On Apr 29, 8:56 pm, Koobee Wublee wrote:

Sure, they all do just like every verse in the bible has at least one
interpreted meaning. <shrug

Again, not my fault you don't know what the words mean. Take a course
in differential geometry or topology some time.

What are these boundary and/or initial conditions?

Depends. Spherical symmetry is an initial condition, for example.

This is utter bullsh*t. You were told of these and don’t even bother
to understand it. <shrug

Can't you do better than abject denial, kooby? Open an ODE textbook
[Boyce & DiPrima] or a PDE textbook [DuChateau & Zachmann] and read
the sections that discuss uniqueness theorems and what is required to
satisfy them.

The solutions yielded by a set of differential equations have no more
extra meaning than the solutions of a quadratic equation such as (x^2
- 3 x + 2 = 0). In this case, x = 1 or 2. Unless you can prove (1 > > 2), the solutions are independent of each other. However, 1 is a
transform from 2, and 2 is a transform from 1. Even if they are a
transform of each other, 1 is still not 2, and 2 is still not 1. This
should be a very basic mathematical concept, but I see that it bothers
and most physicists. Why?

What are you talking about? One is not a "transform" of two - the
numbers are scalars.

Please don't pretend that you have your finger on the pulse of physics
when you don't understand the base concepts involved here.

You have an excuse because you are a multi-
year super senior. How about the professors and PhD’s? Correct me
if I am wrong. The 1st year algebra student should have already
understood this very simple but basic mathematical concept. <shrug

This has never been done. You are dreaming again. <shrug

Progress beyond denial, kooby. You are fooling nobody.

Go back and read the archives.

As I said many times over, the metric cannot be a tensor.

As you were told many times, you are wrong.

It is merely a matrix.

A matrix is just an ordered set of numbers.

A tensor obeys special
properties that happen to be shared with rank 2 matrices.

Isn't learning fun?

That is because the geometry must be invariant. To
describe the geometry, you need the metric and the coordinate system.

To describe the geometry _in a particular coordinate basis_ you need
to project the metric upon a coordinate basis.

Since coordinate system depends on each observer, the metric must vary
with the coordinate system to yield an invariant geometry. This very
basic concept in geometry and methodology of experimentation still
eludes you and all physicists. The nonsense about the metric being
invariant has been around since Ricci’s time. Mysticism is always
wisdom in disguise, is it not?

You wax poetic about a subject you have spent years arguing about but
have never understood.

The word invariance - like tensor - has a
specific mathematical meaning which you simply do not understand.

So what? Black holes or no black holes are serious matters. <shrug

Again...so?

Different solutions have different properties.

I struggle
to understand why you think this is relevant.

What gives you the right to call the Schwarzschild metric more unique
than Schwarzschild’s original metric? Why is Schwarzschild metric not
a transformation of Schwarzschild’s original metric instead?

The modern Schwarzschild solution is preferred because surfaces of
constant r have areas of 4pir^2.

Do you know how to compute the surface area of a surface of a constant
coordinate value yet?

So? Thus, the interior of the sun behaves like an FLRW metric, right?

Is the sun expanding as a function of time? Is the sun homogeneous and
isotropic?

You don’t even know how to transform from Schwarzschild’s original
solution to the Schwarzschild metric. <shrug

http://arxiv.org/pdf/physics/0503095.pdf

Go learn something.

Yes, it is not your fault that you are a multi-year super senior. I
seem to have heard of that excuse many times over though. Since the
metric cannot be a tensor, there is no point of transforming the
metric into something that is still the same metric. <shrug

I will never cease to be amused by your utter certainty that you
understand the subject despite not being able to get even the basic
concepts right, much less compute something.

Since we are clearly revisiting old ground, let us go back to the last
time. It'll make the conversation go faster.
http://groups.google.com/group/sci.physics.relativity/browse_frm/thread/51e774c2099e11b0

You are so sore loser. You have been checkmated many times over and
still refuse to accept defeat. That is also why you are still a super
senior. Is the University of Alaska really that difficult to get a
degree from? I thought they pay you to go to college over there.

