In article .com,
Schoenfeld wrote:

NOTE TO OP: The only relevant contributions to physics and math have
been done by "amateurs". Ironically, one cannot even begin to fathom
how far all have fallen due to the sneering elitist attitude just
demonstrated by Roberts.

Shut up kook

--
Relf's Law? -+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
"Bull**** repeated to the limit of infinity asymptotically approaches
the odour of roses."
Corollary -+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
³It approaches the asymptote faster, the more Œpseduos¹ you throw in
your formulas.²

--
Posted via a free Usenet account from http://www.teranews.com

JanPB wrote:
Koobee Wublee wrote:

Forget about Crothers. I have a more general solution to the field
equations than Crother's C(r). Let review it again.

ds^2 = c^2 (1 - K / R) dt^2 - ((dR/dr)^2 / (1 - K / R)) dr^2 - R^2 ...

Where

** K = Constant
** R(r)

This equation is just as you have claimed that "... coordinate changes
do not affect tensors".

Assuming your metric is correct, it must necessarily be a patch of the
Schwarzschild solution (that's a mathematical fact and is not
debatable).

Well, you wrong. Look at the metric more closely. Through change of
coordinate, it must satisfy G = 0 because the specific form
(Schwarzschild metric) does. Each general form of R as a function of r
represents a complete different metric just as unique as Schwarzschild
metric.

My complaint is that Crothers claims he has a family of solutions. What
he has is a family of diffeomorphic expressions of one and the same
tensor (Schwarzschild's metric).

Read Schwarzschild's orginal paper again, Crothers' is just a more
general form to the original metric discovered by Schwarzschild
himself. It is not the same as the Schwarzschild metric. This is the
best test to find out if you have understood the subject or not. So,
stop pretending you do.

Carrying out the derivatives of R, we have

** dR^2 = (dR/dr)^2 dr^2

The equation I showed earlier becomes

ds^2 = c^2 (1 - K / R) dt^2 - ((dR/dr)^2 / (1 - K / R)) dr^2 - R^2 ...

If (R = sqrt(C)), it yields exactly what Dr. Crother has written down.

If (R = r), it yields the exact equation you are familiar with below
with Schwarzschild metric.

ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / R) - r^2 ...

If (R = r + K), it yields Dr. Rahman's equation below.

ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 ...

With each one rushes out to claim a metric for his own, he is just
building a fat castle in the air. This includes you as well who only
accepts the Schwarzschild metric.

All these expressions (assuming they are correct which I haven't
checked) are the Schwarzschild metric. They are just written
differently. Nothing new can be obtained from these alternate forms by
definition. It's _the same tensor_. So e.g. if one has a singularity,
they all have it. End of story.

Wrong again. Mathematically, each one is a different metric. There
are an infinite number of solutions that can satisfy G = 0. Some
manifest singularities at r 0. Some manifest singularities at r = 0.
Some never manifest singularities.

Given the equation below,

ds^2 = c^2 A(R) dt^2 - B(R) dr^2 - C(R) ...

Solving G = 0, you have A(R) in 2nd derivative, B(R) in 1st derivative,
and C(R) also in 2nd derivative. This means there are 2 integration
constants to A(R), 1 to B(R), and 2 to C(R) associated with the overall
metric in its simplest form that is the Schwarzschild metric. A(R),
B(R), C(R) can all share the same integration constants. Thus, I got

A(R) = A0 (1 + B0 / R)
B(R) = 1 + B0 / R
C(R) = R^2

Where I have still not identified the 2 constants associated with C(R).

With the simplest solution where R(r) = r, B0 must be (- 2 G M / c^2)
to satisfy Newtonian gravity. A0 must be 1 to satisfy SR limit.

For more complicated R(r), these constants don't have to be the ones
you are familiar with. Luckily, the metric pointed out by Dr. Rahman
do share the same constants as Schwarzschild metric.

The singularity theorem is making me laugh. Oops, there, I admit it.

Read e.g. Hawking and Ellis p. 263 and tell us what's wrong with the
proof. "Make me laugh" is not a publishable result :-)

You don't have to worry about me publishing anything related to the
singularity theorem. shrug

Well, then all that talk about solutions without singularities are hot
air, QED.

