Any coordinate system in GR?
Edward Green
formal machinery of GR, treats all coordinate systems equally. Is this
true? Aside from questions about smoothness, it seems to me there is
at least some other implicit condition on permissible coordinate
systems.
Consider a 1-dimensional manifold, a real coordinate x, and the metric
ds = |dx|. Next, stretch this manifold by a postive factor a(x), also
stretching the coordinates (so that points on the manifold are still
labeled by the same numbers). Calling the new, stretched, coordinates
u, the most natural transformation law for the metric would be
ds = a(u)|du|
Now, instead of stretching the manifold, compress the coordinates by
this same function a( ), so that dx = a(u)du. The most natural
transformation law for the metric would again be
ds = a(u)|du|
So, if we allow "any coordinate system", we would apparently be
unable to distinguish the (physical) case of the stretched manifold
from the (unphysical) case of the compressed coordinates.
It may be objected that by compressing the coordinates, we have
implicitly changed physical units, and this is not allowed. But that's
a new rule, isn't it? What's a "unit"? We need additional structure
to specify what we mean by this; we must equip our manifold with some
physics -- a local property establishing the natural scale of the
coordinates. This will insure that we don't arbitrarily distort the
coordinates, and any distortion can be attributed to the manifold.
Am I right or wrong?
Bill Hobba
"Edward Green" <spamsp...@netzero.com> wrote in message
news:1156201991.9...@h48g2000cwc.googlegroups.com...
> The offhand claim is sometimes made that GR, which I take it means the
> formal machinery of GR, treats all coordinate systems equally. Is this
> true? Aside from questions about smoothness, it seems to me there is
> at least some other implicit condition on permissible coordinate
> systems.
>
> Consider a 1-dimensional manifold, a real coordinate x, and the metric
> ds = |dx|. Next, stretch this manifold by a postive factor a(x), also
> stretching the coordinates (so that points on the manifold are still
> labeled by the same numbers). Calling the new, stretched, coordinates
> u, the most natural transformation law for the metric would be
>
> ds = a(u)|du|
>
> Now, instead of stretching the manifold, compress the coordinates by
> this same function a( ), so that dx = a(u)du. The most natural
> transformation law for the metric would again be
>
> ds = a(u)|du|
>
> So, if we allow "any coordinate system", we would apparently be
> unable to distinguish the (physical) case of the stretched manifold
> from the (unphysical) case of the compressed coordinates.
First why is one unphysical and the other not? Secondly since all rulers
are compressed or stretched by the same amount nothing will change. But
that is not really the claim of GR - it does not claim like SR does for
inertial frames that if we take the same experiment and shift it to an other
frame then exactly the same result will occur - shift it to an accelerated
frame and it will be different. Here lies the crucial difference - SR by
being restricted to inertial frames has that property - GR does not. In both
cases the laws of physics are still the same.
GR obeys the principle of general invariance (not to be confused with the
principle of covariance). As pointed out by Kretchmann (and eventually
agreed by Einstein) any law can be put into covariant from so the principle
of general covariance contains no actual physics. Its physics lies in the
fact that when in such a form all absolute terms (ie things like the speed
of light, planks constant etc) remain unchanged and any other terms must be
dynamical - technically this is called general invariance. But you will
find a lot of articles confuse one with the other - but as long as you
understand what is happening no problems will arise - at least I have never
found any. This immediately implies that the metric tensor for example is
not an absolute term so must be dynamical - which is a cornerstone of GR -
in fact all by itself it pretty much implies the EFE's.
For a detailed discussion see
http://www.pitt.edu/~jdnorton/papers/decades.pdf#search=%22general%20covariance%20vs%20invarience%22
Thanks
Bill
Edward Green
> "Edward Green" <spamsp...@netzero.com> wrote ...
<stretching the manifold vs. compressing the coordinate system>
> First why is one unphysical and the other not?
Well, that depends on whether it is "physical" to stretch the manifold.
In a simple minded analogy, the one dimensional manifold might be
represented by a rubber band: stretching the rubber band is a
physically distinct situation from compressing the coordinates -- one
might be able to determine the state of strain by local measurements,
or at least determine the relative strain for two locations. In GR the
"stretch" might correspond to a gravitational field vs. flat spacetime.
Not _everything_ is an artifact of the coordinate system.
> Secondly since all rulers
> are compressed or stretched by the same amount nothing will change.
This might be the case, but I think you assume too much. It may be the
case that "nothing will change" locally -- meaning we cannot tell the
difference by local measurements alone -- but that we can very
definitely tell that something is changing over paths. This is the
situation in GR, I believe. I made my "stretching" dependent on
position BTW -- not the very simplest situation one could imagine -- to
try to avoid the impression that this was solely about an arbitrary
choice of units.
As a silly illustration of principle about a possible physical
difference between distortion of a manifold and distortion of
coordinates, imagine an elastic manifold which turns blue beyond a
certain strain state. Now,eitehr stretching the manifold in strata,
or, on the other, inversely compressing the coordinates, we either see
blue striations or we don't -- physically distinct situations!
> But
> that is not really the claim of GR - it does not claim like SR does for
> inertial frames that if we take the same experiment and shift it to an other
> frame then exactly the same result will occur - shift it to an accelerated
> frame and it will be different. Here lies the crucial difference - SR by
> being restricted to inertial frames has that property - GR does not. In both
> cases the laws of physics are still the same.
>
> GR obeys the principle of general invariance (not to be confused with the
> principle of covariance). As pointed out by Kretchmann (and eventually
> agreed by Einstein) any law can be put into covariant from so the principle
> of general covariance contains no actual physics. Its physics lies in the
> fact that when in such a form all absolute terms (ie things like the speed
> of light, planks constant etc) remain unchanged and any other terms must be
> dynamical - technically this is called general invariance. But you will
> find a lot of articles confuse one with the other - but as long as you
> understand what is happening no problems will arise - at least I have never
> found any. This immediately implies that the metric tensor for example is
> not an absolute term so must be dynamical - which is a cornerstone of GR -
> in fact all by itself it pretty much implies the EFE's.
Well, you have confused me, though I claim I am alive to the kinds of
issues you raise, and you have failed to persuade me, if that was your
purpose, that my question is somehow misguided. However, I thank you
very cordially for your serious answer, and especially your detailed
reference, available free through the miracle of pdf, and I certainly
must be compelled to study it.
That post was rejected by the moderator of s.p.r., BTW, where I only
repaired because that's where the experts are, and the cranks aren't.
The reason given was "mainly mathematical, and containing elementary
errors". I've been led to believe that s.p.r. was waiting with open
arms for the disillusioned amateur to escape the trials and
tribulations of trolls and orcs, provided he only asked respectful
questions. I hardly went ranting about how GR is wrong, only mentioned
a common aside which I suspected was wrong, and if not, could the error
in my ointment please be strained out? I may be wrong -- my usual
refrain -- but I'm not _that_ obviously wrong. Your thoughtful answer
shames this miscreant, but unfortunately does not remove him from his
position.
G. L. Bradford
everything you've been taught about the speed of light and velocity. You are
at once in space and inside a time machine because each distant observable
point of the universe enclosing you has independency of time relative to
you. Generally regarding the hemisphere to your rear you -- relative to
you -- will be mostly retreating in time into histories. Generally regarding
the hemisphere forward you -- relative to you -- will be speeding up in time
into futures of those points you observe.
You cannot travel anywhere whatsoever if you cannot speed up in time --
'relative' time, not 'real' time -- toward whatever point is your
destination. The speed of light limitation upon transmission of information
iron clad guarantees that you will never [observe] yourself at your point of
origin to be current in 'time' with any distant destination, not even if you
are on the same planet with it, much less if you are dealing in differing
points, planets, stars, whatever in the vacuum of space. In order to close
that gap of NON-CURRENCY you must time travel as being identical to space
travel. You must, you will, compress time, shorten it, on a per second per
second basis or you do not close up one silly millisecond (alternatively,
spatially, you do not close up one silly millimeter) of the intervening gap
of non-currency.
And for all the 'relativity' in time you gain forward, you will lose just
that amount of 'relativity' in time rearward. For all the gap (of
non-currency in time) closing you will do ahead, you will open gaps (of
non-currency in time) behind, relative speaking, that is. So if and when you
return to your point of origin you will always have to repeat the order of
physics of time travel (of space travel) precisely.
-------------
You are at a receiving station on the surface of the Earth for orbiting
satellite transmissions. The satellite comes up over the distant horizon of
Earth toward you, or rather toward a point about two hundred kilometers (or
slight fractions of a light second) directly above you. As it surfaces the
distant horizon it is a good deal further away from you than two hundred
kilometers (more fractions of a light second farther away from you then it
will be when it is directly above you). It is constantly transmitting time
signals to your receiving station. The clock aboard that satellite seems to
be behind in time to your clock but seems to be running fast, speeding up in
time per second per second, according to your own clock -- seeming to be on
its way to catching up in time to your clock. As it passes directly above
your position it seems to have caught up in time to you and momentarily
seems to be in sync with your receiving station clock. But what's this?! As
it speeds toward the opposite horizon, rapidly gaining distance from you,
all of sudden it seems to be slowing down in clocking fractions of second in
time. Damnation! Relatively speaking, it is now falling back behind your own
clock in time, now running slower rather than running faster -- or running
in time -- with your own receiving station clock (as it -- the satellite --
now speeds away from you, lengthening out in distance toward the distant
horizon in the opposite direction from which it came).
GLB
GLB
Ken S. Tucker
your question is well posed and penetrating it exceeded
the moderators understanding. Myself I'd be interested
in possible well informed opinions that occasionally
post to s.p.r. , that said I'll try.
Edward Green wrote:
> The offhand claim is sometimes made that GR, which I take it means the
> formal machinery of GR, treats all coordinate systems equally. Is this
> true?
Pardon my apparent evasiveness...
GR treats all Frames of Reference's (FoR's) equally.
AE's law, G_uv = T_uv constrains the metrical
geometry, to solutions of the tensor g_uv satisfying
that law, that is why G_uv=T_uv is a called a field
equation. For example a simplistic solution follows
from G_uv=0 called the Schwarzchild Solution, that
I'm sure you're aware of, that provides metrics like,
g_00 = 1 - 2m/r , g_rr = 1/g_00 etc.
The reason for CS independence is because g_uv
is a tensor and can be properly transformed via
the usual procedure,
g'_ab = (&x^u/&x'_a) (&x^v/&x'_b) g_uv
to any other CS, using light-years, millimetres
polar, elliptical, cylindrical as you please, and
in time, seconds or dog-years, meaning the
physical reality is independent of units.
> Aside from questions about smoothness, it seems to me there is
> at least some other implicit condition on permissible coordinate
> systems.
Yes, the ultimate choice/solution of the metric is
conditioned by the field equation, but you can use
inches or cubits.
> Consider a 1-dimensional manifold, a real coordinate x, and the metric
> ds = |dx|. Next, stretch this manifold by a postive factor a(x), also
> stretching the coordinates (so that points on the manifold are still
> labeled by the same numbers). Calling the new, stretched, coordinates
> u, the most natural transformation law for the metric would be
>
> ds = a(u)|du|
Sure that's a good example, use a tapered elastic.
Lightly stretched mark points against a ruler then
stretch it and you'll end up generating a logrithm
ruler, that's familiar.
> Now, instead of stretching the manifold, compress the coordinates by
> this same function a( ), so that dx = a(u)du. The most natural
> transformation law for the metric would again be
>
> ds = a(u)|du|
>
> So, if we allow "any coordinate system", we would apparently be
> unable to distinguish the (physical) case of the stretched manifold
> from the (unphysical) case of the compressed coordinates.
By "any CS" you can still use feet or metres,
polar or cartesian.
> It may be objected that by compressing the coordinates, we have
> implicitly changed physical units, and this is not allowed. But that's
> a new rule, isn't it? What's a "unit"? We need additional structure
> to specify what we mean by this; we must equip our manifold with some
> physics -- a local property establishing the natural scale of the
> coordinates. This will insure that we don't arbitrarily distort the
> coordinates, and any distortion can be attributed to the manifold.
> Am I right or wrong?
I think you're right when you focus on "unit" specifically
Planck's constant "h" as used here, E = h*f.
Think back in Newtonian terms, light frequency "f" and
energy "E" were unaffected by gravitation.
For example, one could imagine setting up a laser
based cartesian CS, X,Y,Z from distant laser's like
from stars. From the standpoint of Newton, placing
a mass into the X,Y,Z Grid will NOT affect the Grid.
However, light energy is subject to the conservation
of mass-energy law, (in hindsight...duh) and hence
Newtons X,Y,Z Grid needed to be modified to AE's
x,y,z flexible grid, which, as it turns out, flexs the
x,y,z grid to maintain the scalar invariance of "h".
Famous confirmations are the Einstein Shift due
to gravitation and the deflection of light, each mods
the Newtonian Grid to the Einsteinian grid.
Best Regards
Ken S. Tucker
(kst)
Igor
Stretching or compressing the coordinate system just redefines it
mathematically. There's nothing physical there. You can define a
coordinate transformation any way you want to provided that it is
invertible. But that doesn't result in any new physics. It may result
in so-called fictictious forces arising in the new coordinate system
that weren't there previously, but those are never really physical
either. And changing units can indeed be allowed under coordinate
transformations. Think of going from rectangular to polar or vice
versa. Rectangular has both coordinates with length units, and polar
has one with length and the other with angular units. The metric will
always compensate to make up for the difference.
Bill Hobba
> Bill Hobba wrote:
>
>> "Edward Green" <spamsp...@netzero.com> wrote ...
>
> <stretching the manifold vs. compressing the coordinate system>
>
>> First why is one unphysical and the other not?
>
> Well, that depends on whether it is "physical" to stretch the manifold.
Since it is not a material then obviously it is a 'word salad's with no
meaning like saying stew smells like isomorphism.
> In a simple minded analogy, the one dimensional manifold might be
> represented by a rubber band: stretching the rubber band is a
> physically distinct situation from compressing the coordinates -- one
> might be able to determine the state of strain by local measurements,
> or at least determine the relative strain for two locations. In GR the
> "stretch" might correspond to a gravitational field vs. flat spacetime.
> Not _everything_ is an artifact of the coordinate system.
You were on the right track above - stretching is not a property of the
space-time manifold.
Thanks
bill
Edward Green
Igor wrote:
> Stretching or compressing the coordinate system just redefines it
> mathematically. There's nothing physical there.
Of course.
> You can define a
> coordinate transformation any way you want to provided that it is
> invertible. But that doesn't result in any new physics.
Obviously, my dear man.
> It may result
> in so-called fictictious forces arising in the new coordinate system
> that weren't there previously, but those are never really physical
> either.
We are straying from the question.
> And changing units can indeed be allowed under coordinate
> transformations. Think of going from rectangular to polar or vice
> versa. Rectangular has both coordinates with length units, and polar
> has one with length and the other with angular units.
Ok. Good point.
> The metric will always compensate to make up for the difference.
So you claim. I believe I have found the answer to my own question (as
usual), and while I did not state the requirement 100% correctly, I was
on the right track. I happened upon the correct formulation in
Schwarzschild's 1916 paper on the field of a mass point:
"The field equations ... have the fundamental property that
they preserve their form under the substitution of other
arbitrary variables in lieu of x1,x2, x3, x4, as long as the
determinant of the substitution is equal to 1."
http://arxiv.org/PS_cache/physics/pdf/9905/9905030.pdf
If the old coordinates were orthogonal and the new coordinates remained
orthogonal (which requirement I did not state) and we kept the same
units, mine would be a sufficient condition to meet the above
requirement on the named determinate. In general we have more latitude
than that, _but_ we are not free to chose the new coordinates
arbitrarily, even if they are invertible in a neighborhood. The unit of
_volume_ at least must be preserved. In the case of a one-dimensional
manifold this reduces to the requirement to keep the same units.
Are we in agreement now?
JanPB
> [...] I believe I have found the answer to my own question (as
> usual), and while I did not state the requirement 100% correctly, I was
> on the right track. I happened upon the correct formulation in
> Schwarzschild's 1916 paper on the field of a mass point:
>
> "The field equations ... have the fundamental property that
> they preserve their form under the substitution of other
> arbitrary variables in lieu of x1,x2, x3, x4, as long as the
> determinant of the substitution is equal to 1."
>
> http://arxiv.org/PS_cache/physics/pdf/9905/9905030.pdf
It's usually not a good idea to learn stuff from original papers. They
were written for experts and reflect the then-current knowledge. For
example, Schwarzschild didn't know at that time that the determinant
restriction was unnecessary (the equations are invariant under _any_
change of variables).
> If the old coordinates were orthogonal and the new coordinates remained
> orthogonal (which requirement I did not state) and we kept the same
> units, mine would be a sufficient condition to meet the above
> requirement on the named determinate.
Forget the determinants, it's complete mothballs. Read a modern
textbook instead.
> In general we have more latitude
> than that, _but_ we are not free to chose the new coordinates
> arbitrarily, even if they are invertible in a neighborhood. The unit of
> _volume_ at least must be preserved.
No, that's unecessary.
--
Jan Bielawski
JanPB
>
> It's usually not a good idea to learn stuff from original papers.
I meant classic, well-established stuff of course.
--
Jan Bielawski
Koobee Wublee
> Igor wrote:
> > The metric will always compensate to make up for the difference.
>
> So you claim. I believe I have found the answer to my own question (as
> usual), and while I did not state the requirement 100% correctly, I was
> on the right track. I happened upon the correct formulation in
> Schwarzschild's 1916 paper on the field of a mass point:
>
> "The field equations ... have the fundamental property that
> they preserve their form under the substitution of other
> arbitrary variables in lieu of x1,x2, x3, x4, as long as the
> determinant of the substitution is equal to 1."
>
> http://arxiv.org/PS_cache/physics/pdf/9905/9905030.pdf
This is an excellent article. It is very easy to read and understand.
Thanks to Schwarzschild's ingenuity.
> If the old coordinates were orthogonal and the new coordinates remained
> orthogonal (which requirement I did not state) and we kept the same
> units, mine would be a sufficient condition to meet the above
> requirement on the named determinate. In general we have more latitude
> than that, _but_ we are not free to chose the new coordinates
> arbitrarily, even if they are invertible in a neighborhood. The unit of
> _volume_ at least must be preserved. In the case of a one-dimensional
> manifold this reduces to the requirement to keep the same units.
Schwarzschild found a unique solution to the differential equations of
Einstein Field Equations in free space. Hilbert found another one that
he called it Schwarzschild Metric. Recently, Mr. Rahman also a
contributor of this newsgroup presented another solution. The
spacetime with this metric is
ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
The metric above indeed is another solution which anyone can easily
verify because its simplicity. Notice Rahman's metric and
Schwarzschild's original metric do not manifest black holes.
However, since Schwarzschild Metric is much simpler than
Schwarzschild's original solution, Schwarzschild Metric is embraced by
the physics communities today. Mr. Bielawski and Igor have not
understood Schwarzschild's original paper and choose to blindly reject
Schwarzschild's original solution and others.