I don't discuss personal matters with assholes.

On Apr 29, 10:37 pm, Eric Gisse <jowr...@gmail.com> wrote:

On Apr 29, 8:56 pm, Koobee Wublee wrote:
Sure, they all do just like every verse in the bible has at least one
interpreted meaning.  <shrug

Again, not my fault you don't know what the words mean. Take a course
in differential geometry or topology some time.

Again, every single verse in the bible has several different meanings
depending on the politics.  <shrug

What are these boundary and/or initial conditions?

Depends. Spherical symmetry is an initial condition, for example.

They are all spherical!  All use the good old spherically symmetric
polar coordinate system.  Want to try another bullsh*t answer?

This is utter bullsh*t.  You were told of these and don’t even bother
to understand it.  <shrug

Can't you do better than abject denial, kooby? Open an ODE textbook
[Boyce & DiPrima] or a PDE textbook [DuChateau & Zachmann] and read
the sections that discuss uniqueness theorems and what is required to
satisfy them.

There needs no denial in mathematics.  You have failed to see the
logic in it.  You choose to hide behind some book published by some
unknown authors in which I am supposed to waste my time and money to
find it and to buy it.  You got to be kidding me.

The solutions yielded by a set of differential equations have no more
extra meaning than the solutions of a quadratic equation such as (x^2
- 3 x + 2 = 0).  In this case, x = 1 or 2.  Unless you can prove (1 > > > 2), the solutions are independent of each other.  However, 1 is a
transform from 2, and 2 is a transform from 1.  Even if they are a
transform of each other, 1 is still not 2, and 2 is still not 1.  This
should be a very basic mathematical concept, but I see that it bothers
and most physicists.  Why?

What are you talking about? One is not a "transform" of two - the
numbers are scalars.

Hello!  2 = 1 + 1, and 1 = 2 – 1.  These are perfect and valid
transforms.  <shrug

Please don't pretend that you have your finger on the pulse of physics
when you don't understand the base concepts involved here.

Your problem is that I do understand the pulse of physics.
Confronting me, you just fall apart.  All your argument is total
bullsh*t.

You have an excuse because you are a multi-
year super senior.   How about the professors and PhD’s?  Correct me
if I am wrong.  The 1st year algebra student should have already
understood this very simple but basic mathematical concept.  <shrug
This has never been done.  You are dreaming again.  <shrug

Progress beyond denial, kooby. You are fooling nobody.

Go back and read the archives.

You are correct.  I cannot fool anyone and have not attempted and will
not attempt to fool anyone.  <shrug>  However, the archive has no such
record of what you are describing.  You are delusional.

As I said many times over, the metric cannot be a tensor.

As you were told many times, you are wrong.

As I said many times over, the metric cannot be a tensor.

 It is merely a matrix.

A matrix is just an ordered set of numbers.

That is correct.  It also depends on how you group you coordinates.
shrug

A tensor obeys special
properties that happen to be shared with rank 2 matrices.

Yes, but that does not apply to the metric.  <shrug

Isn't learning fun?

Well, in this case for you is lack of learning.  It must really such
to lick up bullsh*t without really understanding the issues involved.
shrug

That is because the geometry must be invariant.  To
describe the geometry, you need the metric and the coordinate system.

To describe the geometry _in a particular coordinate basis_ you need
to project the metric upon a coordinate basis.

Of course, this is what I have been saying.  To describe the geometry,
you need to establish what your choice of coordinate system is and
choose the proper metric.  Thus, the metric is dependent on what
coordinate system you have chosen.  Just how many times do I have to
checkmate you on this issue?

Since coordinate system depends on each observer, the metric must vary
with the coordinate system to yield an invariant geometry.  This very
basic concept in geometry and methodology of experimentation still
eludes you and all physicists.  The nonsense about the metric being
invariant has been around since Ricci’s time.  Mysticism is always
wisdom in disguise, is it not?

You wax poetic about a subject you have spent years arguing about but
have never understood.

