Sagittarius A* - is wikipedia just wrong?
Roland PJ
"This is compatible with, and strong evidence in support of, the
hypothesis that Sagittarius A* is a supermassive black hole. While it
is possible for the observed mass within the observed volume limit to
be distributed among multiple objects, any such objects would undergo
orbital collapse into a single black hole anyway within a few hundred
years at most, a negligible amount of time compared with the lifetime
of the galaxy."
This is surely confusion on the part of the author between the proper
time at Sagittarius A* and observer time on earth.
Sagittarius A* might think it's collapsing in a few hundred years, but
for us on earth that would take an infinite amount of time.
Right?
Regards
Roland
Craig Markwardt
Roland PJ <rola...@gmail.com> writes:
Light from the precise surface of the event horizon would take an
"infinite" time to escape to the distant universe, but emission from
any other point outside the horizon would take a finite amount of
time.
But that is not really relevant. Whether or not the light could
escape in a certain amount of time, the point of the statement in the
article is that the collapse/coalescence timescale based on known
physics is far shorter than the lifetime of the galaxy, thus
increasing the chances of a black hole being present.
CM
harry
"Roland PJ" <rola...@gmail.com> wrote in message
news:1182867330.9...@w5g2000hsg.googlegroups.com...
> http://en.wikipedia.org/wiki/Sagittarius_A%2A
>
> "This is compatible with, and strong evidence in support of, the
> hypothesis that Sagittarius A* is a supermassive black hole. While it
> is possible for the observed mass within the observed volume limit to
> be distributed among multiple objects, any such objects would undergo
> orbital collapse into a single black hole anyway within a few hundred
> years at most, a negligible amount of time compared with the lifetime
> of the galaxy."
Apparently Wikipedia paraphrases:
"Eisenhauer, F., Schödel, R. et al. "A geometric determination of the
distance to the galactic center." The Astrophysical Journal, 597, L121-L124,
(2003)."
Thus, if the above is wrong it could be that the source is wrong about that.
> This is surely confusion on the part of the author between the proper
> time at Sagittarius A* and observer time on earth.
>
> Sagittarius A* might think it's collapsing in a few hundred years, but
> for us on earth that would take an infinite amount of time.
>
> Right?
I guess that it depends on how much the time dilation is at the location of
those objects; certainly the time dilation there is not nearly "infinite"
and perhaps even rather modest. I'm looking forward to see the response of
experts. :-))
Regards,
Harald
Koobee Wublee
> http://en.wikipedia.org/wiki/Sagittarius_A%2A
>
> "This is compatible with, and strong evidence in support of, the
> hypothesis that Sagittarius A* is a supermassive black hole. While it
> is possible for the observed mass within the observed volume limit to
> be distributed among multiple objects, any such objects would undergo
> orbital collapse into a single black hole anyway within a few hundred
> years at most, a negligible amount of time compared with the lifetime
> of the galaxy."
By examining an object orbiting a gravitating body, one can determine
the mass of the gravitating body. Correct me if I am wrong. This is
Newtonian law of gravity.
Now, the logic to prove the existence of a black hole gets really
twisted. If the mass of the gravitating body is confined within the
event horizon, then this gravitating body is a black hole. This, of
course, is according to one of the solutions to the field equations
now commonly referred to as the Schwarzschild metric.
Mathematically, the geometry of such a system is described below where
the mass of the gravitating body is so much larger than the mass of
the orbiting object.
ds^2 = c^2 (1 - 2 U) dt^2 - dr^2 / (1 - 2 U) - r^2 dO^2
Where
** U = G M / c^2 / r
Ironically, Schwarzschild's original solution did not manifest such a
black hole. Thus, mass can concentrate in any confinement without
creating such a black hole. Another solution not as complicated as
Schwarzschild's original solution and almost as simple as Hilbert's
solution (the Schwarzschild metric) is described as follows.
ds^2 = c^2 dt^2 / (1 + 2 U) - (1 + 2 U) dr^2 - (r + R)^2 dO^2
Where
** R = G M / c^2
Historically, because of the complexity to Schwarzschild's original
solution, it prompted Hilbert to find another one simpler. If Hilbert
found the second one, then there would be no issues about black hole
no black hole mumble jumble. <shrug>
Surfer
What do you think of:
"The river model of black holes"?
http://arxiv.org/abs/gr-qc/0411060
"This paper presents an under-appreciated way to conceptualize
stationary black holes, which we call the river model. The river model
is mathematically sound, yet simple enough that the basic picture can
be understood by non-experts.
In the river model, space itself flows like a river through a flat
background, while objects move through the river according to the
rules of special relativity. In a spherical black hole, the river of
space falls into the black hole at the Newtonian escape velocity,
hitting the speed of light at the horizon. Inside the horizon, the
river flows inward faster than light, carrying everything with it. We
show that the river model works also for rotating (Kerr-Newman) black
holes, though with a surprising twist. As in the spherical case, the
river of space can be regarded as moving through a flat background.
However, the river does not spiral inward, as one might have
anticipated, but rather falls inward with no azimuthal swirl at all.
Instead, the river has at each point not only a velocity but also a
rotation, or twist. That is, the river has a Lorentz structure,
characterized by six numbers (velocity and rotation), not just three
(velocity). As an object moves through the river, it changes its
velocity and rotation in response to tidal changes in the velocity and
twist of the river along its path. An explicit expression is given for
the river field, a six-component bivector field that encodes the
velocity and twist of the river at each point, and that encapsulates
all the properties of a stationary rotating black hole."
Eric Gisse
> On Jun 26, 7:15 am, Roland PJ <rolan...@gmail.com> wrote:
>
> >http://en.wikipedia.org/wiki/Sagittarius_A%2A
>
> > "This is compatible with, and strong evidence in support of, the
> > hypothesis that Sagittarius A* is a supermassive black hole. While it
> > is possible for the observed mass within the observed volume limit to
> > be distributed among multiple objects, any such objects would undergo
> > orbital collapse into a single black hole anyway within a few hundred
> > years at most, a negligible amount of time compared with the lifetime
> > of the galaxy."
>
> By examining an object orbiting a gravitating body, one can determine
> the mass of the gravitating body. Correct me if I am wrong. This is
> Newtonian law of gravity.
No. Kepler's laws.
>
> Now, the logic to prove the existence of a black hole gets really
> twisted. If the mass of the gravitating body is confined within the
> event horizon, then this gravitating body is a black hole. This, of
> course, is according to one of the solutions to the field equations
> now commonly referred to as the Schwarzschild metric.
Along with 3 other solutions.
>
> Mathematically, the geometry of such a system is described below where
> the mass of the gravitating body is so much larger than the mass of
> the orbiting object.
>
> ds^2 = c^2 (1 - 2 U) dt^2 - dr^2 / (1 - 2 U) - r^2 dO^2
>
> Where
>
> ** U = G M / c^2 / r
>
> Ironically, Schwarzschild's original solution did not manifest such a
> black hole. Thus, mass can concentrate in any confinement without
> creating such a black hole. Another solution not as complicated as
> Schwarzschild's original solution and almost as simple as Hilbert's
> solution (the Schwarzschild metric) is described as follows.
>
> ds^2 = c^2 dt^2 / (1 + 2 U) - (1 + 2 U) dr^2 - (r + R)^2 dO^2
>
> Where
>
> ** R = G M / c^2
>
> Historically, because of the complexity to Schwarzschild's original
> solution, it prompted Hilbert to find another one simpler. If Hilbert
> found the second one, then there would be no issues about black hole
> no black hole mumble jumble. <shrug>
You fucking moron. They are the _SAME SOLUTION_. Hilbert's solution is
the Schwarzschild solution rescaled. There is no difference between
the solutions you wrote down! Hilbert's solution in the Schwarzschild
coordinate chart is the inverse of the usual Schwarzschild metric.
The physics is the same. You can't wish away black holes by simply
rescaling from r to r - 2GM.
Roland PJ
wrote:
> "Roland PJ" <rolan...@gmail.com> wrote in message
Hi Harald
The time dilation (as viewed from earth) approaches infinity _before_
these objects cross the event horizon (and reaches infinity _as_ the
objects cross the horizon).
So, no external observer (i.e. us) will ever observe any objects
'joining' a black hole (in finite time, which is all we have).
Thanks for the back-link.
Regards
Roland
Roland PJ
<craigm...@REMOVEcow.physics.wisc.edu> wrote:
> Roland PJ <rolan...@gmail.com> writes:
> >http://en.wikipedia.org/wiki/Sagittarius_A%2A
>
> > "This is compatible with, and strong evidence in support of, the
> > hypothesis that Sagittarius A* is a supermassive black hole. While it
> > is possible for the observed mass within the observed volume limit to
> > be distributed among multiple objects, any such objects would undergo
> > orbital collapse into a single black hole anyway within a few hundred
> > years at most, a negligible amount of time compared with the lifetime
> > of the galaxy."
>
> > This is surely confusion on the part of the author between the proper
> > time at Sagittarius A* and observer time on earth.
>
> But that is not really relevant. Whether or not the light could
> escape in a certain amount of time, the point of the statement in the
> article is that the collapse/coalescence timescale based on known
> physics is far shorter than the lifetime of the galaxy, thus
> increasing the chances of a black hole being present.
Craig. Suppose you hopped on-board a space-ship, and managed to stay
alive for the long journey into the centre of the galaxy. You could
take ever-more accurate observations of the objects around Sagitarius
A* as you got closer and closer. But you will never observe objects
that began outside the event horizon to have crossed into the black
hole.
This is the curse of General Relativity. It's just not meaningful to
say that a black hole has formed 'now', because there's no path from
here and now on earth that would let you observe that to have
happened.
And it's particularly confused to use the proper time of the objects
themselves to dictate what has happened 'now'. 'Now' is only relevant
for an observer, and an external observer's time-frame diverges
infinitely from the proper time of the observed object near a 'black
hole'.
Roland
Roland PJ
wrote:
>
> Apparently Wikipedia paraphrases:
> "Eisenhauer, F., Schödel, R. et al. "A geometric determination of the
> distance to the galactic center." The Astrophysical Journal, 597, L121-L124,
> (2003)."
>
The corresponding quote from the above paper is:
"It is also unlikely that the central mass is a binary black hole of
approximately equal masses, since such binary hole would have to have
a separation less than 10 light hours (constrained by the data) and
would coalesce by gravitational radiation in a few hundred years."
They don't appear to specify which observer-frame's time they are
refering to.
So, yes, it could simply be an oversight in the original paper (which
specifically addresses observations from earth).
I need an expert. Any experts out there care to comment? Eric?
Regards
Roland
Eric Gisse
> On Jun 26, 5:55 pm, "harry" <harald.vanlintelButNotT...@epfl.ch>
> wrote:
>
>
>
> > Apparently Wikipedia paraphrases:
> > "Eisenhauer, F., Schödel, R. et al. "A geometric determination of the
> > distance to the galactic center." The Astrophysical Journal, 597, L121-L124,
> > (2003)."
>
> The corresponding quote from the above paper is:
>
> "It is also unlikely that the central mass is a binary black hole of
> approximately equal masses, since such binary hole would have to have
> a separation less than 10 light hours (constrained by the data) and
> would coalesce by gravitational radiation in a few hundred years."
It'd have to be less than 10 light hours, too.
Keep this in mind when considering alternate scenarios like this:
There is about 3 million solar masses stuffed into an area enclosed by
an orbit equivalent to Pluto's.
It is _entirely possible_ that there are multiple black holes. We have
already seen a medium mass black hole hanging around the area, but the
possibility is low. The orbit would have to be _very_ tight to fit
with observation and something like that is going to be spitting
gravitational radiation like crazy.
I bet there could be some severe constraints placed on certain orbital
parameters based upon non-observation of gravitational radiation
through LIGO.
>
> They don't appear to specify which observer-frame's time they are
> refering to.
Earth time. There is a 25,000 year difference.
>
> So, yes, it could simply be an oversight in the original paper (which
> specifically addresses observations from earth).
It really doesn't matter. A few hundred years Sgr A* time, Earth time,
Andromeda time - it doesn't matter. The galaxy is old - I do not
believe it to be reasonable to say that the black hole at the center
formed anytime recently.
Koobee Wublee
> On Jun 26, 9:25 am, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > By examining an object orbiting a gravitating body, one can determine
> > the mass of the gravitating body. Correct me if I am wrong. This is
> > Newtonian law of gravity.
>
> No. Kepler's laws.
Well, in that case, you need to show me how you determine the
gravitating mass using Kepler's laws.
> > Now, the logic to prove the existence of a black hole gets really
> > twisted. If the mass of the gravitating body is confined within the
> > event horizon, then this gravitating body is a black hole. This, of
> > course, is according to one of the solutions to the field equations
> > now commonly referred to as the Schwarzschild metric.
>
> Along with 3 other solutions.
Just 3? Why end at 3? There are in fact an infinite number of them.
> > Mathematically, the geometry of such a system is described below where
> > the mass of the gravitating body is so much larger than the mass of
> > the orbiting object.
>
> > ds^2 = c^2 (1 - 2 U) dt^2 - dr^2 / (1 - 2 U) - r^2 dO^2
>
> > Where
>
> > ** U = G M / c^2 / r
>
> > Ironically, Schwarzschild's original solution did not manifest such a
> > black hole. Thus, mass can concentrate in any confinement without
> > creating such a black hole. Another solution not as complicated as
> > Schwarzschild's original solution and almost as simple as Hilbert's
> > solution (the Schwarzschild metric) is described as follows.
>
> > ds^2 = c^2 dt^2 / (1 + 2 U) - (1 + 2 U) dr^2 - (r + R)^2 dO^2
>
> > Where
>
> > ** R = G M / c^2
>
> > Historically, because of the complexity to Schwarzschild's original
> > solution, it prompted Hilbert to find another one simpler. If Hilbert
> > found the second one, then there would be no issues about black hole
> > no black hole mumble jumble. <shrug>
>
> You fucking moron. They are the _SAME SOLUTION_.
No, the choice of coordinate system is already established in concrete
when these solutions are derived from the field equations. Different
metrics using the same coordinate system describe different
geometries. This is grade school logic. It is missing in you and
other physicists such as Professor Roberts as well as and Mr.
Bielawski and McCullough. <shrug>
> Hilbert's solution is
> the Schwarzschild solution rescaled. There is no difference between
> the solutions you wrote down! Hilbert's solution in the Schwarzschild
> coordinate chart is the inverse of the usual Schwarzschild metric.
You are so full of BS as usual. You have failed grade school logic.
<shrug>
> The physics is the same. You can't wish away black holes by simply
> rescaling from r to r - 2GM.
You cannot simply create a black hole by embracing a particular
metric. <shrug>
Craig Markwardt
Roland PJ <rola...@gmail.com> writes:
And your point is? Since the Wikipedia article never claimed that the
event horizon -- or matter that had passed through the horizon -- had
been detected, your supposition is not relevant.
What the article *did* claim was that the observations require a large
amount of mass to be concentrated into a small volume of space (basis:
Kepler's laws); *and* that multiple bodies would coalesce into a
single body in a short time (I presume the basis of this is the
effects of tidal disruption, collision and gravitational radiation).
So let's review the facts:
1. 100,000 times more dense the densest core collapsed globular;
2. The mass-equivalent of 3.6 million suns (Eisenhauer et al 2003); BUT
2. No appreciable optical emission!
Even ignoring GR, it sounds like the perfect definition of a "black hole."
I.e. the issue of an event horizon never really needed to be brought up.
> This is the curse of General Relativity. It's just not meaningful to
> say that a black hole has formed 'now', because there's no path from
> here and now on earth that would let you observe that to have
> happened.
Since neither the Wikipedia article nor I mentioned the word "now,"
the relevance of your assertion is not clear. Neither the Wikipedia
article, nor the Eisenhauer et al paper it refers to, claim that the
formation of a black hole was detected, nor do they claim that the
formation *could* have been detected.
...
CM
Eric Gisse
> On Jun 26, 2:28 pm, Eric Gisse <jowr...@gmail.com> wrote:
>
> > On Jun 26, 9:25 am, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > > By examining an object orbiting a gravitating body, one can determine
> > > the mass of the gravitating body. Correct me if I am wrong. This is
> > > Newtonian law of gravity.
>
> > No. Kepler's laws.
>
> Well, in that case, you need to show me how you determine the
> gravitating mass using Kepler's laws.
Seriously? You don't know how to do that?
http://en.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion#Kepler.27s_third_law
>
> > > Now, the logic to prove the existence of a black hole gets really
> > > twisted. If the mass of the gravitating body is confined within the
> > > event horizon, then this gravitating body is a black hole. This, of
> > > course, is according to one of the solutions to the field equations
> > > now commonly referred to as the Schwarzschild metric.
>
> > Along with 3 other solutions.
>
> Just 3? Why end at 3? There are in fact an infinite number of them.
*sigh*
Q = J = 0, M =/= 0 : Schwarzschild
Q = 0, J, M =/= 0 : Kerr
Q, J, M =/= 0: Kerr-Newman
There are only 3 solutions for those specific configurations, and no
other black hole solutions exist. The outlier case of the existence of
magnetic charge is folded into the Kerr-Newman solution as well.
You shouldn't argue too hard about uniqueness when Kepler's laws
evades your understanding.
They aren't the same coordinate system, shit for brains. Take the
inverse of the Schwarzschild metric, and transform to the coordinate
system r' = r - R , and you have Hilbert's solution. Therefore, the
two "different" metrics describe the same manifold.
This has been explained to you before.
http://groups.google.com/group/sci.physics.relativity/msg/413dcbe1b5c79241?dmode=source
Since you _still_ haven't figured out the basic truthfulness of this
statement in 3 months, what you need to do is write down the tensor
transformation law and _explicitly_ transform the metric. I have it
down on a piece of paper, it took me all of 5 minutes to do.
What is utterly comical is that it ties in directly to your whining
about how the surface area of constant r,t _must_ be pi*r^2, but from
Hilbert's metric it was calculated to be pi*(r+K)^2. Which you
completely and utterly denied. But if you took the 5 seconds to
remember how r was defined, you would see that it was the same goddamn
result as you expected it to be. Your constant inability to understand
what you are doing trips up your already precarious understanding of
the theory.
Oh, and what is also comical is that you lump me in with 3 folks with
doctorates in physics. Thanks for the complement.
>
> > Hilbert's solution is
> > the Schwarzschild solution rescaled. There is no difference between
> > the solutions you wrote down! Hilbert's solution in the Schwarzschild
> > coordinate chart is the inverse of the usual Schwarzschild metric.
>
> You are so full of BS as usual. You have failed grade school logic.
> <shrug>
Hey, your the one denying mathematics. Not me.
>
> > The physics is the same. You can't wish away black holes by simply
> > rescaling from r to r - 2GM.
>
> You cannot simply create a black hole by embracing a particular
> metric. <shrug>
Y'know, this argument is funny coming from you. You talk so much about
the history of relativity, but apparently you completely _missed_ the
decades-long argument that wasn't settled until the 1960's about
whether the event horizon was a coordinate dependent object or a
global property.
Roland PJ
> The galaxy is old - I do not
> believe it to be reasonable to say that the black hole at the center
> formed anytime recently.
Eric, when you say 'recently', which observer frame are you refering
to?
Also, when you say 'the black hole at the center', do you mean 'the
black hole that we can see at the center'?
Thanks
Roland
Roland PJ
>
>
>
> > On Jun 26, 5:55 pm, "harry" <harald.vanlintelButNotT...@epfl.ch>
> > wrote:
>
> > > Apparently Wikipedia paraphrases:
> > > "Eisenhauer, F., Schödel, R. et al. "A geometric determination of the
> > > distance to the galactic center." The Astrophysical Journal, 597, L121-L124,
> > > (2003)."