As multiple solutions to the vacuum field equations are discovered,
there are actually an infinite number of them. With infinite number of
solutions, it is shaking the very foundation of GR and SR. The house
of cards will soon inevitably collapse. However, refusing to give up
GR and to comfort themselves in false sense of security, they choose to
embrace Voodoo Mathematics. In doing so, they blindly claim all
solutions are indeed the same regardless manifesting black holes,
constant expanding universe, accelerated expanding universe. VOODOO
MATHEMATICS REPRESENTS THE ACHIEVEMENT IN PHYSICS DURING THE LAST 100
YEARS. It is very sad that these clowns are regarded as experts in
their field.
Edward Green
>. The unit of _volume_ at least must be preserved.
>
> No, that's unecessary.
Well sir, I disremember what started me on this tack... maybe I was
trying to figure out what was "physical" in GR. However, I arrived at
the following waystation: Physically distinct (in a specifiable sense)
1-d manifolds with what look like identical representations of a
metric.
Is it a _problem_ if physically distinct situations look identical
given only the variable labels and a representation of the metric? Is
there some missing piece? Did I correctly transform the metrics given
the manipulations of the manifold or coordinate system described? You
would do me a kindness if you were to scan the second two paragraphs in
the post you replied to, and indicate the faults in the argument.
Oh... and learning from original papers... I'm not sure I completely
agree with you. Sometimes you have a conceptual problem, and you find
that the original thinkers were alive to the problem, although it is
glossed over now. At least I think that's happened to me once, so I
can freely generalize. Maybe even twice now. ;-)
JanPB
>
> Schwarzschild found a unique solution to the differential equations of
> Einstein Field Equations in free space. Hilbert found another one that
> he called it Schwarzschild Metric. Recently, Mr. Rahman also a
> contributor of this newsgroup presented another solution. The
> spacetime with this metric is
>
> ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
>
> The metric above indeed is another solution which anyone can easily
> verify because its simplicity. Notice Rahman's metric and
> Schwarzschild's original metric do not manifest black holes.
No, the metric above is equal to Schwarzschild's metric. The form above
is obtained by a coordinate change from the original one, hence the
metric remains the same (tensors do not change under coordinate
changes).
> However, since Schwarzschild Metric is much simpler than
> Schwarzschild's original solution, Schwarzschild Metric is embraced by
> the physics communities today.
It's embraced because it's the same.
> Mr. Bielawski and Igor have not
> understood Schwarzschild's original paper and choose to blindly reject
> Schwarzschild's original solution and others.
There is nothing to reject. One can _prove_ Schwarzschild's metric is
unique. Off the horizon it follows immediately from the particular form
of the Einstein equation in the spherically symmetric case (which is
what we have) and the uniqueness of the extension over the horizon is
slightly more involved but it follows from a similar argument.
> As multiple solutions to the vacuum field equations are discovered,
> there are actually an infinite number of them.
Yes, and 2+2=5.
> With infinite number of
> solutions, it is shaking the very foundation of GR and SR.
Sure. My boots are all torn already.
Give us a break.
--
Jan Bielawski
Ken S. Tucker
Jan, I think you should study the original paper
Ed Green cited, there is no such thing as an
event horizon or Black-Hole's. BTW, Dr. Loinger
and I discussed this at length. In short, the
original Schwarzschild Solution has been
bastardized and mis-understood for simplicity.
The bastardized version became popularized
and embraced by astronomers who now see
BH's under their beds.
Regards
Ken S. Tucker
I.Vecchi
I would say that there are at least two distinct ways to prolong the
solution across the horizon, which yield respectively the black hole
and the white hole solution, describing two different physical
phenomena. They correspond to two distinct choices of the
Eddington-Finkelstein coordinates at the horizon, i.e. to crossing the
horizon on future-directed or on past-directed curves. The extended
solution depends on how space-time is extended beyond the horizon.
IV
Tom Roberts
> I would say that there are at least two distinct ways to prolong the
> solution across the horizon, which yield respectively the black hole
> and the white hole solution, describing two different physical
> phenomena. They correspond to two distinct choices of the
> Eddington-Finkelstein coordinates at the horizon, i.e. to crossing the
> horizon on future-directed or on past-directed curves. The extended
> solution depends on how space-time is extended beyond the horizon.
The Kruskal-Szerkes coordinates show that these two extensions are
merely two aspects of the complete (inextensible) manifold.
Tom Roberts
Koobee Wublee
> Koobee Wublee wrote:
> > Schwarzschild found a unique solution to the differential equations of
> > Einstein Field Equations in free space. Hilbert found another one that
> > he called it Schwarzschild Metric. Recently, Mr. Rahman also a
> > contributor of this newsgroup presented another solution. The
> > spacetime with this metric is
> >
> > ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
> >
> > The metric above indeed is another solution which anyone can easily
> > verify because its simplicity. Notice Rahman's metric and
> > Schwarzschild's original metric do not manifest black holes.
>
> No, the metric above is equal to Schwarzschild's metric. The form above
> is obtained by a coordinate change from the original one, hence the
> metric remains the same (tensors do not change under coordinate
> changes).
** ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
** ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2
K is an integration constant chosen to fit Newtonian result. It is
** K = 2 G M / c^2
For the record, Mr. Bielawski is claiming the above two metrics are the
same despite one manifests a black hole and other one not.
> > However, since Schwarzschild Metric is much simpler than
> > Schwarzschild's original solution, Schwarzschild Metric is embraced by
> > the physics communities today.
>
> It's embraced because it's the same.
You embraced it because of your denial of faulty GR.
> > Mr. Bielawski and Igor have not
> > understood Schwarzschild's original paper and choose to blindly reject
> > Schwarzschild's original solution and others.
>
> There is nothing to reject. One can _prove_ Schwarzschild's metric is
> unique. Off the horizon it follows immediately from the particular form
> of the Einstein equation in the spherically symmetric case (which is
> what we have) and the uniqueness of the extension over the horizon is
> slightly more involved but it follows from a similar argument.
You rejected it because of your denial of faulty GR. You have 90 years
of fun playing with Voodoo Mathematics that gives rise to GR. It is
time to tear it down.
> > As multiple solutions to the vacuum field equations are discovered,
> > there are actually an infinite number of them.
>
> Yes, and 2+2=5.
Only a Voodoo Mathematician like yourself would claim it to be so.
Furthermore, extending the analogy of you claiming all differential
equations having the same solution, we have the following quadraic
equation.
** x^2 - 3 x + 2 = 0
Solving for x, we get (x = 1) or (x = 2). According your silly claim,
all solutions are the same. Thus, 1 = 2.
You really need to go back to kindergarten and study above basic
mathematics instead of embracing Voodoo Mathematics. This is the best
advice I can give you.
> > With infinite number of
> > solutions, it is shaking the very foundation of GR and SR.
>
> Sure. My boots are all torn already.
>
> Give us a break.
If you want a break, you can find a hole on the ground and hide your
head in it, just like an ostrich. In the meantime, the house of cards
called GR and SR, where all pieces are bonded together through Voodoo
Mathematics, is coming down.
Igor
Try to keep up. There's only one Schwarzschild solution. If you can
find any others by merely transforming the coordinates, it won't ever
count as an independent solution. How can it? But you keep claiming
that it is. And that's just plain wrong. You can go ahead and believe
what you want to, even it's wrong. I dare you to find one more
solution that satisfies the conditions of Schwarzschild and that is
truly independent of his original solution. Birkoff proved that it
can't be done.
Igor
Koobee Wublee wrote:
> JanPB wrote:
> > Koobee Wublee wrote:
>
> > > Schwarzschild found a unique solution to the differential equations of
> > > Einstein Field Equations in free space. Hilbert found another one that
> > > he called it Schwarzschild Metric. Recently, Mr. Rahman also a
> > > contributor of this newsgroup presented another solution. The
> > > spacetime with this metric is
> > >
> > > ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
> > >
> > > The metric above indeed is another solution which anyone can easily
> > > verify because its simplicity. Notice Rahman's metric and
> > > Schwarzschild's original metric do not manifest black holes.
> >
> > No, the metric above is equal to Schwarzschild's metric. The form above
> > is obtained by a coordinate change from the original one, hence the
> > metric remains the same (tensors do not change under coordinate
> > changes).
>
> ** ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
> ** ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2
>
> K is an integration constant chosen to fit Newtonian result. It is
>
> ** K = 2 G M / c^2
>
> For the record, Mr. Bielawski is claiming the above two metrics are the
> same despite one manifests a black hole and other one not.
That's just plain wrong. The black hole is still there. You just
refuse to see it. Math has rules and disobeying them is no excuse.
Come back when you've truly learned some of them, especially
transforming a domain.
> > > However, since Schwarzschild Metric is much simpler than
> > > Schwarzschild's original solution, Schwarzschild Metric is embraced by
> > > the physics communities today.
> >
> > It's embraced because it's the same.
>
> You embraced it because of your denial of faulty GR.
You've never shown anything to be faulty, except for your understanding
of the math.
> > > Mr. Bielawski and Igor have not
> > > understood Schwarzschild's original paper and choose to blindly reject
> > > Schwarzschild's original solution and others.
> >
> > There is nothing to reject. One can _prove_ Schwarzschild's metric is
> > unique. Off the horizon it follows immediately from the particular form
> > of the Einstein equation in the spherically symmetric case (which is
> > what we have) and the uniqueness of the extension over the horizon is
> > slightly more involved but it follows from a similar argument.
>
> You rejected it because of your denial of faulty GR. You have 90 years
> of fun playing with Voodoo Mathematics that gives rise to GR. It is
> time to tear it down.
You have a hell of a lot of nerve calling something voodoo when you
can't even play by the proper rules to begin with.
> > > As multiple solutions to the vacuum field equations are discovered,
> > > there are actually an infinite number of them.
> >
> > Yes, and 2+2=5.
>
> Only a Voodoo Mathematician like yourself would claim it to be so.
> Furthermore, extending the analogy of you claiming all differential
> equations having the same solution, we have the following quadraic
> equation.
>
> ** x^2 - 3 x + 2 = 0
>
> Solving for x, we get (x = 1) or (x = 2). According your silly claim,
> all solutions are the same. Thus, 1 = 2.
Who said all the solutions are the same? Nobody. You need to
understand the distinction between the concepts of same and equivalent.
And brushing up on transforming a domain wouldn't hurt either.
Tom Roberts
> ** ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
> ** ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2
>
> K is an integration constant chosen to fit Newtonian result. It is
> ** K = 2 G M / c^2
> For the record, Mr. Bielawski is claiming the above two metrics are the
> same despite one manifests a black hole and other one not.
This is wrong, and both exhibit a black hole. You forgot to specify the
regions of validity of the coordinates: in the first one the horizon is
at r=0 (note the metric components are singular there); the black hole
is the region -K<r<0. While it is labeled "r", that coordinate does
indeed have perfectly valid negative values inside the horizon, and r is
timelike there. From the last term it is clear that r is not "radius" in
this first line element.
Looking just at the last term (solid angle), it is clear that r=-K in
the first line element corresponds to r=0 in the second, and r=0 in the
first corresponds to r=K in the second.
These two line elements do indeed correspond to the same metric,
projected onto different coordinates. <shrug>
Tom Roberts
I.Vecchi
Tom Roberts ha scritto:
Your construct relies on the arbitrary duplication of the the horizon,
inventing two copies of what is actually a single
physical/observational domain. It remains a fact that the extension
across the horizon is not unique and that each extension yields a
distinct physical object.
A black hole is not a white hole. Period.
IV
Edward Green
> "Edward Green" <spamsp...@netzero.com> wrote in message
> news:1156222614....@p79g2000cwp.googlegroups.com...
> > Bill Hobba wrote:
> >
> >> "Edward Green" <spamsp...@netzero.com> wrote ...
> >
> > <stretching the manifold vs. compressing the coordinate system>
> >
> >> First why is one unphysical and the other not?
> >
> > Well, that depends on whether it is "physical" to stretch the manifold.
>
> Since it is not a material then obviously it is a 'word salad's with no
> meaning like saying stew smells like isomorphism.
You are merely being insulting now.
Obviously a manifold is a mathematical abstraction. Obviously any word
normally applied to material objects tentatively when applied to a
mathematical abstraction should be understood to refer to a putative
property of the abstraction analogous to the named material property.
I could equally well scoff that since a manifold is not a material then
it makes no sense assign "curvature" to it. And yet sense can be so
made, as you know.
Of course, in the continuing verbal self-defense your minimax strategy
makes necessary (proving me maximally stupid based on minimal
evidence), I hasten to add that I am _not_ arguing that since a meaning
can be assigned to "curvature" meaning must also be assignable to the
strain or "stretching" of a manifold. I am merely indicating that such
a putative identification is not the silly category error you imply.
In fact, the idea will have meaning to the extent we assign meaning to
it. Modeling the rubber band by a one dimensional manifold, we will
find it natural to assign a pointwise property to the model called
"strain". Is this part of the minimal structure of a differential
manifold, or something added? I don't know. Is there something
analagous to strain assigned to the manifold which is an element of the
model of GR? I don't know.
Since I know that I don't understand the model, I don't know that there
must be a property of the model of GR plausibly described as a local
strain state of spacetime. I am not confident however that there is no
such property is possible merely because Bill Hobba asserts it, because
Bill Hobba denigrates me merely for describing a putative property via
a word applicable to material objects, as if such alone were sufficient
proof of idiocy, when Bill Hobba in fact knows that mathematics freely
borrows terms from the quotidian world.
Your counter argument is falacious.
> > In a simple minded analogy, the one dimensional manifold might be
> > represented by a rubber band: stretching the rubber band is a
> > physically distinct situation from compressing the coordinates -- one
> > might be able to determine the state of strain by local measurements,
> > or at least determine the relative strain for two locations. In GR the
> > "stretch" might correspond to a gravitational field vs. flat spacetime.
> > Not _everything_ is an artifact of the coordinate system.
>
> You were on the right track above - stretching is not a property of the
> space-time manifold.
That's better: a blunt assertion. Maybe you are right, but you will
forgive me if I don't assign your claim 100% Bayesian confidence just
at this moment. You have after all given me no reason at all to
believe you beyond your blunt assertion.
(Leave off the implicit insults in the interest of brevity. Thanks.)
Tom Roberts
> Tom Roberts ha scritto:
>> I.Vecchi wrote:
>>> I would say that there are at least two distinct ways to prolong the
>>> solution across the horizon, which yield respectively the black hole
>>> and the white hole solution, describing two different physical
>>> phenomena. They correspond to two distinct choices of the
>>> Eddington-Finkelstein coordinates at the horizon, i.e. to crossing the
>>> horizon on future-directed or on past-directed curves. The extended
>>> solution depends on how space-time is extended beyond the horizon.
>> The Kruskal-Szerkes coordinates show that these two extensions are
>> merely two aspects of the complete (inextensible) manifold.
>
> Your construct relies on the arbitrary duplication of the the horizon,
Sure. As is easily seen in a Kruskal diagram, and as I said. Here you
seem to be in violent agreement with what I said (but later you get it
wrong).
It is arbitrary which portion of the horizon you consider, but as both
are contained in the manifold it is only the analyst who is subject to
this choice, not the manifold itself (or the physical system for which
it is a model).
> inventing two copies of what is actually a single
> physical/observational domain.
Not really. Those regions of the complete (inextensible) manifold are
DIFFERENT. That is, considered from a given point in spacetime near but
outside the horizon, the past and future horizons are NOT "the same".
This difference between past and future is true in ANY spacetime, of course.
That is, for any point in any manifold of GR, every locus
in the past lightcone of the point is disjoint from every
locus in the future lightcone of the point. This is true
in everyday life -- just think about how different is your
ability to observe events in the past from events in the
future.
> It remains a fact that the extension
> across the horizon is not unique and that each extension yields a
> distinct physical object.
Not "physical object" but rather region of the manifold. Basically you
happened to choose E-F coordinates that do not cover the manifold, but
both sets of E-F coordinates cover the exterior region; which set of E-F
coordinates you choose will determine into which region of the manifold
you can extend. BTW neither choice includes other regions of the
complete (inextensible) manifold, but Kruskal-Szerkes coordinates
include them all.
> A black hole is not a white hole. Period.
Sure. And both are contained in the Kruskal diagram, and in the complete
(inextensible) manifold.
Have you never looked at a Kruskal diagram? -- you seem rather
unknowledgeable about basic aspects of this manifold (yes, singular --
you are discussing different regions of a single manifold). Any
reasonably modern textbook on GR will have it.
Tom Roberts
carlip...@physics.ucdavis.edu
> Tom Roberts ha scritto:
[...]
>> The Kruskal-Szerkes coordinates show that these two extensions are
>> merely two aspects of the complete (inextensible) manifold.
> Your construct relies on the arbitrary duplication of the the horizon,
> inventing two copies of what is actually a single
> physical/observational domain. It remains a fact that the extension
> across the horizon is not unique and that each extension yields a
> distinct physical object.
The Kruskal-Szerkes extension is the unique maximal analytic extension
of the Schwarzschild exterior geometry. Which part do you want to
give up?
Steve Carlip
Koobee Wublee
> Koobee Wublee wrote:
> > ** ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
> > ** ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2
> >
> > K is an integration constant chosen to fit Newtonian result. It is
> > ** K = 2 G M / c^2
> > For the record, Mr. Bielawski is claiming the above two metrics are the
> > same despite one manifests a black hole and other one not.
>
> This is wrong, and both exhibit a black hole. You forgot to specify the
> regions of validity of the coordinates: in the first one the horizon is
> at r=0 (note the metric components are singular there); the black hole
> is the region -K<r<0. While it is labeled "r", that coordinate does
> indeed have perfectly valid negative values inside the horizon, and r is
> timelike there. From the last term it is clear that r is not "radius" in
> this first line element.
Oh, no. Not you too. I should not be surprised from the master of
word salad himself.
You should read the following equations are mutually EXCLUSIVE of each
other. Only one of them can exist at the same time in one world. The
valid range for r for both equations are (r >= 0). At (r = 0), we have
the very center of the gravitating object.
** ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
** ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2
> [...]
>
> These two line elements do indeed correspond to the same metric,
> projected onto different coordinates. <shrug>
Your claim is forever caste in stone (the equivalence in cyberspsace).
Koobee Wublee
> Koobee Wublee wrote:
> > ** ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
> > ** ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2
> >
> > K is an integration constant chosen to fit Newtonian result. It is
> >
> > ** K = 2 G M / c^2
> >
> > For the record, Mr. Bielawski is claiming the above two metrics are the
> > same despite one manifests a black hole and other one not.
>
> That's just plain wrong. The black hole is still there.
The first metric does not manifest a black hole. Show me how you get a
black hole from
** (1 + 2 G M / c^2 / r = 0) where (r >= 0)
> You just
> refuse to see it. Math has rules and disobeying them is no excuse.
> Come back when you've truly learned some of them, especially
> transforming a domain.
Hey, this is very simple math. If you cannot do this, you need to
consider going back to junior high school to brush up on your basic
algebra.
> > > > However, since Schwarzschild Metric is much simpler than
> > > > Schwarzschild's original solution, Schwarzschild Metric is embraced by
> > > > the physics communities today.
> > >
> > > It's embraced because it's the same.
