Are There Unresolver Foundational Issues With GR
Bill Hobba
GR. And to the best of my knowledge there isn't. But an actual
knowledgeable poster thought I was joking. Does anyone know what of any
actual unresolved issues? No cranks please - although I doubt it will stop
anyone.
Thanks
Bill
Eric Gisse
None that I'm aware of, outside the usual noise about singularities
and not-being-quantum.
>
> Thanks
> Bill
Bill Hobba
"Eric Gisse" <jow...@gmail.com> wrote in message
news:1175398394....@y80g2000hsf.googlegroups.com...
Same here - maybe someone like Steve Carlip will give an answer.
Thanks
Bill
>
>>
>> Thanks
>> Bill
>
>
Koobee Wublee
There are no unresolved foundational issues with GR because the
mathematical foundation that GR is based is wrong for at least a
hundred years. Since a hypothesis in physics has to be backed up by
solid mathematics in which physics is still not mathematics, GR's
foundation is actually very faulty. Here are the basics.
1. There are many ways to write the geodesic equation where each way
would yield a different set of connection coefficients. Christoffel's
way is not unique, and the Christoffel symbols are not unique.
2. The covariant derivative was an operator suggested by Ricci based
on the non-unique Christoffel symbols of the second kind. For each
set of connection coefficients to the geodesic equation, there is one
such operator where each such operator is very different from the
other.
3. The Riemann curvature tensor was derived through the covariant
derivative. It was designed by Ricci. There are many but
mathematically different ways to present the Riemann curvature
tensor. Ricci only found one.
4. The Ricci curvature tensor was designed by Ricci's faithful
student Levi-Civita from the Riemann curvature tensor.
5. The Lagrangian to the Einstein-Hilbert action was designed or
cooked up as an alchemist would have done in the medieval times by
Hilbert himself. The main ingredient was none other than the Ricci
curvature tensor.
6. The Einstein field equations are derived by dividing that
Lagrangian Hilbert pulled out of his *ss with each of the elements to
the metric. Since there are 16 elements, there are indeed 16 such
equations.
7. The Schwarzschild metric is one of the infinite numbers of
solutions to the ordinary differential equations (the field equations)
for static spacetime. The creation of a black hole is strictly how
one would embrace one non-unique solution to another. It allows
anyone to play God through a pen and a piece of paper.
The above represent the mathematical foundation of GR. It shows how
the curvature in spacetime is shaped by the mathematical tool invented
by Ricci. Yes, Ricci is the Creator, and Levi-Civita, Hilbert,
Schwarzschild, and Einstein are His prophets.
However, the general concept of the metric being invariant is totally
wrong. The curvature in spacetime must be invariant. This should be
a no-brainer to accept. This invariant geometry is described by the
dot product of the metric itself and the choice of coordinate. If the
choice of coordinate is varying from one observer to another, the
metric must vary with the choice of coordinate system to define an
invariant geometry. This is a concept taught in elementary school
where these kids do understand, but the GRists seem to have great
difficulties grasping. <sigh>
There are more. Need I to continue? These should be enough for one
night.
Thanks for asking.
Eric Gisse
> On Mar 31, 7:26 pm, "Bill Hobba" <rubb...@junk.com> wrote:
>
> > In another post I mentioned there are no unresolved foundational issues with
> > GR. And to the best of my knowledge there isn't. But an actual
> > knowledgeable poster thought I was joking. Does anyone know what of any
> > actual unresolved issues? No cranks please - although I doubt it will stop
> > anyone.
>
> There are no unresolved foundational issues with GR because the
> mathematical foundation that GR is based is wrong for at least a
> hundred years. Since a hypothesis in physics has to be backed up by
> solid mathematics in which physics is still not mathematics, GR's
> foundation is actually very faulty. Here are the basics.
Here comes another Koobee Wublee "shit and run" post.
>
> 1. There are many ways to write the geodesic equation where each way
> would yield a different set of connection coefficients. Christoffel's
> way is not unique, and the Christoffel symbols are not unique.
What is interesting is that you continue to assert this despite having
no evidence that it is true.
I asked you to prove this statement once before, and you just ran off.
It surprises me not in the least to see it again. Isn't it interesting
to see how often you criticize people for not 'showing the math' when
you have never, not even once, shown yours?
>
> 2. The covariant derivative was an operator suggested by Ricci based
> on the non-unique Christoffel symbols of the second kind. For each
> set of connection coefficients to the geodesic equation, there is one
> such operator where each such operator is very different from the
> other.
An even more idiotic conclusion which follows from a dumb premise.
>
> 3. The Riemann curvature tensor was derived through the covariant
> derivative. It was designed by Ricci. There are many but
> mathematically different ways to present the Riemann curvature
> tensor. Ricci only found one.
Show us another, then show that they are different.
>
> 4. The Ricci curvature tensor was designed by Ricci's faithful
> student Levi-Civita from the Riemann curvature tensor.
>
> 5. The Lagrangian to the Einstein-Hilbert action was designed or
> cooked up as an alchemist would have done in the medieval times by
> Hilbert himself. The main ingredient was none other than the Ricci
> curvature tensor.
Have you already forgotten what Bill Hobba was asking about?
>
> 6. The Einstein field equations are derived by dividing that
> Lagrangian Hilbert pulled out of his *ss with each of the elements to
> the metric. Since there are 16 elements, there are indeed 16 such
> equations.
Duh. 4x4 = 16.
At any rate, why is that a problem? There is more than one path to the
field equations.
>
> 7. The Schwarzschild metric is one of the infinite numbers of
> solutions to the ordinary differential equations (the field equations)
> for static spacetime. The creation of a black hole is strictly how
> one would embrace one non-unique solution to another. It allows
> anyone to play God through a pen and a piece of paper.
>
BZZZT, TRY AGAIN.
The field equations are _partial_ differential equations but the
symmetry of the situation allows them to be reduced to ordinary
differential equations, whose solutions are unique if they satisfy the
ODEs as well as the boundary conditions.
> The above represent the mathematical foundation of GR. It shows how
> the curvature in spacetime is shaped by the mathematical tool invented
> by Ricci. Yes, Ricci is the Creator, and Levi-Civita, Hilbert,
> Schwarzschild, and Einstein are His prophets.
You sure do seem to whine a lot about who did what.
>
> However, the general concept of the metric being invariant is totally
> wrong. The curvature in spacetime must be invariant. This should be
> a no-brainer to accept. This invariant geometry is described by the
> dot product of the metric itself and the choice of coordinate. If the
> choice of coordinate is varying from one observer to another, the
> metric must vary with the choice of coordinate system to define an
> invariant geometry. This is a concept taught in elementary school
> where these kids do understand, but the GRists seem to have great
> difficulties grasping. <sigh>
The only one who is having "difficulties grasping" here is YOU. YOU
have been told _countless_ times that while the coordinate
representation of a tensor is obviously coordinate dependent, the
tensor itself is not. Shortly after being told that, it has also been
asked of you to prove the assertion wrong with a counter-example or
some kind of actual mathematical proof.
To date, all you have been able to do is do a coordinate
transformation and say "LOOK! THEY ARE DIFFERENT!" without actually
proving that they are different. Cmon KW, follow your own maxim: SHOW
THE MATH.
For reference, here are some of your previous dumbass claims that you
have never proven.
* The Minkowski metric does not admit a Lagrangian. I explicitly
derived the Lagrangian for you....and you never posted to that thread
again.
* A coordinate transformation can introduce curvature. You kept
asserting it, but you never proven it.
* A tensor is not invariant under a coordinate transformation. You
continue asserting it, but you have never proven it.
* The surface area for space of constant radius for the Schwarzschild
metric is 4pi*r^2. Despite the careful proof to the contrary by JanPB,
you continue to assert that it is.
>
> There are more. Need I to continue? These should be enough for one
> night.
Yes, how about some actual mathematics to prove your dumbass
assertions?
haha like that will ever happen. You don't even have a firm grasp of
multivariable calculus.
>
> Thanks for asking.
Tom Roberts
> 1. There are many ways to write the geodesic equation where each way
> would yield a different set of connection coefficients. Christoffel's
> way is not unique, and the Christoffel symbols are not unique.
Well, yes and no. Yes, the Christoffel symbols are coordinate dependent,
and since coordinates are not unique neither are they. But for a given
coordinate system, in GR they are unique -- consistency with the metric
requires that.
This indeterminacy of the Christoffel symbols is of no concern in GR (or
in differential geometry in general), because such coordinate-dependent
quantities do not represent any physical process (geometrical quantity).
For instance, in a given manifold with metric one can compute geodesic
paths using any coordinate system, and one obtains the same set of
paths. All physical quantities (geometrical quantities), such as paths,
are independent of coordinates -- the variation of the Christoffel
symbols with coordinates is REQUIRED for this to be so.
> 2. The covariant derivative was an operator suggested by Ricci based
> on the non-unique Christoffel symbols of the second kind. For each
> set of connection coefficients to the geodesic equation, there is one
> such operator where each such operator is very different from the
> other.
That is a very old-fashioned and outmoded viewpoint. Today the covariant
derivative is considered an operator in its own right, with an axiomatic
definition that is completely independent of coordinates. When one
applies the modern definition to a specific coordinate system, one
obtains the old formulas, so this approach is equivalent, but avoids the
inherent coordinate dependence of the earlier method -- clearly a good
thing as the old method obviously confused you. You should get a modern
textbook and LEARN about modern methods; I like this one:
Frankel, _The_Geometry_of_Physics_.
> 3. The Riemann curvature tensor was derived through the covariant
> derivative. It was designed by Ricci. There are many but
> mathematically different ways to present the Riemann curvature
> tensor. Ricci only found one.
Not true. The Riemann curvature tensor is unique for any given manifold
with metric. What varies are its PROJECTIONS onto different coordinate
systems (aka components).
> 5. The Lagrangian to the Einstein-Hilbert action was designed or
> cooked up as an alchemist would have done in the medieval times by
> Hilbert himself. The main ingredient was none other than the Ricci
> curvature tensor.
Hmmm. There is no specific algorithm for finding Lagrangians. Medieval
alchemists were not knowledgeable of this sort of thing. Apparently you
aren't, either. <shrug>
> 6. The Einstein field equations are derived by dividing that
> Lagrangian Hilbert pulled out of his *ss with each of the elements to
> the metric.
This is plain and simply false -- there is no such "dividing" at all.
What is done is the _variational_derivative_, a well known and essential
part of Lagrangian mathematics.
> Since there are 16 elements, there are indeed 16 such
> equations.
Well, yes. But they are not all independent....
> 7. The Schwarzschild metric is one of the infinite numbers of
> solutions to the ordinary differential equations (the field equations)
> for static spacetime.
Well, yes. But given the requirements of a static manifold and spherical
symmetry, the Schw. solution is unique. You seem completely mystified by
the solutions to differential equations, and the uniqueness theorems
thereof. More education is necessary before you can understand this
stuff....
> The creation of a black hole is strictly how
> one would embrace one non-unique solution to another. It allows
> anyone to play God through a pen and a piece of paper.
<giggle> Obviously you do not understand the relationship between
theoretical physics and the real world. Nor the theorems on uniqueness
of solutions to differential equations. <shrug>
> The above represent the mathematical foundation of GR.
Not even close. The above represents your personal distortions of your
peculiar mis-perceptions of GR.
> However, the general concept of the metric being invariant is totally
> wrong.
It is a tensor, hence invariant under coordinate changes. <shrug>
> The curvature in spacetime must be invariant. This should be
> a no-brainer to accept.
Hmmm. Curvature is indeed invariant, hence it is represented by a
tensor. After all, coordinate invariance is what tensors were invented
to model.
> This invariant geometry is described by the
> dot product of the metric itself and the choice of coordinate.
Nonsense -- there is no "dot product" involved. Your statement does not
even make sense (one takes dot products between VECTORS, not between a
rank-2 tensor and a coordinate system). As I keep saying, you need to
LEARN what these words and concepts actually mean.
The metric (tensor) is an invariant description of the geometry, in that
all geometrical quantities can be obtained from it. As it is inherently
independent of coordinates, so is this description.
As ought to be obvious, but you clearly missed it, when someone
says "the metric" they mean "the metric TENSOR". Some older
books conflate the meaning of a tensor with its components,
so unwary readers can get confused. That's why you need a
MODERN textbook.
> If the
> choice of coordinate is varying from one observer to another, the
> metric must vary with the choice of coordinate system to define an
> invariant geometry.
You are obviously confusing the metric (tensor) with its projection onto
various coordinate systems -- the COMPONENTS of the metric tensor. The
metric (tensor) does not vary with coordinates, only the COMPONENTS of
the metric (tensor) do so: this variation is specifically defined so the
metric (tensor) itself remains invariant. You SERIOUSLY need to get a
modern textbook and LEARN about modern differential geometry, and modern
physics. <shrug>
Let my arm point to the sun rising directly to the east. My arm
is manifestly completely independent of coordinates. In
coordinates with x=east, y=north, z=up, the components of the
unit vector representing my arm are (1,0,0); in coordinates
with x=south, y=east, z=up, the components of THE SAME UNIT
VECTOR REPRESENTING THE SAME ARM are (0,1,0). The VECTOR has
not changed, only its components have changed with the change
in coordinates. This vector is a rank-1 tensor, and the same
property holds for all tensors. Coordinates are used for a
DESCRIPTION of geometrical and physical quantities, which
cannot possibly depend on the coordinates used to describe
them -- my arm is manifestly completely independent of
coordinates, and the mathematical quantity used to model it
must also be independent of coordinates: THAT'S WHY WE USE
VECTORS AND TENSORS TO MODEL GEOMETRICAL AND PHYSICAL
QUANTITIES.
Tom Roberts
carlip...@physics.ucdavis.edu
> In another post I mentioned there are no unresolved foundational
> issues with GR. And to the best of my knowledge there isn't.
> But an actual knowledgeable poster thought I was joking. Does
> anyone know what of any actual unresolved issues?
It depends on what you count as "foundational," and also on
what scope of issues you are allowing. The main unresolved
issues that might fall under this category -- you can decide
whether they do or not -- include:
1. We don't have a quantum theory of gravity. More generally,
we don't know how to couple general relativity consistently
to quantum matter (though we have good approximations).
2. General relativity predicts the existence of singularities
under a number of physically reasonable circumstances. Some
people consider this a fundamental problem. (Some others
think it comes from issue #1.)
3. We don't yet know whether general relativity predicts "cosmic
censorship" (that is, essentially, that singularities are
generically hidden behind horizons). Is this "foundational"?
It presumably has an answer within the existing framework;
we just don't know what it is.
4. General relativity, like all other standard theories currently
existing in physics, gives predictions that depend on initial
data, but does not tell us what the "right" initial data is
for our universe. If you're doing cosmology, this is a problem
-- it would be nice to have a unique initial condition.
5. Many physicists believe that there should be a single unified
theory that encompasses all fundamental physics. General
relativity isn't one.
6. The Universe apparently has a nonzero cosmological constant.
General relativity allows this, but does not determine its
value; the observed value seems "unnaturally" small, and it
would be good to have an explanation of this. (This is again
tied up with issue #1.)
My own take is that these aren't "foundational issues" in general
relativity, but rather indications that general relativity isn't
the final answer, but needs to be extended in various ways. But
as I said, this depends partly on definitions of terms.
Steve Carlip
Igor
> On Mar 31, 7:26 pm, "Bill Hobba" <rubb...@junk.com> wrote:
>
> > In another post I mentioned there are no unresolved foundational issues with
> > GR. And to the best of my knowledge there isn't. But an actual
> > knowledgeable poster thought I was joking. Does anyone know what of any
> > actual unresolved issues? No cranks please - although I doubt it will stop
> > anyone.
>
> There are no unresolved foundational issues with GR because the
> mathematical foundation that GR is based is wrong for at least a
> hundred years. Since a hypothesis in physics has to be backed up by
> solid mathematics in which physics is still not mathematics, GR's
> foundation is actually very faulty. Here are the basics.
>
> 1. There are many ways to write the geodesic equation where each way
> would yield a different set of connection coefficients. Christoffel's
> way is not unique, and the Christoffel symbols are not unique.
Nobody ever said they were. They're also not tensors.
> 2. The covariant derivative was an operator suggested by Ricci based
> on the non-unique Christoffel symbols of the second kind. For each
> set of connection coefficients to the geodesic equation, there is one
> such operator where each such operator is very different from the
> other.
Once you learn what the word "covariant" actually means, you'll see
just how silly that statement actually is .
> 3. The Riemann curvature tensor was derived through the covariant
> derivative. It was designed by Ricci. There are many but
> mathematically different ways to present the Riemann curvature
> tensor. Ricci only found one.
How else could you define it, without making up your own?
> 4. The Ricci curvature tensor was designed by Ricci's faithful
> student Levi-Civita from the Riemann curvature tensor.
Yeah, so?
> 5. The Lagrangian to the Einstein-Hilbert action was designed or
> cooked up as an alchemist would have done in the medieval times by
> Hilbert himself. The main ingredient was none other than the Ricci
> curvature tensor.
More like the Ricci scalar, but that would be expecting way too much
from you.
> 6. The Einstein field equations are derived by dividing that
> Lagrangian Hilbert pulled out of his *ss with each of the elements to
> the metric. Since there are 16 elements, there are indeed 16 such
> equations.
Actually, there's only ten of them in GR, but thanks for the effort.
> 7. The Schwarzschild metric is one of the infinite numbers of
> solutions to the ordinary differential equations (the field equations)
> for static spacetime. The creation of a black hole is strictly how
> one would embrace one non-unique solution to another. It allows
> anyone to play God through a pen and a piece of paper.
You've obviously never heard of Birkhoff's theorem.
And we'll skip the rest of the insane ignorant ramblings and rantings
inherent in your post. I'll say one thing though, you sure chose the
correct day to post it all.
Eric Gisse
[...]
> You've obviously never heard of Birkhoff's theorem.
Not true - he has _heard_ of it. In fact, this newsgroup was the only
place to ever tell him about it. Except he does not believe it -
something about being "convoluted".
He doesn't believe that vacuum implies a null Ricci tensor [even
though I proved it] nor does he believe in ODE uniqueness theorems
which is even more nuts.
>
> And we'll skip the rest of the insane ignorant ramblings and rantings
> inherent in your post. I'll say one thing though, you sure chose the
> correct day to post it all.
I'm not even bothering to read slashdot and such today.
Bill Hobba
<carlip...@physics.ucdavis.edu> wrote in message
news:euotgl$eq1$1...@skeeter.ucdavis.edu...
> Bill Hobba <rub...@junk.com> wrote:
>> In another post I mentioned there are no unresolved foundational
>> issues with GR. And to the best of my knowledge there isn't.
>> But an actual knowledgeable poster thought I was joking. Does
>> anyone know what of any actual unresolved issues?
>
> It depends on what you count as "foundational," and also on
> what scope of issues you are allowing. The main unresolved
> issues that might fall under this category -- you can decide
> whether they do or not -- include:
>
> 1. We don't have a quantum theory of gravity. More generally,
> we don't know how to couple general relativity consistently
> to quantum matter (though we have good approximations).
> 2. General relativity predicts the existence of singularities
> under a number of physically reasonable circumstances. Some
> people consider this a fundamental problem. (Some others
> think it comes from issue #1.)
I am in the latter camp. When we have a Quantum theory of gravity it will
resolve issues with singularities. Of course this is just an opinion - I
could be wrong.
> 3. We don't yet know whether general relativity predicts "cosmic
> censorship" (that is, essentially, that singularities are
> generically hidden behind horizons). Is this "foundational"?
> It presumably has an answer within the existing framework;
> we just don't know what it is.
> 4. General relativity, like all other standard theories currently
> existing in physics, gives predictions that depend on initial
> data, but does not tell us what the "right" initial data is
> for our universe. If you're doing cosmology, this is a problem
> -- it would be nice to have a unique initial condition.
> 5. Many physicists believe that there should be a single unified
> theory that encompasses all fundamental physics. General
> relativity isn't one.
> 6. The Universe apparently has a nonzero cosmological constant.
> General relativity allows this, but does not determine its
> value; the observed value seems "unnaturally" small, and it
> would be good to have an explanation of this. (This is again
> tied up with issue #1.)
>
> My own take is that these aren't "foundational issues" in general
> relativity, but rather indications that general relativity isn't
> the final answer, but needs to be extended in various ways. But
> as I said, this depends partly on definitions of terms.
Thanks Steve - your response is greatly appreciated. The above is also my
view.
Thanks
Bill
>
> Steve Carlip
>
>
Koobee Wublee
> Koobee Wublee wrote:
> > 1. There are many ways to write the geodesic equation where each way
> > would yield a different set of connection coefficients. Christoffel's
> > way is not unique, and the Christoffel symbols are not unique.
>
> Well, yes and no. Yes, the Christoffel symbols are coordinate dependent,
> and since coordinates are not unique neither are they. But for a given
> coordinate system, in GR they are unique -- consistency with the metric
> requires that.
You don't understand what I was saying. <shrug>
> This indeterminacy of the Christoffel symbols is of no concern in GR (or
> in differential geometry in general), because such coordinate-dependent
> quantities do not represent any physical process (geometrical quantity).
> For instance, in a given manifold with metric one can compute geodesic
> paths using any coordinate system, and one obtains the same set of
> paths. All physical quantities (geometrical quantities), such as paths,
> are independent of coordinates -- the variation of the Christoffel
> symbols with coordinates is REQUIRED for this to be so.
Again, you have failed to understand what I was saying. Try some
mathematical approach instead of relying on hand-waving for a change,
please.
The general geodesic equations are presented independent of a specific
metric. <shrug> Describing an invariant geometry, the metric must be
dependent on the choice of coordinate system.
> > 2. The covariant derivative was an operator suggested by Ricci based
> > on the non-unique Christoffel symbols of the second kind. For each
> > set of connection coefficients to the geodesic equation, there is one
> > such operator where each such operator is very different from the
> > other.
>
> That is a very old-fashioned and outmoded viewpoint.
What is so old-fashioned about any historic viewpoint? It is more
like that you just ran out of arguments and rely on BS to guide you
the way. <Sigh>
> Today the covariant
> derivative is considered an operator in its own right, with an axiomatic
> definition that is completely independent of coordinates.
The covariant derivative involves the Christoffel symbols of the 2nd
kind. Did you not have said the Christoffel symbols of the 2nd kind
are coordinate dependent which they are? So, you are contradicting
yourself once again. <shrug>
> When one
> applies the modern definition to a specific coordinate system, one
> obtains the old formulas, so this approach is equivalent, but avoids the
> inherent coordinate dependence of the earlier method -- clearly a good
> thing as the old method obviously confused you.
The way I interpret what you are saying is that the covariant
derivative just pops out of existence with no connection to what Ricci
had defined it. So, do you want to try the Newtonian law of physics
next?
> You should get a modern
> textbook and LEARN about modern methods; I like this one:
>
> Frankel, _The_Geometry_of_Physics_.
Yes, in time, I shall. However, you really need to brush on your
basics of mathematics. Although physics is not mathematics, physics
should not allow any mathemagics. Please correct me if I am wrong.
> > 3. The Riemann curvature tensor was derived through the covariant
> > derivative. It was designed by Ricci. There are many but
> > mathematically different ways to present the Riemann curvature
> > tensor. Ricci only found one.
>
> Not true. The Riemann curvature tensor is unique for any given manifold
> with metric.
Again, you have not personally derived the Riemann curvature tensor
yourself. Even if you have done it, you are just as blind as Ricci
himself and the rest of the world for that matter. There are many
ways to arrange the term that make up the Riemann curvature tensor.
<shrug>
> What varies are its PROJECTIONS onto different coordinate
> systems (aka components).
Riemann curvature tensors are functions of the Christoffel symbols of
the 2nd kind. Since these symbols are coordinate dependent, Riemann
curvature tensors must be coordinate dependent. Someone is grossly
handicapped with logic here. It is not I. <shrug>
> > 5. The Lagrangian to the Einstein-Hilbert action was designed or
> > cooked up as an alchemist would have done in the medieval times by
> > Hilbert himself. The main ingredient was none other than the Ricci
> > curvature tensor.
>
> Hmmm. There is no specific algorithm for finding Lagrangians.
That is so true, but a Lagrangian must satisfy a few mathematical
axioms. <shrug>
> Medieval alchemists were not knowledgeable of this sort of thing.
This is also true. However, Hilbert's attempt was no different from
the medieval alchemists. <shrug>
> Apparently you aren't, either. <shrug>
You are so wrong. <sigh>
> > 6. The Einstein field equations are derived by dividing that
> > Lagrangian Hilbert pulled out of his *ss with each of the elements to
> > the metric.
>
> This is plain and simply false -- there is no such "dividing" at all.
Apparently you do not know how the field equations are derived.
<shrug> Do you still think the methods that yield the field equations
also yield the geodesic equations?
> What is done is the _variational_derivative_, a well known and essential
> part of Lagrangian mathematics.
The field equations are not derived through the calculus of
variations. <shrug>
> > Since there are 16 elements, there are indeed 16 such
> > equations.
>
> Well, yes. But they are not all independent....
Yes.
> > 7. The Schwarzschild metric is one of the infinite numbers of
> > solutions to the ordinary differential equations (the field equations)
> > for static spacetime.
>
> Well, yes. But given the requirements of a static manifold and spherical
> symmetry, the Schw. solution is unique.
Again, you are so wrong --- misguided by your faith in GR. <shrug>
> You seem completely mystified by
> the solutions to differential equations, and the uniqueness theorems
> thereof.
Not true. There is nothing mystifying about any mathematical
outcomes. <shrug>
> More education is necessary before you can understand this
> stuff....
That is indeed I shall pursue is more education. However, this
subject is so simple. There is no further education necessary that
can contradict the black-and-white indications of the mathematics
involved. <shrug> You need to understand the basics of the
mathematics and NOT USING for word salads for your arguments. <sigh>
> > The creation of a black hole is strictly how
> > one would embrace one non-unique solution to another. It allows
> > anyone to play God through a pen and a piece of paper.
>
> <giggle> Obviously you do not understand the relationship between
> theoretical physics and the real world.
Yes, I do.
> Nor the theorems on uniqueness
> of solutions to differential equations. <shrug>
This I do as well.
> > The above represent the mathematical foundation of GR.
>
> Not even close. The above represents your personal distortions of your
> peculiar mis-perceptions of GR.
It sounds like you have run out of any arguments. <shrug>
> > However, the general concept of the metric being invariant is totally
> > wrong.
>
> It is a tensor, hence invariant under coordinate changes. <shrug>
>
> > The curvature in spacetime must be invariant. This should be
> > a no-brainer to accept.
>
> Hmmm. Curvature is indeed invariant, hence it is represented by a
> tensor.
Unfortunately, there is no way to represent the geometry of curvature
without a specific coordinate system. Your erroneous concept of the
metric does not represent the invariant geometry of the curvature in
space or spacetime. <shrug>
> After all, coordinate invariance is what tensors were invented
> to model.
Someone made a mistake a hundred years ago. It is time to correct
that error.
> > This invariant geometry is described by the
> > dot product of the metric itself and the choice of coordinate.
>
> Nonsense -- there is no "dot product" involved.
Apparently, you have not been reading Mr. Bielawski's posts. <shrug>
> Your statement does not
> even make sense (one takes dot products between VECTORS, not between a
> rank-2 tensor and a coordinate system). As I keep saying, you need to
> LEARN what these words and concepts actually mean.
The 'dot product' I was referring to involves two matrices not two
vectors. You need to learn what I have written in the past. <shrug>
For example, given two n-by-n matrices, [A] and [B], we have the dot
product defined as the following.
[A] * [B] = sum[i = 0 to 3](sum[j = 0 to 3]([A]ij [B]ij))
> The metric (tensor) is an invariant description of the geometry,
This is not what the mathematics of a tensor represents. <shrug>
> in that
> all geometrical quantities can be obtained from it. As it is inherently
> independent of coordinates, so is this description.
You have been so wrong. <shrug>
> As ought to be obvious, but you clearly missed it, when someone
> says "the metric" they mean "the metric TENSOR". Some older
> books conflate the meaning of a tensor with its components,
> so unwary readers can get confused. That's why you need a
> MODERN textbook.
There is only one way to define the metric. It is a set of curvature
correction factors to correct any observations into flat space or
spacetime. <shrug> In doing so, it must be coordinate dependent.
The analogy is a pair of prescribed glasses that corrects only a
particular deficiency in eyesight.
