Re: Gravity and black holes
Oh No
>Hi
>
>I have been wondering about what happens to the gravity from a star when
>it becomes a black hole.
>
>If I have read correctly (and that is not a given) when the star
>collapses an event horizon surrounding it is created, thus it no longer
>can exert a gravitational influence on any other body.
>
No, this is not right. A black hole does exert gravitational influence
on its surroundings.
Regards
--
Charles Francis
moderator sci.physics.foundations.
charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and
braces)
Oh No
>In article <L$0vwXBVWU$KF...@charlesfrancis.wanadoo.co.uk>,
> Oh No <No...@charlesfrancis.wanadoo.co.uk> wrote:
>
>> Thus spake David <post...@REMOVE-TO-REPLYconfidential-counselling.com>
>> >Hi
>> >
>> >I have been wondering about what happens to the gravity from a star when
>> >it becomes a black hole.
>> >
>> >If I have read correctly (and that is not a given) when the star
>> >collapses an event horizon surrounding it is created, thus it no longer
>> >can exert a gravitational influence on any other body.
>> >
>>
>> No, this is not right. A black hole does exert gravitational influence
>> on its surroundings.
>
>not escape because the collapsing "generated" an escape velocity greater
>than the speed of light.
>
>If that is true it must mean that gravity does travel faster than light,
>and so...
>
>Or am I still confused?
The gravity of a black hole (or a planet or a star) can be considered as
a static field - it does not travel. Changes in the gravitational field
travel at light speed.
glird
> Thus spake David <posti...@REMOVE-TO-REPLYconfidential-counselling.com>
>
>
>
> >In article <L$0vwXBVWU$KF...@charlesfrancis.wanadoo.co.uk>,
> > Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
>
> >> Thus spake David <posti...@REMOVE-TO-REPLYconfidential-counselling.com>
> >> >Hi
>
> >> >I have been wondering about what happens to the gravity from a star when
> >> >it becomes a black hole.
>
> >> >If I have read correctly (and that is not a given) when the star
> >> >collapses an event horizon surrounding it is created, thus it no longer
> >> >can exert a gravitational influence on any other body.
>
> >> No, this is not right. A black hole does exert gravitational influence
> >> on its surroundings.
>
> >Thanks, much appreciated - the article I read explained that light could
> >not escape because the collapsing "generated" an escape velocity greater
> >than the speed of light.
>
> >If that is true it must mean that gravity does travel faster than light,
> >and so...
>
> >Or am I still confused?
The article is wrong. The reason light remains within an event-
horizon is that the curvature of its outbound path is sufficient to
bend it either parallel to or back from that spherical level.
> The gravity of a black hole (or a planet or a star) can be considered as
> a static field - it does not travel. Changes in the gravitational field
> travel at light speed.
Two aspects of "gravity" are a gravitational FIELD and a
gravitational FORCE.
The g-field is a density gradient in the material surrounding a
nucleus, whether it is an atomic nucleus or a plant or star or galaxy
etc. As such, the g-field remains at rest relative to the nucleus that
caused it to exist.
The g-force arises inside each atom that is a component of a body
embedded in and partaking of that grad d. As such, it doesn't 'travel"
anywhere other than where the embedded body does.
glird
Eliot Specht
"glird" <gl...@aol.com> wrote in message
news:3bcf7b62-6b4a-473d...@t42g2000yqd.googlegroups.com...
While the collapse of a black hole is fast from the viewpoint of a local
observer, the stong gravity makes the collapse slow down form the viewpoint
of a distant observer. In fact, a distant observer will never see the event
horizon form because the "last photon" to be emitted prior to the formation
of the event horizon takes so long to get out.
As a result, the gravitational field measured by a distant observer comes
from mass which was outside the event horizon.
Eliot Specht
Tom Roberts
> While the collapse of a black hole is fast from the viewpoint of a local
> observer, the stong gravity makes the collapse slow down form the viewpoint
> of a distant observer. In fact, a distant observer will never see the event
> horizon form because the "last photon" to be emitted prior to the formation
> of the event horizon takes so long to get out.
Yes, in principle. Well, sort of -- the horizon can never be "seen". And the
horizon does not really "form", but rather grows outward from a point (it always
moves with local speed c, which is as much a statement about local inertial
frames as it is about "movement" of the horizon).
In practice, to a distant observer such a collapsing object is observationally
indistinguishable from a black hole, just a short time after the final collapse
begins. The start of the final collapse is (in principle at least) visible to
the distant observer, as the surface begins collapsing before the surface is
engulfed by the horizon.
