| Topic: |
Science > Physics |
| User: |
"Starblade Darksquall" |
| Date: |
10 Feb 2004 07:22:36 AM |
| Object: |
Frame dragging? |
How exactly did they determine that, in GR, a moving body would in
fact cause a dragging effect on outside objects? Is this dragging a
real dragging of time and space, or is it merely an expression?
(...Starblade Riven Darksquall...)
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| User: "Jeff Relf" |
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| Title: Re: Frame dragging? |
11 Feb 2004 05:08:32 AM |
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Hi Starblade Darksquall,
Re: Frame dragging,
You ask, " Is this dragging a real dragging of time and space ,
or is it merely an expression ? " ,
General relativity employs two free fall frames of reference,
a local one, where the laws of physics are invariant,
and a distant one which is relative.
The frame dragging would be observed in the distant frame,
not the local one.
To further illustrate this, to the distant frame,
the event horizon of frozen stars
emit almost no radiation, i.e. it seems frozen.
Yet at the event horizon,
the heat at the local frame approaches infinity.
Bizarre, no ?
.
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| User: "Gregory L. Hansen" |
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| Title: Re: Frame dragging? |
10 Feb 2004 12:55:28 PM |
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In article <4aa861fb.0402100522.2e0fa3b5@posting.google.com>,
Starblade Darksquall <Starblade13@Yahoo.com> wrote:
How exactly did they determine that, in GR, a moving body would in
fact cause a dragging effect on outside objects? Is this dragging a
real dragging of time and space, or is it merely an expression?
(...Starblade Riven Darksquall...)
The Kerr metric, the solution for a spherically symmetric mass with
angular momentum. It still hasn't been measured experimentally. The
frame dragging is analogous to a charged particle moving in a magnetic
field. When the particle moves in it gets "dragged" in one direction,
when it moves out it gets "dragged" in the other. And if it just sits
there, e.g. on a thin, stationary shell surrounding a spinning black
hole, it will just sit there and not get spun up to any particular speed.
It's not like a whirlpool. It also wouldn't make you dizzy, since it's
still like free-falling. Free-falling makes people nauseous, not dizzy.
There's shades of Mach here, since it implies angular rotation is not an
absolute, but that your state of zero angular momentum depends on what
the masses in your neighborhood are doing. You could be inside a spinning
spherical shell, a transparant shell so you could see that you're
rotating relative to the stars, but accelerometers you've brought with
you show no acceleration. It's a spacetime thing, not just an expression.
--
Irony: "Small businesses want relief from the flood of spam clogging their
in-boxes, but they fear a proposed national 'Do Not Spam' registry will
make it impossible to use e-mail as a marketing tool."
http://www.bizjournals.com/houston/stories/2003/11/10/newscolumn6.html
.
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| User: "Steve Harris" |
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| Title: Re: Frame dragging? |
10 Feb 2004 10:14:29 PM |
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(Gregory L. Hansen) wrote in message news:<c0b9f0$6s3$4@hood.uits.indiana.edu>...
The Kerr metric, the solution for a spherically symmetric mass with
angular momentum. It still hasn't been measured experimentally. The
frame dragging is analogous to a charged particle moving in a magnetic
field. When the particle moves in it gets "dragged" in one direction,
when it moves out it gets "dragged" in the other. And if it just sits
there, e.g. on a thin, stationary shell surrounding a spinning black
hole, it will just sit there and not get spun up to any particular speed.
It's not like a whirlpool. It also wouldn't make you dizzy, since it's
still like free-falling. Free-falling makes people nauseous, not dizzy.
There's shades of Mach here, since it implies angular rotation is not an
absolute, but that your state of zero angular momentum depends on what
the masses in your neighborhood are doing. You could be inside a spinning
spherical shell, a transparant shell so you could see that you're
rotating relative to the stars, but accelerometers you've brought with
you show no acceleration. It's a spacetime thing, not just an expression.
COMMENT:
I think I disagree. All experts on black holes point out that if you
fall into a rotating one, by the time you hit the event horizon (EH)
you cannot avoid moving angularly at whatever angular rotation the
hole has, since it's really space that is rotating, and being embedded
in space, you can't help doing what it does by the time you hit the
EH, any more than you can escape radially. So at the EH you move at c
in any kind of BH, but in a rotating one you're moving at c not
radially inward, but at an angle. And yes, your on board
accelerometers will show no acceleration (at any point, forget tides),
but that doesn't mean that to *outside viewers* you're not being
angularly accelerated around a rotating hole, just as you're being
downwardly accelerated.