> > If the
> > choice of coordinate is varying from one observer to another, the
> > metric must vary with the choice of coordinate system to define an
> > invariant geometry.
>
> You are obviously confusing the metric (tensor) with its projection onto
> various coordinate systems -- the COMPONENTS of the metric tensor.
This is not true. You are throw excrement because you have run out of
ammunitions. There is no way you can fudge coordinate dependent
elements into an invariant matrix. <shrug> You have been hand-waving
big time.
> The
> metric (tensor) does not vary with coordinates, only the COMPONENTS of
> the metric (tensor) do so: this variation is specifically defined so the
> metric (tensor) itself remains invariant.
If each lens of a pair of prescribed glasses is unique, each pair of
glasses must be uniquely fit only for correcting one particular
deficiency in eyesight. You cannot say all pairs of glasses are the
same. <shrug>
> You SERIOUSLY need to get a
> modern textbook and LEARN about modern differential geometry, and modern
> physics. <shrug>
YOU SERIOUSLY NEED TO GET YOUR COMMON SENSE CHECKED OUT. Seeing a
shrink may not be a bad idea.
> Let my arm point to the sun rising directly to the east. My arm
> is manifestly completely independent of coordinates. In
> coordinates with x=east, y=north, z=up, the components of the
> unit vector representing my arm are (1,0,0); in coordinates
> with x=south, y=east, z=up, the components of THE SAME UNIT
> VECTOR REPRESENTING THE SAME ARM are (0,1,0). The VECTOR has
> not changed, only its components have changed with the change
> in coordinates. This vector is a rank-1 tensor, and the same
> property holds for all tensors. Coordinates are used for a
> DESCRIPTION of geometrical and physical quantities, which
> cannot possibly depend on the coordinates used to describe
> them -- my arm is manifestly completely independent of
> coordinates, and the mathematical quantity used to model it
> must also be independent of coordinates: THAT'S WHY WE USE
> VECTORS AND TENSORS TO MODEL GEOMETRICAL AND PHYSICAL
> QUANTITIES.
Unfortunately what you are describing does not apply to the metric GR
is involved with. <shrug> Each element of the metric in GR must be
coordinate dependent. The whole matrix represented by all these
elements must only be associated with one choice of coordinate system
to describe the invariance in the geometry. Your mind is still in the
Never-Never-Land. <shrug>
Do you not wear any glasses?
JanPB
> On Apr 1, 8:56 am, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
> > Today the covariant
> > derivative is considered an operator in its own right, with an axiomatic
> > definition that is completely independent of coordinates.
>
> The covariant derivative involves the Christoffel symbols of the 2nd
> kind. Did you not have said the Christoffel symbols of the 2nd kind
> are coordinate dependent which they are? So, you are contradicting
> yourself once again. <shrug>
No, you are making a logical mistake here. The fact is that while
Christoffel symbols are coordinate-dependent, that _particular_
combination of them known as _Riemann curvature component_ only
changes according to a rule which governs altering different
coordinate representations of _a single tensor_. You can verify it by
hand although it's quite tedious. The reason this works is that the
Christoffel symbol coordinate-change formula together with the Riemann
curvature component formula produce algebraic cancellations leaving
only the standard tensor-coordinate-change formula behind.
Approaches to curvature developed later are based on people making
clever observations about certain (different) combinations of
Christoffel symbols and general properties of partial differential
operators on vector bundles. These observations - which the original
creators of the field did not make at first - give one a much clearer
understanding of the situation and much more efficient computational
tools.
> > When one
> > applies the modern definition to a specific coordinate system, one
> > obtains the old formulas, so this approach is equivalent, but avoids the
> > inherent coordinate dependence of the earlier method -- clearly a good
> > thing as the old method obviously confused you.
>
> The way I interpret what you are saying is that the covariant
> derivative just pops out of existence with no connection to what Ricci
> had defined it. So, do you want to try the Newtonian law of physics
> next?
No, it's just like vectors, say. They are coordinate-independent but
they still can be expressed as n-tuples of coordinate-dependent
numbers. Covariant derivative does not "pop out" but it's a standard
theorem in differential geometry that covariant derivative is defined
uniquely as the only differential operator from TM to T*M x TM
satisfying the Leibniz rule and compatible with the metric. (You can
find the exact meanings of these words in any decent diff. geom text.)
>
> > > 3. The Riemann curvature tensor was derived through the covariant
> > > derivative. It was designed by Ricci. There are many but
> > > mathematically different ways to present the Riemann curvature
> > > tensor. Ricci only found one.
>
> > Not true. The Riemann curvature tensor is unique for any given manifold
> > with metric.
>
> Again, you have not personally derived the Riemann curvature tensor
> yourself. Even if you have done it, you are just as blind as Ricci
> himself and the rest of the world for that matter. There are many
> ways to arrange the term that make up the Riemann curvature tensor.
> <shrug>
They all describe the same tensor. Just CHECK it as I described
earlier (make sure you've got enough paper).
> > What varies are its PROJECTIONS onto different coordinate
> > systems (aka components).
>
> Riemann curvature tensors are functions of the Christoffel symbols of
> the 2nd kind. Since these symbols are coordinate dependent, Riemann
> curvature tensors must be coordinate dependent. Someone is grossly
> handicapped with logic here. It is not I. <shrug>
Yes, it's you. One more time: the coordinate-dependent Christoffel
symbols combine into components of a coordinate-independent quantity.
It's quite a chore to verify this directly but it can be done. That's
the price you pay for ignoring facts about these objects that were
discovered later.
[...]
--
Jan Bielawski
Koobee Wublee
> On Apr 1, 10:34 pm, "Koobee Wublee" <koobee.wub...@gmail.com> wrote:
> > The covariant derivative involves the Christoffel symbols of the 2nd
> > kind. Did you not have said the Christoffel symbols of the 2nd kind
> > are coordinate dependent which they are? So, you are contradicting
> > yourself once again. <shrug>
>
> No, you are making a logical mistake here. The fact is that while
> Christoffel symbols are coordinate-dependent, that _particular_
> combination of them known as _Riemann curvature component_ only
> changes according to a rule which governs altering different
> coordinate representations of _a single tensor_. You can verify it by
> hand although it's quite tedious. The reason this works is that the
> Christoffel symbol coordinate-change formula together with the Riemann
> curvature component formula produce algebraic cancellations leaving
> only the standard tensor-coordinate-change formula behind.
This is total BS. Check out the specific form of the Riemann
curvature tensor.
http://en.wikipedia.org/wiki/Riemann_tensor
> Approaches to curvature developed later are based on people making
> clever observations about certain (different) combinations of
> Christoffel symbols and general properties of partial differential
> operators on vector bundles. These observations - which the original
> creators of the field did not make at first - give one a much clearer
> understanding of the situation and much more efficient computational
> tools.
The Riemann curvature tensor was designed by Ricci himself. <shrug>
> > The way I interpret what you are saying is that the covariant
> > derivative just pops out of existence with no connection to what Ricci
> > had defined it. So, do you want to try the Newtonian law of physics
> > next?
>
> No, it's just like vectors, say. They are coordinate-independent but
> they still can be expressed as n-tuples of coordinate-dependent
> numbers. Covariant derivative does not "pop out" but it's a standard
> theorem in differential geometry that covariant derivative is defined
> uniquely as the only differential operator from TM to T*M x TM
> satisfying the Leibniz rule and compatible with the metric. (You can
> find the exact meanings of these words in any decent diff. geom text.)
I am not referring to the mechanical functionality of the covariant
derivative. You are barking up the wrong tree once again. <shrug>
> > Again, you have not personally derived the Riemann curvature tensor
> > yourself. Even if you have done it, you are just as blind as Ricci
> > himself and the rest of the world for that matter. There are many
> > ways to arrange the term that make up the Riemann curvature tensor.
> > <shrug>
>
> They all describe the same tensor. Just CHECK it as I described
> earlier (make sure you've got enough paper).
You don't even what I am talking about because you have not personally
derived the Riemann curvature tensor. <shrug>
> > Riemann curvature tensors are functions of the Christoffel symbols of
> > the 2nd kind. Since these symbols are coordinate dependent, Riemann
> > curvature tensors must be coordinate dependent. Someone is grossly
> > handicapped with logic here. It is not I. <shrug>
>
> Yes, it's you. One more time: the coordinate-dependent Christoffel
> symbols combine into components of a coordinate-independent quantity.
> It's quite a chore to verify this directly but it can be done. That's
> the price you pay for ignoring facts about these objects that were
> discovered later.
It does not how many times you present that. It is still BS. Why
don't you get yourself familiarize with the specific presentation of
the Riemann curvature tensor first?
Eric Gisse
[...]
Why is it that you feel you don't have to support anything you say
with actual math or even a reasonable argument?
The Ghost In The Machine
<jow...@gmail.com>
wrote
on 2 Apr 2007 19:05:48 -0700
<1175565948....@p15g2000hsd.googlegroups.com>:
What, and sully one's reputation? Surely you jest! ;-)
--
#191, ewi...@earthlink.net
GNU and improved.
--
Posted via a free Usenet account from http://www.teranews.com
John Anderson
Koobee Wublee
> On Mar 31, 9:27 pm, "Koobee Wublee" <koobee.wub...@gmail.com> wrote:
<yawn> The supposed-to-be educated don't have anything worthwhile to
say. I guess I have some time left to address their apprentices.
> Here comes another Koobee Wublee "shit and run" post.
Not really. I reserve the right not to answer any personal insults.
<shrug>
> > 1. There are many ways to write the geodesic equation where each way
> > would yield a different set of connection coefficients. Christoffel's
> > way is not unique, and the Christoffel symbols are not unique.
>
> What is interesting is that you continue to assert this despite having
> no evidence that it is true.
Don't blame me on your failure in mathematics. <shrug>
> I asked you to prove this statement once before, and you just ran off.
> It surprises me not in the least to see it again. Isn't it interesting
> to see how often you criticize people for not 'showing the math' when
> you have never, not even once, shown yours?
I have addressed them already in my past posts.
> > 2. The covariant derivative was an operator suggested by Ricci based
> > on the non-unique Christoffel symbols of the second kind. For each
> > set of connection coefficients to the geodesic equation, there is one
> > such operator where each such operator is very different from the
> > other.
>
> An even more idiotic conclusion which follows from a dumb premise.
You don't understand what covariant derivative is all about. <shrug>
> > 3. The Riemann curvature tensor was derived through the covariant
> > derivative. It was designed by Ricci. There are many but
> > mathematically different ways to present the Riemann curvature
> > tensor. Ricci only found one.
>
> Show us another, then show that they are different.
If you have the math skills, you can see it yourself. If not, you are
on your own. <shrug>
> > 4. The Ricci curvature tensor was designed by Ricci's faithful
> > student Levi-Civita from the Riemann curvature tensor.
No protest?
> > 5. The Lagrangian to the Einstein-Hilbert action was designed or
> > cooked up as an alchemist would have done in the medieval times by
> > Hilbert himself. The main ingredient was none other than the Ricci
> > curvature tensor.
>
> Have you already forgotten what Bill Hobba was asking about?
No. Have you?
> > 6. The Einstein field equations are derived by dividing that
> > Lagrangian Hilbert pulled out of his *ss with each of the elements to
> > the metric. Since there are 16 elements, there are indeed 16 such
> > equations.
>
> Duh. 4x4 = 16.
Yes. <shrug>
> At any rate, why is that a problem? There is more than one path to the
> field equations.
No, there is only one solid way to derive the field equations.
> > 7. The Schwarzschild metric is one of the infinite numbers of
> > solutions to the ordinary differential equations (the field equations)
> > for static spacetime. The creation of a black hole is strictly how
> > one would embrace one non-unique solution to another. It allows
> > anyone to play God through a pen and a piece of paper.
>
> BZZZT, TRY AGAIN.
7. The Schwarzschild metric is one of the infinite numbers of
solutions to the ordinary differential equations (the field equations)
for static spacetime. The creation of a black hole is strictly how
one would embrace one non-unique solution to another. It allows
anyone to play God through a pen and a piece of paper.
> The field equations are _partial_ differential equations but the
> symmetry of the situation allows them to be reduced to ordinary
> differential equations, whose solutions are unique if they satisfy the
> ODEs as well as the boundary conditions.
'The field equations are _partial_ differential equations' PERIOD.
<shrug>
> > The above represent the mathematical foundation of GR. It shows how
> > the curvature in spacetime is shaped by the mathematical tool invented
> > by Ricci. Yes, Ricci is the Creator, and Levi-Civita, Hilbert,
> > Schwarzschild, and Einstein are His prophets.
>
> You sure do seem to whine a lot about who did what.
This is what history is all about. <shrug>
> > However, the general concept of the metric being invariant is totally
> > wrong. The curvature in spacetime must be invariant. This should be
> > a no-brainer to accept. This invariant geometry is described by the
> > dot product of the metric itself and the choice of coordinate. If the
> > choice of coordinate is varying from one observer to another, the
> > metric must vary with the choice of coordinate system to define an
> > invariant geometry. This is a concept taught in elementary school
> > where these kids do understand, but the GRists seem to have great
> > difficulties grasping. <sigh>
>
> The only one who is having "difficulties grasping" here is YOU.
Wrong. GR concepts and math is so simple to me. It will be so
simple to you too if you understand what I am saying. <shrug>
> YOU
> have been told _countless_ times that while the coordinate
> representation of a tensor is obviously coordinate dependent, the
> tensor itself is not. Shortly after being told that, it has also been
> asked of you to prove the assertion wrong with a counter-example or
> some kind of actual mathematical proof.
You have been told COUNTLESS TIMES by me that holding the metric to be
invariant is absurd and physically invalid.
> To date, all you have been able to do is do a coordinate
> transformation and say "LOOK! THEY ARE DIFFERENT!" without actually
> proving that they are different. Cmon KW, follow your own maxim: SHOW
> THE MATH.
I did. You must be sleeping. <shrug>
> For reference, here are some of your previous dumbass claims that you
> have never proven.
>
> * The Minkowski metric does not admit a Lagrangian. I explicitly
> derived the Lagrangian for you....and you never posted to that thread
> again.
Your derivation is stupid and nonsense. I reserve the right not to
answer any posts so unintelligently put together. You need to get
over with that. <shrug>
> * A coordinate transformation can introduce curvature. You kept
> asserting it, but you never proven it.
I never said that.
> * A tensor is not invariant under a coordinate transformation. You
> continue asserting it, but you have never proven it.
Whatever a tensor is. However, the metric is dependent on the choice
of the coordinate system. <shrug>
> * The surface area for space of constant radius for the Schwarzschild
> metric is 4pi*r^2. Despite the careful proof to the contrary by JanPB,
> you continue to assert that it is.
Yes, the surface area of a sphere to any observer is always 4pi R^2
regardless how much space is curved where R is the observed radius.
<shrug>
> > There are more. Need I to continue? These should be enough for one
> > night.
>
> Yes, how about some actual mathematics to prove your dumbass
> assertions?
Like what?
> haha like that will ever happen. You don't even have a firm grasp of
> multivariable calculus.
I can always surprise you that I promise. <yawn>
Eric Gisse
> On Mar 31, 11:14 pm, "Eric Gisse" <jowr...@gmail.com> wrote:
>
> > On Mar 31, 9:27 pm, "Koobee Wublee" <koobee.wub...@gmail.com> wrote:
>
> <yawn> The supposed-to-be educated don't have anything worthwhile to
> say. I guess I have some time left to address their apprentices.
Yes, retirement....all those hours, so many things to fill them with.
>
> > Here comes another Koobee Wublee "shit and run" post.
>
> Not really. I reserve the right not to answer any personal insults.
> <shrug>
>
> > > 1. There are many ways to write the geodesic equation where each way
> > > would yield a different set of connection coefficients. Christoffel's
> > > way is not unique, and the Christoffel symbols are not unique.
>
> > What is interesting is that you continue to assert this despite having
> > no evidence that it is true.
>
> Don't blame me on your failure in mathematics. <shrug>
Interesting - your failure to disclose the relevant mathematics is
somehow *my* failure.
Ignorance is strength, war is peace, freedom is slavery...
>
> > I asked you to prove this statement once before, and you just ran off.
> > It surprises me not in the least to see it again. Isn't it interesting
> > to see how often you criticize people for not 'showing the math' when
> > you have never, not even once, shown yours?
>
> I have addressed them already in my past posts.
Isn't it interesting how the proofs always end up in 'past posts'?
Perhaps you could show me the 'past posts' that contain the proofs or
you claims - or would you like to shift the blame to me again?
>
> > > 2. The covariant derivative was an operator suggested by Ricci based
> > > on the non-unique Christoffel symbols of the second kind. For each
> > > set of connection coefficients to the geodesic equation, there is one
> > > such operator where each such operator is very different from the
> > > other.
>
> > An even more idiotic conclusion which follows from a dumb premise.
>
> You don't understand what covariant derivative is all about. <shrug>
>
> > > 3. The Riemann curvature tensor was derived through the covariant
> > > derivative. It was designed by Ricci. There are many but
> > > mathematically different ways to present the Riemann curvature
> > > tensor. Ricci only found one.
>
> > Show us another, then show that they are different.
>
> If you have the math skills, you can see it yourself. If not, you are
> on your own. <shrug>
Your failures are not mine. Either you are able to, or you are not.
Since you steadfastly refuse to document your claims, it appears that
you are not.
>
> > > 4. The Ricci curvature tensor was designed by Ricci's faithful
> > > student Levi-Civita from the Riemann curvature tensor.
>
> No protest?
The claim isn't worth arguing about since who developed what has no
influence on the mathematics.
>
> > > 5. The Lagrangian to the Einstein-Hilbert action was designed or
> > > cooked up as an alchemist would have done in the medieval times by
> > > Hilbert himself. The main ingredient was none other than the Ricci
> > > curvature tensor.
>
> > Have you already forgotten what Bill Hobba was asking about?
>
> No. Have you?
>
> > > 6. The Einstein field equations are derived by dividing that
> > > Lagrangian Hilbert pulled out of his *ss with each of the elements to
> > > the metric. Since there are 16 elements, there are indeed 16 such
> > > equations.
>
> > Duh. 4x4 = 16.
>
> Yes. <shrug>
Just making sure you agree since you seem to have trouble with the
idea that there are 16 equations.
>
> > At any rate, why is that a problem? There is more than one path to the
> > field equations.
>
> No, there is only one solid way to derive the field equations.
"Solid" is a fungible concept, especially when the answer does not
change.
>
> > > 7. The Schwarzschild metric is one of the infinite numbers of
> > > solutions to the ordinary differential equations (the field equations)
> > > for static spacetime. The creation of a black hole is strictly how
> > > one would embrace one non-unique solution to another. It allows
> > > anyone to play God through a pen and a piece of paper.
>
> > BZZZT, TRY AGAIN.
>
> 7. The Schwarzschild metric is one of the infinite numbers of
> solutions to the ordinary differential equations (the field equations)
> for static spacetime. The creation of a black hole is strictly how
> one would embrace one non-unique solution to another. It allows
> anyone to play God through a pen and a piece of paper.
ODEs have ONE *unique* solution. This has been explained to you before
- crack open any worthwhile ODE text [Boyce and DiPrima is adequately
deep, as an example] and read it.
The Schwarzschild solution is the *unique* vacuum solution to the
field equations. This has been explained to you before. Open up a
textbook and read about the proof of Birkhoff's theorem.
>
> > The field equations are _partial_ differential equations but the
> > symmetry of the situation allows them to be reduced to ordinary
> > differential equations, whose solutions are unique if they satisfy the
> > ODEs as well as the boundary conditions.
>
> 'The field equations are _partial_ differential equations' PERIOD.
> <shrug>
True, but all those components of the Ricci tensor being equal to zero
via it being a vacuum solution, as well as the spherical symmetry via
constraining the form of the metric anzatz happens to make the
relevant quantities the solutions of a few ODEs rather than PDEs.
You can see an adequate demonstration of this in any proof of
Birkhoff's theorem. Of course, that would require you to go to the
library and _read_ a book on GR which by all accounts you have, and
will, never do.
FYI, you seem to be confused - are the field equations PDEs or ODEs?
Do you know the difference?
>
> > > The above represent the mathematical foundation of GR. It shows how
> > > the curvature in spacetime is shaped by the mathematical tool invented
> > > by Ricci. Yes, Ricci is the Creator, and Levi-Civita, Hilbert,
> > > Schwarzschild, and Einstein are His prophets.
>
> > You sure do seem to whine a lot about who did what.
>
> This is what history is all about. <shrug>
>
> > > However, the general concept of the metric being invariant is totally
> > > wrong. The curvature in spacetime must be invariant. This should be
> > > a no-brainer to accept. This invariant geometry is described by the
> > > dot product of the metric itself and the choice of coordinate. If the
> > > choice of coordinate is varying from one observer to another, the
> > > metric must vary with the choice of coordinate system to define an
> > > invariant geometry. This is a concept taught in elementary school
> > > where these kids do understand, but the GRists seem to have great
> > > difficulties grasping. <sigh>
>
> > The only one who is having "difficulties grasping" here is YOU.
>
> Wrong. GR concepts and math is so simple to me. It will be so
> simple to you too if you understand what I am saying. <shrug>
Perhaps you could spend some time and write down this simple math
rather than argue for months?
>
> > YOU
> > have been told _countless_ times that while the coordinate
> > representation of a tensor is obviously coordinate dependent, the
> > tensor itself is not. Shortly after being told that, it has also been
> > asked of you to prove the assertion wrong with a counter-example or
> > some kind of actual mathematical proof.
>
> You have been told COUNTLESS TIMES by me that holding the metric to be
> invariant is absurd and physically invalid.
Yes, but my statements have the backing of mathematical proofs. Your
statements have the backing of...what, exactly?
I am yet to see you cite one single reference that agrees with your
contrary viewpoint, much less an actual proof of any of them.
>
> > To date, all you have been able to do is do a coordinate
> > transformation and say "LOOK! THEY ARE DIFFERENT!" without actually
> > proving that they are different. Cmon KW, follow your own maxim: SHOW
> > THE MATH.
>
> I did. You must be sleeping. <shrug>
Just because I am typing this in bed does not mean I am sleeping.
I saw your "proof", not only did you transform the line element
incorrectly, you failed to realize it even after it was carefully
explained to you by both myself and JanPB. Your entire proof consisted
of "hey, I did a transformation and the metric looks different! It
must be different!", and of course you couldn't even do the
transformation correctly.
>
> > For reference, here are some of your previous dumbass claims that you
> > have never proven.
>
> > * The Minkowski metric does not admit a Lagrangian. I explicitly
> > derived the Lagrangian for you....and you never posted to that thread
> > again.
>
> Your derivation is stupid and nonsense. I reserve the right not to
> answer any posts so unintelligently put together. You need to get
> over with that. <shrug>
Why is it "stupid and nonsense"? Did you not understand it? Would you
like me to go through it again, slowly?
I derived it straight from the metric without referring to the more
general formula in what was about a third of a page of math on paper.
However, it was derived from the more general formula in about a line
here:
http://groups.google.com/group/sci.physics.relativity/msg/872133b163bba475?dmode=source
Isn't it interesting that there are now two disproofs or your asinine
statement? How many do you require before you admit you are wrong and
drop it?
>
> > * A coordinate transformation can introduce curvature. You kept
> > asserting it, but you never proven it.
>
> I never said that.
Yes, you did.
http://groups.google.com/group/sci.physics.relativity/msg/00d08acc54532fd4?dmode=source
Firefox already had the relevant search phrase in its' history from
the last time you did this song and dance.
>
> > * A tensor is not invariant under a coordinate transformation. You
> > continue asserting it, but you have never proven it.
>
> Whatever a tensor is. However, the metric is dependent on the choice
> of the coordinate system. <shrug>
No, the metric _components_ are dependent. The metric itself is not.
Learn to tell the difference.
>
> > * The surface area for space of constant radius for the Schwarzschild
> > metric is 4pi*r^2. Despite the careful proof to the contrary by JanPB,
> > you continue to assert that it is.
>
> Yes, the surface area of a sphere to any observer is always 4pi R^2
> regardless how much space is curved where R is the observed radius.
> <shrug>
Yet the mathematical proof of a contrary claim hasn't been refuted by
you. Why is that?
Why is it that everyone but you substantiates their arguments with
mathematics? Why is it you are the only one who will see the proof and
ignore the result because it proves you wrong?
>
> > > There are more. Need I to continue? These should be enough for one
> > > night.
>
> > Yes, how about some actual mathematics to prove your dumbass
> > assertions?
>
> Like what?
>
> > haha like that will ever happen. You don't even have a firm grasp of
> > multivariable calculus.
>
> I can always surprise you that I promise. <yawn>
Oh good. Then prove the statement that the Schwarzschild metric has a
surface area of 4pi*r^2 for constant t and r.
Barry
>
> ODEs have ONE *unique* solution. This has been explained to you before
> - crack open any worthwhile ODE text [Boyce and DiPrima is adequately
> deep, as an example] and read it.
>
As everybody knows, I don't know any maths or physics, never having
been to school.
In my ognorance, I thought that ODE's were somewhat analogous to
quadratic equations.
More specifically, I used to think that an "n"th-order ODE would have
"n" linearly independent solutions and that any linear combination of
those solutions would also be a solution.
I really must look at one of those text books one of these days.
Barry
Koobee Wublee
> On Apr 4, 10:23 pm, "Koobee Wublee" <koobee.wub...@gmail.com> wrote:
> > <yawn> The supposed-to-be educated don't have anything worthwhile to
> > say. I guess I have some time left to address their apprentices.
>
> Yes, retirement....all those hours, so many things to fill them with.
I look forward towards my own retirement.
> > Don't blame me on your failure in mathematics. <shrug>
>
> Interesting - your failure to disclose the relevant mathematics is
> somehow *my* failure.
No, your failure to understand the relevant mathematics is not my
fault. <shrug>
> Ignorance is strength, war is peace, freedom is slavery...
This can only be embraced by someone as crooked as Gisse and
incompetence too.
> > I have addressed them already in my past posts.
>
> Isn't it interesting how the proofs always end up in 'past posts'?
Yes, it is called history. <shrug>
> Perhaps you could show me the 'past posts' that contain the proofs or
> you claims - or would you like to shift the blame to me again?
You can find them.
> > If you have the math skills, you can see it yourself. If not, you are
> > on your own. <shrug>
>
> Your failures are not mine. Either you are able to, or you are not.
> Since you steadfastly refuse to document your claims, it appears that
> you are not.
We are talking about you and your mathematical skills. Or rather lack
of it. <shrug>
> > 4. The Ricci curvature tensor was designed by Ricci's faithful
> > student Levi-Civita from the Riemann curvature tensor.
>
> > No protest?
>
> The claim isn't worth arguing about since who developed what has no
> influence on the mathematics.
But this is about history. <shrug>
> Duh. 4x4 = 16.
>
> > Yes. <shrug>
>
> Just making sure you agree since you seem to have trouble with the
> idea that there are 16 equations.
Wrong.
> > No, there is only one solid way to derive the field equations.
>
> "Solid" is a fungible concept, especially when the answer does not
> change.
Your loss, not mine.
> ODEs have ONE *unique* solution. This has been explained to you before
> - crack open any worthwhile ODE text [Boyce and DiPrima is adequately
> deep, as an example] and read it.
Not if they have some dependencies on each other.
> The Schwarzschild solution is the *unique* vacuum solution to the
> field equations. This has been explained to you before. Open up a
> textbook and read about the proof of Birkhoff's theorem.
This has been explained to you numerous times by me on why the
Schwarzschild solution is not unique. <shrug>
> > 'The field equations are _partial_ differential equations' PERIOD.
> > <shrug>
>
> True, but all those components of the Ricci tensor being equal to zero
> via it being a vacuum solution, as well as the spherical symmetry via
> constraining the form of the metric anzatz happens to make the
> relevant quantities the solutions of a few ODEs rather than PDEs.
>
> You can see an adequate demonstration of this in any proof of
> Birkhoff's theorem. Of course, that would require you to go to the
> library and _read_ a book on GR which by all accounts you have, and
> will, never do.
Birkhoff's theorem states that any spherically symmetric solution to
the field equations in vacuum must be static and asymptotically flat.
Spherically symmetric means the geometry in spacetime must be
independent of any angular displacements. Static means the metric is
independent of time. Therefore, it really does not say the
Schwarzschild metric is the only solution. However, there are other
solutions to the field without evoking the Cosmological constant that
would fit an expanding universe. Therefore, Birkhoff's theorem is
wrong on that account. <shrug>
> FYI, you seem to be confused - are the field equations PDEs or ODEs?