Again, you are wrong. There is at least one solution that has the
singularity occuring at r = 0. This means the singularity theorem is a
phantom rock on that fat castle in the air.

tomgee wrote:
Igor wrote:
tomgee wrote:
Igor wrote:
tomgee wrote:
jem wrote:
tomgee wrote:

His idea that space is curved, e.g., or warped, is a conclusion
that is wrong, I believe, and my model offers another conclusion
that is a better explanation that that. For anything to warp, or
bend or fold, it must have the capacity to do so, and he failed
to explain how space could do such physical things when it has
no such capacity.



What makes you think space has the "capacity" to be straight?

You must be assuming that space has some capacity to
do something. That is like saying, "A tree can walk". A
tree does not have such capacity, of course. If you think
space has some capacity or another, tell us what you think
that may be and how it is that space can have the capacity
to do anything.

AE did not tell us how space can have the capacity to do a
single physical act; he just said it does. You can believe him
if you wish, but that only shows you're willing to be cannon
fodder for your leaders without questioning the right or wrong
of it.

But surely trees can walk if you define one of the things that they are
capable of doing as "walking".

Yes, surely, like in Alice in Wonderland, the cards are defined as
walking and talking cards.

You fall into this trap because you
don't understand what things mean in particulat contexts.

That's a trap? What do you call the fix you're in, believing
that fairytales are reality, where you can make trees walk
by simply defining that feature in them?

I didn't say I could make trees walk, you idiot! Learn to read.

Yes, you did. It's up there in B/W, so why deny it?

Ignorance is
no excuse, but it is easier to argue about something you know
absolutely nothing about than actually trying to learn about it.

For you, that's true, no doubt.

I understand what spacetime curvature means and you don't. Keep
wallowing in your own ignorance like a pig.

You understand that spacetime is real, that space curves, and
that trees walk! You're not ignorant - just stupid!

OK. I give up. You're right. There are no spheres or cylinders and
the earth is flat. just as surely as you're doomed to remain ignorant
for eternity.

Schoenfeld wrote:

NOTE TO OP: The only relevant contributions to physics and math have
been done by "amateurs".

That's a lie.

--
Jan Bielawski

Igor wrote:
tomgee wrote:
Igor wrote:
tomgee wrote:
Igor wrote:
tomgee wrote:
jem wrote:
tomgee wrote:

His idea that space is curved, e.g., or warped, is a conclusion
that is wrong, I believe, and my model offers another conclusion
that is a better explanation that that. For anything to warp, or
bend or fold, it must have the capacity to do so, and he failed
to explain how space could do such physical things when it has
no such capacity.



What makes you think space has the "capacity" to be straight?

You must be assuming that space has some capacity to
do something. That is like saying, "A tree can walk". A
tree does not have such capacity, of course. If you think
space has some capacity or another, tell us what you think
that may be and how it is that space can have the capacity
to do anything.

AE did not tell us how space can have the capacity to do a
single physical act; he just said it does. You can believe him
if you wish, but that only shows you're willing to be cannon
fodder for your leaders without questioning the right or wrong
of it.

But surely trees can walk if you define one of the things that they are
capable of doing as "walking".

Yes, surely, like in Alice in Wonderland, the cards are defined as
walking and talking cards.

You fall into this trap because you
don't understand what things mean in particulat contexts.

That's a trap? What do you call the fix you're in, believing
that fairytales are reality, where you can make trees walk
by simply defining that feature in them?

I didn't say I could make trees walk, you idiot! Learn to read.

Yes, you did. It's up there in B/W, so why deny it?

Ignorance is
no excuse, but it is easier to argue about something you know
absolutely nothing about than actually trying to learn about it.

For you, that's true, no doubt.

I understand what spacetime curvature means and you don't. Keep
wallowing in your own ignorance like a pig.

You understand that spacetime is real, that space curves, and
that trees walk! You're not ignorant - just stupid!

OK. I give up. You're right. There are no spheres or cylinders and
the earth is flat. just as surely as you're doomed to remain ignorant
for eternity.

Yes, I'm right that you're stupid in thinking there are "no spheres
or cylinders and the earth is flat". I may remain ignorant for lack
of education, or not, but you're well-educated and stupid, and
there ain't no cure for that.

tomgee wrote:
Igor wrote:

OK. I give up. You're right. There are no spheres or cylinders and
the earth is flat. just as surely as you're doomed to remain ignorant
for eternity.

Yes, I'm right that you're stupid in thinking there are "no spheres
or cylinders and the earth is flat". I may remain ignorant for lack
of education, or not, but you're well-educated and stupid, and
there ain't no cure for that.

It's very good indeed that you just post to the Usenet.