At distances beyond the EH you can avoid doing what space is doing,
but only at the price of actively accelerating. So you have to fire
your rockets not only to keep from being dragged into a rotating hole,
but also to keep from being dragged around radially by frame drag. So
it IS like a whirlpool. It's just that around bodies like the earth,
the angular acceleration is so small compared with the radial one,
that it's hardly noticable. And so far, absent Gravity Probe B,
unmeasurable.
SBH
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| User: "Bruce Bowen" |
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| Title: Re: Frame dragging? |
11 Feb 2004 04:16:16 PM |
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(Steve Harris sbharris@ROMAN9.netcom.com) wrote in message news:<79cf0a8.0402102014.49e75522@posting.google.com>...
glhansen@steel.ucs.indiana.edu (Gregory L. Hansen) wrote in message news:<c0b9f0$6s3$4@hood.uits.indiana.edu>...
The Kerr metric, the solution for a spherically symmetric mass with
angular momentum. It still hasn't been measured experimentally. The
frame dragging is analogous to a charged particle moving in a magnetic
field. When the particle moves in it gets "dragged" in one direction,
when it moves out it gets "dragged" in the other. And if it just sits
there, e.g. on a thin, stationary shell surrounding a spinning black
hole, it will just sit there and not get spun up to any particular speed.
It's not like a whirlpool. It also wouldn't make you dizzy, since it's
still like free-falling. Free-falling makes people nauseous, not dizzy.
There's shades of Mach here, since it implies angular rotation is not an
absolute, but that your state of zero angular momentum depends on what
the masses in your neighborhood are doing. You could be inside a spinning
spherical shell, a transparant shell so you could see that you're
rotating relative to the stars, but accelerometers you've brought with
you show no acceleration. It's a spacetime thing, not just an expression.
COMMENT:
I think I disagree. All experts on black holes point out that if you
fall into a rotating one, by the time you hit the event horizon (EH)
you cannot avoid moving angularly at whatever angular rotation the
hole has, since it's really space that is rotating, and being embedded
in space, you can't help doing what it does by the time you hit the
EH, any more than you can escape radially. So at the EH you move at c
in any kind of BH, but in a rotating one you're moving at c not
radially inward, but at an angle. And yes, your on board
accelerometers will show no acceleration (at any point, forget tides),
but that doesn't mean that to *outside viewers* you're not being
angularly accelerated around a rotating hole, just as you're being
downwardly accelerated.
At distances beyond the EH you can avoid doing what space is doing,
but only at the price of actively accelerating. So you have to fire
your rockets not only to keep from being dragged into a rotating hole,
but also to keep from being dragged around radially by frame drag. So
it IS like a whirlpool.
Here's another metaphore. If you what to think of the situation in
terms of absolute space intead of induction, then frame dragging is
more analogous to the optical "Fizeau Effect", which describes optical
refraction in a moving fluid. This is different than a viscous tug.
Think of the surrounding space as concentric layers of laminar flowing
fluid, with the inside layers moving faster than the outside. A
vertical infalling ray of light would refract spinward in this
situation, but a concentric orbiting ray of light would not be
accelerated either spinward or anti-spinward. To be consistent the
index of refraction needs to increase as well as you go lower into the
gravity well. At the event horizion the index of refraction goes to
infinity. At the static limit the angular fluid flow rate is the same
as the medium wave velocity.
-Bruce bbowen@pppppppppppacbell.nettttttttttttt
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| User: "Bruce Bowen" |
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| Title: Re: Frame dragging? |
11 Feb 2004 03:48:32 PM |
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(Steve Harris sbharris@ROMAN9.netcom.com) wrote in message news:<79cf0a8.0402102014.49e75522@posting.google.com>...
glhansen@steel.ucs.indiana.edu (Gregory L. Hansen) wrote in message news:<c0b9f0$6s3$4@hood.uits.indiana.edu>...
The Kerr metric, the solution for a spherically symmetric mass with
angular momentum. It still hasn't been measured experimentally. The
frame dragging is analogous to a charged particle moving in a magnetic
field. When the particle moves in it gets "dragged" in one direction,
when it moves out it gets "dragged" in the other. And if it just sits
there, e.g. on a thin, stationary shell surrounding a spinning black
hole, it will just sit there and not get spun up to any particular speed.
It's not like a whirlpool. It also wouldn't make you dizzy, since it's
still like free-falling. Free-falling makes people nauseous, not dizzy.
There's shades of Mach here, since it implies angular rotation is not an
absolute, but that your state of zero angular momentum depends on what
the masses in your neighborhood are doing. You could be inside a spinning
spherical shell, a transparant shell so you could see that you're
rotating relative to the stars, but accelerometers you've brought with
you show no acceleration. It's a spacetime thing, not just an expression.