Oh, trying to trick me to do your homework for you.
> Do you know the difference?
Yes.
> > Wrong. GR concepts and math is so simple to me. It will be so
> > simple to you too if you understand what I am saying. <shrug>
>
> Perhaps you could spend some time and write down this simple math
> rather than argue for months?
I did. <shrug>
> > You have been told COUNTLESS TIMES by me that holding the metric to be
> > invariant is absurd and physically invalid.
>
> Yes, but my statements have the backing of mathematical proofs. Your
> statements have the backing of...what, exactly?
No, your statements are BS. <shrug>
> I am yet to see you cite one single reference that agrees with your
> contrary viewpoint, much less an actual proof of any of them.
That is because I am right, and I am the first to point out the errors
made by the forefathers of GR. <shrug> For example, the metric must
be observer dependent.
> > I did. You must be sleeping. <shrug>
>
> Just because I am typing this in bed does not mean I am sleeping.
Close enough. <shrug>
> I saw your "proof", not only did you transform the line element
> incorrectly, you failed to realize it even after it was carefully
> explained to you by both myself and JanPB. Your entire proof consisted
> of "hey, I did a transformation and the metric looks different! It
> must be different!", and of course you couldn't even do the
> transformation correctly.
You have failed to understand what I did. <shrug>
> > Your derivation is stupid and nonsense. I reserve the right not to
> > answer any posts so unintelligently put together. You need to get
> > over with that. <shrug>
>
> Why is it "stupid and nonsense"? Did you not understand it? Would you
> like me to go through it again, slowly?
>
> I derived it straight from the metric without referring to the more
> general formula in what was about a third of a page of math on paper.
> However, it was derived from the more general formula in about a line
> here:
>
> http://groups.google.com/group/sci.physics.relativity/msg/872133b163bba475
What is this? Two buffoon trying to out-design his own version of the
Lagrangian?
> Isn't it interesting that there are now two disproofs or your asinine
> statement? How many do you require before you admit you are wrong and
> drop it?
Not once.
> * A coordinate transformation can introduce curvature. You kept
> asserting it, but you never proven it.
>
> > I never said that.
>
> Yes, you did.
>
> http://groups.google.com/group/sci.physics.relativity/msg/00d08acc54532fd4
Where is the statement saying a coordinate transformation can
introduce curvature?
> Firefox already had the relevant search phrase in its' history from
> the last time you did this song and dance.
Good. Keep it for eternity.
> > Whatever a tensor is. However, the metric is dependent on the choice
> > of the coordinate system. <shrug>
>
> No, the metric _components_ are dependent. The metric itself is not.
That is total BS. If you have a different lens to a pair of glasses,
the prescription is different.
> Learn to tell the difference.
Learn to try not to BS.
> > Yes, the surface area of a sphere to any observer is always 4pi R^2
> > regardless how much space is curved where R is the observed radius.
> > <shrug>
>
> Yet the mathematical proof of a contrary claim hasn't been refuted by
> you. Why is that?
Yes, it did.
> Why is it that everyone but you substantiates their arguments with
> mathematics? Why is it you are the only one who will see the proof and
> ignore the result because it proves you wrong?
You are entitle to hallucinate in your bed.
> > I can always surprise you that I promise. <yawn>
>
> Oh good. Then prove the statement that the Schwarzschild metric has a
> surface area of 4pi*r^2 for constant t and r.
Not now. It is time for you to go to sleep. Make sure you dream of
Koobee Wublee coming to get you.
Eric Gisse
> On Apr 5, 4:41 am, "Eric Gisse" <jowr...@gmail.com> wrote:
>
>
>
> > ODEs have ONE *unique* solution. This has been explained to you before
> > - crack open any worthwhile ODE text [Boyce and DiPrima is adequately
> > deep, as an example] and read it.
>
> As everybody knows, I don't know any maths or physics, never having
> been to school.
>
> In my ognorance, I thought that ODE's were somewhat analogous to
> quadratic equations.
Not really.
The characteristic polynomial for a linear, constant coefficient ODE
is a n'th order polynomial. But that is about it.
>
> More specifically, I used to think that an "n"th-order ODE would have
> "n" linearly independent solutions and that any linear combination of
> those solutions would also be a solution.
That is true only for linear ODEs.
JanPB
> the field equations in vacuum must be static and asymptotically flat.
No, it doesn't say that. You've fallen into the standard "static"
trap. What it says is that the solution _must have coefficients
independent of the "t" variable_. (I guess you never went through
solving the EFE assuming spherical symmetry _only_ or else you'd see
it clearly from one of the relevant ODEs).
Of course independence of the "t" variable (together with the diagonal
form of the metric presumed for the derivation) does translate into
"static" outside the horizon but it doesn't imply static inside (in
fact, it's not static inside as everyone knows).
> Spherically symmetric means the geometry in spacetime must be
> independent of any angular displacements. Static means the metric is
> independent of time. Therefore, it really does not say the
> Schwarzschild metric is the only solution.
It does. You misunderstood the statement of the theorem. I just
noticed Wikipedia also states this incorrectly - is it where you get
your physics from?
> However, there are other
> solutions to the field without evoking the Cosmological constant that
> would fit an expanding universe. Therefore, Birkhoff's theorem is
> wrong on that account. <shrug>
No, it is you who is wrong. It's not a big deal, just study it some
more and don't waste your time on phantoms.
--
Jan Bielawski
Androcles
"JanPB" <fil...@gmail.com> wrote in message news:1175801321....@y80g2000hsf.googlegroups.com...
> On Apr 5, 11:19 am, "Koobee Wublee" <koobee.wub...@gmail.com> wrote:
>>
> [...]
>>
>> Birkhoff's theorem states that any spherically symmetric solution to
>> the field equations in vacuum must be static and asymptotically flat.
>
> No, it doesn't say that.
Hahaha!
> You've fallen into the standard "static"
> trap. What it says is that the solution _must have coefficients
> independent of the "t" variable_.
Oh wow! HAHAHA!
> (I guess
Guessing is always the Polish answer.
Koobee Wublee
> On Apr 5, 11:19 am, "Koobee Wublee" <koobee.wub...@gmail.com> wrote:
> > Birkhoff's theorem states that any spherically symmetric solution to
> > the field equations in vacuum must be static and asymptotically flat.
>
> No, it doesn't say that. You've fallen into the standard "static"
> trap.
I don't get this 'static trap' of yours.
> What it says is that the solution _must have coefficients
> independent of the "t" variable_.
Is that not what 'static' means?
> (I guess you never went through
> solving the EFE assuming spherical symmetry _only_ or else you'd see
> it clearly from one of the relevant ODEs).
Yes, I did.
> Of course independence of the "t" variable (together with the diagonal
> form of the metric presumed for the derivation) does translate into
> "static" outside the horizon but it doesn't imply static inside (in
> fact, it's not static inside as everyone knows).
Inside, outside, then inside? Done with dancing around? Make up your
mind. Do you not know Birkhoff's theorem only applies to vacuum? Is
that outside or inside? Outside or inside of what?
> > Spherically symmetric means the geometry in spacetime must be
> > independent of any angular displacements. Static means the metric is
> > independent of time. Therefore, it really does not say the
> > Schwarzschild metric is the only solution.
>
> It does. You misunderstood the statement of the theorem.
You need to understand the theorem first before making that judgment.
<shrug>
> I just
> noticed Wikipedia also states this incorrectly - is it where you get
> your physics from?
So, just how many schools of physics are there? I suppose you only
endorse one school of interpretations. Does that sound familiar? My
God is always more powerful than someone else's. <shrug>
You must have studied under Reverend Hammond.
> > However, there are other
> > solutions to the field without evoking the Cosmological constant that
> > would fit an expanding universe. Therefore, Birkhoff's theorem is
> > wrong on that account. <shrug>
>
> No, it is you who is wrong. It's not a big deal, just study it some
> more and don't waste your time on phantoms.
Done that. You are still wrong.
Eric Gisse
[...]
> > Perhaps you could show me the 'past posts' that contain the proofs or
> > you claims - or would you like to shift the blame to me again?
>
> You can find them.
If they existed, that is.
[...]
>
> > ODEs have ONE *unique* solution. This has been explained to you before
> > - crack open any worthwhile ODE text [Boyce and DiPrima is adequately
> > deep, as an example] and read it.
>
> Not if they have some dependencies on each other.
I guess you don't know the difference after all.
An ordinary differential equation is an equation involving the
derivatives of a function, the function, and the function's variable.
A partial differential equation is the same - but there is more than
one variable.
The equations that need to be solved are PDEs that reduce to ODEs -
they are functions of _ONE_ variable. You can see this worked out in
an oblique fashion in MTW [page 844] and in more detail in Carrol's
_Spacetime and Geometry_.
Regardless, the hyperbolic PDEs have unique solutions if there is an
adequate amount of initial and/or boundary values. You are wrong
either way.
>
> > The Schwarzschild solution is the *unique* vacuum solution to the
> > field equations. This has been explained to you before. Open up a
> > textbook and read about the proof of Birkhoff's theorem.
>
> This has been explained to you numerous times by me on why the
> Schwarzschild solution is not unique. <shrug>
Except you are wrong. You have too many misconceptions in order to say
such a statement with force, much less prove it.
>
> > > 'The field equations are _partial_ differential equations' PERIOD.
> > > <shrug>
>
> > True, but all those components of the Ricci tensor being equal to zero
> > via it being a vacuum solution, as well as the spherical symmetry via
> > constraining the form of the metric anzatz happens to make the
> > relevant quantities the solutions of a few ODEs rather than PDEs.
>
> > You can see an adequate demonstration of this in any proof of
> > Birkhoff's theorem. Of course, that would require you to go to the
> > library and _read_ a book on GR which by all accounts you have, and
> > will, never do.
>
> Birkhoff's theorem states that any spherically symmetric solution to
> the field equations in vacuum must be static and asymptotically flat.
No, that is not what Birkhoff's theorem states. JanPB adequately
explains it, but I'll add the exact statement of Birkhoff's theorem:
"Let the geometry of a given region of spacetime (1) be spherically
symmetric, and (2) be a solution to the Einstein field equations in
vacuum. Then that geometry is necessarily a piece of the Schwarzschild
geometry". MTW, page 843 - which cites Birkhoff, G.D., 1923
_Relativity and Modern Physics_, Harvard University Press, Cambridge,
Mass.
Compare and contrast what you say [with no reference] vs what was
actually said [with a reference].
> Spherically symmetric means the geometry in spacetime must be
> independent of any angular displacements. Static means the metric is
> independent of time. Therefore, it really does not say the
> Schwarzschild metric is the only solution. However, there are other
> solutions to the field without evoking the Cosmological constant that
> would fit an expanding universe. Therefore, Birkhoff's theorem is
> wrong on that account. <shrug>
You are close...but not quite right. You obviously spent some time in
your life thinking about relativity - too bad you didn't dedicate
enough to getting it _right_.
You understand what spherically symmetric means - good. You slightly
miss the point on static - static means that the metric _components_
are independent of time. It is functionally equivalent to saying the
metric is independent of time, but given your constant
misunderstanding regarding the difference between the metric as an
entity and the components of a metric, it is a distinction worth
making.
The remainder is nonsense - it says _exactly_ that it is the
Schwarzschild solution in the _actual_ statement of Birkhoff's
theorem. Furthermore, both Schwarzschild and Birkhoff did their work
before Einstein included the cosmological constant. Which doesn't even
matter anyway, The analogous Birkhoff's theorem which incorporates the
cosmological constant gives the Schwarzschild-de Sitter metric.
>
> > FYI, you seem to be confused - are the field equations PDEs or ODEs?
>
> Oh, trying to trick me to do your homework for you.
Sorry, I already know the difference.
>
> > Do you know the difference?
>
> Yes.
No you don't.
>
> > > Wrong. GR concepts and math is so simple to me. It will be so
> > > simple to you too if you understand what I am saying. <shrug>
>
> > Perhaps you could spend some time and write down this simple math
> > rather than argue for months?
>
> I did. <shrug>
>
> > > You have been told COUNTLESS TIMES by me that holding the metric to be
> > > invariant is absurd and physically invalid.
>
> > Yes, but my statements have the backing of mathematical proofs. Your
> > statements have the backing of...what, exactly?
>
> No, your statements are BS. <shrug>
Saying so does not make it true. Prove it or admit you are wrong.
>
> > I am yet to see you cite one single reference that agrees with your
> > contrary viewpoint, much less an actual proof of any of them.
>
> That is because I am right, and I am the first to point out the errors
> made by the forefathers of GR. <shrug> For example, the metric must
> be observer dependent.
Once again - you never distinguish between the metric and the metric's
components, which are the projections of the metric along the relevant
coordinate axes.
>
> > > I did. You must be sleeping. <shrug>
>
> > Just because I am typing this in bed does not mean I am sleeping.
>
> Close enough. <shrug>
>
> > I saw your "proof", not only did you transform the line element
> > incorrectly, you failed to realize it even after it was carefully
> > explained to you by both myself and JanPB. Your entire proof consisted
> > of "hey, I did a transformation and the metric looks different! It
> > must be different!", and of course you couldn't even do the
> > transformation correctly.
>
> You have failed to understand what I did. <shrug>
Then explain it.
>
> > > Your derivation is stupid and nonsense. I reserve the right not to
> > > answer any posts so unintelligently put together. You need to get
> > > over with that. <shrug>
>
> > Why is it "stupid and nonsense"? Did you not understand it? Would you
> > like me to go through it again, slowly?
>
> > I derived it straight from the metric without referring to the more
> > general formula in what was about a third of a page of math on paper.
> > However, it was derived from the more general formula in about a line
> > here:
>
> >http://groups.google.com/group/sci.physics.relativity/msg/872133b163b...
>
> What is this? Two buffoon trying to out-design his own version of the
> Lagrangian?
Make a counter-argument or admit you are wrong, rather then insulting
folks who prove you wrong.
>
> > Isn't it interesting that there are now two disproofs or your asinine
> > statement? How many do you require before you admit you are wrong and
> > drop it?
>
> Not once.
>
> > * A coordinate transformation can introduce curvature. You kept
> > asserting it, but you never proven it.
>
> > > I never said that.
>
> > Yes, you did.
>
> >http://groups.google.com/group/sci.physics.relativity/msg/00d08acc545...
>
> Where is the statement saying a coordinate transformation can
> introduce curvature?
"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.
"
I'm surprised you can read an entire post if you have difficulty
reading a paragraph for comprehension.
>
> > Firefox already had the relevant search phrase in its' history from
> > the last time you did this song and dance.
>
> Good. Keep it for eternity.
>
> > > Whatever a tensor is. However, the metric is dependent on the choice
> > > of the coordinate system. <shrug>
>
> > No, the metric _components_ are dependent. The metric itself is not.
>
> That is total BS. If you have a different lens to a pair of glasses,
> the prescription is different.
What is the difference between the metrics represented by the line
elements of ds^2 = -dt^2 + dx^2 + dy^2 + dz^2 and ds^2 = -dt^2 + dr^2
+ r^2(d_theta_^2 + sin(theta)^2*d_phi_^2) ?
>
> > Learn to tell the difference.
>
> Learn to try not to BS.
>
> > > Yes, the surface area of a sphere to any observer is always 4pi R^2
> > > regardless how much space is curved where R is the observed radius.
> > > <shrug>
>
> > Yet the mathematical proof of a contrary claim hasn't been refuted by
> > you. Why is that?
>
> Yes, it did.
Then where is the refutation?
Oh, that's right...more "past posts" that never ever materialize...
>
> > Why is it that everyone but you substantiates their arguments with
> > mathematics? Why is it you are the only one who will see the proof and
> > ignore the result because it proves you wrong?
>
> You are entitle to hallucinate in your bed.
>
> > > I can always surprise you that I promise. <yawn>
>
> > Oh good. Then prove the statement that the Schwarzschild metric has a
> > surface area of 4pi*r^2 for constant t and r.
>
> Not now. It is time for you to go to sleep. Make sure you dream of
> Koobee Wublee coming to get you.
http://www.bjorn-comic.com/art_profile/kirby.htm
JanPB
> On Apr 5, 12:28 pm, "JanPB" <film...@gmail.com> wrote:
>
> > On Apr 5, 11:19 am, "Koobee Wublee" <koobee.wub...@gmail.com> wrote:
> > > Birkhoff's theorem states that any spherically symmetric solution to
> > > the field equations in vacuum must be static and asymptotically flat.
>
> > No, it doesn't say that. You've fallen into the standard "static"
> > trap.
>
> I don't get this 'static trap' of yours.
>
> > What it says is that the solution _must have coefficients
> > independent of the "t" variable_.
>
> Is that not what 'static' means?
Independence of a variable means "static" (ignoring the difference
between "static" and "stationary" for the time being) if that variable
is timelike. In Schwarzschild's chart the coordinate labelled by "t"
is not timelike inside the horizon, so independence of "t" there means
that the region is spatially homogeneous in that direction.
> > (I guess you never went through
> > solving the EFE assuming spherical symmetry _only_ or else you'd see
> > it clearly from one of the relevant ODEs).
>
> Yes, I did.
>
> > Of course independence of the "t" variable (together with the diagonal
> > form of the metric presumed for the derivation) does translate into
> > "static" outside the horizon but it doesn't imply static inside (in
> > fact, it's not static inside as everyone knows).
>
> Inside, outside, then inside? Done with dancing around? Make up your
> mind. Do you not know Birkhoff's theorem only applies to vacuum? Is
> that outside or inside? Outside or inside of what?
I'm just using the standard shortcut: "outside" means "r>2M", "inside"
means "r<2M". Easier to type.
> > > Spherically symmetric means the geometry in spacetime must be
> > > independent of any angular displacements. Static means the metric is
> > > independent of time. Therefore, it really does not say the
> > > Schwarzschild metric is the only solution.
>
> > It does. You misunderstood the statement of the theorem.
>
> You need to understand the theorem first before making that judgment.
> <shrug>
>
> > I just
> > noticed Wikipedia also states this incorrectly - is it where you get
> > your physics from?
>
> So, just how many schools of physics are there? I suppose you only
> endorse one school of interpretations. Does that sound familiar? My
> God is always more powerful than someone else's. <shrug>
This is just a question of mathematics, so interpretations don't enter
into it. Either there are multiple solutions to EFE in the sph. symm.
case or not. The answer is the solution is unique.
> You must have studied under Reverend Hammond.
That would be the sight! :-)
> > > However, there are other
> > > solutions to the field without evoking the Cosmological constant that
> > > would fit an expanding universe. Therefore, Birkhoff's theorem is
> > > wrong on that account. <shrug>
>
> > No, it is you who is wrong. It's not a big deal, just study it some
> > more and don't waste your time on phantoms.
>
> Done that. You are still wrong.
I know that I cannot convince you here - it's impossible to teach a
complex subject on an ASCII BB. Some readers might be interested in
learning more of that stuff though.
--
Jan Bielawski
Tom Roberts
> On Apr 5, 11:19 am, "Koobee Wublee" <koobee.wub...@gmail.com> wrote:
>> Birkhoff's theorem states that any spherically symmetric solution to
>> the field equations in vacuum must be static and asymptotically flat.
>
> No, it doesn't say that. You've fallen into the standard "static"
> trap. What it says is that the solution _must have coefficients
> independent of the "t" variable_.
You're both wrong in important ways: Birkhoff's actual theorem says
nothing about "static" or "asymptotically flat" (and such claims are
clearly wrong). JanPB's statements about specific coordinates are
utterly unwarranted -- this is a geometric theorem and coordinates are
irrelevant.
The best statement of Birkhoff's theorem I have is from Hawking and
Ellis, _The_Large_Scale_Structure_of_Space-Time_, Appendix B:
Any C^2 solution of Einstein's empty space equations which is
spherically symmetric in an open set V, is locally equivalent to
part of the maximally extended Schwarzschild solution in V.
(MTW gives a similar but less precise version; I don't have
Birkhoff's original paper, but it would probably require
translation into modern vocabulary so I follow Hawking and
Ellis.)
[I am surprised they did not state it in terms of a local
isometry between V and some region of the Schw. manifold.
I suspect it is because their proof did not actually do
that, though I believe it can be done.]
Note in particular that inside the horizon this manifold is NOT static.
Eric Gisse said:
> static means that the metric _components_
> are independent of time. It is functionally equivalent to saying the
> metric is independent of time,
Not quite. In GR, for a region V of a manifold to be called "static" means:
a) there is a timelike Killing vector throughout V
and
b) there is a foliation of spacetime throughout V such that spatial
surfaces of constant time are orthogonal to the timelike Killing
vector.
Note that neither the Killing vector nor the foliation need be unique
(e.g. Minkowski spacetime which is static everywhere and everywhen).
(Your statement is essentially equivalent to (a) alone.)
For instance, the Kerr metric (in Boyer-Lindquist coordinates) has
components independent of the time coordinate (outside the horizon,
anyway), yet is not static (it is merely stationary -- you did not
include condition b).
FYI the region V is called "stationary" if (a) holds but not (b).
Tom Roberts
JanPB
> JanPB wrote:
> > On Apr 5, 11:19 am, "Koobee Wublee" <koobee.wub...@gmail.com> wrote:
> >> Birkhoff's theorem states that any spherically symmetric solution to
> >> the field equations in vacuum must be static and asymptotically flat.
>
> > No, it doesn't say that. You've fallen into the standard "static"
> > trap. What it says is that the solution _must have coefficients
> > independent of the "t" variable_.
>
> You're both wrong in important ways: Birkhoff's actual theorem says
> nothing about "static" or "asymptotically flat" (and such claims are
> clearly wrong). JanPB's statements about specific coordinates are
> utterly unwarranted -- this is a geometric theorem and coordinates are
> irrelevant.
I was referring to Schwarzschild's coordinates specifically because
it's easier to see the difference between "static" and "independent of
"t"" this way (I purposely ignored the "static"/"stationary"
distinction - one thing at a time!). Hawking and Ellis are great but
first things first.
--
Jan Bielawski
Eric Gisse
[...]
> Eric Gisse said:
>
> > static means that the metric _components_
> > are independent of time. It is functionally equivalent to saying the
> > metric is independent of time,
>
> Not quite. In GR, for a region V of a manifold to be called "static" means:
> a) there is a timelike Killing vector throughout V
> and
> b) there is a foliation of spacetime throughout V such that spatial
> surfaces of constant time are orthogonal to the timelike Killing
> vector.
> Note that neither the Killing vector nor the foliation need be unique
> (e.g. Minkowski spacetime which is static everywhere and everywhen).
>
> (Your statement is essentially equivalent to (a) alone.)
Mostly because I only knew about a).
>
> For instance, the Kerr metric (in Boyer-Lindquist coordinates) has
> components independent of the time coordinate (outside the horizon,
> anyway), yet is not static (it is merely stationary -- you did not
> include condition b).
>
> FYI the region V is called "stationary" if (a) holds but not (b).
Subtle.
Thanks for the clue transplant.
>
> Tom Roberts
Koobee Wublee
> JanPB wrote:
> > No, it doesn't say that. You've fallen into the standard "static"
> > trap. What it says is that the solution _must have coefficients
> > independent of the "t" variable_.
>
> You're both wrong in important ways: Birkhoff's actual theorem says
> nothing about "static" or "asymptotically flat" (and such claims are
> clearly wrong). JanPB's statements about specific coordinates are
> utterly unwarranted -- this is a geometric theorem and coordinates are
> irrelevant.
So, here is another school of thought on the Birkhoff's theorem, or
rather another interpretation to that obscure theorem.
> The best statement of Birkhoff's theorem I have is from Hawking and
> Ellis, _The_Large_Scale_Structure_of_Space-Time_, Appendix B:
>
> Any C^2 solution of Einstein's empty space equations which is
> spherically symmetric in an open set V, is locally equivalent to
> part of the maximally extended Schwarzschild solution in V.
Is this sentence encrypted by any means?
> (MTW gives a similar but less precise version; I don't have
> Birkhoff's original paper, but it would probably require
> translation into modern vocabulary so I follow Hawking and
> Ellis.)
>
> [I am surprised they did not state it in terms of a local
> isometry between V and some region of the Schw. manifold.
> I suspect it is because their proof did not actually do
> that, though I believe it can be done.]
It sounds like there are as many interpretations to the Birkhoff's
theorem as there are interpreters. <shrug>
I have to conclude the Birkhoff's theorem is not to be taken
seriously.
> Note in particular that inside the horizon this manifold is NOT static.
Did you not have said the field equations do not address anything
inside the horizon? So, why all of a sudden Birkhoff's theorem
applies?
Koobee Wublee
As long as you two do not confine yourselves to mathematics, any of
you two is entitled to his, her, or its philosophical point of view.
<shrug>
Please folks, this is physics not some literature or seminary class
trying to interpret the philosophic meaning of 'static' versus
'stationary'.
Koobee Wublee
> On Apr 5, 5:28 pm, "Koobee Wublee" <koobee.wub...@gmail.com> wrote:
> Independence of a variable means "static" (ignoring the difference
> between "static" and "stationary" for the time being) if that variable
> is timelike.
In our conversation, you have foliaged the word 'static' into
'stationary' and 'non-timelike'. The best BS'ers are the ones who can
come up more words to describe one thing. Was it not one of the
Confucius sayings?
> In Schwarzschild's chart the coordinate labelled by "t"
> is not timelike inside the horizon, so independence of "t" there means
> that the region is spatially homogeneous in that direction.
You still allow yourself to be trapped in the mental black hole.
<shrug> In the meantime, we are discussing about solutions in vacuum.
> > Inside, outside, then inside? Done with dancing around? Make up your
> > mind. Do you not know Birkhoff's theorem only applies to vacuum? Is
> > that outside or inside? Outside or inside of what?
>
> I'm just using the standard shortcut: "outside" means "r>2M", "inside"
> means "r<2M". Easier to type.
Inside does not apply with any mathematics we have. So, how the heck
can you apply the Birkhoff's theorem inside the event horizon? You do
not have any mathematics to back you back in this case.
> > So, just how many schools of physics are there? I suppose you only
> > endorse one school of interpretations. Does that sound familiar? My
> > God is always more powerful than someone else's. <shrug>
>
> This is just a question of mathematics, so interpretations don't enter
> into it.
May I remind you that you have not addressed any mathematics!
> Either there are multiple solutions to EFE in the sph. symm.
> case or not. The answer is the solution is unique.
Mechanically, indeed there are infinite number of independent and
unique solutions. Philosophically, you have interpreted all solutions
are identical due to your misconception that the metric is invariant.
<shrug>
> > You must have studied under Reverend Hammond.
>
> That would be the sight! :-)
Indeed. :-)
> > Done that. You are still wrong.
>
> I know that I cannot convince you here - it's impossible to teach a
> complex subject on an ASCII BB. Some readers might be interested in
> learning more of that stuff though.
The outcome of a physical hypothesis can only be decided in the arena
of mathematics.
Ps. I know you have backed away from how an observer would observe
the surface area of a sphere in curved space. You and Mr. McCullough
were wrong about that one. Please tell your student/admirer Gisse
about this one. So, he can have a good-night sleep. Thank you.
JanPB
> On Apr 5, 5:48 pm, "JanPB" <film...@gmail.com> wrote:
>
> > On Apr 5, 5:28 pm, "Koobee Wublee" <koobee.wub...@gmail.com> wrote:
> > Independence of a variable means "static" (ignoring the difference
> > between "static" and "stationary" for the time being) if that variable
> > is timelike.
>
> In our conversation, you have foliaged the word 'static' into
> 'stationary' and 'non-timelike'. The best BS'ers are the ones who can
> come up more words to describe one thing. Was it not one of the
> Confucius sayings?
I just prefer to focus on one issue at a time. The issue at hand was
that deriving Schwarzschild's solution yielded particular metric
coefficients not depending on the "t" variable which you concluded
meant the metric was necessarily static. All I said was that this
implication follows only for locations outside the horizon (r>2M)
where the letter "t" denotes a timelike coordinate.
> > In Schwarzschild's chart the coordinate labelled by "t"
> > is not timelike inside the horizon, so independence of "t" there means
> > that the region is spatially homogeneous in that direction.
>
> You still allow yourself to be trapped in the mental black hole.
> <shrug> In the meantime, we are discussing about solutions in vacuum.
This is pure rhetoric.
