Spacetime Questions for Mr. Bielawski, Mr. McCullough, Dr. Roberts, and Others Interested
Koobee Wublee
ds^2 = c^2 dt^2 - dq1^2 - dq2^2 - dq3^2
The metric is Minkowski. It is diagonal.
{ c^2, -1, -1, -1 }
Does this mean the described spacetime is always flat?
Now, we have another spaceimte as described below.
ds^2 = c^2 (1 - 2 U) dt^2 - dq1^2 / (1 - 2 U) - q1^2 cos^2(q3)
dq2^2 - q1^2 dq3^2
Where
** U = G M / c^2 / q1
The metric is Schwarzschild. It is diagonal as well.
{ c^2 (1 - 2 U), - 1 / (1 - 2 U), q1^2 cos^2(q3), q1^2 }
Does this mean the described spacetime is always curved if (M > 0)?
Dirk Van de moortel
"Koobee Wublee" <koobee...@gmail.com> wrote in message news:1162232475....@e64g2000cwd.googlegroups.com...
What don't you first try to understand elementary physics?
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/NewPotential.html
Dirk Vdm
Androcles
"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
in message news:Zyr1h.157081$2x5.2...@phobos.telenet-ops.be...
[anip]
| What don't you first try to understand elementary physics?
I tink you mean elemtary fisicks, what don't you?
How old are you?
sal
>
> "Koobee Wublee" <koobee...@gmail.com> wrote in message
> news:1162232475....@e64g2000cwd.googlegroups.com...
>> Let's say we have the following spacetime.
>>
>> ds^2 = c^2 dt^2 - dq1^2 - dq2^2 - dq3^2
>>
>> The metric is Minkowski. It is diagonal.
>>
>> { c^2, -1, -1, -1 }
>>
>> Does this mean the described spacetime is always flat?
[SAL replying to K-B here]
Define "always". Do you mean for all time, or for all coordinate systems,
or everywhere in space, or on the entire manifold, or what?
"flat" is a coordinate-independent property, so at any particular point
either the manifold is flat or it isn't. "Sometimes", "usually",
"always", and "never" are qualifiers one doesn't typically use with such
an absolute state of affairs.
Since the curvature depends on the second partials of the metric, and the
metric you just described is constant, then in the region covered by that
metric the manifold is flat.
>> Now, we have another spaceimte as described below.
>>
>> ds^2 = c^2 (1 - 2 U) dt^2 - dq1^2 / (1 - 2 U) - q1^2 cos^2(q3) dq2^2 -
>> q1^2 dq3^2
>>
>> Where
>>
>> ** U = G M / c^2 / q1
>>
>> The metric is Schwarzschild. It is diagonal as well.
The Schwarzschild metric leads to a spacetime with nonzero curvature, as
you can show for yourself by plug-and-chugging the terms of the S. metric
into the formula for the Riemann tensor which you can find in lots of
books or on the web.
>>
>> { c^2 (1 - 2 U), - 1 / (1 - 2 U), q1^2 cos^2(q3), q1^2 }
>>
>> Does this mean the described spacetime is always curved if (M > 0)?
>
> What don't you first try to understand elementary physics?
> http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/NewPotential.html
Thank you, Dirk, that's a good one. Did dear old K-B ever retract that
statement, or modify it in any way (or, for that matter, deny saying it)?
>
> Dirk Vdm
--
Nospam becomes physicsinsights to fix the email
Igor
Koobee Wublee wrote:
> Let's say we have the following spacetime.
>
> ds^2 = c^2 dt^2 - dq1^2 - dq2^2 - dq3^2
>
> The metric is Minkowski. It is diagonal.
>
> { c^2, -1, -1, -1 }
>
> Does this mean the described spacetime is always flat?
Any spacetime that admits a Minkowski metric globally is indeed flat.
Not necessarily true locally.
>
> Now, we have another spaceimte as described below.
>
> ds^2 = c^2 (1 - 2 U) dt^2 - dq1^2 / (1 - 2 U) - q1^2 cos^2(q3)
> dq2^2 - q1^2 dq3^2
>
> Where
>
> ** U = G M / c^2 / q1
>
> The metric is Schwarzschild. It is diagonal as well.
>
> { c^2 (1 - 2 U), - 1 / (1 - 2 U), q1^2 cos^2(q3), q1^2 }
>
> Does this mean the described spacetime is always curved if (M > 0)?
Non-trivial versions of the Schwarzchild metric (M > 0) have globally a
non-vanishing curvature tensor. Although this may not be true for the
Ricci tensor and it's scalar, and indeed this is the case for
Schwarzchild. Ricci tensor vanishes.
Dirk Van de moortel
"sal" <pragm...@nospam.org> wrote in message news:pan.2006.10.30....@nospam.org...
Do real crackpots ever retract?
Does Androcles ever retract?
They rather dig and die ;-)
Dirk Vdm
Tom Roberts
> Let's say we have the following spacetime.
> ds^2 = c^2 dt^2 - dq1^2 - dq2^2 - dq3^2
> The metric is Minkowski. It is diagonal.
> { c^2, -1, -1, -1 }
> Does this mean the described spacetime is always flat?
Assuming you mean that the coordinates {t,q1,q2,q3} can be applied
globally, then yes. That is, the manifold is flat everywhere and everywhen.
(Igor said "Not necessarily true locally", which is wrong
-- the Riemann curvature tensor is identically zero at
every point of the manifold.)
Note that while the line element above is Minkowski, the manifold itself
need not be Minkowski spacetime. In particular, a manifold with topology
RxSxR^2 admits that line element (i.e. space is a "cylinder"); but, of
course, then the above coordinates cannot be applied globally.
> Now, we have another spaceimte as described below.
> ds^2 = c^2 (1 - 2 U) dt^2 - dq1^2 / (1 - 2 U) - q1^2 cos^2(q3)
> dq2^2 - q1^2 dq3^2
> Where
> ** U = G M / c^2 / q1
> The metric is Schwarzschild. It is diagonal as well.
> { c^2 (1 - 2 U), - 1 / (1 - 2 U), q1^2 cos^2(q3), q1^2 }
> Does this mean the described spacetime is always curved if (M > 0)?
Yes, in every region for which those coordinates are valid. Note that
unlike the previous case they cannot cover the entire manifold. But we
know enough about this manifold to know that it indeed has nonzero
Riemann curvature tensor at every point of the manifold (including
regions not covered by the above coordinates). There is also a
singularity at q1=0 which must be deleted from the manifold (be careful
which set of the coordinates you use -- this of course applies to the
sets in which q1 is timelike).
Note that the Ricci curvature tensor and scalar are both identically
zero at every point of the manifold. But when we say a manifold "is
curved" we mean that Riemann is nonzero.
Tom Roberts
JanPB
> Let's say we have the following spacetime.
>
> ds^2 = c^2 dt^2 - dq1^2 - dq2^2 - dq3^2
>
> The metric is Minkowski. It is diagonal.
>
> { c^2, -1, -1, -1 }
>
> Does this mean the described spacetime is always flat?
Yes, since it follows that Riemann=0. The topology could be different
than R^4, e.g. replace any (or all) of the factors in RxRxRxR with the
circle S^1, as in the flat torus:
S^1 x S^1 x S^1 x S^1.
> Now, we have another spaceimte as described below.
>
> ds^2 = c^2 (1 - 2 U) dt^2 - dq1^2 / (1 - 2 U) - q1^2 cos^2(q3)
> dq2^2 - q1^2 dq3^2
>
> Where
>
> ** U = G M / c^2 / q1
>
> The metric is Schwarzschild. It is diagonal as well.
>
> { c^2 (1 - 2 U), - 1 / (1 - 2 U), q1^2 cos^2(q3), q1^2 }
>
> Does this mean the described spacetime is always curved if (M > 0)?
Calculate the Riemann tensor. It's nonzero if M>0 so this spacetime is
curved.
--
Jan Bielawski
Koobee Wublee
> Let's say we have the following spacetime.
>
> ds^2 = c^2 dt^2 - dq1^2 - dq2^2 - dq3^2
>
> The metric is Minkowski. It is diagonal.
>
> { c^2, -1, -1, -1 }
>
> Does this mean the described spacetime is always flat?
>
> Now, we have another spaceimte as described below.
>
> ds^2 = c^2 (1 - 2 U) dt^2 - dq1^2 / (1 - 2 U) - q1^2 cos^2(q3)
> dq2^2 - q1^2 dq3^2
>
> Where
>
> ** U = G M / c^2 / q1
>
> The metric is Schwarzschild. It is diagonal as well.
>
> { c^2 (1 - 2 U), - 1 / (1 - 2 U), q1^2 cos^2(q3), q1^2 }
>
> Does this mean the described spacetime is always curved if (M > 0)?
On Oct 30, 12:48 pm, sal <pragmat...@nospam.org> wrote:
> Since the curvature depends on the second partials of the metric, and the
> metric [Minkowski] you just described is constant, then in the region
> covered by that metric the manifold is flat.
> The Schwarzschild metric leads to a spacetime with nonzero curvature, as
> you can show for yourself by plug-and-chugging the terms of the S. metric
> into the formula for the Riemann tensor which you can find in lots of
> books or on the web.
On Oct 30, 1:01 pm, "Igor" <thoov...@excite.com> wrote:
> Any spacetime that admits a Minkowski metric globally is indeed flat.
>
> Non-trivial versions of the Schwarzchild metric (M > 0) have globally a
> non-vanishing curvature tensor. Although this may not be true for the
> Ricci tensor and it's scalar, and indeed this is the case for
> Schwarzchild. Ricci tensor vanishes.
On Oct 30, 2:07 pm, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
> [...] yes. That is, the manifold is flat everywhere and everywhen.
> Yes, [Schwarzschild metric indicates a curvature in spacetime] [...]
> unlike the previous case they cannot cover the entire manifold. But we
> know enough about this manifold to know that it indeed has nonzero
> Riemann curvature tensor at every point of the manifold (including
> regions not covered by the above coordinates). There is also a
> singularity at q1=0 which must be deleted from the manifold (be careful
> which set of the coordinates you use -- this of course applies to the
> sets in which q1 is timelike).
>
> Note that the Ricci curvature tensor and scalar are both identically
> zero at every point of the manifold. But when we say a manifold "is
> curved" we mean that Riemann is nonzero.
On Oct 30, 5:47 pm, "JanPB" <film...@gmail.com> wrote:
> Yes, since it [Minkowski metric] follows that Riemann=0. [...]
> Calculate the Riemann tensor [for the Schwarzschild metric]. It's
> nonzero if M>0 so this spacetime is curved.
Basically, these notable correspondences above agree with the following
with various dingleberries (degrees of BS attached).
** Minkowski metric describes flat spacetime.
** Schwarzschild metric describes curved spacetime.
This appears to be what everyone is taught. However, one must also
examine the choice of coordinate system. The metric along cannot tell
how the spacetime is curved. One must also know the choice of
coordinate system to complete the analysis. For examples,
1) If we have a Minkowski metric and a coordinate system of the
following,
** t = (x^2 + y^2 + z^2) t' / R^2
** q1 = R^2 / x
** q2 = R^2 / y
** q3 = R^2 / z
Where
** (x, y, z) = Euclidean
** t' = Greenwich time
** R > 0
Then the spacetime has to be curved. The Minkowski metric along with
the choice of coordinate system above describes a curved spacetime.
Linear metric times curved coordinate system yields curved spacetime.
2) If we have a Schwarzschild metric and a coordinate system of the
following.
** t = G0(r, h, p) t'
** q1 = G1(r, h, p)
** q2 = G2(r, h, p)
** q3 = G3(r, h, p)
Where the above equations satisfy as solutions to the following
differential equations
** (1 - 2 U) dt^2 = dt'^2
** dq1^2 / (1 - 2 U) = dr^2
** dq1^2 cos^2(q3) dq2^2 = r^2 cos^2(p) dh^2
** dq1^2 dq3^2 = r^2 dp^2
Where
** U = G M / c^2 / q1
** r, h, p = polar coordinate describing radial position, longitude,
and latitude respectively
** t' = Greenwich time
Then the spacetime has to be flat. The Schwarzschild metric along with
the choice of coordinate system above describes a flat spacetime.
Curved metric times inversely curved coordinate system yields flat
spacetime.
Misuse of Riemann tensor or Ricci tensor is a tell-tale sign of not
understanding tensor calculus and/or the application of tensor calculus
to real world physics problems. It is indeed very lonely to be the
only person [Koobee Wublee] since Riemann to understand the curvature
of space as well as the curvature of spacetime. <shrug>
Sal further wrote:
> http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/NewPotential.html
>
> Dirk, that's a good one. Did dear old K-B ever retract that
> statement, or modify it in any way (or, for that matter, deny saying it)?
There is no need to retract that statement. It is very correct. I
choose to define
Total energy = Kinetic energy - Potential energy
Where
** Kinetic energy >= 0
** Potential energy >= 0
I did this to emphasize the non-existence of either the kinetic energy
or the potential energy. Thus, gravitational potential has to be zero
or positive as well. Dr. Roberts at first did not understand that.
However, after only a few days, he finally understood that as a valid
description but after giving off a few ranting remarks. Check out the
rest of the postings in that thread.
Eric Gisse
Koobee Wublee wrote:
[...]
>
> This appears to be what everyone is taught. However, one must also
> examine the choice of coordinate system. The metric along cannot tell
> how the spacetime is curved. One must also know the choice of
> coordinate system to complete the analysis.
Idiot of the day, right here. This is basic, basic shit you are fucking
up. A coordinate transformation does NOT change the physics in a theory
that is BASED upon covariance.
It is obvious that you already know this because you refrained from
computing the curvature scalar.
Where the hell did you learn GR? Can you name the book, or did you just
absorb it through your numerous physics classes? Oh, thats right...no
physics classes. Just engineering classes.