COMMENT:
I think I disagree. All experts on black holes point out that if you
fall into a rotating one, by the time you hit the event horizon (EH)
you cannot avoid moving angularly at whatever angular rotation the
hole has, since it's really space that is rotating, and being embedded
in space, you can't help doing what it does by the time you hit the
EH, any more than you can escape radially. So at the EH you move at c
in any kind of BH, but in a rotating one you're moving at c not
radially inward, but at an angle. And yes, your on board
accelerometers will show no acceleration (at any point, forget tides),
but that doesn't mean that to *outside viewers* you're not being
angularly accelerated around a rotating hole, just as you're being
downwardly accelerated.
At distances beyond the EH you can avoid doing what space is doing,
but only at the price of actively accelerating. So you have to fire
your rockets not only to keep from being dragged into a rotating hole,
but also to keep from being dragged around radially by frame drag. So
it IS like a whirlpool.
No it's not. Space is not viscous!! A stationary plumb will hang
vertically over a rotating black hole (outside the static limit). It
will not list spinward or anti-spinward. A FALLING plumb will receive
and induced spinward gravitomagnetic acceleration. The falling plumb
will of course not "feel" this due to the principle of equivalence.
Until you get into stong non-linear gravity regions the situation is
qualitively exactly the same as a classic negative charge orbiting a
more massive, positively charged spinning object.
The reason an object must orbit spinward inside the static limit is
because the RADIAL force on a STATIONARY object diverges at the static
limit (this is "trivial" to show by plugging the values into the Kerr
metric), hence it MUST start to fall, in which case it immediately
gets an induced spinward orbit.
Think of a figure skater over the equator and parallel to the surface
(i.e., horizontal, or "standing" on a wall, or rotating on a spit). If
she extends her arms the inside arm will be torqued spinward and the
outside arm will be torqued anti-spinward. There is a specific
initial rotation (dependent on location) at which, when she extends
her arms nothing changes. This is the local inertial reference for
non-rotation. It is NOT stationary with respect to the fixed stars.
She is philosophically like a planetary gear with the BH being the sun
gear and the universe being the ring gear.
Another fascinating feature is that, due to time dilation, the speed
of light (relative to the outside world) goes to zero at the event
horizon, but the rotation rate DOES NOT!!! This is totally cool! One
way to view the static limit is as the point were the rotation rate
exceeds the local speed of light. ("Rotation Rate" must be understood
somewhat metaphorically because there is nothing physically there
(other than space, except perhaps in a limiting neutron star) that is
physically rotating. To an outside observer, at the event horizon the
speed of light is zero while the rotation rate is finite. So to an
outside observer everything that crosses the event horizon does so at
a fixed angular rotation frequency.
In a very real sense gravity literally does not care about the speed
of light, which is why things like black holes, ergospheres and
singularities exist in the first place.
-Bruce bbowen@pppppppppppppppacbell.nnnnnnnnnnnnnet
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| User: "Edward Green" |
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| Title: Re: Frame dragging? |
15 Feb 2004 06:40:47 AM |
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(Bruce Bowen) wrote in message news:<b824a8a0.0402111348.6a1c30e9@posting.google.com>...
sbharris@ix.netcom.com (Steve Harris sbharris@ROMAN9.netcom.com) wrote in message news:<79cf0a8.0402102014.49e75522@posting.google.com>...
....
At distances beyond the EH you can avoid doing what space is doing,
but only at the price of actively accelerating. So you have to fire
your rockets not only to keep from being dragged into a rotating hole,
but also to keep from being dragged around radially by frame drag. So
it IS like a whirlpool.
Furthering my riff to you about this, I think you can always "avoid
what space is doing" to exactly the same extent: that is, some, and
not at all.
For example, standing next to me, you have the freedom to move
relatively to me in an open velocity ball of radius c. Standing on
the other side of a black hole horizon from me, or over a distant
event horizon in an expanding universe, you retain this same freedom
of movement -- locally. But globally space can arrange so we can
never meet again. Space proposes our fated geodesic, always giving us
some room, but never complete freedom.
No it's not. Space is not viscous!!
That was my lay-reaction: there is a real sense in GR in which we are
locally trapped in space and dragged along with it -- with a remaining
local freedom of movement -- but not, as you indicate, a viscous
sense. Yet there are at least three test cases where worldlines are
fatefully controlled by space: objects over an horizon in an expanding
universe cannot return to us, objects over an horizon in a collapsing
part of the universe (i.e., a black hole) cannot return to us , and
.... subject of current thread ... objects over a different horizon
next to a massive rotating object cannot avoid moving _laterally_ wrt
to us.