> > > Inside, outside, then inside? Done with dancing around? Make up your
> > > mind. Do you not know Birkhoff's theorem only applies to vacuum? Is
> > > that outside or inside? Outside or inside of what?
>
> > I'm just using the standard shortcut: "outside" means "r>2M", "inside"
> > means "r<2M". Easier to type.
>
> Inside does not apply with any mathematics we have.
Of course it does.
> So, how the heck
> can you apply the Birkhoff's theorem inside the event horizon? You do
> not have any mathematics to back you back in this case.
I've just told you. The derivation of the solution works both inside
and outside and in either case one of the Einstein equations says that
the metric coefficients (in the presumed diagonal form of the metric)
do not depend on the "t" variable. Examining the signs of the
coefficients this implies that the metric is static outside and
spatially homogeneous inside.
> > > So, just how many schools of physics are there? I suppose you only
> > > endorse one school of interpretations. Does that sound familiar? My
> > > God is always more powerful than someone else's. <shrug>
>
> > This is just a question of mathematics, so interpretations don't enter
> > into it.
>
> May I remind you that you have not addressed any mathematics!
I'll do it in my next post then. I was only addressing the general
flow of logic in these posts as you had this false implication at the
forefront: [Birkhoff]=>[static].
> > Either there are multiple solutions to EFE in the sph. symm.
> > case or not. The answer is the solution is unique.
>
> Mechanically, indeed there are infinite number of independent and
> unique solutions. Philosophically, you have interpreted all solutions
> are identical due to your misconception that the metric is invariant.
> <shrug>
No, they are all identical in the sense of being physically
indistinguishable by any means.
> > > You must have studied under Reverend Hammond.
>
> > That would be the sight! :-)
>
> Indeed. :-)
>
> > > Done that. You are still wrong.
>
> > I know that I cannot convince you here - it's impossible to teach a
> > complex subject on an ASCII BB. Some readers might be interested in
> > learning more of that stuff though.
>
> The outcome of a physical hypothesis can only be decided in the arena
> of mathematics.
>
> Ps. I know you have backed away from how an observer would observe
> the surface area of a sphere in curved space.
That's total news to me. Surface areas are integrals of 2D volume
forms, and these are given in terms of the metric - there is no wiggle
room here, everything is set.
> You and Mr. McCullough
> were wrong about that one. Please tell your student/admirer Gisse
> about this one. So, he can have a good-night sleep. Thank you.
More rhetoric. In general n-dim. volume = Integral(dvol), where "dvol"
is the volume form (the Hodge dual of 1). No one can "back away" from
it.
--
Jan Bielawski
JanPB
> On Apr 5, 6:40 pm, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
>
> > JanPB wrote:
> > > No, it doesn't say that. You've fallen into the standard "static"
> > > trap. What it says is that the solution _must have coefficients
> > > independent of the "t" variable_.
>
> > You're both wrong in important ways: Birkhoff's actual theorem says
> > nothing about "static" or "asymptotically flat" (and such claims are
> > clearly wrong). JanPB's statements about specific coordinates are
> > utterly unwarranted -- this is a geometric theorem and coordinates are
> > irrelevant.
>
> So, here is another school of thought on the Birkhoff's theorem, or
> rather another interpretation to that obscure theorem.
I knew this would confuse you. That's why I wanted to focus on the
meaning of "t"-independence first.
> > The best statement of Birkhoff's theorem I have is from Hawking and
> > Ellis, _The_Large_Scale_Structure_of_Space-Time_, Appendix B:
>
> > Any C^2 solution of Einstein's empty space equations which is
> > spherically symmetric in an open set V, is locally equivalent to
> > part of the maximally extended Schwarzschild solution in V.
>
> Is this sentence encrypted by any means?
C^2 means twice continuously differentiable, open set is a universal
technical requirement for domains over which limits (as in calculus)
are most conveniently defined, locally equivalent means isometric over
an open subset, isometric means dot-product-preserving.
> > (MTW gives a similar but less precise version; I don't have
> > Birkhoff's original paper, but it would probably require
> > translation into modern vocabulary so I follow Hawking and
> > Ellis.)
>
> > [I am surprised they did not state it in terms of a local
> > isometry between V and some region of the Schw. manifold.
> > I suspect it is because their proof did not actually do
> > that, though I believe it can be done.]
>
> It sounds like there are as many interpretations to the Birkhoff's
> theorem as there are interpreters. <shrug>
>
> I have to conclude the Birkhoff's theorem is not to be taken
> seriously.
>
> > Note in particular that inside the horizon this manifold is NOT static.
>
> Did you not have said the field equations do not address anything
> inside the horizon? So, why all of a sudden Birkhoff's theorem
> applies?
It always applied there. Mathematical details to follow - they are not
too bad actually. But I agree that many sources get the statement of
the theorem wrong. Sloppiness, I guess.
--
Jan Bielawski
JanPB
> On Apr 6, 12:20 am, "Koobee Wublee" <koobee.wub...@gmail.com> wrote:
>
> > May I remind you that you have not addressed any mathematics!
>
> I'll do it in my next post then. I was only addressing the general
> flow of logic in these posts as you had this false implication at the
> forefront: [Birkhoff]=>[static].
Here is the relevant mathematics. You'll see precisely how in the
Schwarzschild case the independence of metric coefficients of t
follows from the EFE, not staticity everywhere.
We seek a spherically symmetric metric of signature 2 satisfying the
Einstein equation in vacuum:
R_ab = 0
The assumption of symmetry allows us to assume WLOG that the metric
has the general form like so:
ds^2 = f(r,t) dt^2 + g(r,t) dr^2 + h(r,t) dr dt + k(r,t) dO^2,
where r is a coordinate labelling the spheres of symmetry and:
dO^2 = dtheta^2 + sin^2(theta) dphi^2, as usual.
It turns out that even this reasonably nice general form results in an
unmanageable computational mess so one simplifies the general form
further by a local change of coordinates which cleverly removes the
"dr dt" term and also relabels the spheres of symmetry by their areas
(divided by 4pi). (I can supply more details on this final coordinate
change if you wish.) This results in the following two possible
simplified, diagonal, forms consistent with the given signature
constraint:
either:
ds^2 = -A^2 dt^2 + B^2 dr^2 + r^2 dO^2 [1]
or:
ds^2 = +A^2 dt^2 - B^2 dr^2 + r^2 dO^2 [2]
where A(r,t) and B(r,t) are some functions which are never zero (or
else the metric would not be rank-4).
NB: this local coordinate change happens to rely on dividing by
certain function, so any solutions for metric coefficients we may
obtain by proceeding from these two general forms may not be defined
over the spacetime events for which the function we divided by is
zero.
[In fact (getting slightly ahead of myself here) this is precisely
what happens: the function we divide by is zero for r=2M, hence the
(in)famous exclusion of the locus r=2M from the domain of the metric
coefficients in the Schwarzschild form. There is of course no reason
to presume this is an actual physical exclusion as it potentially
results from the man-made work-saving transformation. So whether it's
a true singularity must be verified separately. It turns out - as is
well known - that the metric is well-defined for r=2M, it is merely
non-diagonalizable there hence the forms [1] and [2] break down there
- it's just a case of being too greedy/lazy.]
OK, so we have those two forms potentially satisfying the EFE. The
maximal domain possible (unless the EFE themselves somehow end up
restricting it) is:
-infty < t < +infty,
0 < r < +infty,
0 < theta < 2pi,
0 < phi < pi.
I'm going to skip the exact details of deriving the equations below
(we can go into it later if necessary). I'm using Cartan's method of
moving frames, and after computing the connection and curvature forms
the Einstein equation R_ab=0 reduces to the following four PDEs:
R_tt = 0
R_rr = 0
R_theta theta = 0
R_rt = 0
...with the remaining equations being either identities or equivalent
to those four ones. (For example, R_phi phi = R_theta theta.)
The exact forms of the four left-hand sides are, few pages of
calculations later, respectively (excuse the mess, it's the fourth,
simplest, equation that's interesting to us):
(1/AB)(-d/dt((1/A)*dB/dt) + d/dr((1/B)*dA/dr) + (2/rB)*dA/dr) = 0
(1/AB)( d/dt((1/A)*dB/dt) - d/dr((1/B)*dA/dr)) - (2/rB)*d/dr(1/B) = 0
-+(1/rAB^2)*dA/dr -+ (1/rB)*d/dr(1/B) + (1/r^2)*(1 -+ 1/B^2) = 0
(2/rAB^2)*dB/dt = 0
...where "-+" means use the "-" sign for the first general form of the
diagonal metric [1], use "+" for the second [2]. Also "d" obviously
denotes partial derivatives.
Note the fourth equation R_rt = 0 says then (cancelling the factor in
front):
dB/dt = 0
...in other words, B(r,t) is independent of t, both in case [1] and
[2]. This means that B is only a function of r which means that the
first two PDEs simplify quite a bit:
(1/AB)( d/dr((1/B)*dA/dr) + (2/rB)*dA/dr) = 0
(1/AB)(-d/dr((1/B)*dA/dr)) - (2/rB)*d/dr(1/B) = 0
This is actually a pair of _ODEs_ (r is the only variable in them)
which remarkably simplify to the following (denoting the r-derivatives
by primes now):
A'/A + B'/B = 0
i.e.,
(ln |A|)' + (ln |B|)' = 0
hence:
|AB| = C (a positive constant of integration)
i.e.,
AB = C (an arbitary nonzero constant of integration).
The important thing as far as Birkhoff is concerned is that this means
that:
B = constant/A
...in other words B(r,t) is independent of t as well, in both [1] and
[2].
Plugging A(r) and B(r)=1/A(r) into the remaining (third) PDE yields
another _ordinary_ differential equation whose exact form and details
of the solution are less interesting to us at this point. Suffice to
say it yields the well-known Schwarzschild form of the metric
components:
ds^2 = -(1 - 2M/r) dt^2 + 1/(1 - 2M/r) dr^2 + r^2 dO^2 (case
[1])
...with the domain restricted to r > 2M, and:
ds^2 = (2M/r - 1) dt^2 - 1/(2M/r - 1) dr^2 + r^2 dO^2 (case
[2])
...with the domain restricted to 0 < r < 2M.
[As I mentioned earlier, the case of r=2M must be verified separately,
as logically at this point we don't know if the exclusion is real or
merely due to our coordinate change in the beginning.]
A quick examination of these two formulas shows that they can be
written as a single equation by a simple sign switch:
ds^2 = -(1 - 2M/r) dt^2 + 1/(1 - 2M/r) dr^2 + r^2 dO^2
...with the domain restricted to r>0 AND r=/=2M.
We note that this solution is static for r>2M and spatially
homogeneous for r<2M. It is also unique over this domain as it's a
solution to an ODE system.
Incidentally, since the coordinate denoted by "r" inside the horizon
is timelike, it's instructive to switch around the letters "r" and "t"
in case [2] because of our strong mental habit of thinking of "t" as
denoting time:
ds^2 = -1/(2M/t - 1) dt^2 + (2M/t - 1) dr^2 + t^2 dO^2,
...with r arbitrary and t > 0 only (we "fall off" spacetime at t=0).
This is what the metric inside "really" is.
--
Jan Bielawski
Androcles
"JanPB" <fil...@gmail.com> wrote in message news:1175891499....@o5g2000hsb.googlegroups.com...
> We seek a spherically symmetric metric of signature 2 satisfying the
> Einstein equation
Another week, another post full of idiocies about another book full of
idiocies.
Koobee Wublee
> We seek a spherically symmetric metric of signature 2 satisfying the
> Einstein equation in vacuum:
>
> R_ab = 0
Or (G_ab = 0)
> The assumption of symmetry allows us to assume WLOG that the metric
> has the general form like so:
WLOG?
> ds^2 = f(r,t) dt^2 + g(r,t) dr^2 + h(r,t) dr dt + k(r,t) dO^2,
>
> where r is a coordinate labelling the spheres of symmetry and:
>
> dO^2 = dtheta^2 + sin^2(theta) dphi^2, as usual.
>
> It turns out that even this reasonably nice general form results in an
> unmanageable computational mess so one simplifies the general form
> further by a local change of coordinates which cleverly removes the
> "dr dt" term and also relabels the spheres of symmetry by their areas
> (divided by 4pi). (I can supply more details on this final coordinate
> change if you wish.) This results in the following two possible
> simplified, diagonal, forms consistent with the given signature
> constraint:
>
> either:
>
> ds^2 = -A^2 dt^2 + B^2 dr^2 + r^2 dO^2 [1]
>
> or:
>
> ds^2 = +A^2 dt^2 - B^2 dr^2 + r^2 dO^2 [2]
If you are describing the same invariant spacetime, ds^2, then your
dr, dt must be different due to your coordinate transformation.
ds^2 = A^2 dt'^2 + B^2 dr'^2 + C^2 dO'^2 [3]
I don't agree with (r^2 dO^2).
> where A(r,t) and B(r,t) are some functions which are never zero (or
> else the metric would not be rank-4).
Rank-4?
> NB: this local coordinate change happens to rely on dividing by
> certain function, so any solutions for metric coefficients we may
> obtain by proceeding from these two general forms may not be defined
> over the spacetime events for which the function we divided by is
> zero.
This transformation is not necessary a division transformation. Why
do you think it is so?
> [In fact (getting slightly ahead of myself here) this is precisely
> what happens: the function we divide by is zero for r=2M, hence the
> (in)famous exclusion of the locus r=2M from the domain of the metric
> coefficients in the Schwarzschild form. There is of course no reason
> to presume this is an actual physical exclusion as it potentially
> results from the man-made work-saving transformation. So whether it's
> a true singularity must be verified separately. It turns out - as is
> well known - that the metric is well-defined for r=2M, it is merely
> non-diagonalizable there hence the forms [1] and [2] break down there
> - it's just a case of being too greedy/lazy.]
This makes no sense. It has the signature of Archimedes Plutonium.
Are you the real person behind this handle?
> OK, so we have those two forms potentially satisfying the EFE. The
> maximal domain possible (unless the EFE themselves somehow end up
> restricting it) is:
>
> -infty < t < +infty,
> 0 < r < +infty,
> 0 < theta < 2pi,
> 0 < phi < pi.
>
> I'm going to skip the exact details of deriving the equations below
> (we can go into it later if necessary). I'm using Cartan's method of
> moving frames, and after computing the connection and curvature forms
> the Einstein equation R_ab=0 reduces to the following four PDEs:
>
> R_tt = 0
> R_rr = 0
> R_theta theta = 0
> R_rt = 0
No, with (3), you only get
** Rt't' = 0
** Rr'r' = 0
** RO'O' = 0
Recall
ds^2 = A^2 dt'^2 + B^2 dr'^2 + C^2 dO'^2 [3]
You have reduced the spacetime from 4 variables to 3, remember?
> ...with the remaining equations being either identities or equivalent
> to those four ones. (For example, R_phi phi = R_theta theta.)
>
> The exact forms of the four left-hand sides are, few pages of
> calculations later, respectively (excuse the mess, it's the fourth,
> simplest, equation that's interesting to us):
>
> (1/AB)(-d/dt((1/A)*dB/dt) + d/dr((1/B)*dA/dr) + (2/rB)*dA/dr) = 0
> (1/AB)( d/dt((1/A)*dB/dt) - d/dr((1/B)*dA/dr)) - (2/rB)*d/dr(1/B) = 0
> -+(1/rAB^2)*dA/dr -+ (1/rB)*d/dr(1/B) + (1/r^2)*(1 -+ 1/B^2) = 0
> (2/rAB^2)*dB/dt = 0
There is no way you can get ((2/rAB^2)*dB/dt=0) from (3).
Actually, with time derivative, these equations are a lot more
complicated than you have shown above.
> ...where "-+" means use the "-" sign for the first general form of the
> diagonal metric [1], use "+" for the second [2]. Also "d" obviously
> denotes partial derivatives.
>
> Note the fourth equation R_rt = 0 says then (cancelling the factor in
> front):
>
> dB/dt = 0
Therefore showing (dB/dt = 0) is just BS.
> ...in other words, B(r,t) is independent of t, both in case [1] and
> [2]. This means that B is only a function of r which means that the
> first two PDEs simplify quite a bit:
>
> (1/AB)( d/dr((1/B)*dA/dr) + (2/rB)*dA/dr) = 0
> (1/AB)(-d/dr((1/B)*dA/dr)) - (2/rB)*d/dr(1/B) = 0
>
> This is actually a pair of _ODEs_ (r is the only variable in them)
> which remarkably simplify to the following (denoting the r-derivatives
> by primes now):
>
> A'/A + B'/B = 0
Bullsh*t. Your differential equations involve a second derivative of
A! Where is it?
> i.e.,
>
> (ln |A|)' + (ln |B|)' = 0
>
> hence:
>
> |AB| = C (a positive constant of integration)
>
> i.e.,
>
> AB = C (an arbitary nonzero constant of integration).
Don't you feel like a prophet if you know the answer before hand?
> The important thing as far as Birkhoff is concerned is that this means
> that:
>
> B = constant/A
>
> ...in other words B(r,t) is independent of t as well, in both [1] and
> [2].
Only with your mathematical errors, that is so.
[The rest snipped due to the obvious mathematical error.]
Also, you have conveniently assumed (C^2 = r^2). See (3).
That is how the Birkhoff's theorem as the author of Wikipedia defined
it.
Assuming (C^2 = r^2) which translates to spherically symmetric. For
every C^2, there is a new set of metric that satisfies the field
equations. Have you finally understood Mr. Crothers?
Assuming (static) which translates to all elements of the metric
independent of time.
Then, you end up with much, much, much, and much simpler equations
described below.
** (r / B^2) (dB/dr) - (1 / B) + 1 = 0
** (r / A / B) (dA/dr) + (1 / B) - 1 = 0
These are actually not Ricci tensor equations but rather Einstein
tensor equations. With all these simplifications, they are all the
same.
For the spacetime described below,
ds^2 = A dt^2 - B dr^2 - C dO^2
Where (C = r^2)
You will see that you can easily solve for B to be the following.
** B = 1 / (1 + K / r)
Where (K = integration constant)
And A follows.
** A = K' (1 + K / r)
Where (K' = another integration constant)
For (C = (r + K)^2),
You get
** A = K' / (1 + K / r)
** B = 1 + K / r
In this case, you don't get a black hole.
For (C = ((r^3 + K^3)^(1/3) - K)^2),
You end up with Schwarzschild's original solution which does not
manifest any black holes either.
There are infinite numbers of solutions more.
Eric Gisse
> On Apr 6, 1:31 pm, "JanPB" <film...@gmail.com> wrote:
>
> > We seek a spherically symmetric metric of signature 2 satisfying the
> > Einstein equation in vacuum:
>
> > R_ab = 0
>
> Or (G_ab = 0)
>
> > The assumption of symmetry allows us to assume WLOG that the metric
> > has the general form like so:
>
> WLOG?
"Without loss of generality"
It is a phrase you should be used to.
>
>
>
> > ds^2 = f(r,t) dt^2 + g(r,t) dr^2 + h(r,t) dr dt + k(r,t) dO^2,
>
> > where r is a coordinate labelling the spheres of symmetry and:
>
> > dO^2 = dtheta^2 + sin^2(theta) dphi^2, as usual.
>
> > It turns out that even this reasonably nice general form results in an
> > unmanageable computational mess so one simplifies the general form
> > further by a local change of coordinates which cleverly removes the
> > "dr dt" term and also relabels the spheres of symmetry by their areas
> > (divided by 4pi). (I can supply more details on this final coordinate
> > change if you wish.) This results in the following two possible
> > simplified, diagonal, forms consistent with the given signature
> > constraint:
>
> > either:
>
> > ds^2 = -A^2 dt^2 + B^2 dr^2 + r^2 dO^2 [1]
>
> > or:
>
> > ds^2 = +A^2 dt^2 - B^2 dr^2 + r^2 dO^2 [2]
>
> If you are describing the same invariant spacetime, ds^2, then your
> dr, dt must be different due to your coordinate transformation.
They aren't coordinate transformations - they are representations of
the same manifold just with different signatures - remember the
diag(-1,1,1,1) and the diag(1,-1,-1.-1) representations of Minkowski
space?
>
> ds^2 = A^2 dt'^2 + B^2 dr'^2 + C^2 dO'^2 [3]
>
> I don't agree with (r^2 dO^2).
Why? ds^2 is a squared distance - it needs to have units of squared
distance.
>
> > where A(r,t) and B(r,t) are some functions which are never zero (or
> > else the metric would not be rank-4).
>
> Rank-4?
>
> > NB: this local coordinate change happens to rely on dividing by
> > certain function, so any solutions for metric coefficients we may
> > obtain by proceeding from these two general forms may not be defined
> > over the spacetime events for which the function we divided by is
> > zero.
>
> This transformation is not necessary a division transformation. Why
> do you think it is so?
>
> > [In fact (getting slightly ahead of myself here) this is precisely
> > what happens: the function we divide by is zero for r=2M, hence the
> > (in)famous exclusion of the locus r=2M from the domain of the metric
> > coefficients in the Schwarzschild form. There is of course no reason
> > to presume this is an actual physical exclusion as it potentially
> > results from the man-made work-saving transformation. So whether it's
> > a true singularity must be verified separately. It turns out - as is
> > well known - that the metric is well-defined for r=2M, it is merely
> > non-diagonalizable there hence the forms [1] and [2] break down there
> > - it's just a case of being too greedy/lazy.]
>
> This makes no sense. It has the signature of Archimedes Plutonium.
> Are you the real person behind this handle?
Of course it doesn't make sense - you are confused by C^n notation for
being n-times differentiable and you don't know what WLOG means.
What he is saying is that you can't represent the manifold at r=2GM
with a metric that has no cross terms.
>
>
>
> > OK, so we have those two forms potentially satisfying the EFE. The
> > maximal domain possible (unless the EFE themselves somehow end up
> > restricting it) is:
>
> > -infty < t < +infty,
> > 0 < r < +infty,
> > 0 < theta < 2pi,
> > 0 < phi < pi.
>
> > I'm going to skip the exact details of deriving the equations below
> > (we can go into it later if necessary). I'm using Cartan's method of
> > moving frames, and after computing the connection and curvature forms
> > the Einstein equation R_ab=0 reduces to the following four PDEs:
>
> > R_tt = 0
> > R_rr = 0
> > R_theta theta = 0
> > R_rt = 0
>
> No, with (3), you only get
>
> ** Rt't' = 0
> ** Rr'r' = 0
> ** RO'O' = 0
>
> Recall
>
> ds^2 = A^2 dt'^2 + B^2 dr'^2 + C^2 dO'^2 [3]
>
> You have reduced the spacetime from 4 variables to 3, remember?
It isn't 3 variables - the variables are _still_ (t,r,theta,phi).
Remember the definition of dO he wrote down for you?
>
> > ...with the remaining equations being either identities or equivalent
> > to those four ones. (For example, R_phi phi = R_theta theta.)
>
> > The exact forms of the four left-hand sides are, few pages of
> > calculations later, respectively (excuse the mess, it's the fourth,
> > simplest, equation that's interesting to us):
>
> > (1/AB)(-d/dt((1/A)*dB/dt) + d/dr((1/B)*dA/dr) + (2/rB)*dA/dr) = 0
> > (1/AB)( d/dt((1/A)*dB/dt) - d/dr((1/B)*dA/dr)) - (2/rB)*d/dr(1/B) = 0
> > -+(1/rAB^2)*dA/dr -+ (1/rB)*d/dr(1/B) + (1/r^2)*(1 -+ 1/B^2) = 0
> > (2/rAB^2)*dB/dt = 0
>
> There is no way you can get ((2/rAB^2)*dB/dt=0) from (3).
Why not?
You have the metric anzatz - go derive the R_tr equation. You will
find that he is, in fact, right if you do it correctly.
>
> Actually, with time derivative, these equations are a lot more
> complicated than you have shown above.
No, they are not.
>
> > ...where "-+" means use the "-" sign for the first general form of the
> > diagonal metric [1], use "+" for the second [2]. Also "d" obviously
> > denotes partial derivatives.
>
> > Note the fourth equation R_rt = 0 says then (cancelling the factor in
> > front):
>
> > dB/dt = 0
>
> Therefore showing (dB/dt = 0) is just BS.
*sigh*
It is the only way for the R_tr equation to hold for all t,r that
exclude trivial A,B. I am amazed that you have trouble with something
as basic as *this*.
>
> > ...in other words, B(r,t) is independent of t, both in case [1] and
> > [2]. This means that B is only a function of r which means that the
> > first two PDEs simplify quite a bit:
>
> > (1/AB)( d/dr((1/B)*dA/dr) + (2/rB)*dA/dr) = 0
> > (1/AB)(-d/dr((1/B)*dA/dr)) - (2/rB)*d/dr(1/B) = 0
>
> > This is actually a pair of _ODEs_ (r is the only variable in them)
> > which remarkably simplify to the following (denoting the r-derivatives
> > by primes now):
>
> > A'/A + B'/B = 0
>
> Bullsh*t. Your differential equations involve a second derivative of
> A! Where is it?
It went away. He integrated both equations with respect to r.
How can someone claim, with a straight face, that all of differential
geometry is wrong when that someone struggles with regular one
dimensional integration?
>
> > i.e.,
>
> > (ln |A|)' + (ln |B|)' = 0
>
> > hence:
>
> > |AB| = C (a positive constant of integration)
>
> > i.e.,
>
> > AB = C (an arbitary nonzero constant of integration).
>
> Don't you feel like a prophet if you know the answer before hand?
Don't you feel like a jackass for insulting someone who obviously
spent a lot of time writing down something you should already know,
which you will not understand anyway?
Integrate the equations - you get ln(AB) = C. Which is equivalent to
AB = C, where C in both cases are _completely arbitrary_ constants. Is
that enough detail, or would you like it done in even more baby steps?
I guess JanPB was entirely unreasonable in expecting that the guru of
Riemann be capable of understanding the concept of "integration".
>
> > The important thing as far as Birkhoff is concerned is that this means
> > that:
>
> > B = constant/A
>
> > ...in other words B(r,t) is independent of t as well, in both [1] and
> > [2].
>
> Only with your mathematical errors, that is so.
*laughs*
>
> [The rest snipped due to the obvious mathematical error.]
>
> Also, you have conveniently assumed (C^2 = r^2). See (3).
YOU ARE THE ONE WHO WROTE (3). THE FUCKUP IS YOURS.
>
> That is how the Birkhoff's theorem as the author of Wikipedia defined
> it.
Who cares? You don't have the education to differentiate between good
and bad yet - if ever.
>
> Assuming (C^2 = r^2) which translates to spherically symmetric. For
> every C^2, there is a new set of metric that satisfies the field
> equations. Have you finally understood Mr. Crothers?
Hellooo he never wrote C^2 - you were the one who introduced it. Is
your head that far up your ass that you have to invent completely
nonsensical arguments?
[snip remaining junk which is based upon a false premise]
Eric Gisse
wrote:
> "JanPB" <film...@gmail.com> wrote in messagenews:1175891499....@o5g2000hsb.googlegroups.com...
> > We seek a spherically symmetric metric of signature 2 satisfying the
> > Einstein equation
>
> Another week, another post full of idiocies about another book full of
> idiocies.
Then why participate in this newsgroup?
JanPB
wrote:
> "JanPB" <film...@gmail.com> wrote in messagenews:1175891499....@o5g2000hsb.googlegroups.com...
> > We seek a spherically symmetric metric of signature 2 satisfying the
> > Einstein equation
>
> Another week, another post full of idiocies about another book full of
> idiocies.
Wait a moment, which book? I was just reading David Parlett's "History
of Card Games" but I guess that's not what you meant?
--
Jan Bielawski
Koobee Wublee
> On Apr 6, 4:53 pm, "Koobee Wublee" <koobee.wub...@gmail.com> wrote:
> > WLOG?
>
> "Without loss of generality"
Boy, you need know how to read Mr. Bielawski's mind.
> It is a phrase you should be used to.
No, I won't.
It is about an hour and a half since Mr. Bielawski posted his last
post. It is certain that he is trying to find another mathemagic
trick for a show. In the meantime, you are here to keep the rest
entertained. That is OK.
> > > ds^2 = +A^2 dt^2 - B^2 dr^2 + r^2 dO^2 [2]
>
> > If you are describing the same invariant spacetime, ds^2, then your
> > dr, dt must be different due to your coordinate transformation.
>
> They aren't coordinate transformations - they are representations of
> the same manifold just with different signatures - remember the
> diag(-1,1,1,1) and the diag(1,-1,-1.-1) representations of Minkowski
> space?