>
> > The corresponding quote from the above paper is:
>
> > "It is also unlikely that the central mass is a binary black hole of
> > approximately equal masses, since such binary hole would have to have
> > a separation less than 10 light hours (constrained by the data) and
> > would coalesce by gravitational radiation in a few hundred years."
>
>
> > refering to.
>
> Earth time. There is a 25,000 year difference.
>
OK. I get it (I think). What we see now on earth, is information that
has travelled at light speed (it's light) taking 25,000 years to reach
us. 100 years doesn't seem very long compared to this.
But, on the other hand, if we wait 100 years on earth, and take
another look, _we_ still won't notice that much has changed.
Roland
Roland PJ
<craigm...@REMOVEcow.physics.wisc.edu> wrote:
> Roland PJ <rolan...@gmail.com> writes:
>
> What the article *did* claim was that the observations require a large
> amount of mass to be concentrated into a small volume of space (basis:
> Kepler's laws); *and* that multiple bodies would coalesce into a
> single body in a short time (I presume the basis of this is the
> effects of tidal disruption, collision and gravitational radiation).
> So let's review the facts:
> 1. 100,000 times more dense the densest core collapsed globular;
> 2. The mass-equivalent of 3.6 million suns (Eisenhauer et al 2003); BUT
> 2. No appreciable optical emission!
> Even ignoring GR, it sounds like the perfect definition of a "black hole."
> I.e. the issue of an event horizon never really needed to be brought up.
I agree with all of this. Whether or not the 'event horizon' actually
exists 'now' is the frame of any observer is somewhat irrelevant - the
immense concentration of mass, and hence gravitation is indisputable.
>
> Since neither the Wikipedia article nor I mentioned the word "now,"
> the relevance of your assertion is not clear. Neither the Wikipedia
> article, nor the Eisenhauer et al paper it refers to, claim that the
> formation of a black hole was detected, nor do they claim that the
> formation *could* have been detected.
I'll quote again:
"It is also unlikely that the central mass is a binary black hole of
approximately equal masses..."
The present tense "is" denotes "now" in the frame of some observer
(presumably the frame of the authors when they wrote the paper?). Do
you have another way to interpret this?
Regards
Roland
Eric Gisse
This planet is ~5 billion years old.
Which do you think came first? The galaxy or the planet?
Consider the binary pulsar system PSR 1913+16. It is scheduled for
decay in 300 million years +/- a bit. The binary pulsars are about a
solar mass each, and have a closest approach of a few solar radii. The
_total mass_ at the center of the galaxy is at _LEAST_ 2.5 [I have
heard it from anywhere from 2 to 4 being cited as serious numbers]
million solar masses.
http://www.mpe.mpg.de/ir/GC/res_mass.php?lang=en
It is one black hole.
>
> Roland
Roland PJ
> On Jun 28, 1:04 am, Roland PJ <rolan...@gmail.com> wrote:
>
>
> > But, on the other hand, if we wait 100 years on earth, and take
> > another look, _we_ still won't notice that much has changed.
>
> This planet is ~5 billion years old.
>
> Which do you think came first? The galaxy or the planet?
Eric, when you say 'first', which reference frame are you using?
I think today on earth will come 'before' event horizon formation, or
binary black hole coalescence, for all reference frames external to
the black hole(s). Right?
> Consider the binary pulsar system PSR 1913+16. It is scheduled for
> decay in 300 million years +/- a bit. The binary pulsars are about a
> solar mass each, and have a closest approach of a few solar radii. The
> _total mass_ at the center of the galaxy is at _LEAST_ 2.5 [I have
> heard it from anywhere from 2 to 4 being cited as serious numbers]
> million solar masses.
>
> http://www.mpe.mpg.de/ir/GC/res_mass.php?lang=en
>
> It is one black hole.
When you say 'is', do you mean 'now' in the reference frame of Eric
Gisse, when he wrote this?
If that's what you mean, then I think you are wrong.
Regards
Roland
Koobee Wublee
> On Jun 27, 10:51 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > Well, in that case, you need to show me how you determine the
> > gravitating mass using Kepler's laws.
>
> Seriously? You don't know how to do that?
Yeah seriously, you don't know both Kepler's laws of Newton's law of
gravity. <shrug>
> http://en.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion#Kepler...
Oh yeah! I am still laughing.
> Q = J = 0, M =/= 0 : Schwarzschild
> Q = 0, J, M =/= 0 : Kerr
> Q, J, M =/= 0: Kerr-Newman
>
> There are only 3 solutions for those specific configurations, and no
> other black hole solutions exist. The outlier case of the existence of
> magnetic charge is folded into the Kerr-Newman solution as well.
One more such solution that manifests a black hole is the following.
ds^2 = c^2 (1 - U)^2 dt^2 - dr^2 / (1 - U)^2 / (1 - 2 U)^2
- r^2 dO^2 / (1 - 2 U)^2
Where
** U = G M / c^2 / r
And of course, there are still an infinite others out there. There
are also an infinite numbers out that do not manifest any black
holes. <shrug>
> You shouldn't argue too hard about uniqueness when Kepler's laws
> evades your understanding.
Well, I won't argue about Kelper's laws anymore. You need to
understand them first.
> > No, the choice of coordinate system is already established in concrete
> > when these solutions are derived from the field equations. Different
> > metrics using the same coordinate system describe different
> > geometries. This is grade school logic. It is missing in you and
> > other physicists such as Professor Roberts as well as and Mr.
> > Bielawski and McCullough. <shrug>
>
> They aren't the same coordinate system, shit for brains. Take the
> inverse of the Schwarzschild metric, and transform to the coordinate
> system r' = r - R , and you have Hilbert's solution. Therefore, the
> two "different" metrics describe the same manifold.
This is your mistake. No one is talking about transformation in the
sense as what you are doing here. <shrug>
> This has been explained to you before.
>
> http://groups.google.com/group/sci.physics.relativity/msg/413dcbe1b5c...
And I have explained to you many times before that what I have done is
a transformation of what you are thinking.
> Since you _still_ haven't figured out the basic truthfulness of this
> statement in 3 months, what you need to do is write down the tensor
> transformation law and _explicitly_ transform the metric. I have it
> down on a piece of paper, it took me all of 5 minutes to do.
Done that. <shrug>
> What is utterly comical is that it ties in directly to your whining
> about how the surface area of constant r,t _must_ be pi*r^2, but from
> Hilbert's metric it was calculated to be pi*(r+K)^2. Which you
> completely and utterly denied. But if you took the 5 seconds to
> remember how r was defined, you would see that it was the same goddamn
> result as you expected it to be. Your constant inability to understand
> what you are doing trips up your already precarious understanding of
> the theory.
Again, I stand by what I have said before. As an observer using a
coordinate system of (t, r, O), he is going to observe the surface
area of a sphere to be (4 pi R^2) where R is the radius determined or
measured from his coordinate system. This is so regardless what the
metric is.
> Oh, and what is also comical is that you lump me in with 3 folks with
> doctorates in physics. Thanks for the complement.
I did not. You must be very mistaken and have misunderstood grossly.
Remember that you are a super, super senior who is still struggling to
get that some BS in art degree from the university for snowmen.
<shrug>
> > You are so full of BS as usual. You have failed grade school logic.
> > <shrug>
>
> Hey, your the one denying mathematics. Not me.
On the contrary, you are the one denying mathematics. I have shown
you many times over the mathematics involved therein. <shrug>
> > You cannot simply create a black hole by embracing a particular
> > metric. <shrug>
>
> Y'know, this argument is funny coming from you. You talk so much about
> the history of relativity, but apparently you completely _missed_ the
> decades-long argument that wasn't settled until the 1960's about
> whether the event horizon was a coordinate dependent object or a
> global property.
Given the Schwarzschild metric, the event horizon is invariant. This
is no brainer because it is the property of the geometry itself.
Remember that the geometry must be invariant regardless of any
observers. On the other hand, it is impossible to completely describe
the geometry without using a set of coordinate system. Doing so, an
appropriate metric must be utilized. Using another set of coordinate
system, you must also provide its appropriate metric to agree on the
same geometry. Again, this logic should be so basic, but you and
other physicists have so much trouble grasping. <shrug>
Craig Markwardt
Roland PJ <rola...@gmail.com> writes:
> >
> > Since neither the Wikipedia article nor I mentioned the word "now,"
> > the relevance of your assertion is not clear. Neither the Wikipedia
> > article, nor the Eisenhauer et al paper it refers to, claim that the
> > formation of a black hole was detected, nor do they claim that the
> > formation *could* have been detected.
>
> I'll quote again:
>
> "It is also unlikely that the central mass is a binary black hole of
> approximately equal masses..."
>
> The present tense "is" denotes "now" in the frame of some observer
> (presumably the frame of the authors when they wrote the paper?). Do
> you have another way to interpret this?
You really are missing the point.
The point is that the dynamical lifetime of two compact orbitting
bodies -- any bodies -- is far shorter than the present age of the
galaxy. Thus the chances of having multiple bodies within such a
small volume are *a priori* very small. The question of "now" is
irrelevant, since the determination of whether there are one or two
bodies within a certain volume can be answered in a finite amount of
time by a distant observer, independent of the presence of a
relativistic event horizon.
The "black hole" aspect only enters when one considers that there are
3 million suns worth of mass in a region from which no light is being
emitted. Well, it enters in other ways as well, since there is no
known theoretical or observational basis that any body other than
black hole could have 1.5 million solar masses.
Craig
Eric Gisse
> On Jun 28, 12:51 am, Eric Gisse <jowr...@gmail.com> wrote:
>
> > On Jun 27, 10:51 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > > Well, in that case, you need to show me how you determine the
> > > gravitating mass using Kepler's laws.
>
> > Seriously? You don't know how to do that?
>
> Yeah seriously, you don't know both Kepler's laws of Newton's law of
> gravity. <shrug>
Kepler came first, dipshit.
Kepler formulated his laws from the data from Tycho. Newton only
formulated his inverse square law _after_ Kepler and Tycho.
>
> >http://en.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion#Kepler...
>
> Oh yeah! I am still laughing.
>
> > Q = J = 0, M =/= 0 : Schwarzschild
> > Q = 0, J, M =/= 0 : Kerr
> > Q, J, M =/= 0: Kerr-Newman
>
> > There are only 3 solutions for those specific configurations, and no
> > other black hole solutions exist. The outlier case of the existence of
> > magnetic charge is folded into the Kerr-Newman solution as well.
>
> One more such solution that manifests a black hole is the following.
>
> ds^2 = c^2 (1 - U)^2 dt^2 - dr^2 / (1 - U)^2 / (1 - 2 U)^2
> - r^2 dO^2 / (1 - 2 U)^2
dr^2 / (1 - U)^2 / (1 - 2 U)^2 = dr^2, moron.
>
> Where
>
> ** U = G M / c^2 / r
>
> And of course, there are still an infinite others out there. There
> are also an infinite numbers out that do not manifest any black
> holes. <shrug>
1) You haven't demonstrated that the metric this line element
represents is a solution to the vacuum field equations.
2) You haven't explained the meanings of the coordinate variables and
their domains.
3) You haven't demonstrated that there is an event horizon, and that
it is a global phenomena rather than an aspect of the coordinates.
Pulling a metric out of your butt just isn't acceptable.
>
> > You shouldn't argue too hard about uniqueness when Kepler's laws
> > evades your understanding.
>
> Well, I won't argue about Kelper's laws anymore. You need to
> understand them first.
No, YOU need to understand them first. You asked how to determine the
mass of an orbited body through Kepler's laws, and I gave you a link
that had the goddamn formula.
>
> > > No, the choice of coordinate system is already established in concrete
> > > when these solutions are derived from the field equations. Different
> > > metrics using the same coordinate system describe different
> > > geometries. This is grade school logic. It is missing in you and
> > > other physicists such as Professor Roberts as well as and Mr.
> > > Bielawski and McCullough. <shrug>
>
> > They aren't the same coordinate system, shit for brains. Take the
> > inverse of the Schwarzschild metric, and transform to the coordinate
> > system r' = r - R , and you have Hilbert's solution. Therefore, the
> > two "different" metrics describe the same manifold.
>
> This is your mistake. No one is talking about transformation in the
> sense as what you are doing here. <shrug>
Actually, it is yours. That is my entire point. That is always the
point.
>
> > This has been explained to you before.
>
> >http://groups.google.com/group/sci.physics.relativity/msg/413dcbe1b5c...
>
> And I have explained to you many times before that what I have done is
> a transformation of what you are thinking.
DUH
Any two metrics that can be related by a coordinate transformation are
the _SAME_. Welcome to introductory tensor analysis - Enjoy your stay.
>
> > Since you _still_ haven't figured out the basic truthfulness of this
> > statement in 3 months, what you need to do is write down the tensor
> > transformation law and _explicitly_ transform the metric. I have it
> > down on a piece of paper, it took me all of 5 minutes to do.
>
> Done that. <shrug>
OK. Write down the tensor transformation law.
>
> > What is utterly comical is that it ties in directly to your whining
> > about how the surface area of constant r,t _must_ be pi*r^2, but from
> > Hilbert's metric it was calculated to be pi*(r+K)^2. Which you
> > completely and utterly denied. But if you took the 5 seconds to
> > remember how r was defined, you would see that it was the same goddamn
> > result as you expected it to be. Your constant inability to understand
> > what you are doing trips up your already precarious understanding of
> > the theory.
>
> Again, I stand by what I have said before. As an observer using a
> coordinate system of (t, r, O), he is going to observe the surface
> area of a sphere to be (4 pi R^2) where R is the radius determined or
> measured from his coordinate system. This is so regardless what the
> metric is.
This is most definitely not true, as proven by explicit calculation.
It isn't 4piR^2 for r' = r - K. It isn't for r' = exp(r). It isn't for
r' = r + K.
>
> > Oh, and what is also comical is that you lump me in with 3 folks with
> > doctorates in physics. Thanks for the complement.
>
> I did not. You must be very mistaken and have misunderstood grossly.
> Remember that you are a super, super senior who is still struggling to
> get that some BS in art degree from the university for snowmen.
> <shrug>
I am 8 classes [5 humanities, 3 physics] away from a full Bachelors of
Science in Physics. As usual, you say something stupid then punctuate
it with <shrug>.
Where was _YOUR_ alma mater? What was _YOUR_ degree in? It sure as
hell wasn't in physics.
>
> > > You are so full of BS as usual. You have failed grade school logic.
> > > <shrug>
>
> > Hey, your the one denying mathematics. Not me.
>
> On the contrary, you are the one denying mathematics. I have shown
> you many times over the mathematics involved therein. <shrug>
Writing down a line element and screaming "DON'T YOU SEE?!!?!?" is not
mathematics. That is the whole sum of your mathematics contributions
to date - you have never went further than writing down a line element
you think is relevant to your problem.
You have not once calculated one of the curvature quantities that you
claim prove what you were saying. Not once have you calculated area
using your metrics. Not once have you even written down an integral to
support your statements.
Whenever confronted with your dishonest behavior, all you can do is
meekly bleat that the poster should "search my past postings" in the
vain hope that there exists something, somewhere that supports what
you are saying.
>
> > > You cannot simply create a black hole by embracing a particular
> > > metric. <shrug>
>
> > Y'know, this argument is funny coming from you. You talk so much about
> > the history of relativity, but apparently you completely _missed_ the
> > decades-long argument that wasn't settled until the 1960's about
> > whether the event horizon was a coordinate dependent object or a
> > global property.
>
> Given the Schwarzschild metric, the event horizon is invariant. This
> is no brainer because it is the property of the geometry itself.
Is it? You can only say that with the certainty of hindsight because
you are certainly incapable of proving it. I could write the
Schwarzschild metric in an exotic set of coordinates and you would
have _NO WAY_ of telling me whether it *was* the Schwarzschild metric,
much less where the event horizon is.
> Remember that the geometry must be invariant regardless of any
> observers. On the other hand, it is impossible to completely describe
> the geometry without using a set of coordinate system. Doing so, an
> appropriate metric must be utilized. Using another set of coordinate
> system, you must also provide its appropriate metric to agree on the
> same geometry. Again, this logic should be so basic, but you and
> other physicists have so much trouble grasping. <shrug>
That was the exact debate that raged on until the 1960s until Kruskal
found the coordinates that were named after him. As usual, you are
ignorant about the history of the subjects you argue.
Eric Gisse
[...]
>
> If that's what you mean, then I think you are wrong.
>From the reference frame of having your head up your ass?
>
> Regards
> Roland
Joseph Lazio
RP> On Jun 26, 5:55 pm, Craig Markwardt
RP> <craigm...@REMOVEcow.physics.wisc.edu> wrote:
>> Roland PJ <rolan...@gmail.com> writes:
>> >http://en.wikipedia.org/wiki/Sagittarius_A%2A
>>
>>> "This is compatible with, and strong evidence in support of, the
>>> hypothesis that Sagittarius A* is a supermassive black
>>> hole. While it is possible for the observed mass within the
>>> observed volume limit to be distributed among multiple objects,
>>> any such objects would undergo orbital collapse into a single
>>> black hole anyway within a few hundred years at most, [...].
>>
>>> This is surely confusion on the part of the author between the
>>> proper time at Sagittarius A* and observer time on earth.
>> But that is not really relevant. Whether or not the light could
>> escape in a certain amount of time, the point of the statement in
>> the article is that the collapse/coalescence timescale based on
>> known physics is far shorter than the lifetime of the galaxy, thus
>> increasing the chances of a black hole being present.
RP> Suppose you hopped on-board a space-ship, and managed to stay
RP> alive for the long journey into the centre of the galaxy. You
RP> could take ever-more accurate observations of the objects around
RP> Sagitarius A* as you got closer and closer. But you will never
RP> observe objects that began outside the event horizon to have
RP> crossed into the black hole.
Define "observe."
This exact problem is treated in _Gravitation_ (MTW). They show that
the luminosity of an object falling into a black hole dims
exponentially fast. For an object falling into Sgr A* (with a mass of
about 4E6 solar masses), the e-folding time is about 100 s. After
about 10 min., the infalling object will be only 0.1% as bright as it
initially was. Note that time here is defined in terms of a distant,
external observer. In a very short time, the gravitational redshift
from the black hole will cause any infalling object to disappear from
view.
--
Lt. Lazio, HTML police | e-mail: jla...@patriot.net
No means no, stop rape. | http://patriot.net/%7Ejlazio/
sci.astro FAQ at http://sciastro.astronomy.net/sci.astro.html
Koobee Wublee
> On Jun 28, 8:45 am, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > Yeah seriously, you don't know both Kepler's laws of Newton's law of
> > gravity. <shrug>
>
> Kepler came first, dipshit.
>
> Kepler formulated his laws from the data from Tycho. Newton only
> formulated his inverse square law _after_ Kepler and Tycho.
Stop steering the subject away. You are the one claiming that the
mass of the gravitating body can be determined just by examining the
orbit using only Kepler's laws.
> > Oh yeah! I am still laughing.
At your ignorance. It is no wonder you are a super senior for several
years already. <shrug>
> > One more such solution that manifests a black hole is the following.
>
> > ds^2 = c^2 (1 - U)^2 dt^2 - dr^2 / (1 - U)^2 / (1 - 2 U)^2
> > - r^2 dO^2 / (1 - 2 U)^2
>
> dr^2 / (1 - U)^2 / (1 - 2 U)^2 = dr^2, moron.
You are just too ignorant. <shrug>
> > Where
>
> > ** U = G M / c^2 / r
>
> > And of course, there are still an infinite others out there. There
> > are also an infinite numbers out that do not manifest any black
> > holes. <shrug>
>
> 1) You haven't demonstrated that the metric this line element
> represents is a solution to the vacuum field equations.
So, it is. <shrug>
> 2) You haven't explained the meanings of the coordinate variables and
> their domains.
Need I to? It should be so self-evident. <shrug>
> 3) You haven't demonstrated that there is an event horizon, and that
> it is a global phenomena rather than an aspect of the coordinates.
I never claimed the event horizon is an aspect of the coordinates.