The last, transverse, case is deliciously suggestive of a
generalization. The horizons associated with expanding and collapsing
spaces ... or should I say spacetime geometries ... may be more
commonly considered by (relative) simplicity of the solutions. The
Kerr metric is somewhat more complicated and exotic, and something
very much like this happens _across_ a line joining two observers, and
acquires a new name. Now, it would be nice to be able to dispense
with special cases and understand this as a general feature of GR:
there is _some_ sense in which patches of space enjoy a relative
velocity wrt to each other in GR, though we may only be driven to
consider this as a special phenomenon when this presumptive relative
velocity exceeds c.
A stationary plumb will hang
vertically over a rotating black hole (outside the static limit). It
will not list spinward or anti-spinward. A FALLING plumb will receive
and induced spinward gravitomagnetic acceleration. <...>
Are you _quite_ sure a hanging plumb will hang vertically before
crossing the ergosphere? Abstract considerations of continuity of
solutions suggest otherwise.
The reason an object must orbit spinward inside the static limit is
because the RADIAL force on a STATIONARY object diverges at the static
limit <...> hence it MUST start to fall, in which case it immediately
gets an induced spinward orbit.
So something there is divergent at the static limit, a blow to my
"continuity" argument. I remain skeptical. If the radial force
diverges, then why isn't this the same as the event horizon, beyond
which an object cannot be retrieved. Are you certain it's not the
transverse component of force required to keep the bob vertical which
diverges at the stationary limit?
Here's a solution I'd like to see: two infinite parallel sheets of
relatively moving matter. This might nicely separate "frame drag"
from rotation.
Think of a figure skater over the equator and parallel to the surface
(i.e., horizontal, or "standing" on a wall, or rotating on a spit). If
she extends her arms the inside arm will be torqued spinward and the
outside arm will be torqued anti-spinward. There is a specific
initial rotation (dependent on location) at which, when she extends
her arms nothing changes. This is the local inertial reference for
non-rotation. It is NOT stationary with respect to the fixed stars.
She is philosophically like a planetary gear with the BH being the sun
gear and the universe being the ring gear.
Neat. You could also say, from the POV of an object _emerging_ from
the hole, that it is being "frame dragged" by the rest of the
universe!
Another fascinating feature is that, due to time dilation, the speed
of light (relative to the outside world) goes to zero at the event
horizon, but the rotation rate DOES NOT!!! This is totally cool! One
way to view the static limit is as the point were the rotation rate
exceeds the local speed of light.
That was about my idea -- making an analogy with limits where the
recession velocity of distant parts of the universe exceeds the speed
of light.
("Rotation Rate" must be understood
somewhat metaphorically because there is nothing physically there
(other than space, except perhaps in a limiting neutron star) that is
physically rotating.
Ooohhhh. :-)
Is it then consonant with GR to have a material object -- say the
surface of a sufficiently massive neutron star -- spinning faster than
c relative to nearby material objects?
I was recently speculating about such things -- given the apparently
acknowledged possibility in GR of writing down solutions where "X
moves with some v > c wrt Y", for some reasonable sense of relative
velocity "v" in context -- of writing down solutions which would make
such effects much more up close and personal than "distant parts of
the universe", and demonstrate a possibility for ftl travel within GR
which didn't require wormholes. cf:
In a very real sense gravity literally does not care about the speed
of light, which is why things like black holes, ergospheres and
singularities exist in the first place.
<snipped from above>
The falling plumb
will of course not "feel" this due to the principle of equivalence.
Until you get into stong non-linear gravity regions the situation is
qualitively exactly the same as a classic negative charge orbiting a
more massive, positively charged spinning object.
Given the close relation of magnetism to spin and angular momentum
this (quantitative?) analogy is equally suggestive in both directions.
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| User: "Bruce Bowen" |
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| Title: Re: Frame dragging? |
15 Feb 2004 03:20:01 PM |
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(Edward Green) wrote in message news:<2a0cceff.0402150440.66ae3e53@posting.google.com>...
brucebo@my-deja.com (Bruce Bowen) wrote in message news:<b824a8a0.0402111348.6a1c30e9@posting.google.com>...
A stationary plumb will hang
vertically over a rotating black hole (outside the static limit). It
will not list spinward or anti-spinward. A FALLING plumb will receive
and induced spinward gravitomagnetic acceleration. <...>
Are you _quite_ sure a hanging plumb will hang vertically before
crossing the ergosphere? Abstract considerations of continuity of
solutions suggest otherwise.