Yes, I know. <shrug>
> > ds^2 = A^2 dt'^2 + B^2 dr'^2 + C^2 dO'^2 [3]
>
> > I don't agree with (r^2 dO^2).
>
> Why? ds^2 is a squared distance - it needs to have units of squared
> distance.
Right. I still don't agree with (r^2 dO^2). It should be more
general such as (C^2 dO^2) where C is a function of r.
> > > [In fact (getting slightly ahead of myself here) this is precisely
> > > what happens: the function we divide by is zero for r=2M, hence the
> > > (in)famous exclusion of the locus r=2M from the domain of the metric
> > > coefficients in the Schwarzschild form. There is of course no reason
> > > to presume this is an actual physical exclusion as it potentially
> > > results from the man-made work-saving transformation. So whether it's
> > > a true singularity must be verified separately. It turns out - as is
> > > well known - that the metric is well-defined for r=2M, it is merely
> > > non-diagonalizable there hence the forms [1] and [2] break down there
> > > - it's just a case of being too greedy/lazy.]
>
> > This makes no sense. It has the signature of Archimedes Plutonium.
> > Are you the real person behind this handle?
>
> Of course it doesn't make sense - you are confused by C^n notation for
> being n-times differentiable and you don't know what WLOG means.
I am not laughing. Can you find something better to say while the
magician is trying to come up with another mathemagic trick?
> What he is saying is that you can't represent the manifold at r=2GM
> with a metric that has no cross terms.
Oh, that's it! What does that mean? Notice I am still not laughing.
> > Recall
>
> > ds^2 = A^2 dt'^2 + B^2 dr'^2 + C^2 dO'^2 [3]
>
> > You have reduced the spacetime from 4 variables to 3, remember?
>
> It isn't 3 variables - the variables are _still_ (t,r,theta,phi).
Not anymore.
> Remember the definition of dO he wrote down for you?
Yes.
> > > The exact forms of the four left-hand sides are, few pages of
> > > calculations later, respectively (excuse the mess, it's the fourth,
> > > simplest, equation that's interesting to us):
>
> > > (1/AB)(-d/dt((1/A)*dB/dt) + d/dr((1/B)*dA/dr) + (2/rB)*dA/dr) = 0
> > > (1/AB)( d/dt((1/A)*dB/dt) - d/dr((1/B)*dA/dr)) - (2/rB)*d/dr(1/B) = 0
> > > -+(1/rAB^2)*dA/dr -+ (1/rB)*d/dr(1/B) + (1/r^2)*(1 -+ 1/B^2) = 0
> > > (2/rAB^2)*dB/dt = 0
>
> > There is no way you can get ((2/rAB^2)*dB/dt=0) from (3).
>
> Why not?
Why yes?
> You have the metric anzatz - go derive the R_tr equation. You will
> find that he is, in fact, right if you do it correctly.
Through a mathemagic trik?
> > Actually, with time derivative, these equations are a lot more
> > complicated than you have shown above.
>
> No, they are not.
Says someone who has not gone through them. <shrug>
> > Therefore showing (dB/dt = 0) is just BS.
>
> *sigh*
>
> It is the only way for the R_tr equation to hold for all t,r that
> exclude trivial A,B. I am amazed that you have trouble with something
> as basic as *this*.
<sigh> You are drowning in BS, and you don't know it.
> > > (1/AB)( d/dr((1/B)*dA/dr) + (2/rB)*dA/dr) = 0
> > > (1/AB)(-d/dr((1/B)*dA/dr)) - (2/rB)*d/dr(1/B) = 0
>
> > > This is actually a pair of _ODEs_ (r is the only variable in them)
> > > which remarkably simplify to the following (denoting the r-derivatives
> > > by primes now):
>
> > > A'/A + B'/B = 0
>
> > Bullsh*t. Your differential equations involve a second derivative of
> > A! Where is it?
>
> It went away. He integrated both equations with respect to r.
Come back after you have understood integration.
> How can someone claim, with a straight face, that all of differential
> geometry is wrong when that someone struggles with regular one
> dimensional integration?
Easily if it is wrong which it is. <shrug>
> > Don't you feel like a prophet if you know the answer before hand?
>
> Don't you feel like a jackass for insulting someone who obviously
> spent a lot of time writing down something you should already know,
> which you will not understand anyway?
No.
> Integrate the equations - you get ln(AB) = C. Which is equivalent to
> AB = C, where C in both cases are _completely arbitrary_ constants. Is
> that enough detail, or would you like it done in even more baby steps?
Yes, but the differential equations don't indicate that result.
<shrug>
> I guess JanPB was entirely unreasonable in expecting that the guru of
> Riemann be capable of understanding the concept of "integration".
Correct me if I am wrong. It is up to him to decide on that one.
> > > The important thing as far as Birkhoff is concerned is that this means
> > > that:
>
> > > B = constant/A
>
> > > ...in other words B(r,t) is independent of t as well, in both [1] and
> > > [2].
>
> > Only with your mathematical errors, that is so.
>
> *laughs*
I am still not laughing.
> > Also, you have conveniently assumed (C^2 = r^2). See (3).
>
> YOU ARE THE ONE WHO WROTE (3). THE FUCKUP IS YOURS.
Yes, I wrote that one as a correction. <shrug>
> > That is how the Birkhoff's theorem as the author of Wikipedia defined
> > it.
>
> Who cares? You don't have the education to differentiate between good
> and bad yet - if ever.
I do, and I did.
> > Assuming (C^2 = r^2) which translates to spherically symmetric. For
> > every C^2, there is a new set of metric that satisfies the field
> > equations. Have you finally understood Mr. Crothers?
>
> Hellooo he never wrote C^2
I know.
> - you were the one who introduced it.
Yes, as a correction.
> Is
> your head that far up your ass that you have to invent completely
> nonsensical arguments?
No.
Any other questions?
Eric Gisse
> On Apr 6, 7:55 pm, "Eric Gisse" <jowr...@gmail.com> wrote:
>
> > On Apr 6, 4:53 pm, "Koobee Wublee" <koobee.wub...@gmail.com> wrote:
> > > WLOG?
>
> > "Without loss of generality"
>
> Boy, you need know how to read Mr. Bielawski's mind.
>
> > It is a phrase you should be used to.
>
> No, I won't.
>
> It is about an hour and a half since Mr. Bielawski posted his last
> post. It is certain that he is trying to find another mathemagic
> trick for a show. In the meantime, you are here to keep the rest
> entertained. That is OK.
>
> > > > ds^2 = +A^2 dt^2 - B^2 dr^2 + r^2 dO^2 [2]
>
> > > If you are describing the same invariant spacetime, ds^2, then your
> > > dr, dt must be different due to your coordinate transformation.
>
> > They aren't coordinate transformations - they are representations of
> > the same manifold just with different signatures - remember the
> > diag(-1,1,1,1) and the diag(1,-1,-1.-1) representations of Minkowski
> > space?
>
> Yes, I know. <shrug>
If you actually knew you would not have not made that comment.
>
> > > ds^2 = A^2 dt'^2 + B^2 dr'^2 + C^2 dO'^2 [3]
>
> > > I don't agree with (r^2 dO^2).
>
> > Why? ds^2 is a squared distance - it needs to have units of squared
> > distance.
>
> Right. I still don't agree with (r^2 dO^2). It should be more
> general such as (C^2 dO^2) where C is a function of r.
Then use C(r) in place of r^2. The argument that r^2*dO^2 appears in
the metric is an argument from spherical symmetry - since you have
such issue with it, start off with the most general metric and work
from there.
>
>
>
> > > > [In fact (getting slightly ahead of myself here) this is precisely
> > > > what happens: the function we divide by is zero for r=2M, hence the
> > > > (in)famous exclusion of the locus r=2M from the domain of the metric
> > > > coefficients in the Schwarzschild form. There is of course no reason
> > > > to presume this is an actual physical exclusion as it potentially
> > > > results from the man-made work-saving transformation. So whether it's
> > > > a true singularity must be verified separately. It turns out - as is
> > > > well known - that the metric is well-defined for r=2M, it is merely
> > > > non-diagonalizable there hence the forms [1] and [2] break down there
> > > > - it's just a case of being too greedy/lazy.]
>
> > > This makes no sense. It has the signature of Archimedes Plutonium.
> > > Are you the real person behind this handle?
>
> > Of course it doesn't make sense - you are confused by C^n notation for
> > being n-times differentiable and you don't know what WLOG means.
>
> I am not laughing. Can you find something better to say while the
> magician is trying to come up with another mathemagic trick?
>
> > What he is saying is that you can't represent the manifold at r=2GM
> > with a metric that has no cross terms.
>
> Oh, that's it! What does that mean? Notice I am still not laughing.
Defining "cross terms" is left as an exercise for those who have
enough math to understand.
>
> > > Recall
>
> > > ds^2 = A^2 dt'^2 + B^2 dr'^2 + C^2 dO'^2 [3]
>
> > > You have reduced the spacetime from 4 variables to 3, remember?
>
> > It isn't 3 variables - the variables are _still_ (t,r,theta,phi).
>
> Not anymore.
In what fantasy world?
The metric is manifestly a function of t,r,theta,phi - is your reading
disability so strong that you cannot read what is in front of your
face?
>
> > Remember the definition of dO he wrote down for you?
>
> Yes.
>
> > > > The exact forms of the four left-hand sides are, few pages of
> > > > calculations later, respectively (excuse the mess, it's the fourth,
> > > > simplest, equation that's interesting to us):
>
> > > > (1/AB)(-d/dt((1/A)*dB/dt) + d/dr((1/B)*dA/dr) + (2/rB)*dA/dr) = 0
> > > > (1/AB)( d/dt((1/A)*dB/dt) - d/dr((1/B)*dA/dr)) - (2/rB)*d/dr(1/B) = 0
> > > > -+(1/rAB^2)*dA/dr -+ (1/rB)*d/dr(1/B) + (1/r^2)*(1 -+ 1/B^2) = 0
> > > > (2/rAB^2)*dB/dt = 0
>
> > > There is no way you can get ((2/rAB^2)*dB/dt=0) from (3).
>
> > Why not?
>
> Why yes?
>
> > You have the metric anzatz - go derive the R_tr equation. You will
> > find that he is, in fact, right if you do it correctly.
>
> Through a mathemagic trik?
It is only magic to those who do not understand the process.
If you know how to obtain the Ricci tensor from a given metric - work
out the R_tr equation. It will be equal to 0 via being a vacuum
solution. If you *don't* know how to....well that's your problem and
you shouldn't be making the claims you are making.
>
> > > Actually, with time derivative, these equations are a lot more
> > > complicated than you have shown above.
>
> > No, they are not.
>
> Says someone who has not gone through them. <shrug>
Actually I have.
>
> > > Therefore showing (dB/dt = 0) is just BS.
>
> > *sigh*
>
> > It is the only way for the R_tr equation to hold for all t,r that
> > exclude trivial A,B. I am amazed that you have trouble with something
> > as basic as *this*.
>
> <sigh> You are drowning in BS, and you don't know it.
These tools have served me well thus far. Not my problem you are
incapable of understanding their usage.
>
> > > > (1/AB)( d/dr((1/B)*dA/dr) + (2/rB)*dA/dr) = 0
> > > > (1/AB)(-d/dr((1/B)*dA/dr)) - (2/rB)*d/dr(1/B) = 0
>
> > > > This is actually a pair of _ODEs_ (r is the only variable in them)
> > > > which remarkably simplify to the following (denoting the r-derivatives
> > > > by primes now):
>
> > > > A'/A + B'/B = 0
>
> > > Bullsh*t. Your differential equations involve a second derivative of
> > > A! Where is it?
>
> > It went away. He integrated both equations with respect to r.
>
> Come back after you have understood integration.
Well then what do _you_ get when you integrate both equations with
respect to r? Did you even try before crying "BS" ?
>
> > How can someone claim, with a straight face, that all of differential
> > geometry is wrong when that someone struggles with regular one
> > dimensional integration?
>
> Easily if it is wrong which it is. <shrug>
Did you even check?
This can be as interactive as you want to be - why not start off with
the metric anzatz _he_ uses and calculate the relevant equations
yourself. Or is that too much work to ask of you?
[...]
Androcles
"JanPB" <fil...@gmail.com> wrote in message news:1175916481.5...@y80g2000hsf.googlegroups.com...
You tell me, they are your words, idiot. Have a pleasant divide-by-zero, Hawking is as good at it as Einstein was. He even wrote "A Brief History of Time", which is briefer than the history of card games but funnier.
Koobee Wublee
> On Apr 6, 9:08 pm, "Koobee Wublee" <koobee.wub...@gmail.com> wrote:
> They aren't coordinate transformations - they are representations of
> the same manifold just with different signatures - remember the
> diag(-1,1,1,1) and the diag(1,-1,-1.-1) representations of Minkowski
> space?
>
> > Yes, I know. <shrug>
>
> If you actually knew you would not have not made that comment.
No, I comment was totally irrelevant to that! <shrug>
> > Right. I still don't agree with (r^2 dO^2). It should be more
> > general such as (C^2 dO^2) where C is a function of r.
>
> Then use C(r) in place of r^2. The argument that r^2*dO^2 appears in
> the metric is an argument from spherical symmetry - since you have
> such issue with it, start off with the most general metric and work
> from there.
This was Mr. Bielawski's convention. I personally would use C(r). As
long as it is C(r) in the general case and not r^2, I have no issues.
<shrug>
I want to correct myself that 'spherically symmetric' means A, B, and
C are independent of any angular displacement.
ds^2 = A dt^2 - B dr^2 - C dO^2
> Defining "cross terms" is left as an exercise for those who have
> enough math to understand.
Whatever! It is more like a call to another mathemagic trick.
> > > > ds^2 = A^2 dt'^2 + B^2 dr'^2 + C^2 dO'^2 [3]
>
> > > > You have reduced the spacetime from 4 variables to 3, remember?
>
> > > It isn't 3 variables - the variables are _still_ (t,r,theta,phi).
>
> > Not anymore.
>
> In what fantasy world?
Yours, of course.
> The metric is manifestly a function of t,r,theta,phi - is your reading
> disability so strong that you cannot read what is in front of your
> face?
No. Try to find the problem in front of a mirror.
> > Through a mathemagic trik?
>
> It is only magic to those who do not understand the process.
The trick is exposed. <shrug>
> If you know how to obtain the Ricci tensor from a given metric - work
> out the R_tr equation. It will be equal to 0 via being a vacuum
> solution. If you *don't* know how to....well that's your problem and
> you shouldn't be making the claims you are making.
To do so, it involves more assumptions. <shrug>
> > Says someone who has not gone through them. <shrug>
>
> Actually I have.
I don't mean in your dreams, please.
> > <sigh> You are drowning in BS, and you don't know it.
>
> These tools have served me well thus far.
I can see that. It makes you look awful.
> Not my problem you are
> incapable of understanding their usage.
I do. However if you don't which is the case, you are just drowning
in your own BS and won't know it. <shrug>
> > Come back after you have understood integration.
>
> Well then what do _you_ get when you integrate both equations with
> respect to r? Did you even try before crying "BS" ?
Try to look up the answer in a textbook. <shrug>
> > Easily if it is wrong which it is. <shrug>
>
> Did you even check?
Yes. <shrug>
> This can be as interactive as you want to be - why not start off with
> the metric anzatz _he_ uses and calculate the relevant equations
> yourself. Or is that too much work to ask of you?
I have done that. Good night!
Eric Gisse
[...]
>
> > This can be as interactive as you want to be - why not start off with
> > the metric anzatz _he_ uses and calculate the relevant equations
> > yourself. Or is that too much work to ask of you?
>
> I have done that. Good night!
Good.
Show your work.
JanPB
[...]
Something went wrong with my response, it looks like it never showed
up. Fortunately, I'm in the habit of writing posts in my text editor
as I lost a few posts before. So I'm reposting it now (and apologies
if this thing does end up posted twice).
On Apr 6, 5:53 pm, "Koobee Wublee" <koobee.wub...@gmail.com> wrote:
> On Apr 6, 1:31 pm, "JanPB" <film...@gmail.com> wrote:
>
> > We seek a spherically symmetric metric of signature 2 satisfying the
> > Einstein equation in vacuum:
>
> > R_ab = 0
>
> Or (G_ab = 0)
>
> > The assumption of symmetry allows us to assume WLOG that the metric
> > has the general form like so:
>
> WLOG?
"Without loss of generality". This abbreviation is rarely seen in
books but is otherwise very common.
> > ds^2 = f(r,t) dt^2 + g(r,t) dr^2 + h(r,t) dr dt + k(r,t) dO^2,
>
> > where r is a coordinate labelling the spheres of symmetry and:
>
> > dO^2 = dtheta^2 + sin^2(theta) dphi^2, as usual.
>
> > It turns out that even this reasonably nice general form results in an
> > unmanageable computational mess so one simplifies the general form
> > further by a local change of coordinates which cleverly removes the
> > "dr dt" term and also relabels the spheres of symmetry by their areas
> > (divided by 4pi). (I can supply more details on this final coordinate
> > change if you wish.) This results in the following two possible
> > simplified, diagonal, forms consistent with the given signature
> > constraint:
>
> > either:
>
> > ds^2 = -A^2 dt^2 + B^2 dr^2 + r^2 dO^2 [1]
>
> > or:
>
> > ds^2 = +A^2 dt^2 - B^2 dr^2 + r^2 dO^2 [2]
>
> If you are describing the same invariant spacetime, ds^2, then your
> dr, dt must be different due to your coordinate transformation.
Yes, I have recycled the letters "r" and "t" from the first formula
(the one with coefficients f, g, h, k) into the formulas [1] and [2].
BTW, did you notice the nonzero cross term "dr dt"? _This_ is the
source of the computational complexity, not k(r,t) which is trivially
replaced by r^2 (here I'm recycling "r" as usual).
> ds^2 = A^2 dt'^2 + B^2 dr'^2 + C^2 dO'^2 [3]
>
> I don't agree with (r^2 dO^2).
It's simply switching from r to r', where:
r' = sqrt(k(r,t))
> > where A(r,t) and B(r,t) are some functions which are never zero (or
> > else the metric would not be rank-4).
>
> Rank-4?
Brush up on your linear algebra. Here it's synonymous with
"nondegenerate".
> > NB: this local coordinate change happens to rely on dividing by
> > certain function, so any solutions for metric coefficients we may
> > obtain by proceeding from these two general forms may not be defined
> > over the spacetime events for which the function we divided by is
> > zero.
>
> This transformation is not necessary a division transformation. Why
> do you think it is so?
It's not _just_ a division - I said it only relied on it. The entire
procedure involves cooking up an integrating factor. Here is a quickie
summary. We have:
ds^2 = f(r,t) dt^2 + g(r,t) dr^2 + h(r,t) dt dr + k(r,t) dO^2 (*)
...and we want to get rid of the cross-term involving dt dr. We
consider the 1-form:
f dt + (h/2) dr
We know there exists a function F(r,t) (an "integrating factor") which
makes this form into a _closed_ form when multiplied by F:
d( F(f dt + (h/2) dr) ) = 0
...which means that locally there exists another function G(r,t) such
that:
F(f dt + (h/2) dr) = dG (**)
Turns out if we use this G in the following change of variables:
t' = G(r,t)
r = r
theta = theta
phi = phi
...then the crossterm dt dr drops out. Specifically, from (**) we
have:
dt'^2 = F^2 ( f dt + (h/2) dr )^2 =
= F^2 ( f^2 dt^2 + fh dt dr + (h^2/4) dr^2)
...which means (collecting terms):
f^2 dt^2 + fh dt dr = (1/F^2) dt'^2 - (h^2/4) dr^2
...and so DIVIDING BY f (this is the step I was referring to in my
previous post):
f dt^2 + h dt dr = U dt'^2 + V dr^2
...where I substituted U(r,t) and V(r,t) for the actual algebraic
expressions.
Now the reason it works is that when one plugs the LHS above into the
expression (*) for ds^2, we cancel out the unwanted cross-term like
so:
ds^2 = f dt^2 + g dr^2 + h dt dr + k dO^2 =
= U dt'^2 + (V + g) dr^2 + k dO^2
...and we have a diagonal form.(2/rAB^2)*dB/dt
> > [In fact (getting slightly ahead of myself here) this is precisely
> > what happens: the function we divide by is zero for r=2M, hence the
> > (in)famous exclusion of the locus r=2M from the domain of the metric
> > coefficients in the Schwarzschild form. There is of course no reason
> > to presume this is an actual physical exclusion as it potentially
> > results from the man-made work-saving transformation. So whether it's
> > a true singularity must be verified separately. It turns out - as is
> > well known - that the metric is well-defined for r=2M, it is merely
> > non-diagonalizable there hence the forms [1] and [2] break down there
> > - it's just a case of being too greedy/lazy.]
>
> This makes no sense. It has the signature of Archimedes Plutonium.
> Are you the real person behind this handle?
I am quite real, yes. Do you know Archimedes' original name? I think
he had like 5 name changes? The paragraph you quoted above simply
describes what happens for those r and t for which the function f(r,t)
equals zero (we've just divided by f, remember? so there is a
possibility we threw out a portion of the domain just out of
convenience).
> > OK, so we have those two forms potentially satisfying the EFE. The
> > maximal domain possible (unless the EFE themselves somehow end up
> > restricting it) is:
>
> > -infty < t < +infty,
> > 0 < r < +infty,
> > 0 < theta < 2pi,
> > 0 < phi < pi.
>
> > I'm going to skip the exact details of deriving the equations below
> > (we can go into it later if necessary). I'm using Cartan's method of
> > moving frames, and after computing the connection and curvature forms
> > the Einstein equation R_ab=0 reduces to the following four PDEs:
>
> > R_tt = 0
> > R_rr = 0
> > R_theta theta = 0
> > R_rt = 0
>
> No, with (3), you only get
>
> ** Rt't' = 0
> ** Rr'r' = 0
> ** RO'O' = 0
You get three equations if you assume A and B are functions of r only
(i.e. independent of t), IOW if you assume a static metric from the
beginning. Here we are interested in Birkhoff's theorem so we only
assume spherical symmetry. Because A and B are thus functions of r and
t, the rt-component of Ricci curvature is NOT identically zero, and is
in fact equal to (2/rAB^2)*dB/dt. It's just a calculation - I typed up
the details behind this calculation in TeX and posted them here:
http://www.mastersofcinema.org/jan/sch.pdf
TeX's deafult typeface is rather ugly but it beats ASCII anytime :-)
> Recall
>
> ds^2 = A^2 dt'^2 + B^2 dr'^2 + C^2 dO'^2 [3]
>
> You have reduced the spacetime from 4 variables to 3, remember?
I haven't reduced anything in this sense. There are still 4 variables
t,r,theta,phi. The Ricci rt-component just happens to involve dB/dt,
so it's obviously identically zero if you presume B is a function of r
only. But R_rt is not identically zero in general. Just compute (or
read that PDF doc) and see for yourself.
> > ...with the remaining equations being either identities or equivalent
> > to those four ones. (For example, R_phi phi = R_theta theta.)
>
> > The exact forms of the four left-hand sides are, few pages of
> > calculations later, respectively (excuse the mess, it's the fourth,
> > simplest, equation that's interesting to us):
>
> > (1/AB)(-d/dt((1/A)*dB/dt) + d/dr((1/B)*dA/dr) + (2/rB)*dA/dr) = 0
> > (1/AB)( d/dt((1/A)*dB/dt) - d/dr((1/B)*dA/dr)) - (2/rB)*d/dr(1/B) = 0
> > -+(1/rAB^2)*dA/dr -+ (1/rB)*d/dr(1/B) + (1/r^2)*(1 -+ 1/B^2) = 0
> > (2/rAB^2)*dB/dt = 0
>
> There is no way you can get ((2/rAB^2)*dB/dt=0) from (3).
That's right. You don't get the fourth equation out of the remaining
three. It's just a fourth independent equation (there are 10 of them
in general, 6 of which are either equivalent to the 4 above or are
identities 0=0).
> Actually, with time derivative, these equations are a lot more
> complicated than you have shown above.
That's because you use Christoffel symbols for your calculations. I
use Cartan's connection forms which are based on orthonormal moving
frames instead of coordinate frames. This technology is well-worth
learning as it offers significant simplification of the calculations.
At the "bottom" both Christoffel and Cartan-based quantities are the
same of course, but Cartan's approach makes clever use of certain
hidden SKEW-symmetries (rather than Christoffel symmetries in the
lower indices) which means you have less quantities to deal with: for
example, there are only 6 independent entries in a 4x4 skew-symmetric
matrix vs. 10 entries in a 4x4 symmetric matrix. Etc. MTW has a
chapter on Cartan's calculus and books like Wald use this technique
extensively.
> > ...where "-+" means use the "-" sign for the first general form of the
> > diagonal metric [1], use "+" for the second [2]. Also "d" obviously
> > denotes partial derivatives.
>
> > Note the fourth equation R_rt = 0 says then (cancelling the factor in
> > front):
>
> > dB/dt = 0
>
> Therefore showing (dB/dt = 0) is just BS.
You have to work harder than that :-) The fourth equation R_rt = 0
says that dB/dt=0. It just says what it says.
> > ...in other words, B(r,t) is independent of t, both in case [1] and
> > [2]. This means that B is only a function of r which means that the
> > first two PDEs simplify quite a bit:
>
> > (1/AB)( d/dr((1/B)*dA/dr) + (2/rB)*dA/dr) = 0
> > (1/AB)(-d/dr((1/B)*dA/dr)) - (2/rB)*d/dr(1/B) = 0
>
> > This is actually a pair of _ODEs_ (r is the only variable in them)
> > which remarkably simplify to the following (denoting the r-derivatives
> > by primes now):
>
> > A'/A + B'/B = 0
>
> Bullsh*t. Your differential equations involve a second derivative of
> A! Where is it?
Just multiply through 1/AB in the first equation and then add both
equations together. The first term cancels out. Then note that d/dr(1/
B) = -B'/B^2 and you obtain the simple equation in A and B.
> > i.e.,
>
> > (ln |A|)' + (ln |B|)' = 0
>
> > hence:
>
> > |AB| = C (a positive constant of integration)
>
> > i.e.,
>
> > AB = C (an arbitary nonzero constant of integration).
>
> Don't you feel like a prophet if you know the answer before hand?
It's just following the math - no room wiggle. When I was writing this
I didn't know exactly where it lead (besides the general expectation
of seeing Schwarzschild's solution in the end somewhere).
> > The important thing as far as Birkhoff is concerned is that this means
> > that:
>
> > B = constant/A
>
> > ...in other words B(r,t) is independent of t as well, in both [1] and
> > [2].
>
> Only with your mathematical errors, that is so.
>
> [The rest snipped due to the obvious mathematical error.]
Conveniently pulling a rabbit out of the hat? At least tell us what
mathematical errors lead to B=constant/A?
> Also, you have conveniently assumed (C^2 = r^2). See (3).
This is a trivial coordinate change. I'm beginning to suspect where
you (and Crothers) get this idea of metric being coordinate-dependent.
It's the problem of physical units, isn't it? You both think that
changing units of measurement is the same as changing coordinates,
right? So when I change my unit of length - so you both think - the
numerical values of, say, vectors lengths change also, hence the
values of dot products change, and these values ARE the metric, so the
metric changes too! Am I getting close?
> That is how the Birkhoff's theorem as the author of Wikipedia defined
> it.
>
> Assuming (C^2 = r^2) which translates to spherically symmetric. For
> every C^2, there is a new set of metric that satisfies the field
> equations. Have you finally understood Mr. Crothers?
C^2 makes no difference. Mr. Crothers is a confused soul.
> Assuming (static) which translates to all elements of the metric
> independent of time.
This whole exchange started with you knocking Birkhoff's theorem and
how it implied the metric had to be static. So don't assume now that
the metric is static.
> Then, you end up with much, much, much, and much simpler equations
> described below.
>
> ** (r / B^2) (dB/dr) - (1 / B) + 1 = 0
> ** (r / A / B) (dA/dr) + (1 / B) - 1 = 0
Well, wonderful, but again - we are discussing how Birkhoff says
spherical symmetry in Schwarzschild coordinates implies independence
of t (which you mistook for static). We cannot assume that which we
want to prove!
> These are actually not Ricci tensor equations but rather Einstein
> tensor equations. With all these simplifications, they are all the
> same.