A fair conclusion is that you do not understand anything.  That is why
you remain a multi-year super senior at the University of Alaska.
shrug

The word invariance - like tensor - has a
specific mathematical meaning which you simply do not understand.

I have used it properly.  If you don’t understand what I mean, you are
the one who does not understand the meaning of invariance.  <shrug

So what?  Black holes or no black holes are serious matters.  <shrug

Again...so?

Again, black holes or no black holes are serious matters.  <shrug

Different solutions have different properties.

This is very correct.

I struggle
to understand why you think this is relevant.

Your struggle perplexes me.  As you have said a moment again that
different solutions have different properties, and yet you do not
understand the significance of a solution manifesting black holes and
another one that does not.  I cannot help you on this one.  Try some
psychological help.

What gives you the right to call the Schwarzschild metric more unique
than Schwarzschild’s original metric?  Why is Schwarzschild metric not
a transformation of Schwarzschild’s original metric instead?

The modern Schwarzschild solution is preferred because surfaces of
constant r have areas of 4pir^2.

Preferred is not a scientific methodology.  <shrug

Do you know how to compute the surface area of a surface of a constant
coordinate value yet?

Yes, always.  I have shown you the answer many times over.  <shrug

So?  Thus, the interior of the sun behaves like an FLRW metric, right?

Is the sun expanding as a function of time? Is the sun homogeneous and
isotropic?

You tell me.  It depends on your FLRW metric.

You don’t even know how to transform from Schwarzschild’s original
solution to the Schwarzschild metric.  <shrug

http://arxiv.org/pdf/physics/0503095.pdf

Go learn something.

That paper does not tell me jack sh*t other than what I have known
about Schwarzschild’s original solution and the Schwarzschild metric
(Hilbert’s solution).  <shrug

Yes, it is not your fault that you are a multi-year super senior.  I
seem to have heard of that excuse many times over though.  Since the
metric cannot be a tensor, there is no point of transforming the
metric into something that is still the same metric.  <shrug

I will never cease to be amused by your utter certainty that you
understand the subject despite not being able to get even the basic
concepts right, much less compute something.

Since we are clearly revisiting old ground, let us go back to the last
time. It'll make the conversation go faster.

 http://groups.google.com/group/sci.physics.relativity/browse_frm/thre....

Holy cow!  I don’t even remember that post of mine.  It is a great
post, is it not?  I am still amazed that I can express my point so
well.  It just buries all the clowns in which you know who they are.
It really makes my day.

You are so sore loser.  You have been checkmated many times over and
still refuse to accept defeat.  That is also why you are still a super
senior.  Is the University of Alaska really that difficult to get a
degree from?  I thought they pay you to go to college over there.

I don't discuss personal matters with assholes.

Likewise, but being a sore loser who refuses the logical argument I
presented is hardly an issue of personal matters.  Are you running out
of books to hide from?  I thought you are sitting on piles after piles
of books, no?

On Apr 29, 10:37 pm, Eric Gisse <jowr...@gmail.com> wrote:

On Apr 29, 8:56 pm, Koobee Wublee wrote:
Sure, they all do just like every verse in the bible has at least one
interpreted meaning.  <shrug

Again, not my fault you don't know what the words mean. Take a course
in differential geometry or topology some time.

Again, every single verse in the bible has several different meanings
depending on the politics.  <shrug

What are these boundary and/or initial conditions?

Depends. Spherical symmetry is an initial condition, for example.

They are all spherical!  All use the good old spherically symmetric
polar coordinate system.  Want to try another bullsh*t answer?

This is utter bullsh*t.  You were told of these and don’t even bother
to understand it.  <shrug

Can't you do better than abject denial, kooby? Open an ODE textbook
[Boyce & DiPrima] or a PDE textbook [DuChateau & Zachmann] and read
the sections that discuss uniqueness theorems and what is required to
satisfy them.

There needs no denial in mathematics.  You have failed to see the
logic in it.  You choose to hide behind some book published by some
unknown authors in which I am supposed to waste my time and money to
find it and to buy it.  You got to be kidding me.