Quite sure. You have the metric, you have the initial position and
direction (4-velocity) (dt!=0, dr=0, d_theta=0, d_phi=0), and you have
the Euler-Lagrange equations. What more do you need? :) Actually
it's much simpler than I imply. You're only interested is the space
tract direction at time 0 when you release the plumb. In this case,
dr/dt, dtheta/dt and dphi/dt are also initially at zero, so it is
adequate to show that at time 0, d^2r/dt^2 !=0 while d^2theta/dt^2 and
d^2phy/dt^2 =0.
The reason an object must orbit spinward inside the static limit is
because the RADIAL force on a STATIONARY object diverges at the static
limit <...> hence it MUST start to fall, in which case it immediately
gets an induced spinward orbit.
So something there is divergent at the static limit, a blow to my
"continuity" argument. I remain skeptical. If the radial force
diverges, then why isn't this the same as the event horizon, beyond
which an object cannot be retrieved. Are you certain it's not the
transverse component of force required to keep the bob vertical which
diverges at the stationary limit?
It only diverges for a stationary object, not a moving (spinward)
object. This is very easy to show. At the static limit if you plug
in dr=0, dtheta=0 and dphi=0 you get ds/dt=0 or conversely,
dt/ds=infinity.
-Bruce
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| User: "Bruce Bowen" |
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| Title: Re: Frame dragging? |
01 Mar 2004 12:25:29 PM |
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(Edward Green) wrote in message news:<2a0cceff.0402150440.66ae3e53@posting.google.com>...
brucebo@my-deja.com (Bruce Bowen) wrote in message news:<b824a8a0.0402111348.6a1c30e9@posting.google.com>...
The reason an object must orbit spinward inside the static limit is
because the RADIAL force on a STATIONARY object diverges at the static
limit <...> hence it MUST start to fall, in which case it immediately
gets an induced spinward orbit.
So something there is divergent at the static limit, a blow to my
"continuity" argument. I remain skeptical. If the radial force
diverges, then why isn't this the same as the event horizon, beyond
which an object cannot be retrieved.
In some ways it is. For a Schwarzchild BH, the event horizon is at r =
2M. For a Kerr BH the ***Static Limit*** is at 2M (at the
equator)!!! Where "M" is the total mass/energy, including rotational
kinetic energy. The actual event horizon is closer in, at M +
sqrt(M^2 - a^2) where "a" is the angular momentum.
This line of argument brings to mind a recent mental query. If you
cross the static limit with initially NO angular velocity, can you get
up enough speed in time to still leave the BH (ergosphere)???? Or do
you have to dip into the ergosphere with some initial spinward
velocity in order to safely traverse it and re-emerge? Interesting
homework problem. :)
Another interesting feature:
The equatorial spinward "photon sphere" (unstable circular orbit for
a photon) is farther out than the anti-spinward photon sphere.
-Bruce
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| User: "Bruce Bowen" |
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| Title: Re: Frame dragging? |
01 Mar 2004 12:30:03 PM |
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Oops, I meant to say the "anti-spinward" photon sphere has a larger
orbit than the spinward photon sphere.
Another interesting feature:
The equatorial spinward "photon sphere" (unstable circular orbit
for a photon) is farther out than the anti-spinward photon sphere.
-Bruce
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| User: "Gregory L. Hansen" |
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| Title: Re: Frame dragging? |
11 Feb 2004 09:03:45 AM |
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In article <79cf0a8.0402102014.49e75522@posting.google.com>,
Steve Harris <sbharris@ix.netcom.com> wrote:
glhansen@steel.ucs.indiana.edu (Gregory L. Hansen) wrote in message
news:<c0b9f0$6s3$4@hood.uits.indiana.edu>...
The Kerr metric, the solution for a spherically symmetric mass with
angular momentum. It still hasn't been measured experimentally. The
frame dragging is analogous to a charged particle moving in a magnetic
field. When the particle moves in it gets "dragged" in one direction,
when it moves out it gets "dragged" in the other. And if it just sits
there, e.g. on a thin, stationary shell surrounding a spinning black
hole, it will just sit there and not get spun up to any particular speed.
It's not like a whirlpool. It also wouldn't make you dizzy, since it's
still like free-falling. Free-falling makes people nauseous, not dizzy.
There's shades of Mach here, since it implies angular rotation is not an
absolute, but that your state of zero angular momentum depends on what
the masses in your neighborhood are doing. You could be inside a spinning
spherical shell, a transparant shell so you could see that you're
rotating relative to the stars, but accelerometers you've brought with
you show no acceleration. It's a spacetime thing, not just an expression.