<shrug>
> Pulling a metric out of your butt just isn't acceptable.
That I did not. <shrug>
> > Well, I won't argue about Kelper's laws anymore. You need to
> > understand them first.
>
> No, YOU need to understand them first. You asked how to determine the
> mass of an orbited body through Kepler's laws, and I gave you a link
> that had the goddamn formula.
Kepler's three laws do not contain any mass information. <shrug>
> > This is your mistake. No one is talking about transformation in the
> > sense as what you are doing here. <shrug>
>
> Actually, it is yours. That is my entire point. That is always the
> point.
You are so mentally challenged. Again, that is why you remain a super
senior. <shrug>
> > And I have explained to you many times before that what I have done is [not]
> > a transformation of what you are thinking.
>
> DUH
I meant [not]. <shrug>
> Any two metrics that can be related by a coordinate transformation are
> the _SAME_. Welcome to introductory tensor analysis - Enjoy your stay.
This is not true. Remember in order to get to the field equations you
must have already on a particular set of coordinate system. If you
want to change the coordinate system, you need to start over again.
Go back to deriving the Riemann curvature tensor. <shrug>
> > Done that. <shrug>
>
> OK. Write down the tensor transformation law.
Look it up in your textbook.
> > Again, I stand by what I have said before. As an observer using a
> > coordinate system of (t, r, O), he is going to observe the surface
> > area of a sphere to be (4 pi R^2) where R is the radius determined or
> > measured from his coordinate system. This is so regardless what the
> > metric is.
>
> This is most definitely not true, as proven by explicit calculation.
> It isn't 4piR^2 for r' = r - K. It isn't for r' = exp(r). It isn't for
> r' = r + K.
You are still clueless. You need to go back to study the meaning of
coordinate system according to an observer. <shrug>
That disease of Mr. McCullough seems to be really contagious. First,
it infected Mr. Bielawski. Then, you. After I came down hard on
those two gentlemen, they backed off from that. It looks like you are
the one holding the bag of the garbage started with Mr. McCullough.
> I am 8 classes [5 humanities, 3 physics] away from a full Bachelors of
> Science in Physics. As usual, you say something stupid then punctuate
> it with <shrug>.
We will see you back as a super senior again. <shrug>
> Where was _YOUR_ alma mater? What was _YOUR_ degree in? It sure as
> hell wasn't in physics.
It does not concern you. <shrug>
> > On the contrary, you are the one denying mathematics. I have shown
> > you many times over the mathematics involved therein. <shrug>
>
> Writing down a line element and screaming "DON'T YOU SEE?!!?!?" is not
> mathematics.
Well, you have to do your diligence to show the metric I have given is
indeed a solution to the field equations. Remember that you are the
one still trying to learn GR. <shrug>
> That is the whole sum of your mathematics contributions
> to date - you have never went further than writing down a line element
> you think is relevant to your problem.
You got to be joking. In a newsgroup like thiat?
> You have not once calculated one of the curvature quantities that you
> claim prove what you were saying. Not once have you calculated area
> using your metrics. Not once have you even written down an integral to
> support your statements.
Actually I did in a more general way. Just look it up. Or have your
buddy moortel do that for you.
> Whenever confronted with your dishonest behavior, all you can do is
> meekly bleat that the poster should "search my past postings" in the
> vain hope that there exists something, somewhere that supports what
> you are saying.
I am telling the truth though. You are just too ignorant and too lazy
to do any work. Again, it is no wonder that you are still a super
senior. <shrug>
> > Given the Schwarzschild metric, the event horizon is invariant. This
> > is no brainer because it is the property of the geometry itself.
>
> Is it? You can only say that with the certainty of hindsight because
> you are certainly incapable of proving it. I could write the
> Schwarzschild metric in an exotic set of coordinates and you would
> have _NO WAY_ of telling me whether it *was* the Schwarzschild metric,
> much less where the event horizon is.
No, no, no. The Schwarzschild metric is only valid with the
spherically symmetric polar coordinate system. Any of its sister
solutions to the same field equations are also only valid to the same
set of coordinate system. Since each metric or solution is different
from each other, using the same set of coordinate system each metric
must describe a different and very independent geometry. I have no
faith that you can understand this.
> > Remember that the geometry must be invariant regardless of any
> > observers. On the other hand, it is impossible to completely describe
> > the geometry without using a set of coordinate system. Doing so, an
> > appropriate metric must be utilized. Using another set of coordinate
> > system, you must also provide its appropriate metric to agree on the
> > same geometry. Again, this logic should be so basic, but you and
> > other physicists have so much trouble grasping. <shrug>
>
> That was the exact debate that raged on until the 1960s until Kruskal
> found the coordinates that were named after him. As usual, you are
> ignorant about the history of the subjects you argue.
Who cares about another set of coordinate system? I certainly don't.
Craig Markwardt
Joseph Lazio <jla...@adams.patriot.net> writes:
Joe, "Roland PJ" is disputing the wikipedia article, which claims
there "is" a single black hole in the galactic center, on the basis
that today, we could never observe the event horizon of the black hole
because of infinite time dilation.
However, the argument for a single black hole is based on the fact
that the dynamical lifetime of such a compact and massive binary
system would be very short due to gravitational radiation. Nothing
about the time dilation of the event horizon or the "blackness" of a
black hole is involved. For this reason, "Roland PJ's" argument
doesn't really connect.
Craig
--
--------------------------------------------------------------------------
Craig B. Markwardt, Ph.D. EMAIL: crai...@REMOVEcow.physics.wisc.edu
Astrophysics, IDL, Finance, Derivatives | Remove "net" for better response
--------------------------------------------------------------------------
Eric Gisse
> On Jun 28, 1:51 pm, Eric Gisse <jowr...@gmail.com> wrote:
>
> > On Jun 28, 8:45 am, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > > Yeah seriously, you don't know both Kepler's laws of Newton's law of
> > > gravity. <shrug>
>
> > Kepler came first, dipshit.
>
> > Kepler formulated his laws from the data from Tycho. Newton only
> > formulated his inverse square law _after_ Kepler and Tycho.
>
> Stop steering the subject away. You are the one claiming that the
> mass of the gravitating body can be determined just by examining the
> orbit using only Kepler's laws.
Kepler's 3rd law, specifically.
>
> > > Oh yeah! I am still laughing.
>
> At your ignorance. It is no wonder you are a super senior for several
> years already. <shrug>
I was only a senior as of the end of the spring semester. I was pretty
close at the end of fall though - 5 credits.
>
> > > One more such solution that manifests a black hole is the following.
>
> > > ds^2 = c^2 (1 - U)^2 dt^2 - dr^2 / (1 - U)^2 / (1 - 2 U)^2
> > > - r^2 dO^2 / (1 - 2 U)^2
>
> > dr^2 / (1 - U)^2 / (1 - 2 U)^2 = dr^2, moron.
>
> You are just too ignorant. <shrug>
>
> > > Where
>
> > > ** U = G M / c^2 / r
>
> > > And of course, there are still an infinite others out there. There
> > > are also an infinite numbers out that do not manifest any black
> > > holes. <shrug>
>
> > 1) You haven't demonstrated that the metric this line element
> > represents is a solution to the vacuum field equations.
>
> So, it is. <shrug>
Assertion is different from demonstration. You _assert_ that it is,
but you have not _shown_ that it is.
>
> > 2) You haven't explained the meanings of the coordinate variables and
> > their domains.
>
> Need I to? It should be so self-evident. <shrug>
Of course. What is the domain of r?
>
> > 3) You haven't demonstrated that there is an event horizon, and that
> > it is a global phenomena rather than an aspect of the coordinates.
>
> I never claimed the event horizon is an aspect of the coordinates.
> <shrug>
Can't have a black hole without an event horizon.
>
> > Pulling a metric out of your butt just isn't acceptable.
>
> That I did not. <shrug>
Then where did it come from?
>
> > > Well, I won't argue about Kelper's laws anymore. You need to
> > > understand them first.
>
> > No, YOU need to understand them first. You asked how to determine the
> > mass of an orbited body through Kepler's laws, and I gave you a link
> > that had the goddamn formula.
>
> Kepler's three laws do not contain any mass information. <shrug>
http://en.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion#Kepler.27s_third_law
[...]
>
> > Any two metrics that can be related by a coordinate transformation are
> > the _SAME_. Welcome to introductory tensor analysis - Enjoy your stay.
>
> This is not true. Remember in order to get to the field equations you
> must have already on a particular set of coordinate system. If you
> want to change the coordinate system, you need to start over again.
Yes, it is true.
It is _basic_ tensor analysis. This is the same mathematics that is
used in stress analysis in classical physics, as well as the covariant
formalism of classical electrodynamics.
> Go back to deriving the Riemann curvature tensor. <shrug>
You continue to misunderstand the geometric meaning of a tensor. If I
can write down a coordinate transformation that relates two tensors in
different coordinate representations, they are the same tensor.
Christ, this is basic stuff that is proven in explicit detail.
>
> > > Done that. <shrug>
>
> > OK. Write down the tensor transformation law.
>
> Look it up in your textbook.
Or you could have simply said "I don't know", the effect would be the
same.
I'll make it _very_ simple for you: (???) * T * (???) = T, where T is
the tensor.
What are the (???) terms?
>
> > > Again, I stand by what I have said before. As an observer using a
> > > coordinate system of (t, r, O), he is going to observe the surface
> > > area of a sphere to be (4 pi R^2) where R is the radius determined or
> > > measured from his coordinate system. This is so regardless what the
> > > metric is.
>
> > This is most definitely not true, as proven by explicit calculation.
> > It isn't 4piR^2 for r' = r - K. It isn't for r' = exp(r). It isn't for
> > r' = r + K.
>
> You are still clueless. You need to go back to study the meaning of
> coordinate system according to an observer. <shrug>
How about giving me a literature reference that supports your point of
view?
Or, better yet, have your testicles drop and engage me in actual
debate using the very mathematics you claim you accept.
>
> That disease of Mr. McCullough seems to be really contagious. First,
> it infected Mr. Bielawski. Then, you. After I came down hard on
> those two gentlemen, they backed off from that. It looks like you are
> the one holding the bag of the garbage started with Mr. McCullough.
...because I'm the only one who can put up with your intellectual
dishonesty and general bullshit.
>
> > I am 8 classes [5 humanities, 3 physics] away from a full Bachelors of
> > Science in Physics. As usual, you say something stupid then punctuate
> > it with <shrug>.
>
> We will see you back as a super senior again. <shrug>
How much do you want to bet?
>
> > Where was _YOUR_ alma mater? What was _YOUR_ degree in? It sure as
> > hell wasn't in physics.
>
> It does not concern you. <shrug>
What, you can dish it out but you can't take it? What are you afraid
of me finding out? That maybe you don't actually have any background
in physics or mathematics and have been bullshitting all along?
>
> > > On the contrary, you are the one denying mathematics. I have shown
> > > you many times over the mathematics involved therein. <shrug>
>
> > Writing down a line element and screaming "DON'T YOU SEE?!!?!?" is not
> > mathematics.
>
> Well, you have to do your diligence to show the metric I have given is
> indeed a solution to the field equations. Remember that you are the
> one still trying to learn GR. <shrug>
Pretty hard to do that when you horribly mangled writing it down - you
have dr^2 / (1 - U)^2 / (1 - 2 U)^2 which is either dr^2 or dr^2 / (1-
U)^4, depending how much you felt like abusing ACSII convention.
>
> > That is the whole sum of your mathematics contributions
> > to date - you have never went further than writing down a line element
> > you think is relevant to your problem.
>
> You got to be joking. In a newsgroup like thiat?
You think highly enough to post in it year after year. Why not support
your arguments?
>
> > You have not once calculated one of the curvature quantities that you
> > claim prove what you were saying. Not once have you calculated area
> > using your metrics. Not once have you even written down an integral to
> > support your statements.
>
> Actually I did in a more general way. Just look it up. Or have your
> buddy moortel do that for you.
I got a better idea - YOU look it up.
>
> > Whenever confronted with your dishonest behavior, all you can do is
> > meekly bleat that the poster should "search my past postings" in the
> > vain hope that there exists something, somewhere that supports what
> > you are saying.
>
> I am telling the truth though. You are just too ignorant and too lazy
> to do any work. Again, it is no wonder that you are still a super
> senior. <shrug>
Lazy yes, ignorant no.
>
> > > Given the Schwarzschild metric, the event horizon is invariant. This
> > > is no brainer because it is the property of the geometry itself.
>
> > Is it? You can only say that with the certainty of hindsight because
> > you are certainly incapable of proving it. I could write the
> > Schwarzschild metric in an exotic set of coordinates and you would
> > have _NO WAY_ of telling me whether it *was* the Schwarzschild metric,
> > much less where the event horizon is.
>
> No, no, no. The Schwarzschild metric is only valid with the
> spherically symmetric polar coordinate system. Any of its sister
> solutions to the same field equations are also only valid to the same
> set of coordinate system. Since each metric or solution is different
> from each other, using the same set of coordinate system each metric
> must describe a different and very independent geometry. I have no
> faith that you can understand this.
That's because it is bullshit that is unsupported by simple tensor
analysis. A coordinate change does _NOT_ change the fact it is a
solution to the field equations.
>
> > > Remember that the geometry must be invariant regardless of any
> > > observers. On the other hand, it is impossible to completely describe
> > > the geometry without using a set of coordinate system. Doing so, an
> > > appropriate metric must be utilized. Using another set of coordinate
> > > system, you must also provide its appropriate metric to agree on the
> > > same geometry. Again, this logic should be so basic, but you and
> > > other physicists have so much trouble grasping. <shrug>
>
> > That was the exact debate that raged on until the 1960s until Kruskal
> > found the coordinates that were named after him. As usual, you are
> > ignorant about the history of the subjects you argue.
>
> Who cares about another set of coordinate system? I certainly don't.
That's the point. It proves you wrong, and you just don't care.
George Dishman
> On Jun 28, 11:58 am, Eric Gisse <jowr...@gmail.com> wrote:
>> On Jun 28, 1:04 am, Roland PJ <rolan...@gmail.com> wrote:
>>
>>
>> > But, on the other hand, if we wait 100 years on earth, and take
>> > another look, _we_ still won't notice that much has changed.
>>
>> This planet is ~5 billion years old.
>>
>> Which do you think came first? The galaxy or the planet?
>
> Eric, when you say 'first', which reference frame are you using?
>
> I think today on earth will come 'before' event horizon formation, or
> binary black hole coalescence, for all reference frames external to
> the black hole(s). Right?
I think you are missing the point. Your argument
is that we don't see an event horizon but a
surface which is grossly red-shifted because the
infalling material takes an infinite time to
reach the horizon in our frame. A simple black
hole would have a spherical surface of that type.
Imagine two such objects in a tight binary
configuration. How long would it take them to
spiral together to the point where they touch
and merge into a single object of dumbbell shape?
They can "coalesce" into a single object in a
short time in our frame regardless of how long it
takes an event horizon to form.
HTH
George
Jeff Root
Joseph Lazio said:
> This exact problem is treated in _Gravitation_ (MTW). They
> show that the luminosity of an object falling into a black
> hole dims exponentially fast. For an object falling into
> Sgr A* (with a mass of about 4E6 solar masses), the e-folding
> time is about 100 s. After about 10 min., the infalling
> object will be only 0.1% as bright as it initially was.
> Note that time here is defined in terms of a distant,
> external observer. In a very short time, the gravitational
> redshift from the black hole will cause any infalling object
> to disappear from view.
Joe,
I haven't done any of the math, but I'm surprised that the
luminosity declines as slowly as you say. Not that the
difference between your numbers and my WAG is great: only
an order of magnitude or two. But my guess would be that
after a couple of seconds, you will see no more photons of
detectable energy. So if I jump into a black hole while
you watch from a safe distance with your super-duper-ultra-
mega-hyper-wideband telescope, you would see light from me
go from bright visible to dim radio in a couple of seconds.
However... You're talking about an object falling into a
black hole of 4 million solar masses. And I'm probably
thinking of falling into the black hole of a collapsed star
of only 4 solar masses. I'd disappear more quickly there,
wouldn't I, because of the far smaller radius?
-- Jeff, in Minneapolis
Koobee Wublee
> On Jun 28, 9:42 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > Stop steering the subject away. You are the one claiming that the
> > mass of the gravitating body can be determined just by examining the
> > orbit using only Kepler's laws.
>
> Kepler's 3rd law, specifically.
Kepler's third law does not address any mass involved. You need to
study Kepler's laws for a change. <shrug>
> > At your ignorance. It is no wonder you are a super senior for several
> > years already. <shrug>
>
> I was only a senior as of the end of the spring semester. I was pretty
> close at the end of fall though - 5 credits.
So, this is an excuse of a super, super senior. Yes, psychologically
you can forget yourself in everything and picture yourself in high
esteem if you ever try hard enough. <shrug>
> 1) You haven't demonstrated that the metric this line element
> represents is a solution to the vacuum field equations.
>
> > So, it is. <shrug>
>
> Assertion is different from demonstration. You _assert_ that it is,
> but you have not _shown_ that it is.
Not in any newsgroup. If you do your diligence, you will find out
that it is. <shrug>
> 2) You haven't explained the meanings of the coordinate variables and
> their domains.
>
> > Need I to? It should be so self-evident. <shrug>
>
> Of course. What is the domain of r?
Zero to infinity, of course as always.
> Can't have a black hole without an event horizon.
I agree. So, what is your objection?
> Pulling a metric out of your butt just isn't acceptable.
>
> > That I did not. <shrug>
>
> Then where did it come from?
Through solving the field equations, hello!!! <shrug>
> > Kepler's three laws do not contain any mass information. <shrug>
>
> http://en.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion#Kepler...
Only when referring to the Newtonian law of gravity that it does, and
Kepler's laws are so much older than the Newtonian law of gravity.
<shrug>
> > This is not true. Remember in order to get to the field equations you
> > must have already on a particular set of coordinate system. If you
> > want to change the coordinate system, you need to start over again.
>
> It is _basic_ tensor analysis. This is the same mathematics that is
> used in stress analysis in classical physics, as well as the covariant
> formalism of classical electrodynamics.
The application is drastically different. Now, shall go into fluid
mechanics? <shrug>
> > Go back to deriving the Riemann curvature tensor. <shrug>
>
> You continue to misunderstand the geometric meaning of a tensor. If I
> can write down a coordinate transformation that relates two tensors in
> different coordinate representations, they are the same tensor.
> Christ, this is basic stuff that is proven in explicit detail.
The concept of a tensor is a misnomer. <shrug>
> I'll make it _very_ simple for you: (???) * T * (???) = T, where T is
> the tensor.
>
> What are the (???) terms?
1 <shrug>
> > You are still clueless. You need to go back to study the meaning of
> > coordinate system according to an observer. <shrug>
>
> How about giving me a literature reference that supports your point of
> view?
Go read my past posts. They are free. <shrug>
> Or, better yet, have your testicles drop and engage me in actual
> debate using the very mathematics you claim you accept.
I don't do homosexual activities. <shrug>
> > That disease of Mr. McCullough seems to be really contagious. First,
> > it infected Mr. Bielawski. Then, you. After I came down hard on
> > those two gentlemen, they backed off from that. It looks like you are
> > the one holding the bag of the garbage started with Mr. McCullough.
>
> ...because I'm the only one who can put up with your intellectual
> dishonesty and general bullshit.
That is because you are the most ignorant of the bunch. <shrug>
> > We will see you back as a super senior again. <shrug>
>
> How much do you want to bet?
Emperor's clothes.
> What, you can dish it out but you can't take it? What are you afraid
> of me finding out? That maybe you don't actually have any background
> in physics or mathematics and have been bullshitting all along?
Hey, you are the one who wants to play with my testicles, remember.