The solutions yielded by a set of differential equations have no more
extra meaning than the solutions of a quadratic equation such as (x^2
- 3 x + 2 = 0).  In this case, x = 1 or 2.  Unless you can prove (1 > > > 2), the solutions are independent of each other.  However, 1 is a
transform from 2, and 2 is a transform from 1.  Even if they are a
transform of each other, 1 is still not 2, and 2 is still not 1.  This
should be a very basic mathematical concept, but I see that it bothers
and most physicists.  Why?

What are you talking about? One is not a "transform" of two - the
numbers are scalars.

Hello!  2 = 1 + 1, and 1 = 2 – 1.  These are perfect and valid
transforms.  <shrug

Please don't pretend that you have your finger on the pulse of physics
when you don't understand the base concepts involved here.

Your problem is that I do understand the pulse of physics.
Confronting me, you just fall apart.  All your argument is total
bullsh*t.

You have an excuse because you are a multi-
year super senior.   How about the professors and PhD’s?  Correct me
if I am wrong.  The 1st year algebra student should have already
understood this very simple but basic mathematical concept.  <shrug
This has never been done.  You are dreaming again.  <shrug

Progress beyond denial, kooby. You are fooling nobody.

Go back and read the archives.

You are correct.  I cannot fool anyone and have not attempted and will
not attempt to fool anyone.  <shrug>  However, the archive has no such
record of what you are describing.  You are delusional.

As I said many times over, the metric cannot be a tensor.

As you were told many times, you are wrong.

As I said many times over, the metric cannot be a tensor.

 It is merely a matrix.

A matrix is just an ordered set of numbers.

That is correct.  It also depends on how you group you coordinates.
shrug

A tensor obeys special
properties that happen to be shared with rank 2 matrices.

Yes, but that does not apply to the metric.  <shrug

Isn't learning fun?

Well, in this case for you is lack of learning.  It must really such
to lick up bullsh*t without really understanding the issues involved.
shrug

That is because the geometry must be invariant.  To
describe the geometry, you need the metric and the coordinate system.

To describe the geometry _in a particular coordinate basis_ you need
to project the metric upon a coordinate basis.

Of course, this is what I have been saying.  To describe the geometry,
you need to establish what your choice of coordinate system is and
choose the proper metric.  Thus, the metric is dependent on what
coordinate system you have chosen.  Just how many times do I have to
checkmate you on this issue?

Since coordinate system depends on each observer, the metric must vary
with the coordinate system to yield an invariant geometry.  This very
basic concept in geometry and methodology of experimentation still
eludes you and all physicists.  The nonsense about the metric being
invariant has been around since Ricci’s time.  Mysticism is always
wisdom in disguise, is it not?

You wax poetic about a subject you have spent years arguing about but
have never understood.

A fair conclusion is that you do not understand anything.  That is why
you remain a multi-year super senior at the University of Alaska.
shrug

The word invariance - like tensor - has a
specific mathematical meaning which you simply do not understand.

I have used it properly.  If you don’t understand what I mean, you are
the one who does not understand the meaning of invariance.  <shrug

So what?  Black holes or no black holes are serious matters.  <shrug

Again...so?

Again, black holes or no black holes are serious matters.  <shrug

Different solutions have different properties.

This is very correct.

I struggle
to understand why you think this is relevant.

Your struggle perplexes me.  As you have said a moment again that
different solutions have different properties, and yet you do not
understand the significance of a solution manifesting black holes and
another one that does not.  I cannot help you on this one.  Try some
psychological help.

What gives you the right to call the Schwarzschild metric more unique
than Schwarzschild’s original metric?  Why is Schwarzschild metric not
a transformation of Schwarzschild’s original metric instead?

The modern Schwarzschild solution is preferred because surfaces of
constant r have areas of 4pir^2.

Preferred is not a scientific methodology.  <shrug

Do you know how to compute the surface area of a surface of a constant
coordinate value yet?

Yes, always.  I have shown you the answer many times over.  <shrug

So?  Thus, the interior of the sun behaves like an FLRW metric, right?