> >
> > You embraced it because of your denial of faulty GR.
>
> You've never shown anything to be faulty, except for your understanding
> of the math.
I have shown them through out many posts in the past year or so. Every
post is backed up by rigorous mathematical analysis.
> > You rejected it because of your denial of faulty GR. You have 90 years
> > of fun playing with Voodoo Mathematics that gives rise to GR. It is
> > time to tear it down.
>
> You have a hell of a lot of nerve calling something voodoo when you
> can't even play by the proper rules to begin with.
Oh, I play within the rules of mathematics. Since Mr. Bielawski fails
at simple numeric operation, he needs to go back to kindergarten. If
you still have not learnt proper arithmetic, maybe you will study them
with my 2.5 year old twins in the future. As far as you go, you need
to go back to junior high to study the basic algebra.
> Who said all the solutions are the same? Nobody. You need to
> understand the distinction between the concepts of same and equivalent.
> And brushing up on transforming a domain wouldn't hurt either.
Each metric is merely a unique and independent solution to a set of
differential equations called Einstein Field Equations. These are
merely differential equations. Since you, Mr. Bielawski, and Dr.
Roberts are all claiming all solutions are the same. To be fair, you
are saying all solutions to a set of differential equations are the
same.
> > As multiple solutions to the vacuum field equations are discovered,
> > there are actually an infinite number of them. With infinite number of
> > solutions, it is shaking the very foundation of GR and SR. The house
> > of cards will soon inevitably collapse. However, refusing to give up
> > GR and to comfort themselves in false sense of security, they choose to
> > embrace Voodoo Mathematics. In doing so, they blindly claim all
> > solutions are indeed the same regardless manifesting black holes,
> > constant expanding universe, accelerated expanding universe. VOODOO
> > MATHEMATICS REPRESENTS THE ACHIEVEMENT IN PHYSICS DURING THE LAST 100
> > YEARS. It is very sad that these clowns are regarded as experts in
> > their field.
>
> Try to keep up. There's only one Schwarzschild solution. If you can
> find any others by merely transforming the coordinates, it won't ever
> count as an independent solution. How can it?
All solutions to a set of differential equations are related but
independent of each other. Why is it a surprise to be able to obtain
them through a theorem I showed you guys?
> But you keep claiming that it is. And that's just plain wrong.
The rules of mathematics show so.
> You can go ahead and believe
> what you want to, even it's wrong. I dare you to find one more
> solution that satisfies the conditions of Schwarzschild and that is
> truly independent of his original solution.
What do you mean by the conditions of Schwarzschild? Do you
Schwarzschild Metric as discovered by Hilbert or Schwarzschild's
original metric?
> Birkoff proved that it can't be done.
Birkoff assumed that Schwarzschild Metric is the only solution to
derive that theorem. His assumption is blatantly wrong. His theorem
is thus total rubbish.
Sorcerer
"Koobee Wublee" <koobee...@gmail.com> wrote in message
news:1156547523.8...@m73g2000cwd.googlegroups.com...
For the record,
http://www.androcles01.pwp.blueyonder.co.uk/DominoEffect.GIF
Hence anyone using c has to be a lunatic with a peanut for a neuron.
Androcles
JanPB
> Igor wrote:
> > Koobee Wublee wrote:
>
> > > ** ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
> > > ** ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2
> > >
> > > K is an integration constant chosen to fit Newtonian result. It is
> > >
> > > ** K = 2 G M / c^2
> > >
> > > For the record, Mr. Bielawski is claiming the above two metrics are the
> > > same despite one manifests a black hole and other one not.
> >
> > That's just plain wrong. The black hole is still there.
>
> The first metric does not manifest a black hole. Show me how you get a
> black hole from
>
> ** (1 + 2 G M / c^2 / r = 0) where (r >= 0)
At r = -2GM/c^2. Your restriction r>0 is wrong, it should read
r>-2GM/c^2.
The reason for it this is that the first form of the metric is obtained
from the second (for which r>0) by means of the coordinate change:
r = u + K
t = t
theta = theta
phi = phi
...where you are recycling the letter "r" in the first form where I
have used "u". Look at the formula r=u+K. It's patently obvious that
when r changes between 0 and infinity then u changes between -K and
infinity. Tell me, is this not true?
> > Who said all the solutions are the same? Nobody. You need to
> > understand the distinction between the concepts of same and equivalent.
> > And brushing up on transforming a domain wouldn't hurt either.
>
> Each metric is merely a unique and independent solution to a set of
> differential equations called Einstein Field Equations. These are
> merely differential equations. Since you, Mr. Bielawski, and Dr.
> Roberts are all claiming all solutions are the same. To be fair, you
> are saying all solutions to a set of differential equations are the
> same.
These are *tensor* PDEs, not scalar. The particular form of the
equations therefore changes depending on the coordinate system used.
The solutions of these _particular_ equations (the actual PDEs you
solve) are then *components* of the tensor you seek. That tensor in the
spherically symmetric vacuum case is unique although - obviously - a
particular coordinate representation will look different and will have
correspondingly transformed domain compared to some other coordinate
representation.
> > Try to keep up. There's only one Schwarzschild solution. If you can
> > find any others by merely transforming the coordinates, it won't ever
> > count as an independent solution. How can it?
>
> All solutions to a set of differential equations are related but
> independent of each other.
To a *scalar* PDE, yes (with technical qualifications). But you have
here a non-zero-rank *tensor* PDE which means for any given coordinate
system it's a PDE for *tensor components* in that system. These
components are going to look different in different coordinates but the
*tensor*, that is the actual *solution* they describe, is the same. Of
course it's possible to have multiple *tensor solutions* in general but
this is not what happens in the spherically symmetric vacuum case.
> > You can go ahead and believe
> > what you want to, even it's wrong. I dare you to find one more
> > solution that satisfies the conditions of Schwarzschild and that is
> > truly independent of his original solution.
>
> What do you mean by the conditions of Schwarzschild? Do you
> Schwarzschild Metric as discovered by Hilbert or Schwarzschild's
> original metric?
I assume he meant spherically symmetric and vacuum.
> > Birkoff proved that it can't be done.
>
> Birkoff assumed that Schwarzschild Metric is the only solution to
> derive that theorem. His assumption is blatantly wrong. His theorem
> is thus total rubbish.
What utter nonsense. Birkhoff only assumed the two things I mentioned:
spherical symmetry and vacuum (and signature 2 and other obviosities).
You don't assume what you want to prove for goodness sake. You simply
write down the Einstein equation and out comes the *unique* tensor
because in this case this equation reduces to an ODE, so the solution
must be unique once the boundary conditions are pinned down. It's an
elementary fact about ODEs in general (look up Picard-Lindeloef
theorem).
--
Jan Bielawski
JanPB
>
> Jan, I think you should study the original paper
> Ed Green cited, there is no such thing as an
> event horizon or Black-Hole's. BTW, Dr. Loinger
> and I discussed this at length. In short, the
> original Schwarzschild Solution has been
> bastardized and mis-understood for simplicity.
No, the "original" solution is the only one. This follows from basic
ODE theory and the definition of tensor.
> The bastardized version became popularized
> and embraced by astronomers who now see
> BH's under their beds.
There is no other version, bastardized or not. How many solutions do
you have to the following ODE:
f'(x) = f(x)
...given the initial condition f(0) = 1 ? I can see f(t) = exp(t). Is
there any other?
How about another ODE, just slightly more complicated:
-2 f(x) f'(x) + 1/x * (1 - f(x)^2) = 0
...given the initial condition f(1) = 0 ?
--
Jan Bielawski
I.Vecchi
Tom Roberts ha scritto:
> I.Vecchi wrote:
> > Your construct relies on the arbitrary duplication of the the horizon,
>
> Sure. As is easily seen in a Kruskal diagram, and as I said. Here you
> seem to be in violent agreement with what I said (but later you get it
> wrong).
>
> It is arbitrary which portion of the horizon you consider, but as both
> are contained in the manifold it is only the analyst who is subject to
> this choice, not the manifold itself (or the physical system for which
> it is a model).
Are you saying that for every black hole in the universe there is a
corresponding white hole?
>
> > inventing two copies of what is actually a single
> > physical/observational domain.
>
> Not really. Those regions of the complete (inextensible) manifold are
> DIFFERENT. That is, considered from a given point in spacetime near but
> outside the horizon, the past and future horizons are NOT "the same".
> This difference between past and future is true in ANY spacetime, of course.
>
> That is, for any point in any manifold of GR, every locus
> in the past lightcone of the point is disjoint from every
> locus in the future lightcone of the point. This is true
> in everyday life -- just think about how different is your
> ability to observe events in the past from events in the
> future.
>
>
> > It remains a fact that the extension
> > across the horizon is not unique and that each extension yields a
> > distinct physical object.
>
> Not "physical object" but rather region of the manifold.
Which corresponds to a physical/observational object. Otherwise we are
talking about nothing.
> Basically you
> happened to choose E-F coordinates that do not cover the manifold, but
> both sets of E-F coordinates cover the exterior region; which set of E-F
> coordinates you choose will determine into which region of the manifold
> you can extend.
Yes, which set of coordinates I choose will determine it . And
correspondily yield the black hole or the white hole solution.
> BTW neither choice includes other regions of the
> complete (inextensible) manifold, but Kruskal-Szerkes coordinates
> include them all.
Yes, it's mathematical construct useful for didactic purposes.
>
>
> > A black hole is not a white hole. Period.
>
> Sure. And both are contained in the Kruskal diagram, and in the complete
> (inextensible) manifold.
>
> Have you never looked at a Kruskal diagram? -- you seem rather
> unknowledgeable about basic aspects of this manifold (yes, singular --
> you are discussing different regions of a single manifold). Any
> reasonably modern textbook on GR will have it.
The point here is its relevance to the topic of this discussion. As for
extending my knowledge, which may indeed be limited, vigorous (and,
ideally, polite) online discussion appears to be a fruitful approach.
Beside the above, I surmise that there are other space-time extensions
across the horizon corresponding to hybrid white hole/black hole
solutions (*). This would be impossible according to your argument,
right?
Cheers,
IV
(*) In the corresponding charts the metric gets singular at one point
on the horizon, but that does not make it more unphysical than, say,
the Schwarzschild solution.
I.Vecchi
carlip...@physics.ucdavis.edu ha scritto:
> The Kruskal-Szerkes extension is the unique maximal analytic extension
> of the Schwarzschild exterior geometry.
Nice. Is there a proof for that? I mean a mathematically rigorous one,
where the hypotheses are cleary stated, so that one can weigh their
physical relevance. A reference would be appreciated.
> Which part do you want to
> give up?
In any given case, the one that does not fit observation.
You are warmly welcome to provide your feedback on the point I raise
about "hybrid solutions" in my reply to Tom Roberts.
IV
Edward Green
Hi Jan,
I don't see you ever got back to me on my humble request to show me
where I went wrong in my heuristic. It may be, as Hobba suggested,
that there is nothing corresponding to a dilation or strain of the
manifold in GR. If that's true, then that removes the conflict between
the explicit form of the metric trying to compensate for dilation and
also for an arbitrary change of units in different regions of the
manifold. Maybe that's the answer to my question.
Anyway, since you are discussing black holes/white holes, let me jump
ahead. I'd like to know: do black holes and white holes represent
regions of a single complete inextensible manifold corresponding to the
Schwarzschild metric, or are they result of something like a sign
choice -- disconnected solutions?
This seems like a rather important point. If the solutions are
disconnected, then it is less damaging to discard one of the solutions
as unphysical. If they are different branches of the same analytic
extension, then discarding one is a much more serious theoretical
problem: analytic extensions are what we obtain by creeping across the
surface of the solution, so to speak, and saying "well, we have a
solution up to here, what is required to extend the solution a little
further, assuming the field equations to be satisfied". An arbitrary
truncation of such a process because we obtain unpalatable results is
tantamount to saying the theory is broken somewhere.
(Theories are allowed to be broken somewhere, and still be beautiful
and useful).
Edward Green
Sorry, I missed your reply at first.
> Basically s.p.r. has no expertise in GR so although
> your question is well posed and penetrating it exceeded
> the moderators understanding. Myself I'd be interested
> in possible well informed opinions that occasionally
> post to s.p.r. , that said I'll try.
You are a gentleman, as always.
> > The offhand claim is sometimes made that GR, which I take it means the
> > formal machinery of GR, treats all coordinate systems equally. Is this
> > true?
>
> Pardon my apparent evasiveness...
And mine. :-)
[snip gracious mini tutorial in GR]
Ken, I seem to be subject to a mini-curse: nobody quite recognizes what
I'm asking; and finally, given enough perserverence, I may come up with
answers, still unrecognizable. Not recognizing the question,
respondants are either sympathetic or insulting (or silent) according
to their own personality, but not, from my point of view, on target.
I asked how the metric -- particularly the explicit coordinate
representation of the metric -- adapts to dilations in spacetime.
Hobba's answer is, there is nothing corresponding to a dilation in
spacetime. I'm not sure if I believe this or not. What about
gravitaional time dilation? This phrase tends to indicate that a
second as measured by a standard physical clock recorded over _here_
may not correspond to a second over _there_. Physical clocks and
rulers, composed of standard arrangement of atoms, give what we might
consider a natural local metric to spacetime. If we have reason to
believe -- without the possibility of direct comparison -- that these
metrics would not agree in different regions, then it seems plausible
to say that there is a relative dilation of spacetime beween the
regions.
How does the formal representation of a metric handle this? Is the
change in a time-like variable measured in local seconds prefixed with
a different coefficient in region one than in region two? Or are local
seconds always prefixed with the same constant coefficient in a given
representation of a metric, as are, say, local centimeters. That
doesn't seem right, otherwise all spacetimes would look like flat
spacetime as seen through the lens of the explicit metric!
Schwarschild's metric, e.g.
http://en.wikipedia.org/wiki/Schwarzschild_metric
trivially has the property that the prefix to dt is a function of r.
The "dt" implicitly represents time as measured by a appropriate
standard physical clock. So evidently not all such clocks are equal
with respect to the metric -- time is "stretched" in some regions
relative to others.
Now, I can rephrase my question. Suppose we replaced "t" with t' =
a(t)t , a( ) a positive increasing function. This can be interpreted
as changing the units of time from "natural clock seconds", and not in
a uniform way. How does the explicit form of the metric handle this?
In this case, a possible answer seems trivial: we simply insert the
inverse of the positive factor "a(t)"... better, a^-1(t') ... in the
formal metric, and everything is fine.
This is the point where I see a possible problem. We now have a
function of variables labeled t',r,O (omega) which corresponds to the
same spacetime as a different function of variables labeled t,r,O .
So, what's the problem? We've just changed varables. The problem is,
suppose we could demonstrate a second _different spacetime_ with the
same variable labels t',r,O and the same formal metric. Is this a
problem?
I can't carry through my threat here, with the Schwarschild metric, but
that was the point of my toy one dimensional manifold -- I tried to
motivate such a possibility.
OK, Ken, here is my current postulation for an answer: you _can't_
reconstruct a physical spacetime simply from a set of variable labels
and an explicit form of a metric on such labels: there is an element
missing. That element is a perscription of the relationship between
the local differentials of the variables and standard local physical
processes -- "so many wavelengths of a certain spectral transition",
and so forth. If this perscription is the same everywhere, then it can
simply be stated as "choice of units", as in, centimeters and seconds;
if the perscription varies, then the additional information will be
more complicated. But there is additional structure required to
complete the model.
Comment: the "standard local clock" may need clarification.
Comment: my internal pendulum is again swinging towards supposing my
question was close to being well-posed or at least well-posable, and
that therefore those who merely dismissed it as containing elementary
errors would have done better to remain silent, than open their mouth
and remove all doubt.
> GR treats all Frames of Reference's (FoR's) equally.
> AE's law, G_uv = T_uv constrains the metrical
> geometry, to solutions of the tensor g_uv satisfying
> that law, that is why G_uv=T_uv is a called a field
> equation. For example a simplistic solution follows
> from G_uv=0 called the Schwarzchild Solution, that
> I'm sure you're aware of, that provides metrics like,
>
> g_00 = 1 - 2m/r , g_rr = 1/g_00 etc.
>
> The reason for CS independence is because g_uv
> is a tensor and can be properly transformed via
> the usual procedure,
>
> g'_ab = (&x^u/&x'_a) (&x^v/&x'_b) g_uv
>
> to any other CS, using light-years, millimetres
> polar, elliptical, cylindrical as you please, and
> in time, seconds or dog-years, meaning the
> physical reality is independent of units.
Of course. I'm thinking however that the units must be stated, which
is sufficiently trivial as to be invisible, and that if we allow
arbitrary point to point fluctuations in the units, we may need some
additional non-trivial statements to complete the specification, beyond
the explicit form of the metric.
As always, I could be wrong.
Thanks, Ken.
Sorcerer
| Ken, I seem to be subject to a mini-curse: nobody quite recognizes what
| I'm asking; and finally, given enough perserverence, I may come up with
| answers, still unrecognizable. Not recognizing the question,
| respondants are either sympathetic or insulting (or silent) according
| to their own personality, but not, from my point of view, on target.
|
| I asked how the metric -- particularly the explicit coordinate
| representation of the metric -- adapts to dilations in spacetime.
Edward, I seem to be subject to a mini-curse: nobody quite recognizes what
I'm asking; and finally, given enough perserverence, I may come up with
answers, still unrecognizable. Not recognizing the question,
respondants are either sympathetic or insulting (or silent) according
to their own personality, but not, from my point of view, on target.
I asked how the egg -- particularly the explicit ovum representation
of the bright green flying elephant -- adapts to contours in black
holes.
Androcles
Ken S. Tucker
Edward Green wrote:
> Ken S. Tucker wrote:
> I asked how the metric -- particularly the explicit coordinate
> representation of the metric -- adapts to dilations in spacetime.
> Hobba's answer is, there is nothing corresponding to a dilation in
> spacetime. I'm not sure if I believe this or not. What about
> gravitational time dilation? This phrase tends to indicate that a
> second as measured by a standard physical clock recorded over _here_
> may not correspond to a second over _there_. Physical clocks and
> rulers, composed of standard arrangement of atoms, give what we might
> consider a natural local metric to spacetime. If we have reason to
> believe -- without the possibility of direct comparison -- that these
> metrics would not agree in different regions, then it seems plausible
> to say that there is a relative dilation of spacetime beween the
> regions.
Yes, I agree.
> How does the formal representation of a metric handle this?
Well first we need to agree that survey's are done
using light, radar and laser's, by astronomers, and
astrophysicist's. In the case of Newton, the "metric"
was assumed to be equalivalent to a Cartesian CS,
and I repeat, light is unaffected in a Cartesian CS,
(metric) because the light-ray paths define the axes.
The so-called "spacetime-field" is defined by the
velocity (direction and speed) of light-rays.
Subsequent to SR and QT, conservation of the
energy of light needed to be included in the
method of surveying astronomical calculations,
primarily using the Einstein shift and to a lesser
degree the deflection of light.
Thus the old Cartesian metric was subtly altered,
accounting for the effects mentioned above, and
I think an outstanding proof is given by a more
accurate description of Mercury's orbit and the
deflection of light.
