2. Now,

v^2 = u*[1 - u^2] = u - u^3

Now, the derivative of that is just d(v^2)/du = 1 - 3u^2. Second derivatie will be negative, so this is a local maximum. That will be zero at

u = 1/sqrt(3). So our maximum occurs at

(1 - x)^2 = 1/sqrt(3) --> x = R/r = 1 - 1/(4throot(3)), or

r ~ 4.2R.

Plugging u at the maximum back into the v equation, we find

v^2 = 2/3 * 1/sqrt(3), or v ~ 0.62c

So, to summarize, my E = mc^2 logic says that the maximum velocity will occur at just over 4 Schwarzchild radii, with the peak velocity reaching
62% c.

Note the square (and 4th) root of 3 did come into this.


-Richard

publius
2006-Aug-11, 01:39 AM
Let's see, Rs for a solar mass black hole is about 3km ~ 1.8 miles. So 4.2Rs is only about 12.6km or 7.6 miles from the center. That's very close.

So Jeff, if it makes it any easier to swallow, an object falling into a stellar mass black hole will appear to accelerate until it gets very close to the center. It will then *quickly* deccelerate to zero over those last few km, with its clock rapidly slowing, and its light rapidly redshifting during that same distance as well.

So, basically, the decceleration will occur in the blur of the redshift to oblivion. It will be moving at 0.62c when that redshift to oblivion occurs.

ETA, at 4.2Rs, the coordinate speed of light will be (1 - x)c ~ 0.76c. So in a local frame, the free fall velocity is 62/76c ~ 0.82c

-Richard

Ken G
2006-Aug-11, 01:52 AM
You remember the thread about the guy whose name escapes who made a PR release that he had discovered a "repulsive" solution for a gravitational source approaching a test mass at high speed?

Yes, I think we agreed it was a coordinatization issue, so is very much what you are talking about now as well.


Well, this falls right out of the above reasoning, and is actually nothing all that spectacular at all, just gravity behaving like GR says it does.


The reson is appears to slow down is because light is slowing down or "space-time is becomming denser", however you like to think of it.
Or the rulers are getting longer relative to the space they're embedded in, that's how I like to think of it. It's the opposite of the Big Bang, because in the latter the gravity is weakening.


IIRC, he pulled out a factor v/c = Sqt(3)/2 when things turned "repulsive" -- I wonder if that will fall out of the above E=mc^2 playing.
That would be interesting, but you have gone farther into the math already, and I will trust your result.

publius
2006-Aug-11, 02:10 AM
Ken,

Well, the square root of 3 did fall out, but it wasn't over 2. The peak velocity was 0.62c at 4.2Rs. But this is for 0 initial velocity at infinity, and things might be different when you start adding initial velocity. I may try that. It may be as simple as just starting out with some E = y0*mc^2, where y0 is the gamma factor at infinity. Just add initial kinetic energy to the mix, rather than all rest energy.

And I caution I'm not completely confident in this E = mc^2 reasoning. What this is doing is holding total energy starting at infinity constant. That is, an object's motion through time becomes motion through space, as the "direction" of time (and space) "warps". Now, I know that is the GR view of where the kinetic energy comes from.

What I'm doing is equating that with the gamma factor, holding E constant as c slows down in the well, which says that as the motion through time slows, the difference in energy has to show up as velocity.

And, if you'll look, you'll see that even though the velocity peaks, the kinetic energy part continues to increase. At the event horizon, it says all of the original energy is now kinetic energy. But, since time has stopped, non-zero kinetic energy translates to 0 velocity! :)

But again, I'm not 100% confident the above is correct. It looks right, but there may be something else going on the gamma and E=mc^2 didn't take into account.

-Richard

RussT
2006-Aug-11, 02:17 AM
OK. We are in a stationary frame a large distance away from the black hole, watching something free-fall in a straight line radial trajectory, no angular momentum to worry about.

This may be the main problem in the first place.

I find it very hard to imagine that there are any black holes that are not spinning. Also, from this perspective, from a far away reference frame, all the matter being frozen at the event horizon "forever', would just keep accumulating and building up forever, which after billions of years would certainly be a viewing quandry!

Also, Massive galactic black holes are certainly spinning, so everything approaching the accretion disc is forced into an angular momentum that sets up a very complex situation with the ergoshere above the accretion disc going down to the accretion disc itself.

I saw a couple of the Fraser stories several weeks ago (that I realized later I should have saved). One dealt with this field I am referencing accretion disc up to the ergoshere and the plasma/electical/magnetic fields and formed what they wound up calling a 'helicopter rotor effect'. The other was an ultra violet (I think) view of the galactic center and showed where the massive black hole was and 4 other numbered bright light spots in the very near proximity to it.