[...]
sal
> On Oct 30, 10:21 am, "Koobee Wublee" <koobee.wub...@gmail.com> wrote:
>
>
> On Oct 30, 12:48 pm, sal <pragmat...@nospam.org> wrote:
>
>> Since the curvature depends on the second partials of the metric, and
>> the metric [Minkowski] you just described is constant, then in the
>> region covered by that metric the manifold is flat.
Yup, and I'd go on to add that a tensor which is zero in any coordinate
system is zero in all of them.
And I'd also add that the standard meaning of the term "curved" in a forum
devoted to relativity is that the Riemann tensor is nonzero. But perhaps
K-W has decided to redefine "curvature" just so he can claim everyone else
is "wrong".
[ snip ]
> ** Minkowski metric describes flat spacetime. ** Schwarzschild metric
> describes curved spacetime.
>
> This appears to be what everyone is taught. However, one must also
> examine the choice of coordinate system.
Do tell.
> The metric along cannot tell
> how the spacetime is curved. One must also know the choice of
> coordinate system to complete the analysis. For examples,
>
> 1) If we have a Minkowski metric and a coordinate system of the
> following,
>
> ** t = (x^2 + y^2 + z^2) t' / R^2
> ** q1 = R^2 / x
> ** q2 = R^2 / y
> ** q3 = R^2 / z
>
> Where
>
> ** (x, y, z) = Euclidean
> ** t' = Greenwich time
> ** R > 0
>
> Then the spacetime has to be curved. The Minkowski metric along with
> the choice of coordinate system above describes a curved spacetime.
> Linear metric times curved coordinate system yields curved spacetime.
Why don't you prove this? Go ahead and compute the Riemann tensor in
terms of your curvilinear coordinates with the flat metric. See what you
get. Post the result.
While you're at it you might want to explain how the Riemann tensor can be
nonzero in one coordinate system and zero in another coordinate system.
Or are you defining "curvature" as something other than a nonzero Riemann
tensor? Like, "The space is curved if the bullshit coefficient is
nonzero, and that's defined as the value Koobee-Wublee picked out of his
nose, and it's nonzero in the case given above".
> Misuse of Riemann tensor or Ricci tensor is a tell-tale sign of not
> understanding tensor calculus and/or the application of tensor calculus
> to real world physics problems. It is indeed very lonely to be the only
> person [Koobee Wublee] since Riemann to understand the curvature of
> space as well as the curvature of spacetime. <shrug>
Godamighty. Did you really write this? You're as bad as Androcles.
> Sal further wrote:
>
>> http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/NewPotential.html
>>
>> Dirk, that's a good one. Did dear old K-B ever retract that statement,
>> or modify it in any way (or, for that matter, deny saying it)?
>
> There is no need to retract that statement. It is very correct. I choose
> to define
>
> Total energy = Kinetic energy - Potential energy
Most people call that the Lagrangian rather than the total energy.
Unfortunately as you add energy to a system that value doesn't necessarily
increase, which makes calling it the "total energy" somewhat awkward. The
Lagrangian also has the unfortunate property of changing its value as time
goes by in some systems which are commonly called "conservative", such as
an orbiting planet, a bouncing ball, or a mass attached to a spring
anchored at the other end; this, also, makes calling it the "total
energy", which one might expect to remain constant in a "conservative"
system, somewhat awkward.
But then, you may be defining "potential energy" as something which the
system must lose in order to become more ... ahem ... energetic. But in
that case you're certainly not talking about what anyone else means by
"energy".
>
> Where
>
> ** Kinetic energy >= 0
> ** Potential energy >= 0
>
> I did this to ...
Oh who cares why you did it? Redefining common words, like saying "I'm
using "up" to mean "toward the center of the earth"", and then pretending
you've done something clever and that anyone who finds your posts
confusing as a result must be stupid, is sophomoric.
Communication is difficult enough when everyone is working to use standard
terms correctly. If you want to work to intentionally violate common
usage rules, go right ahead, but please don't think it shows how smart you
are.
Dirk Van de moortel
"sal" <pragm...@nospam.org> wrote in message news:pan.2006.10.31....@nospam.org...
> On Mon, 30 Oct 2006 22:54:19 -0800, Koobee Wublee wrote:
>
>> On Oct 30, 10:21 am, "Koobee Wublee" <koobee.wub...@gmail.com> wrote:
>>
>
>>
>> On Oct 30, 12:48 pm, sal <pragmat...@nospam.org> wrote:
>>
>>> Since the curvature depends on the second partials of the metric, and
>>> the metric [Minkowski] you just described is constant, then in the
>>> region covered by that metric the manifold is flat.
>
> Yup, and I'd go on to add that a tensor which is zero in any coordinate
> system is zero in all of them.
>
> And I'd also add that the standard meaning of the term "curved" in a forum
> devoted to relativity is that the Riemann tensor is nonzero. But perhaps
> K-W has decided to redefine "curvature" just so he can claim everyone else
> is "wrong".
Koobee Afarensis also thinks that the surface of a sphere
is flat:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/FlatSphere.html
Dirk Vdm
Eric Gisse
Dirk Van de moortel wrote:
Where did this moron happen to think he learned GR from?
Given his massive boner for calculus of variations and Lagrangians, you
would think he actually knows what he is talking about. But apparently
he manages to fuck up everything yet retaining that smug sense of
superiority that feels great to quench.
>
> Dirk Vdm
Dirk Van de moortel
"Eric Gisse" <jow...@gmail.com> wrote in message news:1162289768.5...@m73g2000cwd.googlegroups.com...
>
> Dirk Van de moortel wrote:
>> "sal" <pragm...@nospam.org> wrote in message news:pan.2006.10.31....@nospam.org...
>> > On Mon, 30 Oct 2006 22:54:19 -0800, Koobee Wublee wrote:
>> >
>> >> On Oct 30, 10:21 am, "Koobee Wublee" <koobee.wub...@gmail.com> wrote:
>> >>
>> >
>> >>
>> >> On Oct 30, 12:48 pm, sal <pragmat...@nospam.org> wrote:
>> >>
>> >>> Since the curvature depends on the second partials of the metric, and
>> >>> the metric [Minkowski] you just described is constant, then in the
>> >>> region covered by that metric the manifold is flat.
>> >
>> > Yup, and I'd go on to add that a tensor which is zero in any coordinate
>> > system is zero in all of them.
>> >
>> > And I'd also add that the standard meaning of the term "curved" in a forum
>> > devoted to relativity is that the Riemann tensor is nonzero. But perhaps
>> > K-W has decided to redefine "curvature" just so he can claim everyone else
>> > is "wrong".
>>
>> Koobee Afarensis also thinks that the surface of a sphere
>> is flat:
>> http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/FlatSphere.html
>
> Where did this moron happen to think he learned GR from?
Beats me.
>
> Given his massive boner for calculus of variations and Lagrangians, you
> would think he actually knows what he is talking about. But apparently
> he manages to fuck up everything yet retaining that smug sense of
> superiority that feels great to quench.
He even manages to fuck up a simple Lorentz transformation:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/LorentzTale.html
I never understood why people take the trouble to actually *talk*
to this imbecile :-)
Dirk Vdm
Eric Gisse
Dirk Van de moortel wrote:
> "Eric Gisse" <jow...@gmail.com> wrote in message news:1162289768.5...@m73g2000cwd.googlegroups.com...
> >
> > Dirk Van de moortel wrote:
> >> "sal" <pragm...@nospam.org> wrote in message news:pan.2006.10.31....@nospam.org...
> >> > On Mon, 30 Oct 2006 22:54:19 -0800, Koobee Wublee wrote:
> >> >
> >> >> On Oct 30, 10:21 am, "Koobee Wublee" <koobee.wub...@gmail.com> wrote:
> >> >>
> >> >
> >> >>
> >> >> On Oct 30, 12:48 pm, sal <pragmat...@nospam.org> wrote:
> >> >>
> >> >>> Since the curvature depends on the second partials of the metric, and
> >> >>> the metric [Minkowski] you just described is constant, then in the
> >> >>> region covered by that metric the manifold is flat.
> >> >
> >> > Yup, and I'd go on to add that a tensor which is zero in any coordinate
> >> > system is zero in all of them.
> >> >
> >> > And I'd also add that the standard meaning of the term "curved" in a forum
> >> > devoted to relativity is that the Riemann tensor is nonzero. But perhaps
> >> > K-W has decided to redefine "curvature" just so he can claim everyone else
> >> > is "wrong".
> >>
> >> Koobee Afarensis also thinks that the surface of a sphere
> >> is flat:
> >> http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/FlatSphere.html
> >
> > Where did this moron happen to think he learned GR from?
>
> Beats me.
I need someone new to bug.
Henri has been reduced to one liners about my intelligence or simply
not replying. Androcles, while psychotic, might have actually managed
to make a killfile work against me [why me and not you? zuh?]. Ken Seto
doesn't look like he will be entertaining for too much longer.
Plus it is funny when people make cracks about my education while being
completely retarded.
>
> >
> > Given his massive boner for calculus of variations and Lagrangians, you
> > would think he actually knows what he is talking about. But apparently
> > he manages to fuck up everything yet retaining that smug sense of
> > superiority that feels great to quench.
>
> He even manages to fuck up a simple Lorentz transformation:
> http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/LorentzTale.html
I think this latest episode is even worse. He just demonstrated a
fantastic level of misunderstanding. I mean...that is a HUGE fuckup.
Someone completely unfamiliar with GR and its' principles would be
excused because that *is* a subtle point. But it is a fundamental point
that he should understand....yet doesn't.
Apparently understanding covariance is hard...
>
> I never understood why people take the trouble to actually *talk*
> to this imbecile :-)
Same reason you do - entertainment.
The idiots aren't directly teaching me anything, but the indirect
benefits are large. At least for me. The responses the Folks With
Clue(tm) produce tend to be educational and/or entertaining, and the
arguments the cranks make are [sometimes!] subtle enough to require
some thinking. Plus they provide a great benchmark - 'at least I'm not
*that* bad!', 'yea I fucked that up but at least I can admit it', etc.
>
> Dirk Vdm
Dirk Van de moortel
"Eric Gisse" <jow...@gmail.com> wrote in message news:1162291709.6...@k70g2000cwa.googlegroups.com...
He never had me in his killfile, at least not for much longer
than a few days. He just pretended. During his Great Plonk
Period, I got dozens of hits from his IP address on my fumble
page immediately after I announced a new entry.
Finally after he got sick of being unable to 'defend' himself,
and perhaps because I kept telling him how I loved being
in his killfile for this very reason, he 'removed me' from his
alleged list. He is so easy to manipulate :-)
Don't be mistaken about people who advertise their
plonking. Most of the time it's just a lie ;-)
> Ken Seto
> doesn't look like he will be entertaining for too much longer.
>
> Plus it is funny when people make cracks about my education while being
> completely retarded.
Yes.
And you might think that they are driven by jealousy.
They aren't. Too stupid to be jealous.
>
>>
>> >
>> > Given his massive boner for calculus of variations and Lagrangians, you
>> > would think he actually knows what he is talking about. But apparently
>> > he manages to fuck up everything yet retaining that smug sense of
>> > superiority that feels great to quench.
>>
>> He even manages to fuck up a simple Lorentz transformation:
>> http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/LorentzTale.html
>
> I think this latest episode is even worse. He just demonstrated a
> fantastic level of misunderstanding. I mean...that is a HUGE fuckup.
> Someone completely unfamiliar with GR and its' principles would be
> excused because that *is* a subtle point. But it is a fundamental point
> that he should understand....yet doesn't.
He shouldn't even bother looking at advanced stuff like this.
Surely he must realize that he can't even cope with the basics.
>
> Apparently understanding covariance is hard...
yes, specially for someone like this who doesn't even understand
how coordinates are used in a simple Lotrentz transformation ;-)
>
>>
>> I never understood why people take the trouble to actually *talk*
>> to this imbecile :-)
>
> Same reason you do - entertainment.
yes, but some seem to take him So Seriously - that is what
I don't understand.
>
> The idiots aren't directly teaching me anything, but the indirect
> benefits are large. At least for me. The responses the Folks With
> Clue(tm) produce tend to be educational and/or entertaining, and the
> arguments the cranks make are [sometimes!] subtle enough to require
> some thinking. Plus they provide a great benchmark - 'at least I'm not
> *that* bad!', 'yea I fucked that up but at least I can admit it', etc.
:-)
Keep up the good work and enjoy!
Dirk Vdm
sal
>
> Koobee Wublee wrote:
>> Let's say we have the following spacetime.
>>
>> ds^2 = c^2 dt^2 - dq1^2 - dq2^2 - dq3^2
>>
>> The metric is Minkowski. It is diagonal.
>>
>> { c^2, -1, -1, -1 }
>>
>> Does this mean the described spacetime is always flat?
>
> Any spacetime that admits a Minkowski metric globally is indeed flat. Not
> necessarily true locally.
If it admits a Minkowski metric on an open neighborhood of some point then
it's flat on that neighborhood.
"Locally flat" coordinates at a point where the curvature is nonzero only
have the Minkowski metric (and zero connection coefficients) at a single
point. In other words, if you're orbiting in free-fall, your
natural coordinate system is "locally flat". However, nearby stationary
objects will appear to accelerate (slowly!) due to tidal forces; the tidal
forces can't be eliminated by a coordinate transformation.
>
>
>> Now, we have another spaceimte as described below.
>>
>> ds^2 = c^2 (1 - 2 U) dt^2 - dq1^2 / (1 - 2 U) - q1^2 cos^2(q3) dq2^2 -
>> q1^2 dq3^2
>>
>> Where
>>
>> ** U = G M / c^2 / q1
>>
>> The metric is Schwarzschild. It is diagonal as well.
>>
>> { c^2 (1 - 2 U), - 1 / (1 - 2 U), q1^2 cos^2(q3), q1^2 }
>>
>> Does this mean the described spacetime is always curved if (M > 0)?