The altered metric is different from Cartesian,
and changes thoses predictions w.r.t Newton.
I've studied your post below, but I'll hold here
pending a mutual understanding, you know, it's
the kinda of basics I grew up with.
Cheers
Ken S. Tucker
Ken S. Tucker
What Newton assumed was,
r= sqrt(x^2 + y^2 +z^2)
but in GR the "real" R = r - m
where m =1.47 kms in the case of the Sun.
Most treatise begin by defining "r" that way.
Schwarzschild emphasized that difference and obtained
a singularity only at the very center of a mass, and that
would assume an infinite density, which is physically
impossible. But it's a good exercise.
Ken
Bill Hobba
> Bill Hobba wrote:
>
>> "Edward Green" <spamsp...@netzero.com> wrote in message
>> news:1156222614....@p79g2000cwp.googlegroups.com...
>> > Bill Hobba wrote:
>> >
>> >> "Edward Green" <spamsp...@netzero.com> wrote ...
>> >
>> > <stretching the manifold vs. compressing the coordinate system>
>> >
>> >> First why is one unphysical and the other not?
>> >
>> > Well, that depends on whether it is "physical" to stretch the manifold.
>>
>> Since it is not a material then obviously it is a 'word salad's with no
>> meaning like saying stew smells like isomorphism.
>
> You are merely being insulting now.
If you thought that then I apologize. You have done nothing to deserve
insults.
>
> Obviously a manifold is a mathematical abstraction. Obviously any word
> normally applied to material objects tentatively when applied to a
> mathematical abstraction should be understood to refer to a putative
> property of the abstraction analogous to the named material property.
That is not so obvious to me. Have you considered your musings are better
suited to a philosophy forum?
Thanks
Bill
Tom Roberts
> Tom Roberts ha scritto:
>> are contained in the manifold it is only the analyst who is subject to
>> this choice, not the manifold itself (or the physical system for which
>> it is a model).
>
> Are you saying that for every black hole in the universe there is a
> corresponding white hole?
Not at all! But I am pointing out that in the Schwarzschild manifold
there is both a black hole and a white hole. That manifold _IS_ what we
are discussing. It is quite clear that the universe we inhabit is not
described by the Schw. manifold.
>>> It remains a fact that the extension
>>> across the horizon is not unique and that each extension yields a
>>> distinct physical object.
>> Not "physical object" but rather region of the manifold.
>
> Which corresponds to a physical/observational object. Otherwise we are
> talking about nothing.
I don't know what you mean by "a physical/observational object". But it
IS clear that a region of the manifold is not any sort of "object" at
all. <shrug>
One might call it an "observational domain" I suppose, but "object" is
quite definitely not applicable.
> Beside the above, I surmise that there are other space-time extensions
> across the horizon corresponding to hybrid white hole/black hole
> solutions (*). This would be impossible according to your argument,
> right?
The geodesically complete extension of the Schwarzschild charts is
unique, given by the Kruskal chart.
Tom Roberts
Tom Roberts
> Now, I can rephrase my question. Suppose we replaced "t" with t' =
> a(t)t , a( ) a positive increasing function. This can be interpreted
> as changing the units of time from "natural clock seconds", and not in
> a uniform way. How does the explicit form of the metric handle this?
The metric itself is of course unaffected.
How the metric components behave is given by the usual coordinate
transform of the components of a rank-2 covariant tensor.
> In this case, a possible answer seems trivial: we simply insert the
> inverse of the positive factor "a(t)"... better, a^-1(t') ... in the
> formal metric, and everything is fine.
No. that is not correct.
g_i'j' = (dx^i/dx^i') (dx^j/dx^j') g_ij
Here primed indices are for the primed coordinates, and unprimed are for
unprimed coords. Basically you forgot to account for how the basis
vectors of the coordinate system change because of the transform.
[Note that a^-1(t') will occur in dx^t/dx^t'.]
> OK, Ken, here is my current postulation for an answer: you _can't_
> reconstruct a physical spacetime simply from a set of variable labels
> and an explicit form of a metric on such labels: there is an element
> missing. That element is a perscription of the relationship between
> the local differentials of the variables and standard local physical
> processes -- "so many wavelengths of a certain spectral transition",
> and so forth.
Yes. Coordinates by themselves are essentially useless in physics. One
needs a metric, and in particular the metric components in terms of
those coordinates. That permits one to compute distances between points
labeled by the coordinates, and from that everything else follows.
> If this perscription is the same everywhere, then it can
> simply be stated as "choice of units", as in, centimeters and seconds;
> if the perscription varies, then the additional information will be
> more complicated. But there is additional structure required to
> complete the model.
Yes. This "additional structure" is called a metric.
> I'm thinking however that the units must be stated, which
> is sufficiently trivial as to be invisible, and that if we allow
> arbitrary point to point fluctuations in the units, we may need some
> additional non-trivial statements to complete the specification, beyond
> the explicit form of the metric.
The metric components and the coordinate differentials must include
units. It is a matter of personal choice whether the units are put into
the metric components or into the coordinate differentials; in any case
ds^2 must have units (length)^2.
BTW Ken Tucker in the past has been particularly confused about the
relationship among coordinates, units, and metric components....
Tom Roberts
Koobee Wublee
JanPB wrote:
> Koobee Wublee wrote:
>> ** ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
>> ** ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2
>>
>> K is an integration constant chosen to fit Newtonian result. It is
>>
>> ** K = 2 G M / c^2
>
> At r = -2GM/c^2. Your restriction r>0 is wrong, it should read
> r>-2GM/c^2.
>
> The reason for it this is that the first form of the metric is obtained
> from the second (for which r>0) by means of the coordinate change:
How do you know which of the two is the first form? And why?
> ...where you are recycling the letter "r" in the first form where I
> have used "u". Look at the formula r=u+K. It's patently obvious that
> when r changes between 0 and infinity then u changes between -K and
> infinity. Tell me, is this not true?
No.
JanPB
> JanPB wrote:
> > Koobee Wublee wrote:
>
> >> ** ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
> >> ** ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2
> >>
> >> K is an integration constant chosen to fit Newtonian result. It is
> >>
> >> ** K = 2 G M / c^2
> >
> > At r = -2GM/c^2. Your restriction r>0 is wrong, it should read
> > r>-2GM/c^2.
> >
> > The reason for it this is that the first form of the metric is obtained
> > from the second (for which r>0) by means of the coordinate change:
>
> How do you know which of the two is the first form? And why?
If one starts - as is usual - with a spherically symmetric
(t,r,theta,phi)-space with r>0, etc., then any spherically symmetric
metric can be locally written in the two forms we had before (not the
two forms you wrote above - the two forms with the sign switch). One
solves the Einstein equations which describe a metric (uniquely,
because the equations happen to be ODEs in this case) described by
components which happen to blow up at r=2m in that basis. This is your
second equation.
If one started instead with a spherically symmetric
(t,r,theta,phi)-space with labelling set so that r>-K (K - some number)
then you'd end up with the solution that looks like your first
equation. Same thing, different chart.
> > ...where you are recycling the letter "r" in the first form where I
> > have used "u". Look at the formula r=u+K. It's patently obvious that
> > when r changes between 0 and infinity then u changes between -K and
> > infinity. Tell me, is this not true?
>
> No.
Are you sure? This is aritmetic now. One more time: assuming r=u+K if r
has values between 0 and infinity then u has values between what and
what?
--
Jan Bielawski
Sorcerer
"JanPB" <fil...@gmail.com> wrote in message
news:1156659448.0...@m73g2000cwd.googlegroups.com...
err... -1 and 3? I used ter be gud at ritmatec 'n' speling.
| --
| Jan Bielawski
|
JanPB
Not sure what you're alluding to.
For the record, the second equation I wrote:
-2 f(x) f'(x) + 1/x * (1 - f(x)^2) = 0
...is what the Einstein field equations reduce to (in the case that
turns out to be the exterior).
The solution is the familiar:
f(x) = sqrt(1 + C/x)
I set up the initial condition at random where normally one chooses the
constant C so that the metric is Newtonian for large x, namely C = -2m.
My point was that at this stage this is really the ODE theory, hence
the uniqueness.
--
Jan Bielawski
Koobee Wublee
JanPB wrote:
> Koobee Wublee wrote:
>> ** ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
>> ** ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2
>>
>> K is an integration constant chosen to fit Newtonian result. It is
>>
>> ** K = 2 G M / c^2
>>
>
> If one starts - as is usual - with a spherically symmetric
> (t,r,theta,phi)-space with r>0, etc., then any spherically symmetric
> metric can be locally written in the two forms we had before (not the
> two forms you wrote above - the two forms with the sign switch). One
> solves the Einstein equations which describe a metric (uniquely,
> because the equations happen to be ODEs in this case) described by
> components which happen to blow up at r=2m in that basis. This is your
> second equation.
Of course, r >=0. This is spherically symmetric polar coordinate.
> If one started instead with a spherically symmetric
> (t,r,theta,phi)-space with labelling set so that r>-K (K - some number)
> then you'd end up with the solution that looks like your first
> equation. Same thing, different chart.
Is there any reason for you to avoid my simple question?
> > > ...where you are recycling the letter "r" in the first form where I
> > > have used "u". Look at the formula r=u+K. It's patently obvious that
> > > when r changes between 0 and infinity then u changes between -K and
> > > infinity. Tell me, is this not true?
> >
> > No.
>
> Are you sure? This is aritmetic now. One more time: assuming r=u+K if r
> has values between 0 and infinity then u has values between what and
> what?
Very sure. In both equations, r >= 0.
Ken S. Tucker
I perfer other approaches but I'll use your's here.
f(x) ~ (1-m/x) , x*f(x) = x - m.
In more standard notation, that is written,
R = r - m
as I posted. Where Newton says the light particle
would be at "r" Einstein predicted it would be at "R",
with the caveat that we're in 3D. When translating
to 4D the actual quantity becomes R(4D) = r - 2m.
> My point was that at this stage this is really the ODE theory, hence
> the uniqueness.
What is ODE?
FrediFizzx
news:Sg9Ig.20724$gY6....@newssvr11.news.prodigy.com...
Tom, please don't get sloppy with terminology. It is ds^2 must have
*dimensions* of length^2. However when c = 1, length = time so it can
also have dimensions of time^2.
FrediFizzx
Quantum Vacuum Charge papers;
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.pdf
or postscript
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.ps
http://www.arxiv.org/abs/physics/0601110
http://www.vacuum-physics.com
JanPB
That's only if K=42.
--
Jan Bielawski
JanPB
Not sure what the Einstein equation has to do with R = r - m at all...
> > My point was that at this stage this is really the ODE theory, hence
> > the uniqueness.
>
> What is ODE?
Ordinary differential equations.
--
Jan Bielawski
JanPB
> JanPB wrote:
> > Koobee Wublee wrote:
> >> ** ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
> >> ** ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2
> >>
> >> K is an integration constant chosen to fit Newtonian result. It is
> >>
> >> ** K = 2 G M / c^2
> >>
> >> How do you know which of the two is the first form? And why?
> >
> > If one starts - as is usual - with a spherically symmetric
> > (t,r,theta,phi)-space with r>0, etc., then any spherically symmetric
> > metric can be locally written in the two forms we had before (not the
> > two forms you wrote above - the two forms with the sign switch). One
> > solves the Einstein equations which describe a metric (uniquely,
> > because the equations happen to be ODEs in this case) described by
> > components which happen to blow up at r=2m in that basis. This is your
> > second equation.
>
> Of course, r >=0. This is spherically symmetric polar coordinate.
Yes, although there is nothing magical about 0. It's just a convention
everyone uses by analogy with the planar polar coordinates. One can
assign the number 42 to the centre of symmetry.
> > If one started instead with a spherically symmetric
> > (t,r,theta,phi)-space with labelling set so that r>-K (K - some number)
> > then you'd end up with the solution that looks like your first
> > equation. Same thing, different chart.
>
> Is there any reason for you to avoid my simple question?
I though I'd just answered it. I assume it was "How do you know which
of the two is the first form?" Did you mean something else? I assumed
you meant "how does one know to which form r>0 applies to?".
> > > > ...where you are recycling the letter "r" in the first form where I
> > > > have used "u". Look at the formula r=u+K. It's patently obvious that
> > > > when r changes between 0 and infinity then u changes between -K and
> > > > infinity. Tell me, is this not true?
> > >
> > > No.
> >
> > Are you sure? This is aritmetic now. One more time: assuming r=u+K if r
> > has values between 0 and infinity then u has values between what and
> > what?
>
> Very sure. In both equations, r >= 0.
No. The first equation follows from the second when you change the
coordinates as r=u+K. So for clarity you should not use the same letter
"r" there. I used "u". So if in the second equation the range of r is 0
through infinity it means that the range of u=r-K is -K through
infinity. Elementary.
--
Jan Bielawski
I.Vecchi
> I.Vecchi wrote:
...
> > Are you saying that for every black hole in the universe there is a
> > corresponding white hole?
>
> Not at all! But I am pointing out that in the Schwarzschild manifold
> there is both a black hole and a white hole. That manifold _IS_ what we
> are discussing.
What we are discussing is "the uniqueness of the extension over the
horizon" and the relevant selection criteria . What I am saying is that
there are at least two ways to extend the solution over the horizon,
correspoinding to two different physical situations, the black hole and
the white hole.
> It is quite clear that the universe we inhabit is not
> described by the Schw. manifold.
Yes, as I said, it is a physically irrelevant mathematical construct,
whose physical irrelevance reflects the irrelevance of the criteria it
fulfills.
> >>> It remains a fact that the extension
> >>> across the horizon is not unique and that each extension yields a
> >>> distinct physical object.
> >> Not "physical object" but rather region of the manifold.
> >
> > Which corresponds to a physical/observational object. Otherwise we are
> > talking about nothing.
>
> I don't know what you mean by "a physical/observational object". But it
> IS clear that a region of the manifold is not any sort of "object" at
> all. <shrug>
>
> One might call it an "observational domain" I suppose, but "object" is
> quite definitely not applicable.
You may call it as you like.
>
> > Beside the above, I surmise that there are other space-time extensions
> > across the horizon corresponding to hybrid white hole/black hole
> > solutions (*). This would be impossible according to your argument,
> > right?
>
> The geodesically complete extension of the Schwarzschild charts is
> unique, given by the Kruskal chart.
The question is whether the requirement of geodesic completeness (which
in this setting is obtained by shifting trouble to infinity) is
appriopriate, i.e. physically relevant.
I surmise that there are geodesically incomplete extensions,
corresponding to hybrid solutions which may be physically relevant.
I am not keen on throwing away interesting extensions/solutions just
because they do not not comply with some arbitrary uniqueness
criterion.
Cheers.
IV
Sorcerer
Yes, fuckhead. r>=0. c= 0/0.
The problem in both equations is c which is the velocity of light from
A to A in time t'A-tA.
http://www.androcles01.pwp.blueyonder.co.uk/DominoEffect.GIF
Androcles
Daryl McCullough
>The question is whether the requirement of geodesic completeness (which
>in this setting is obtained by shifting trouble to infinity) is
>appriopriate, i.e. physically relevant.
The answer is definitely "yes". Consider an observer in freefall near the
event horizon. Using his local coordinates, there is *nothing* to prevent
him from reaching and passing the event horizon in a finite amount of
proper time, because for him, spacetime near the event horizon is
approximately *flat*. Anything other than demanding geodesic completeness
would violate the equivalence principle, I think.
>I surmise that there are geodesically incomplete extensions,
>corresponding to hybrid solutions which may be physically relevant.
>
>I am not keen on throwing away interesting extensions/solutions just
>because they do not not comply with some arbitrary uniqueness
>criterion.
I don't think that there is anything arbitrary about the criteria.
--
Daryl McCullough
Ithaca, NY
Daryl McCullough
>JanPB wrote:
>> Koobee Wublee wrote:
>>> ** ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
>>> ** ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2
>Of course, r >=0. This is spherically symmetric polar coordinate.
What about the first metric implies that the range of r is between
0 and infinity? Are you thinking that the use of the name "r" implies
such a range?
You do realize, don't you, that the first metric, with range
r > -K is exactly the same as the second metric, with range r > 0?
Or, the other way around, the first metric with range r > 0 is
exactly the same as the second metric with range r > K>?
Daryl McCullough
>>> ** ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
>>> ** ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2
>> The reason for it this is that the first form of the metric is obtained
>> from the second (for which r>0) by means of the coordinate change:
>
>How do you know which of the two is the first form? And why?
If you compute the area of a sphere of radius r and constant t using
the top metric, you will find that it is
4 pi (r+K)^2
which implies that the effective radius is r+K, not r.
>> ...where you are recycling the letter "r" in the first form where I
>> have used "u". Look at the formula r=u+K. It's patently obvious that
>> when r changes between 0 and infinity then u changes between -K and
>> infinity. Tell me, is this not true?
>
>No.
Are you saying that it's not true that if we start with
ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2
and change variables to u = r-K, then we obtain
ds^2 = c^2 dt^2 / (1 + K / u) - (1 + K / u) du^2 - (u + K)^2 dO^2
Or are you saying that it's not true that under the variable change
u = r-K, the range r=0 to infinity is mapped to the range u=-K to
infinity?
Sorcerer
"Daryl McCullough" <stevend...@yahoo.com> wrote in message
news:ecs3o...@drn.newsguy.com...
| Koobee Wublee says...
|
| >>> ** ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
| >>> ** ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2
|
| >> The reason for it this is that the first form of the metric is obtained
| >> from the second (for which r>0) by means of the coordinate change:
| >
| >How do you know which of the two is the first form? And why?
|
| If you compute the area of a sphere of radius r and constant t using
| the top metric,
Metric.
A nonnegative function g(x,y) describing the "distance" between neighboring
points for a given set.
No metric can contain c or c^2
http://www.androcles01.pwp.blueyonder.co.uk/DominoEffect.GIF
Androcles
Daryl McCullough
>I asked how the metric -- particularly the explicit coordinate
>representation of the metric -- adapts to dilations in spacetime.
>Hobba's answer is, there is nothing corresponding to a dilation in
>spacetime. I'm not sure if I believe this or not. What about
>gravitaional time dilation?
Ed, the usual way of understanding gravitational time dilation
is that it is about the *relationship* between two different
coordinate systems---the coordinate system of an inertial
observer, and the acceleration of a noninertial observer. So
it's not really a property of *spacetime*, but of particular
coordinates on spacetime. The same effect (gravitational
time dilation) takes place inside an accelerating rocket
in flat spacetime.
Of course, spacetime curvature is relevant because if spacetime
is curved, then there *are* no extended inertial coordinate systems,
so we're forced to use noninertial coordinates.
>This phrase tends to indicate that a second as measured by a
>standard physical clock recorded over _here_ may not correspond
>to a second over _there_.
>Physical clocks and rulers, composed
>of standard arrangement of atoms, give what we might
>consider a natural local metric to spacetime. If we have reason to
>believe -- without the possibility of direct comparison -- that these
>metrics would not agree in different regions, then it seems plausible
>to say that there is a relative dilation of spacetime beween the
>regions.