COMMENT:
I think I disagree. All experts on black holes point out that if you
fall into a rotating one, by the time you hit the event horizon (EH)
you cannot avoid moving angularly at whatever angular rotation the
hole has, since it's really space that is rotating, and being embedded
in space, you can't help doing what it does by the time you hit the
EH, any more than you can escape radially. So at the EH you move at c
in any kind of BH, but in a rotating one you're moving at c not
radially inward, but at an angle. And yes, your on board
accelerometers will show no acceleration (at any point, forget tides),
but that doesn't mean that to *outside viewers* you're not being
angularly accelerated around a rotating hole, just as you're being
downwardly accelerated.
At distances beyond the EH you can avoid doing what space is doing,
but only at the price of actively accelerating. So you have to fire
your rockets not only to keep from being dragged into a rotating hole,
but also to keep from being dragged around radially by frame drag. So
it IS like a whirlpool. It's just that around bodies like the earth,
the angular acceleration is so small compared with the radial one,
that it's hardly noticable. And so far, absent Gravity Probe B,
unmeasurable.
SBH
http://scienceworld.wolfram.com/physics/KerrBlackHole.html
Specialize to the equator, divide by (dt)^2, and note that there's no
angular acceleration terms. Solve the Kerr metric for d(phi)/dt. The
only time dependence will involve dr/dt and d(tau)/dt. If, at constant r,
you change your angular speed (e.g. come to rest with respect to the
stars) you'll stay there because there's no d^2(phi)/dt^2 term. If you
calculate a d^2(phi)/dt^2 by taking the time derivative of that
expression, it will be in terms of changing radius, changing mass,
changing angular momentum, but not in terms of angular speed.
--
"The average person, during a single day, deposits in his or her underwear
an amount of fecal bacteria equal to the weight of a quarter of a peanut."
-- Dr. Robert Buckman, Human Wildlife, p119.
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| User: "Steve Harris" |
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| Title: Re: Frame dragging? |
11 Feb 2004 03:50:20 PM |
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(Gregory L. Hansen) wrote in message news:<c0dg8h$uc9$3@hood.uits.indiana.edu>...
COMMENT:
I think I disagree. All experts on black holes point out that if you
fall into a rotating one, by the time you hit the event horizon (EH)
you cannot avoid moving angularly at whatever angular rotation the
hole has, since it's really space that is rotating, and being embedded
in space, you can't help doing what it does by the time you hit the
EH, any more than you can escape radially. So at the EH you move at c
in any kind of BH, but in a rotating one you're moving at c not
radially inward, but at an angle. And yes, your on board
accelerometers will show no acceleration (at any point, forget tides),
but that doesn't mean that to *outside viewers* you're not being
angularly accelerated around a rotating hole, just as you're being
downwardly accelerated.
At distances beyond the EH you can avoid doing what space is doing,
but only at the price of actively accelerating. So you have to fire
your rockets not only to keep from being dragged into a rotating hole,
but also to keep from being dragged around radially by frame drag. So
it IS like a whirlpool. It's just that around bodies like the earth,
the angular acceleration is so small compared with the radial one,
that it's hardly noticable. And so far, absent Gravity Probe B,
unmeasurable.
SBH
http://scienceworld.wolfram.com/physics/KerrBlackHole.html
Specialize to the equator, divide by (dt)^2, and note that there's no
angular acceleration terms. Solve the Kerr metric for d(phi)/dt. The
only time dependence will involve dr/dt and d(tau)/dt. If, at constant r,
you change your angular speed (e.g. come to rest with respect to the
stars) you'll stay there because there's no d^2(phi)/dt^2 term. If you
calculate a d^2(phi)/dt^2 by taking the time derivative of that
expression, it will be in terms of changing radius, changing mass,
changing angular momentum, but not in terms of angular speed.
Look, just google "black hole stationary limit". Kerr holes have an
ergosphere defined by a stationary limit. Inside that limit, but
before you reach the event horizon, you cannot remain stationary WRT
the rest of the non-rotating universe. Equations for you minimal
angular rotation rate as a function of radial r, are given in several
sources.
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| User: "Edward Green" |
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| Title: Re: Frame dragging? |
14 Feb 2004 09:48:35 AM |
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(Steve Harris sbharris@ROMAN9.netcom.com) wrote in message news:<79cf0a8.0402111350.1f565361@posting.google.com>...
First of all, let me say how delighting I am to read your not entirely
expert speculations. We both know that to really understand something
like this, a minimum prerequisite to to grok the mathematics. But in
this, I think we are both laymen, so, continuing peer-to-peer ...