> As a result, the gravitational field measured by a distant observer comes
> from mass which was outside the event horizon.
The gravitational field does not "come" from mass. Your statement is seriously
over-simplistic. Note that for a spherical object that collapses in a
spherically symmetric manner [#], the gravitational field outside its surface is
static and remains constant regardless of whether or not it collapses
completely. The gravitational field at the location of the observer does not
depend on the radius of such an object, as long as that radius is smaller than
the radial distance of the observer from its center; this holds for a spherical
surface just under the observer's feet, all the way to complete collapse into a
black hole (the observer must of course remain at constant radial distance).
[#] I choose this simple situation, because it is simple
enough for the answer to be known rigorously within GR.
Tom Roberts
glird
>
>< Note that for a spherical object that collapses in a spherically symmetric manner [#], the gravitational field outside its surface is static and remains constant regardless of whether or not it collapses completely. [snip]
[#] I choose this simple situation, because it is simple enough for
the answer to be known rigorously within GR. >
Consider a spherical object moving along the X axis at v. Suppose
it deforms by phi(v)q in X and by phi(v) in Y and Z. For any value of
q, there are only two values of phi(v) for which the object will
deform asymmetrically. One of them is phi(v) = 1. THAT is its value in
the LTE.
Note, then, that "reasons of symmetry" are ruled out by phi(v) = 1!
Similarly, since the space and time of STR and GR deform
asymmetrically, the space-time continuum cannot be homogeneous.
Therefore Einstein's assertion that his 1905 paper's "equations must
be linear on account of the properties of homogeneity which we
attribute to space and time" requires that phi(v) = q = 1; which rules
out the LTE!!
glird
Tom Roberts
> Consider a spherical object moving along the X axis at v. Suppose
> it deforms by phi(v)q in X and by phi(v) in Y and Z.
Then in GR you must include the mechanism of deformation in the energy-momentum
tensor. That gives you a completely different problem to study, and it's not
obvious how to compare two such different situations....
> For any value of
> q, there are only two values of phi(v) for which the object will
> deform asymmetrically.
Huh? If q != 1 that deformation is not spherically symmetric, for any nonzero
phi(v).
There are many different types of symmetry. You must be more
specific. For instance, for any nonzero values of q and phi(v)
that deformation is cylindrically symmetric with axis along X.
For q=1 is it also spherically symmetric.
> One of them is phi(v) = 1. THAT is its value in
> the LTE.
Huh?? Lorentz transforms do not involve the deformation of any object. They
describe variations in geometrical projections, not deformations.
Just THINK about it: how could an observer moving past an object
possibly affect the object itself? How could choice of
coordinates possibly affect any physical phenomenon, including
shapes of objects?
Looking at an object cannot possibly "deform" the object, and that's what
different inertial frames are -- different ways of looking at THE SAME world.
Lorentz transforms merely relate measurements in one inertial frame to those in
another such frame.
> Note, then, that "reasons of symmetry" are ruled out by phi(v) = 1!
Huh? You are confused. See above.
> Similarly, since the space and time of STR and GR deform
> asymmetrically, the space-time continuum cannot be homogeneous.
You are confused. In neither SR nor GR do space and time "deform", symmetrically
or asymmetrically. But yes, in GR the spacetime manifold is not homogeneous
(except for unphysical situations like the Minkowski spacetime of SR).
In SR, observers at rest in different inertial frames see different PROJECTIONS
of the world as their personal views of space and time. To each observer, space
is isotropic and homogeneous and time is homogeneous. But they see space and
time DIFFERENTLY. That is, for each observer the points that form the locus of
3-space at a given instant in time are different for the two observers, but each
sees their locus as both isotropic and homogeneous. Similar remarks apply to time.
IOW: These two observers are foliating the single 4-d spacetime
manifold into DIFFERENT submanifolds of (3-d) space and time.
Here's an example: consider 3-d Euclidean space, and construct
Cartesian coordinates {x,y,z} on it. The 2-d projection onto
x=0 is isotropic and homogeneous. Now construct Cartesian
coordinates {x',y',z'} rotated relative to the first set. The
2-d projection onto x'=0 is isotropic and homogeneous. These
two projections are QUITE different, yet are both isotropic and
homogeneous. A similar thing happens for those observers making
3-d projections onto t=0 and t'=0 -- both 3-d submanifolds are
isotropic and homogeneous, yet they are different.