Let e1 and e2 be two successive ticks of a clock (assuming that
you have an old-fashioned clock that ticks, if they make those
anymore). There are at least three different ways to compute the
"time between ticks": (1) Use the metric, and compute
Integral from e1 to e2 of square-root |g_ij dx^i dx^j|
along a geodesic path connecting e1 and e2. (2) Look at
the elapsed time shown on the clock. (3) Use a particular
coordinate system, and use t2 - t1 as the time between
ticks (where t2 is the time of e2 in that coordinate
system, and t1 is the time of e1).
The usual assumption is that methods (1) and (2) will
give the same answer. Method (3) can give a wildly
different answer, depending on the coordinate system.
Two time intervals that are the same "duration" according
to methods (1) and (2) might have different durations
according to method (3).
Now, the usual interpretation of such a discrepancy is
to treat the *coordinate* notion of "duration" as not
physically meaningful.
>How does the formal representation of a metric handle this? Is the
>change in a time-like variable measured in local seconds prefixed with
>a different coefficient in region one than in region two? Or are local
>seconds always prefixed with the same constant coefficient in a given
>representation of a metric, as are, say, local centimeters. That
>doesn't seem right, otherwise all spacetimes would look like flat
>spacetime as seen through the lens of the explicit metric!
But that's *exactly* the modern interpretation of General
Relativity: all spacetimes look flat if you look at them locally.
You can divide up spacetime into tiny little regions (for example,
consider filling spacetime with observers in rocket ships). Each
region is approximately like a little piece of flat spacetime.
Each region can be described (approximately) using good old
inertial cartesian coordinates, and all the laws of physics
will (approximately) hold in these coordinates as if there
were no gravity, no curved spacetime.
Curvature only comes into play when you consider how to glue
all these little regions together into a coherent whole. If
spacetime is flat, then you can partition it into equal-sized
rectangular regions that match neatly on the borders. If spacetime
is curved, then the regions will have overlaps or gaps and won't
fit together so smoothly.
For a lower-dimensional analogy, you can imagine dividing the
surface of the Earth into little regions that are say 100 miles
by 100 miles. You can draw a map of such a region on a flat
piece of paper without worrying about the fact that the Earth
is curved. However, if you try to fit all the little maps together,
you will find that they *can't* be fit together without overlaps
or gaps.
>Schwarschild's metric, e.g.
>
>http://en.wikipedia.org/wiki/Schwarzschild_metric
>
>trivially has the property that the prefix to dt is a function of r.
>The "dt" implicitly represents time as measured by a appropriate
>standard physical clock.
No, it really *doesn't* mean that. What a physical clock measures
is the *whole* expression
ds^2 = -(1-2m/r) dt^2 + 1/(1-2m/r) dr^2 + r^2 dOmega^2
If the clock travels from the point with coordinates
(t1,r1,theta1,phi1) to the point with coordinates
(t2,r2,theta2,phi2) then the elapsed time on the
clock T will approximately satisfy
T^2 = (1 - 2m/r_average) (t2 - t1)^2
- 1/(1 - 2m/r_average) (r2 - r1)^2
- r_average^2 (theta2 - theta1)^2
- r_average^2 sin^2(theta_average) (phi2 - phi1)^2
where r_average = (r2 + r1)/2 and theta_average = (theta2 + theta1)/2.
There is no direct way to measure the Schwarzchild coordinate
t using clocks.
Tom Roberts
> Tom Roberts ha scritto:
>> there is both a black hole and a white hole. That manifold _IS_ what we
>> are discussing.
>
> What we are discussing is "the uniqueness of the extension over the
> horizon" and the relevant selection criteria .
Yes. Extension of the exterior region of the Schw. manifold.
The complete extension is known, and is called the "Kruskal extension",
and is covered by Kruskal-Szerkes coordinates. You should get a GR
textbook and learn about them. It astounds me that you keep trying to
argue from ignorance of what is already known about this manifold.
> What I am saying is that
> there are at least two ways to extend the solution over the horizon,
> correspoinding to two different physical situations, the black hole and
> the white hole.
Knowledge of the complete (inextensible) manifold shows that your "two
ways" are merely different regions of a single manifold. I have said
this several times, and don't know how to say it differently. <shrug>
The two sets of E-F coords. merely cover different regions of the
(single) manifold; both include the exterior region, and neither
includes the "other" exterior region of the manifold.
>> The geodesically complete extension of the Schwarzschild charts is
>> unique, given by the Kruskal chart.
>
> The question is whether the requirement of geodesic completeness (which
> in this setting is obtained by shifting trouble to infinity) is
> appriopriate, i.e. physically relevant.
Of course it is relevant! If the manifold is not geodesically complete
(for timelike paths), then objects would simply "leave the manifold",
which is tantamount to "leaving the universe". As the manifold is
intended to be a model of the entire universe, this does not make sense.
> I surmise that there are geodesically incomplete extensions,
> corresponding to hybrid solutions which may be physically relevant.
You think objects can arbitrarily leave the universe? With the
arbitrariness being your personal whim (of which incomplete extension
you select)?
> I am not keen on throwing away interesting extensions/solutions just
> because they do not not comply with some arbitrary uniqueness
> criterion.
The requirement is not necessarily uniqueness (though that is highly
desirable), but for the Schw. manifold the RESULT OF EXTENSIVE ANALYSIS
is that the extension is unique.
Tom Roberts
Edward Green
> "Edward Green" <spamsp...@netzero.com> wrote in message
> news:1156542691.4...@75g2000cwc.googlegroups.com...
> > You are merely being insulting now.
>
> If you thought that then I apologize. You have done nothing to deserve
> insults.
Thank you.
> >
> > Obviously a manifold is a mathematical abstraction. Obviously any word
> > normally applied to material objects tentatively when applied to a
> > mathematical abstraction should be understood to refer to a putative
> > property of the abstraction analogous to the named material property.
>
> That is not so obvious to me.
Possibly it should be. If I ask you if a manifold can be twisted, does
it mean I can't distinguish mathematical from physical objects, or does
it mean I want to know whether any property of the manifold is
analogous to twisting a material body? As a courtesy, please assume
enough background on my part to suppose the latter.
> Have you considered your musings are better suited to a philosophy forum?
Sigh. I guess it just comes naturally then.
Igor
Koobee Wublee wrote:
> Tom Roberts wrote:
> > Koobee Wublee wrote:
>
> > > ** ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
> > > ** ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2
> > >
> > > ** K = 2 G M / c^2
> > > same despite one manifests a black hole and other one not.
> >
> > This is wrong, and both exhibit a black hole. You forgot to specify the
> > regions of validity of the coordinates: in the first one the horizon is
> > at r=0 (note the metric components are singular there); the black hole
> > is the region -K<r<0. While it is labeled "r", that coordinate does
> > indeed have perfectly valid negative values inside the horizon, and r is
> > timelike there. From the last term it is clear that r is not "radius" in
> > this first line element.
>
> Oh, no. Not you too. I should not be surprised from the master of
> word salad himself.
>
> You should read the following equations are mutually EXCLUSIVE of each
> other. Only one of them can exist at the same time in one world. The
> valid range for r for both equations are (r >= 0).
And that's where it all goes wrong. Again you need to learn to
transform a domain. Also you need to understand that just because you
call it r, it may not represent an actual polar radial coordinate.
>At (r = 0), we have
> the very center of the gravitating object.
>
> ** ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
> ** ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2
>
> > [...]
> >
> > These two line elements do indeed correspond to the same metric,
> > projected onto different coordinates. <shrug>
>
> Your claim is forever caste in stone (the equivalence in cyberspsace).
So is your ignorance and unwillingness to learn.
Igor
Koobee Wublee wrote:
> Igor wrote:
> > Koobee Wublee wrote:
>
> > > ** ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
> > > ** ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2
> > >
> > > K is an integration constant chosen to fit Newtonian result. It is
> > >
> > > ** K = 2 G M / c^2
> > >
> > > For the record, Mr. Bielawski is claiming the above two metrics are the
> > > same despite one manifests a black hole and other one not.
> >
>
> The first metric does not manifest a black hole. Show me how you get a
> black hole from
>
> ** (1 + 2 G M / c^2 / r = 0) where (r >= 0)
Actually r >= -K, and it is not a valid polar radial coordinate. You
just keep failing to transform the domain properly.
> > You just
> > refuse to see it. Math has rules and disobeying them is no excuse.
> > Come back when you've truly learned some of them, especially
> > transforming a domain.
>
> Hey, this is very simple math. If you cannot do this, you need to
> consider going back to junior high school to brush up on your basic
> algebra.
It is very simple math and apparently, you've failed to understand it.
What part of transforming a domain are you still failing to grasp?
> > > > > However, since Schwarzschild Metric is much simpler than
> > > > > Schwarzschild's original solution, Schwarzschild Metric is embraced by
> > > > > the physics communities today.
> > > >
> > > > It's embraced because it's the same.
> > >
> > > You embraced it because of your denial of faulty GR.
> >
> > You've never shown anything to be faulty, except for your understanding
> > of the math.
>
> I have shown them through out many posts in the past year or so. Every
> post is backed up by rigorous mathematical analysis.
No they have not been. Some of us have repeatedly pointed out where
you went wrong, but you keep making the same stupid mistakes. We have
no responsibility for you unwillingness to learn.
> > > You rejected it because of your denial of faulty GR. You have 90 years
> > > of fun playing with Voodoo Mathematics that gives rise to GR. It is
> > > time to tear it down.
> >
> > You have a hell of a lot of nerve calling something voodoo when you
> > can't even play by the proper rules to begin with.
>
> Oh, I play within the rules of mathematics. Since Mr. Bielawski fails
> at simple numeric operation, he needs to go back to kindergarten. If
> you still have not learnt proper arithmetic, maybe you will study them
> with my 2.5 year old twins in the future. As far as you go, you need
> to go back to junior high to study the basic algebra.
Argument from ignorance will never work.
> > Who said all the solutions are the same? Nobody. You need to
> > understand the distinction between the concepts of same and equivalent.
> > And brushing up on transforming a domain wouldn't hurt either.
>
> Each metric is merely a unique and independent solution to a set of
> differential equations called Einstein Field Equations. These are
> merely differential equations. Since you, Mr. Bielawski, and Dr.
> Roberts are all claiming all solutions are the same. To be fair, you
> are saying all solutions to a set of differential equations are the
> same.
Nobody ever said they were the same. Look at them. They have
different forms, because you are using a different variable. But one
is reachable from the other by way of a coordinate transformation, so
the two are said to be EQUIVALENT. But when you perform a
transformation, you also need to transform the domain. You keep
failing to do that and then you call us uneducated. And the fact that
you refuse to recognise your own little mistakes makes it worse. A
true scholar will recognise his mistakes and learn from them, but you
don't want to do that.
> > > As multiple solutions to the vacuum field equations are discovered,
> > > there are actually an infinite number of them. With infinite number of
> > > solutions, it is shaking the very foundation of GR and SR. The house
> > > of cards will soon inevitably collapse. However, refusing to give up
> > > GR and to comfort themselves in false sense of security, they choose to
> > > embrace Voodoo Mathematics. In doing so, they blindly claim all
> > > solutions are indeed the same regardless manifesting black holes,
> > > constant expanding universe, accelerated expanding universe. VOODOO
> > > MATHEMATICS REPRESENTS THE ACHIEVEMENT IN PHYSICS DURING THE LAST 100
> > > YEARS. It is very sad that these clowns are regarded as experts in
> > > their field.
> >
> > Try to keep up. There's only one Schwarzschild solution. If you can
> > find any others by merely transforming the coordinates, it won't ever
> > count as an independent solution. How can it?
>
> All solutions to a set of differential equations are related but
> independent of each other. Why is it a surprise to be able to obtain
> them through a theorem I showed you guys?
What about Birkoff's theorem?
> > But you keep claiming that it is. And that's just plain wrong.
>
> The rules of mathematics show so.
Which you keep showing that you're completely ignorant of. Learn how
to transform domains.
> > You can go ahead and believe
> > what you want to, even it's wrong. I dare you to find one more
> > solution that satisfies the conditions of Schwarzschild and that is
> > truly independent of his original solution.
>
> What do you mean by the conditions of Schwarzschild? Do you
> Schwarzschild Metric as discovered by Hilbert or Schwarzschild's
> original metric?
They're equivalent also.
> > Birkoff proved that it can't be done.
>
> Birkoff assumed that Schwarzschild Metric is the only solution to
> derive that theorem. His assumption is blatantly wrong. His theorem
> is thus total rubbish.
He assumed no such thing. But he proved that there are no spherically
symmetric solutions that are not equivalent to Schwarzschild.
Igor
Still ignoring the transformation of the domain? If so, you may be
hopeless.
Igor
Now I know you're hopeless.
I.Vecchi
Daryl McCullough wrote:
> I.Vecchi says...
>
> >The question is whether the requirement of geodesic completeness (which
> >in this setting is obtained by shifting trouble to infinity) is
> >appriopriate, i.e. physically relevant.
>
> The answer is definitely "yes". Consider an observer in freefall near the
> event horizon. Using his local coordinates, there is *nothing* to prevent
> him from reaching and passing the event horizon in a finite amount of
> proper time, because for him, spacetime near the event horizon is
> approximately *flat*.
Yes, but the problem is to define what is the event horizont for him.
What does it mean PHYSICALLY that he has crossed the event horizon?
Let me go back for a moment to the Agata and Bruno example ([1]). For
Agata, Bruno never crosses the horizon . She cannot associate a time
T_o to Bruno stepping over the horizon. For an external observer Bruno
NEVER crosses the horizon.
Let's consider Bruno perspective then. You are right that nothing
prevents him to reach the horizon, the problem is the the horizon's
position in his perspective becomes arbitrary. If his proper time at
the horizon had a physically determined value he should be able to
look at his clock just before talking the plunge (e.g. switching off
the rocket that keeps him over the horizon) calculate and say: "In ten
minutes I will have crossed the horizont", and that would mean that in
his proper time 10 minutes would pass and then he'd look at his clock
and know he's over . The problem is that in order to calculate that
proper time he needs to know on which KS chart he is. Now, as we have
seen ([2]), the correspondence between the KS chart and the
Schwarzschild chart is undetermined, so he will not be able to do
that. The question is then, is there any chart independent way (i.e.
independent from the space-time measurement model that defines a chart)
to calculate Bruno's proper time across the horizon? And my tentative
answer is, no there is not. Actually, it seems to me that different
charts correspond to different proper times that may be attributed to
Bruno and that there is no way to associate a unique chart to Bruno
based on the Schwarzschild chart alone (i.e. based on information that
is available this side of the horizont).
>Anything other than demanding geodesic completeness
> would violate the equivalence principle, I think.
Would it? I would like to inspect a rigorous proof that it would, and
examine the role that proper time plays in it. I am not saying that no
such proof exists, though indeed i doubt it, but it seems to me that
the arguments being thrown around are riddled with implicit assumption
that may or may not be warranted. It may be worthwhile to make them
explicit and scrutinise them.
Besides, note that neither KS is geodesically complete. It's just that
in it the only obstruction to geodesic completeness is the curvature
singularity at the origin. That does not violate the equivalence
principle , as far as I know.
> >I surmise that there are geodesically incomplete extensions,
> >corresponding to hybrid solutions which may be physically relevant.
> >
> >I am not keen on throwing away interesting extensions/solutions just
> >because they do not not comply with some arbitrary uniqueness
> >criterion.
>
> I don't think that there is anything arbitrary about the criteria.
Good for you.
Cheers,
IV
PS I am reading a paper by Christian Fronsdal
(http://arxiv.org/abs/gr-qc/0508048) built around "an observer who is
aware of a limited portion of space", where "the horizon recedes as it
is approached and has no physical reality". It's good deconstructivist
stuff, highlighting some of the issues we are discussing.
[1]
http://groups.google.com/group/sci.physics.relativity/msg/9ca04b30b02cc1ef
[2]
http://groups.google.com/group/sci.physics.relativity/msg/8a5ffdd9c5944461
JanPB
> Daryl McCullough wrote:
> > I.Vecchi says...
> >
> > >The question is whether the requirement of geodesic completeness (which
> > >in this setting is obtained by shifting trouble to infinity) is
> > >appriopriate, i.e. physically relevant.
> >
> > The answer is definitely "yes". Consider an observer in freefall near the
> > event horizon. Using his local coordinates, there is *nothing* to prevent
> > him from reaching and passing the event horizon in a finite amount of
> > proper time, because for him, spacetime near the event horizon is
> > approximately *flat*.
>
> Yes, but the problem is to define what is the event horizont for him.
> What does it mean PHYSICALLY that he has crossed the event horizon?
>
> Let me go back for a moment to the Agata and Bruno example ([1]). For
> Agata, Bruno never crosses the horizon . She cannot associate a time
> T_o to Bruno stepping over the horizon. For an external observer Bruno
> NEVER crosses the horizon.
Except Agata makes this conclusion based either on her visual
(literally) observation - which is subject to light delay, or based on
her Schwarzschild simultaneity notion - which we already know is very
odd.
> Let's consider Bruno perspective then. You are right that nothing
> prevents him to reach the horizon, the problem is the the horizon's
> position in his perspective becomes arbitrary. If his proper time at
> the horizon had a physically determined value he should be able to
> look at his clock just before talking the plunge (e.g. switching off
> the rocket that keeps him over the horizon) calculate and say: "In ten
> minutes I will have crossed the horizont", and that would mean that in
> his proper time 10 minutes would pass and then he'd look at his clock
> and know he's over . The problem is that in order to calculate that
> proper time he needs to know on which KS chart he is.
What do you mean by "on which KS chart he is"? (There is only one.) Do
you mean Daryl's A and B time shifts? Bruno could calculate the time to
reach the horizon using only the exterior Schwarzschild chart. If just
before beginning his free fall he sits above the horizon at r=r0>2m,
then:
time to the horizon = lim_{r->2m+} integral_r0^r ds
(integrate along the free-fall geodesic).
Adding the constant A doesn't change this integral.
> Now, as we have
> seen ([2]), the correspondence between the KS chart and the
> Schwarzschild chart is undetermined, so he will not be able to do
> that.
It is determined off the horizon and then one needs to prove that this
diffeomorphism extends uniquely over the horizon.
> The question is then, is there any chart independent way (i.e.
> independent from the space-time measurement model that defines a chart)
> to calculate Bruno's proper time across the horizon?
The integral above is chart-independent.
--
Jan Bielawski
Koobee Wublee
Daryl McCullough wrote:
> Koobee Wublee says...
> > ** ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
> > ** ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2
>
> > Of course, r >=0. This is spherically symmetric polar coordinate.
>
> What about the first metric implies that the range of r is between
> 0 and infinity? Are you thinking that the use of the name "r" implies
> such a range?
Yes.
> You do realize, don't you, that the first metric, with range
> r > -K is exactly the same as the second metric, with range r > 0?
Yes.
> Or, the other way around, the first metric with range r > 0 is
> exactly the same as the second metric with range r > K>?
Yes.
Do you realize these two metrics are mutually independent of each
other? Do you realize these two metrics are independent of each other?
Do you realize both of these metrics have their origins at (r = 0)?