I think I disagree. All experts on black holes point out that if you
fall into a rotating one, by the time you hit the event horizon (EH)
you cannot avoid moving angularly at whatever angular rotation the
hole has, since it's really space that is rotating, and being embedded
in space, you can't help doing what it does by the time you hit the
EH, any more than you can escape radially. So at the EH you move at c
in any kind of BH, but in a rotating one you're moving at c not
radially inward, but at an angle. And yes, your on board
accelerometers will show no acceleration (at any point, forget tides),
but that doesn't mean that to *outside viewers* you're not being
angularly accelerated around a rotating hole, just as you're being
downwardly accelerated.
Regarding being dragged by space: "it's really space that is rotating,
and being embedded in space, you can't help doing what it does". I
think something like this is true, but we must be careful in
expressing it. One _might_ think that this idea violated special
covariance.
(Pause while Steve realizes Ed has picked up a new phrase on the
street).
Let's, just for the sake of contrapositive argument, say that space
had a preferred local rest frame into which we tended to settle, a la
Aristotle. Then, if space was moving past us at .7 c (relative to the
local space of a distant observer) we would eventually, in the absence
of a continuously applied force, be dragged along at that rate also.
Now _that_ would be frame dragging, I tell you!
But, let's suppose again for the sake of argument, that we are again
in the generally relativistic world (locally specially covariant), and
that we _still_ are able to assign some meaning to a claim "the
relative velocity of our space to that of a distant observer is .7 c".
If we were now at initially at rest wrt to that distant observer
(assuming for the sake of argument we have in mind a well defined
sense of "relative velocity" over distant separations), we will remain
at rest wrt that observer: space does not "drag" us along.
Now, let's take things one step further: suppose, in whatever sense we
were able to make of this measure above, that our local space is now
moving at 1.5c wrt to that distant observer: say, receding. _Now_ we
can no longer remain at rest wrt that obsever: in fact, our possible
recessional velocities wrt that observer now lie in the open inverval
(0.5, 2.5). The threshold for being able to remain at rest wrt to the
distant observer is evidently crossed at a "relative space velocity"
(rsv) of c.
Ok. We are now, armed with the ad hoc concept of rsv, in a position
to make sweeping speculo-generalizations tying together ordinary black
holes, Kerr black holes, cosmology and Valentine's day!
First we explore some more speculo-results:
rsv < c : a distant observer may remain stationary wrt us, but there
are still observable consequences to the rsv. Moving in free fall
from our location to that distant location a test body will tend to
accelerate: the rocket will only be able to remain at rest wrt us by
constantly firing rocket motors.
rsv > c : a distant observer can no longer remain stationary wrt us
by any amount of rocket firing.
rsv || separation : test body moving away from us will accelerate
along the line of sepration. When rsv > c (assuming rsv points away
from us), the body will no longer be able to return itself or a signal
by any method.
rsv |_ separation : test body moving away from us will accelerate
athwart our line of separation. When rsv > c, the body will no longer
be able to avoid perpendicular motion by any method.
Suddenly, a lot of things seem likely to fall together:
In cosmology, distant parts of space are receding from us. When the
rsv exceeds c, an horizon is crossed, and we lose all knowledge of
parts of the universe outside it.
In a black hole, cosmological geometry is reversed, and nearby parts
of space are receding _inwardly_ from us. When the rsv exceeds c, an
horizon is crossed, and we lose all knowledge of of parts of the
universe inside it.
In a rotating black hole, nearby part of space are moving _across_ our
path. When the (perpedicular) rsv exceeds c, an horizon is crossed,
and bodies inside it are no longer able to avoid moving across our
path. In this case, however, communication with space outside the
horizon is not lost, and escape may be possible by following the
advice given to a bather in the ocean: swim across the current.
In any of these cases, before an horizon is crossed it is possible to
resist acceleration relative to a distant observer by sufficient
application of reactive thrust.
It seems, following the guidance of our still steaming
speculo-generalitive insight, that "frame-dragging" belongs in the
great class of misleading slogans. It suggests by common usage that
there is something special going on in rotating cases not going on in
rotationally static ones. Indeed there is, but _not_ something which
seems to deserve the exclusive use of the name "frame dragging": what
is special is that in the rotating case the frames are being dragged
not only in the line of seperation, but across it. What is outside
our ordinary experience and leads us to coin a special phrase is
"sideways gravity". Whatever "frame dragging" is, it seems likely to
best describe a range of normal cases, not a special case.