> > Well, you have to do your diligence to show the metric I have given is
> > indeed a solution to the field equations. Remember that you are the
> > one still trying to learn GR. <shrug>
>
> Pretty hard to do that when you horribly mangled writing it down - you
> have dr^2 / (1 - U)^2 / (1 - 2 U)^2 which is either dr^2 or dr^2 / (1-
> U)^4, depending how much you felt like abusing ACSII convention.
Well, sorry about that. Your concern is genuinely taken. I meant the
following.
(dr^2 / (1 - U)^2) / (1 - 2 U)^2
> > I am telling the truth though. You are just too ignorant and too lazy
> > to do any work. Again, it is no wonder that you are still a super
> > senior. <shrug>
>
> Lazy yes, ignorant no.
Both.
> > No, no, no. The Schwarzschild metric is only valid with the
> > spherically symmetric polar coordinate system. Any of its sister
> > solutions to the same field equations are also only valid to the same
> > set of coordinate system. Since each metric or solution is different
> > from each other, using the same set of coordinate system each metric
> > must describe a different and very independent geometry. I have no
> > faith that you can understand this.
>
> That's because it is bullshit that is unsupported by simple tensor
> analysis. A coordinate change does _NOT_ change the fact it is a
> solution to the field equations.
There is no coordinate change in the sense that you are portraying it
to be. <shrug>
> > Who cares about another set of coordinate system? I certainly don't.
>
> That's the point. It proves you wrong, and you just don't care.
No, it doesn't. To work with another set of coordinate system, you
need to go back to the Lagranagian of the Eisntein-Hilbert action.
<shrug>
Eric Gisse
> On Jun 29, 1:36 am, Eric Gisse <jowr...@gmail.com> wrote:
>
> > On Jun 28, 9:42 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > > Stop steering the subject away. You are the one claiming that the
> > > mass of the gravitating body can be determined just by examining the
> > > orbit using only Kepler's laws.
>
> > Kepler's 3rd law, specifically.
>
> Kepler's third law does not address any mass involved. You need to
> study Kepler's laws for a change. <shrug>
Wrong, as usual. You can't even argue classical mechanics without
fucking up.
Pg. 135, _Mechanics_, 3rd. ed, Symon
Pg. 101, _Classical Mechanics_, 3rd. ed, Goldstein
http://en.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion#Kepler.27s_third_law
They all say the SAME GODDAMN THING:
\tau^2 = 4 pi^2 a^3 / G M
Here you are claiming you have studied Kepler's laws:
http://groups.google.com/group/sci.physics.relativity/msg/6d2b40a0776c8cf2?dmode=source
Lying? Stupid? Which is it..?
>
> > > At your ignorance. It is no wonder you are a super senior for several
> > > years already. <shrug>
>
> > I was only a senior as of the end of the spring semester. I was pretty
> > close at the end of fall though - 5 credits.
>
> So, this is an excuse of a super, super senior. Yes, psychologically
> you can forget yourself in everything and picture yourself in high
> esteem if you ever try hard enough. <shrug>
You say that, yet you post under a pseudonym...whose the one with low
self esteem again?
>
> > 1) You haven't demonstrated that the metric this line element
> > represents is a solution to the vacuum field equations.
>
> > > So, it is. <shrug>
>
> > Assertion is different from demonstration. You _assert_ that it is,
> > but you have not _shown_ that it is.
>
> Not in any newsgroup. If you do your diligence, you will find out
> that it is. <shrug>
http://img503.imageshack.us/img503/9130/negatoryup1.png
Apparently the one true guru of Riemann doesn't quite understand that
a vacuum solution required R_uv = 0.
[...]
Koobee Wublee
> On Jun 29, 8:38 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > Kepler's third law does not address any mass involved. You need to
> > study Kepler's laws for a change. <shrug>
>
> Wrong, as usual. You can't even argue classical mechanics without
> fucking up.
>
> Pg. 135, _Mechanics_, 3rd. ed, Symon
> Pg. 101, _Classical Mechanics_, 3rd. ed, Goldstein
> http://en.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion#Kepler...
Again, Kelper's laws came before Newton's law of gravity. Therefore,
you can never predict the amount of the gravitating mass under
Kepler's laws. <shrug>
> They all say the SAME GODDAMN THING:
>
> \tau^2 = 4 pi^2 a^3 / G M
This is Newton's law of gravity. <shrug>
> Here you are claiming you have studied Kepler's laws:
>
> http://groups.google.com/group/sci.physics.relativity/msg/6d2b40a0776...
>
> Lying? Stupid? Which is it..?
As I said, you need to study Kepler's laws as well as Newton's law of
gravity. Well, your idol Einstein did. After finally understood
Newton's law of gravity he came up with the equivalence principle to
brag about his feat. Well, Einstein was almost 300 years too late.
http://en.wikipedia.org/wiki/Galileo
> > So, this is an excuse of a super, super senior. Yes, psychologically
> > you can forget yourself in everything and picture yourself in high
> > esteem if you ever try hard enough. <shrug>
>
> You say that, yet you post under a pseudonym...whose the one with low
> self esteem again?
How do I know Eric Gisse is indeed your name? What is your social
security number? What is your passport number?
> > Not in any newsgroup. If you do your diligence, you will find out
> > that it is. <shrug>
>
> http://img503.imageshack.us/img503/9130/negatoryup1.png
>
> Apparently the one true guru of Riemann doesn't quite understand that
> a vacuum solution required R_uv = 0.
You are confusing R with r in your calculation, and I actually made a
mistake in calculation. Nobody is perfect. So, here is the correct
solution to the field equations using the common spherically symmetric
polar coordinate system. Just like the Schwarzschild metric, it also
exhibits black holes.
ds^2 = (1 - 2 G M / R) dt^2 - (dR/dr)^2 dr^2 / (1 - 2 G M / R) - R^2
dO^2
Where
** R = r / (1 - G M / r / 2)
** c^2 = 1
** A / B / C = A / (B C)
The equation above can be written as the following.
ds^2 = (1 - G M / r)^2 dt^2 - dr^2 / (1 - G M / r / 2)^4 - r^2 dO^2 /
(1 - G M / r / 2)^2
The event horizon occurs at (G M) not (2 G M).
At this point, we have gone far enough that I have no choice but to
finish it. Good job. You sucker me into doing your homework for you.
Eric Gisse
> On Jun 30, 1:46 am, Eric Gisse <jowr...@gmail.com> wrote:
>
> > On Jun 29, 8:38 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > > Kepler's third law does not address any mass involved. You need to
> > > study Kepler's laws for a change. <shrug>
>
> > Wrong, as usual. You can't even argue classical mechanics without
> > fucking up.
>
> > Pg. 135, _Mechanics_, 3rd. ed, Symon
> > Pg. 101, _Classical Mechanics_, 3rd. ed, Goldstein
> >http://en.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion#Kepler...
>
> Again, Kelper's laws came before Newton's law of gravity. Therefore,
> you can never predict the amount of the gravitating mass under
> Kepler's laws. <shrug>
Familiarize yourself with physics as formulated sometime in the past
400 years. Kepler's laws are 100% consistent with Newtonian
gravitation [in fact, Newton used Kepler's laws to derive the inverse
square law] and are contained wholly within. Which you would know if
you would read any of the references I gave you.
>
> > They all say the SAME GODDAMN THING:
>
> > \tau^2 = 4 pi^2 a^3 / G M
>
> This is Newton's law of gravity. <shrug>
No, it is not. It is Kepler's 3rd law, which is derived _from_
Newton's law of gravity.
>
> > Here you are claiming you have studied Kepler's laws:
>
> >http://groups.google.com/group/sci.physics.relativity/msg/6d2b40a0776...
>
> > Lying? Stupid? Which is it..?
>
> As I said, you need to study Kepler's laws as well as Newton's law of
> gravity. Well, your idol Einstein did. After finally understood
> Newton's law of gravity he came up with the equivalence principle to
> brag about his feat. Well, Einstein was almost 300 years too late.
I have studied both already. Thats why I can immediately call you upon
your ignorant bullshit rather than let it get swept under the rug.
>
> http://en.wikipedia.org/wiki/Galileo
>
> > > So, this is an excuse of a super, super senior. Yes, psychologically
> > > you can forget yourself in everything and picture yourself in high
> > > esteem if you ever try hard enough. <shrug>
>
> > You say that, yet you post under a pseudonym...whose the one with low
> > self esteem again?
>
> How do I know Eric Gisse is indeed your name? What is your social
> security number? What is your passport number?
:)
>
> > > Not in any newsgroup. If you do your diligence, you will find out
> > > that it is. <shrug>
>
> >http://img503.imageshack.us/img503/9130/negatoryup1.png
>
> > Apparently the one true guru of Riemann doesn't quite understand that
> > a vacuum solution required R_uv = 0.
>
> You are confusing R with r in your calculation, and I actually made a
> mistake in calculation. Nobody is perfect. So, here is the correct
> solution to the field equations using the common spherically symmetric
> polar coordinate system. Just like the Schwarzschild metric, it also
> exhibits black holes.
Coordinate labels are irrelevant. I felt like using R instead of r. I
could call it Q and nothing would change. I find it odd that you make
the elementary error of assuming that the label determines the
physics.
>
> ds^2 = (1 - 2 G M / R) dt^2 - (dR/dr)^2 dr^2 / (1 - 2 G M / R) - R^2
> dO^2
>
> Where
>
> ** R = r / (1 - G M / r / 2)
> ** c^2 = 1
> ** A / B / C = A / (B C)
Not a solution to the vacuum field equations, either. Oh guru of
Riemann, what part of "R_uv = 0 for vacuum" do you NOT understand?
http://img244.imageshack.us/img244/444/negatorypart2vr0.png
Isn't it a BITCH when your bullshit can be checked in less than 5
minutes of work?
>
> The equation above can be written as the following.
>
> ds^2 = (1 - G M / r)^2 dt^2 - dr^2 / (1 - G M / r / 2)^4 - r^2 dO^2 /
> (1 - G M / r / 2)^2
>
> The event horizon occurs at (G M) not (2 G M).
>
> At this point, we have gone far enough that I have no choice but to
> finish it. Good job. You sucker me into doing your homework for you.
Spending a few minutes entering your line element into maple and
typing grcalc(R(dn,dn)) then grdisplay(_) is barely homework.
Why does the one, true guru of Riemann struggle to give me an _actual_
solution to the vacuum field equations?
Koobee Wublee
> On Jun 30, 12:30 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > Again, Kelper's laws came before Newton's law of gravity. Therefore,
> > you can never predict the amount of the gravitating mass under
> > Kepler's laws. <shrug>
>
> Familiarize yourself with physics as formulated sometime in the past
> 400 years.
Yes, done that. <shrug>
> Kepler's laws are 100% consistent with Newtonian
> gravitation
This is not true. Kepler's laws do not deal with three bodies. On
the other hand, Newton's law of gravity predicted the existence of the
planets Neptune, Uranus, and Pluto. <shrug>
> [in fact, Newton used Kepler's laws to derive the inverse
> square law] and are contained wholly within.
No, Newton solely used Galileo's principles of equivalence and
relativity to derive the law of gravity.
> Which you would know if
> you would read any of the references I gave you.
I have other references. <shrug>
> \tau^2 = 4 pi^2 a^3 / G M
>
> > This is Newton's law of gravity. <shrug>
>
> No, it is not. It is Kepler's 3rd law, which is derived _from_
> Newton's law of gravity.
When Kepler died, Newton was not even born yet. There is no way that
Kepler's 3rd law being derived from Newton's law of gravity. However,
only after Newton's law of gravity, you can derive Kepler's 1st and
3rd laws. Our original discussion was that only through the Newtonian
law of gravity we can determine the gravitating mass. Remember?
Thus, Kepler's 3rd law does not apply here. <shrug>
> > As I said, you need to study Kepler's laws as well as Newton's law of
> > gravity. Well, your idol Einstein did. After finally understood
> > Newton's law of gravity he came up with the equivalence principle to
> > brag about his feat. Well, Einstein was almost 300 years too late.
>
> I have studied both already. Thats why I can immediately call you upon
> your ignorant bullshit rather than let it get swept under the rug.
You action have shown you are both ignorant in Kepler's laws as well
as Newton's law of gravity. <shrug>
> > You are confusing R with r in your calculation, and I actually made a
> > mistake in calculation. Nobody is perfect. So, here is the correct
> > solution to the field equations using the common spherically symmetric
> > polar coordinate system. Just like the Schwarzschild metric, it also
> > exhibits black holes.
>
> Coordinate labels are irrelevant.
Again, only after you have settled on your choice of coordinate
system, you can derive the Riemann curvature tensor, the Ricci
curvature tensor, the Ricci curvature scalar, the Lagrangian to the
Einstein-Hilbert Action whatever meaningful it is, the Einstein field
equations, and finally the metric itself.
> I felt like using R instead of r.
You can use whatever you want. In this case, R = r. <shrug>
> I
> could call it Q and nothing would change. I find it odd that you make
> the elementary error of assuming that the label determines the
> physics.
No, I did not. I did not even make the mistake of using the metric to
describe the geometry as yourself and the physicists since Ricci's
time. <shrug>
> > ds^2 = (1 - 2 G M / R) dt^2 - (dR/dr)^2 dr^2 / (1 - 2 G M / R) - R^2
> > dO^2
>
> > Where
>
> > ** R = r / (1 - G M / r / 2)
> > ** c^2 = 1
> > ** A / B / C = A / (B C)
>
> Not a solution to the vacuum field equations, either. Oh guru of
> Riemann, what part of "R_uv = 0 for vacuum" do you NOT understand?
>
> http://img244.imageshack.us/img244/444/negatorypart2vr0.png
>
> Isn't it a BITCH when your bullshit can be checked in less than 5
> minutes of work?
Well, you made a mistake copying down the equation. <shrug>
> > The equation above can be written as the following.
>
> > ds^2 = (1 - G M / r)^2 dt^2 - dr^2 / (1 - G M / r / 2)^4 - r^2 dO^2 /
> > (1 - G M / r / 2)^2
The dr^2 term has the denominator term to the order of 4 not 2 as you
have entered in your software. That is provided your software is
checked out. <shrug>
> > The event horizon occurs at (G M) not (2 G M).
>
> > At this point, we have gone far enough that I have no choice but to
> > finish it. Good job. You sucker me into doing your homework for you.
>
> Spending a few minutes entering your line element into maple and
> typing grcalc(R(dn,dn)) then grdisplay(_) is barely homework.
>
> Why does the one, true guru of Riemann struggle to give me an _actual_
> solution to the vacuum field equations?
After this one, there are infinite numbers of solutions.
Eric Gisse
> On Jun 30, 5:40 pm, Eric Gisse <jowr...@gmail.com> wrote:
>
> > On Jun 30, 12:30 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > > Again, Kelper's laws came before Newton's law of gravity. Therefore,
> > > you can never predict the amount of the gravitating mass under
> > > Kepler's laws. <shrug>
>
> > Familiarize yourself with physics as formulated sometime in the past
> > 400 years.
>
> Yes, done that. <shrug>
>
> > Kepler's laws are 100% consistent with Newtonian
> > gravitation
>
> This is not true. Kepler's laws do not deal with three bodies. On
> the other hand, Newton's law of gravity predicted the existence of the
> planets Neptune, Uranus, and Pluto. <shrug>
Lets see, is this intellectual dishonesty or stupidity? Kepler's laws
follow _FROM_ Newton's law of gravitation, not the other way around.
>
> > [in fact, Newton used Kepler's laws to derive the inverse
> > square law] and are contained wholly within.
>
> No, Newton solely used Galileo's principles of equivalence and
> relativity to derive the law of gravity.
Two textbooks on classical mechanics and Wikipedia explicitly disagree
with you.
>
> > Which you would know if
> > you would read any of the references I gave you.
>
> I have other references. <shrug>
Such as?
>
> > \tau^2 = 4 pi^2 a^3 / G M
>
> > > This is Newton's law of gravity. <shrug>
>
> > No, it is not. It is Kepler's 3rd law, which is derived _from_
> > Newton's law of gravity.
>
> When Kepler died, Newton was not even born yet. There is no way that
> Kepler's 3rd law being derived from Newton's law of gravity. However,
> only after Newton's law of gravity, you can derive Kepler's 1st and
> 3rd laws. Our original discussion was that only through the Newtonian
> law of gravity we can determine the gravitating mass. Remember?
> Thus, Kepler's 3rd law does not apply here. <shrug>
No, the original discussion was how to find the gravitating mass and I
said "Kepler's 3rd law". Then you alternated between being
intellectually dishonest and outright stupid.
>
> > > As I said, you need to study Kepler's laws as well as Newton's law of
> > > gravity. Well, your idol Einstein did. After finally understood
> > > Newton's law of gravity he came up with the equivalence principle to
> > > brag about his feat. Well, Einstein was almost 300 years too late.
>
> > I have studied both already. Thats why I can immediately call you upon
> > your ignorant bullshit rather than let it get swept under the rug.
>
> You action have shown you are both ignorant in Kepler's laws as well
> as Newton's law of gravity. <shrug>
Go ahead. Find one literature reference that disagrees with me. You
say you have other references - put up or shut up.
>
> > > You are confusing R with r in your calculation, and I actually made a
> > > mistake in calculation. Nobody is perfect. So, here is the correct
> > > solution to the field equations using the common spherically symmetric
> > > polar coordinate system. Just like the Schwarzschild metric, it also
> > > exhibits black holes.
>
> > Coordinate labels are irrelevant.
>
> Again, only after you have settled on your choice of coordinate
> system, you can derive the Riemann curvature tensor, the Ricci
> curvature tensor, the Ricci curvature scalar, the Lagrangian to the
> Einstein-Hilbert Action whatever meaningful it is, the Einstein field
> equations, and finally the metric itself.
Coordinate _LABELS_ are irrelevant. That is distinct from saying the
_COORDINATES_ are irrelevant.
Furthermore, the Riemann curvature tensor and its contractions are
invariant by virtue of being tensorial quantities. So is the action,
since it is constructed from metric and its derivatives.
>
> > I felt like using R instead of r.
>
> You can use whatever you want. In this case, R = r. <shrug>
>
> > I
> > could call it Q and nothing would change. I find it odd that you make
> > the elementary error of assuming that the label determines the
> > physics.
>
> No, I did not. I did not even make the mistake of using the metric to
> describe the geometry as yourself and the physicists since Ricci's
> time. <shrug>
The metric _IS_ the geometry. I need no other information to
completely specify how particles will behave on that manifold.
>
>
>
> > > ds^2 = (1 - 2 G M / R) dt^2 - (dR/dr)^2 dr^2 / (1 - 2 G M / R) - R^2
> > > dO^2
>
> > > Where
>
> > > ** R = r / (1 - G M / r / 2)
> > > ** c^2 = 1
> > > ** A / B / C = A / (B C)
>
> > Not a solution to the vacuum field equations, either. Oh guru of
> > Riemann, what part of "R_uv = 0 for vacuum" do you NOT understand?
>
> >http://img244.imageshack.us/img244/444/negatorypart2vr0.png
>
> > Isn't it a BITCH when your bullshit can be checked in less than 5
> > minutes of work?
>
> Well, you made a mistake copying down the equation. <shrug>
Ruh roh.
>
> > > The equation above can be written as the following.
>
> > > ds^2 = (1 - G M / r)^2 dt^2 - dr^2 / (1 - G M / r / 2)^4 - r^2 dO^2 /
> > > (1 - G M / r / 2)^2
>
> The dr^2 term has the denominator term to the order of 4 not 2 as you
> have entered in your software. That is provided your software is
> checked out. <shrug>
A quick re-roll into grtensor showed that you are, in fact, correct.
It does satisfy R_uv = 0. Furthermore, a few minutes of computation
revealed that it is the Schwarzschild metric in different coordinates.
>
> > > The event horizon occurs at (G M) not (2 G M).
But if you plug R=GM into the coordinate transform you used, you find
out that the event horizon is in fact at r = 2GM as expected.