Do you realize both of these two metrics are solutions to Einstein
Field Equations in spherically symmetric vacuum?
Koobee Wublee
> Koobee Wublee says...
>> ** ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
>> ** ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2
>
>> The reason for it this is that the first form of the metric is obtained
>> from the second (for which r>0) by means of the coordinate change:
>
> >How do you know which of the two is the first form? And why?
>
> If you compute the area of a sphere of radius r and constant t using
> the top metric, you will find that it is
>
> 4 pi (r+K)^2
>
> which implies that the effective radius is r+K, not r.
Wrong. The surface area of a sphere is always (4 pi r^2) where r is
the radius of the sphere.
How do you know which of the two is the first form? And why?
> >> ...where you are recycling the letter "r" in the first form where I
> >> have used "u". Look at the formula r=u+K. It's patently obvious that
> >> when r changes between 0 and infinity then u changes between -K and
> >> infinity. Tell me, is this not true?
> >
> >No.
>
> Are you saying that it's not true that if we start with
>
> ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2
>
> and change variables to u = r-K, then we obtain
>
> ds^2 = c^2 dt^2 / (1 + K / u) - (1 + K / u) du^2 - (u + K)^2 dO^2
Yes, this is true. You can also start with the following trial
equation (Safartti's term).
ds^2 = c^2 (1 - K / u) dt^2 - du^2 / (1 - K / u) - u^2 dO^2
Setting (u = r + K), you get
ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
You can also start with the following trial equation.
ds^2 = c^2 dt^2 / (1 + K / u) - (1 + K / u) du^2 - (u + K)^2 dO^2
Setting (u = r - K), you get
ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2
You can also start with the following trial equation.
ds^2 = c^2 dt^2 / (1 + K / u) - (1 + K / u) du^2 - (u + K)^2 dO^2
Setting (u = K^2 / r), you get
ds^2 = c^2 dt^2 / (1 + r / K) - (1 + r / K) (K^4 / r^4) dr^2 - (K^4 /
r^2) (1 + r / K)^2 dO^2
If the following is a valid metric,
ds^2 = c^2 A(r) dt^2 - B(r) dr^2 - C(r)^2 dO^2
Then, the following is also a valid metric.
ds^2 = c^2 A(u) dt^2 - B(u) (du/dr)^2 dr^2 - C(r)^2 dO^2
Koobee Wublee
JanPB wrote:
> Koobee Wublee wrote:
> > JanPB wrote:
> > > Koobee Wublee wrote:
> > >> ** ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
> > >> ** ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2
> > >>
> > >> K is an integration constant chosen to fit Newtonian result. It is
> > >>
> > >> ** K = 2 G M / c^2
> > >>
> > >> How do you know which of the two is the first form? And why?
> > >
> > > If one starts - as is usual - with a spherically symmetric
> > > (t,r,theta,phi)-space with r>0, etc., then any spherically symmetric
> > > metric can be locally written in the two forms we had before (not the
> > > two forms you wrote above - the two forms with the sign switch). One
> > > solves the Einstein equations which describe a metric (uniquely,
> > > because the equations happen to be ODEs in this case) described by
> > > components which happen to blow up at r=2m in that basis. This is your
> > > second equation.
> >
> > Of course, r >=0. This is spherically symmetric polar coordinate.
>
> Yes, although there is nothing magical about 0. It's just a convention
> everyone uses by analogy with the planar polar coordinates. One can
> assign the number 42 to the centre of symmetry.
With the following spacetime,
ds^2 = c^2 dt^2 / (1 + r / K) - (1 + r / K) (K^4 / r^4) dr^2 - (K^4 /
r^2) (1 + r / K)^2 dO^2
How do you specify a distance 2 AU from the center of the sun?
> > > If one started instead with a spherically symmetric
> > > (t,r,theta,phi)-space with labelling set so that r>-K (K - some number)
> > > then you'd end up with the solution that looks like your first
> > > equation. Same thing, different chart.
> >
> > Is there any reason for you to avoid my simple question?
>
> I though I'd just answered it. I assume it was "How do you know which
> of the two is the first form?" Did you mean something else? I assumed
> you meant "how does one know to which form r>0 applies to?".
You did not answer my question. I am expecting a 'first' or 'second'
answer following by an expalantion on why the answer is 'first' or
'second'.
JanPB
> Daryl McCullough wrote:
> > Koobee Wublee says...
>
> >> ** ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
> >> ** ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2
> >
> >> The reason for it this is that the first form of the metric is obtained
> >> from the second (for which r>0) by means of the coordinate change:
> >
> > >How do you know which of the two is the first form? And why?
> >
> > If you compute the area of a sphere of radius r and constant t using
> > the top metric, you will find that it is
> >
> > 4 pi (r+K)^2
> >
> > which implies that the effective radius is r+K, not r.
>
> Wrong. The surface area of a sphere is always (4 pi r^2) where r is
> the radius of the sphere.
No. (This is too funny.) The surface area is determined by the metric.
Let's compute then this area precisely using the top metric. Fix r = C
= const. and t = D = const. - this describes the sphere corresponding
to the fixed value of your "r" coordinate equal to C.
Using your top form and inserting r=C and t=D, we get dr = dt = 0, and:
dsigma^2 = -(C+K)^2 dtheta^2 - (C+K)^2 sin^2(theta) dphi^2
...so the volume 2-form on the sphere is:
w = sqrt(g) dtheta /\ dphi
...where:
[ -(C+K)^2 0 ]
g = det [ ] = (C+K)^4 sin^2(theta)
[ 0 -(C+K)^2 sin^2(theta) ]
...hence:
w = (C+K)^2 sin(theta) dtheta /\ dphi
...and we integrate over the sphere:
Area = integral(volume form) =
integral_{-pi}^{pi} integral_0^{pi} (C+K)^2 sin(theta)
dtheta dphi =
= (C+K)^2 integral_{-pi}^{pi} [-cos(theta)]_0^{pi} dphi =
= (C+K)^2 * 2 * integral_{-pi}^{pi} dphi =
= (C+K)^2 * 2 * 2pi = 4 pi (C+K)^2.
QED.
You can change and substitute every which way and you'll never change
the tensor. That's part of what it means to be a tensor. Plain vectors
are the same way.
--
Jan Bielawski
Daryl McCullough
>
>Daryl McCullough wrote:
>> Koobee Wublee says...
>
>>> ** ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
>>> ** ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2
>> If you compute the area of a sphere of radius r and constant t using
>> the top metric, you will find that it is
>>
>> 4 pi (r+K)^2
>>
>> which implies that the effective radius is r+K, not r.
>
>Wrong. The surface area of a sphere is always (4 pi r^2) where r is
>the radius of the sphere.
No, that's only true for *Euclidean* geometry.
The formula for area is given by (in the special case of
a diagonal metric)
Integral from theta=0 to theta = pi
Integral from phi=0 to phi=2pi
square-root(|g_{theta,theta} g_{phi,phi}|) dtheta dphi
In the case of the metric
ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
the important term is
- (r + K)^2 dO^2
= - (r + K)^2 dtheta^2 - (r + K)^2 sin^2(theta) dphi^2
The metric component g_theta, theta is just the term multiplying
dtheta^2, and the component g_phi,phi is the term multiplying
dphi^2. So we see
|g_theta,theta| = (r+K)^2
|g_phi,phi| = (r+K)^2 sin^2(theta)
So the area is given by
Integral from theta=0 to theta = pi
Integral from phi=0 to phi=2pi
(r+K)^2 sin(theta) dtheta dphi
Integrating over phi just gives a factor of 2pi, so we have
Area = (r+K)^2 2pi Integral from theta=0 to pi of sin(theta) dtheta
Since the integral of sin(theta) is -cos(theta), we get
Area = (r+K)^2 2pi [-cos(pi) - -cos(0)]
= (r+K)^2 2pi [-(-1) - (-1)]
= 4pi (r+K)^2
>How do you know which of the two is the first form? And why?
Didn't you just ask that question? Didn't I just answer it?
If you compute the area of a sphere of radius r and constant t using
the top metric, you will find that it is 4 pi (r+K)^2
which implies that the effective radius is r+K, not r.
--
Daryl McCullough
Ithaca, NY
Daryl McCullough
>> > ** ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
>> > ** ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2
>Do you realize these two metrics are mutually independent of each
>other?
No, they are not. If you can transform from one metric to another
via a coordinate change, then that means that they are physically
the *same* metric.
Daryl McCullough
>Daryl McCullough wrote:
>> I.Vecchi says...
>>
>> >The question is whether the requirement of geodesic completeness (which
>> >in this setting is obtained by shifting trouble to infinity) is
>> >appriopriate, i.e. physically relevant.
>>
>> The answer is definitely "yes". Consider an observer in freefall near the
>> event horizon. Using his local coordinates, there is *nothing* to prevent
>> him from reaching and passing the event horizon in a finite amount of
>> proper time, because for him, spacetime near the event horizon is
>> approximately *flat*.
>
>Yes, but the problem is to define what is the event horizont for him.
>What does it mean PHYSICALLY that he has crossed the event horizon?
Using Schwarzchild coordinates, you can compute the proper time for
a freefalling observer as a function of his radius r. The exact
computation depends on the initial conditions, but for one particular
geodesic, the relationship is simple:
r(tau) = 3/2 (2m)^{1/3} (tau_0 - tau))^{2/3}
where tau_0 is a constant depending on the initial conditions.
For this geodesic, the observer passes the event horizon
at a proper time tau_1 given by
tau_1 = tau_0 - (2/3)^{3/2} 2m
and reaches r=0 at proper time
tau = tau_0
So what it means physically for the observer to pass the
event horizon is just that his proper time is greater than
tau_1. As I said, from the point of view of the freefalling
observer, there is no physical reason for his proper time
to stop before reaching tau_1 (when he crosses the event
horizon) and tau_0 (when he reaches r=0).
>Let me go back for a moment to the Agata and Bruno example ([1]). For
>Agata, Bruno never crosses the horizon . She cannot associate a time
>T_o to Bruno stepping over the horizon. For an external observer Bruno
>NEVER crosses the horizon.
Yes, that shows that Agata's coordinate system is incomplete. There
are physically meaningful events that are given no coordinate whatsoever.
The same thing happens in flat spacetime with accelerated observers.
If you are on board a rocket ship with constant proper acceleration g,
then times and distances as measured by clocks and rulers on the ship
will be related to times and distances as measured by inertial clocks
that are at rest relative to the ship's initial reference frame as
follows:
x = X cosh(gT)
t = X sinh(gT)
where (x,t) are the coordinates as measured by the inertial observers,
and (X,T) are the coordinates as measured by the accelerated observers.
Note that the point (x=1 light year, t=1 year) is given no coordinates
*at* *all* in the accelerated coordinate system: there is no values of
X and T such that X cosh(gT) = 1 and X sinh(gT) = 1.
From the point of view of the accelerated
observer, the point (x=1, t=1) can be approached as T--> infinity,
but it can never be reached. That just means that the accelerated
coordinates are incomplete.
The same thing is true of the Schwarzchild coordinates. They are
incomplete in exactly the same way. There are valid points on
the manifold that are given no coordinates whatsoever in the
Schwarzchild coordinate system.
>Let's consider Bruno perspective then. You are right that nothing
>prevents him to reach the horizon, the problem is the the horizon's
>position in his perspective becomes arbitrary.
Yes, from his point of view, there is nothing special about the
horizon. However, Agata can calculate the proper time tau_1 at
which Bruno will hit the event horizon, and she can make the
prediction that Bruno's clock will never advance past tau_1.
Bruno can watch his clock and prove that Agata is wrong (although
his message gloating about it will never reach Agata).
>If his proper time at
>the horizon had a physically determined value
It does. Agata can calculate it.
>he should be able to look at his clock just before talking the
>plunge (e.g. switching off the rocket that keeps him over the
>horizon) calculate and say: "In ten
>minutes I will have crossed the horizon",
Yes, and at 1 second past ten minutes, assuming he hasn't
hit the singularity, he'll know that he has crossed the
event horizon.
>and that would mean that in his proper time 10 minutes would
>pass and then he'd look at his clock and know he's over. The
>problem is that in order to calculate that proper time he
>needs to know on which KS chart he is.
The calculation can be done using KS, or using Schwarzchild
coordinates. It's not hard using Schwarzchild coordinates.
From the metric and the geodesic equation, you can prove
that there are two constants of motion for radial geodesics:
(1-2m/r) V^t = P
(1-2m/r) (V^t)^2 - 1/(1-2m/r) (V^r)^2 = E
where V^t = dt/dtau and V^r = dr/dtau. Using
the first equation in the second, we find
P/(1-2m/r) - (V^r)^2/(1-2m/r) = E
So
V^r = - square-root(P - E(1-2m/r))
dr/dtau = - square-root(P - E(1-2m/r))
(where I chose the minus sign because the observer is falling, so
dr/dtau is negative). This gives tau as a function of r:
dtau/dr = - 1/square-root(P - E(1-2m/r))
tau = tau_0 - integral of 1/square-root(P - E(1-2m/r)) dr
The integral is a perfectly ordinary integral, with no singularities
except at r=0. So the value of tau at which r=2m is computable.
>Now, as we have seen ([2]), the correspondence between the KS chart
>and the Schwarzschild chart is undetermined, so he will not be able
>to do that.
The relationship between tau and r is perfectly well determined.
The relationship between tau and t is not completely determined
but that's not relevant to the question of what the value of tau
is when the infalling observer hits the event horizon.
>The question is then, is there any chart independent way (i.e.
>independent from the space-time measurement model that defines a chart)
>to calculate Bruno's proper time across the horizon?
Yes. I just gave it to you. If you know Bruno's initial radius
and the initial time on Bruno's clock, then you can easily
compute the time on Bruno's clock when he reaches the event
horizon.
>>Anything other than demanding geodesic completeness
>> would violate the equivalence principle, I think.
>
>Would it? I would like to inspect a rigorous proof that it would,
I don't know what a rigorous proof would look like, but suppose
that we compute (as shown above) that Bruno will hit the event
horizon at proper time tau = 10 minutes. Bruno, working in his
local coordinate system can see no physical reason that his clock
would stop running before reaching tau=10 minutes.
>Besides, note that neither KS is geodesically complete. It's just that
>in it the only obstruction to geodesic completeness is the curvature
>singularity at the origin. That does not violate the equivalence
>principle, as far as I know.
The singularity at r=0 is present in *all* coordinate systems,
including the coordinate system of a freefalling observer. In
contrast, the "singularity" at the event horizon vanishes
completely when you change coordinate systems. So there is
a big difference there.
Tom S.
> Let e1 and e2 be two successive ticks of a clock (assuming that
> you have an old-fashioned clock that ticks, if they make those
> anymore). There are at least three different ways to compute the
> "time between ticks": (1) Use the metric, and compute
> Integral from e1 to e2 of square-root |g_ij dx^i dx^j|
> along a geodesic path connecting e1 and e2. (2) Look at
> the elapsed time shown on the clock. (3) Use a particular
> coordinate system, and use t2 - t1 as the time between
> ticks (where t2 is the time of e2 in that coordinate
> system, and t1 is the time of e1).
>
> The usual assumption is that methods (1) and (2) will
> give the same answer.
Maybe I'm misinterpreting what you're saying. But I don't think that (1)
and (2) are usually assumed to yield the same result.
For example, let e1 and e2 be ticks of a clock sitting at rest on the
surface of a neutron star. Then the elapsed time between e1 and e2 as shown
on this clock is your method (2).
Let another clock be launched radially outward from the surface of the star
coincident with the fixed clock at event e1 with an initial speed chosen
such that it freefalls out and back, returning to the surface at event e2.
This clock has traveled along a geodesic between e1 and e2, so it's elapsed
time corresponds to method (1).
But (1) will not agree with (2).
In fact, you can arrange a scenario where two different geodesic paths
connect the same two events e1 and e2 such that the ''lengths'' of the two
geodesics between e1 and e2 are different. So, method (1) doesn't
necessarily lead to a unique result.
Tom
Bill Hobba
> Bill Hobba wrote:
>
>> "Edward Green" <spamsp...@netzero.com> wrote in message
>> news:1156542691.4...@75g2000cwc.googlegroups.com...
>
>> > You are merely being insulting now.
>>
>> If you thought that then I apologize. You have done nothing to deserve
>> insults.
>
> Thank you.
>
>> >
>> > Obviously a manifold is a mathematical abstraction. Obviously any word
>> > normally applied to material objects tentatively when applied to a
>> > mathematical abstraction should be understood to refer to a putative
>> > property of the abstraction analogous to the named material property.
>>
>> That is not so obvious to me.
>
> Possibly it should be. If I ask you if a manifold can be twisted, does
> it mean I can't distinguish mathematical from physical objects, or does
> it mean I want to know whether any property of the manifold is
> analogous to twisting a material body?
No because twist is a well defined topological property with a well accepted
meaning in physics and mathematics. Stress however does not have such a
generaly accepted interpretation. Physics is not a game of definitions.
Unless you are uber experienced then in any endeavor it is best to stick to
the standard usage of words before venturing your own.
Thanks
Bill
Koobee Wublee
No, as you have written the spacetime is its incremental discrete form,
you are actually referring to its change of line element of spacetime
which translates to velocity. This (r + K) term only affect the
velocity. Nice try but no cigar.
Is there any reason to avoid my comment other than this puny
introduction from my previous post?
In case if you have a short memory, here it is
...
Yes, this is true. You can also start with the following trial
equation (Safartti's term).
ds^2 = c^2 (1 - K / u) dt^2 - du^2 / (1 - K / u) - u^2 dO^2
Setting (u = r + K), you get
ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
You can also start with the following trial equation.
ds^2 = c^2 dt^2 / (1 + K / u) - (1 + K / u) du^2 - (u + K)^2 dO^2
Setting (u = r - K), you get
ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2
You can also start with the following trial equation.
ds^2 = c^2 dt^2 / (1 + K / u) - (1 + K / u) du^2 - (u + K)^2 dO^2
Koobee Wublee
JanPB wrote:
> Koobee Wublee wrote:
> > Daryl McCullough wrote:
> > > Koobee Wublee says...
> >
> > >> ** ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
> > >> ** ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2
> > >
> > >> The reason for it this is that the first form of the metric is obtained
> > >> from the second (for which r>0) by means of the coordinate change:
> > >
> > > >How do you know which of the two is the first form? And why?
> > >
> > > If you compute the area of a sphere of radius r and constant t using
> > > the top metric, you will find that it is
> > >
> > > 4 pi (r+K)^2
> > >
> > > which implies that the effective radius is r+K, not r.
> >
> > Wrong. The surface area of a sphere is always (4 pi r^2) where r is
> > the radius of the sphere.
>
> No. (This is too funny.) The surface area is determined by the metric.
>
The metric specifies how the 2nd derivative of spatial components are
going to behave in which there is no definitive saying on the
integrated result. The surface area of a sphere in this case still
remains to be (4 pi R^2). However, the trajectory is a different
matter.
This is not funny! This is serious stuff. You clowns trying to make
this issue into funny matter to confuse everyone else is absolutely
sinful.