Disclaimer: all the above speculo-generalization is of course posited
on an effective map to the math of GR. The writers subjective
Bayesian confidence that he is at least wrong, and maybe not entirely
wrong, is 80%.
Comment: I still don't understand (even on this level) how it is that
space seems not merely to have relative velocity, but relative
_acceleration_, even when it is not going anywhere. If space merely
had relative velocity, then, given an initial correction for a probe
moving through a relative velocity (rsv) difference of .7c, you would
expect it to just hang there, inertially. But this seems never to be
the case: as an object falls from infinity to the surface of the
Earth, it is not sufficient simply to brake at a given height; it must
continue braking, or else continue falling.
I notice this appears to conflict with the paragraph beginning "But,
let's suppose ...". Oh well. Work in progress.
At distances beyond the EH you can avoid doing what space is doing,
but only at the price of actively accelerating. So you have to fire
your rockets not only to keep from being dragged into a rotating hole,
but also to keep from being dragged around radially by frame drag. So
it IS like a whirlpool. It's just that around bodies like the earth,
the angular acceleration is so small compared with the radial one,
that it's hardly noticable. And so far, absent Gravity Probe B,
unmeasurable.
Look, just google "black hole stationary limit". Kerr holes have an
ergosphere defined by a stationary limit. Inside that limit, but
before you reach the event horizon, you cannot remain stationary WRT
the rest of the non-rotating universe. Equations for you minimal
angular rotation rate as a function of radial r, are given in several
sources.
.
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| User: "Gregory L. Hansen" |
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| Title: Re: Frame dragging? |
11 Feb 2004 08:13:04 PM |
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In article <79cf0a8.0402111350.1f565361@posting.google.com>,
Steve Harris <sbharris@ix.netcom.com> wrote:
glhansen@steel.ucs.indiana.edu (Gregory L. Hansen) wrote in message
news:<c0dg8h$uc9$3@hood.uits.indiana.edu>...
COMMENT:
I think I disagree. All experts on black holes point out that if you
fall into a rotating one, by the time you hit the event horizon (EH)
you cannot avoid moving angularly at whatever angular rotation the
hole has, since it's really space that is rotating, and being embedded
in space, you can't help doing what it does by the time you hit the
EH, any more than you can escape radially. So at the EH you move at c
in any kind of BH, but in a rotating one you're moving at c not
radially inward, but at an angle. And yes, your on board
accelerometers will show no acceleration (at any point, forget tides),
but that doesn't mean that to *outside viewers* you're not being
angularly accelerated around a rotating hole, just as you're being
downwardly accelerated.
At distances beyond the EH you can avoid doing what space is doing,
but only at the price of actively accelerating. So you have to fire
your rockets not only to keep from being dragged into a rotating hole,
but also to keep from being dragged around radially by frame drag. So
it IS like a whirlpool. It's just that around bodies like the earth,
the angular acceleration is so small compared with the radial one,
that it's hardly noticable. And so far, absent Gravity Probe B,
unmeasurable.
SBH
http://scienceworld.wolfram.com/physics/KerrBlackHole.html
Specialize to the equator, divide by (dt)^2, and note that there's no
angular acceleration terms. Solve the Kerr metric for d(phi)/dt. The
only time dependence will involve dr/dt and d(tau)/dt. If, at constant r,
you change your angular speed (e.g. come to rest with respect to the
stars) you'll stay there because there's no d^2(phi)/dt^2 term. If you
calculate a d^2(phi)/dt^2 by taking the time derivative of that
expression, it will be in terms of changing radius, changing mass,
changing angular momentum, but not in terms of angular speed.
Look, just google "black hole stationary limit". Kerr holes have an
ergosphere defined by a stationary limit. Inside that limit, but
before you reach the event horizon, you cannot remain stationary WRT
the rest of the non-rotating universe. Equations for you minimal
angular rotation rate as a function of radial r, are given in several
sources.
I vaguely remember reading something about that. But black holes have a
lot of atypical cases. The "black hole stationary limit" does not apply
when you're sufficiently far from a black hole, when you're interested in
a neutron star, doesn't apply to the Gravity Probe B experiment around
Earth, for instances.
--
"Don't try to teach a pig how to sing. You'll waste your time and annoy
the pig."
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| User: "Sam Wormley" |
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| Title: Re: Frame dragging? |
10 Feb 2004 11:50:16 AM |
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Starblade Darksquall wrote:
How exactly did they determine that, in GR, a moving body would in
fact cause a dragging effect on outside objects? Is this dragging a
real dragging of time and space, or is it merely an expression?
(...Starblade Riven Darksquall...)
See: http://www.google.com/search?q=%22Frame+Dragging%22
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