>
> Non-trivial versions of the Schwarzchild metric (M > 0) have globally a
> non-vanishing curvature tensor. Although this may not be true for the
> Ricci tensor and it's scalar, and indeed this is the case for
> Schwarzchild. Ricci tensor vanishes.
--
Nospam becomes physicsinsights to fix the email
I can be also contacted through http://www.physicsinsights.org
sal
>
> Koobee Wublee wrote:
>
> [...]
>
>
>> This appears to be what everyone is taught. However, one must also
>> examine the choice of coordinate system. The metric along cannot tell
>> how the spacetime is curved. One must also know the choice of
>> coordinate system to complete the analysis.
>
> Idiot of the day, right here. This is basic, basic shit you are fucking
> up. A coordinate transformation does NOT change the physics in a theory
> that is BASED upon covariance.
>
> It is obvious that you already know this because you refrained from
> computing the curvature scalar.
I'd have guessed it meant he didn't know how to compute the curvature
scalar, actually.
> Where the hell did you learn GR? Can you name the book, or did you just
> absorb it through your numerous physics classes? Oh, thats right...no
> physics classes. Just engineering classes.
>
> [...]
--
Tom Roberts
> ** Minkowski metric describes flat spacetime.
Yes. One of them. And it is an incomplete description.
> ** Schwarzschild metric describes curved spacetime.
Yes. One of them, not every one as your phrasing might suggest. And it
is an incomplete description.
> This appears to be what everyone is taught.
Well, yes. And it is also what any COMPETENT person _COMPUTES_. That, of
course, is why it is taught. <shrug>
> However, one must also
> examine the choice of coordinate system.
No. The Riemann curvature tensor is utterly and completely independent
of coordinates. As is the metric tensor. As is any other tensor. <shrug>
> The metric along cannot tell
> how the spacetime is curved.
Not true. Give the metric tensor on an open region of the manifold, one
can COMPUTE the Riemann curvature tensor throughout that region.
> [... further nonsense riddled throughout with errors]
As I have said so often: you must LEARN the basics. Essentially
everything you said in this post is wrong. Most especially the part
about you being the only person "since Riemann" to understand such
curvatures.
> [...] I choose to define
> Total energy = Kinetic energy - Potential energy
How many legs does a horse have if we call its tail a leg? -- FOUR --
calling a tail a leg does not make it one.
Tom Roberts
Igor
You cannot change the curvature properties of a manifold merely by a
coordinate transformation. Riemann curvature is a tensor. If it's
zero everywhere, it remains zero everywhere regardless of what
coordinate system you use. Likewise, if it has non-zero curvature,
this cannot be transformed away by an invertible transformation. This
is basic tensor analysis.
carlip...@physics.ucdavis.edu
[...]
> This appears to be what everyone is taught. However, one must also
> examine the choice of coordinate system. The metric along cannot tell
> how the spacetime is curved. One must also know the choice of
> coordinate system to complete the analysis.
If I describe a plane with polar coordinates, does it become curved?
If I describe the Earth using a Mercator projection, does it become
flat?
A coordinate system is a human-made choice of how to label points.
Do you really believe that curvature depends on your particular
choice of how to label points, and that it will change if I choose
a different labeling system?
Steve Carlip
Igor
Tom Roberts wrote:
, then yes. That is, the manifold is flat everywhere and everywhen.
>
> (Igor said "Not necessarily true locally", which is wrong
> -- the Riemann curvature tensor is identically zero at
> every point of the manifold.)
>
Poor choice of words on my part. What I actually meant to say is that
a manifold that admits a Minkowski metric locally is not necessarily
flat.
Koobee Wublee
On Oct 31, 10:12 am, carlip-nos...@physics.ucdavis.edu wrote:
> If I describe a plane with polar coordinates, does it become curved?
No, a plane is still flat. Polar coordinate describes curved
projections. To properly describe the flat plane, you need the metric
associated with the polar coordinate.
** (r, h) = Polar coordinate in two dimensions
** diag(1, r^2) = The metric.
> If I describe the Earth using a Mercator projection, does it become
> flat?
No, the earth is spherical. Mercator projection is flat. To properly
describe the surface of the spherical earth, you need the metric
associated with the Mercator projection.
> A coordinate system is a human-made choice of how to label points.
> Do you really believe that curvature depends on your particular
> choice of how to label points, and that it will change if I choose
> a different labeling system?
No, the curvature of space or spacetime exists independent of man-made
labeling points. However, man-made labeling points do not properly
describe the curvature of space or spacetime. That is why we need a
metric to complete the description of this curvature in space or
spacetime. Since there are an infinite number of choices in coordinate
system, to describe the same space or spacetime, there needs to be a
unique metric associated with each coordinate system to do so. Here
are another two examples.
In a flat spacetime, we can describe it as follows.
ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2
Where
** (x, y, z) = Euclidean coordinate
** diag(c^2, -1, -1, -1) = the metric
We can also write the same spacetime as follows.
ds^2 = c^2 dt^2 - dr^2 - r^2 cos^2(p) dh^2 - r^2 dp^2
Where
** (r, h, p) = Spherically symmetrical polar coordinate
** diag(c^2, -1, - r^2 cos^2(p), - r^2) = the metric
Clearly, the two metrics are different. It is not surprising because
the choice of coordinate is different. However, both equations
describe the same spacetime through different metric.
Thus, a metric along cannot determine if spacetime is curved or not.
One needs to throw in the associated coordinate system to complete the
analysis.
Koobee Wublee
On Oct 31, 9:07 am, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
> Koobee Wublee wrote:
> [...] it is also what any COMPETENT person _COMPUTES_. That, of
> course, is why it is taught. <shrug>
A better explanation is that the teacher does not understand the
subject very well.
> No. The Riemann curvature tensor is utterly and completely independent
> of coordinates. As is the metric tensor. As is any other tensor. <shrug>
The Riemann curvature tensor is different if I use the rectangular
coordinate or the spherically symmetric polar coordinate. Thus, it is
coordinate dependent. You are utterly wrong.
> > The metric along cannot tell
> > how the spacetime is curved.Not true.
>
> Give the metric tensor on an open region of the manifold, one
> can COMPUTE the Riemann curvature tensor throughout that region.
Any curvature tensor is useless if you don't specify what coordinate
system you have.
> As I have said so often: you must LEARN the basics. Essentially
> everything you said in this post is wrong.
It appears to be wrong to you because of your incomplete understanding
of the curvature of space or spacetime. You need to start with the
basics by learning from my posts here.
> Most especially the part about you being the only person "since
> Riemann" to understand such curvatures.
But that is the truth. Just by looking at your postings here, they
fully demonstrate a lack of understanding in basic structure of
spacetime. As I said, it is very lonely to be the only person to have
fully understood the concept of the curvature in space or spacetime
since Riemann. <shrug>
> > [...] I choose to define
> > Total energy = Kinetic energy - Potential energy
>
> How many legs does a horse have if we call its tail a leg?
> -- FOUR -- calling a tail a leg does not make it one.
You are getting ridiculous.
Koobee Wublee
> You cannot change the curvature properties of a manifold merely by a
> coordinate transformation. Riemann curvature is a tensor. If it's
> zero everywhere, it remains zero everywhere regardless of what
> coordinate system you use. Likewise, if it has non-zero curvature,
> this cannot be transformed away by an invertible transformation. This
> is basic tensor analysis.
We have the following spacetime described by the Schwarzschild metric
in spherically symmetric polar coordinate.
ds^2 = c^2 (1 - 2 U) dt^2 - dr^2 / (1 - 2 U) - r^2 cos^2(p)
dh^2 - r^2 dp^2
Where
** U = G M / c^2 / r
** r, h, p = radial distance, longitude, latitude
Of course, the metric is
[ c^2 (1 - 2 U), 0, 0, 0 ]
[ 0, -1 / (1 - 2 U), 0, 0 ]
[ 0, 0, - r^2 cos^2(p), 0 ]
[ 0, 0, 0, - r^2 ]
The same spacetime above can be written into the following using the
Euclidean coordinate system.
ds^2 = c^2 (1 - 2 U) - dx^2 - dy^2 - dz^2 - 2 U (x dx + y dy
+ z dz)^2 / (1 - 2 U) / r^2
Where
** U = G M / c^2 / r
** r^2 = x^2 + y^2 + z^2
Of course, the metric is
[ c^2 (1 - 2 U), 0, 0, 0 ]
[ 0, -XX, -XY, -XZ ]
[ 0, -XY, -YY, -YZ ]
[ 0, -XZ, -YZ, -ZZ ]
Where
** XX = 1 + 2 x^2 U / (1 - 2 U)^2 / r^2
** YY = 1 + 2 y^2 U / (1 - 2 U)^2 / r^2
** ZZ = 1 + 2 z^2 U / (1 - 2 U)^2 / r^2
** XY = 2 x y U / (1 - 2 U) / r^2
** YZ = 2 y z U / (1 - 2 U) / r^2
** XZ = 2 x z U / (1 - 2 U) / r^2
Clearly, the metric for each case is very different. Whether if the
curvature tensor is zero or not, it bears no counter argument to any of
my postings.
If you are still confused, don't feel so bad. All others still don't
understand this. Can you feel my frustration when I am the only one
since Riemann to be able to understand this subject fully?
Dirk Van de moortel
"Koobee Wublee" <koobee...@gmail.com> wrote in message news:1162329665....@e3g2000cwe.googlegroups.com...
We see your frustration.
Being such a genius must be a real burden.
The responsibility must be crushing.
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/SmellHere.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/TwoMetrics.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/DiffGeoAero.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/FlatSphere.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/LorentzTale.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/SRBogus.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/ReasonLaws.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/NewLagrangian.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/LosingIt.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/AerospaceRelativity.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/NewPotential.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/WhatWrong.html
Dirk Vdm
sal
>
>
> On Oct 31, 9:07 am, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
>> Koobee Wublee wrote:
>
>> [...] it is also what any COMPETENT person _COMPUTES_. That, of course,
>> is why it is taught. <shrug>
>
> A better explanation is that the teacher does not understand the subject
> very well.
>
>> No. The Riemann curvature tensor is utterly and completely independent
>> of coordinates. As is the metric tensor. As is any other tensor. <shrug>
>
> The Riemann curvature tensor is different if I use the rectangular
> coordinate or the spherically symmetric polar coordinate. Thus, it is
> coordinate dependent. You are utterly wrong.
Dirk! Dirk! You've gotta stick this one in the fumbles list!
This guy has no clue what a "tensor" is, and doesn't even have a clue that
he doesn't have a clue.
Igor
The metric is not changed, just it's components have changed. Just as
the components of the curvature tensor will change under such a
relabelling of the coordinates. But the overall tensor doesn't change.
That's a fundamental principle of tensor analysis.
> If you are still confused, don't feel so bad. All others still don't
> understand this. Can you feel my frustration when I am the only one
> since Riemann to be able to understand this subject fully?
You're seriously confused and extremely deluded if you keep insisting
that you can transform curvature away. That's all there is to it.
Eric Gisse
Koobee Wublee wrote:
[...]
> Thus, a metric along cannot determine if spacetime is curved or not.
> One needs to throw in the associated coordinate system to complete the
> analysis.
hahahhahahahahahahahahahahha
Paul B. Andersen
>
> On Oct 31, 9:07 am, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
>> As I have said so often: you must LEARN the basics. Essentially
>> everything you said in this post is wrong.
>
> It appears to be wrong to you because of your incomplete understanding
> of the curvature of space or spacetime. You need to start with the
> basics by learning from my posts here.
>
>> Most especially the part about you being the only person "since
>> Riemann" to understand such curvatures.
>
> But that is the truth. Just by looking at your postings here, they
> fully demonstrate a lack of understanding in basic structure of
> spacetime. As I said, it is very lonely to be the only person to have
> fully understood the concept of the curvature in space or spacetime
> since Riemann. <shrug>
And he is dead serious! :-)
The hallmark of a crank is that he believes himself to
be the only person who has understood what the rest
of the world have failed to understand.
But Koobee Wublee is an advanced crank.
The more mundane cranks like Henri Wilson, Androcles and Ken Seto
fail to understand elementary math and physics,
Koobee Wublee fails to understand Riemannian geometry and GR.
Paul
Dirk Van de moortel
"It is very lonely at the top":
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/LonelyTop.html
A stunning beauty.
A Fox.
Dirk Vdm
Koobee Wublee
> Koobee Wublee wrote:
What are these components you are referring to?
As I have pointed through a series of simple mathematics, we have
** Curvature of spacetime is invariant.
** Coordinate is a choice of the observer.
** Since the observer cannot fully observe the curvature in spacetime,
he needs help with a metric attached to his choice of coordinate system
to see clearly just like a pair of prescribed glasses to correct eye
sights.
> > If you are still confused, don't feel so bad. All others still don't
> > understand this. Can you feel my frustration when I am the only one
> > since Riemann to be able to understand this subject fully?
>
> You're seriously confused and extremely deluded if you keep insisting
> that you can transform curvature away. That's all there is to it.
Hey, I do insist on the truth. It is indeed lonely to be the only one
fully understands the subject of curved spacetime since Riemann.
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/LonelyTop.html
Koobee Wublee
On Oct 31, 2:22 pm, "Paul B. Andersen"
<paul.b.ander...@hiadeletethis.no> wrote:
> The hallmark of a crank is that he believes himself to
> be the only person who has understood what the rest
> of the world have failed to understand.
Not quite, a hallmark of a crank is not to be consistent. He embraces
the principle of Consistency, and yet he fouls up the whole landscape
with droppings that falsifies the very principle of Consistency.
The principle of Consistency states that the laws of physics must be
the same at any point in space as well as any moment in time.
Galileo's very familiar principle of Relativity is a subset.
> But Koobee Wublee is an advanced crank.
Another hallmark of a crank is to call a guru an advanced crank.