> Therefore Einstein's assertion that his 1905 paper's "equations must
> be linear on account of the properties of homogeneity which we
> attribute to space and time" requires that phi(v) = q = 1; which rules
> out the LTE!!
I repeat: you are confused, but don't seem to know it. That is not at all what
SR says (GR is irrelevant here, as Lorentz transforms do not apply in it).
The symmetry for motion along X/X' requires that the
transforms relating Y and Z to Y' and Z' be the same. But
symmetry puts no constraint on the relationship between that
and the transform relating X to X' -- the motion itself
INHERENTLY breaks spherical symmetry. Einstein's paper shows
that the motion need not break homogeneity (see my Euclidean
example above).
Tom Roberts
glird
glird wrote:
< Consider a spherical object moving along the X axis at v. Suppose
it deforms by phi(v)q in X and by phi(v) in Y and Z. >
Tom: <Then in GR you must include the mechanism of deformation in the
energy-momentum tensor. That gives you a completely different problem
to study, and it's not obvious how to compare two such different
situations....>
In the LTE of SR the mechanism was spelled out by Lorentz in his
1904 paper.
glird: <For any value of q, there are only two values of phi(v)
for which the object will deform asymmetrically. >
Tom: <Huh? If q != 1 that deformation is not spherically symmetric,
for any nonzero phi(v). >
RIGHT!!! I should have said, There are only two values
for which the object will deform asymmetrically.
Tom: <There are many different types of symmetry.
You must be more specific. >
So must YOU. To me, in order for a deformation to be symmetrical,
the general shape of the object must remain constant regardless of its
size. for example, if it is an oval that expands for any reason, it
mut remain an oval with the same proportion between its long and short
radii no matter how large or small it ends up.
Tom: <For instance, for any nonzero values of q and phi(v)
that deformation is cylindrically symmetric with axis along
X.
For q=1 is it also spherically symmetric. >
For phi(v) = 1, then q = sqrt(1-v^2/c^2); which means that the
deformed object IS cylindrical. BUT!! if v changes, then the length
of the cylinder changes in X but NOT in Y or Z. The deformation is
therefore Asymmetrical.
For q = 1, then phi)v = 1/sqrt(1-v^2/c^2); which means that the
deformed object is cylindrical. BUT if v changes, then the length of
the cylinder changes in Y and Z but NOT in X! The deformation is
asymmetrical again!!
glird wrote: <One of them is phi(v) = 1. THAT is its value in the
LTE.
Tom: <Huh?? Lorentz transforms do not involve the deformation of any
object. They describe variations in geometrical projections, not
deformations. >
You are old enough to remember that prior to circa 1980, when
Wheeler invented the {uncaused} "rotations" of the X and t axes of
moving systems, the LTE not only involved deformations of moving
objects, they were based on and imposed them.
Tom: <Just THINK about it: how could an observer moving past an object
possibly affect the object itself? >
Think about this, Tom. A few years before Wheeler invented
rotations as the cause of the projected length and rate changes I was
talking to a friend who was a Math Professor at Rutgers -- which is
only a few miles from Princeton where Wheeler lived. When my friend
said something about the mass of an object being a function of its
velocity, I scoffed, "It is impossible for the quantity of matter to
be a function of who is looking at it". The Professor, who smilingly
agreed, probably asked his fellow profs at Rutgers that question;
which may have gone on to Princeton etc etc etc.
Tom: <How could choice of coordinates possibly affect any
physical phenomenon, including shapes of objects? .
Thank you for asking that question in the way that you did, Tom.
It shows me that you are one of the myriad contributors to these
newsgroups who doesn't understand Einstein's 1905 STR paper
nor the LTE he failed to derive.
Tom: <Looking at an object cannot possibly "deform" the object, and
that's what different inertial frames are -- different ways of looking
at THE SAME world.
Lorentz transforms merely relate measurements in one inertial frame
to those in another such frame. >
The key to your failure to understand how the LTE work is the word
"merely". Delete it, and we have,
"Lorentz transforms relate measurements in one inertial frame to
those in another such frame". The new key word is "measurements".
Put that into your head and rewrite your
statement like this:
Looking at an object cannot possibly "deform" the object, although
that's what different inertial frames do -- they use different ways of
MEASURING the SAME world. The Lorentz transforms relate measurements
by one inertial esynched frame to those a differently moving such
frame would obtain for a given event.