>
> For the spacetime described below,
>
> ds^2 = A dt^2 - B dr^2 - C dO^2
>
> Where (C = r^2)
>
> You will see that you can easily solve for B to be the following.
>
> ** B = 1 / (1 + K / r)
>
> Where (K = integration constant)
>
> And A follows.
>
> ** A = K' (1 + K / r)
>
> Where (K' = another integration constant)
>
> For (C = (r + K)^2),
>
> You get
>
> ** A = K' / (1 + K / r)
> ** B = 1 + K / r
>
> In this case, you don't get a black hole.
You do - you simply neglected to use the correct domain (you changed
coordinates but kept the domain the same - this is like saying "0
Celsius equals 0 Kelvin, therefore absolute zero doesn't exist").
You've changed the domain of r from r>0 to r>-K.
> For (C = ((r^3 + K^3)^(1/3) - K)^2),
>
> You end up with Schwarzschild's original solution which does not
> manifest any black holes either.
>
> There are infinite numbers of solutions more.
You've just altered the domain to exclude the black hole. This is not
a different solution, it's just a subset of the good old Schwarzschild
with a restriction of the radial coordinate. See:
http://groups.google.com/group/sci.physics.relativity/msg/068e4e33d3fecf0d?hl=en&
--
Jan Bielawski
Eric Gisse
[...]
I don't think he will ever reverse his position, to be honest. He has
spent too many years on USENET claiming stupid shit to suddenly turn
around and admit 'oh man - that was stupid!'. Your efforts, though,
aren't wasted. I have appreciated seeing the derivation of the
Schwarzschild solution through a different way. I am completely
unfamiliar with Cartan's method.
>
> > Recall
>
> > ds^2 = A^2 dt'^2 + B^2 dr'^2 + C^2 dO'^2 [3]
>
> > You have reduced the spacetime from 4 variables to 3, remember?
>
> I haven't reduced anything in this sense. There are still 4 variables
> t,r,theta,phi. The Ricci rt-component just happens to involve dB/dt,
> so it's obviously identically zero if you presume B is a function of r
> only. But R_rt is not identically zero in general. Just compute (or
> read that PDF doc) and see for yourself.
I asked him to do this.
It might be worth asking if he still thinks T_uv = 0 does not imply
that R_uv = 0. I showed him, straight from the field equations, awhile
back that it is true. The reason I did that is because he could never
get past the first step of Birkhoff's theorem - understanding that a
vacuum solution implies that R_uv = 0 - which he always called
"convoluted" for whatever reason.
At any rate it is still unclear to me why he thinks you reduced the
metric from four to three variables despite having an explicit
definition for dO^2.
[...]
> > Actually, with time derivative, these equations are a lot more
> > complicated than you have shown above.
>
> That's because you use Christoffel symbols for your calculations. I
> use Cartan's connection forms which are based on orthonormal moving
> frames instead of coordinate frames. This technology is well-worth
> learning as it offers significant simplification of the calculations.
> At the "bottom" both Christoffel and Cartan-based quantities are the
> same of course, but Cartan's approach makes clever use of certain
> hidden SKEW-symmetries (rather than Christoffel symmetries in the
> lower indices) which means you have less quantities to deal with: for
> example, there are only 6 independent entries in a 4x4 skew-symmetric
> matrix vs. 10 entries in a 4x4 symmetric matrix. Etc. MTW has a
> chapter on Cartan's calculus and books like Wald use this technique
> extensively.
That explains why you never wrote down the symbols... ^_^
[...]
>
> > Assuming (C^2 = r^2) which translates to spherically symmetric. For
> > every C^2, there is a new set of metric that satisfies the field
> > equations. Have you finally understood Mr. Crothers?
>
> C^2 makes no difference. Mr. Crothers is a confused soul.
>
I wonder - do you think ol' KW and Crothers are the same fellow [or
far more related than normal]? It would neither be the first or last
time for a crank author to reference his own work extensively on
USENET while posting under a pseudonym.
At any rate - I told KW to assume C(r) instead of r^2. He said he did
it, and now I'm just waiting for him to show the work he claimed he
did.
[...]
JanPB
> On Apr 8, 10:45 am, "JanPB" <film...@gmail.com> wrote:
>
> [...]
>
> I don't think he will ever reverse his position, to be honest. He has
> spent too many years on USENET claiming stupid shit to suddenly turn
> around and admit 'oh man - that was stupid!'. Your efforts, though,
> aren't wasted.
To tell you the truth I rarely post detailed technical stuff unless
I've done it earlier just for myself (to clear up something) or the
stuff is just very easy. So last year I decided once and for all to
derive the spherically symmetric solution without presuming anything
else, just to convince myself the logic was fine. One reason was that
all texts that I know of tend to gloss over the important details and
carelessly write stuff that's plain wrong. Perhaps in another post
I'll go over the relevant fragments from the standard textbooks :-) Of
course the logic works fine and I simply wrote it down in that TeX
note. So it's easy for me now to reuse it.
It's small details, yes, but it's enough to derail people like
Crothers & Co. At the core they are probably well-meaning but they
don't read those texts aggressively enough to _really question_ what
they read, and instead accept the first false excuse that seems to
"resolve" an incorrect statement (like "spherically symmetric implies
static" in Ohanian & Ruffini, say).
> I have appreciated seeing the derivation of the
> Schwarzschild solution through a different way. I am completely
> unfamiliar with Cartan's method.
MTW provides probably the quickest explanation. I think Wald has an
introduction too (at about 200 mph!). Many standard diff. geo. texts
like Spivak's vol. II do it in great detail, although math texts tend
to ignore the indefinite metric case which brings an extra twist or
two into the proceedings (plus it allows null frames - besides
orthonormal - which are also very interesting, Cartan's method applies
to both with some modifications).
[...]
> That explains why you never wrote down the symbols... ^_^
...not even connection and Riemann curvature forms. Too much typing -
it's all in that PDF file. I only wrote down the 4 non-trivial
components (out of 10 possible) of the equation Ricci=0.
> I wonder - do you think ol' KW and Crothers are the same fellow [or
> far more related than normal]? It would neither be the first or last
> time for a crank author to reference his own work extensively on
> USENET while posting under a pseudonym.
Everything is possible although Mr. C. seems to have no problems with
properly identifying himself on his web site so I don't know why he'd
bother with all that hide-and-seek here?
> At any rate - I told KW to assume C(r) instead of r^2. He said he did
> it, and now I'm just waiting for him to show the work he claimed he
> did.
I'll take a look but I think leaving arbitrary C(r,t) in place again
would complicate the derivation beyond anything sane, just like the
non-diagonal form.
--
Jan Bielawski
JanPB
Just a typo correction:
> ...and we have a diagonal form.(2/rAB^2)*dB/dt
It should just read:
> ...and we have a diagonal form.
(I don't know where the junk after the period came from - some random
paste looks like.)
--
Jan Bielawski
John Park
> On Apr 6, 5:53 pm, "Koobee Wublee" <koobee.wub...@gmail.com> wrote:
>
> [...]
>>
>> > Einstein equation in vacuum:
>>
>> > R_ab = 0
>>
>> Or (G_ab = 0)
>>
>> > The assumption of symmetry allows us to assume WLOG that the metric
>> > has the general form like so:
>>
>>
>> > where r is a coordinate labelling the spheres of symmetry and:
>>
>> > dO^2 = dtheta^2 + sin^2(theta) dphi^2, as usual.
>>
>> > It turns out that even this reasonably nice general form results in an
>> > unmanageable computational mess so one simplifies the general form
>> > further by a local change of coordinates which cleverly removes the
>> > "dr dt" term and also relabels the spheres of symmetry by their areas
>> > (divided by 4pi). (I can supply more details on this final coordinate
>> > change if you wish.) This results in the following two possible
>> > simplified, diagonal, forms consistent with the given signature
>> > constraint:
>>
>> > either:
>>
>> > ds^2 = -A^2 dt^2 + B^2 dr^2 + r^2 dO^2 [1]
>>
>> > or:
>>
>> > ds^2 = +A^2 dt^2 - B^2 dr^2 + r^2 dO^2 [2]
> (the one with coefficients f, g, h, k) into the formulas [1] and [2].
> BTW, did you notice the nonzero cross term "dr dt"? _This_ is the
> source of the computational complexity, not k(r,t) which is trivially
> replaced by r^2 (here I'm recycling "r" as usual).
>
>> ds^2 = A^2 dt'^2 + B^2 dr'^2 + C^2 dO'^2 [3]
>>
>> I don't agree with (r^2 dO^2).
>
> It's simply switching from r to r', where:
>
> r' = sqrt(k(r,t))
>
I don't understand. Wouldn't this make r' time-dependent?
Thanks,
John Park
JanPB
r' is a new independent variable, it doesn't depend on other
variables. The formula above is a portion of a change of variables
from one independent set to another. It involves t because the
"original" labeling of the spheres may have been t-dependent, so to
"undo" this dependence, the change of variables may also involve t.
--
Jan Bielawski
Androcles
"JanPB" <fil...@gmail.com> wrote in message news:1176178763.0...@o5g2000hsb.googlegroups.com...
HAHAHA!
r' is an independent variable that is dependent on sqrt(k(r,t)).
Light bulb changing is dependent on sqrt(k(step-ladder, JanPB^2))
but independent of the Pole Bielawski, call an electrician.
How many Poles does it take to change a light bulb?
JanPB
wrote:
> "JanPB" <film...@gmail.com> wrote in messagenews:1176178763.0...@o5g2000hsb.googlegroups.com...
Independent variable is one thing, coordinate change is another.
--
Jan Bielawski
Androcles
"JanPB" <fil...@gmail.com> wrote in message news:1176232221.9...@b75g2000hsg.googlegroups.com...
Did you snip something I wrote to hide your embarrassment for
being a stupid Pole who makes contradictory statements?
Oh yes, here it is:
Light bulb changing is dependent on sqrt(k(step-ladder, JanPB^2))
but independent of the Pole Bielawski, call an electrician.
How many Poles does it take to change a light bulb?
Try to answer the question, you might learn what "dependence"
means.
(Hahaha)^2
Koobee Wublee
I have been gone for over a week. Coming back, I am very surprised
that you would pile on your BS and present it as a gourmet dish.
> > > ds^2 = +A^2 dt^2 - B^2 dr^2 + r^2 dO^2 [2]
>
> > If you are describing the same invariant spacetime, ds^2, then your
> > dr, dt must be different due to your coordinate transformation.
>
> Yes, I have recycled the letters "r" and "t" from the first formula
> (the one with coefficients f, g, h, k) into the formulas [1] and [2].
> BTW, did you notice the nonzero cross term "dr dt"? _This_ is the
> source of the computational complexity, not k(r,t) which is trivially
> replaced by r^2 (here I'm recycling "r" as usual).
What happened to the cross terms (dO dt)? You have assumed t0o much
to arrive at a solution that is thoroughly dependent on the assumption
you have made, and yet you don't realize it. <shrug>
> > ds^2 = A^2 dt'^2 + B^2 dr'^2 + C^2 dO'^2 [3]
>
> > I don't agree with (r^2 dO^2).
>
> It's simply switching from r to r', where:
>
> r' = sqrt(k(r,t))
>
> > > where A(r,t) and B(r,t) are some functions which are never zero (or
> > > else the metric would not be rank-4).
>
> > Rank-4?
>
> Brush up on your linear algebra. Here it's synonymous with
> "nondegenerate".
>
> > > NB: this local coordinate change happens to rely on dividing by
> > > certain function, so any solutions for metric coefficients we may
> > > obtain by proceeding from these two general forms may not be defined
> > > over the spacetime events for which the function we divided by is
> > > zero.
>
> > This transformation is not necessary a division transformation. Why
> > do you think it is so?
>
> It's not _just_ a division - I said it only relied on it. The entire
> procedure involves cooking up an integrating factor. Here is a quickie
> summary. We have:
>
> ds^2 = f(r,t) dt^2 + g(r,t) dr^2 + h(r,t) dt dr + k(r,t) dO^2 (*)
>
> ...and we want to get rid of the cross-term involving dt dr. We
> consider the 1-form:
>
> f dt + (h/2) dr
More assumptions. You cannot make these assumptions and claiming that
you have not made any assumptions to derive at the Birkhoff's
theorem. That is why in Birkhoff'theorem it lays out the following
assumptions before the proof begins.
** If static, independent of t
** If spherically symmetric, independent of O
> [...]
So, if you use some other connection coefficients that are so abstract
in nature compared with Christoffel's which are more tangible you end
up with something exponentially simpler. Yeah, right. Give me a
break. There is nothing as realistic as the Christoffel symbols.
<shrug>
> [...]
>
> Conveniently pulling a rabbit out of the hat? At least tell us what
> mathematical errors lead to B=constant/A?
We will leave that in a future discussion. I am very tired tonight.
> > Also, you have conveniently assumed (C^2 = r^2). See (3).
>
> This is a trivial coordinate change.
This is not coordinate change.
> I'm beginning to suspect where
> you (and Crothers) get this idea of metric being coordinate-dependent.
I don't recall Mr. Crothers is saying the metric is coordinate
dependent. As far as I know, I am the only one so far.
> It's the problem of physical units, isn't it?
No.
> You both think that
> changing units of measurement is the same as changing coordinates,
> right?
Not I.
> So when I change my unit of length - so you both think - the
> numerical values of, say, vectors lengths change also, hence the
> values of dot products change, and these values ARE the metric, so the
> metric changes too! Am I getting close?
No. That is the problem here. There is no coordinate change. We are
still talking about using the same spherically symmetric coordinate
system.
> > That is how the Birkhoff's theorem as the author of Wikipedia defined
> > it.
>
> > Assuming (C^2 = r^2) which translates to spherically symmetric. For
> > every C^2, there is a new set of metric that satisfies the field
> > equations. Have you finally understood Mr. Crothers?
>
> C^2 makes no difference. Mr. Crothers is a confused soul.
Perhaps, he is. However, you are even more confused than Mr. Crothers
is.
> > Assuming (static) which translates to all elements of the metric
> > independent of time.
>
> This whole exchange started with you knocking Birkhoff's theorem and
> how it implied the metric had to be static. So don't assume now that
> the metric is static.
I still maintain the assumptions of static and spherically
symmetricity must be established before going further with Birkhoff's
theorem. After making these assumptions, the theorem appears to be
correct with the Schwarzschild metric, or Schwarzschild's original
metric, or even the metric Mr. Crothers is championing over the
Schwarzschild metric. However, there are just other metrics that
would invalidate Birkhoff's theorem. I have given some of these
metrics that do so in the past.
I have not alter any domains. I just presented more independent
solutions where each one is just as valid as the Schwarzschild metric
in which you have worshipped like a God. In the meantime, the
coordinate of choice is still the familiar spherically symmetric polar
coordinate system for each of the solutions I have presented so far
--- no domain change nor change of coordinate system, just independent
solutons to the field equations.
JanPB
> On Apr 8, 11:45 am, "JanPB" <film...@gmail.com> wrote:
>
> > On Apr 6, 5:53 pm, "Koobee Wublee" <koobee.wub...@gmail.com> wrote:
>
> I have been gone for over a week.
As you can see, I've been delayed as well!
> Coming back, I am very surprised
> that you would pile on your BS and present it as a gourmet dish.
And I am surprised that you *still* defend your position despite my
explicit disproof of it. So I guess we're even :-)
> > > > ds^2 = +A^2 dt^2 - B^2 dr^2 + r^2 dO^2 [2]
>
> > > If you are describing the same invariant spacetime, ds^2, then your
> > > dr, dt must be different due to your coordinate transformation.
>
> > Yes, I have recycled the letters "r" and "t" from the first formula
> > (the one with coefficients f, g, h, k) into the formulas [1] and [2].
> > BTW, did you notice the nonzero cross term "dr dt"? _This_ is the
> > source of the computational complexity, not k(r,t) which is trivially
> > replaced by r^2 (here I'm recycling "r" as usual).
>
> What happened to the cross terms (dO dt)?
Can be removed by spherical symmetry (I am assuming you mean "dtheta
dt" and "dphi dt" when you write "dO dt"?)
> You have assumed t0o much
> to arrive at a solution that is thoroughly dependent on the assumption
> you have made, and yet you don't realize it. <shrug>
The assumptions made are:
1. spherical symmetry,
2. nondegenerate metric,
3. signature 2 metric,
4. ...and various other standard unstated assumptions, like the
smoothness of all the relevant objects, etc.
> > > ds^2 = A^2 dt'^2 + B^2 dr'^2 + C^2 dO'^2 [3]
>
> > > I don't agree with (r^2 dO^2).
>
> > It's simply switching from r to r', where:
>
> > r' = sqrt(k(r,t))
>
> > > > where A(r,t) and B(r,t) are some functions which are never zero (or
> > > > else the metric would not be rank-4).
>
> > > Rank-4?
>
> > Brush up on your linear algebra. Here it's synonymous with
> > "nondegenerate".
>
> > > > NB: this local coordinate change happens to rely on dividing by
> > > > certain function, so any solutions for metric coefficients we may
> > > > obtain by proceeding from these two general forms may not be defined
> > > > over the spacetime events for which the function we divided by is
> > > > zero.
>
> > > This transformation is not necessary a division transformation. Why
> > > do you think it is so?
>
> > It's not _just_ a division - I said it only relied on it. The entire
> > procedure involves cooking up an integrating factor. Here is a quickie
> > summary. We have:
>
> > ds^2 = f(r,t) dt^2 + g(r,t) dr^2 + h(r,t) dt dr + k(r,t) dO^2 (*)
>
> > ...and we want to get rid of the cross-term involving dt dr. We
> > consider the 1-form:
>
> > f dt + (h/2) dr
>
> More assumptions.
This is just a construction of an auxiliary function - no additional
assumptions are being made.
> You cannot make these assumptions and claiming that
> you have not made any assumptions to derive at the Birkhoff's
> theorem.
I repeat, the only assumptions made are 1-4 above.
> That is why in Birkhoff'theorem it lays out the following
> assumptions before the proof begins.
>
> ** If static, independent of t
This is a conclusion of the theorem, not the assumption.
> ** If spherically symmetric, independent of O
And this is just a general fact about the form of the metric in a
spherically symmetric spacetime.
Yes, of course - both Christoffel and Cartan are the same calculation,
only compartmentalised differently (more efficiently for Cartan).
> Give me a
> break. There is nothing as realistic as the Christoffel symbols.
> <shrug>
Both are very realistic - in fact they are in a sense almost
identical: both Christoffel coefficients and Cartan connection forms
are components of the rate of change of certain basis vectors wrt to
other basis vectors. The difference is that in Christoffel's setup you
take _coordinate vectors_ as your basis while in Cartan's approach you
take _orthonormal vectors_ as the basis (i.e., a non-holonomic frame).
So both approaches are tangible.
The advantage of Cartan's setup is that with respect to an orthonormal
basis the entries of the matrix g_ij are obviously constant (namely,
+1, -1, and 0). Because of that Cartan's connection forms are skew-
symmetric in their lower indices and this is the fact responsible for
the significant reduction of the wasteful computational effort which
is present in Christoffel-based computations. If you need Christoffel
coefficients at the end, you can always compute them from the Cartan's
form after the main calculation is done.
> > Conveniently pulling a rabbit out of the hat? At least tell us what
> > mathematical errors lead to B=constant/A?
>
> We will leave that in a future discussion. I am very tired tonight.
>
> > > Also, you have conveniently assumed (C^2 = r^2). See (3).
>
> > This is a trivial coordinate change.
>
> This is not coordinate change.
t' = t
r' = sqrt(C(r,t))
theta' = theta
phi' = phi
...is a coordinate change that removes C(r,t) which is therefore
redundant. C(r,t) is simply man-made labelling of the spheres of
symmetry. No physics is changed by my relabelling the spheres.
> > I'm beginning to suspect where
> > you (and Crothers) get this idea of metric being coordinate-dependent.
>
> I don't recall Mr. Crothers is saying the metric is coordinate
> dependent. As far as I know, I am the only one so far.
>
> > It's the problem of physical units, isn't it?
>
> No.
>
> > You both think that
> > changing units of measurement is the same as changing coordinates,
> > right?
>
> Not I.
>
> > So when I change my unit of length - so you both think - the
> > numerical values of, say, vectors lengths change also, hence the
> > values of dot products change, and these values ARE the metric, so the
> > metric changes too! Am I getting close?
>
> No. That is the problem here. There is no coordinate change. We are
> still talking about using the same spherically symmetric coordinate
> system.
OK, so if it's not the unit problem then why do you claim that metric
changes when coordinates change? For example, take a vector. Its
length is, say, 17 units. I change the coordinates. The same vectors
must still have the same length, 17 units (physics cannot depend on
man-made labels attached to objects).
Your matrix of g_ij does change because your coordinate basis vectors
change with coordinate change (g_ij represent the _changed_
coefficients necessary to reconstruct the _unchanged_ number 17 from
the _changed_ coordinate vectors).
When people say "metric", they mean by this the thing that assigns
actual lengths and angles to vectors and vector pairs. And lengths and
angles are properties of vectors and vector pairs _themselves_, not of
their coordinate labels. When I take a stick in my hand, it has a
definite numerically quantified length and the metric is the abstract
model that defines this assignment - coordinates do not enter into it
as the stick clearly has the length 17 regardless of any coordinates.
Coordinates are only a computational tool, just like analytic geometry
is a better computational tool than the old Greek geometry that used
just geometric constructions.
Your coordinate-dependent "metric" is not a model for physical lengths
and angles. As to what model it is OF is anyone's guess.
> > > That is how the Birkhoff's theorem as the author of Wikipedia defined
> > > it.
>
> > > Assuming (C^2 = r^2) which translates to spherically symmetric. For
> > > every C^2, there is a new set of metric that satisfies the field
> > > equations. Have you finally understood Mr. Crothers?
>
> > C^2 makes no difference. Mr. Crothers is a confused soul.
>
> Perhaps, he is. However, you are even more confused than Mr. Crothers
> is.
>
> > > Assuming (static) which translates to all elements of the metric
> > > independent of time.
>
> > This whole exchange started with you knocking Birkhoff's theorem and
> > how it implied the metric had to be static. So don't assume now that
> > the metric is static.
>
> I still maintain the assumptions of static and spherically
> symmetricity must be established before going further with Birkhoff's
> theorem.
You don't need static - where in my PDF note did you see this
assumption?
> After making these assumptions, the theorem appears to be
> correct with the Schwarzschild metric, or Schwarzschild's original
> metric, or even the metric Mr. Crothers is championing over the
> Schwarzschild metric. However, there are just other metrics that
> would invalidate Birkhoff's theorem. I have given some of these
> metrics that do so in the past.
They are all isometric to (a subset of) the standard Schwarzschild
solution.
> > > Then, you end up with much, much, much, and much simpler equations
> > > described below.
>
> > > ** (r / B^2) (dB/dr) - (1 / B) + 1 = 0
> > > ** (r / A / B) (dA/dr) + (1 / B) - 1 = 0
Yes, it's much simpler if you assume a static metric from the start.
That's not what we are supposed to be doing though.
You've changed the coordinates by r'=r-K. This means the domain r>0 in
the new coordinates is r'>-K. It's just like, say, switching from the
Cartesians to polars: a unit disc (say) is described in Cartesians by:
-1 <= x <= +1
-sqrt(1 - x^2) <= y <= +sqrt(1 - x^2)
Switching to polars: x=r cos(theta), y=r sin(theta) changes the
domain:
0 < r <= 1
0 <= theta < 2 pi.
Your "different" solution is exactly like saying that unit disc in
polars is described by:
-1 <= r <= +1
-sqrt(1 - r^2) <= theta <= +sqrt(1 - r^2).
...i.e., patent nonsense. And your case is even simpler than that
because all you change is ONE coordinate (r), not two.
> I just presented more independent
> solutions where each one is just as valid as the Schwarzschild metric
> in which you have worshipped like a God.
What's up with this "worshipping" business. You must have caught
whatever bug Pentcho is trying to recover from. It's just a
mathematical theorem that certain differential equations have unique
solutions.
> In the meantime, the
> coordinate of choice is still the familiar spherically symmetric polar
> coordinate system for each of the solutions I have presented so far
> --- no domain change nor change of coordinate system, just independent
> solutons to the field equations.
Your "independent" solution is obtained from Schwarzschild by plugging
in r'=r-K - hence it's the same solution.
--
Jan Bielawski
Androcles
"JanPB" <fil...@gmail.com> wrote in message news:1177368104.4...@e65g2000hsc.googlegroups.com...
Theorems without axioms are "worshipped". You must have caught
Alzheimers.
Koobee Wublee
> On Apr 18, 12:04 am, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > What happened to the cross terms (dO dt)?
>
> Can be removed by spherical symmetry (I am assuming you mean "dtheta
> dt" and "dphi dt" when you write "dO dt"?)
No, you have made this assumption. Spherically symmetry means all the
elements of the metric are independent of O. (a) <shrug>
> > You have assumed t0o much
> > to arrive at a solution that is thoroughly dependent on the assumption
> > you have made, and yet you don't realize it. <shrug>
>
> The assumptions made are:
>
> 1. spherical symmetry,
> 2. nondegenerate metric,
> 3. signature 2 metric,
> 4. ...and various other standard unstated assumptions, like the
> smoothness of all the relevant objects, etc.
Plus the one above and more to come.
> f dt + (h/2) dr
>
> > More assumptions.
>
> This is just a construction of an auxiliary function - no additional
> assumptions are being made.
No, it is another assumption. (b)
> > You cannot make these assumptions and claiming that
> > you have not made any assumptions to derive at the Birkhoff's
> > theorem.
>
> I repeat, the only assumptions made are 1-4 above.
You have made two more assumptions (a) and (b) so far.
> > That is why in Birkhoff'theorem it lays out the following
> > assumptions before the proof begins.
>
> > ** If static, independent of t
>
> This is a conclusion of the theorem, not the assumption.
If you don't agree with Widinpedia, that is your problem only.
> > ** If spherically symmetric, independent of O
>
> And this is just a general fact about the form of the metric in a
> spherically symmetric spacetime.
So, why is (dO dt = zero)?
> > So, if you use some other connection coefficients that are so abstract
> > in nature compared with Christoffel's which are more tangible you end
> > up with something exponentially simpler. Yeah, right.
>
> Yes, of course - both Christoffel and Cartan are the same calculation,
> only compartmentalised differently (more efficiently for Cartan).
In a set of real geodesic equations, you have to use the Christoffel
symbols of the 2nd kind. <shrug>
> > Give me a
> > break. There is nothing as realistic as the Christoffel symbols.
> > <shrug>
>
> Both are very realistic - in fact they are in a sense almost
> identical: both Christoffel coefficients and Cartan connection forms
> are components of the rate of change of certain basis vectors wrt to
> other basis vectors. The difference is that in Christoffel's setup you
> take _coordinate vectors_ as your basis while in Cartan's approach you
> take _orthonormal vectors_ as the basis (i.e., a non-holonomic frame).
> So both approaches are tangible.
You are very incorrect about the Christoffel symbols. <shrug>
> The advantage of Cartan's setup is that with respect to an orthonormal
> basis the entries of the matrix g_ij are obviously constant (namely,
> +1, -1, and 0). Because of that Cartan's connection forms are skew-
> symmetric in their lower indices and this is the fact responsible for
> the significant reduction of the wasteful computational effort which
> is present in Christoffel-based computations. If you need Christoffel
> coefficients at the end, you can always compute them from the Cartan's
> form after the main calculation is done.
With the way the geodesic equations are written, there is no advantage
in using holonomic bases. With Riemann curvature tensor, I have been
using holonomic bases. Thus, what you are saying just still does not
gain any simplicity in mathematics.
> > This is not coordinate change.
>
> t' = t
> r' = sqrt(C(r,t))
> theta' = theta
> phi' = phi
>
> ...is a coordinate change that removes C(r,t) which is therefore
> redundant. C(r,t) is simply man-made labelling of the spheres of
> symmetry. No physics is changed by my relabelling the spheres.
Oh, no. C can take many forms. All C's are independent of each
other. There is no coordinate transformation. The choice of
coordinate WAS ALREADY decided when the field equations WERE
ESTABLISHED. You need to stick with that coordinate system. <shrug>
> > No. That is the problem here. There is no coordinate change. We are
> > still talking about using the same spherically symmetric coordinate
> > system.
>
> OK, so if it's not the unit problem then why do you claim that metric
> changes when coordinates change?