--
Jan Bielawski

JanPB wrote:
Koobee Wublee wrote:
JanPB wrote:
Koobee Wublee wrote:

If you cannot see other solution other than the Schwarzschild metric,
please don't whine at me. Go cry on your buddy JanPB's shoulder
instead. You need to grow up. Gee!

Sorry dude, the uniqueness of Schwarzschild follows immediately from
its method of derivation (even Schwarzschild mentions it already in
that December 1915 letter to Einstein: "Es gibt nur ein
Linienelement...") - see again http://mastersofcinema.org/jan/t.pdf

You need to read that paper you again yourself. That metric is not
Schwarzschild metric. I am very convinced that you don't understand
that. shrug

Words are cheap. Proof? I can prove my claims. Can you?

Oh, yes, I as well. You don't even bothet to read and understand that
paper you provided.

Not only that, Schwarzschild was (obviously) aware that coordinate
system choice does not affect the metric tensor (obviously) because
when he was writing this letter the theory was not 100% finished and it
was assumed the coordinate systems had to satisfy det=1. Nevertheless,
he uses a coordinate system that does NOT satisfy this condition merely
because the metric looks, as he puts it, "am schoensten" (nicest) in
them - and proceeds to write the well-known form. So he knew it didn't
matter back then, in 1915, he knew the solution was unique - something
that Crothers & Co. still don't understand in the 21st century. They
all should go back to school and study some differential geometry 101,
it's really basic stuff. I recommend the first two volumes of Spivak,
it's back in print.

It sounds like you are jealous of Crothers et al because they are able
to understand this subject.

Reread what I wrote. Where do you see jealousy in it? I'm simply saying
they don't know the basics which were clear even to Schwarzschild in
1915 yet they attempt an investigation of a rather subtle manifold
geometry.

The evidence is all in your hot-air like criticism of Crothers. Face
it. He read and understood Schwarzschild's oringinal paper. He saw a
more general solution to that metric discovered by Schwarzschild in
which the Schwarzschild metric is just a special case. And that, you
are jealous of Dr. Crothers' capability. You are a whiner.

It's kind of embarassing that this baloney got published (who was the
reviewer - that's what I'd like to know).

Because you don't understand it, it does not make the work baloney.
shrug

Rhetoric. The differential geometry 101 fact remains that you cannot
change a tensor by a coordinate switch.

Read and understand Schwarzschild's paper. You will understand the
product of two tensors can be the product of another two other tensors.
Do you even understand matrices? Personally, I doubt that you do.

Tom Roberts wrote:
JanPB wrote:
the uniqueness of Schwarzschild follows immediately from
its method of derivation

Not really. But in the vacuum exterior region where the metric is both
static and spherically symmetric, it does follow from Birkhoff's
theorem. There is a reason this is not known as "Schwarzschild's
theorem", and the reason is generality: Schwarzschild found a single
line element, Birkhoff _proved_ there are no others.

But didn't Schwarzschild's method prove uniqueness given staticity and
symmetry? I mean you solve the PDEs and they force 1-r/2m (etc.) on
you. Birkhoff removed the staticity assumption. BTW, did you know
Birkhoff's theorem was proved 2 years earlier by a Norwegian physicist?
I didn't: http://www.arxiv.org/abs/physics/0508163

Birkhoff's theorem does not require the spherically symmetric solution
to be Schwarzschild.

Of course not. I never said it did. (1)

Yes, but Dr. Roberts did.

In fact, it can be any spherically symmetric
solutions. There are an infinite number of them.

Prove it. (2)

Your statement (2) is contradicting your statement (1). So, as GR has
demonstrated, it is a dynamic theory capable of explaining any
observation because it yields an infinite number of solutions. Is that
affecting you too that you are able to agree and disagree a logic
statement at the same time?

It is time to tear down your fat castle in the air. It is not unique.
GR is BS.

Prove it.

Read what I have written you. I have shown the field equations do
yield an infinite number of metrics. Still in denial?

wrote:

Indeed. from Einstein's 1905 paper on SR, (Dover) The Principle of
Relativity, p.63, we find Einstein claiming the KE (W) of the moving
electron to be (\gamma - 1)mc^2, where m is the rest mass of the
electron and \gamma = 1 / [1 - v^2/c^2]^(1/2).

\gamma mc^2 is the total energy and mc^2 is the rest energy of the
electron, that is, the equivalent energy of the mass for the electron
in the rest frame of the electron. In other words

E = mc^2.