> The more mundane cranks like Henri Wilson, Androcles and Ken Seto
> fail to understand elementary math and physics,
> Koobee Wublee fails to understand Riemannian geometry and GR.
Yet, another hallmark of a crank is to accuse a guru of not
understanding the said subjects.
JanPB
Oh puhleeze. Tensors are independent of coordinates. End of story. They
are frequently specified by writing down their components wrt a
coordinate system but that a different thing. The tensor so specified
doesn't know anything about coordinates because it's just a bundle
section.
An obvious special case is a vector field. Say you have this vector
field defined by specifying its components in the Cartesian system:
X = d/dx + d/dy
(a vector field "pointing northeast").
Now let's change to the polar coordinates:
x = r cos(theta)
y = r sin(theta)
...therefore, the components of X change as:
X = (cos(theta)+sin(theta)) d/dr + (cos(theta)-sin(theta))/r d/dtheta
And you claim that this is a different tensor????
--
Jan Bielawski
Eric Gisse
Koobee Wublee wrote:
> On Oct 31, 2:22 pm, "Paul B. Andersen"
> <paul.b.ander...@hiadeletethis.no> wrote:
>
> > The hallmark of a crank is that he believes himself to
> > be the only person who has understood what the rest
> > of the world have failed to understand.
>
> Not quite, a hallmark of a crank is not to be consistent. He embraces
> the principle of Consistency, and yet he fouls up the whole landscape
> with droppings that falsifies the very principle of Consistency.
Henri Wilson is very consistent in his denial of reality. He is still a
crank.
Redefining what a crank is doesn't change the nature of what a crank
is. You are a crank.
>
> The principle of Consistency states that the laws of physics must be
> the same at any point in space as well as any moment in time.
> Galileo's very familiar principle of Relativity is a subset.
>
> > But Koobee Wublee is an advanced crank.
>
> Another hallmark of a crank is to call a guru an advanced crank.
>
> > The more mundane cranks like Henri Wilson, Androcles and Ken Seto
> > fail to understand elementary math and physics,
> > Koobee Wublee fails to understand Riemannian geometry and GR.
>
> Yet, another hallmark of a crank is to accuse a guru of not
> understanding the said subjects.
By what figment of your imagination are you a "guru" ?
You exhibit incredibly basic misunderstandings about simple tensor
analysis.
Yet you are so sure you are right, you won't even prove all us
naysayers wrong by doing the one thing that would prove you right:
computing the curvature scalar. Why have you refrained from computing
the curvature scalar for all your examples?
Koobee Wublee
> Koobee Wublee wrote:
> > Hey, I do insist on the truth. It is indeed lonely to be the only one
> > fully understands the subject of curved spacetime since Riemann.
>
> Oh puhleeze. Tensors are independent of coordinates. End of story. They
> are frequently specified by writing down their components wrt a
> coordinate system but that a different thing. The tensor so specified
> doesn't know anything about coordinates because it's just a bundle
> section.
You are very confused. The following discussion should clear it up.
> An obvious special case is a vector field. Say you have this vector
> field defined by specifying its components in the Cartesian system:
>
> X = d/dx + d/dy
>
> (a vector field "pointing northeast").
What you have written down has a coordinate equivalence of
<d/dx, d/dy>
And a metric of
<1, 1>
> Now let's change to the polar coordinates:
>
> x = r cos(theta)
> y = r sin(theta)
>
> ...therefore, the components of X change as:
>
> X = (cos(theta)+sin(theta)) d/dr + (cos(theta)-sin(theta))/r d/dtheta
Now, the coordinate equivalence is
<d/dr, d/dtheta>
I did not check your math. Assuming it is correct, the metric is
<cos(theta) + sin(theta), (cos(theta) - sin(theta)) / r>
> And you claim that this is a different tensor????
X = X, of course
However, the choice of coordinate system is different. Not
surprisingly, the metric is different as well.
Well, now you have to agree with me that I am the only one who is able
to understand the curvature of space or spacetime since Riemann. It is
getting very lonely.
Eric Gisse
Koobee Wublee wrote:
[...]
> Well, now you have to agree with me that I am the only one who is able
> to understand the curvature of space or spacetime since Riemann. It is
> getting very lonely.
So when are you going to compute the curvature scalar and show us how
incredibly wrong we are?
JanPB
> On Oct 31, 5:49 pm, "JanPB" <film...@gmail.com> wrote:
> > Koobee Wublee wrote:
>
> > > Hey, I do insist on the truth. It is indeed lonely to be the only one
> > > fully understands the subject of curved spacetime since Riemann.
> >
> > Oh puhleeze. Tensors are independent of coordinates. End of story. They
> > are frequently specified by writing down their components wrt a
> > coordinate system but that a different thing. The tensor so specified
> > doesn't know anything about coordinates because it's just a bundle
> > section.
>
> You are very confused. The following discussion should clear it up.
>
> > An obvious special case is a vector field. Say you have this vector
> > field defined by specifying its components in the Cartesian system:
> >
> > X = d/dx + d/dy
> >
> > (a vector field "pointing northeast").
>
> What you have written down has a coordinate equivalence of
>
> <d/dx, d/dy>
>
> And a metric of
>
> <1, 1>
No, I wrote a vector field X = d/dx + d/dy. I have no "coordinate
equivalence" (there is no such thing). And I haven't specified ANY
metric. Perhaps you should reread what I wrote: a plane (2D linear
space R^2) with a vector field on it. No metric.
> > Now let's change to the polar coordinates:
> >
> > x = r cos(theta)
> > y = r sin(theta)
> >
> > ...therefore, the components of X change as:
> >
> > X = (cos(theta)+sin(theta)) d/dr + (cos(theta)-sin(theta))/r d/dtheta
>
> Now, the coordinate equivalence is
>
> <d/dr, d/dtheta>
It's a coordinate basis, not any "equivalence". Don't use your private
terminology, we are not clairvoyant around here.
> I did not check your math. Assuming it is correct, the metric is
>
> <cos(theta) + sin(theta), (cos(theta) - sin(theta)) / r>
>
> > And you claim that this is a different tensor????
>
> X = X, of course
>
> However, the choice of coordinate system is different. Not
> surprisingly, the metric is different as well.
What metric? I have not specified ANY metric. The question is: did X
change when I switched from Cartesians to polars? Yes/no?
> Well, now you have to agree with me that I am the only one who is able
> to understand the curvature of space or spacetime since Riemann. It is
> getting very lonely.
Obviously I disagree, you not only don't understand curvature, you need
to learn manifold basics first.
--
Jan Bielawski
JanPB
I think I can sort of (barely) understand where he got his ideas from.
He thinks the fact that coordinate systems are used to compute things
from tensors implies coordinates actually determine tensors.
--
Jan Bielawski
badd...@yahoo.com
Koobee Wublee wrote:
> Let's say we have the following spacetime.
>
> ds^2 = c^2 dt^2 - dq1^2 - dq2^2 - dq3^2
>
> The metric is Minkowski. It is diagonal.
>
> { c^2, -1, -1, -1 }
>
> Does this mean the described spacetime is always flat?
You are confused. In curved spacetime, the metric can always
be transformed into the Minkowski metric _at a point_. What
distinguishes flat spacetime from curved spacetime is whether
or not you can construct a _global_ coordinate system that is
flat (vanishing Riemann tensor).
>
> Now, we have another spaceimte as described below.
>
> ds^2 = c^2 (1 - 2 U) dt^2 - dq1^2 / (1 - 2 U) - q1^2 cos^2(q3)
> dq2^2 - q1^2 dq3^2
>
> Where
>
Mike
Koobee Wublee wrote:
> Thus, a metric along cannot determine if spacetime is curved or not.
> One needs to throw in the associated coordinate system to complete the
> analysis.
Koobee, Koobee, Koobee, Koobee, Koobee....
At this time I can do nothing to stir up some controversy to your
benefit. Math is not philosophy. unfortunately. If I were to try this
time I would earn the most prominent placement in Fumbles pyramid.
Koobee, Koobee, Koobee.... in simple words, curvature is the amount by
which a geometric object deviates from being flat. Obviously, this
property of geometric objects is independent of the coordinate system
used, naturally.
Furthermore, do you think that the distance between two points in
Euclidean space depends on the coordinate system used?
Mike
Mike
Do you know of any implications of the Nash embedding theorem on
General Relativity?
It appears that as a consequence of the theorem and its C^1
generalization by Kuiper that General Relativity is a special case of
some (unknown to this date) gravity theory with a flat spacetime
manifold and a Euclidean metric.
Mike
>
> Steve Carlip
Tom Roberts
> On Oct 31, 9:07 am, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
>> The Riemann curvature tensor is utterly and completely independent
>> of coordinates. As is the metric tensor. As is any other tensor. <shrug>
>
> The Riemann curvature tensor is different if I use the rectangular
> coordinate or the spherically symmetric polar coordinate. Thus, it is
> coordinate dependent.
I repeat: YOU NEED TO LEARN THE BASICS. <shrug>
The Riemann curvature TENSOR does not change regardless of which
coordinate system one uses to compute it. The COMPONENTS of the tensor
relative to the coordinate system do change for different coordinate
systems. As I keep saying and you keep ignoring: there is a BIG
DIFFERENCE between a tensor and its components, and when you ignore the
difference you confuse yourself.
This has been known for more than a century. Yes, some older physics
textbooks confuse the issue.
(I'm giving up, as you obviously have no interest in learning anything.)
Tom Roberts
Dirk Van de moortel
"Tom Roberts" <tjrobe...@sbcglobal.net> wrote in message news:3F42h.24209$e66....@newssvr13.news.prodigy.com...
The only interest he has, is in making you lose your
time & temper :-)
Dirk Vdm
Koobee Wublee
> Koobee Wublee wrote:
> > On Oct 31, 5:49 pm, "JanPB" <film...@gmail.com> wrote:
> > > Koobee Wublee wrote:
>
> > > > Hey, I do insist on the truth. It is indeed lonely to be the only one
> > > > fully understands the subject of curved spacetime since Riemann.
>
> > > Oh puhleeze. Tensors are independent of coordinates. End of story. They
> > > are frequently specified by writing down their components wrt a
> > > coordinate system but that a different thing. The tensor so specified
> > > doesn't know anything about coordinates because it's just a bundle
> > > section.
>
> > You are very confused. The following discussion should clear it up.
>
> > > An obvious special case is a vector field. Say you have this vector
> > > field defined by specifying its components in the Cartesian system:
>
> > > X = d/dx + d/dy
>
> > > (a vector field "pointing northeast").
>
> > What you have written down has a coordinate equivalence of
>
> > <d/dx, d/dy>
>
> > And a metric of
>
> > <1, 1>
>
> No, I wrote a vector field X = d/dx + d/dy. I have no "coordinate
> equivalence" (there is no such thing). And I haven't specified ANY
> metric. Perhaps you should reread what I wrote: a plane (2D linear
> space R^2) with a vector field on it. No metric.
You wrote down
X = d/dx + d/dy
And this is the same as
X = <1, 1> * <d/dx, d/dy> = d/dx + d/dy
If you don't understand dot product, then our discussion should end.
You need to learn what dot product is and then inner product.
<1, 1> clearly is the metric.
<d/dx, d/dy> clearly is your very choice of coordinate.
> > > Now let's change to the polar coordinates:
>
> > > x = r cos(theta)
> > > y = r sin(theta)
>
> > > ...therefore, the components of X change as:
>
> > > X = (cos(theta)+sin(theta)) d/dr + (cos(theta)-sin(theta))/r d/dtheta
>
> > Now, the coordinate equivalence is
>
> > <d/dr, d/dtheta>
>
> It's a coordinate basis, not any "equivalence". Don't use your private
> terminology, we are not clairvoyant around here.
You wrote down
X = (cos(theta) + sin(theta)) d/dr + ((cos(theta) - sin(theta)) / r)
d/dtheta
And the above equation can be written as
X = <cos(theta) + sin(theta), (cos(theta) - sin(theta)) / r> * <d/dr,
d/dtheta>
Again, X is a dot product of a metric and a coordinate vector.
> > I did not check your math. Assuming it is correct, the metric is
>
> > <cos(theta) + sin(theta), (cos(theta) - sin(theta)) / r>
>
> > > And you claim that this is a different tensor????
>
> > X = X, of course
>
> > However, the choice of coordinate system is different. Not
> > surprisingly, the metric is different as well.
>
> What metric? I have not specified ANY metric. The question is: did X
> change when I switched from Cartesians to polars? Yes/no?
Your metric whether you try to deny it or not is
<cos(theta) + sin(theta), (cos(theta) - sin(theta)) / r>
The corresponding coordinate system as observer's choice is
<d/dr, d/dtheta>
THE METRIC IS A FUNCTION OF THE OBSERVER'S CHOICE OF COORDINATE
SYSTEM. SINCE THE CURVATURE OF SPACE IS INDEPENDENT OF ANY OBSERVERS,
AS THE OBSERVER'S CHOICE OF COORDINATE SYSTEM VARIES FROM ONE
OBSERVER TO ANOTHER, THE METRIC HAS TO VARY WITH EACH OBSERVER AS WELL.
THUS, THE METRIC HAS TO BE COORDINATE DEPENDENT. THIS IS THE ONLY WAY
TO GUARANTEE A CURVATURE OF SPACE INDPENDENT OF ANY OBSERVER AND HIS
CHOICE OF COORDINATE SYSETM.
> > Well, now you have to agree with me that I am the only one who is able
> > to understand the curvature of space or spacetime since Riemann. It is
> > getting very lonely.
>
> Obviously I disagree, you not only don't understand curvature, you need
> to learn manifold basics first.
You will agree after you have learned your dot product.
Koobee Wublee
On Nov 1, 9:17 am, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
> The Riemann curvature TENSOR does not change regardless of which
> coordinate system one uses to compute it. The COMPONENTS of the tensor
> relative to the coordinate system do change for different coordinate
> systems. As I keep saying and you keep ignoring: there is a BIG
> DIFFERENCE between a tensor and its components, and when you ignore the
> difference you confuse yourself.