I do not think that this means there is no or very little matter going into the black hole, but it does set up a very intersting scenario, as to what is actually happening at the accretion disc/ergoshere.

If someone remembers the 2 articles I am refering to, please let me know so I can find them again.

One other thing...afaik, once anything crosses the event horizon, it is thought that it is automatically accelerated to the speed of light in the black hole. In other words, space inside a black hole is traveling at the speed of light. Now, I'm not questioning that space inside a black hole is traveling at the speed of light, I just would like to know why we think that it is.

Thanks

Tensor
2006-Aug-11, 02:21 AM
But I would argue the most "natural" global coordinate system to use is one stationary a large distance away, as that should give us an idea of what we would see watching stuff like this.

-Richard

Actually, Richard, I would argue that using a coordinate system describing both would be more natural. If you use either Eddington-Finkelstein or Kruskal-Szekeres as your global coordinates, both the distant observer and the infalling observer can be described at the same time.
-

publius
2006-Aug-11, 02:35 AM
Tensor,

I think the Eddington-Finkelstein coordinates are the one where distance goes to infinity as you approach the horizon? Well, I see they make use of that so-called "tortoise" radial coordinate


-Richard

Jeff Root
2006-Aug-11, 02:47 AM
I wrote this for a different thread, but I'll post it here.




To another observer
farther away, though, the observer close to the black hole just
falls in, and the black hole expands slightly as he does.
By the phrase "expands slightly" do you mean that the size of the
event horizon increases which also means the strength of the
gravitational field grows?
Yes. Directly dependant on the mass of the infalling object,
of course. Negligible for anything less than planet-size, but
theoretically there.



From what I have read we the distant observer would never see the
participant fall into the event horizon and now from what you
state neither will the participant see themself fall in and so
how can the blackhole grow if no one ever sees anything enter it?
I think that the slowing has been described incorrectly, giving
a totally wrong idea of what relativity actually predicts.

You watch me with your hypertelescope as I fall toward a
comparatively large (maybe 40 solar masses) stellar-mass black
hole at ever-increasing speed. You have to adjust the controls
faster and faster to compensate for increasing redshift.
I am carrying a clock which we can both read. At one second
before noon, I'm approaching the BH at terrifying speed. The
near side of event horizon is moving away from me slightly,
but more distant parts are curving around me, so I can only
see a small circle of sky. Nothing special happens when the
clock reads noon, but I know that you will never see anything
that happens to me from that instant on, because I just crossed
your event horizon (according to calculations done ahead of the
trip, which we all agree are correct). I am still approaching
the BH at terrifying speed when the clock reads noon, and the
horizon is still between me and the black hole's center. One
tenth of a second later, when the clock reads 0.1 second after
noon, I am spaghettified, very close to the singularity.

Some time later the light from these events reaches you, and
you see my clock read one second before noon. I am redshifted
into the infrared and getting smaller and smaller with distance,
but your hypertelescope can still follow me, although the image
is getting grainy. One-half second later I'm redshifted into
the microwave, and you can just barely read my clock. It says
0.8 second before noon, so the clock has slowed significantly,
but I am falling away from you faster than ever. Three-tenths
of a second later I'm shifted into the radio portion of the
spectrum, and my clock reads 0.7 second before noon. A tenth
of a second after that and the signal is so weak that you can't
see me anymore. I'm gone.



bigsplit: What does the acronym "GRB" mean?
Gamma-ray burst or gamma-ray burster.

-- Jeff, in Minneapolis

Squashed
2006-Aug-11, 03:15 AM
...If you use either Eddington-Finkelstein or Kruskal-Szekeres ...

Just for reference I'll give the two links you gave me a while back: Eddington-Finkelstein (http://en.wikipedia.org/wiki/Eddington_coordinates) or Kruskal-Szekeres (http://en.wikipedia.org/wiki/Kruskal-Szekeres_coordinates).

publius
2006-Aug-11, 06:35 AM
I won't post the derivation unless someone is interested, but adding an initial velocity into this gives v/c = Sqrt(2/3) ~ 0.82. You'll note that is the same as the intial turning point speed in local coordinates. I was wondering if it would work out that way.

Now, about that "repulsive gravity" guy. That was a Dr. Felber, and he came up with v/c = 1/sqrt(3) (not sqrt(3)/2 as I thought I remembered).

So these are different by a factor of the square root of two, which, assuming he knew what he was doing, makes me less confident in the above E=mc^2 logic -- there may be something else that comes into play.

What is interesting to me here is the Felber number is smaller, which would seem to indicate the free-fall turning point occurs sooner than I came up with, at a larger radius.

-Richard