To get the above equation, E = m c^2

You have to assume

m = m0 / sqrt(1 - v^2 / c^2)

To get the above equation for mass, you have to assume

p = m v = m0 v / sqrt(1 - v^2 / c^2)

To get the above equation for observed momentum, you have to assume

E = m c^2 = m0 c^2 / sqrt(1 - v^2 / c^2)

With circular reason like that, how can you say Einstein proved (E = m
c^2)?

Koobee Wublee wrote:

The evidence is all in your hot-air like criticism of Crothers. Face
it. He read and understood Schwarzschild's oringinal paper. He saw a
more general solution to that metric discovered by Schwarzschild in
which the Schwarzschild metric is just a special case. And that, you
are jealous of Dr. Crothers' capability. You are a whiner.

This is total nonsense. Given the symmetry, the solution to G=0 is
unique. No amount of laborious coordinate changes will ever produce
anything but that unique tensor. If you clam otherwise then you should
be prepared to show a mistake in Birkhoff's derivation.

It's kind of embarassing that this baloney got published (who was the
reviewer - that's what I'd like to know).

Because you don't understand it, it does not make the work baloney.
shrug

Rhetoric. The differential geometry 101 fact remains that you cannot
change a tensor by a coordinate switch.

Read and understand Schwarzschild's paper. You will understand the
product of two tensors can be the product of another two other tensors.

No matter what manipulations you do to the tensor, it won't change as
long as it's required to remain symmetric and satisfy G=0.

Do you even understand matrices? Personally, I doubt that you do.

Oh give me a break. Not even worth a comment.

Tom Roberts wrote:
JanPB wrote:
the uniqueness of Schwarzschild follows immediately from
its method of derivation

Not really. But in the vacuum exterior region where the metric is both
static and spherically symmetric, it does follow from Birkhoff's
theorem. There is a reason this is not known as "Schwarzschild's
theorem", and the reason is generality: Schwarzschild found a single
line element, Birkhoff _proved_ there are no others.

But didn't Schwarzschild's method prove uniqueness given staticity and
symmetry? I mean you solve the PDEs and they force 1-r/2m (etc.) on
you. Birkhoff removed the staticity assumption. BTW, did you know
Birkhoff's theorem was proved 2 years earlier by a Norwegian physicist?
I didn't: http://www.arxiv.org/abs/physics/0508163

Birkhoff's theorem does not require the spherically symmetric solution
to be Schwarzschild.

Of course not. I never said it did. (1)

Yes, but Dr. Roberts did.

I was just being sloppy with my terminology - when I said originally
"the uniqueness of Schwarzschild follows immediately from its method of
derivation" I meant given Schwarzschild's own assumptions (i.e.,
staticity was included) - which is not modern usage. Tom corrected me
on that.

In fact, it can be any spherically symmetric
solutions. There are an infinite number of them.

Prove it. (2)

Your statement (2) is contradicting your statement (1).

See above.

So, as GR has
demonstrated, it is a dynamic theory capable of explaining any
observation because it yields an infinite number of solutions. Is that
affecting you too that you are able to agree and disagree a logic
statement at the same time?

You misunderstood the exchange. It was just my carelessness when I
conflated the two meanings of the phrase "the uniqueness of
Schwarzschild follows immediately from its method of derivation": 1.
derivation that assumes only the symmetry, 2. original derivation by
Schwarzschild. When you type quickly, things like this happen
sometimes.

It is time to tear down your fat castle in the air. It is not unique.
GR is BS.

Prove it.

Read what I have written you. I have shown the field equations do
yield an infinite number of metrics. Still in denial?

No denial, it's just a fact that if you assume spherical symmetry then
the metric tensor is forced on you uniquely. It's really quite easy
(although a bit messy, it's less work using Cartan's frames rather than
Christoffel symbols).

--
Jan Bielawski

JanPB wrote:
Schoenfeld wrote:

NOTE TO OP: The only relevant contributions to physics and math have
been done by "amateurs".

That's a lie.

How many PhD's did Einstein have in 1905?
How many PhD's did Faraday have when he proposed the electric field of
force?


How many PhD's did the elite of the times, those who were calling
others laymen and telling them not to study physics because they "have
no chance", have when they were talking about the dynamics of the
electrical fluid (pre-faraday)?

The only lie here is the nonsense ordained idiots would have people
believe (i.e. string theory/loopy quantum gravity/unobservable dark
matter halo's). The worst part is they tell others "not to study"
because "they have no chance" when in reality they are the only ones
with a chance.




--
Jan Bielawski