I have not addressed the Riemann curvature tensor. All I have said so
far is that metric must be coordinate dependent. Each observer must
have his own metric and his coordinate system describing the curvature
of space or spacetime which is observer independent. Let's go back
to the example of the following familiar spacetime segment.
ds^2 = c^2 (1 - 2 U) dt^2 - dr^2 / (1 - 2 U) - r^2 cos^2(p)
dh^2 - r^2 dh^2
Where
** U = G M / c^2 / r
The metric is
[c^2 (1 - 2 U), 0, 0, 0]
[0, - 1 / (1 / 2 U), 0, 0]
[0, 0, -r^2 cos^2(p), 0]
[0, 0, 0, -r^2]
Is the metric a function of r?
YES.
Is r a choice of coordinate?
YES.
The very example tells you exactly what my point is right in front of
your very face. If you still don't see this, hey, you do not deserve
that phd.
Eric Gisse
Koobee Wublee wrote:
> On Nov 1, 9:17 am, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
>
> > The Riemann curvature TENSOR does not change regardless of which
> > coordinate system one uses to compute it. The COMPONENTS of the tensor
> > relative to the coordinate system do change for different coordinate
> > systems. As I keep saying and you keep ignoring: there is a BIG
> > DIFFERENCE between a tensor and its components, and when you ignore the
> > difference you confuse yourself.
>
> I have not addressed the Riemann curvature tensor. All I have said so
> far is that metric must be coordinate dependent.
...and you are WRRRROOONNNNGGGGGG.
This is *BASIC* differential geometry. A metric is *NOT* changed by an
invertable coordinate transformation.
Why have you refused to compute the curvature scalars for the metrics
you supposedly converted from curved to flat and flat to curved? That
is *all* you have to do to prove how wrong the naysayers are, but you
won't even do that.
Is it that you don't know how? Do I need to do it for you?
[...]
Koobee Wublee
On Nov 1, 7:17 am, "Mike" <elea...@yahoo.gr> wrote:
> Koobee, Koobee, Koobee.... in simple words, curvature is the amount by
> which a geometric object deviates from being flat. Obviously, this
> property of geometric objects is independent of the coordinate system
> used, naturally.
Yes, the curvature is independent of the coordinate system. Yes, the
curvature is independent of any observer. However, when comes an
observer using his own coordinate system, he is going to measure a
distorted amount because he cannot perceive a curvature in space or
spacetime. He sees everything straight despite bending photons. Since
the mathematics is laid out using the observer's coordinate, as the
coordinate changes from observer to observer, the metric must change
with each coordinate to accommodate an invariant curvature in
spacetime. Mathematically, we have
ds^2 = Q1^ij q1_ij = Q2^ij q2_ij = Q3^ij q3_ij
Where
** ds^2 = Invariant from observer to observer
** q1, q2, q3 = Coordinate system of observer 1, 2, 3
** Q1, Q2, Q3 = Metric 1, 2, 3 of each observer
If you can understand this very simple concept, you are better than the
professors and the phds.
Eric Gisse
Koobee Wublee wrote:
[...]
>
> THE METRIC IS A FUNCTION OF THE OBSERVER'S CHOICE OF COORDINATE
> SYSTEM. SINCE THE CURVATURE OF SPACE IS INDEPENDENT OF ANY OBSERVERS,
> AS THE OBSERVER'S CHOICE OF COORDINATE SYSTEM VARIES FROM ONE
> OBSERVER TO ANOTHER, THE METRIC HAS TO VARY WITH EACH OBSERVER AS WELL.
> THUS, THE METRIC HAS TO BE COORDINATE DEPENDENT. THIS IS THE ONLY WAY
> TO GUARANTEE A CURVATURE OF SPACE INDPENDENT OF ANY OBSERVER AND HIS
> CHOICE OF COORDINATE SYSETM.
You are so sure that you don't even bother proving it!
Igor
You're the expert on tensors. Why don't you tell me? If you don't
know what the word "components" refers to, you're dead in the water.
Why did you start such a discussion in the first place if you don't
know the basics of the subject?
> As I have pointed through a series of simple mathematics, we have
>
> ** Curvature of spacetime is invariant.
That's correct, but that's not what you've been saying previously. So
now you just contradicted yourself. You need to get your story
straight.
> ** Coordinate is a choice of the observer.
Correct again, but...
> ** Since the observer cannot fully observe the curvature in spacetime,
> he needs help with a metric attached to his choice of coordinate system
> to see clearly just like a pair of prescribed glasses to correct eye
> sights.
But the metric either cooresponds to curved spacetime or a flat
spacetime and they're completely mutually exclusive. Changing your
coordinates won't change curvature. You've already said it's
invariant. You need to learn what that word means.
> > > If you are still confused, don't feel so bad. All others still don't
> > > understand this. Can you feel my frustration when I am the only one
> > > since Riemann to be able to understand this subject fully?
> >
> > You're seriously confused and extremely deluded if you keep insisting
> > that you can transform curvature away. That's all there is to it.
>
> Hey, I do insist on the truth. It is indeed lonely to be the only one
> fully understands the subject of curved spacetime since Riemann.
You're the one that keeps flip-flopping on the concept of invariant
curvature. So you're as about as confused as a person can get.
Igor
Koobee Wublee wrote:
> On Nov 1, 7:17 am, "Mike" <elea...@yahoo.gr> wrote:
>
> > Koobee, Koobee, Koobee.... in simple words, curvature is the amount by
> > which a geometric object deviates from being flat. Obviously, this
> > property of geometric objects is independent of the coordinate system
> > used, naturally.
>
> Yes, the curvature is independent of the coordinate system.
Now stop right there and say no other stupid things and you've got it
correct for a change. Case closed.
Koobee Wublee
On Nov 1, 10:57 am, "Igor" <thoov...@excite.com> wrote:
> Koobee Wublee wrote:
> > Yes, the curvature is independent of the coordinate system.
In order for the above statement to be true, the metric must be a
function of the choice of coordinate system. If the metric in
invariant, then the curvature of space or spacetime will change with
the choice of coordinate system.
> Now stop right there and say no other stupid things and you've got it
> correct for a change. Case closed.
What you have learned is wrong. That is why you are such confused.
Better read on.
Igor
Ignorance is curable, stupidity is not. When people repeatedly point
out your errors, but you keep refusing to see them, which side do you
think you fall on?
sal
>
> THE METRIC IS A FUNCTION OF THE OBSERVER'S CHOICE OF COORDINATE SYSTEM.
> SINCE THE CURVATURE OF SPACE IS INDEPENDENT OF ANY OBSERVERS, AS THE
> OBSERVER'S CHOICE OF COORDINATE SYSTEM VARIES FROM ONE OBSERVER TO
> ANOTHER, THE METRIC HAS TO VARY WITH EACH OBSERVER AS WELL.
> THUS, THE METRIC HAS TO BE COORDINATE DEPENDENT. THIS IS THE ONLY WAY
> TO GUARANTEE A CURVATURE OF SPACE INDPENDENT OF ANY OBSERVER AND HIS
> CHOICE OF COORDINATE SYSETM.
Omigod will you look at that, he puts his ignorance in CAPS just so
nobody can overlook it! And he clarifies the statement of it to
the point where there can be no question over his confusion level.
Does KW think the metric TENSOR is a different kind of entity from the
curvature TENSOR?
Does he think the metric tensor is and fundamentally
different from every other tensor?
Does he think there are two different kinds of tensors, those that are
invariant and those that aren't? Or does he think the metric tensor isn't
a tensor? Or does he think the curvature tensor isn't a tensor?
Does he think tensors aren't invariant?
Does he think?
--
Nospam becomes physicsinsights to fix the email
JanPB
Nothing you write makes sense:
1. You originally wrote "What you have written down has a coordinate
equivalence of <d/dx, d/dy> and a metric of <1, 1>", and this is what I
objected to.
2. "<1, 1> * <d/dx, d/dy>" is not any "dot product" - it seems you need
to brush up on basic linear algebra as well.
3. "You need to learn what dot product is and then inner product" -
says the guy who refers to
"<1, 1> * <d/dx, d/dy>" as a "dot product". Unbelievable.
> <1, 1> clearly is the metric.
> <d/dx, d/dy> clearly is your very choice of coordinate.
>
> > > > Now let's change to the polar coordinates:
> >
> > > > x = r cos(theta)
> > > > y = r sin(theta)
> >
> > > > ...therefore, the components of X change as:
> >
> > > > X = (cos(theta)+sin(theta)) d/dr + (cos(theta)-sin(theta))/r d/dtheta
> >
> > > Now, the coordinate equivalence is
> >
> > > <d/dr, d/dtheta>
> >
> > It's a coordinate basis, not any "equivalence". Don't use your private
> > terminology, we are not clairvoyant around here.
>
> You wrote down
>
> X = (cos(theta) + sin(theta)) d/dr + ((cos(theta) - sin(theta)) / r)
> d/dtheta
>
> And the above equation can be written as
>
> X = <cos(theta) + sin(theta), (cos(theta) - sin(theta)) / r> * <d/dr,
> d/dtheta>
>
> Again, X is a dot product of a metric and a coordinate vector.
This is not a dot product. In fact this object has nothing to do with
any metric.
> > > I did not check your math. Assuming it is correct, the metric is
> >
> > > <cos(theta) + sin(theta), (cos(theta) - sin(theta)) / r>
> >
> > > > And you claim that this is a different tensor????
> >
> > > X = X, of course
> >
> > > However, the choice of coordinate system is different. Not
> > > surprisingly, the metric is different as well.
> >
> > What metric? I have not specified ANY metric. The question is: did X
> > change when I switched from Cartesians to polars? Yes/no?
>
> Your metric whether you try to deny it or not is
>
> <cos(theta) + sin(theta), (cos(theta) - sin(theta)) / r>
No, there is no metric in my question. Just a 2D linear space R^2 with
two coordinate systems on it, plus a vector field.
> The corresponding coordinate system as observer's choice is
>
> <d/dr, d/dtheta>
>
> THE METRIC IS A FUNCTION OF THE OBSERVER'S CHOICE OF COORDINATE
> SYSTEM.
No. Metric - just like curvature or a vector field - is a tensor, hence
a coordinate-independent object. Coordinates are typically used to pin
down or define a particular tensor, but the tensor so defined does not
depend on coordinates.
> SINCE THE CURVATURE OF SPACE IS INDEPENDENT OF ANY OBSERVERS,
Nor the metric. Nor any tensor whatsoever.
> AS THE OBSERVER'S CHOICE OF COORDINATE SYSTEM VARIES FROM ONE
> OBSERVER TO ANOTHER, THE METRIC HAS TO VARY WITH EACH OBSERVER AS WELL.
No. Tensors do not depend on coordinates. (Their components do, of
course.)
> THUS, THE METRIC HAS TO BE COORDINATE DEPENDENT.
False hypothesis, false conclusion.
> THIS IS THE ONLY WAY
> TO GUARANTEE A CURVATURE OF SPACE INDPENDENT OF ANY OBSERVER AND HIS
> CHOICE OF COORDINATE SYSETM.
No.
> > > Well, now you have to agree with me that I am the only one who is able
> > > to understand the curvature of space or spacetime since Riemann. It is
> > > getting very lonely.
> >
> > Obviously I disagree, you not only don't understand curvature, you need
> > to learn manifold basics first.
>
> You will agree after you have learned your dot product.
You referered to "<1, 1> * <d/dx, d/dy>" as a "dot product", so it is
you who has a problem here.
--
Jan Bielawski
JanPB
> On Nov 1, 10:57 am, "Igor" <thoov...@excite.com> wrote:
> > Koobee Wublee wrote:
>
> > > Yes, the curvature is independent of the coordinate system.
>
> In order for the above statement to be true, the metric must be a
> function of the choice of coordinate system.
Not the metric, the _metric components_. The metric doesn't change
under coordinate changes. It's a tensor.
--
Jan Bielawski
sal
> Koobee Wublee wrote:
>> On Oct 31, 9:50 pm, "JanPB" <film...@gmail.com> wrote:
>> > Koobee Wublee wrote:
>> > [ snip a bunch of nonsense and empty posturing ]
>> > > Well, now you have to agree with me that I am the only one who is
>> > > able to understand the curvature of space or spacetime since
>> > > Riemann. It is getting very lonely.
>> >
>> > Obviously I disagree, you not only don't understand curvature, you
>> > need to learn manifold basics first.
>>
>> You will agree after you have learned your dot product.
>
> You referered to "<1, 1> * <d/dx, d/dy>" as a "dot product", so it is you
> who has a problem here.
Oh, stop, this is getting too funny.
Off hand I'd say KW has no idea what you're talking about.
Koobee Wublee
I have to say it does not make sense to you because you do not
understand what dot product is.
> 1. You originally wrote "What you have written down has a coordinate
> equivalence of <d/dx, d/dy> and a metric of <1, 1>", and this is what I
> objected to.
That indicates you do not understand what dot product is.
> 2. "<1, 1> * <d/dx, d/dy>" is not any "dot product" - it seems you need
> to brush up on basic linear algebra as well.
You definitely do not know what dot product is.
> 3. "You need to learn what dot product is and then inner product" -
> says the guy who refers to
> "<1, 1> * <d/dx, d/dy>" as a "dot product". Unbelievable.
Yes, unbelievable and sad, all that education wasted without knowing
what dot product is.
[...rest snipped due to Mr. Bielawski's lack of understanding of what
dot product is]
Dirk Van de moortel
"sal" <pragm...@nospam.org> wrote in message news:pan.2006.11.01....@nospam.org...
It doesn't really matter whether he has an idea.