Geometry = Metric * Coordinate
If Geometry is invariant, Metric must vary with the choice of
Coordinate. <shrug>
> For example, take a vector. Its
> length is, say, 17 units. I change the coordinates. The same vectors
> must still have the same length, 17 units (physics cannot depend on
> man-made labels attached to objects).
You are saying Geometry = Geometry. Of course, it is. <shrug>
> Your matrix of g_ij does change because your coordinate basis vectors
> change with coordinate change (g_ij represent the _changed_
> coefficients necessary to reconstruct the _unchanged_ number 17 from
> the _changed_ coordinate vectors).
What are you talking about?
> When people say "metric", they mean by this the thing that assigns
> actual lengths and angles to vectors and vector pairs. And lengths and
> angles are properties of vectors and vector pairs _themselves_, not of
> their coordinate labels. When I take a stick in my hand, it has a
> definite numerically quantified length and the metric is the abstract
> model that defines this assignment - coordinates do not enter into it
> as the stick clearly has the length 17 regardless of any coordinates.
In the past 100 years, when people talk about 'metric', they have been
referring to A, B, and C of the following equation.
ds^2 = A dt^2 - B dr^2 - C dO^2
> Coordinates are only a computational tool, just like analytic geometry
> is a better computational tool than the old Greek geometry that used
> just geometric constructions.
>
> Your coordinate-dependent "metric" is not a model for physical lengths
> and angles. As to what model it is OF is anyone's guess.
Once again, Geometry = Metric * Coordinate
Where
** Geometry = invariant
What you have been talking about the metric is Geometry. However, it
is invalid when applied to the field equations, Riemann curvature
tensors, etc. You are very confused. <shrug>
> > After making these assumptions, the theorem appears to be
> > correct with the Schwarzschild metric, or Schwarzschild's original
> > metric, or even the metric Mr. Crothers is championing over the
> > Schwarzschild metric. However, there are just other metrics that
> > would invalidate Birkhoff's theorem. I have given some of these
> > metrics that do so in the past.
>
> They are all isometric to (a subset of) the standard Schwarzschild
> solution.
The Birkhoff's theorem has a minor impact on the course of GR,
anyway. In time, I will prove it wrong through other solutions to the
field equations where the coordinate of choice is already
established. Stop conjuring up another coordinate system. Your
matheMagic trick is exposed.
> > I have not alter any domains.
>
> You've changed the coordinates by r'=r-K. This means the domain r>0 in
> the new coordinates is r'>-K. It's just like, say, switching from the
> Cartesians to polars: a unit disc (say) is described in Cartesians by:
How can I? I am bound to the same choice of coordinate system that
established the field equations. You are very unreasonable here.
> -1 <= x <= +1
> -sqrt(1 - x^2) <= y <= +sqrt(1 - x^2)
>
> Switching to polars: x=r cos(theta), y=r sin(theta) changes the
> domain:
>
> 0 < r <= 1
> 0 <= theta < 2 pi.
>
> Your "different" solution is exactly like saying that unit disc in
> polars is described by:
>
> -1 <= r <= +1
> -sqrt(1 - r^2) <= theta <= +sqrt(1 - r^2).
>
> ...i.e., patent nonsense. And your case is even simpler than that
> because all you change is ONE coordinate (r), not two.
Please stop this nonsense. The choice of coordinate system is already
established as the commonly utilized spherically symmetric polar
coordinate system.
Face the mathematics. You are looking at an infinite numbers of
solutions to the field equations. Stop burying your head in the hole
on the ground.
> > I just presented more independent
> > solutions where each one is just as valid as the Schwarzschild metric
> > in which you have worshipped like a God.
>
> What's up with this "worshipping" business. You must have caught
> whatever bug Pentcho is trying to recover from. It's just a
> mathematical theorem that certain differential equations have unique
> solutions.
I can assure you I did not catch any bug from Pentcho.
> > In the meantime, the
> > coordinate of choice is still the familiar spherically symmetric polar
> > coordinate system for each of the solutions I have presented so far
> > --- no domain change nor change of coordinate system, just independent
> > solutons to the field equations.
>
> Your "independent" solution is obtained from Schwarzschild by plugging
> in r'=r-K - hence it's the same solution.
Would it make you feel better if my independent solution is truly
independent without plugging (r' = r - K)? Claiming so, my point of
view is still very much valid.
Don Stockbauer
Eric Gisse
> On Apr 23, 3:41 pm, JanPB <film...@gmail.com> wrote:
>
> > On Apr 18, 12:04 am, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > > What happened to the cross terms (dO dt)?
>
> > Can be removed by spherical symmetry (I am assuming you mean "dtheta
> > dt" and "dphi dt" when you write "dO dt"?)
>
> No, you have made this assumption. Spherically symmetry means all the
> elements of the metric are independent of O. (a) <shrug>
Combined with the assumption of "static", there are no dt crossterms.
[...]
> If you don't agree with Widinpedia, that is your problem only.
Try a source that isn't Wikipedia.
>
> > > ** If spherically symmetric, independent of O
>
> > And this is just a general fact about the form of the metric in a
> > spherically symmetric spacetime.
>
> So, why is (dO dt = zero)?
Static metric + spherically symmetry.
Spherically symmetric means I can move all around surfaces of constant
r without having anything change. If there are dt crossterms, that
means slices of constant r,t _will_ be different.
>
> > > So, if you use some other connection coefficients that are so abstract
> > > in nature compared with Christoffel's which are more tangible you end
> > > up with something exponentially simpler. Yeah, right.
>
> > Yes, of course - both Christoffel and Cartan are the same calculation,
> > only compartmentalised differently (more efficiently for Cartan).
>
> In a set of real geodesic equations, you have to use the Christoffel
> symbols of the 2nd kind. <shrug>
Or you can use the calculus of variations method. They are _all_
equivalent.
>
> > > Give me a
> > > break. There is nothing as realistic as the Christoffel symbols.
> > > <shrug>
>
> > Both are very realistic - in fact they are in a sense almost
> > identical: both Christoffel coefficients and Cartan connection forms
> > are components of the rate of change of certain basis vectors wrt to
> > other basis vectors. The difference is that in Christoffel's setup you
> > take _coordinate vectors_ as your basis while in Cartan's approach you
> > take _orthonormal vectors_ as the basis (i.e., a non-holonomic frame).
> > So both approaches are tangible.
>
> You are very incorrect about the Christoffel symbols. <shrug>
Naturally.
Perhaps you would be kind enough to explain what the Christoffel
symbols represent then?
[...]
Jan, why did you have to tell him about Cartan forms? Christ.
>
> > ...is a coordinate change that removes C(r,t) which is therefore
> > redundant. C(r,t) is simply man-made labelling of the spheres of
> > symmetry. No physics is changed by my relabelling the spheres.
>
> Oh, no. C can take many forms. All C's are independent of each
> other. There is no coordinate transformation. The choice of
> coordinate WAS ALREADY decided when the field equations WERE
> ESTABLISHED. You need to stick with that coordinate system. <shrug>
No, he doesn't. You are falling into the trap of "the label determines
the physics".
>
> > > No. That is the problem here. There is no coordinate change. We are
> > > still talking about using the same spherically symmetric coordinate
> > > system.
>
> > OK, so if it's not the unit problem then why do you claim that metric
> > changes when coordinates change?
>
> Geometry = Metric * Coordinate
That isn't true in any sense. The metric _IS_ the geometry.
>
> If Geometry is invariant, Metric must vary with the choice of
> Coordinate. <shrug>
What's the difference between the geometries represented by the given
line elements?
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
>
> > For example, take a vector. Its
> > length is, say, 17 units. I change the coordinates. The same vectors
> > must still have the same length, 17 units (physics cannot depend on
> > man-made labels attached to objects).
>
> You are saying Geometry = Geometry. Of course, it is. <shrug>
>
> > Your matrix of g_ij does change because your coordinate basis vectors
> > change with coordinate change (g_ij represent the _changed_
> > coefficients necessary to reconstruct the _unchanged_ number 17 from
> > the _changed_ coordinate vectors).
>
> What are you talking about?
Coordinate transformations change the representation of your basis
vectors.
Why are you being confused by this?
>
> > When people say "metric", they mean by this the thing that assigns
> > actual lengths and angles to vectors and vector pairs. And lengths and
> > angles are properties of vectors and vector pairs _themselves_, not of
> > their coordinate labels. When I take a stick in my hand, it has a
> > definite numerically quantified length and the metric is the abstract
> > model that defines this assignment - coordinates do not enter into it
> > as the stick clearly has the length 17 regardless of any coordinates.
>
> In the past 100 years, when people talk about 'metric', they have been
> referring to A, B, and C of the following equation.
>
> ds^2 = A dt^2 - B dr^2 - C dO^2
Um, no. Not even close.
I can crack open any of my relativity texts, or my differential
geometry text. None of them will say this.
What set of knowledge are _you_ working from?
>
> > Coordinates are only a computational tool, just like analytic geometry
> > is a better computational tool than the old Greek geometry that used
> > just geometric constructions.
>
> > Your coordinate-dependent "metric" is not a model for physical lengths
> > and angles. As to what model it is OF is anyone's guess.
>
> Once again, Geometry = Metric * Coordinate
Oh, you really do think it works like that?
Wow.
>
> Where
>
> ** Geometry = invariant
>
> What you have been talking about the metric is Geometry. However, it
> is invalid when applied to the field equations, Riemann curvature
> tensors, etc. You are very confused. <shrug>
The metric _IS_ the geometry, you confused little man.
>
> > > After making these assumptions, the theorem appears to be
> > > correct with the Schwarzschild metric, or Schwarzschild's original
> > > metric, or even the metric Mr. Crothers is championing over the
> > > Schwarzschild metric. However, there are just other metrics that
> > > would invalidate Birkhoff's theorem. I have given some of these
> > > metrics that do so in the past.
>
> > They are all isometric to (a subset of) the standard Schwarzschild
> > solution.
>
> The Birkhoff's theorem has a minor impact on the course of GR,
> anyway. In time, I will prove it wrong through other solutions to the
> field equations where the coordinate of choice is already
> established. Stop conjuring up another coordinate system. Your
> matheMagic trick is exposed.
Minor? Establishing the uniqueness of the Schwarzschild solution is
actually quite important.
You will never "prove" anything when you quite obviously don't know
what you are talking about.
>
> > > I have not alter any domains.
>
> > You've changed the coordinates by r'=r-K. This means the domain r>0 in
> > the new coordinates is r'>-K. It's just like, say, switching from the
> > Cartesians to polars: a unit disc (say) is described in Cartesians by:
>
> How can I? I am bound to the same choice of coordinate system that
> established the field equations. You are very unreasonable here.
No. You are not bound to a particular coordinate system.
>
>
>
> > -1 <= x <= +1
> > -sqrt(1 - x^2) <= y <= +sqrt(1 - x^2)
>
> > Switching to polars: x=r cos(theta), y=r sin(theta) changes the
> > domain:
>
> > 0 < r <= 1
> > 0 <= theta < 2 pi.
>
> > Your "different" solution is exactly like saying that unit disc in
> > polars is described by:
>
> > -1 <= r <= +1
> > -sqrt(1 - r^2) <= theta <= +sqrt(1 - r^2).
>
> > ...i.e., patent nonsense. And your case is even simpler than that
> > because all you change is ONE coordinate (r), not two.
>
> Please stop this nonsense. The choice of coordinate system is already
> established as the commonly utilized spherically symmetric polar
> coordinate system.
Hey, look! Tripping over the label-determines-the-physics trap again!
>
> Face the mathematics. You are looking at an infinite numbers of
> solutions to the field equations. Stop burying your head in the hole
> on the ground.
Only one geometry, but an infinite number of representations. That is
what you have been told countless times already.
[snip remaining]
Koobee Wublee
> On Apr 26, 9:50 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > No, you have made this assumption. Spherically symmetry means all the
> > elements of the metric are independent of O. (a) <shrug>
>
> Combined with the assumption of "static", there are no dt crossterms.
You need to tell Mr. Bielawski that, and you don't know what we are
discussing. <shrug>
> [...]
>
> > If you don't agree with Widinpedia, that is your problem only.
>
> Try a source that isn't Wikipedia.
You first.
> > > > ** If spherically symmetric, independent of O
>
> > > And this is just a general fact about the form of the metric in a
> > > spherically symmetric spacetime.
>
> > So, why is (dO dt = zero)?
>
> Static metric + spherically symmetry.
You need to tell Mr. Bielawski that.
> Spherically symmetric means I can move all around surfaces of constant
> r without having anything change. If there are dt crossterms, that
> means slices of constant r,t _will_ be different.
You have no clues to what we are discussing.
> > In a set of real geodesic equations, you have to use the Christoffel
> > symbols of the 2nd kind. <shrug>
>
> Or you can use the calculus of variations method. They are _all_
> equivalent.
Christoffel symbols are the result of the calculus of variations.
<shrug>
> > You are very incorrect about the Christoffel symbols. <shrug>
>
> Naturally.
>
> Perhaps you would be kind enough to explain what the Christoffel
> symbols represent then?
Read your textbook.
> [...]
>
> Jan, why did you have to tell him about Cartan forms? Christ.
He had to impress someone.
> > Oh, no. C can take many forms. All C's are independent of each
> > other. There is no coordinate transformation. The choice of
> > coordinate WAS ALREADY decided when the field equations WERE
> > ESTABLISHED. You need to stick with that coordinate system. <shrug>
>
> No, he doesn't. You are falling into the trap of "the label determines
> the physics".
No, you have not falling into that trap. You are trapped in a well.
> > Geometry = Metric * Coordinate
>
> That isn't true in any sense. The metric _IS_ the geometry.
Metric is an interpretation of the choice of coordinate that describes
the actual geometry. Metric is not Geometry.
> > If Geometry is invariant, Metric must vary with the choice of
> > Coordinate. <shrug>
>
> What's the difference between the geometries represented by the given
> line elements?
>
> 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
There is only one geometry to the above equations. However, there are
4 different metrics where each is represented by a coordinate system
to describe that one and only one geometry. <shrug>
> Coordinate transformations change the representation of your basis
> vectors.
So?
> Why are you being confused by this?
No.
> > In the past 100 years, when people talk about 'metric', they have been
> > referring to A, B, and C of the following equation.
>
> > ds^2 = A dt^2 - B dr^2 - C dO^2
>
> Um, no. Not even close.
Read you textbook.
> I can crack open any of my relativity texts, or my differential
> geometry text. None of them will say this.
All say that.
> What set of knowledge are _you_ working from?
Your textbooks.
> > Once again, Geometry = Metric * Coordinate
>
> Oh, you really do think it works like that?
Yes.
> Wow.
Wow, how simple can this be.
> > Where
>
> > ** Geometry = invariant
>
> > What you have been talking about the metric is Geometry. However, it
> > is invalid when applied to the field equations, Riemann curvature
> > tensors, etc. You are very confused. <shrug>
>
> The metric _IS_ the geometry, you confused little man.
The metric cannot be the geometry. You are still trapped in well.
<shrug>
> > The Birkhoff's theorem has a minor impact on the course of GR,
> > anyway. In time, I will prove it wrong through other solutions to the
> > field equations where the coordinate of choice is already
> > established. Stop conjuring up another coordinate system. Your
> > matheMagic trick is exposed.
>
> Minor? Establishing the uniqueness of the Schwarzschild solution is
> actually quite important.
Schwarzschild solution is just one of the infinite numbers of them out
there. There are also an infinite numbers of solutions that easily
prove Birkhoff's Theorem wrong. <shrug>
> You will never "prove" anything when you quite obviously don't know
> what you are talking about.
Yes, I do know what I am talking about.
> > How can I? I am bound to the same choice of coordinate system that
> > established the field equations. You are very unreasonable here.
>
> No. You are not bound to a particular coordinate system.
Yes, I am. If not, there requires a new set of field equations.
> > Please stop this nonsense. The choice of coordinate system is already
> > established as the commonly utilized spherically symmetric polar
> > coordinate system.
>
> Hey, look! Tripping over the label-determines-the-physics trap again!
Hey, think. Trapped in a well, you need to come up with ways to get
out of it.
> > Face the mathematics. You are looking at an infinite numbers of
> > solutions to the field equations. Stop burying your head in the hole
> > on the ground.
>
> Only one geometry, but an infinite number of representations.
That is the real world, but not according to the field equations.
These equations yield an infinite numbers of solutions (metrics) using
the same set of coordinate system. According to the field equations,
there can co-exist an infinite numbers of geometry. That is the silly
nature in the field equations and GR in general. <shrug>
> That is
> what you have been told countless times already.
And what has been told to me is still wrong every time.
JanPB
> On Apr 27, 4:35 am, Eric Gisse <jowr...@gmail.com> wrote:
>
> > Jan, why did you have to tell him about Cartan forms? Christ.
>
> He had to impress someone.
What an idea. The reason was that several people last year objected -
correctly - to some of the logical details in the derivation of
Schwarzschild's solution as presented in some textbooks. Based on that
some people reached an incorrect conclusion that the entire derivation
is logically faulty or invalid. So I simply wrote down the derivation
step by step, choosing Cartan's method to compute the Riemann tensor
simply because it involves less computational effort. The 9-page PDF
note I posted would have probably been twice as long had I used
Christoffel-based approach.
But that's not important. It's clear now that we cannot continue until
we resolve the basic problem at the foundations of this business:
> > > Geometry = Metric * Coordinate
>
> > That isn't true in any sense. The metric _IS_ the geometry.
>
> Metric is an interpretation of the choice of coordinate that describes
> the actual geometry. Metric is not Geometry.
You have never explained to us what made you reach this conclusion.
Suppose I write two expressions for the 2D plane:
(1) ds^2 = dx^2 + dy^2 in Cartesian coordinates,
(2) ds^2 = dr^2 + r^2 d theta^2 in polar coordinates.
In your answer to Eric you say that these two would represent the same
geometry but two different metrics - correct?
If yes, then I'd like to see how (1) differs from (2) _physically_.
They both assign the same lengths and the same angles to the same
vectors and vector pairs. Since there is nothing _else_ that
physically determines a geometry, in what physical way is (1)
different than (2)?
--
Jan Bielawski
Koobee Wublee
> On Apr 27, 10:35 am, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > Geometry = Metric * Coordinate
>
> > Metric is an interpretation of the choice of coordinate that describes
> > the actual geometry. Metric is not Geometry.
>
> You have never explained to us what made you reach this conclusion.
> Suppose I write two expressions for the 2D plane:
>
> (1) ds^2 = dx^2 + dy^2 in Cartesian coordinates,
> (2) ds^2 = dr^2 + r^2 d theta^2 in polar coordinates.
>
> In your answer to Eric you say that these two would represent the same
> geometry but two different metrics - correct?
Yes.
> If yes, then I'd like to see how (1) differs from (2) _physically_.
The geometry is the same. The choice of coordinate systems is
different. The metrics are different. This is so simple. Just what
is confusing you?
> They both assign the same lengths and the same angles to the same
> vectors and vector pairs.
Yes.
> Since there is nothing _else_ that
> physically determines a geometry, in what physical way is (1)
> different than (2)?
A better question is to ask what physically DESCRIBES a geometry since
nothing can change or determine the geometry. If you ask that
question, I will answer.
JanPB
> On Apr 27, 2:39 pm, JanPB <film...@gmail.com> wrote:
>
> > On Apr 27, 10:35 am, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > > Geometry = Metric * Coordinate
>
> > > Metric is an interpretation of the choice of coordinate that describes
> > > the actual geometry. Metric is not Geometry.
>
> > You have never explained to us what made you reach this conclusion.
> > Suppose I write two expressions for the 2D plane:
>
> > (1) ds^2 = dx^2 + dy^2 in Cartesian coordinates,
> > (2) ds^2 = dr^2 + r^2 d theta^2 in polar coordinates.
>
> > In your answer to Eric you say that these two would represent the same
> > geometry but two different metrics - correct?
>
> Yes.
>
> > If yes, then I'd like to see how (1) differs from (2) _physically_.
>
> The geometry is the same. The choice of coordinate systems is
> different. The metrics are different. This is so simple. Just what
> is confusing you?
Your terminology is non-standard - when people say "metric" they
normally mean the assignment of lengths and angles to vectors.
> > They both assign the same lengths and the same angles to the same
> > vectors and vector pairs.
>
> Yes.
Good.
> > Since there is nothing _else_ that
> > physically determines a geometry, in what physical way is (1)
> > different than (2)?
>
> A better question is to ask what physically DESCRIBES a geometry since
> nothing can change or determine the geometry. If you ask that
> question, I will answer.
It's not just a DESCRIPTION - if that was the case, nobody would
bother you :-) The problem with your claims is that you insist on an
_actual physical difference_ between certain two expressions for ds^2
differing only by a coordinate change - you said in one such case, for
example, that one of the ds^2 exhibited a black hole while the other
one didn't.
Bottom line:
I'd like to know how two functions assigning the same lengths and
angles to the same vectors can STILL be considered as different
solutions to Einstein's equation. What distinguishes them, physically?
--
Jan Bielawski
Koobee Wublee
> On Apr 27, 4:02 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > A better question is to ask what physically DESCRIBES a geometry since
> > nothing can change or determine the geometry. If you ask that
> > question, I will answer.
>
> It's not just a DESCRIPTION - if that was the case, nobody would
> bother you :-)
When talking about an invariant object or a feature, we can only
describe. We cannot change it.
> The problem with your claims is that you insist on an
> _actual physical difference_ between certain two expressions for ds^2
> differing only by a coordinate change - you said in one such case, for
> example, that one of the ds^2 exhibited a black hole while the other
> one didn't.
No, I did not say that. I said since the field equations are only
valid for a set of coordinate system that we initially have chosen,
any solutions found must be conformed to this coordinate system and
nothing else. If there are two solutions, then each metric must
represent a different geometry since both metrics employ the same
coordinate system. With the field equations able to predict the
existence of so many independent metrics at the same time, the field
equations must be wrong, silly, and absurd.
The analogy is like solving the following quadratic equation.
x^2 - 4 x + 3 = 0
The solutions are
** x = 1
** x = 3
To say (x = 1 = 3) is wrong, silly, and absurd. To insist on (x = 1 =
3) is barbaric, irrespnosible and sinister.
> Bottom line:
> I'd like to know how two functions assigning the same lengths and
> angles to the same vectors can STILL be considered as different
> solutions to Einstein's equation.
They can't, but they do exist. Thus, the field equations are just
wrong, silly, and absurd.
> What distinguishes them, physically?
The metric with identical coordinate system.
Tom Roberts
> On Apr 27, 2:39 pm, JanPB <film...@gmail.com> wrote:
>> (1) ds^2 = dx^2 + dy^2 in Cartesian coordinates,
>> (2) ds^2 = dr^2 + r^2 d theta^2 in polar coordinates.
>> In your answer to Eric you say that these two would represent the same
>> geometry but two different metrics - correct?
> Yes.
> different. The metrics are different.
That is not how these words are used, and is not what they mean.
Those are line elements, not "metrics". From the line element one can
easily read out the COMPONENTS of the metric when the metric is
PROJECTED onto the coordinates used in the line element.
> This is so simple. Just what
> is confusing you?
YOU are the one who is "confused", by your insistence on using
non-standard terminology, and letting your MISUSE of words color your
perception of the underlying mathematics.
for (1) the components of the metric are:
g_xx = 1
g_yy = 1
g_xy = g_yx = 0
for (2) the components of the metric are:
g_rr = 1
g_tt = r^2 (here t stands for theta)
g_rt = g_tr = 0
For (1) the metric is:
g = g_xx dx X dx + g_yy dy X dy
where dx and dy are the corresponding 1-forms, and X stands for the
tensor product.
For (2) the metric is:
g = g_rr dr X dr + g_tt dt X dt
A modest amount of calculus shows that these two metrics are the same.
This is no different from using i and j as unit vectors along the x and
y axes, and then saying that the vector v (with components v_x and v_y) is:
v = v_x i + v_y j
except that g is a rank-2 tensor.
When we say "geometry", we normally mean the metric. That is completely
independent of coordinates, as geometry MUST be.
Get off your high horse and LEARN what differential geometry actually
is, and how its terminology is actually used. Your high horse is down in
a very deep canyon, anyway.
Tom Roberts
Eric Gisse
> On Apr 27, 4:35 am, Eric Gisse <jowr...@gmail.com> wrote:
>
> > On Apr 26, 9:50 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > > No, you have made this assumption. Spherically symmetry means all the
> > > elements of the metric are independent of O. (a) <shrug>
>
> > Combined with the assumption of "static", there are no dt crossterms.
>
> You need to tell Mr. Bielawski that, and you don't know what we are
> discussing. <shrug>
>
> > [...]
>
> > > If you don't agree with Widinpedia, that is your problem only.
>
> > Try a source that isn't Wikipedia.
>
> You first.
"Differential Geometry of Curves and Surfaces" - M. Do Camro
The metric has the name "first fundamental form" here, though.
"Gravitation" - Misner, Thorne, Wheeler
"Spacetime and Geometry" - S. Carroll
[...]
>
> Christoffel symbols are the result of the calculus of variations.
> <shrug>
True but calculating them directly through calculus of variations is
much easier if your metric has certain symmetries.
>
> > > You are very incorrect about the Christoffel symbols. <shrug>
>
> > Naturally.
>
> > Perhaps you would be kind enough to explain what the Christoffel
> > symbols represent then?
>
> Read your textbook.
No, it does not work that way.
If you say I'm wrong, you are supposed to explain _how_ I am wrong
rather than say "read your textbook" which says the exact same thing
anyway.
>
> > [...]
>
> > Jan, why did you have to tell him about Cartan forms? Christ.
>
> He had to impress someone.
>
> > > Oh, no. C can take many forms. All C's are independent of each
> > > other. There is no coordinate transformation. The choice of
> > > coordinate WAS ALREADY decided when the field equations WERE
> > > ESTABLISHED. You need to stick with that coordinate system. <shrug>
>
> > No, he doesn't. You are falling into the trap of "the label determines
> > the physics".
>
> No, you have not falling into that trap. You are trapped in a well.
>
> > > Geometry = Metric * Coordinate
>
> > That isn't true in any sense. The metric _IS_ the geometry.
>
> Metric is an interpretation of the choice of coordinate that describes
> the actual geometry. Metric is not Geometry.
Oh, but it is - if I have the metric I have all the information I need
in order to describe everything that occurs on a specified manifold.
Using only the metric or quantities derived from it, I can obtain
lengths and angles [this is actually how the metric is defined],
curvature, equations of motion that describe the shortest path...etc
I need no additional information. The metric is the geometry.
>
> > > If Geometry is invariant, Metric must vary with the choice of
> > > Coordinate. <shrug>
>
> > What's the difference between the geometries represented by the given
> > line elements?
>
> > 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
>
> There is only one geometry to the above equations. However, there are
> 4 different metrics where each is represented by a coordinate system
> to describe that one and only one geometry. <shrug>
If they all represent the same thing, they aren't all that different
now are they?
That is the point folks have been trying to hammer home for god knows
how long now.
>
> > Coordinate transformations change the representation of your basis
> > vectors.
>
> So?
>
> > Why are you being confused by this?
>
> No.
>
> > > In the past 100 years, when people talk about 'metric', they have been
> > > referring to A, B, and C of the following equation.
>
> > > ds^2 = A dt^2 - B dr^2 - C dO^2
>
> > Um, no. Not even close.
>
> Read you textbook.
Which one?
MTW, which disagrees with you?
Carroll, which disagrees with you?
Do Camro, which disagrees with you?
>
> > I can crack open any of my relativity texts, or my differential
> > geometry text. None of them will say this.
>
> All say that.
No. They don't.
Not only do you attempt to do what I was saying you were doing, namely
defining the physics through coordinate labels, but you ignore
dx^idx^j crossterms that can't always be shoved away.
>
> > What set of knowledge are _you_ working from?
>
> Your textbooks.
Oh, really?
What text are you referencing?
[...]
>
> Schwarzschild solution is just one of the infinite numbers of them out
> there. There are also an infinite numbers of solutions that easily
> prove Birkhoff's Theorem wrong. <shrug>
Then write down two solutions that aren't isomorphic to Schwarzschild.
Since you are so sure there are an infinite number of different
solutions, asking for two of them shouldn't be an insurmountable task
for you.
[....]
Eric Gisse
[...]
>
> Get off your high horse and LEARN what differential geometry actually
> is, and how its terminology is actually used. Your high horse is down in
> a very deep canyon, anyway.
Have you ever seen him explain where he learned this stuff?