Glird: <Note, then, that "reasons of symmetry" are ruled out 'by phi
(v) = 1! >
Tom: <Huh? You are confused. See above. >
> Similarly, since the space and time of STR and GR deform
> asymmetrically, the space-time continuum cannot be
> homogeneous.
Tom: <You are confused. In neither SR nor GR do space and time
"deform", symmetrically or asymmetrically. But yes, in GR the
spacetime manifold is not homogeneous (except for unphysical
situations like the Minkowski spacetime of SR)..
I'm glad you understand that the space-time invented by Minkowski is
an "unphysical" mathematical abstraction that doesn't physically exist
in the real world.
Tom: <In {WHEELER'S} SR, observers at rest in different inertial
frames see different PROJECTIONS of the world as their personal views
of space and time. To each observer, space is isotropic and
homogeneous and time is homogeneous. But they see space and time
DIFFERENTLY. That is, for each observer the points that form the locus
of 3-space at a given instant in time are different for the two
observers, but each sees their locus as both isotropic and
homogeneous. Similar remarks apply to time. IOW: These two observers
are foliating {FOLDING?} 4-d spacetime manifold into DIFFERENT
submanifolds of (3-d) space and time. Here's an example:
Consider 3-d Euclidean space, and construct Cartesian coordinates
{x,y,z} on it. The 2-d projection onto x=0 is isotropic and
homogeneous. Now construct Cartesian coordinates {x',y',z'} rotated
relative to the first set. The 2-d projection onto x'=0 is isotropic
and homogeneous. These two projections are QUITE different, yet are
both isotropic and homogeneous. A similar thing happens for those
observers making 3-d projections onto t=0 and t'=0 -- both 3-d
submanifolds are isotropic and homogeneous, yet they are different. >
From steps 1 and 3 we have a Cartesian co-ordinate system (x,y,z)
and a Cartesian cs (x',y',z') whose axes are rotated relative to the
first one.
NOW TRY THIS: Rotate the axes of cs 2 back enough that they are
COINCIDENT with those of cs 1; and THEN let them measure given events.
Since you stipulated that that these are Cartesian systems, the
transformations will be the same Euclidean ones Newton used. If you
want to understand how the non-Euclidean, non-Cartesian co-ordinate
systems of the LTE work, try this:
We have from modified steps 1 and 3, a non-Euclidean co-ordinate
system (x,y,z) and a non-Cartesian cs (x',y',z') whose axes are
rotated relative to the first one.
Now rotate the axes of cs 2 back enough that they are COINCIDENT
with those of cs 1; and THEN let them measure given events.
For ANY value of phi(v) the following transformations will hold
good:
x' = phi(v)b(x-vt), y' = phi(v)y; z' = phi(v)z;
and t' = phi(v)b(t-vx/c^2)
in which b = 1/(c^2-v^2)^.5.
There are an infinity of groups that satisfy those equations. The
LTE Group, in which phi(v) = 1 this is invisible, is only one of
them.
The Galilean group of Cartesian-Euclidean classical physics do not
fit, thus are ruled out by the above transformations -- which are
those Einstein DID almost-correctly derive in his 1905 paper.
glird: <Therefore Einstein's assertion that his 1905 paper's
"equations must be linear on account of the properties of homogeneity
which we attribute to space and time" requires that phi(v) = q = 1;
which rules out the LTE!! >
Tom: <I repeat: you are confused, but don't seem to know it. That is
not at all what SR says (GR is irrelevant here, as Lorentz transforms
do not apply in it).
The symmetry for motion along X/X' requires that the
transforms relating Y and Z to Y' and Z' be the same. >
"Reasons of symmetry" presumably require "that the transforms
relating Y and Z to Y' and Z' be the same" AS THOSE RELATING
Y' and Z' to Y and Z. Put simply, they require that the inverse
transformations have the same form as the direct ones.
Einstein, and presumably all of his followers including Tom Roberts,
mistakenly thought that the only value of phi(v) that allowed that to
happen is
"phi(v) = 1".
Tom: <But symmetry puts no constraint on the relationship between that
and the transform relating X to X' -- the motion itself
INHERENTLY breaks spherical symmetry. Einstein's paper shows that the
motion need not break homogeneity (see my Euclidean example above). >
Your Euclidean example, all by itself, PROVES that you don't
understand the LTE nor the STR either.