What matters is that he gets lots of attention :-)
Dirk Vdm
sal
>
>
> On Nov 1, 12:59 pm, "JanPB" <film...@gmail.com> wrote:
>> Koobee Wublee wrote:
>> > On Oct 31, 9:50 pm, "JanPB" <film...@gmail.com> wrote:
>> > > Koobee Wublee wrote:
>> > > > On Oct 31, 5:49 pm, "JanPB" <film...@gmail.com> wrote:
>> > > > [ ... ]
You, on the other hand, have no clue what a **vector** is.
And you don't even know that you don't know.
sal
You're probably right, and I should get out of this thread and stop
feeding his ego. But on the other hand I've found it extremely
enlightening -- I had mistakenly thought that K-W knew a _lot_ more than
he actually knows.
Koobee Wublee
On Nov 1, 1:01 pm, "JanPB" <film...@gmail.com> wrote:
> Koobee Wublee wrote:
> > Yes, the curvature is independent of the coordinate system.
>
> > In order for the above statement to be true, the metric must be a
> > function of the choice of coordinate system.
>
> Not the metric, the _metric components_. The metric doesn't change
> under coordinate changes. It's a tensor.
Consider the following 2-dimensional space.
ds^2 = dx^2 + dy^2
Where
** (x, y) = Euclidean coordinate
The metric is
[1, 0]
[0, 1]
Now, the following also represents the same 2-dimensional space.
ds^2 = dr^2 + r^2 dh^2
Where
** (r, h) = Symmetrical polar coordinate
The metric is
[1, 0]
[0, r^2]
So, you argue for the metrics to be the same but just the elements that
is different. The determinant of the 1st metric is 1. The determinant
of the 2nd metric is r^2. The determinants are different. They cannot
be the same metric. You are totally wrong.
The next subject after this would be the tensor in which the metric is
dependent on the choice of coordinate system. Although a metric
dependency on choice of an observer's coordinate system does not
affect the geodesic equations, the tensor based on coordinate
independency is the very foundation in which the field equations are
based on. I am going to rock it sooner or later. Starting to sweat
bullets yet?
As I said before, I am the only one since Riemann to have thoroughly
understood the curvature of space or spacetime, and this is no
under-statement.
Mike
Koobee Wublee wrote:
> On Nov 1, 7:17 am, "Mike" <elea...@yahoo.gr> wrote:
>
> > Koobee, Koobee, Koobee.... in simple words, curvature is the amount by
> > which a geometric object deviates from being flat. Obviously, this
> > property of geometric objects is independent of the coordinate system
> > used, naturally.
>
> Yes, the curvature is independent of the coordinate system. Yes, the
> curvature is independent of any observer. However, when comes an
> observer using his own coordinate system, he is going to measure a
> distorted amount because he cannot perceive a curvature in space or
> spacetime. He sees everything straight despite bending photons. Since
> the mathematics is laid out using the observer's coordinate, as the
> coordinate changes from observer to observer, the metric must change
> with each coordinate to accommodate an invariant curvature in
> spacetime.
I think this dispute is silly. I will not attack KB as several other
posters did and seem to do out of a habbit in the ngs. KB says the
metric changes and he really means the components of the metric.
> Mathematically, we have
>
> ds^2 = Q1^ij q1_ij = Q2^ij q2_ij = Q3^ij q3_ij
>
> Where
>
> ** ds^2 = Invariant from observer to observer
> ** q1, q2, q3 = Coordinate system of observer 1, 2, 3
> ** Q1, Q2, Q3 = Metric 1, 2, 3 of each observer
>
> If you can understand this very simple concept, you are better than the
> professors and the phds.
I will try to correct everybody here, including Ph.D's, professors,
lunatics, cranks, crackpots, idtiots, etc. by saying the following:
Although carvuture, segment length, etc. do not depend on the choice of
coordinates, but depend oin the metric and manifold only, the metric
components (and units) depend on the choice of coordinate system.
Thus, the explicit form of the metric depends on the coordinate system
used, whether Cartesian, polar, etc., and thus, in loose language
(usually used by engineers or by other para-physical professions (c)
Mike) one could say that the metric is coordinate dependent although
the measurements are not.
This is really another conflict between strict mathematical formalism
and mathematical aesthetics of common sense KB promotes. One argument
against the stand of KB is that there can be no progress in math
outside of a formalistic framework but only satisfaction of the same
form obtained by virtual reality sexual encounters as opposed to hard,
physical, plain and rigorous sex.
Mike
Dirk Van de moortel
Yes, well... I used to think he knew at least slightly more than he
seems to fail to know now. A bit like Ken Tucker, you know.
And perhaps he even does. As I said, it does't matter, he has
found out that he gets a *lot* more ego-food this way, and
*that* is what these mental cases are after :-)
Dirk Vdm
Dirk Van de moortel
"Mike" <ele...@yahoo.gr> wrote in message news:1162418213.3...@e64g2000cwd.googlegroups.com...
Why don't you first try to calculate how many days it takes
to dig a hole that fits you?
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/BrainHoles.html
and then tell us where the hell you "stydied" phyiscs?
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/PhysicsHell.html
Dirk Vdm
Mike
Dirk Van de moortel wrote:
I made a wrong calculation once and I admitted my mistake and corrected
it. Attacking someone who admits a mistake and corrects it is unethical
behavior.
Your postings are proof that you have no ethics. As a consequence you
are no physicist because all of them I know have an ethics standard you
do not have.
This is the reason that you are not employed as a physicist and you
spent 365/24/7 trying to be a man in these ngs.
have you tried one of those pumps? maybe that will help you leave us
alone idiot.
Mike
>
> Dirk Vdm
Mike
Dirk Van de moortel wrote:
Do you have yet another apology to make to your fellow physicists in
Africa?
Mike
>
> Dirk Vdm
Hexenmeister
"JanPB" <fil...@gmail.com> wrote in message
news:1162414740....@e3g2000cwe.googlegroups.com...
| Nothing you write makes sense:
|
| 1. You originally wrote "What you have written down has a coordinate
| equivalence of <d/dx, d/dy> and a metric of <1, 1>", and this is what I
| objected to.
| 2. "<1, 1> * <d/dx, d/dy>" is not any "dot product" - it seems you need
| to brush up on basic linear algebra as well.
| 3. "You need to learn what dot product is and then inner product" -
| says the guy who refers to
| "<1, 1> * <d/dx, d/dy>" as a "dot product". Unbelievable.
| any metric.
| two coordinate systems on it, plus a vector field.
| No.
| You referered to "<1, 1> * <d/dx, d/dy>" as a "dot product", so it is
| you who has a problem here.
| --
| Jan Bielawski
"No... <your response>"
See if you can say "Yes" for a change, Bilewacky.
Androcles
Hexenmeister
| Not <snip> doesn't <snip>
Construct a sentence without "not" in it, Bilewacky.
Eric Gisse
Koobee Wublee wrote:
> On Nov 1, 10:57 am, "Igor" <thoov...@excite.com> wrote:
> > Koobee Wublee wrote:
>
> > > Yes, the curvature is independent of the coordinate system.
>
> In order for the above statement to be true, the metric must be a
> function of the choice of coordinate system. If the metric in
> invariant, then the curvature of space or spacetime will change with
> the choice of coordinate system.
NO, it WON'T.
Open a book that touches on differential geometry.
>
> > Now stop right there and say no other stupid things and you've got it
> > correct for a change. Case closed.
>
> What you have learned is wrong. That is why you are such confused.
> Better read on.
So you know differential geometry better than Misner, Thorne, and
Wheeler? You know differential geometry better than Sean Carroll? You
know differential geometry better than Ray d'Inverno? Better than Wald,
do Camro?
You must obviously think you do because I can recall FIVE books off the
top of my head that explicitly disagree with your "conclusion".
Eric Gisse
Who gives a shit about the determinant? If you have no cross terms,
that is simply the trace of the metric in a particular coordinate
system.
What is the CURVATURE SCALAR? When are you going to compute the
CURVATURE SCALAR?
>
> The next subject after this would be the tensor in which the metric is
> dependent on the choice of coordinate system. Although a metric
> dependency on choice of an observer's coordinate system does not
> affect the geodesic equations, the tensor based on coordinate
> independency is the very foundation in which the field equations are
> based on. I am going to rock it sooner or later. Starting to sweat
> bullets yet?
Oh, so a change of coordinates doesn't change the geodesic equation but
it changes the curvature scalar?
>
> As I said before, I am the only one since Riemann to have thoroughly
> understood the curvature of space or spacetime, and this is no
> under-statement.
Really?
Then why don't you compute the curvature scalar? Do I have to do it for
you?
Dirk Van de moortel
"Mike" <ele...@yahoo.gr> wrote in message news:1162419247.0...@h48g2000cwc.googlegroups.com...
Take the ones in Alaska as well :-)
How did your wife enjoy her holiday, where did she go again?
- Alaska.
No it's ok, I'll ask her myself.
:-)
Dirk Vdm
Hexenmeister
"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
in message news:Wz82h.161151$Ln3.2...@phobos.telenet-ops.be...
So anxious to be heard, yet never listened to.
What matters is that the fucking cowardly psychopath Dork Van de merde
gets no attention :-)
BTW, How are you getting on with Nicky Vroom?
Did you tell him about the French Concorde crash because the stupid frogs
left shit on the runway?
http://video.google.co.uk/videoplay?docid=-4825616503797900495&q=Concorde
At least the frogs are not so stupid as to get their arses kicked
anywhere but Trafalgar and Waterloo. Hitler stomped on Waterloo
and nobody noticed.
In Britain "loo" (a derivation of Waterloo) is common slang for a potty.
HAHAHAHAHA!
You are funnier than Wilson Rabbidge!
Hexenmeister
|
| Dirk Van de moortel wrote:
If you quit answering the local village dog tord it might go away, Mike.
All it seeks is attention, not even Nicky Vroom buys its shit for all
its best efforts.
Androcles
sal
Aha. Some things are becoming clearer about K-W. Here, he is not just
being a complete idiot and spouting gibberish. He is, rather, making a
sophomoric error: He is confusing linear operators with bilinear
operators.
The metric, after all, _looks_ like a linear operator on a vector space,
and in fact when applied to a single vector, in its "one-up/one-down" form
it _is_ a linear operator. However, the transformation rules for a
bilinear operator (in this case, a rank 2 tensor with both indices "down")
are subtly different from the transformation rules for a linear operator,
and this is the point K-W is apparently ignorant of: The determinant of
(the matrix of) a linear operator is invariant under a change of basis;
the determinant of (the matrix of) a bilinear operator is not.
It's obvious if you look at the transformations used to map the components
from one basis to another, but I doubt very much that K-W even knows what
the transformation rules are for bilinear operators. In fact, after
reading this thread, I very much doubt that he's ever gotten past freshman
linear algebra ... which doesn't normally include a treatment of such
things. In short, he doesn't know what a tensor is.
(For the benefit of any lurkers, bilinear operators transform by
multiplying on one side by the (reverse) change of basis matrix and on the
other side by its transpose. Linear operators transform by multiplying on
one side by the (reverse) change of basis matrix and on the other side by
its inverse. A subtle but important difference.)
> What is the CURVATURE SCALAR? When are you going to compute the
> CURVATURE SCALAR?
He can't. That's obvious.
JanPB
"<1, 1> * <d/dx, d/dy>" is not a dot product. You said it was and it is
nonsense to claim that.
> [...rest snipped due to Mr. Bielawski's lack of understanding of what
> dot product is]
"<1, 1> * <d/dx, d/dy>" is not a dot product. Never was, never shall
be.
You still haven't answered the question about the vector field X: is it
or is it not the same vector field after the coordinate switch?
--
Jan Bielawski
JanPB
Read my lips: there is no such thing as "determinant of a metric". It
is linear maps - not metrics - that possess determinants.
> The next subject after this would be the tensor in which the metric is
> dependent on the choice of coordinate system. Although a metric
> dependency on choice of an observer's coordinate system does not
> affect the geodesic equations, the tensor based on coordinate
> independency is the very foundation in which the field equations are
> based on. I am going to rock it sooner or later. Starting to sweat
> bullets yet?
Surely you jest.
> As I said before, I am the only one since Riemann to have thoroughly
> understood the curvature of space or spacetime, and this is no
> under-statement.
If it's no under-statement then what is it? An over-statement, I
suppose? Yes, I agree. A massive one.
Simply put, it's 100% baloney. You don't even know linear algebra
basics. Forget differential geometry and curvature.
--
Jan Bielawski
badd...@yahoo.com
Koobee Wublee wrote:
> On Oct 31, 10:12 am, carlip-nos...@physics.ucdavis.edu wrote:
>
> > If I describe a plane with polar coordinates, does it become curved?
>
> No, a plane is still flat. Polar coordinate describes curved
> projections. To properly describe the flat plane, you need the metric
> associated with the polar coordinate.
>
*snip*
> In a flat spacetime, we can describe it as follows.
>
> ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2
>
You are confusing the metric with the line element and further
confusing a particular representation with the tensor equations.
The metric, g_ij, is a rank 2 tensor. The line element, ds^2, is
a scalar (rank 0 tensor) contraction of the metric with the
coordinates,
ds^2 = g_ij dx^i dx^j. ds^2 and g_ij are invariants. For a
particular choice of representation (coordinate system), the
components are not invariants, but transform covariantly so that
the tensors remain invariant. Physicists typically refer to both
the metric tensor and the line element as the metric, assuming
that the meaning is clear from context. It is up to the reader
to know if he doesn't understand what he is reading.
*snip*
> Clearly, the two metrics are different.
The metrics (line elements in this case) are not different.
Your choice of representation is different. The line element
is still ds^2 = g_ij dx^i dx^j.
*snip*
> Thus, a metric along cannot determine if spacetime is curved or not.
The vanishing of the Riemann tensor is both necessary and
sufficient to determine if there is curvature and it may be
calculated from the second derivatives of the metric tensor.