Is the sun expanding as a function of time? Is the sun homogeneous and
isotropic?

You tell me.  It depends on your FLRW metric.

You don’t even know how to transform from Schwarzschild’s original
solution to the Schwarzschild metric.  <shrug

http://arxiv.org/pdf/physics/0503095.pdf

Go learn something.

That paper does not tell me jack sh*t other than what I have known
about Schwarzschild’s original solution and the Schwarzschild metric
(Hilbert’s solution).  <shrug

hanson wrote:

NONE of the presenters, nowhere in the entire presentation,

So, perhaps we should fire all the professionals and put all the

This would have great social impact... The public would no longer

Thanks for the laughs all your precious Einstein Dingleberries,

On Apr 29, 10:15 pm, Koobee Wublee wrote:

Again, every single verse in the bible has several different meanings
depending on the politics. <shrug

Way to go kooby - don't bother learning the technical terms of the
field, just fart and call it a religion.

That saves a lot of effort
that could be spent learning.

They are all spherical! All use the good old spherically symmetric
polar coordinate system. Want to try another bullsh*t answer?

Yea, expecting you to understand spherical symmetry was a bit
optimistic.

There needs no denial in mathematics. You have failed to see the
logic in it. You choose to hide behind some book published by some
unknown authors in which I am supposed to waste my time and money to
find it and to buy it. You got to be kidding me.

Or go to a library, or search for some other books that cover ODE &
PDE uniqueness theorems. Or stamp your feet because nobody will spoon
feed you an education. Whichever you decide!

Hello! 2 = 1 + 1, and 1 = 2 – 1. These are perfect and valid
transforms. <shrug

Congratulations! You have no idea what the word "transform" means.
I'll add it to the list.

Your problem is that I do understand the pulse of physics.
Confronting me, you just fall apart. All your argument is total
bullsh*t.

Have you figured out how to do an integral yet?

You are correct. I cannot fool anyone and have not attempted and will
not attempt to fool anyone. <shrug> However, the archive has no such
record of what you are describing. You are delusional.

JanPB explained it to you a year ago...

http://groups.google.com/group/sci.physics.relativity/msg/413dcbe1b5c...

Mentioned to you again 3 months later.

http://groups.google.com/group/sci.physics.relativity/msg/7c4b0cb7052...

Here I give you the explicit coordinate transformation.

http://groups.google.com/group/sci.physics.relativity/msg/5e4cf198adb...

Denial is not just a river in Egypt, now is it?

As I said many times over, the metric cannot be a tensor.

You think your proclamations are relevant because....?

Yes, but that does not apply to the metric. <shrug

Prove it.

Of course, this is what I have been saying. To describe the geometry,
you need to establish what your choice of coordinate system is and
choose the proper metric. Thus, the metric is dependent on what
coordinate system you have chosen. Just how many times do I have to
checkmate you on this issue?

You are deluded if you think you are "checkmating" me. This isn't a
physics issue, this is a "kooby doesn't get math" issue.

Your struggle perplexes me. As you have said a moment again that
different solutions have different properties, and yet you do not
understand the significance of a solution manifesting black holes and
another one that does not. I cannot help you on this one. Try some
psychological help.

In F = ma you can have solutions that are oscillatory in time or
solutions that are quadratic in time depending what you put in for F.
Explain how that is ok in F = ma but not ok in G_uv = kT_uv.

Preferred is not a scientific methodology. <shrug

You asked why, and you were told why. Don't stamp your feet because
you don't like what you were told.

Yes, always. I have shown you the answer many times over. <shrug

Yes - you _show_ the answer but you cannot derive the answer. Plus you
always show the same answer, even when it is wrong!

You tell me. It depends on your FLRW metric.

Your idea, your problem.

That paper does not tell me jack sh*t other than what I have known
about Schwarzschild’s original solution and the Schwarzschild metric
(Hilbert’s solution). <shrug

It gives you the coordinate mapping between the two. Don't cry because
you can't read for comprehension.