JanPB
> JanPB wrote:
> > Koobee Wublee wrote:
> > > Daryl McCullough wrote:
> > > > Koobee Wublee says...
> > >
> > > >> ** ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
> > > >> ** ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2
> > > >
> > > >> The reason for it this is that the first form of the metric is obtained
> > > >> from the second (for which r>0) by means of the coordinate change:
> > > >
> > > > >How do you know which of the two is the first form? And why?
> > > >
> > > > If you compute the area of a sphere of radius r and constant t using
> > > > the top metric, you will find that it is
> > > >
> > > > 4 pi (r+K)^2
> > > >
> > > > which implies that the effective radius is r+K, not r.
> > >
> > > Wrong. The surface area of a sphere is always (4 pi r^2) where r is
> > > the radius of the sphere.
> >
> > No. (This is too funny.) The surface area is determined by the metric.
> >
> > [...]
>
> The metric specifies how the 2nd derivative of spatial components are
> going to behave in which there is no definitive saying on the
> integrated result.
Words words words.
> The surface area of a sphere in this case still
> remains to be (4 pi R^2).
It's 4 pi (R+K)^2 - I have just CALCULATED it from YOUR OWN metric. Do
you know calculus?
--
Jan Bielawski
Koobee Wublee
JanPB wrote:
> Koobee Wublee wrote:
> > JanPB wrote:
> > > Koobee Wublee wrote:
> > > > Daryl McCullough wrote:
> > > > > Koobee Wublee says...
> > > >
> > > > >> ** ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
> > > > >> ** ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2
> > > > >
> > > > >> The reason for it this is that the first form of the metric is obtained
> > > > >> from the second (for which r>0) by means of the coordinate change:
> > > > >
> > > > > >How do you know which of the two is the first form? And why?
> > > > >
> > > > > If you compute the area of a sphere of radius r and constant t using
> > > > > the top metric, you will find that it is
> > > > >
> > > > > 4 pi (r+K)^2
> > > > >
> > > > > which implies that the effective radius is r+K, not r.
> > > >
> > > > Wrong. The surface area of a sphere is always (4 pi r^2) where r is
> > > > the radius of the sphere.
> > >
> > > No. (This is too funny.) The surface area is determined by the metric.
> > >
> > > [...]
> >
> > The metric specifies how the 2nd derivative of spatial components are
> > going to behave in which there is no definitive saying on the
> > integrated result.
>
> Words words words.
And mathematics between the words.
> > The surface area of a sphere in this case still
> > remains to be (4 pi R^2).
>
> It's 4 pi (R+K)^2 - I have just CALCULATED it from YOUR OWN metric. Do
> you know calculus?
I do. Your assumption is very faulty in the very start. The surface
area of a sphere remains (4 pi R^2). Your calculation is based on
something that only applies to velocity terms which is grossly
inapproppriate and faulty in nature. So, please act like an adult for
a change and stop whining about spilled milk.
JanPB
What "mathematics"? After I CALCULATED the sphere area you said "The
metric specifies how the 2nd derivative of spatial components are going
to behave in which there is no definitive saying on the integrated
result" - which is poetry and has nothing to do with the area of the
sphere in question.
> > > The surface area of a sphere in this case still
> > > remains to be (4 pi R^2).
> >
> > It's 4 pi (R+K)^2 - I have just CALCULATED it from YOUR OWN metric. Do
> > you know calculus?
>
> I do. Your assumption is very faulty in the very start. The surface
> area of a sphere remains (4 pi R^2).
No, it doesn't. I have just CALCULATED it from YOUR OWN metric. If you
want to claim otherwise, you have to find an error in my calculation
(you can't).
> Your calculation is based on
> something that only applies to velocity terms which is grossly
> inapproppriate and faulty in nature.
There is no velocity there. There is just a metric you yourself wrote -
all I've done is calculated the area of the sphere r=const. according
to your metric.
> So, please act like an adult for a change and stop whining about spilled milk.
I prefer facts to rethoric.
--
Jan Bielawski
Daryl McCullough
>"Daryl McCullough" <stevend...@yahoo.com> wrote
>> Let e1 and e2 be two successive ticks of a clock (assuming that
>> you have an old-fashioned clock that ticks, if they make those
>> anymore). There are at least three different ways to compute the
>> "time between ticks": (1) Use the metric, and compute
>> Integral from e1 to e2 of square-root |g_ij dx^i dx^j|
>> along a geodesic path connecting e1 and e2. (2) Look at
>> the elapsed time shown on the clock. (3) Use a particular
>> coordinate system, and use t2 - t1 as the time between
>> ticks (where t2 is the time of e2 in that coordinate
>> system, and t1 is the time of e1).
>>
>> The usual assumption is that methods (1) and (2) will
>> give the same answer.
>
>Maybe I'm misinterpreting what you're saying. But I don't think that (1)
>and (2) are usually assumed to yield the same result.
>
>For example, let e1 and e2 be ticks of a clock sitting at rest on the
>surface of a neutron star. Then the elapsed time between e1 and e2 as shown
>on this clock is your method (2).
>
>Let another clock be launched radially outward from the surface of the star
>coincident with the fixed clock at event e1 with an initial speed chosen
>such that it freefalls out and back, returning to the surface at event e2.
>This clock has traveled along a geodesic between e1 and e2, so it's elapsed
>time corresponds to method (1).
>
>But (1) will not agree with (2).
That's true. I think, though, that if e1 and e2 are very close together
in time, then the difference between the result of an inertial clock and
a noninertial clock becomes negligible.
Daryl McCullough
>I do. Your assumption is very faulty in the very start. The surface
>area of a sphere remains (4 pi R^2).
In *Euclidean* geometry, but not in curved space. Distances
have to be computed using the metric. Areas are computed using
distances.
In curved 2-D, if you have two coordinates x and y (assumed to be
orthogonal) and you look at the rectangle with corners
(x,y)
(x+delta_x,y)
(x+delta_x,y+delta_y)
(x,y+delta_y)
what is the area of the rectangle? Well, the length of
the sides are *not* delta_x and delta_y. To compute
lengths in curved space, you have to use the metric.
The length of the sides are square-root(|g_xx|) delta_x
and square-root(|g_yy| delta_y). The area of that rectangle is
square-root(|g_xx| |g_yy|) delta_x delta_y
The area of a sphere is calculated using exactly this
principle. In the case of a sphere, instead of using
x and y, you use theta and phi. So the area of a tiny
patch on the surface of a sphere is given by
square-root(|g_theta,theta| |g_phi,phi|) delta_theta delta_theta
In Euclidean geometry with the usual coordinates,
g_theta,theta = R^2
g_phi,phi = R^2 sin^2(theta)
So the area of the little patch is
square-root(R^2 * R^2 sin^2(theta)) delta_theta delta_theta
= R^2 sin(theta) delta_theta delta_phi
>Your calculation is based on something that only applies to
>velocity terms
You are confused about the meaning of "metric". The metric
tells how to measure proper distances and proper times. It
doesn't have anything to do with velocity.
Once again, look at the surface of a sphere. Suppose
you have two points on the sphere. The first point has
coordinates (theta=pi/4, phi=0). The second point has
coordinates (theta=pi/4, phi=pi). What is the distance
between these two points along the curve theta=pi/4?
To compute distances, you use the metric:
ds^2 = g_uv dx^u dx^v
Since the only coordinate that is varying is phi, this
simplifies to
ds^2 = g_phi,phi dphi^2
So
ds = square-root(g_phi,phi) dphi
g_phi,phi for spherical coordinates is R^2 sin^2(theta). So
ds = R sin(theta) dphi
In our case, theta = pi/4, so sin(theta) = 1/square-root(2).
So
ds= R/square-root(2) dphi
Integrating from phi=0 to phi=pi gives
s = R/square-root(2) pi
The metric is primarily about calculating *distances*
and *times*.
Daryl McCullough
>Daryl McCullough wrote:
>> >How do you know which of the two is the first form? And why?
>>
>> Didn't you just ask that question? Didn't I just answer it?
>> If you compute the area of a sphere of radius r and constant t using
>> the top metric, you will find that it is 4 pi (r+K)^2
>> which implies that the effective radius is r+K, not r.
>
>No, as you have written the spacetime is its incremental discrete form,
>you are actually referring to its change of line element of spacetime
>which translates to velocity. This (r + K) term only affect the
>velocity. Nice try but no cigar.
We're not talking about velocity. Do you understand what a metric is?
It doesn't have anything to do with velocity. The metric is used
for computing *distances*. To compute the area, you have to know
distances. To compute distances, you have to use the metric.
So you have to use the metric to compute areas.
>Is there any reason to avoid my comment other than this puny
>introduction from my previous post?
I answered several times: The area of a sphere using your
metric is given by 4pi(r+K)^2, thus showing that the relevant
radius is r+K, not r.
--
Daryl McCullough
Ithaca, NY
Daryl McCullough
>JanPB wrote:
>> > The surface area of a sphere in this case still
>> > remains to be (4 pi R^2).
>>
>> It's 4 pi (R+K)^2 - I have just CALCULATED it from YOUR OWN metric. Do
>> you know calculus?
>
>I do. Your assumption is very faulty in the very start. The surface
>area of a sphere remains (4 pi R^2).
At this point, the disagreement has nothing to do with physics
or relativity, but is a purely mathematical question.
Given a two-dimensional surface described by coordinates u and v,
what is the area enclosed by the region
0 < u < pi
0 < v < 2 pi
Can you answer that question without knowing anything about
the coordinates u and v? Of course not. If u and v are cartesian
coordinates, then the area will be
2 pi^2
If the surface is a flat 2-D plane, and u is the radial coordinate
and v is the angle (in radians) then the area will be
pi^2
If the surface is the surface of a sphere of radius R in flat
Euclidean 3D space, and u and v are the colatitude and
azimuth, respectively, then the area is
4 pi R^2
If the surface is the surface of a cylinder of radius R,
and u is the distance along the axis and v is the angle
around the circumference, then the area is
2 pi^2
You can't compute the area without knowing more information
about the coordinates u and v. Where does that additional
information come from?
In Riemannian geometry, that additional information is
found in the *metric*. The distinction between spherical
coordinates and rectangular coordinates is given by the
metric coefficients.
You seem to think that the information is conveyed by the
*names* of the coordinate. If the names are r and theta,
then we are using polar coordinates, and if the names are
theta and phi, then we are using spherical coordinates, and
if the names are x and y, then we are using Cartesian
coordinates. However, that's just a convention. You can
name a coordinate anything you like. You have to look at
the geometry (which is described by the metric).
Ken S. Tucker
JanPB wrote:
> Ken S. Tucker wrote:
> > JanPB wrote:
> > > Ken S. Tucker wrote:
> > > > JanPB wrote:
> > > > > Ken S. Tucker wrote:
> > > > > >
> > > > > > Jan, I think you should study the original paper
> > > > > > Ed Green cited, there is no such thing as an
> > > > > > event horizon or Black-Hole's. BTW, Dr. Loinger
> > > > > > and I discussed this at length. In short, the
> > > > > > original Schwarzschild Solution has been
> > > > > > bastardized and mis-understood for simplicity.
> > > > >
> > > > > No, the "original" solution is the only one. This follows from basic
> > > > > ODE theory and the definition of tensor.
> > > > >
> > > > > > The bastardized version became popularized
> > > > > > and embraced by astronomers who now see
> > > > > > BH's under their beds.
> > > > >
> > > > > There is no other version, bastardized or not. How many solutions do
> > > > > you have to the following ODE:
> > > > >
> > > > > f'(x) = f(x)
> > > > >
> > > > > ...given the initial condition f(0) = 1 ? I can see f(t) = exp(t). Is
> > > > > there any other?
> > > > >
> > > > > How about another ODE, just slightly more complicated:
> > > > >
> > > > > -2 f(x) f'(x) + 1/x * (1 - f(x)^2) = 0
> > > > >
> > > > > ...given the initial condition f(1) = 0 ?
> > > > >
> > > > > --
> > > > > Jan Bielawski
> > > >
> > > > What Newton assumed was,
> > > >
> > > > r= sqrt(x^2 + y^2 +z^2)
> > > >
> > > > but in GR the "real" R = r - m
> > > > where m =1.47 kms in the case of the Sun.
> > > >
> > > > Most treatise begin by defining "r" that way.
> > > > Schwarzschild emphasized that difference and obtained
> > > > a singularity only at the very center of a mass, and that
> > > > would assume an infinite density, which is physically
> > > > impossible. But it's a good exercise.
> > >
> > > Not sure what you're alluding to.
> > >
> > > For the record, the second equation I wrote:
> > >
> > > -2 f(x) f'(x) + 1/x * (1 - f(x)^2) = 0
> > >
> > > ...is what the Einstein field equations reduce to (in the case that
> > > turns out to be the exterior).
> > >
> > > The solution is the familiar:
> > >
> > > f(x) = sqrt(1 + C/x)
> > >
> > > I set up the initial condition at random where normally one chooses the
> > > constant C so that the metric is Newtonian for large x, namely C = -2m.
> >
> > I perfer other approaches but I'll use your's here.
> >
> > f(x) ~ (1-m/x) , x*f(x) = x - m.
> >
> > In more standard notation, that is written,
> >
> > R = r - m
> >
> > as I posted. Where Newton says the light particle
> > would be at "r" Einstein predicted it would be at "R",
> > with the caveat that we're in 3D. When translating
> > to 4D the actual quantity becomes R(4D) = r - 2m.
>
> Not sure what the Einstein equation has to do with R = r - m at all...
That's the basis of the Schwarz. Solution, Weinberg's
"Grav&Cosmo" pg. 181. It is how energy is *stored*
in the spacetime field.
> > > My point was that at this stage this is really the ODE theory, hence
> > > the uniqueness.
> >
> > What is ODE?
>
> Ordinary differential equations.
Thanks
Ken
I.Vecchi
have taken me a substantially larger amount of time, effort and wit to
reach whatever understanding of the issues I may have now.
One more consideration. A clock is, by its very nature, a massive,
extended object. Beyond the horizon no information can be tranferred
from the bottom to the top of the clock. It is not clear to me that any
clock would keep working under such conditions. So, thinking about it,
I am no longer sure that the statement that "nothing special" happens
at the horizon is accurate. Clocks may stop working there. Actually,
this is not a purely speculative statement, since the effect of a very
strong gravitational field could be tested on physical clocks also in
the exterior domain. If physical clocks stop, what happens to time?
Cheers,
IV
carlip...@physics.ucdavis.edu
> carlip...@physics.ucdavis.edu ha scritto:
>> The Kruskal-Szerkes extension is the unique maximal analytic extension
>> of the Schwarzschild exterior geometry.
> Nice. Is there a proof for that? I mean a mathematically rigorous one,
> where the hypotheses are cleary stated, so that one can weigh their
> physical relevance. A reference would be appreciated.
One reference are Hawking and Ellis, _The Large Scale Structure of
Space-Time_, p. 155. No proof is given. If you look at De Felice
and Clarke, _Relativity on curved manifolds_, section 10.4, you will
find a careful discussion of maximal analytic extensions, along with
a statement that if such an extension exists, it is unique. This is
not my specialty, and I don't know off hand where to find the exact
proofs, but I am certainly inclined to believe these authors.
>> Which part do you want to give up?
> In any given case, the one that does not fit observation.
A black hole formed from collapsing matter is not, of course, described
by the full Kruskal-Szekeres extension. That's because the spacetime
is not empty -- the region containing matter can't be described by a
solution of the vacuum field equations. If you solve the equations
for collapsing matter, you'll find a portion of regions I and II of
the Kruskal-Szekeres extension (the "black hole" part) glued to a
non-vacuum solution, whose nature depends on exactly what sort of
collapsing matter you're describing.
Is this all you're worried about?
Steve Carlip
I.Vecchi
I.Vecchi ha scritto:
> One more consideration. A clock is, by its very nature, a massive,
> extended object. Beyond the horizon no information can be tranferred
> from the bottom to the top of the clock. It is not clear to me that any
> clock would keep working under such conditions. So, thinking about it,
> I am no longer sure that the statement that "nothing special" happens
> at the horizon is accurate. Clocks may stop working there. Actually,
> this is not a purely speculative statement, since the effect of a very
> strong gravitational field could be tested on physical clocks also in
> the exterior domain. If physical clocks stop, what happens to time?
Nonsense,
IV
Koobee Wublee
JanPB wrote:
> No, it doesn't. I have just CALCULATED it from YOUR OWN metric. If you
> want to claim otherwise, you have to find an error in my calculation
> (you can't).
The metric indicates a distortion in space. Thus, after computing the
area in curved space, the radius does not necessarily follow the
computation in flat space. Your mistake is to mix curved space with
flat space. In the meantime, you still owe me answers on the
following.
With the following spacetime,
ds^2 = c^2 dt^2 / (1 + r / K) - (1 + r / K) (K^4 / r^4) dr^2 - (K^4 /
r^2) (1 + r / K)^2 dO^2
How do you specify a distance 2 AU from the center of the sun?
Given the following two independent spacetime where K is an integration
constant,
** ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
** ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2
How do you know which of the two is the first form? And why?
Or better yet, let's write the above equations into a more general
form,
** ds^2 = c^2 dt^2 / (1 + K / (r + Q)) - (1 + K / (r + Q)) dr^2 - (r +
Q)^2 dO^2
where (K = 2 G M / c^2), (Q = any constant)
How do you know Q has to be (- K) and not some other number?
I.Vecchi
Thanks for the feedback and references, which I will look up. A
"careful discussio of maximal analytic extensions" sounds like
something I should read.
IV
Aluminium Holocene Holodeck Zoroaster
the area of its greatcircles is then piDD/4.
coordinatewisely, I can only do tripolars!
> You seem to think that the information is conveyed by the
> *names* of the coordinate. If the names are r and theta,
> then we are using polar coordinates, and if the names are
> theta and phi, then we are using spherical coordinates, and
> if the names are x and y, then we are using Cartesian
thus:
[How Big Is Infinity?]
nowhere nohow noway anywise as big as yo' momma....
anywho, the Hindoos are never going to agree with that assumption.
> >>> Assuming that the universe is finite (which at this point is a
thus:
he is Right. I mean,
Why are their no left trigona?
it seems a lot of this is wrapped-up in the projections
to the ordinary plain, which provides the canonical right angle,
made on the diameter of a circle with compasses. the above dyscussion
on affine geometry may be the best approach.
three mathematical dimensions are very special;
four are also special, but they aren't "right" in the same way:
the four altitudes of a regular tetrahedron all meet pairwise
at the same angle, not ninety degrees (arcos 2/3 ?);
the altitudes of a general tetrahedron do not meet, at all,
as you might expect by analogy to the trigon. also,
no more than four lines can meet pairwise at the same angle,
in this static (and 3D) sense.
Hamilton's vector terminology was deconstructed
by Gibbs into "inner & outer products," which seems
to give rise to co- and contra-variance in these "bigger"
(somehow) spaces. (I just found a good text on this,
vis-a-vu electromagnetism, stressing that
4D in this area is really always 3D + 1D, in spite
of any ellision or hype over or from Michelson-Morley-Miller-Minkowski,
or decategorizations (in modern term; Minkowski was added
to the problem-set in Memorium & by Mystaque,
by special order of the Department of Einsteinmania,
the Musical .-))
below, you are comparing a pure (and unit) vector
with a pure scalar (also unit, and negative) e.g.