This is what happens when you play fast and loose with coordinates.
>
> > > At this point, we have gone far enough that I have no choice but to
> > > finish it. Good job. You sucker me into doing your homework for you.
>
> > Spending a few minutes entering your line element into maple and
> > typing grcalc(R(dn,dn)) then grdisplay(_) is barely homework.
>
> > Why does the one, true guru of Riemann struggle to give me an _actual_
> > solution to the vacuum field equations?
>
> After this one, there are infinite numbers of solutions.
...that describe the same solution. Remember Birkhoff's theorem?
Apply the coordinate transform r = 2*R^2/(2*R-G*M) to Schwarzschild,
and the "new" metric is obtained.
ds^2 = dx^2 + dy^2 + dz^2
ds^2 = dr^2 + r^2[ d\theta^2 + sin(\theta^2)d\phi^2 ]
ds^2 = da^2 + db^2 + dc^2
ds^2 = dr^2 + r^2d\theta^2 + dz^2
Same manifold, different projections of the metric on different
coordinate axes.
What's old is new. You are making the same mistakes as you were making
in April.
http://groups.google.com/group/sci.physics.relativity/msg/1418eb64933c3545?dmode=source
You are now gradually changing the topics of our discussion. Obviously
if one has two different "metrics" as you call them (meaning, two
different matrices of g_ij's) written wrt one and the same coordinate
system, then they represent different geometries.
This was never under the discussion.
What we are debating is your claim that there are infinitely many
solutions to EFE in the spherically symmetric vacuum case. You gave
several examples of different line elements claiming they were
physically different solutions. These line elements differed by a
coordinate change - so what you said above does not apply. Nobody here
ever claimed than in a _fixed coordinate system_ different line
elements yielded the same geometry.
Koobee Wublee
On Jun 30, 11:02 pm, Eric Gisse <jowr...@gmail.com> wrote:
> On Jun 30, 9:01 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > This is not true. Kepler's laws do not deal with three bodies. On
> > the other hand, Newton's law of gravity predicted the existence of the
> > planets Neptune, Uranus, and Pluto. <shrug>
>
> Lets see, is this intellectual dishonesty or stupidity? Kepler's laws
> follow _FROM_ Newton's law of gravitation, not the other way around.
During Kepler's time, Newtonian law of gravity was not yet
discovered. There is no way in hell that you can determine the
gravitating mass through any of Kepler's laws. What you have posted
is Newton's law PERIOD.
> > No, Newton solely used Galileo's principles of equivalence and
> > relativity to derive the law of gravity.
>
> Two textbooks on classical mechanics and Wikipedia explicitly disagree
> with you.
So, they are wrong. <shrug>
> > When Kepler died, Newton was not even born yet. There is no way that
> > Kepler's 3rd law being derived from Newton's law of gravity. However,
> > only after Newton's law of gravity, you can derive Kepler's 1st and
> > 3rd laws. Our original discussion was that only through the Newtonian
> > law of gravity we can determine the gravitating mass. Remember?
> > Thus, Kepler's 3rd law does not apply here. <shrug>
>
> No, the original discussion was how to find the gravitating mass and I
> said "Kepler's 3rd law". Then you alternated between being
> intellectually dishonest and outright stupid.
No, the original discussion was how to determine the gravitating mass,
and I said "Newton's law of gravity". <shrug>
> > You action have shown you are both ignorant in Kepler's laws as well
> > as Newton's law of gravity. <shrug>
>
> Go ahead. Find one literature reference that disagrees with me. You
> say you have other references - put up or shut up.
We know Kepler's laws cannot be right. Therefore, you are totally
wrong along with all these infinite numbers of textbooks you possess.
You should consider returning them for a full refund on each. <shrug>
> > Again, only after you have settled on your choice of coordinate
> > system, you can derive the Riemann curvature tensor, the Ricci
> > curvature tensor, the Ricci curvature scalar, the Lagrangian to the
> > Einstein-Hilbert Action whatever meaningful it is, the Einstein field
> > equations, and finally the metric itself.
>
> Coordinate _LABELS_ are irrelevant. That is distinct from saying the
> _COORDINATES_ are irrelevant.
So, I challenge you to describe a geometry without using any
coordinate system. <shrug>
> Furthermore, the Riemann curvature tensor and its contractions are
> invariant by virtue of being tensorial quantities.
This is totally wrong. Since all your textbooks and physicists have
been wrong, I very humbly describe myself as the only one since
Riemann to have understood the curvature business. <shrug>
> So is the action,
> since it is constructed from metric and its derivatives.
Since the geometry must be invariant, the metric describing this
geometry must be only unique to the choice of coordinate system. In
doing so, the so called "tensors" which are merely matrices
constructed from the metrics and their derivatives must also be unique
to the initial choice of coordinate system. Just what logic do you
not understand here?
> > No, I did not. I did not even make the mistake of using the metric to
> > describe the geometry as yourself and the physicists since Ricci's
> > time. <shrug>
>
> The metric _IS_ the geometry.
The metric cannot be the geometry because you cannot describe the
geometry without any coordinate system. <shrug>
> I need no other information to
> completely specify how particles will behave on that manifold.
Without a coordinate system, you cannot specify how fast the particles
are moving. <shrug>
> > ds^2 = (1 - G M / r)^2 dt^2 - dr^2 / (1 - G M / r / 2)^4 - r^2 dO^2 /
> > (1 - G M / r / 2)^2
>
> > The dr^2 term has the denominator term to the order of 4 not 2 as you
> > have entered in your software. That is provided your software is
> > checked out. <shrug>
>
> A quick re-roll into grtensor showed that you are, in fact, correct.
> It does satisfy R_uv = 0.
Yes, I told you so.
> Furthermore, a few minutes of computation
> revealed that it is the Schwarzschild metric in different coordinates.
What coordinate system would that be? Remember one shows the event
horizon at (G M) and the other one at (2 G M).
> > The event horizon occurs at (G M) not (2 G M).
>
> But if you plug R=GM into the coordinate transform you used, you find
> out that the event horizon is in fact at r = 2GM as expected.
There is no coordinate transformation as what you are leading to.
<shrug>
The choice of coordinate system is already cast in concrete that is
the common spherically symmetric coordinate system.
> This is what happens when you play fast and loose with coordinates.
Yes, you have lost the original choice of coordinate system that give
rise to the Riemann curvature tensor, the Ricci curvature tensor, the
Ricci scalar, the field equations, and finally the metric itself.
> > After this one, there are infinite numbers of solutions.
>
> ...that describe the same solution. Remember Birkhoff's theorem?
Birkhoff's theorem is wrong. We have discussed that already.
> Apply the coordinate transform r = 2*R^2/(2*R-G*M) to Schwarzschild,
> and the "new" metric is obtained.
>
> ds^2 = dx^2 + dy^2 + dz^2
The coordinate system is (x, y, z), and the metric is
[1, 0, 0]
[0, 1, 0]
[0, 0, 1]
> ds^2 = dr^2 + r^2[ d\theta^2 + sin(\theta^2)d\phi^2 ]
The coordinate system is (r, \phi, \theta), and the metric is
[1, 0, 0]
[0, r^2 sin^2(\theta), 0]
[0, 0, r^2]
> ds^2 = da^2 + db^2 + dc^2
The coordinate system is (a, b, c), and the metric is
[1, 0, 0]
[0, 1, 0]
[0, 0, 1]
> ds^2 = dr^2 + r^2d\theta^2 + dz^2
The coordinate system is (r, \theta, z), and the metric is
[1, 0, 0]
[0, r^2, 0]
[0, 0, 1]
Notice all these metrics are different. If they all describe the same
geometry, then the metric is unique for each choice of coordinate
system as I have trying to tell you. <shrug>
> Same manifold, different projections of the metric on different
> coordinate axes.
The sentence above can only mean what I have been saying.
> What's old is new. You are making the same mistakes as you were making
> in April.
>
> http://groups.google.com/group/sci.physics.relativity/msg/1418eb64933...
This is not a mistake. I stand by as what I wrote.
> You are now gradually changing the topics of our discussion. Obviously
> if one has two different "metrics" as you call them (meaning, two
> different matrices of g_ij's) written wrt one and the same coordinate
> system, then they represent different geometries.
Well, I have not changed the topic, and this is exactly what I have
been talking about.
> This was never under the discussion.
You are a liar.
> What we are debating is your claim that there are infinitely many
> solutions to EFE in the spherically symmetric vacuum case.
Yes, my claim is the only valid one that still eludes all physicists.
That is why I have very humbly stated that I am the only one to have
understand this curvature business since Riemann. <shrug>
> You gave
> several examples of different line elements claiming they were
> physically different solutions.
Yes, you have verified them.
> These line elements differed by a
> coordinate change
They seem to be coordinate change to you because they are all
solutions that satisfy the field equations. All solutions to the
field equations have such unique commonality, and you should construed
them as all describing the same geometry.
> - so what you said above does not apply.
You are wrong, here.
> Nobody here
> ever claimed than in a _fixed coordinate system_ different line
> elements yielded the same geometry.
No, you have. The purpose of your discussion is to matheMagically
change the coordinate system to allow different metrics to describe
the same geometry. You are the biggest liar. <shrug>
Eric Gisse
> On Jun 30, 11:02 pm, Eric Gisse <jowr...@gmail.com> wrote:
>
> > On Jun 30, 9:01 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > > This is not true. Kepler's laws do not deal with three bodies. On
> > > the other hand, Newton's law of gravity predicted the existence of the
> > > planets Neptune, Uranus, and Pluto. <shrug>
>
> > Lets see, is this intellectual dishonesty or stupidity? Kepler's laws
> > follow _FROM_ Newton's law of gravitation, not the other way around.
>
> During Kepler's time, Newtonian law of gravity was not yet
> discovered. There is no way in hell that you can determine the
> gravitating mass through any of Kepler's laws. What you have posted
> is Newton's law PERIOD.
Consult any classical mechanics text published EVER. Your brand of
history is not accepted.
Isn't it odd that practically every opinion you hold is at odds with
conventional wisdom in a technical field?
>
> > > No, Newton solely used Galileo's principles of equivalence and
> > > relativity to derive the law of gravity.
>
> > Two textbooks on classical mechanics and Wikipedia explicitly disagree
> > with you.
>
> So, they are wrong. <shrug>
Not only does Goldstein and Symon make the same identification, but so
does Landau & Lifshitz.
The physics community has equated the original formulation of Kepler's
laws with their more exact counterparts derived via Newtonian
gravitation. There is no point in whining because there is nothing you
can do to change things.
I find it far more illuminating that you disagree with every aspect of
physics.
>
> > > When Kepler died, Newton was not even born yet. There is no way that
> > > Kepler's 3rd law being derived from Newton's law of gravity. However,
> > > only after Newton's law of gravity, you can derive Kepler's 1st and
> > > 3rd laws. Our original discussion was that only through the Newtonian
> > > law of gravity we can determine the gravitating mass. Remember?
> > > Thus, Kepler's 3rd law does not apply here. <shrug>
>
> > No, the original discussion was how to find the gravitating mass and I
> > said "Kepler's 3rd law". Then you alternated between being
> > intellectually dishonest and outright stupid.
>
> No, the original discussion was how to determine the gravitating mass,
> and I said "Newton's law of gravity". <shrug>
>From which Kepler's 3rd law, the specific tool you would use, is
derived.
>
> > > You action have shown you are both ignorant in Kepler's laws as well
> > > as Newton's law of gravity. <shrug>
>
> > Go ahead. Find one literature reference that disagrees with me. You
> > say you have other references - put up or shut up.
>
> We know Kepler's laws cannot be right. Therefore, you are totally
> wrong along with all these infinite numbers of textbooks you possess.
> You should consider returning them for a full refund on each. <shrug>
>
> > > Again, only after you have settled on your choice of coordinate
> > > system, you can derive the Riemann curvature tensor, the Ricci
> > > curvature tensor, the Ricci curvature scalar, the Lagrangian to the
> > > Einstein-Hilbert Action whatever meaningful it is, the Einstein field
> > > equations, and finally the metric itself.
>
> > Coordinate _LABELS_ are irrelevant. That is distinct from saying the
> > _COORDINATES_ are irrelevant.
>
> So, I challenge you to describe a geometry without using any
> coordinate system. <shrug>
Try reading for comprehension.
It does _NOT_ matter if you call the Cartesian x, y, z coordinates a,
b, c or x_1, x_2, or x_3. Your failure to understand this means you
don't even understand the Lagrangian formalism of classical mechanics
because there is no way you could grasp the meaning of generalized
coordinates with this mental fixation on labels.
>
> > Furthermore, the Riemann curvature tensor and its contractions are
> > invariant by virtue of being tensorial quantities.
>
> This is totally wrong. Since all your textbooks and physicists have
> been wrong, I very humbly describe myself as the only one since
> Riemann to have understood the curvature business. <shrug>
Yet you struggle to compute anything or come up with even one example
that supports any of your assertions. Isn't that interesting?
>
> > So is the action,
> > since it is constructed from metric and its derivatives.
>
> Since the geometry must be invariant, the metric describing this
> geometry must be only unique to the choice of coordinate system. In
> doing so, the so called "tensors" which are merely matrices
> constructed from the metrics and their derivatives must also be unique
> to the initial choice of coordinate system. Just what logic do you
> not understand here?
Tensors aren't matrices, dipshit. How many times does that need to be
explained to you before it finally sinks in?
>
> > > No, I did not. I did not even make the mistake of using the metric to
> > > describe the geometry as yourself and the physicists since Ricci's
> > > time. <shrug>
>
> > The metric _IS_ the geometry.
>
> The metric cannot be the geometry because you cannot describe the
> geometry without any coordinate system. <shrug>
A change in coordinates simply projects the metric components upon a
different set of basis vectors. The metric is unchanged. If you would
get over that "tensor = matrix" bullshit, you might actually learn
something.
>
> > I need no other information to
> > completely specify how particles will behave on that manifold.
>
> Without a coordinate system, you cannot specify how fast the particles
> are moving. <shrug>
Yet a coordinate change does not change how the particles move. What
does that tell you?
>
> > > ds^2 = (1 - G M / r)^2 dt^2 - dr^2 / (1 - G M / r / 2)^4 - r^2 dO^2 /
> > > (1 - G M / r / 2)^2
>
> > > The dr^2 term has the denominator term to the order of 4 not 2 as you
> > > have entered in your software. That is provided your software is
> > > checked out. <shrug>
>
> > A quick re-roll into grtensor showed that you are, in fact, correct.
> > It does satisfy R_uv = 0.
>
> Yes, I told you so.
...and how do you know? Did you actually solve the field equations by
hand [unlikely, given how you couldn't understand the proof of
Birkhoff's theorem] or did you simply use a coordinate transformation?
>
> > Furthermore, a few minutes of computation
> > revealed that it is the Schwarzschild metric in different coordinates.
>
> What coordinate system would that be? Remember one shows the event
> horizon at (G M) and the other one at (2 G M).
The one I explicitly explained in the very post you replied to, which
you apparently missed?
Here it is again: r(R) = 2*R^2/(2*R-G*M)
Apply the tensor transformation law, or just use the differentials in
the Schwarzschild line element. Plug R = GM into r(R), and you will
find r(GM) = 2GM. Different coordinate systems give different labels
for the same thing.
>
> > > The event horizon occurs at (G M) not (2 G M).
>
> > But if you plug R=GM into the coordinate transform you used, you find
> > out that the event horizon is in fact at r = 2GM as expected.
>
> There is no coordinate transformation as what you are leading to.
MORON.
Plug r = 2*R^2/(2*R-G*M) into the Schwarzschild line element and you
will get EXACTLY the line element you gave me.
> <shrug>
Nothing like shrugging your shoulders after saying something
fantastically stupid to remind everyone that you are, in fact, a
fucking moron.
>
> The choice of coordinate system is already cast in concrete that is
> the common spherically symmetric coordinate system.
No, it isn't. You chose a DIFFERENT radial coordinate. There isn't one
particular radial coordinate - I could define a _new_ radial
coordinate R = exp(r), and it will retain the spherical symmetry.
You either do not understand what is meant by spherical symmetry, or
you are being deliberately stupid. Which is it?
>
> > This is what happens when you play fast and loose with coordinates.
>
> Yes, you have lost the original choice of coordinate system that give
> rise to the Riemann curvature tensor, the Ricci curvature tensor, the
> Ricci scalar, the field equations, and finally the metric itself.
Hey, idiot. The coordinates don't matter. Basic tensor analysis. Any
result you give me which satisfies the same field equations and
boundary conditions will be the same result in a different coordinate
system.
>
> > > After this one, there are infinite numbers of solutions.
>
> > ...that describe the same solution. Remember Birkhoff's theorem?
>
> Birkhoff's theorem is wrong. We have discussed that already.
More like, a half dozen folks explained to you why you are so
amazingly wrong and you ran away because you couldn't defend your half-
assed position.
You were asked to provide two different solutions in the same
coordinate system that satisfied the same boundary conditions. You
couldn't even do that - you handed us two solutions in _DIFFERENT_
coordinate systems that were related by a simple coordinate
transformation. Then you ran away.
>
> > Apply the coordinate transform r = 2*R^2/(2*R-G*M) to Schwarzschild,
> > and the "new" metric is obtained.
Somehow you managed to miss this.
[Euclid]
*sigh*
Same metric, different coordinate systems. 3 examples, all of them
represent Euclidean geometry in different coordinate systems. Welcome
to tensor analysis - it would do you some good to learn about it.
>
> > Same manifold, different projections of the metric on different
> > coordinate axes.
>
> The sentence above can only mean what I have been saying.
Yet somehow you haven't managed to display even a glimmer of
understanding.
>
> > What's old is new. You are making the same mistakes as you were making
> > in April.
>
> >http://groups.google.com/group/sci.physics.relativity/msg/1418eb64933...
>
> This is not a mistake. I stand by as what I wrote.
You didn't even read it, moron. Everything you wrote below was in
response to what JanPB was telling you in April.
[snip remaining stupidity]
Koobee Wublee
> On Jul 1, 8:17 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > During Kepler's time, Newtonian law of gravity was not yet
> > discovered. There is no way in hell that you can determine the
> > gravitating mass through any of Kepler's laws. What you have posted
> > is Newton's law PERIOD.
>
> Consult any classical mechanics text published EVER. Your brand of
> history is not accepted.
I did, and Newton was born after Kepler's death.
> Isn't it odd that practically every opinion you hold is at odds with
> conventional wisdom in a technical field?
No, just some. <shrug>
> > So, they are wrong. <shrug>
>
> Not only does Goldstein and Symon make the same identification, but so
> does Landau & Lifshitz.
So, we have two wrong books. <shrug>
> The physics community has equated the original formulation of Kepler's
> laws with their more exact counterparts derived via Newtonian
> gravitation.
So.
> There is no point in whining because there is nothing you
> can do to change things.
Since Kepler's laws are derived from Newton's, it is solely Newton's
law of gravity that can determine the gravitating mass. <shrug>
> I find it far more illuminating that you disagree with every aspect of
> physics.
Big deal. <shrug>
> > So, I challenge you to describe a geometry without using any
> > coordinate system. <shrug>
>
> Try reading for comprehension.
OK, done just that.
> It does _NOT_ matter if you call the Cartesian x, y, z coordinates a,
> b, c or x_1, x_2, or x_3. Your failure to understand this means you
> don't even understand the Lagrangian formalism of classical mechanics
> because there is no way you could grasp the meaning of generalized
> coordinates with this mental fixation on labels.
You are the one who is trapped in this mental fixation thing on
labels. <shrug>
I have already explained to you numerous times that
Geometry = Metric * Coordinate
Where
** Geometry = Invariant
** Metric = Function of Coordinate
> > This is totally wrong. Since all your textbooks and physicists have
> > been wrong, I very humbly describe myself as the only one since
> > Riemann to have understood the curvature business. <shrug>
>
> Yet you struggle to compute anything or come up with even one example
> that supports any of your assertions. Isn't that interesting?