Fortunately for all of us, other than that it too requires that
successive clocks be esynched via Einstein's postulated method, the
mathematics of General Relativity is entirely different than the
restricted theory's LTE.
glird
Tom Roberts
> On Jan 4, 12:32 pm, Tom Roberts <tjrob...@sbcglobal.net> wrote:
> glird wrote:
> < Consider a spherical object moving along the X axis at v. Suppose
> it deforms by phi(v)q in X and by phi(v) in Y and Z. >
>
> Tom: <Then in GR you must include the mechanism of deformation in the
> energy-momentum tensor. That gives you a completely different problem
> to study, and it's not obvious how to compare two such different
> situations....>
>
> In the LTE of SR the mechanism was spelled out by Lorentz in his
> 1904 paper.
That paper is NOT Special Relativity. It is the basis of Lorentz Ether Theory,
which is a DIFFERENT theory from SR, but "happens to be" experimentally
indistinguishable from it.
Note he made a mistake in his transform of charge density.
When you say "an object is deformed", that is an ACTIVE process that acts ON THE
OBJECT ITSELF, and requires some mechanism and some energy. That is NOT what
happens in SR and GR due to relative velocity.
From now on I omit GR and stick to SR.
> For phi(v) = 1, then q = sqrt(1-v^2/c^2); which means that the
> deformed object IS cylindrical. BUT!! if v changes, then the length
> of the cylinder changes in X but NOT in Y or Z. The deformation is
> therefore Asymmetrical.
You are confused. And remain confused. I repeat: in SR, no object ever "deforms"
due to a relative velocity between observer and object. It is the RELATIONSHIP
BETWEEN OBSERVER AND OBJECT that varies with their relative velocity, not the
object itself.
Observers are just LOOKING AT OBJECTS, and have no power
whatsoever to "change" or "modify" or "deform" the object.
When you look at a friend from far away, she LOOKS very
small to you, but SHE can attest to the fact that she HERSELF
did not shrink. That is the same idea as what I am saying
here.
> Tom: <Huh?? Lorentz transforms do not involve the deformation of any
> object. They describe variations in geometrical projections, not
> deformations. >
>
> You are old enough to remember that prior to circa 1980, when
> Wheeler invented the {uncaused} "rotations" of the X and t axes of
> moving systems, the LTE not only involved deformations of moving
> objects, they were based on and imposed them.
Not true. Go back and READ those old references. They are discussions HOW AN
OBSERVER sees the object, not how the object itself "deforms".
It seems you have read your own PERSONAL BIASES into other peoples' writings.
DON'T DO THAT. You need to improve the care and accuracy with which you read.
> Tom: <Just THINK about it: how could an observer moving past an object
> possibly affect the object itself? >
You need to do that. You remain confused. You also need to improve the care and
accuracy with which you read.
> Think about this, Tom. A few years before Wheeler invented
> rotations as the cause of the projected length and rate changes [...]
Wheeler did not consider them "causes", and he did not invent them -- Minkowski
did so ~ 1909. Wheeler just explored their consequences further than previously.
He also introduced coordinate-independent language and methods.
> Tom: <How could choice of coordinates possibly affect any
> physical phenomenon, including shapes of objects? .
>
> Thank you for asking that question in the way that you did, Tom.
> It shows me that you are one of the myriad contributors to these
> newsgroups who doesn't understand Einstein's 1905 STR paper
> nor the LTE he failed to derive.
YOU are the one confused, not me. OBJECTS do not vary, but rather an observer's
MEASUREMENTS of an object can vary with relative velocity of the object being
measured. Neither object nor observer are "changed" in any way due to different
inertial motions, but the RELATIONSHIP between them does vary.
> Tom: <In {WHEELER'S} SR, [...]
SR is not "Wheeler's". It is a physical theory invented by Einstein to which
many later physicists have contributed. Yes, Wheeler is among them.
> IOW: These two observers
> are foliating {FOLDING?}
Foliating. Learn the vocabulary or forever remain confused.
> Your Euclidean example, all by itself, PROVES that you don't
> understand the LTE nor the STR either.
No. It merely proves you do not understand what I am saying. Everything I have
said in this thread is part of the standard physics curriculum. There are
thousands of physicists who can attest to that, but very few of them bother to
read this newsgroup, and even fewer bother with threads like this.
You have A LOT of rather serious misconceptions about relativity and its
history. You will remain confused until you admit that to yourself, and go back
and STUDY.
[This is going nowhere. I probably won't respond until you
actually learn something.]
Tom Roberts