> > > > 0 + 1 * i + 0 * j + 0 * k and (- 1) + 0 * i + 0 * j + 0 * k.
> > > The angle between these is 90 degrees when plotted in
> > > Cartesian coordinates.
> > > Their projections in 3-space: (0,1,0) and (-1,0,0) are also
> > > at 90 degrees. Do you think otherwise?
> > This evidence of projections doesn't give us the required condition of
> > rectangularity
> Usually, the angle between two functions isn't as important as being
> able to prove (in some circumstances) that the angle is 90 degrees. The
> formula for the coefficients of the Fourier Transform relies on the
> fact that the following functions are pairwise orthogonal:
> 1, sin x, sin(2x), sin(3x), ..., cos x, cos(2x), cos(3x), ...
> Here, f*g = integral(f(x)g(x), x=-Pi..Pi).
thus:
the brachistocrhone problem actually helped Leibniz and Bernoulli
to establish "the" calculus. I'm sure, Newton had to weigh-in
with his cannonballs, but, Who cares?... I mean,
all he did was algebraize Kepler's orbital constraints
-- and didn't cite his contemporary ... I mean, Galileo didn't, either
--
although evidence points to his stealing the 2nd-power law
from what's-his-face (no-one knows, since Sir Duh destroyed all
of his portraits at the Society, which wrote a whole Philosphic Tract,
to obfuscate the basic idea....
Bernoulli's paper is fairly elementary,
even in French.
thus:
to be featured in the next movie,
"Harry Potter's New Crusades and
the 'Public' Charter Schools: Faith-based Initiatives
in the New Millennium CCE: Come the Rapture,
No Child Left Behind!:"
http://larouchepub.com/other/2006/3333uk_scoop_soc.html
--it takes some to jitterbug!
http://members.tripod.com/~american_almanac
http://www.21stcenturysciencetech.com/2006_articles/Amplitude.W05.pdf
http://www.rwgrayprojects.com/synergetics/plates/figs/plate01.html
http://larouchepub.com/other/2006/3322_ethanol_no_science.html
http://www.wlym.com/pdf/iclc/howthenation.pdf
Daryl McCullough
>JanPB wrote:
>
>> No, it doesn't. I have just CALCULATED it from YOUR OWN metric. If you
>> want to claim otherwise, you have to find an error in my calculation
>> (you can't).
>
>The metric indicates a distortion in space. Thus, after computing the
>area in curved space, the radius does not necessarily follow the
>computation in flat space. Your mistake is to mix curved space with
>flat space.
How about telling the truth, Koobee? You had no idea
how to calculate area from the metric prior to Jan's
explaining it to you. Rather than honestly saying
"Thanks for explaining that to me, Jan", you instead
pretended that *Jan* was the one who was confused.
Jan didn't make a mistake---*YOU* did.
Koobee Wublee
The cows are
** Mr. Bielawski, a known crackpot and a master Voodoo mathematician
** Mr. McCullogh, also a known crackpot and a Voodoo mathematician
want-to-be
Given the following metrics,
** ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
** ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2
The want-to-be suggested the following.
"Daryl McCullough" <stevend...@yahoo.com> wrote in message
news:ecs3o...@drn.newsguy.com...
>
> If you compute the area of a sphere of radius r and constant t using
> the top metric, you will find that it is
>
> 4 pi (r+K)^2
>
> which implies that the effective radius is r+K, not r.
The master smelled blood and came in for the kill by adding the
following.
"JanPB" <fil...@gmail.com> wrote in message
news:1156719174.9...@75g2000cwc.googlegroups.com...
>
> The surface area is determined by the metric.
>
> Let's compute then this area precisely using the top metric. Fix r = C
> = const. and t = D = const. - this describes the sphere corresponding
> to the fixed value of your "r" coordinate equal to C.
>
> Using your top form and inserting r=C and t=D, we get dr = dt = 0, and:
>
> dsigma^2 = -(C+K)^2 dtheta^2 - (C+K)^2 sin^2(theta) dphi^2
>
> ...so the volume 2-form on the sphere is:
>
> w = sqrt(g) dtheta /\ dphi
>
> ...where:
>
> [ -(C+K)^2 0 ]
> g = det [ ] = (C+K)^4 sin^2(theta)
> [ 0 -(C+K)^2 sin^2(theta) ]
>
> ...hence:
>
> w = (C+K)^2 sin(theta) dtheta /\ dphi
>
> ...and we integrate over the sphere:
>
> Area = integral(volume form) =
>
> integral_{-pi}^{pi} integral_0^{pi} (C+K)^2 sin(theta)
> dtheta dphi =
>
> = (C+K)^2 integral_{-pi}^{pi} [-cos(theta)]_0^{pi} dphi =
>
> = (C+K)^2 * 2 * integral_{-pi}^{pi} dphi =
>
> = (C+K)^2 * 2 * 2pi = 4 pi (C+K)^2.
>
> QED.
After a little fastening up, the want-to-be boldly suggests the
following.
"Daryl McCullough" <stevend...@yahoo.com> wrote in message
news:ecuq5...@drn.newsguy.com...
> tells how to measure proper distances and proper times. It
> doesn't have anything to do with velocity.
>
> Once again, look at the surface of a sphere. Suppose
> you have two points on the sphere. The first point has
> coordinates (theta=pi/4, phi=0). The second point has
> coordinates (theta=pi/4, phi=pi). What is the distance
> between these two points along the curve theta=pi/4?
>
> To compute distances, you use the metric:
>
> ds^2 = g_uv dx^u dx^v
>
> Since the only coordinate that is varying is phi, this
> simplifies to
>
> ds^2 = g_phi,phi dphi^2
>
> So
>
> ds = square-root(g_phi,phi) dphi
>
> g_phi,phi for spherical coordinates is R^2 sin^2(theta). So
>
> ds = R sin(theta) dphi
>
> In our case, theta = pi/4, so sin(theta) = 1/square-root(2).
> So
>
> ds= R/square-root(2) dphi
>
> Integrating from phi=0 to phi=pi gives
>
> s = R/square-root(2) pi
>
> The metric is primarily about calculating *distances*
> and *times*.
And after more fluttering, the want-to-be felt ever more confident also
came in for the kill.
"Daryl McCullough" <stevend...@yahoo.com> wrote in message
news:ecvpj...@drn.newsguy.com...
>
> How about telling the truth, Koobee? You had no idea
> how to calculate area from the metric prior to Jan's
> explaining it to you. Rather than honestly saying
> "Thanks for explaining that to me, Jan", you instead
> pretended that *Jan* was the one who was confused.
>
> Jan [the master] didn't make a mistake---*YOU* did.
These two gentlemen fail to understand what the metric actually
represents. It is thus necessary to explain what the metric really
represents. Given the ubiquitously familiar Schwarzschild metric as
described in the very second equation, dt, dr, and dO (longitudinal and
latitudinal information) are the observed quantities. In GR, the
observed quantities are what actually are. The role of the metric is
to measure how much distortion in the observed quantities from the
non-distorted ones. Thus, what the want-to-be suggested regarding the
surface area of a sphere in a distorted space where (r = 0) resides
right at the center of the gravitating body is totally wrong. The
observed surface area of such sphere is still always (4 pi R^2) not (4
pi (R + K)^2) where R is the observed radius.
Mr. McCullough needs to return his degree and ask for a complete refund
for his tuition because of this embarrassment. Mr. Bielawski after
receiving a somewhat higher degree in math should stay where he is that
is watching second-rated films for a living. Minds practicing Voodoo
math need to allocate wasteful time to vent their nonsense.
In the meantime, the Voodoo mathematical minds still need to answer and
address the following.
"Koobee Wublee" <koobee...@gmail.com> wrote in message
news:1156798554.8...@m73g2000cwd.googlegroups.com...
>
> With the following spacetime,
>
> ds^2 = c^2 dt^2 / (1 + r / K) - (1 + r / K) (K^4 / r^4) dr^2 - (K^4 /
> r^2) (1 + r / K)^2 dO^2
>
> How do you specify a distance 2 AU from the center of the sun?
>
> Given the following two independent spacetime where K is an integration
> constant,
>
> ** ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
> ** ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2
>
> How do you know which of the two is the first form? And why?
>
> Or better yet, let's write the above equations into a more general
> form,
>
> ** ds^2 = c^2 dt^2 / (1 + K / (r + Q)) - (1 + K / (r + Q)) dr^2 - (r +
> Q)^2 dO^2
>
> where (K = 2 G M / c^2), (Q = any constant)
>
> How do you know Q has to be (- K) and not some other number?
And
"Koobee Wublee" <koobee...@gmail.com> wrote in message
news:1156742081....@p79g2000cwp.googlegroups.com...
>
> You can also start with the following trial
> equation (Safartti's term).
>
> ds^2 = c^2 (1 - K / u) dt^2 - du^2 / (1 - K / u) - u^2 dO^2
>
> Setting (u = r + K), you get
>
> ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
>
> You can also start with the following trial equation.
>
> ds^2 = c^2 dt^2 / (1 + K / u) - (1 + K / u) du^2 - (u + K)^2 dO^2
>
> Setting (u = r - K), you get
>
> ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2
>
> You can also start with the following trial equation.
>
> ds^2 = c^2 dt^2 / (1 + K / u) - (1 + K / u) du^2 - (u + K)^2 dO^2
>
> Setting (u = K^2 / r), you get
>
> ds^2 = c^2 dt^2 / (1 + r / K) - (1 + r / K) (K^4 / r^4) dr^2 - (K^4 /
> r^2) (1 + r / K)^2 dO^2
>
Tom Roberts
> One more consideration. A clock is, by its very nature, a massive,
> extended object. Beyond the horizon no information can be tranferred
> from the bottom to the top of the clock.
Yes a clock has finite height. Let me assume it remains a clock as it
passes the horizon (that is, it is not ripped apart by an attempt to
hover the upper part of the clock above the horizon while slowly
"dipping" the lower part into the horizon). As soon as the lower edge of
the clock reaches the horizon, the horizon zooms past the clock at the
speed of light, so there is no issue of transferring information from
inside to outside the horizon -- the top will be inside as soon as any
such information can reach it.
The horizon always moves with speed c relative to ANY locally inertial
frame valid at the horizon. All such frames are infalling of course.
Yes, this means that in principle an outgoing light pulse could remain
at the horizon forever; in practice this is an unstable equilibrium.
> It is not clear to me that any
> clock would keep working under such conditions.
As long as the clock is not ripped apart, it is not possible to know
when it passes the horizon by any local measurement; the operation of
the clock is, of course, completely local. This is true for any
freefalling clock, and for any clock that is accelerated within its
specification.
Tom Roberts
Sorcerer
| After a few days of fastening these cows, it is time for slaughter.
ROFL!
According to Bielawski, Relativity works because
An error in Relativity "would be like Stephen Hawking dividing by zero or
something equally trivial." -- Bielawski.
It's WAY too simple-minded.-- Bielawski.
"would have been caught immediately by the AdP reviewer." -- Bielawski.
It is boring to find out how far Bielawski can be pushed into
irrelevancies and nonsense.
news:1127968741.0...@o13g2000cwo.googlegroups.com
Actually, it is like Albert Einstein dividing by zero:
http://www.androcles01.pwp.blueyonder.co.uk/DominoEffect.GIF
Androcles.
Daryl McCullough
[stuff deleted]
I'm sorry if what I wrote went over your head. I tried to make
things as simple as possible, but I see that you don't understand
what a "metric" is, so we should back up a little.
If x and y are Cartesian coordinates, then we can
easily compute the length of a line segment in terms
of the coordinates of its endpoints as follows:
If the line segment runs from (x0,y0) to (x1,y1),
then the length is given by the Pythagorean theorem
as follows:
L = square-root(L_x^2 + L_y^2)
where L_x = (x1 - x0) and L_y = (y1 - y0).
Now, suppose that instead of a straight line,
we are trying to compute the length of a *curve*.
We can approximate the length of the curve by
breaking it up into tiny segments, using
Pythagoras to approximate the length of each
segment, and then summing the lengths to get
the length of the entire curve. In the limit
as the number of segments becomes very large,
the result will be the length of the curve.
Let (x_i,y_i) be the starting point of segment
number i. The endpoint of segment number i will
be the starting point of segment number i+1:
(x_{i+1}, y_{i+1}). The length of each segment
will be approximated by
L_i = square-root(delta-x_i^2 + delta-y_i^2)
where delta-x_i = (x_{i+1} - x_i) and
delta-y_i = (y_{i+1} - y_i). So we can
approximate the length of the entire curve
by the sum
L = sum over i of L_i
= sum over i of square-root(delta-x_i^2 + delta-y_i^2)
In the limit as the number of segments gets very large, this
becomes an integral:
L = integral along the curve of square-root(dx^2 + dy^2)
The combination dx^2 + dy^2 is called the "line element" and
is usually written ds^2. So for Euclidean geometry with cartesian
coordinates, we have
ds^2 = dx^2 + dy^2
Now, what happens when we use non-Cartesian coordinates? For
example, what if, instead of using x and y, we use r and theta,
defined indirectly via
x = r cos(theta)
y = r sin(theta)
Well, what we need to do is to rewrite the line element ds^2
in terms of new coordinates r and theta. We can easily compute:
dx = dr cos(theta) - r sin(theta) dtheta
dy = dr sin(theta) + r cos(theta) dtheta
ds^2 = dx^2 + dy^2
= dr^2 cos^2(theta) - 2 dr dtheta cos(theta) sin(theta)
+ r^2 sin^2(theta) dtheta^2 + dr^2 sin^2(theta)
+ 2 dr dtheta cos(theta) sin(theta)
+ r^2 cos^2(theta) dtheta^2
= (dr^2 + r^2 dtheta^2) (sin^2(theta) + cos^2(theta))
= dr^2 + r^2 dtheta^2
So in order to compute the length of a curve using r and theta,
we use:
L = Integral along the curve of square-root(dr^2 + r^2 dtheta^2)
If we want to compute L using any coordinates whatsoever, then we
can use the metric g to write it as
L = Integral along the curve of
square-root(sum over j, k of g_jk dx^j dx^k)
That's the *fundamental* use of the metric: to compute lengths
of curves. The metric doesn't have anything, a priori, to do
with gravity or relativity or velocity.
Sorcerer
| Koobee Wublee says...
|
| [stuff deleted]
No, he didn't say that at all, you did.
You relativists are so shitheaded you are compelled
to snip what you have no answer for.
Androcles
JanPB
> "Koobee Wublee" <koobee...@gmail.com> wrote in message
> news:1156826811....@b28g2000cwb.googlegroups.com...
> | After a few days of fastening these cows, it is time for slaughter.
>
> ROFL!
> According to Bielawski, Relativity works because
>
>
> An error in Relativity "would be like Stephen Hawking dividing by zero or
> something equally trivial." -- Bielawski.
>
> It's WAY too simple-minded.-- Bielawski.
>
> "would have been caught immediately by the AdP reviewer." -- Bielawski.
>
> It is boring to find out how far Bielawski can be pushed into
> irrelevancies and nonsense.
Just to keep it incontext - I meant specifically mathematical errors in
special relativity (i.e., the type of errors you complained about).
These errors are out.
--
Jan Bielawski
JanPB
Why not scroll up the screen then? :-)
--
Jan Bielawski
Sorcerer
Such trite comment displays your troll nature and fanatical
lunatic belief in division by zero.
http://www.androcles01.pwp.blueyonder.co.uk/DominoEffect.GIF
Any so-called "metric" containing a reversing velocity has to
be the product of lunacy, all you guys are debating is how
many angels can dance on the head of a pin.
Androcles.
Sorcerer
You are way too simple-minded.
http://www.androcles01.pwp.blueyonder.co.uk/DominoEffect.GIF
Relativity doesn't work because Einstein divided by zero,
never mind the other equally trivial blunders, and the AdP
reviewer was a shithead like you.
Androcles
carlip...@physics.ucdavis.edu
> These two gentlemen fail to understand what the metric actually
> represents. It is thus necessary to explain what the metric really
> represents. Given the ubiquitously familiar Schwarzschild metric as
> described in the very second equation, dt, dr, and dO (longitudinal
> and latitudinal information) are the observed quantities.
No, they aren't. As a simple example, compare the Schwarzschild
metric in isotropic coordinates and Schwarzschild coordinates (see
http://relativity.livingreviews.org/open?pubNo=lrr-2000-5&page=node9.html
for the explicit expressions). Both describe the same spacetime, and
both have a radial coordinate called r. But at a given point, the
numerical value of r_Schwazrschild will be different from the value of
r_isotropic.
Clearly they can't both be "the observed quantities." So how do you
propose to decide which one is "right"? (Note: an answer of the form,
"I saw the Schwarzschild form first" is not passing.)
For a *physical* example of the difference, see Bodenner and Will,
"Deflection of light to second order: A tool for illustrating principles
of general relativity," American Journal of Physics, Vol. 71, No. 8,
pp. 770–773, August 2003. You will learn that the coordinate values
for the deflection of light by the Sun differ depending on which "r"
you use, and that the correct result can only be obtained by using
"measurable, coordinate-independent quantities."
[...]
> In the meantime, the Voodoo mathematical minds still need to answer and
> address the following.
> "Koobee Wublee" <koobee...@gmail.com> wrote in message
> news:1156798554.8...@m73g2000cwd.googlegroups.com...
>> With the following spacetime,
>> ds^2 = c^2 dt^2 / (1 + r / K) - (1 + r / K) (K^4 / r^4) dr^2 - (K^4 /
>> r^2) (1 + r / K)^2 dO^2
>> How do you specify a distance 2 AU from the center of the sun?
First of all, you can't use this metric to answer that question --
the Sun is not a vacuum, and the metric you've written down is only
valid in empty space. To answer the question as stated, you need
first to give an interior metric that "glues" on to the exterior
metric you've written down. The answer will depend on the details
of that interior metric.
You can, however, answer the question, "How do you specify a distance
2 AU from the edge of the Sun." To answer that, you need to take
the interval ds at fixed time and fixed angle (so dt=0 and dO=0) and
integrate ds from the edge of the Sun to an arbitrary value r. That
integral will give you the physical ("proper") distance D, as measured
by an observer at rest. Inverting the answer, you can find r as a
function of D. Take that expression and set D = 2 AU.
This is fairly basic general relativity. It's true that beginners
sometimes get coordinate differences and physical quantities mixed up,
but if you take a decent course in the subject, or actually read a
decent book (and work through the exercises, etc.), you should be able
to correct that kind of mistake pretty quickly.
(For example, see the discussion on page 191 of Weinberg's text about
the problem of whether a prediction "really refers to an objective
physical measurement or whether it has folded into it arbitrary
subjective elements dependent on our choice of coordinate system";
or, for the meaning of Schwarzschild coordinates, Wald's careful
discussion in section 6.1 of his text. And by all means, read the
paper by Bodenner and Will that I cited above, which is about almost
precisely this issue.)
Steve Carlip