I gave you another solution to the field equations, did I not? Want
another one? Does it look like I am struggling? As I said, there are
infinite numbers of solutions to the field equations including the
ones that describe accelerated expansion of our universe without
provoking the Cosmological constant. <shrug>
> > Since the geometry must be invariant, the metric describing this
> > geometry must be only unique to the choice of coordinate system. In
> > doing so, the so called "tensors" which are merely matrices
> > constructed from the metrics and their derivatives must also be unique
> > to the initial choice of coordinate system. Just what logic do you
> > not understand here?
>
> Tensors aren't matrices, dipshit.
Then, the metric cannot be tensors.
> How many times does that need to be
> explained to you before it finally sinks in?
Never, because what you are saying is just wrong. <shrug>
> > The metric cannot be the geometry because you cannot describe the
> > geometry without any coordinate system. <shrug>
>
> A change in coordinates simply projects the metric components upon a
> different set of basis vectors. The metric is unchanged. If you would
> get over that "tensor = matrix" bullshit, you might actually learn
> something.
Well, whatever "tensor" is I don't care. However, "metric" is merely
"matrix". Treat the metric as a matrix throughout all stages of
mathematical manipulations, and the result is the same. <shrug>
> > Without a coordinate system, you cannot specify how fast the particles
> > are moving. <shrug>
>
> Yet a coordinate change does not change how the particles move. What
> does that tell you?
That is right, but what is your point? I never said changing the
coordinate will affect the motion of a particle. <shrug>
> ...and how do you know? Did you actually solve the field equations by
> hand
Yes, I did.
> [unlikely, given how you couldn't understand the proof of
> Birkhoff's theorem]
Birkhoff's theorem is just wrong by inferring to the observed
accelerated expansion in our universe. Again, this is also identified
from the solutions to the field equations without provoking the
Cosmological constant. <shrug>
> or did you simply use a coordinate transformation?
No, I did not. <shrug>
> > What coordinate system would that be? Remember one shows the event
> > horizon at (G M) and the other one at (2 G M).
>
> The one I explicitly explained in the very post you replied to, which
> you apparently missed?
>
> Here it is again: r(R) = 2*R^2/(2*R-G*M)
Why? Have I not presented you with a solution other than the
Schwarzschild metric where it also manifest black holes except with
half of the event horizon?
ds^2 = (1 - K / r)^2 dt^2 - dr^2 / (1 - K / r / 2)^4 - r^2 dO^2 / (1 -
K / r / 2)^2
What is your motivation to write
r(R) = R / (1 - K / R / 2)?
Have you not read what I wrote? I have already specified my choice of
coordinate system as (r, O).
You are the one doing the coordinate transformation. Not I. <shrug>
> Apply the tensor transformation law,
Tensor transformation law?
> or just use the differentials in
> the Schwarzschild line element. Plug R = GM into r(R), and you will
> find r(GM) = 2GM. Different coordinate systems give different labels
> for the same thing.
You have way too vivid of imagination. <shrug>
> > There is no coordinate transformation as what you are leading to.
>
> Plug r = 2*R^2/(2*R-G*M) into the Schwarzschild line element and you
> will get EXACTLY the line element you gave me.
So? Again, I have already specified my choice of coordinate system to
be (r, O), and the metric exhibits a black hole with an event horizon
of only (G M / c^2).
Where
** G = 0.667E-10 (whatever units)
** M = Gravitating mass
** c = 3E8 m/sec
<shrug>
> Nothing like shrugging your shoulders after saying something
> fantastically stupid to remind everyone that you are, in fact, a
> fucking moron.
Dr. Roberts (excuse me), Professor Roberts, has shrugged more often
than I have. <shrug>
> > The choice of coordinate system is already cast in concrete that is
> > the common spherically symmetric coordinate system.
>
> No, it isn't. You chose a DIFFERENT radial coordinate. There isn't one
> particular radial coordinate - I could define a _new_ radial
> coordinate R = exp(r), and it will retain the spherical symmetry.
>
> You either do not understand what is meant by spherical symmetry, or
> you are being deliberately stupid. Which is it?
I am the observer here. I have the choice of using my own coordinate
system. In doing so, I have cast it in concrete. You cannot change
that. <shrug>
> > Yes, you have lost the original choice of coordinate system that give
> > rise to the Riemann curvature tensor, the Ricci curvature tensor, the
> > Ricci scalar, the field equations, and finally the metric itself.
>
> Hey, idiot. The coordinates don't matter.
Again, the invariant geometry = the metric * the coordinate.
> Basic tensor analysis.
If you insist, physicists have applied the mathematics incorrectly in
the past 100 years or so. <shrug>
> Any
> result you give me which satisfies the same field equations and
> boundary conditions will be the same result in a different coordinate
> system.
Again, this cannot be true. Any metric or solution I give you is only
for the coordinate that I have already chosen. Thus, each metric or
solution I find represents a different geometry. <shrug>
> > Birkhoff's theorem is wrong. We have discussed that already.
>
> More like, a half dozen folks explained to you why you are so
> amazingly wrong and you ran away because you couldn't defend your half-
> assed position.
They have not discovered another solution to the field equations that
would allow an accelerated expansion in our universe without invoking
the Cosmological constant, but I did. <shrug>
> You were asked to provide two different solutions in the same
> coordinate system that satisfied the same boundary conditions. You
> couldn't even do that - you handed us two solutions in _DIFFERENT_
> coordinate systems that were related by a simple coordinate
> transformation. Then you ran away.
No, I just refused to do your homework for you. <shrug>
> Apply the coordinate transform r = 2*R^2/(2*R-G*M) to Schwarzschild,
> and the "new" metric is obtained.
>
> Somehow you managed to miss this.
No, I thought that was very stupid. <shrug>
>
> [Euclid]
>
> *sigh*
<shrug>
> Same metric, different coordinate systems.
I have already told you my choice of coordinate system is the common
spherically symmetric polar coordinate system already. There are no
other coordinate systems I have referred to. Can you read what I
wrote? Can you read at all? <shrug>
> 3 examples, all of them
> represent Euclidean geometry in different coordinate systems.
Yes, same geometry, different metrics, different coordinate systems.
<shrug>
> Welcome
> to tensor analysis - it would do you some good to learn about it.
You and the physicists need to stop mis-apply the mathematics.
<shrug>
> > The sentence above can only mean what I have been saying.
>
> Yet somehow you haven't managed to display even a glimmer of
> understanding.
Your opinion is worthless to me. <shrug>
> You didn't even read it, moron. Everything you wrote below was in
> response to what JanPB was telling you in April.
Mr. Bielawski was just wrong. <shrug>
> [snip remaining stupidity]
Why not the rest?
Eric Gisse
[...]
> Since Kepler's laws are derived from Newton's, it is solely Newton's
> law of gravity that can determine the gravitating mass. <shrug>
Congratulations, you finally learned what I have been telling you for
a week now.
[...]
>
> I have already explained to you numerous times that
>
> Geometry = Metric * Coordinate
No. Go study some differential geometry.
The geometry is the metric.
>
> Where
>
> ** Geometry = Invariant
> ** Metric = Function of Coordinate
>
> > > This is totally wrong. Since all your textbooks and physicists have
> > > been wrong, I very humbly describe myself as the only one since
> > > Riemann to have understood the curvature business. <shrug>
>
> > Yet you struggle to compute anything or come up with even one example
> > that supports any of your assertions. Isn't that interesting?
>
> I gave you another solution to the field equations, did I not? Want
> another one? Does it look like I am struggling? As I said, there are
> infinite numbers of solutions to the field equations including the
> ones that describe accelerated expansion of our universe without
> provoking the Cosmological constant. <shrug>
No, moron. You gave me the same solution in different coordinates.
That is not another solution - it is the same, and _ONLY_, solution.
You still don't even understand the most basic fact about Birkhoff's
theorem: it is solved assuming VACUUM. FRW cosmologies that have
accelerated expansion are NOT in vacuum - they assume a perfect fluid
instead of vacuum.
And yes, you do struggle. You claimed that you could introduce
curvature through a coordinate transformation - yet you could never
prove it. You claimed there are infinite solutions to the field
equations for the same initial and boundary conditions - yet all you
have been able to do is hand me variations on the same solution.
On, and on, and ON. You don't know what you are talking about, and it
is hilariously obvious.
>
> > > Since the geometry must be invariant, the metric describing this
> > > geometry must be only unique to the choice of coordinate system. In
> > > doing so, the so called "tensors" which are merely matrices
> > > constructed from the metrics and their derivatives must also be unique
> > > to the initial choice of coordinate system. Just what logic do you
> > > not understand here?
>
> > Tensors aren't matrices, dipshit.
>
> Then, the metric cannot be tensors.
The metric transforms like a TENSOR, not a matrix. Since you couldn't
even tell me how a tensor transforms, your not understanding what a
tensor is doesn't come as too much of a surprise.
>
> > How many times does that need to be
> > explained to you before it finally sinks in?
>
> Never, because what you are saying is just wrong. <shrug>
Why, oh guru of Riemann?
You have no literature to back you up. Not one textbook, journal
article, or web page. You don't have any education in either physics
or mathematics, as witnessed by your constant fuckups with both.
>
> > > The metric cannot be the geometry because you cannot describe the
> > > geometry without any coordinate system. <shrug>
>
> > A change in coordinates simply projects the metric components upon a
> > different set of basis vectors. The metric is unchanged. If you would
> > get over that "tensor = matrix" bullshit, you might actually learn
> > something.
>
> Well, whatever "tensor" is I don't care. However, "metric" is merely
> "matrix". Treat the metric as a matrix throughout all stages of
> mathematical manipulations, and the result is the same. <shrug>
That means you never did the damn manipulations even once in your
life. The metric transforms like a tensor, not a matrix.
>
> > > Without a coordinate system, you cannot specify how fast the particles
> > > are moving. <shrug>
>
> > Yet a coordinate change does not change how the particles move. What
> > does that tell you?
>
> That is right, but what is your point? I never said changing the
> coordinate will affect the motion of a particle. <shrug>
Look up "covariance".
>
> > ...and how do you know? Did you actually solve the field equations by
> > hand
>
> Yes, I did.
>
> > [unlikely, given how you couldn't understand the proof of
> > Birkhoff's theorem]
>
> Birkhoff's theorem is just wrong by inferring to the observed
> accelerated expansion in our universe. Again, this is also identified
> from the solutions to the field equations without provoking the
> Cosmological constant. <shrug>
Hey moron - Birkhoff is in vacuum, accelerated expansion is not. Don't
critique what you do not understand.
>
> > or did you simply use a coordinate transformation?
>
> No, I did not. <shrug>
>
> > > What coordinate system would that be? Remember one shows the event
> > > horizon at (G M) and the other one at (2 G M).
>
> > The one I explicitly explained in the very post you replied to, which
> > you apparently missed?
>
> > Here it is again: r(R) = 2*R^2/(2*R-G*M)
>
> Why? Have I not presented you with a solution other than the
> Schwarzschild metric where it also manifest black holes except with
> half of the event horizon?
No! You haven't!
THEY ARE THE SAME SOLUTION IN DIFFERENT COORDINATES.
>
> ds^2 = (1 - K / r)^2 dt^2 - dr^2 / (1 - K / r / 2)^4 - r^2 dO^2 / (1 -
> K / r / 2)^2
>
> What is your motivation to write
>
> r(R) = R / (1 - K / R / 2)?
The coordinates \theta and \phi are not functions of r or R. Compare g_
\theta\theta in Schwarzschild to what you have here. Solve for r as a
function of R. Apply the tensor transformation law to Schwarzschild.
>
> Have you not read what I wrote? I have already specified my choice of
> coordinate system as (r, O).
No, that is not correct. Your coordinate system is (t,r,\theta\,\phi)
- it appears you don't regard t as a coordinate and have forgotten the
meaning of \omega [what you call O].
As I already explained to you at least a half dozen times here, the
labels for the coordinates are irrelevant. I could change your
coordinate system to (a,b,c,d) and _nothing would change_ because
there is no physics in the labels.
>
> You are the one doing the coordinate transformation. Not I. <shrug>
Of course I am, because you didn't bother checking.
Didn't you _just_ explain that two metric in the same coordinate
system describe the same geometry?
>
> > Apply the tensor transformation law,
>
> Tensor transformation law?
Forgotten already? Never learned? No matter.
It is transpose(A) * T * A , where A is the Jacobian.
>
> > or just use the differentials in
> > the Schwarzschild line element. Plug R = GM into r(R), and you will
> > find r(GM) = 2GM. Different coordinate systems give different labels
> > for the same thing.
>
> You have way too vivid of imagination. <shrug>
Why don't you check?
>
> > > There is no coordinate transformation as what you are leading to.
>
> > Plug r = 2*R^2/(2*R-G*M) into the Schwarzschild line element and you
> > will get EXACTLY the line element you gave me.
>
> So? Again, I have already specified my choice of coordinate system to
> be (r, O), and the metric exhibits a black hole with an event horizon
> of only (G M / c^2).
Again, the label doesn't matter. I could call it q and it /would not
matter/, there is no physics in the coordinate label. Just in the
coordinate itself - if you studied classical mechanics to the point of
understanding generalized coordinates, you would be far better
prepared to understand this instead of tripping over it constantly.
Why don't I calculate the area? For concreteness, it will be the
surface area of a sphere of constant r,t.
For constant r,t the dr and dt terms drop out [you do understand this,
right?]. This is my induced metric. Its' determinant [you do know how
to take a determinant of a matrix, right?] will be r^4 * sin(\theta)^2
* (1 - K / 2 / r)^-4.
The quantity I need is the square root of the determinant. So, det(|
g|) = r^2 * sin(\theta) * (1 - K / 2 / r)^-2.
So, the area then is r^2 * (1 - K / 2 / r)^-2 * int(int(sin(\theta),
\theta=0...2pi),\phi=0...pi). Which evaluates directly to [you can
calculate that integral, right?] to 4 * \pi * r^2 * (1 - K / 2 /
r)^-2.
Since the "true" radial coordinate is defined through the surface area
of a sphere, something is fairly amiss. But we _know_ the surface area
of a sphere in Schwarzschild coordinates is 4 * pi * r^2. Plug in r(R)
= 2*R^2/(2*R-G*M)...and you get exactly 4 * \pi * R^2 * (1 - K / 2 /
R)^-2, which is the surface area in your "other" geometry.
What the FUCK does that tell you?
>
> Where
>
> ** G = 0.667E-10 (whatever units)
> ** M = Gravitating mass
> ** c = 3E8 m/sec
>
> <shrug>
>
> > Nothing like shrugging your shoulders after saying something
> > fantastically stupid to remind everyone that you are, in fact, a
> > fucking moron.
>
> Dr. Roberts (excuse me), Professor Roberts, has shrugged more often
> than I have. <shrug>
Try finding your own style rather than blindly copying someone
else's.
>
> > > The choice of coordinate system is already cast in concrete that is
> > > the common spherically symmetric coordinate system.
>
> > No, it isn't. You chose a DIFFERENT radial coordinate. There isn't one
> > particular radial coordinate - I could define a _new_ radial
> > coordinate R = exp(r), and it will retain the spherical symmetry.
>
> > You either do not understand what is meant by spherical symmetry, or
> > you are being deliberately stupid. Which is it?
>
> I am the observer here. I have the choice of using my own coordinate
> system. In doing so, I have cast it in concrete. You cannot change
> that. <shrug>
You haven't fixed anything in concrete, as I have just shown.
>
> > > Yes, you have lost the original choice of coordinate system that give
> > > rise to the Riemann curvature tensor, the Ricci curvature tensor, the
> > > Ricci scalar, the field equations, and finally the metric itself.
>
> > Hey, idiot. The coordinates don't matter.
>
> Again, the invariant geometry = the metric * the coordinate.
Not only is that stupid because it is a childishly way of trying to
express a complicated subject, but it is wrong. Coordinates are not
invariant.
>
> > Basic tensor analysis.
>
> If you insist, physicists have applied the mathematics incorrectly in
> the past 100 years or so. <shrug>
Then why does it work? Why is it internally self consistent? Why is it
someone totally uneducated in the craft managed to find an "error" in
the craft that none of the professionals have seen in over a century?
>
> > Any
> > result you give me which satisfies the same field equations and
> > boundary conditions will be the same result in a different coordinate
> > system.
>
> Again, this cannot be true. Any metric or solution I give you is only
> for the coordinate that I have already chosen. Thus, each metric or
> solution I find represents a different geometry. <shrug>
Why is it I can relate two "different" geometries with a coordinate
transformation? What does that say to you?
Remember my examples of Euclid in different coordinates? Think about
it.
>
> > > Birkhoff's theorem is wrong. We have discussed that already.
>
> > More like, a half dozen folks explained to you why you are so
> > amazingly wrong and you ran away because you couldn't defend your half-
> > assed position.
>
> They have not discovered another solution to the field equations that
> would allow an accelerated expansion in our universe without invoking
> the Cosmological constant, but I did. <shrug>
Hey look, more changing goalposts. Birkhoff's theorem is under
discussion, not the cosmological constant.
>
> > You were asked to provide two different solutions in the same
> > coordinate system that satisfied the same boundary conditions. You
> > couldn't even do that - you handed us two solutions in _DIFFERENT_
> > coordinate systems that were related by a simple coordinate
> > transformation. Then you ran away.
>
> No, I just refused to do your homework for you. <shrug>
I'm not the one posting about how all physicists and mathematicians
got basic tensor analysis wrong.
>
> > Apply the coordinate transform r = 2*R^2/(2*R-G*M) to Schwarzschild,
> > and the "new" metric is obtained.
>
> > Somehow you managed to miss this.
>
> No, I thought that was very stupid. <shrug>
Was it more stupid than saying a metric cannot have a Lagrangian? More
stupid than saying you can spot curvature by inspecting the
coordinates? More stupid than saying you can introduce curvature
through a coordinate transformation?
>
>
>
> > [Euclid]
>
> > *sigh*
>
> <shrug>
>
> > Same metric, different coordinate systems.
>
> I have already told you my choice of coordinate system is the common
> spherically symmetric polar coordinate system already. There are no
> other coordinate systems I have referred to. Can you read what I
> wrote? Can you read at all? <shrug>
Except it isn't.
You chose a DIFFERENT radial coordinate. There isn't one particular
radial coordinate - I could define a _new_ radial coordinate R =
exp(r), and it will retain the spherical symmetry. Your "r" _is_ a
radial coordinate, just not the usual one.
As explicitly shown. It doesn't reduce to Minkowski space in spherical
coordinates, nor does it have the proper area.
You either do not understand what is meant by spherical symmetry, or
you are being deliberately stupid. Which is it?
>
> > 3 examples, all of them
> > represent Euclidean geometry in different coordinate systems.
>
> Yes, same geometry, different metrics, different coordinate systems.
> <shrug>
Very good. You understand that much.
Now think about it a little more. I can construct transformations
[TENSOR transformations] that can relate all those representations.
What does that tell you?
>
> > Welcome
> > to tensor analysis - it would do you some good to learn about it.
>
> You and the physicists need to stop mis-apply the mathematics.
> <shrug>
You don't even know how to transform a tensor, much less have the
ability to make wild claims like "all physicists are wrong".
>
> > > The sentence above can only mean what I have been saying.
>
> > Yet somehow you haven't managed to display even a glimmer of
> > understanding.
>
> Your opinion is worthless to me. <shrug>
Yet you keep responding.
[...]
Koobee Wublee
> On Jul 1, 11:21 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > Since Kepler's laws are derived from Newton's, it is solely Newton's
> > law of gravity that can determine the gravitating mass. <shrug>
>
> Congratulations, you finally learned what I have been telling you for
> a week now.