2 R_abcd = d_c d_b g_ad + d_a d_d g_cb
- d_b d_d g_ac - d_a d_c g_db
Mike
badd...@yahoo.com wrote:
> Koobee Wublee wrote:
> > On Oct 31, 10:12 am, carlip-nos...@physics.ucdavis.edu wrote:
> >
> > > If I describe a plane with polar coordinates, does it become curved?
> >
> > No, a plane is still flat. Polar coordinate describes curved
> > projections. To properly describe the flat plane, you need the metric
> > associated with the polar coordinate.
> >
>
> *snip*
>
> > In a flat spacetime, we can describe it as follows.
> >
> > ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2
> >
>
> You are confusing the metric with the line element and further
> confusing a particular representation with the tensor equations.
> The metric, g_ij, is a rank 2 tensor. The line element, ds^2, is
> a scalar (rank 0 tensor) contraction of the metric with the
> coordinates,
> ds^2 = g_ij dx^i dx^j. ds^2 and g_ij are invariants. For a
> particular choice of representation (coordinate system), the
> components are not invariants, but transform covariantly so that
> the tensors remain invariant. Physicists typically refer to both
> the metric tensor and the line element as the metric, assuming
> that the meaning is clear from context. It is up to the reader
> to know if he doesn't understand what he is reading.
That pretty much sums it up. Honestly, you should publish something
along these lines in AJP so that professors clear up the confusion
upfront and students like K-W see the distinction.
IMO it is all a matter of convention and rigorous math and K-W has a
point assuming the dinstinctions are not clear in some texts.
Mike
Eric Gisse
If you actually understand what is being written, why you just sitting
on the sidelines while nodding your head in agreement rather than
actually telling this moron how wrong he is?
>
> IMO it is all a matter of convention and rigorous math and K-W has a
> point assuming the dinstinctions are not clear in some texts.
The distinctions are made ABUNDANTLY clear in every textbook I have
ever seen. Any textbook that even uses the word "covariance" WILL
explain this point. It isn't the fault of the textbook authors that KW
is a flaming moron with delusions of competency. He proclaims himself a
"guru" in differential geometry yet he doesn't even have a grasp of
linear algebra.
carlip...@physics.ucdavis.edu
> On Oct 31, 10:12 am, carlip-nos...@physics.ucdavis.edu wrote:
>> If I describe a plane with polar coordinates, does it become curved?
> No, a plane is still flat. Polar coordinate describes curved
> projections. To properly describe the flat plane, you need the metric
> associated with the polar coordinate.
> ** (r, h) = Polar coordinate in two dimensions
> ** diag(1, r^2) = The metric.
>> If I describe the Earth using a Mercator projection, does it become
>> flat?
> No, the earth is spherical. Mercator projection is flat. To properly
> describe the surface of the spherical earth, you need the metric
> associated with the Mercator projection.
>> A coordinate system is a human-made choice of how to label points.
>> Do you really believe that curvature depends on your particular
>> choice of how to label points, and that it will change if I choose
>> a different labeling system?
> No, the curvature of space or spacetime exists independent of man-made
> labeling points. However, man-made labeling points do not properly
> describe the curvature of space or spacetime. That is why we need a
> metric to complete the description of this curvature in space or
> spacetime. Since there are an infinite number of choices in coordinate
> system, to describe the same space or spacetime, there needs to be a
> unique metric associated with each coordinate system to do so.
This is wrong, but it's a common and understandable error for someone
who hasn't studied differential geometry. When you change coordinates,
what changes is not the metric, but the *components* of the metric.
There's a difference.
As a simple analogy, think of a vector on an ordinary flat plane, pointing
along the x axis. To make the arithmetic simple, suppose the vector has
length 5. Geometrically, that vector can be represented as an arrow; it
exists independent of any coordinate system. If you want to describe its
*components* in your coordinate system, you can represent them as a pair
(0,5).
Now rotate you axes 30 degrees, while leaving the vector -- the arrow --
unchanged. In your new coordinates, the components of the vector will be
(3,4). But the vector hasn't changed; all that has changed is your
arbitrary choice of how to describe it.
In the same way, the metric tensor does not depend on a choice of coordinates.
Its *components* in a particular coordinate system depend on which particular
coordinate system you choose, but that's not because the metric is different,
it's because you've chosen a different way to describe it. That's what tensor
analysis is all about.
Now, if you want to know whether a space (or a spacetime) is curved, the
object you look at is the curvature tensor. This, like the metric -- and
like the vector on the plane -- has an objective existence, independent of
your choice of how to describe it. Its *components*, on the other hand,
depend on what coordinates you use. But one of the things that goes along
with the objective existence of the curvature tensor is that if it has nonzero
components in one coordinate system, it will have nonzero components in all
coordinate systems. So the question, "Is this spacetime curved" has a real,
objective answer that does not depend on what coordinates you use to describe
the curvature or the metric.
[...]
> Thus, a metric along cannot determine if spacetime is curved or not.
> One needs to throw in the associated coordinate system to complete the
> analysis.
No. Compute the curvature tensor of the metric, in any coordinate system
you like. If the curvature tensor has nonzero components in any coordinate
system, it has nonzero compnents in all coordinate systems.
Steve Carlip
Mike
My initial responce to him:
I'm not going to call him a moron because my sense is that K-W is a
nice person who is just confused about some specific notions of
differential geometry. I think Roberts, badd and Carlip explained to
him his problem and he must now concede. If he does not concede and
continuous to insist then he may act as a moron but still I'm not going
to call him names because he is not insulting other people, he is a
polite individual.
Mike
Koobee Wublee
> Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > ...the curvature of space or spacetime exists independent of man-made
> > labeling points. However, man-made labeling points do not properly
> > describe the curvature of space or spacetime. That is why we need a
> > metric to complete the description of this curvature in space or
> > spacetime. Since there are an infinite number of choices in coordinate
> > system, to describe the same space or spacetime, there needs to be a
> > unique metric associated with each coordinate system to do so.
>
> This is wrong, but it's a common and understandable error for someone
> who hasn't studied differential geometry. When you change coordinates,
> what changes is not the metric, but the *components* of the metric.
> There's a difference.
We do agree that the real (or actual) space which is commonly called
the proper space must be observer independent, don't we? Assuming
so, we have
ds^2 = gI_ij dqI^i dqI^j = gJ_ij dqJ^i dqJ^j
Where
** ds = Actual segment of real space
** gI = Metric associated with observer I's choice of coordinate, qI
** gJ = Metric associated with observer J's choice of coordinate, qJ
** i, j = 1, 2, 3 that represent the 3 spatial dimensions
Since ds^2 is invariant, that is very audacious to say (gI = gJ) when
(qI != qJ).
> As a simple analogy, think of a vector on an ordinary flat plane, pointing
> along the x axis. To make the arithmetic simple, suppose the vector has
> length 5. Geometrically, that vector can be represented as an arrow; it
> exists independent of any coordinate system. If you want to describe its
> *components* in your coordinate system, you can represent them as a pair
> (0,5).
>
> Now rotate you axes 30 degrees, while leaving the vector -- the arrow --
> unchanged. In your new coordinates, the components of the vector will be
> (3,4). But the vector hasn't changed; all that has changed is your
> arbitrary choice of how to describe it.
It appears to be a good analogy. Unfortunately, in this case, (gI =
gJ) because in each case it is the same scenario of flat space. Given
the same flat space in 2 dimensions, the following example would be
better.
ds^2 = dx^2 + dy^2 = dr^2 + r^2 dtheta^2 = Flat space
Now, we have
** gI = [1, 0], dqI^2 = [dx dx, dx dy]
** gI = [0, 1], dqI^2 = [dy dx, dy dy]
** gJ = [1, 0], dqJ^2 = [dr dr, dr dtheta]
** gJ = [0, r^2], dqj^2 = [dtheta dr, dtheta dtheta]
How can you justify (gI = gJ) since obviously (dqI^2 != dqJ^2)?
> In the same way, the metric tensor does not depend on a choice of coordinates.
> Its *components* in a particular coordinate system depend on which particular
> coordinate system you choose, but that's not because the metric is different,
> it's because you've chosen a different way to describe it. That's what tensor
> analysis is all about.
Is not gJ (a metric) a function of r while gI is an identity matrix
which is constant?
I believe the tensor analysis is wrong, but that is another chapter of
discussion. In the meantime, we have to iron out the very fundamental
concept of curved space.
> Now, if you want to know whether a space (or a spacetime) is curved, the
> object you look at is the curvature tensor. This, like the metric -- and
> like the vector on the plane -- has an objective existence, independent of
> your choice of how to describe it. Its *components*, on the other hand,
> depend on what coordinates you use. But one of the things that goes along
> with the objective existence of the curvature tensor is that if it has nonzero
> components in one coordinate system, it will have nonzero components in all
> coordinate systems. So the question, "Is this spacetime curved" has a real,
> objective answer that does not depend on what coordinates you use to describe
> the curvature or the metric.
To gage how space is curved requires a man-made gage. In this case, it
is the mathematical structure of the Riemann curvature tensor designed
by Ricci himself. At this moment, I'd rather not go there because we
have not agreed on the very basic concept of the curvature in space
yet.
Koobee Wublee
On Nov 2, 4:40 pm, carlip-nos...@physics.ucdavis.edu wrote:
> Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > ...the curvature of space or spacetime exists independent of man-made
> > labeling points. However, man-made labeling points do not properly
> > describe the curvature of space or spacetime. That is why we need a
> > metric to complete the description of this curvature in space or
> > spacetime. Since there are an infinite number of choices in coordinate
> > system, to describe the same space or spacetime, there needs to be a
> > unique metric associated with each coordinate system to do so.
>
> This is wrong, but it's a common and understandable error for someone
> who hasn't studied differential geometry. When you change coordinates,
> what changes is not the metric, but the *components* of the metric.
> There's a difference.
We do agree that the real (or actual) space must be observer
independent, don't we? Assuming so, we have
ds^2 = gI_ij dqI^i dqI^j = gJ_ij dqJ^i dqJ^j
Where
** ds = Actual segment of real space
** gI = Metric associated with observer I's choice of coordinate, qI
** gJ = Metric associated with observer J's choice of coordinate, qJ
** i, j = 1, 2, 3 that represent the 3 spatial dimensions
Since ds^2 is invariant, that is very audacious to say (gI = gJ) when
(qI != qJ).
> As a simple analogy, think of a vector on an ordinary flat plane, pointing
> along the x axis. To make the arithmetic simple, suppose the vector has
> length 5. Geometrically, that vector can be represented as an arrow; it
> exists independent of any coordinate system. If you want to describe its
> *components* in your coordinate system, you can represent them as a pair
> (0,5).
>
> Now rotate you axes 30 degrees, while leaving the vector -- the arrow --
> unchanged. In your new coordinates, the components of the vector will be
> (3,4). But the vector hasn't changed; all that has changed is your
> arbitrary choice of how to describe it.
It appears to be a good analogy. Unfortunately, in this case, (gI =
gJ) because in each case it is the same scenario of flat space. Given
the same flat space in 2 dimensions, the following example would be
better.
ds^2 = dx^2 + dy^2 = dr^2 + r^2 dtheta^2 = Flat space
Now, we have
** gI = [1, 0], dqI^2 = [dx dx, dx dy]
** gI = [0, 1], dqI^2 = [dy dx, dy dy]
** gJ = [1, 0], dqJ^2 = [dr dr, dr dtheta]
** gJ = [0, r^2], dqj^2 = [dtheta dr, dtheta dtheta]
How can you justify (gI = gJ) since obviously (dqI^2 != dqJ^2)?
> In the same way, the metric tensor does not depend on a choice of coordinates.
> Its *components* in a particular coordinate system depend on which particular
> coordinate system you choose, but that's not because the metric is different,
> it's because you've chosen a different way to describe it. That's what tensor
> analysis is all about.
Is not gJ (a metric) a function of r while gI is an identity matrix
which is constant?
I believe the tensor analysis is wrong, but that is another chapter of
discussion. In the meantime, we have to iron out the very fundamental
concept of curved space.
> Now, if you want to know whether a space (or a spacetime) is curved, the
> object you look at is the curvature tensor. This, like the metric -- and
> like the vector on the plane -- has an objective existence, independent of
> your choice of how to describe it. Its *components*, on the other hand,
> depend on what coordinates you use. But one of the things that goes along
> with the objective existence of the curvature tensor is that if it has nonzero
> components in one coordinate system, it will have nonzero components in all
> coordinate systems. So the question, "Is this spacetime curved" has a real,
> objective answer that does not depend on what coordinates you use to describe
> the curvature or the metric.
To gage how space is curved requires a man-made gage. In this case, it
Eric Gisse
Koobee Wublee wrote:
[...snip ten thousand misconceptions...]
A) Nobody ever claimed the _form_ of the metric is
coordinate-independent.
B) COMPUTE THE FUCKING CURVATURE SCALAR LIKE I HAVE TOLD YOU A DOZEN
TIMES ALREADY.
JanPB
>
> We do agree that the real (or actual) space must be observer
> independent, don't we? Assuming so, we have
>
> ds^2 = gI_ij dqI^i dqI^j = gJ_ij dqJ^i dqJ^j
>
> Where
>
> ** ds = Actual segment of real space
> ** gI = Metric associated with observer I's choice of coordinate, qI
> ** gJ = Metric associated with observer J's choice of coordinate, qJ
> ** i, j = 1, 2, 3 that represent the 3 spatial dimensions
OK.
> Since ds^2 is invariant, that is very audacious to say (gI = gJ) when
> (qI != qJ).
Nobody claims that gI_ij = gJ_ij. They are different in general since
these are components of the metric wrt two different coordinate
systems. The metric tensor itself is gI_ij dqI^i dqI^j which _is_ equal
to gJ_ij dqJ^i dqJ^j.