>
> Tom Roberts
Koobee Wublee
> Koobee Wublee wrote:
> > The geometry is the same. The choice of coordinate systems is
> > different. The metrics are different.
>
> That is not how these words are used, and is not what they mean.
>
> Those are line elements, not "metrics". From the line element one can
> easily read out the COMPONENTS of the metric when the metric is
> PROJECTED onto the coordinates used in the line element.
Given the line element (squared) below,
ds^2 = A dt^2 - B dr^2 - C dO^2
(A, B, C) are elements to the metric, and (dt, dr, dO) are the
coordinate system.
Do you disagree that (A, B, C) are elements to the metric?
> > This is so simple. Just what
> > is confusing you?
>
> YOU are the one who is "confused", by your insistence on using
> non-standard terminology, and letting your MISUSE of words color your
> perception of the underlying mathematics.
No, I am not confused. I laid out exactly what I meant through simple
mathematics that anyone with any education can understand. <shrug>
> for (1) the components of the metric are:
> g_xx = 1
> g_yy = 1
> g_xy = g_yx = 0
>
> for (2) the components of the metric are:
> g_rr = 1
> g_tt = r^2 (here t stands for theta)
> g_rt = g_tr = 0
This is exactly what I have been trying to tell you. What is that you
are confused about?
Notice that these two metrics are very different because the choice of
coordinate systems is very different as well. However, the geometry
or the line element as you have bullied it to be remains the same as
interpreted by each of these two metrics using each of the choice of
coordinate systems.
> For (1) the metric is:
> g = g_xx dx X dx + g_yy dy X dy
> where dx and dy are the corresponding 1-forms, and X stands for the
> tensor product.
No, you need to review your metrices.
g1 = [g_xx, g_xy] = [1, 0]
[g_yx, g_yy] [0, 1]
> For (2) the metric is:
> g = g_rr dr X dr + g_tt dt X dt
No.
g2 = [g_rr, g_rt] = [1, 0 ]
[g_tr, g_tt] [0, r^2]
> A modest amount of calculus shows that these two metrics are the same.
No, your matheMagic trick is caught. g1 is not the same as g2.
You have showed geometry = geometry or line element = line element
like a magician. We are talking about the metrics here. Stick to the
metric.
> When we say "geometry", we normally mean the metric.
However, mathematically when you talk about the metric you don't mean
"geometry". This is how a very absurd hypothesis like GR can be
widely accepted. <shrug>
> That is completely
> independent of coordinates, as geometry MUST be.
Again, mathematically used in the past 100 years, the metric is
coordinate dependent.
> Get off your high horse and LEARN what differential geometry actually
> is, and how its terminology is actually used. Your high horse is down in
> a very deep canyon, anyway.
I don't ride horses. If you do, that explains your childish behavior
here. <shrug>
Eric Gisse
[...]
So where exactly did you learn this amusing interpretation?
JanPB
> On Apr 27, 4:37 pm, JanPB <film...@gmail.com> wrote:
>
> > On Apr 27, 4:02 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > > A better question is to ask what physically DESCRIBES a geometry since
> > > nothing can change or determine the geometry. If you ask that
> > > question, I will answer.
>
> > It's not just a DESCRIPTION - if that was the case, nobody would
> > bother you :-)
>
> When talking about an invariant object or a feature, we can only
> describe. We cannot change it.
OK. But you didn't answer the question that's really basic here:
how two functions assigning the same lengths and the same angles to
the same vectors can STILL be considered physically different. What
distinguishes them, physically?
[I am skipping the rest for one moment]
--
Jan Bielawski
Koobee Wublee
> On Apr 27, 9:35 am, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > You first.
>
> "Differential Geometry of Curves and Surfaces" - M. Do Camro
>
> The metric has the name "first fundamental form" here, though.
>
> "Gravitation" - Misner, Thorne, Wheeler
> "Spacetime and Geometry" - S. Carroll
Keep trying.
> > Christoffel symbols are the result of the calculus of variations.
> > <shrug>
>
> True but calculating them directly through calculus of variations is
> much easier if your metric has certain symmetries.
The metric does not have to be symmetric. That was what Christoffel's
position when he derived them. Later on, we found out the metric must
be symmetric. That means there are more than one way to represent the
geodesic equations. In doing so, there are more than one way to write
down the covariant derivative. Settling on just one and call it the
one only is just silly and absurd.
> No, it does not work that way.
>
> If you say I'm wrong, you are supposed to explain _how_ I am wrong
> rather than say "read your textbook" which says the exact same thing
> anyway.
You are incomprehensible.
> > Metric is an interpretation of the choice of coordinate that describes
> > the actual geometry. Metric is not Geometry.
>
> Oh, but it is - if I have the metric I have all the information I need
> in order to describe everything that occurs on a specified manifold.
You still have to know what coordinate system your metric is abided
to. <shrug>
> Using only the metric or quantities derived from it, I can obtain
> lengths and angles [this is actually how the metric is defined],
> curvature, equations of motion that describe the shortest path...etc
But your lengths and angle are exactly what the initial choice of
coordinate system. However, you can now transform your coordinate
system to another. <shrug>
> I need no additional information. The metric is the geometry.
No, the metric is only an interpretation to the geometry via the
choice of coordinate system. You cannot argue against the very simple
mathematics. <shrug>
> > There is only one geometry to the above equations. However, there are
> > 4 different metrics where each is represented by a coordinate system
> > to describe that one and only one geometry. <shrug>
>
> If they all represent the same thing, they aren't all that different
> now are they?
Yes.
> That is the point folks have been trying to hammer home for god knows
> how long now.
No, the silly folks are saying if the geometry is the same, then the
metrics must be the same regardless the coordinate system. They are
so wrong and absurd.
> > Read you textbook.
>
> Which one?
Which one do you have?
> MTW, which disagrees with you?
> Carroll, which disagrees with you?
> Do Camro, which disagrees with you?
As I said, it is lonely to be the only one who understands Riemann's
original concept of curvature in space. <shrug> It is ridiculous
simple, and yet it is made so complicated and wrong by the likes of
Ricci, Levi-Civita, etc.
> Not only do you attempt to do what I was saying you were doing, namely
> defining the physics through coordinate labels, but you ignore
> dx^idx^j crossterms that can't always be shoved away.
It is getting late. I don't understand what you are saying, and I
don't have the energy to interpret what you are saying. So, talk to
yourself more.
> > Your textbooks.
>
> Oh, really?
>
> What text are you referencing?
Which text do you have?
> > Schwarzschild solution is just one of the infinite numbers of them out
> > there. There are also an infinite numbers of solutions that easily
> > prove Birkhoff's Theorem wrong. <shrug>
>
> Then write down two solutions that aren't isomorphic to Schwarzschild.
Just look through my previous posts.
> Since you are so sure there are an infinite number of different
> solutions, asking for two of them shouldn't be an insurmountable task
> for you.
Oh, I am very, very sure that there are an infinite number of
solutions to the field equations that all reference to the same set of
coordinate system. The field equations thus represent an infinite
number of independent geometry or universe. The field equations are
just BS to the max. The fat castle in air known as GR is crumbling.
Koobee Wublee
This is a stupid question. Before I answer this stupid question, you
have to answer the following question first.
Given a quadratic equation of the following,
x^2 - 3 x + 2 = 0
The solutions are (x = 2) or (x = 1).
What make all solutions equivalent? What makes (x = 1 = 2)?
JanPB
> On Apr 27, 11:11 pm, JanPB <film...@gmail.com> wrote:
>
> > On Apr 27, 5:52 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > OK. But you didn't answer the question that's really basic here:
> > how two functions assigning the same lengths and the same angles to
> > the same vectors can STILL be considered physically different. What
> > distinguishes them, physically?
>
> This is a stupid question.
No, this is the root of your confusion. You claim that two different
expressions of the same geometry (what you call "two different
metrics") are to be treated as physically different.
So I'm asking a simple question: HOW can a person - some observer in
space or whatever - tell the difference?
I'm not at all surprised you refuse to answer it.
> Before I answer this stupid question, you
> have to answer the following question first.
>
> Given a quadratic equation of the following,
>
> x^2 - 3 x + 2 = 0
>
> The solutions are (x = 2) or (x = 1).
>
> What make all solutions equivalent? What makes (x = 1 = 2)?
This is of course irrelevant. Here you have an equation for points in
the parameter space R^1 - not a manifold. Einstein's equation OTOH is
an equation for an unknown function between two manifolds.
--
Jan Bielawski
Eric Gisse
> On Apr 27, 7:21 pm, Eric Gisse <jowr...@gmail.com> wrote:
>
> > On Apr 27, 9:35 am, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > > You first.
>
> > "Differential Geometry of Curves and Surfaces" - M. Do Camro
>
> > The metric has the name "first fundamental form" here, though.
>
> > "Gravitation" - Misner, Thorne, Wheeler
> > "Spacetime and Geometry" - S. Carroll
>
> Keep trying.
At least I have actual references.
>
> > > Christoffel symbols are the result of the calculus of variations.
> > > <shrug>
>
> > True but calculating them directly through calculus of variations is
> > much easier if your metric has certain symmetries.
>
> The metric does not have to be symmetric. That was what Christoffel's
> position when he derived them. Later on, we found out the metric must
> be symmetric. That means there are more than one way to represent the
> geodesic equations. In doing so, there are more than one way to write
> down the covariant derivative. Settling on just one and call it the
> one only is just silly and absurd.
Notice what I said - going through calculus of variations is an
_easier_ way if your metric has certain symmetries.
You don't have to take umbrage with everything I say.
>
> > No, it does not work that way.
>
> > If you say I'm wrong, you are supposed to explain _how_ I am wrong
> > rather than say "read your textbook" which says the exact same thing
> > anyway.
>
> You are incomprehensible.
Here, I'll write it in a way that even babelfish can understand:
Show me why I am wrong instead of saying I am wrong.
>
> > > Metric is an interpretation of the choice of coordinate that describes
> > > the actual geometry. Metric is not Geometry.
>
> > Oh, but it is - if I have the metric I have all the information I need
> > in order to describe everything that occurs on a specified manifold.
>
> You still have to know what coordinate system your metric is abided
> to. <shrug>
True but irrelevant - the result is simply the metric projected on the
relevant basis vectors.
>
> > Using only the metric or quantities derived from it, I can obtain
> > lengths and angles [this is actually how the metric is defined],
> > curvature, equations of motion that describe the shortest path...etc
>
> But your lengths and angle are exactly what the initial choice of
> coordinate system. However, you can now transform your coordinate
> system to another. <shrug>
>
> > I need no additional information. The metric is the geometry.
>
> No, the metric is only an interpretation to the geometry via the
> choice of coordinate system. You cannot argue against the very simple
> mathematics. <shrug>
You need a representation in order to perform computations. The point
is the representation of the geometry _does not matter_.
>
> > > There is only one geometry to the above equations. However, there are
> > > 4 different metrics where each is represented by a coordinate system
> > > to describe that one and only one geometry. <shrug>
>
> > If they all represent the same thing, they aren't all that different
> > now are they?
>
> Yes.
>
> > That is the point folks have been trying to hammer home for god knows
> > how long now.
>
> No, the silly folks are saying if the geometry is the same, then the
> metrics must be the same regardless the coordinate system. They are
> so wrong and absurd.
Of course. But that isn't what they are saying.
Folks are saying if the geometry is the same, then all the metrics
which are related by an invertable coordinate transformation are
representing the _same geometry_.
>
> > > Read you textbook.
>
> > Which one?
>
> Which one do you have?
>
> > MTW, which disagrees with you?
> > Carroll, which disagrees with you?
> > Do Camro, which disagrees with you?
>
> As I said, it is lonely to be the only one who understands Riemann's
> original concept of curvature in space. <shrug> It is ridiculous
> simple, and yet it is made so complicated and wrong by the likes of
> Ricci, Levi-Civita, etc.
So where did you learn this stuff?
>
> > Not only do you attempt to do what I was saying you were doing, namely
> > defining the physics through coordinate labels, but you ignore
> > dx^idx^j crossterms that can't always be shoved away.
>
> It is getting late. I don't understand what you are saying, and I
> don't have the energy to interpret what you are saying. So, talk to
> yourself more.
That's what it comes down to - you don't understand what is being
explained to you, so you interpret that to mean you are right and
everyone else on the face of the earth is wrong. That is a frequent
occurance on this newsgroup.
>
> > > Your textbooks.
>
> > Oh, really?
>
> > What text are you referencing?
>
> Which text do you have?
>
> > > Schwarzschild solution is just one of the infinite numbers of them out
> > > there. There are also an infinite numbers of solutions that easily
> > > prove Birkhoff's Theorem wrong. <shrug>
>
> > Then write down two solutions that aren't isomorphic to Schwarzschild.
>
> Just look through my previous posts.
Hah. This old standby. "Do my research for me".
Either you know them or you don't.
>
> > Since you are so sure there are an infinite number of different
> > solutions, asking for two of them shouldn't be an insurmountable task
> > for you.
>
> Oh, I am very, very sure that there are an infinite number of
> solutions to the field equations that all reference to the same set of
> coordinate system. The field equations thus represent an infinite
> number of independent geometry or universe. The field equations are
> just BS to the max. The fat castle in air known as GR is crumbling.
Yes, your postings on USENET sure are taking the world by storm.
All it takes to prove you right is the derivation of two non-
isomorphic solutions. That's all. Are you going to expend the tiny
amount of effort required to make your claims a bit less insane, or
are you going to continue posting to USENET about it?
bz
news:1177741152....@o5g2000hsb.googlegroups.com:
Your so called solutions actually represent the intersection of the curve
x^2-3x+2=0
with the locus of points (the line)
y=0.
There is only ONE unique solution consisting of the pair of points (2,0)
and (1,0).
> What make all solutions equivalent? What makes (x = 1 = 2)?
Meaningless question.
The 'solution' consists of two points (2,0) and (1,0).
--
bz
please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.
bz+...@ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap
Koobee Wublee
> On Apr 27, 11:19 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> OK. But you didn't answer the question that's really basic here:
> how two functions assigning the same lengths and the same angles to
> the same vectors can STILL be considered physically different. What
> distinguishes them, physically?
>
> > This is a stupid question.
>
> No, this is the root of your confusion. You claim that two different
> expressions of the same geometry (what you call "two different
> metrics") are to be treated as physically different.
This question remains the most stupid as I have seriously
encountered. It is not that I refuse to answer that question. I am
still appalled by your lack of understanding in the very fundamental
algebra. Sometimes, I am even debating with myself if I should
continue with this pointless discussion with you on the most basic of
the mathematics involved.
> So I'm asking a simple question: HOW can a person - some observer in
> space or whatever - tell the difference?
Your question has no relevance to our discussion. I am not claiming
two different expressions for the same geometry. I am claiming two
different metrics using the same coordinate system must each
represents a different geometry. <shrug>
> I'm not at all surprised you refuse to answer it.
>
> > Before I answer this stupid question, you
> > have to answer the following question first.
>
> > Given a quadratic equation of the following,
>
> > x^2 - 3 x + 2 = 0
>
> > The solutions are (x = 2) or (x = 1).
>
> > What make all solutions equivalent? What makes (x = 1 = 2)?
>
> This is of course irrelevant. Here you have an equation for points in
> the parameter space R^1 - not a manifold. Einstein's equation OTOH is
> an equation for an unknown function between two manifolds.
This is very relevant. Your question is like asking the above. Or
better yet let's throw in the units. Given the following,
x^2 - 3 dollars * x + 2 dollars^2 = 0
What are the solutions?
Your stupid question is like asking how (x = 1 dollar) and (x = 2
dollars) can be in co-existence at the same time. In your twisted
logic, you somehow conjured up another unit say 'gollar' where (1
gollar = 0.50 dollar). Therefore, continuing with your fouled
mathematics, you are claiming (x = 1 dollar) and (x = 2 gollars).
Since (x = 1 dollar = 2 gollars), therefore the above quadratic
equation really has only one solution. This is called matheMagic.
You really need to go back to the basics of algebra. I cannot help
you on this one.
JanPB
> On Apr 28, 12:14 am, JanPB <film...@gmail.com> wrote:
>
> > On Apr 27, 11:19 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > OK. But you didn't answer the question that's really basic here:
> > how two functions assigning the same lengths and the same angles to
> > the same vectors can STILL be considered physically different. What
> > distinguishes them, physically?
>
> > > This is a stupid question.
>
> > No, this is the root of your confusion. You claim that two different
> > expressions of the same geometry (what you call "two different
> > metrics") are to be treated as physically different.
>
> This question remains the most stupid as I have seriously
> encountered. It is not that I refuse to answer that question. I am
> still appalled by your lack of understanding in the very fundamental
> algebra. Sometimes, I am even debating with myself if I should
> continue with this pointless discussion with you on the most basic of
> the mathematics involved.
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.
Here is the problem. Since you keep insisting on existence of
different solutions besides Schwarzschild's I'm simply asking (for the
third time) HOW are they different? IOW, what can an observer
physically do to tell these "infinitely many" solutions of yours
apart?
You've agreed that geometry is determined by an assignment of lengths
and angles to tangent vectors. So obviously there must be something
ELSE in your mind besides lengths and angles that is physically
detectable and is implied by the EFE. WHAT IS IT? Is this some kind of
a secret?
> > So I'm asking a simple question: HOW can a person - some observer in
> > space or whatever - tell the difference?
>
> Your question has no relevance to our discussion.
Just answer it, please. (Of course I think it's not only relevant,
it's the crux of the matter, that's why I'm asking it.)
> I am not claiming
> two different expressions for the same geometry. I am claiming two
> different metrics using the same coordinate system must each
> represents a different geometry. <shrug>
Of course. But that was never under the discussion. That is not what
Crothers claims. What you and Crothers have been saying was that there
was more than one solution to the EFE in the spherically symmetric
vacuum case.
Each such solution is a function assigning lengths and angles to
vectors. So you and Crothers claim that there is more than one such
function satisfying EFE. And yet ALL examples he and you produced so
far were of the following form: ONE AND THE SAME FUNCTION assigning
lengths and angles to vectors, ONLY WRITTEN IN DIFFERENT COORDINATES.
Same function assigning lengths and angles to vectors. Same solution.
> > I'm not at all surprised you refuse to answer it.
>
> > > Before I answer this stupid question, you
> > > have to answer the following question first.
>
> > > Given a quadratic equation of the following,
>
> > > x^2 - 3 x + 2 = 0
>
> > > The solutions are (x = 2) or (x = 1).
>
> > > What make all solutions equivalent? What makes (x = 1 = 2)?
>
> > This is of course irrelevant. Here you have an equation for points in
> > the parameter space R^1 - not a manifold. Einstein's equation OTOH is
> > an equation for an unknown function between two manifolds.
>
> This is very relevant. Your question is like asking the above.
It's not. I've explained why.
> Or
> better yet let's throw in the units. Given the following,
>
> x^2 - 3 dollars * x + 2 dollars^2 = 0
>
> What are the solutions?
>
> Your stupid question is like asking how (x = 1 dollar) and (x = 2
> dollars) can be in co-existence at the same time. In your twisted
> logic, you somehow conjured up another unit say 'gollar' where (1
> gollar = 0.50 dollar). Therefore, continuing with your fouled
> mathematics, you are claiming (x = 1 dollar) and (x = 2 gollars).
> Since (x = 1 dollar = 2 gollars), therefore the above quadratic
> equation really has only one solution. This is called matheMagic.
>
> You really need to go back to the basics of algebra. I cannot help
> you on this one.
Nonsense.
--
Jan Bielawski
Eric Gisse
> On Apr 28, 12:14 am, JanPB <film...@gmail.com> wrote:
>
> > On Apr 27, 11:19 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > OK. But you didn't answer the question that's really basic here:
> > how two functions assigning the same lengths and the same angles to
> > the same vectors can STILL be considered physically different. What
> > distinguishes them, physically?
>
> > > This is a stupid question.
>
> > No, this is the root of your confusion. You claim that two different
> > expressions of the same geometry (what you call "two different
> > metrics") are to be treated as physically different.
>
> This question remains the most stupid as I have seriously
> encountered. It is not that I refuse to answer that question. I am
> still appalled by your lack of understanding in the very fundamental
> algebra. Sometimes, I am even debating with myself if I should
> continue with this pointless discussion with you on the most basic of
> the mathematics involved.
>
> > So I'm asking a simple question: HOW can a person - some observer in
> > space or whatever - tell the difference?
>
> Your question has no relevance to our discussion. I am not claiming
> two different expressions for the same geometry. I am claiming two
> different metrics using the same coordinate system must each
> represents a different geometry. <shrug>
Then show us two different metrics using the same coordinate system
that represent different geometries while being solutions to the
Einstein field equations in vacuum with spherical symmetry.
[...]
Koobee Wublee
> On Apr 28, 8:51 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > This question remains the most stupid as I have seriously
> > encountered. It is not that I refuse to answer that question. I am
> > still appalled by your lack of understanding in the very fundamental
> > algebra. Sometimes, I am even debating with myself if I should
> > continue with this pointless discussion with you on the most basic of
> > the mathematics involved.
>
> 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.
Yes, 2 different metrics using the same coordinate system represent 2
different geometries. <shrug>
> This was never under the discussion.
Bullsh*t! This is exactly what I have been talking about. This is so
obvious, and I am so surprise that you have so much trouble to
understand this.
> 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.
You have been speaking with a forked tongue here. These very
different and thus independent line elements share the same coordinate
system. They are not coordinate transform of the other. What you
call coordinate transformation is merely a mathematical tool and trick
but certainly no matheMagic that allows one to find other solutions if
one is found.
> Here is the problem. Since you keep insisting on existence of
> different solutions besides Schwarzschild's I'm simply asking (for the
> third time) HOW are they different? IOW, what can an observer
> physically do to tell these "infinitely many" solutions of yours
> apart?
For the third time, I am answering that they are very different just
as the mathematics that represent them is very different.
> You've agreed that geometry is determined by an assignment of lengths
> and angles to tangent vectors. So obviously there must be something
> ELSE in your mind besides lengths and angles that is physically
> detectable and is implied by the EFE. WHAT IS IT? Is this some kind of
> a secret?
The geometry cannot be determined, shaped, modified, or annihilated.
It can only be observed and interpreted. So, I have no clues as to
what you are claiming of my agreement with whatever.
> So I'm asking a simple question: HOW can a person - some observer in
> space or whatever - tell the difference?
>
> > Your question has no relevance to our discussion.
>
> Just answer it, please. (Of course I think it's not only relevant,
> it's the crux of the matter, that's why I'm asking it.)
Through the metric.
> > I am not claiming
> > two different expressions for the same geometry. I am claiming two
> > different metrics using the same coordinate system must each
> > represents a different geometry. <shrug>
>
> Of course. But that was never under the discussion.
Again, this is exactly what my point is. It is pitiful that you just
start to understand what I have been saying.
> That is not what Crothers claims.
I have never said that was what Mr. Crothers was claiming. He had his
own agenda. He wanted to promote his metric out of the infinite
numbers of them. My point is that no one can say the metric he
discovers after solving the field equations represents the actual
geometry of our universe. The hypothesis known as GR is totally BS.
> What you and Crothers have been saying was that there
> was more than one solution to the EFE in the spherically symmetric
> vacuum case.
No, I have been saying that there are more than one independent
solution to the field equations regardless of it being spherically
symmetric or not.
> Each such solution is a function assigning lengths and angles to
> vectors. So you and Crothers claim that there is more than one such
> function satisfying EFE.
Yes.
> And yet ALL examples he and you produced so
> far were of the following form: ONE AND THE SAME FUNCTION assigning
> lengths and angles to vectors, ONLY WRITTEN IN DIFFERENT COORDINATES.
Back to the solutions to that quadratic equation, (x = 2) is merely a
coordinate transformation from (x = 1). That does not make (x = 1)
and (x = 2) identical.
> Same function assigning lengths and angles to vectors. Same solution.
Your logic is very faulty. You don't understand the basics of
algebra. <shrug>
> > Given a quadratic equation of the following,
>
> > x^2 - 3 x + 2 = 0
>
> > The solutions are (x = 2) or (x = 1).
>
> > What make all solutions equivalent? What makes (x = 1 = 2)?
>
> This is of course irrelevant. Here you have an equation for points in
> the parameter space R^1 - not a manifold. Einstein's equation OTOH is
> an equation for an unknown function between two manifolds.
>
> > This is very relevant. Your question is like asking the above.
>
> It's not. I've explained why.
You are in denial. <shrug>
> > Or
> > better yet let's throw in the units. Given the following,
>
> > x^2 - 3 dollars * x + 2 dollars^2 = 0
>
> > What are the solutions?
>
> > Your stupid question is like asking how (x = 1 dollar) and (x = 2
> > dollars) can be in co-existence at the same time. In your twisted
> > logic, you somehow conjured up another unit say 'gollar' where (1
> > gollar = 0.50 dollar). Therefore, continuing with your fouled
> > mathematics, you are claiming (x = 1 dollar) and (x = 2 gollars).
> > Since (x = 1 dollar = 2 gollars), therefore the above quadratic
> > equation really has only one solution. This is called matheMagic.
>
> > You really need to go back to the basics of algebra. I cannot help
> > you on this one.
>
> Nonsense.
It is up to you to go back to understand the basics of algebra. I
cannot make you. You need to make that initiative yourself. Good
night, Mr. Bielawski!
JanPB
That's a good question too. Please do that, Koobee:
(i) a fixed coordinate system,
(ii) two different sets of g_ij's written with respect to that fixed
coordinate system, both sets spherically symmetric and satisfying the
vaccum EFE.
--
Jan Bielawski
Eric Gisse
[...]
While your ignorance makes discussions with you incredibly difficult,
it is truly your alternative vocabulary that makes things the most
difficult.
Koobee Wublee
> On Apr 28, 9:19 pm, Eric Gisse <jowr...@gmail.com> wrote:
> > Then show us two different metrics using the same coordinate system
> > that represent different geometries while being solutions to the
> > Einstein field equations in vacuum with spherical symmetry.
http://groups.google.com/group/sci.physics.relativity/msg/a510f9508677596d?hl=en&
> That's a good question too. Please do that, Koobee:
>
> (i) a fixed coordinate system,
What do you mean by a fixed coordinate system? Aren't all coordinate
system by definition fixed?
> (ii) two different sets of g_ij's written with respect to that fixed
> coordinate system, both sets spherically symmetric and satisfying the
> vaccum EFE.
Is the following what you are looking for? These two g_ij are
different and spherically symmetric. Both are solutions to the vacuum
EFE where you have no justification to prefer one over the other.
Now, there are an infinite numbers more of them.
http://groups.google.com/group/sci.physics.relativity/msg/a510f9508677596d?hl=en&
Eric Gisse
> On Apr 29, 9:09 am, JanPB <film...@gmail.com> wrote:
>
> > On Apr 28, 9:19 pm, Eric Gisse <jowr...@gmail.com> wrote:
> > > Then show us two different metrics using the same coordinate system
> > > that represent different geometries while being solutions to the
> > > Einstein field equations in vacuum with spherical symmetry.
>
Does not qualify. There is an explicit coordinate change that
transforms the Schwarzschild metric to that metric. Therefore they
represent the same geometry.
Notice how I said _different_ geometries?
>
> > That's a good question too. Please do that, Koobee:
>
> > (i) a fixed coordinate system,
>
> What do you mean by a fixed coordinate system? Aren't all coordinate
> system by definition fixed?
hahahahahhaahahahahah
Have you ever had a classical mechanics course worth the name?
Rotating coordinate systems are a standard fixture.
>
> > (ii) two different sets of g_ij's written with respect to that fixed
> > coordinate system, both sets spherically symmetric and satisfying the
> > vaccum EFE.
>
> Is the following what you are looking for? These two g_ij are
> different and spherically symmetric. Both are solutions to the vacuum
> EFE where you have no justification to prefer one over the other.
> Now, there are an infinite numbers more of them.
>
> http://groups.google.com/group/sci.physics.relativity/msg/a510f950867...
*sigh*
You are representing the same geometry in two different coordinate
systems. That was exactly _not_ what was being requested of you. Since
you are once again having reading difficulties I will be more explicit
in the request.
Write down two line elements corresponding to two metrics that are
constructed from the _same_ coordinate system with the _same_
coordinate ranges that satisfy spherical symmetry, possess a timelike
killing vector at positive spatial infinity [An essential component of
Birkhoff's theorem], and are solutions to the Einstein field equations
in vacuum.