That is not what you are saying. You have claimed the gravitating
mass can only solely determined through the Newtonian law of gravity,
but you claimed so solely from Kepler's 3rd law. Kepler's laws only
work for some special circumstance. Ultimately, it is Newton's law of
gravity that describes these phenomena. <shrug>
> > I have already explained to you numerous times that
>
> > Geometry = Metric * Coordinate
>
> No. Go study some differential geometry.
I did, and "Geometry = Metric * Coordinate" <shrug>
> The geometry is the metric.
It makes no sense to consider the metric as the geometry. <shrug>
> > Where
>
> > ** Geometry = Invariant
> > ** Metric = Function of Coordinate
> > I gave you another solution to the field equations, did I not? Want
> > another one? Does it look like I am struggling? As I said, there are
> > infinite numbers of solutions to the field equations including the
> > ones that describe accelerated expansion of our universe without
> > provoking the Cosmological constant. <shrug>
>
> No, moron. You gave me the same solution in different coordinates.
> That is not another solution - it is the same, and _ONLY_, solution.
I gave you solutions to the field equations that can only be specified
with the same set of spherically symmetric polar coordinate system.
Therefore, each metric must describe different geometry using the same
set of coordinate system.
> You still don't even understand the most basic fact about Birkhoff's
> theorem: it is solved assuming VACUUM. FRW cosmologies that have
> accelerated expansion are NOT in vacuum - they assume a perfect fluid
> instead of vacuum.
I get you don't understand Birkhoff's theorem. It assumes the
Schwarzschild metric is the only solution and establishes the solution
must be spherically symmetric and asymptotically flat. <shrug>
> And yes, you do struggle. You claimed that you could introduce
> curvature through a coordinate transformation
No, I did not introduce curvature through a coordinate
transformation. Remember the geometry is invariant. <shrug>
> - yet you could never prove it.
The fact that the geometry must be invariant is a fundamental
principle, and you have so much trouble understanding that. <shrug>
> You claimed there are infinite solutions to the field
> equations for the same initial and boundary conditions
The choice of coordinate system must be firmly established before we
can proceed to get to the solutions of the field equations. <shrug>
> - yet all you
> have been able to do is hand me variations on the same solution.
No, I have shown you a number of solutions already. <shrug>
> On, and on, and ON. You don't know what you are talking about, and it
> is hilariously obvious.
It is hilarious that you kept hallucinating all these solutions are
indeed one. This is so because you don't understand the subject.
This is so because you do not understand the fundamental principle
that the geometry must be invariant, and to describe the geometry
fully you need to reference it to a set of coordinate system. <shrug>
> > Then, the metric cannot be tensors.
>
> The metric transforms like a TENSOR, not a matrix.
Your math is very poor. The metric transforms like a matrix. <shrug>
> Since you couldn't
> even tell me how a tensor transforms, your not understanding what a
> tensor is doesn't come as too much of a surprise.
I don't give a damn about tensors at this moment because the metric is
a matrix. <shrug>
> How many times does that need to be
> explained to you before it finally sinks in?
>
> > Never, because what you are saying is just wrong. <shrug>
>
> Why, oh guru of Riemann?
Because what you are saying is just wrong. <shrug>
> You have no literature to back you up. Not one textbook, journal
> article, or web page.
Has it get to your head that the physicists have been mis-applying
mathematics to differential geometry for almost 100 years?
> You don't have any education in either physics
> or mathematics, as witnessed by your constant fuckups with both.
This is your ASSumption of me. Who gives a sh*t? <shrug>
> > Well, whatever "tensor" is I don't care. However, "metric" is merely
> > "matrix". Treat the metric as a matrix throughout all stages of
> > mathematical manipulations, and the result is the same. <shrug>
>
> That means you never did the damn manipulations even once in your
> life. The metric transforms like a tensor, not a matrix.
I have been showing you the metric a merely a matrix. Any mathematics
treating the metric as a matrix is sufficient for this application.
<shrug>
> > > Yet a coordinate change does not change how the particles move. What
> > > does that tell you?
>
> > That is right, but what is your point? I never said changing the
> > coordinate will affect the motion of a particle. <shrug>
>
> Look up "covariance".
Yes, I did. I never said changing the coordinate will affect the
motion of a particle. <shrug>
> > Birkhoff's theorem is just wrong by inferring to the observed
> > accelerated expansion in our universe. Again, this is also identified
> > from the solutions to the field equations without provoking the
> > Cosmological constant. <shrug>
>
> Hey moron - Birkhoff is in vacuum, accelerated expansion is not. Don't
> critique what you do not understand.
That is a very good excuse to keep a foul theorem alive I suppose.
<shrug>
> > Why? Have I not presented you with a solution other than the
> > Schwarzschild metric where it also manifest black holes except with
> > half of the event horizon?
>
> No! You haven't!
Yes, I have. Here is once again.
ds^2 = (1 - K / r)^2 dt^2 - dr^2 / (1 - K / r / 2)^4 - r^2 dO^2 / (1 -
K / r / 2)^2
Where
** (t, R, O) = Choice of coordinate system
> THEY ARE THE SAME SOLUTION IN DIFFERENT COORDINATES.
THEY ARE DIFFERENT SOLUTIONS IN THE ASAME COORDINATE SYSTEM.
> > ds^2 = (1 - K / r)^2 dt^2 - dr^2 / (1 - K / r / 2)^4 - r^2 dO^2 / (1 -
> > K / r / 2)^2
>
> > What is your motivation to write
>
> > r(R) = R / (1 - K / R / 2)?
>
> The coordinates \theta and \phi are not functions of r or R. Compare g_
> \theta\theta in Schwarzschild to what you have here. Solve for r as a
> function of R. Apply the tensor transformation law to Schwarzschild.
Why Schwarzschild? Fudging everything into the Schwarzschild result
is not doing any serious physics. <shrug>
> > Have you not read what I wrote? I have already specified my choice of
> > coordinate system as (r, O).
>
> No, that is not correct. Your coordinate system is (t,r,\theta\,\phi)
Fine, picky.
> - it appears you don't regard t as a coordinate and have forgotten the
> meaning of \omega [what you call O].
No, I have not. It is much simpler to write it as 'O' for the purpose
of any informal discussions. It is a good idea, and I did not
initiate this. Mr. Bielawski did. <shrug>
> As I already explained to you at least a half dozen times here, the
> labels for the coordinates are irrelevant.
And you are wrong because there is no way you can describe the
geometry without a set of coordinate system. <shrug>
> I could change your
> coordinate system to (a,b,c,d) and _nothing would change_ because
> there is no physics in the labels.
Yes, that is right. However, going to a different coordinate system,
you must also use the appropriate metric for each set of coordinate
systems. <shrug>
> > You are the one doing the coordinate transformation. Not I. <shrug>
>
> Of course I am, because you didn't bother checking.
I understand exactly what you are doing, and you are just wrong. You
don't even understand the subject matter. <shrug>
> Didn't you _just_ explain that two metric in the same coordinate
> system describe the same geometry?
No, play attention. Two different metrics referencing to the same
coordinate system conclusively describe two different geometries. Two
different metrics referencing to a set of coordinate system each have
the chance of describing the same geometry. <shrug>
> > Tensor transformation law?
>
> Forgotten already? Never learned? No matter.
>
> It is transpose(A) * T * A , where A is the Jacobian.
<Yawn> <shrug>
> > You have way too vivid of imagination. <shrug>
>
> Why don't you check?
I checked. You are still wrong.
> > So? Again, I have already specified my choice of coordinate system to
> > be (r, O), and the metric exhibits a black hole with an event horizon
> > of only (G M / c^2).
>
> Again, the label doesn't matter. I could call it q and it /would not
> matter/, there is no physics in the coordinate label. Just in the
> coordinate itself - if you studied classical mechanics to the point of
> understanding generalized coordinates, you would be far better
> prepared to understand this instead of tripping over it constantly.
The choice of coordinate system does matter to describe the geometry.
The labels does not matter if you also throw in the appropriate metric
that goes along with this coordinate system. <shrug>
> Why don't I calculate the area? For concreteness, it will be the
> surface area of a sphere of constant r,t.
>
> For constant r,t the dr and dt terms drop out [you do understand this,
> right?]. This is my induced metric. Its' determinant [you do know how
> to take a determinant of a matrix, right?] will be r^4 * sin(\theta)^2
> * (1 - K / 2 / r)^-4.
>
> The quantity I need is the square root of the determinant. So, det(|
> g|) = r^2 * sin(\theta) * (1 - K / 2 / r)^-2.
>
> So, the area then is r^2 * (1 - K / 2 / r)^-2 * int(int(sin(\theta),
> \theta=0...2pi),\phi=0...pi). Which evaluates directly to [you can
> calculate that integral, right?] to 4 * \pi * r^2 * (1 - K / 2 /
> r)^-2.
>
> Since the "true" radial coordinate is defined through the surface area
> of a sphere, something is fairly amiss. But we _know_ the surface area
> of a sphere in Schwarzschild coordinates is 4 * pi * r^2. Plug in r(R)
> = 2*R^2/(2*R-G*M)...and you get exactly 4 * \pi * R^2 * (1 - K / 2 /
> R)^-2, which is the surface area in your "other" geometry.
The surface area as described by the choice of coordinate system I
have settled with is as follows.
4 PI r^2
> What the FUCK does that tell you?
You don't understand geometry and basic, fundamental principles.
<shrug>
> > Dr. Roberts (excuse me), Professor Roberts, has shrugged more often
> > than I have. <shrug>
>
> Try finding your own style rather than blindly copying someone
> else's.
I don't have a style. I learn well. <shrug>
> > I am the observer here. I have the choice of using my own coordinate
> > system. In doing so, I have cast it in concrete. You cannot change
> > that. <shrug>
>
> You haven't fixed anything in concrete, as I have just shown.
Well, you don't understand what geometry is. <shrug>
> > Again, the invariant geometry = the metric * the coordinate.
>
> Not only is that stupid because it is a childishly way of trying to
> express a complicated subject, but it is wrong.
<CORRECT> Geometry = Metric * Coordinate
Why is it wrong? Why is it chidish?
Now the following is wrong, absurd, silly, and childish mistakes.
<WRONG> Geometry = Metric
> Coordinates are not invariant.
I never said it is. <shrug>
> > If you insist, physicists have applied the mathematics incorrectly in
> > the past 100 years or so. <shrug>
>
> Then why does it work?
No, it does not work.
> Why is it internally self consistent?
No, it is not internally self consistent.
> Why is it
> someone totally uneducated in the craft managed to find an "error" in
> the craft that none of the professionals have seen in over a century?
That is because I am educated in and have thoroughly understood the
craft. That is why I can find errors and lots of them where they have
eluded physicists for almost 100 years. <shrug>
> > Again, this cannot be true. Any metric or solution I give you is only
> > for the coordinate that I have already chosen. Thus, each metric or
> > solution I find represents a different geometry. <shrug>
>
> Why is it I can relate two "different" geometries with a coordinate
> transformation?
Because there are infinite solutions to the field equations. <shrug>
> What does that say to you?
All solutions to the field equations are related but not the same.
Each metric is a solution to the same field equations. <shrug>
> Remember my examples of Euclid in different coordinates? Think about
> it.
OK, I did. What I am saying still stands.
> > They have not discovered another solution to the field equations that
> > would allow an accelerated expansion in our universe without invoking
> > the Cosmological constant, but I did. <shrug>
>
> Hey look, more changing goalposts. Birkhoff's theorem is under
> discussion, not the cosmological constant.
The Cosmological constant basically implies the Birkhoff's theorem
wrong.
> > No, I just refused to do your homework for you. <shrug>
>
> I'm not the one posting about how all physicists and mathematicians
> got basic tensor analysis wrong.
That is correct. I am. <shrug>
> Was it more stupid than saying a metric cannot have a Lagrangian?
No, a metric cannot have a Lagrangian. <shrug>
> More
> stupid than saying you can spot curvature by inspecting the
> coordinates?
No, to describe the curvature, you must start with a set of coordinate
system and use the appropriate metric associated with this choice of
coordinate system. <shrug>
> More stupid than saying you can introduce curvature
> through a coordinate transformation?
No, because the geometry is invariant. You cannot change the geometry
through a coordinate transformation. <shrug>
> > I have already told you my choice of coordinate system is the common
> > spherically symmetric polar coordinate system already. There are no
> > other coordinate systems I have referred to. Can you read what I
> > wrote? Can you read at all? <shrug>
>
> Except it isn't.
>
> You chose a DIFFERENT radial coordinate.
How? I have already settled on the common spherically symmetric polar
coordinate system. Do you know what that is?
> There isn't one particular
> radial coordinate - I could define a _new_ radial coordinate R =
> exp(r), and it will retain the spherical symmetry. Your "r" _is_ a
> radial coordinate, just not the usual one.
My 'r' is always the usual radial coordinate. <shrug>
> As explicitly shown. It doesn't reduce to Minkowski space in spherical
> coordinates, nor does it have the proper area.
At (r = infinity), the metric becomes Minkowkski. The area of the
sphere is always (4 PI r^2). <shrug>
What is this proper area you are describing?
> You either do not understand what is meant by spherical symmetry, or
> you are being deliberately stupid. Which is it?
Neither. <shrug>
> > Yes, same geometry, different metrics, different coordinate systems.
> > <shrug>
>
> Very good. You understand that much.
I understand very well in this subject.
> Now think about it a little more. I can construct transformations
> [TENSOR transformations] that can relate all those representations.
> What does that tell you?
Geometry must be invariant. Coordinate must be chosen before
describing the geometry. In choosing a coordinate system, you must
find the appropriate metric associated with this coordinate system.
Finally, the field equations yield an infinite numbers of solutions.
Given the same coordinate system, each solution or metric describes a
very different geometry. The field equations describe a infinity
numbers of geometries.
> > You and the physicists need to stop mis-apply the mathematics.
> > <shrug>
>
> You don't even know how to transform a tensor, much less have the
> ability to make wild claims like "all physicists are wrong".
As I said transformation of tensors do not apply here. All physicists
are wrong.
> > Your opinion is worthless to me. <shrug>
>
> Yet you keep responding.
Yes, your observation is indeed correct in this case. <shrug>
Eric Gisse
> On Jul 2, 2:20 am, Eric Gisse <jowr...@gmail.com> wrote:
>
> > On Jul 1, 11:21 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > > Since Kepler's laws are derived from Newton's, it is solely Newton's
> > > law of gravity that can determine the gravitating mass. <shrug>
>
> > Congratulations, you finally learned what I have been telling you for
> > a week now.
>
> That is not what you are saying. You have claimed the gravitating
> mass can only solely determined through the Newtonian law of gravity,
> but you claimed so solely from Kepler's 3rd law. Kepler's laws only
> work for some special circumstance. Ultimately, it is Newton's law of
> gravity that describes these phenomena. <shrug>
Nope. Read what I said. I never said Kepler's 3rd law was the only
way, just that it is how I would do it.
It is well understood that Kepler's laws are approximations. For
example, Kepler's 3rd law is only valid when the main body is much
more massive than the orbiting body.
>
> > > I have already explained to you numerous times that
>
> > > Geometry = Metric * Coordinate
>
> > No. Go study some differential geometry.
>
> I did, and "Geometry = Metric * Coordinate" <shrug>
Really? What book?
>
> > The geometry is the metric.
>
> It makes no sense to consider the metric as the geometry. <shrug>
Why? What information do I need other than the metric?
> > > Where
>
> > > ** Geometry = Invariant
> > > ** Metric = Function of Coordinate
> > > I gave you another solution to the field equations, did I not? Want
> > > another one? Does it look like I am struggling? As I said, there are
> > > infinite numbers of solutions to the field equations including the
> > > ones that describe accelerated expansion of our universe without
> > > provoking the Cosmological constant. <shrug>
>
> > No, moron. You gave me the same solution in different coordinates.
> > That is not another solution - it is the same, and _ONLY_, solution.
>
> I gave you solutions to the field equations that can only be specified
> with the same set of spherically symmetric polar coordinate system.
> Therefore, each metric must describe different geometry using the same
> set of coordinate system.
>
> > You still don't even understand the most basic fact about Birkhoff's
> > theorem: it is solved assuming VACUUM. FRW cosmologies that have
> > accelerated expansion are NOT in vacuum - they assume a perfect fluid
> > instead of vacuum.
>
> I get you don't understand Birkhoff's theorem. It assumes the
> Schwarzschild metric is the only solution and establishes the solution
> must be spherically symmetric and asymptotically flat. <shrug>
>
No, it doesn't. Birkhoff's theorem was explained in excruciating
detail to you. All that is assumed is spherical symmetry and a time-
like killing vector at infinity.
> > And yes, you do struggle. You claimed that you could introduce
> > curvature through a coordinate transformation
>
> No, I did not introduce curvature through a coordinate
> transformation. Remember the geometry is invariant. <shrug>
But you claimed you could.
"The metric is Minkowski but the choice of coordinate system is
hopelessly non-linear. Thus, the only logical conclusion is that the
described spacetime must be curved despite the metric being
Minkowski."
http://groups.google.com/group/sci.physics.relativity/msg/00d08acc54532fd4?dmode=source
Do you retract this statement?
>
> > - yet you could never prove it.
>
> The fact that the geometry must be invariant is a fundamental
> principle, and you have so much trouble understanding that. <shrug>
>
> > You claimed there are infinite solutions to the field
> > equations for the same initial and boundary conditions
>
> The choice of coordinate system must be firmly established before we
> can proceed to get to the solutions of the field equations. <shrug>
No, it doesn't.
The popular derivation of the Schwarzschild solution goes through a
half dozen coordinate changes.
>
> > - yet all you
> > have been able to do is hand me variations on the same solution.
>
> No, I have shown you a number of solutions already. <shrug>
Yes, and they are all the _same_ solution.
>
> > On, and on, and ON. You don't know what you are talking about, and it
> > is hilariously obvious.
>
> It is hilarious that you kept hallucinating all these solutions are
> indeed one. This is so because you don't understand the subject.
> This is so because you do not understand the fundamental principle
> that the geometry must be invariant, and to describe the geometry
> fully you need to reference it to a set of coordinate system. <shrug>
>
> > > Then, the metric cannot be tensors.
>
> > The metric transforms like a TENSOR, not a matrix.
>
> Your math is very poor. The metric transforms like a matrix. <shrug>
Prove it.
>
> > Since you couldn't
> > even tell me how a tensor transforms, your not understanding what a
> > tensor is doesn't come as too much of a surprise.
>
> I don't give a damn about tensors at this moment because the metric is
> a matrix. <shrug>
One representation is a matrix, but that does not mean a rank 2 tensor
has all the properties of matrices.
>
> > How many times does that need to be
> > explained to you before it finally sinks in?
>
> > > Never, because what you are saying is just wrong. <shrug>
>
> > Why, oh guru of Riemann?
>
> Because what you are saying is just wrong. <shrug>
>
> > You have no literature to back you up. Not one textbook, journal
> > article, or web page.
>
> Has it get to your head that the physicists have been mis-applying
> mathematics to differential geometry for almost 100 years?
How do you know?
>
> > You don't have any education in either physics
> > or mathematics, as witnessed by your constant fuckups with both.
>
> This is your ASSumption of me. Who gives a sh*t? <shrug>
Correct the assumption then. What is your background in physics and
mathematics?
>
> > > Well, whatever "tensor" is I don't care. However, "metric" is merely
> > > "matrix". Treat the metric as a matrix throughout all stages of
> > > mathematical manipulations, and the result is the same. <shrug>
>
> > That means you never did the damn manipulations even once in your
> > life. The metric transforms like a tensor, not a matrix.
>
> I have been showing you the metric a merely a matrix. Any mathematics
> treating the metric as a matrix is sufficient for this application.
> <shrug>
Then explain how to transform the Euclidean metric tensor into
spherical coordinates. You see, I get the correct answer [ds^2 = dr^2
+ r^2 [ sin(\theta)^2*d\phi^2 + d\theta^2] by using the tensor
transformation law. Can you explain why that is the case since the
tensor transformation law is different from the matrix transformation
law?
>
> > > > Yet a coordinate change does not change how the particles move. What
> > > > does that tell you?
>
> > > That is right, but what is your point? I never said changing the
> > > coordinate will affect the motion of a particle. <shrug>
>
> > Look up "covariance".
>
> Yes, I did. I never said changing the coordinate will affect the
> motion of a particle. <shrug>
Very good. You understand that a change of coordinates won't do
anything to the physics. Keep that in mind as you read on.
>
> > > Birkhoff's theorem is just wrong by inferring to the observed
> > > accelerated expansion in our universe. Again, this is also identified
> > > from the solutions to the field equations without provoking the
> > > Cosmological constant. <shrug>
>
> > Hey moron - Birkhoff is in vacuum, accelerated expansion is not. Don't
> > critique what you do not understand.
>
> That is a very good excuse to keep a foul theorem alive I suppose.
> <shrug>
>
> > > Why? Have I not presented you with a solution other than the
> > > Schwarzschild metric where it also manifest black holes except with
> > > half of the event horizon?
>
> > No! You haven't!
>
> Yes, I have. Here is once again.
>
> ds^2 = (1 - K / r)^2 dt^2 - dr^2 / (1 - K / r / 2)^4 - r^2 dO^2 / (1 -
> K / r / 2)^2
>
> Where
>
> ** (t, R, O) = Choice of coordinate system
>
> > THEY ARE THE SAME SOLUTION IN DIFFERENT COORDINATES.
>
> THEY ARE DIFFERENT SOLUTIONS IN THE ASAME COORDINATE SYSTEM.
No, they aren't. Your radial coordinate is not the one you claim it
is.
>
> > > ds^2 = (1 - K / r)^2 dt^2 - dr^2 / (1 - K / r / 2)^4 - r^2 dO^2 / (1 -
> > > K / r / 2)^2
>
> > > What is your motivation to write
>
> > > r(R) = R / (1 - K / R / 2)?
>
> > The coordinates \theta and \phi are not functions of r or R. Compare g_
> > \theta\theta in Schwarzschild to what you have here. Solve for r as a
> > function of R. Apply the tensor transformation law to Schwarzschild.
>
> Why Schwarzschild? Fudging everything into the Schwarzschild result
> is not doing any serious physics. <shrug>
Birkhoff's theorem, for one. Birkhoff's theorem would be wrong if I
were unable to do what I just did.
>
> > > Have you not read what I wrote? I have already specified my choice of
> > > coordinate system as (r, O).
>
> > No, that is not correct. Your coordinate system is (t,r,\theta\,\phi)
>
> Fine, picky.
>
> > - it appears you don't regard t as a coordinate and have forgotten the
> > meaning of \omega [what you call O].
>
> No, I have not. It is much simpler to write it as 'O' for the purpose
> of any informal discussions. It is a good idea, and I did not
> initiate this. Mr. Bielawski did. <shrug>
>
> > As I already explained to you at least a half dozen times here, the
> > labels for the coordinates are irrelevant.
>
> And you are wrong because there is no way you can describe the
> geometry without a set of coordinate system. <shrug>
That's right - you can't actually describe the manifold without
picking a coordinate system. But that isn't what I'm saying. What I am
saying is what you _call_ the coordinates does not matter. There is no
physical significance in saying "r" - the meaning you attach to it is
derived through other properties. I could call it "q", and it'd work
just fine.
>
> > I could change your
> > coordinate system to (a,b,c,d) and _nothing would change_ because
> > there is no physics in the labels.
>
> Yes, that is right. However, going to a different coordinate system,
> you must also use the appropriate metric for each set of coordinate
> systems. <shrug>
That's right - you transform the metric components to the new
coordinate system.
>
> > > You are the one doing the coordinate transformation. Not I. <shrug>
>
> > Of course I am, because you didn't bother checking.
>
> I understand exactly what you are doing, and you are just wrong. You
> don't even understand the subject matter. <shrug>
Why is it wrong? It is basic tensor analysis.
If you hand me the line element ds^2 = dr^2 + r^2 [ sin(\theta)^2*d
\phi^2 + d\theta^2] and I can find a coordinate transformation to it
from ds^2 = dx^2 + dy^2 + dz^2, they represent the _same_ manifold.
That is exactly what I did between your line element and the
Schwarzschild line element. It isn't that complicated.
>
> > Didn't you _just_ explain that two metric in the same coordinate
> > system describe the same geometry?
>
> No, play attention. Two different metrics referencing to the same
> coordinate system conclusively describe two different geometries. Two
> different metrics referencing to a set of coordinate system each have
> the chance of describing the same geometry. <shrug>
True, but that doesn't apply since the same coordinate system isn't
being used!
>
> > > Tensor transformation law?
>
> > Forgotten already? Never learned? No matter.
>
> > It is transpose(A) * T * A , where A is the Jacobian.
>
> <Yawn> <shrug>
>
> > > You have way too vivid of imagination. <shrug>
>
> > Why don't you check?
>
> I checked. You are still wrong.
>
> > > So? Again, I have already specified my choice of coordinate system to
> > > be (r, O), and the metric exhibits a black hole with an event horizon
> > > of only (G M / c^2).
>
> > Again, the label doesn't matter. I could call it q and it /would not
> > matter/, there is no physics in the coordinate label. Just in the
> > coordinate itself - if you studied classical mechanics to the point of
> > understanding generalized coordinates, you would be far better
> > prepared to understand this instead of tripping over it constantly.
>
> The choice of coordinate system does matter to describe the geometry.
Yep.
> The labels does not matter if you also throw in the appropriate metric
> that goes along with this coordinate system. <shrug>
Yep.
I'm failing to see how you can disagree with me.
>
> > Why don't I calculate the area? For concreteness, it will be the
> > surface area of a sphere of constant r,t.
>
> > For constant r,t the dr and dt terms drop out [you do understand this,
> > right?]. This is my induced metric. Its' determinant [you do know how
> > to take a determinant of a matrix, right?] will be r^4 * sin(\theta)^2
> > * (1 - K / 2 / r)^-4.
>
> > The quantity I need is the square root of the determinant. So, det(|
> > g|) = r^2 * sin(\theta) * (1 - K / 2 / r)^-2.
>
> > So, the area then is r^2 * (1 - K / 2 / r)^-2 * int(int(sin(\theta),
> > \theta=0...2pi),\phi=0...pi). Which evaluates directly to [you can
> > calculate that integral, right?] to 4 * \pi * r^2 * (1 - K / 2 /
> > r)^-2.
>
> > Since the "true" radial coordinate is defined through the surface area
> > of a sphere, something is fairly amiss. But we _know_ the surface area
> > of a sphere in Schwarzschild coordinates is 4 * pi * r^2. Plug in r(R)
> > = 2*R^2/(2*R-G*M)...and you get exactly 4 * \pi * R^2 * (1 - K / 2 /
> > R)^-2, which is the surface area in your "other" geometry.
>
> The surface area as described by the choice of coordinate system I
> have settled with is as follows.
>
> 4 PI r^2
Really? Why does what you claim disagree with the calculation?
Since you can't even figure out how to transform the metric to a
different coordinate system, I find your claim highly dubious.
>
> > Why is it
> > someone totally uneducated in the craft managed to find an "error" in
> > the craft that none of the professionals have seen in over a century?
>
> That is because I am educated in and have thoroughly understood the
> craft. That is why I can find errors and lots of them where they have
> eluded physicists for almost 100 years. <shrug>
Really? Where were you educated? What did you learn from?
>
> > > Again, this cannot be true. Any metric or solution I give you is only
> > > for the coordinate that I have already chosen. Thus, each metric or
> > > solution I find represents a different geometry. <shrug>
>
> > Why is it I can relate two "different" geometries with a coordinate
> > transformation?
>
> Because there are infinite solutions to the field equations. <shrug>
How come what I just showed you is exactly consistent with what I have
been saying all along? Two "different" metrics that satisfy the same
boundary and initial conditions are related by a coordinate
transformation.
>
> > What does that say to you?
>
> All solutions to the field equations are related but not the same.
> Each metric is a solution to the same field equations. <shrug>
The set of all metrics that are related by a coordinate transformation
are a solution to the same field equations.
>
> > Remember my examples of Euclid in different coordinates? Think about
> > it.
>
> OK, I did. What I am saying still stands.
>
> > > They have not discovered another solution to the field equations that
> > > would allow an accelerated expansion in our universe without invoking
> > > the Cosmological constant, but I did. <shrug>
>
> > Hey look, more changing goalposts. Birkhoff's theorem is under
> > discussion, not the cosmological constant.
>
> The Cosmological constant basically implies the Birkhoff's theorem
> wrong.
Nope. Birkhoff's theorem doesn't assume a cosmological constant.
There is an equivalent of Birkhoff's theorem for the case of a
cosmological constant - it yields the de Sittter-Schwarzschild metric.
>
> > > No, I just refused to do your homework for you. <shrug>
>
> > I'm not the one posting about how all physicists and mathematicians
> > got basic tensor analysis wrong.
>
> That is correct. I am. <shrug>
Then where did you learn this stuff?
>
> > Was it more stupid than saying a metric cannot have a Lagrangian?
>
> No, a metric cannot have a Lagrangian. <shrug>
http://groups.google.com/group/sci.physics.relativity/msg/1289007518ac4f67?dmode=source
Yet another discussion you simply fled because you couldn't argue with
the mathematics.
>
> > More
> > stupid than saying you can spot curvature by inspecting the
> > coordinates?
>
> No, to describe the curvature, you must start with a set of coordinate
> system and use the appropriate metric associated with this choice of
> coordinate system. <shrug>
Forgotten already?
"The metric is Minkowski but the choice of coordinate system is
hopelessly non-linear. Thus, the only logical conclusion is that the
described spacetime must be curved despite the metric being
Minkowski."
http://groups.google.com/group/sci.physics.relativity/msg/00d08acc54532fd4?dmode=source
>
> > More stupid than saying you can introduce curvature
> > through a coordinate transformation?
>
> No, because the geometry is invariant. You cannot change the geometry
> through a coordinate transformation. <shrug>
"The metric is Minkowski but the choice of coordinate system is
hopelessly non-linear. Thus, the only logical conclusion is that the
described spacetime must be curved despite the metric being
Minkowski."
http://groups.google.com/group/sci.physics.relativity/msg/00d08acc54532fd4?dmode=source
>
> > > I have already told you my choice of coordinate system is the common
> > > spherically symmetric polar coordinate system already. There are no
> > > other coordinate systems I have referred to. Can you read what I
> > > wrote? Can you read at all? <shrug>
>
> > Except it isn't.
>
> > You chose a DIFFERENT radial coordinate.
>
> How? I have already settled on the common spherically symmetric polar
> coordinate system. Do you know what that is?
The usual radial coordinate is defined such that it represents spheres
of area 4*\pi*r^2. Yours does not qualify. Take the asymptotic limit
as r --> \infty, and you will see that it doesn't reduce to Minkowski
space.
>
> > There isn't one particular
> > radial coordinate - I could define a _new_ radial coordinate R =
> > exp(r), and it will retain the spherical symmetry. Your "r" _is_ a
> > radial coordinate, just not the usual one.
>
> My 'r' is always the usual radial coordinate. <shrug>
>
> > As explicitly shown. It doesn't reduce to Minkowski space in spherical
> > coordinates, nor does it have the proper area.
>
> At (r = infinity), the metric becomes Minkowkski. The area of the
> sphere is always (4 PI r^2). <shrug>
Neither of these are true. I proved the latter explicitly, and the
former is obvious through inspection. You could power expand the
metric components as a formal proof, but the answer won't change.
>
> What is this proper area you are describing?
4 * \pi * r^2
>
> > You either do not understand what is meant by spherical symmetry, or
> > you are being deliberately stupid. Which is it?
>
> Neither. <shrug>
>
> > > Yes, same geometry, different metrics, different coordinate systems.
> > > <shrug>
>
> > Very good. You understand that much.
>
> I understand very well in this subject.
>
> > Now think about it a little more. I can construct transformations
> > [TENSOR transformations] that can relate all those representations.
> > What does that tell you?
>
> Geometry must be invariant. Coordinate must be chosen before
> describing the geometry. In choosing a coordinate system, you must
> find the appropriate metric associated with this coordinate system.
> Finally, the field equations yield an infinite numbers of solutions.
> Given the same coordinate system, each solution or metric describes a
> very different geometry. The field equations describe a infinity
> numbers of geometries.
>
> > > You and the physicists need to stop mis-apply the mathematics.
> > > <shrug>
>
> > You don't even know how to transform a tensor, much less have the
> > ability to make wild claims like "all physicists are wrong".
>
> As I said transformation of tensors do not apply here. All physicists
> are wrong.
So where did you learn the material? If all physicists and
mathematicians are wrong, there isn't even one piece of literature for
you to learn the material from.
In reality, your understanding got corrupted early and rather than
fixing the problem you simply bleat over and over that ALL
mathematicians and physicists have been getting it wrong for over a
century.
Koobee Wublee
> On Jul 2, 12:37 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > That is not what you are saying. You have claimed the gravitating
> > mass can only solely determined through the Newtonian law of gravity,
> > but you claimed so solely from Kepler's 3rd law. Kepler's laws only
> > work for some special circumstance. Ultimately, it is Newton's law of
> > gravity that describes these phenomena. <shrug>
>
> Nope. Read what I said. I never said Kepler's 3rd law was the only
> way, just that it is how I would do it.
>
> It is well understood that Kepler's laws are approximations. For
> example, Kepler's 3rd law is only valid when the main body is much
> more massive than the orbiting body.
http://groups.google.com/group/sci.physics.relativity/msg/7971142029f6543c?hl=en&
I said:
"By examining an object orbiting a gravitating body, one can determine
the mass of the gravitating body. Correct me if I am wrong. This is
Newtonian law of gravity."
Then, you corrected me:
"No. Kepler's laws."
Therefore, according to you, it is not Newton's law of gravity that
determines the gravitating mass, but now you are claiming Kepler's law
is derived from Newton's law of gravity. You are contradicting
yourself as usual. <shrug>
> > I did, and "Geometry = Metric * Coordinate" <shrug>
>
> Really? What book?
Right here.
> > It makes no sense to consider the metric as the geometry. <shrug>
>
> Why? What information do I need other than the metric?
Yes, the choice of coordinate system. <shrug>
> > I get you don't understand Birkhoff's theorem. It assumes the
> > Schwarzschild metric is the only solution and establishes the solution
> > must be spherically symmetric and asymptotically flat. <shrug>
>
> No, it doesn't. Birkhoff's theorem was explained in excruciating
> detail to you. All that is assumed is spherical symmetry and a time-
> like killing vector at infinity.
Not all the solutions to the field equations obey the Birkhoff's
theorem. Your silly worship of the Birkhoff's theorem is plainly
wrong.
> > No, I did not introduce curvature through a coordinate
> > transformation. Remember the geometry is invariant. <shrug>
>
> But you claimed you could.
No, I never did. Remember the geometry is invariant. <shrug>
> "The metric is Minkowski but the choice of coordinate system is
> hopelessly non-linear. Thus, the only logical conclusion is that the
> described spacetime must be curved despite the metric being
> Minkowski."
That is correct. The Minkowski metric and choice of nonlinear
coordinate system results in describing curved geometry. <shrug>
> http://groups.google.com/group/sci.physics.relativity/msg/00d08acc545...
>
> Do you retract this statement?
No.
> > The choice of coordinate system must be firmly established before we
> > can proceed to get to the solutions of the field equations. <shrug>
>
> No, it doesn't.
>
> The popular derivation of the Schwarzschild solution goes through a
> half dozen coordinate changes.
No, you don't know the math involved. <shrug>
> > No, I have shown you a number of solutions already. <shrug>
>
> Yes, and they are all the _same_ solution.
Again, the coordinate system is the same. Each solution must
represent a different geometry. <shrug>
> > Your math is very poor. The metric transforms like a matrix. <shrug>
>
> Prove it.
Get a book on matrices. <shrug>
> > I don't give a damn about tensors at this moment because the metric is
> > a matrix. <shrug>
>
> One representation is a matrix, but that does not mean a rank 2 tensor
> has all the properties of matrices.
The metric has all the properties of a matrix. <shrug>
> > Has it get to your head that the physicists have been mis-applying
> > mathematics to differential geometry for almost 100 years?
>
> How do you know?
This is what we have been discussing. So, you don't understand what I
am saying, and I really don't care. There is no point in continuing.
Go to argue against yourself.
Eric Gisse
[...]
As expected, you run away when cornered.
Koobee Wublee
> As expected, you run away when cornered.
No, I did not run away. I do not wish to engage in meaningless
discussions.
In the post below, I said "Has it get to your head that the physicists
have been mis-applying mathematics to differential geometry for almost
100 years?"
http://groups.google.com/group/sci.physics.relativity/msg/36f0301fc6e8dcc4?hl=en&
You replied in the post below where you said "How do you know?"
http://groups.google.com/group/sci.physics.relativity/msg/ac7fceaa6d0ce678?hl=en&
Well, you obviously do not try to understand my point of view. You
don't really care. You are indeed a troll. As an agent of the
politically correct to silence any logical thinking, how much do they
pay you to do this?
If nothing, you must be the most evil person in the world. If so, you
cannot stop the ones who think more logically than you brain-washed.
I was brain-washed once, and I vowed not to be so again. The
revolution has begun, and it will be successful if we can get to the
young ones before they are brain-wahsed. One day we will be laughing
at the ones like your kind who worship the nitwit Einstein. <shrug>
Eric Gisse
> On Jul 3, 12:58 pm, Eric Gisse <jowr...@gmail.com> wrote:
>
> > As expected, you run away when cornered.
>
> No, I did not run away. I do not wish to engage in meaningless
> discussions.
Good. Then have a meaningful discussion instead of dodging questions.
Explain how come the calculated area using your line element is
something other than 4*pi*r^2. I would prefer a counter with
_MATHEMATICS_ rather than pure assertion.
Explain how come I can, using a simple coordinate transformation,
express your line element in Schwarzschild coordinates. Try to retain
self-consistency with what you said about the Euclid examples.
Explain how come the surface area computed in Schwarzschild
coordinates transforms into the surface area derived previously?
Explain where you learned general relativity and differential geometry
if all physicists and mathematicians have gotten it wrong. While
explaining, try to fit in a rational explanation for why you are
unable to transform the metric from one coordinate system to another.
>
> In the post below, I said "Has it get to your head that the physicists
> have been mis-applying mathematics to differential geometry for almost
> 100 years?"
>
> http://groups.google.com/group/sci.physics.relativity/msg/36f0301fc6e...
>
> You replied in the post below where you said "How do you know?"
Is there something difficult you find about that question?
Why is it that you are unwilling or unable to explain how you came to
the conclusion that an entire century's worth of scientists _GOT IT
WRONG_ in the most basic fashion?
>
> http://groups.google.com/group/sci.physics.relativity/msg/ac7fceaa6d0...
>
> Well, you obviously do not try to understand my point of view. You
> don't really care. You are indeed a troll. As an agent of the
> politically correct to silence any logical thinking, how much do they
> pay you to do this?
I have asked you, many times, to provide the references from which you
learned what you know. You have consistently been either unwilling or
unable to provide them.
>
> If nothing, you must be the most evil person in the world. If so, you
> cannot stop the ones who think more logically than you brain-washed.
> I was brain-washed once, and I vowed not to be so again. The
> revolution has begun, and it will be successful if we can get to the
> young ones before they are brain-wahsed. One day we will be laughing
> at the ones like your kind who worship the nitwit Einstein. <shrug>
The only Einstein-worshipper I see here is you. You promote him to the
status of a religious figure whose words are treated as gospel and
whose predictions are taken on faith despite neither of those being
true in the least.