--
Jan Bielawski
Koobee Wublee
> Koobee Wublee wrote:
>
> > We do agree that the real (or actual) space must be observer
> > independent, don't we? Assuming so, we have
>
> > ds^2 = gI_ij dqI^i dqI^j = gJ_ij dqJ^i dqJ^j
>
> > Where
>
> > ** ds = Actual segment of real space
> > ** gI = Metric associated with observer I's choice of coordinate, qI
> > ** gJ = Metric associated with observer J's choice of coordinate, qJ
> > ** i, j = 1, 2, 3 that represent the 3 spatial dimensions
>
> OK.
Good! Excellent!
> > Since ds^2 is invariant, that is very audacious to say (gI = gJ) when
> > (qI != qJ).Nobody claims that gI_ij = gJ_ij. They are different in general since
> these are components of the metric wrt two different coordinate
> systems. The metric tensor itself is gI_ij dqI^i dqI^j which _is_ equal
> to gJ_ij dqJ^i dqJ^j.
gI_ij dqI^i dqI^j = gJ_ij dqJ^i dqJ^j = ds^2 = Scalar
No, you are absolutely confused. (gI_ij dqI^i dqI^j) nor (gJ_ij dqJ^i
dqJ^j) are tensors. They are scalars.
Still, the metrics are gI_ij and gJ_ij WHICH ARE TENSORS and are the
victims of the errors created by tensor analyses.
If (dqI^i dqI^j != dqJ^i dqJ^j) due to freedom of choices in different
coordinate systems, then (gI_ij != gJ_ij) to keep ds^2 invariant from
observer's choice of coordinate system to observer's choice of
coordinate system.
Just how hard can this be?
After swallowing this hard, we can now move on to address the fallacy
in tensor analysis.
Yes, after seeing the light, you can stop sweating bullets now. The
conclusion is inevitable. The road to GR only points to absurdity
whether you accept the "Checkmate" move now or later, but it will
happen.
Tom Roberts
> ds^2 = gI_ij dqI^i dqI^j = gJ_ij dqJ^i dqJ^j
> Where
> ** ds = Actual segment of real space
> ** gI = Metric associated with observer I's choice of coordinate, qI
> ** gJ = Metric associated with observer J's choice of coordinate, qJ
> ** i, j = 1, 2, 3 that represent the 3 spatial dimensions
This is supposed to be GR, and i,j={0,1,2,3} representing the 4
spaceTIME coordinates. But this all works in a manifold of any dimension.
> Since ds^2 is invariant, that is very audacious to say (gI = gJ) when
> (qI != qJ).
Your last claim is wrong, and is the crux of your confusion.
Yes indeed, gI = gJ, where that is a SINGLE equation BETWEEN THE TENSORS
-- in a textbook both gI and gJ would be bold, indicating they are
tensors. But gI_ij != gJ_ij, where that is a SET of 16 equations BETWEEN
THEIR COMPONENTS (i.e. each of those 16 equations is a relationship
between real numbers) -- in a textbook neither the gI_ij nor the gJ_ij
would be in bold, as they are all real numbers, not tensors.
Moreover, one can explicitly write:
gI = gI_ij DxI^i DxI^j = gJ = gJ_uv DxJ^u DxJ^v
where to distinguish the exterior derivative "D" from your (real)
differential "d" I have used the unusual notation of capitalizing the
"D" rather than putting it in bold (which I cannot do in this ASCII
medium). The {DxI^i} are the basis 1-forms used for expanding gI, and
the {DxJ^u} are the basis 1-forms used for expanding gJ, but still gI = gJ.
[This is just a generalization of the usual way of
writing a vector in terms of its components:
V = V^i e_i
Where the V and the e are both bold (indicating
they are vectors), and the V^i are not bold because
vector components are real numbers. The {e_i} are
of course the basis vectors.]
Note that gI = gJ is an invariant equation BETWEEN TENSORS, and that the
{gI_ij} and the {gJ_uv} are not themselves invariant -- only the
specific combination above using BOTH the components and the BASIS is
invariant.
Please, go LEARN THE BASICS. You are very confused.
Tom Roberts
JanPB
> On Nov 2, 11:16 pm, "JanPB" <film...@gmail.com> wrote:
> > Koobee Wublee wrote:
> >
> > > We do agree that the real (or actual) space must be observer
> > > independent, don't we? Assuming so, we have
> >
> > > ds^2 = gI_ij dqI^i dqI^j = gJ_ij dqJ^i dqJ^j
> >
> > > Where
> >
> > > ** ds = Actual segment of real space
> > > ** gI = Metric associated with observer I's choice of coordinate, qI
> > > ** gJ = Metric associated with observer J's choice of coordinate, qJ
> > > ** i, j = 1, 2, 3 that represent the 3 spatial dimensions
> >
> > OK.
>
> Good! Excellent!
>
> > > Since ds^2 is invariant, that is very audacious to say (gI = gJ) when
> > > (qI != qJ).Nobody claims that gI_ij = gJ_ij. They are different in general since
> > these are components of the metric wrt two different coordinate
> > systems. The metric tensor itself is gI_ij dqI^i dqI^j which _is_ equal
> > to gJ_ij dqJ^i dqJ^j.
>
> gI_ij dqI^i dqI^j = gJ_ij dqJ^i dqJ^j = ds^2 = Scalar
>
> No, you are absolutely confused. (gI_ij dqI^i dqI^j) nor (gJ_ij dqJ^i
> dqJ^j) are tensors. They are scalars.
No, they are tensors. I know what you are trying to get at but you are
using a thoroughly antiquated notation from the time calculus was not
an airt tight science yet. Here is the breakdown of the symbol
gI_ij dqI^i dqI^j:
gI_ij => a function (one for each pair of i,j with i<=j),
dqI^i => a covector (aka. 1-form, aka. (1,0)-tensor), namely the
exterior derivative of the coordinate function qI^i (one for each i),
dqI^i dqI^j => a symmetric tensor product of two such tensors (one
such term for each pair i,j with i<=j), where "symmetric tensor
product" means "1/2 (dqI^i X dqI^j + dqI^j X dqI^i), where "X" denotes
the standard tensor product. Each such term is thus a (2,0)-tensor,
...hence gI_ij dqI^i dqI^j is a (2,0)-tensor (being a linear
combination of same).
> Still, the metrics are gI_ij and gJ_ij WHICH ARE TENSORS and are the
> victims of the errors created by tensor analyses.
No, these are the components of these tensors. They change with the
coordinates.
> If (dqI^i dqI^j != dqJ^i dqJ^j) due to freedom of choices in different
> coordinate systems, then (gI_ij != gJ_ij) to keep ds^2 invariant from
> observer's choice of coordinate system to observer's choice of
> coordinate system.
>
> Just how hard can this be?
Nobody is disagreeing with you when you say that gI_ij != gJ_ij. But
these functions are components of the metric tensor (hence they change
with coordinates) which doesn't change with coordinates.
> After swallowing this hard, we can now move on to address the fallacy
> in tensor analysis.
No, first you must learn the basics of linear algebra. I suggest you
stay away from differential geometry for now. This is all very basic
stuff.
--
Jan Bielawski
Tom Roberts
> The metric tensor itself is gI_ij dqI^i dqI^j
Your thought is correct, but your notation is ambiguous -- that looks
just like ds^2. See my other post where I discussed this in detail (I
realize you know all this).
Tom Roberts
Ken S. Tucker
Steve wrote, "That's what tensor analysis is all about."
What Steve wrote is true only *locally* as he set forth,
his use of the word "all" is at issue due to the *global*
nature of non-euclidian geometry, a good review is in
Weinberg's G&C pg.4, and one can read how the
developement of tensor analysis followed from the legal
need for accurate surveys of a globe, the Earth, hence
the term "global".
For example, a cartesian CS is impossible on a globe,
and only certain CS's are "global" as specified by the
curvature in the surveyors 2D world, and so longitude
and latitude was popularized.
In a broad sense, Earth's surface became described
by a *global field equation* to which a chosen metric
was in accord with, like lat. and long. but not cartesian.
Of course you can still use a cartesian CS in your
back vegetable garden :-).
Both SR and GR place strict specifcations on permitted
metrics that are true *universally*, using the term
universally to be true in all of space and time in place
of the archaic term "globally" that is a bit misnomer.
In that sense, the allowable metrics themselves (as KW
argues) are determined by our choice of field equations.
Regards
Ken S. Tucker
Koobee Wublee
On Nov 3, 9:02 am, "JanPB" <film...@gmail.com> wrote:
> Koobee Wublee wrote:
> > gI_ij dqI^i dqI^j = gJ_ij dqJ^i dqJ^j = ds^2 = Scalar
>
> > No, you are absolutely confused. (gI_ij dqI^i dqI^j) nor (gJ_ij dqJ^i
> > dqJ^j) are tensors. They are scalars.
>
> No, they are tensors. I know what you are trying to get at but you are
> using a thoroughly antiquated notation from the time calculus was not
> an airt tight science yet. Here is the breakdown of the symbol
> gI_ij dqI^i dqI^j:
>
> gI_ij => a function (one for each pair of i,j with i<=j),
> dqI^i => a covector (aka. 1-form, aka. (1,0)-tensor), namely the
> exterior derivative of the coordinate function qI^i (one for each i),
> dqI^i dqI^j => a symmetric tensor product of two such tensors (one
> such term for each pair i,j with i<=j), where "symmetric tensor
> product" means "1/2 (dqI^i X dqI^j + dqI^j X dqI^i), where "X" denotes
> the standard tensor product. Each such term is thus a (2,0)-tensor,
> ...hence gI_ij dqI^i dqI^j is a (2,0)-tensor (being a linear
> combination of same).
As I have already pointed out, you don't even understand what a dot
product is. The above example should be correctly written as follows.
ds^2 = gI_ij dqI^i dqI^j = SUM[i = 1 to 3](SUM[j = 1 to 3](gI_ij dqI_i
dqJ_j)) = A number
A number is a scalar.
> > Still, the metrics are gI_ij and gJ_ij WHICH ARE TENSORS and are the
> > victims of the errors created by tensor analyses.
>
> No, these are the components of these tensors. They change with the
> coordinates.
I meant gI and gJ are tensors. They should change with choice of
coordinate system.
> > If (dqI^i dqI^j != dqJ^i dqJ^j) due to freedom of choices in different
> > coordinate systems, then (gI_ij != gJ_ij) to keep ds^2 invariant from
> > observer's choice of coordinate system to observer's choice of
> > coordinate system.
>
> > Just how hard can this be?
>
> Nobody is disagreeing with you when you say that gI_ij != gJ_ij. But
> these functions are components of the metric tensor (hence they change
> with coordinates) which doesn't change with coordinates.
Let's try (gI != gJ).
> > After swallowing this hard, we can now move on to address the fallacy
> > in tensor analysis.No, first you must learn the basics of linear algebra. I suggest you
> stay away from differential geometry for now. This is all very basic
> stuff.
I suggest you to learn what a dot product is.
Koobee Wublee
> Koobee Wublee wrote:
> > ds^2 = gI_ij dqI^i dqI^j = gJ_ij dqJ^i dqJ^j
> > Where
> > ** ds = Actual segment of real space
> > ** gI = Metric associated with observer I's choice of coordinate, qI
> > ** gJ = Metric associated with observer J's choice of coordinate, qJ
> > ** i, j = 1, 2, 3 that represent the 3 spatial dimensions
>
> This is supposed to be GR, and i,j={0,1,2,3} representing the 4
> spaceTIME coordinates. But this all works in a manifold of any dimension.
Yes, you are right. I chose to work only with the 3 spatial dimensions
for simplicity in concept. We can easily add another dimension to it
later on.
> > Since ds^2 is invariant, that is very audacious to say (gI = gJ) when
> > (qI != qJ).
>
> Yes indeed, gI = gJ, where that is a SINGLE equation BETWEEN THE TENSORS
> -- in a textbook both gI and gJ would be bold, indicating they are
> tensors. But gI_ij != gJ_ij, where that is a SET of 16 equations BETWEEN
> THEIR COMPONENTS (i.e. each of those 16 equations is a relationship
> between real numbers) -- in a textbook neither the gI_ij nor the gJ_ij
> would be in bold, as they are all real numbers, not tensors.
If (gI_ij != gJ_ij), how is that (gI = gJ)?
> Moreover, one can explicitly write:
>
> gI = gI_ij DxI^i DxI^j = gJ = gJ_uv DxJ^u DxJ^v
No, you are totally confused and wrong. For example, we have
gI = [5, 8, 3]
gI = [2, 7, 9]
gI = [4, 1, 6]
Where
** gI_11 = 5
** gI_12 = 8
** gI_13 = 3
** gI_21 = 2
** gI_22 = 7
** gI_23 = 9
** gI_22 = 7
** gI_31 = 4
** gI_32 = 1
** gI_33 = 6
> [...]
Still confused with the elements and the matrix?
> Please, go LEARN THE BASICS. You are very confused.
You need to learn the basics of matrices.
Mike
[snip]
>
> You need to learn the basics of matrices.
That is definitevily a red herring here. Your original assertion based
on which the responses were provoked was:
"Thus, a metric along cannot determine if spacetime is curved or not.
One needs to throw in the associated coordinate system to complete the
analysis. "
As can be seen from this post of yours:
This statement of yours as indicated clearly by badd was an outcome of
loose use of terminology, i.e. calling a segment length a metric. See:
But as it was shown to you by Carlip in:
all you have to do is compute the curvature tensor and if it has
non-zero components then spacetime is curved. The choice of coordinate
system is irrelevant.
Therefore, your original question was answered completely and if you
still insist it is either because you do not understand it or because
you have something else in mind that you do not state very clearly and
you will have to be more rigorous and explicit about it.
Mike
Koobee Wublee
On Nov 3, 1:31 pm, "Mike" <elea...@yahoo.gr> wrote:
> Your original assertion based
> on which the responses were provoked was:
>
> "Thus, a metric along cannot determine if spacetime is curved or not.
> One needs to throw in the associated coordinate system to complete the
> analysis. "
>
> As can be seen from this post of yours:
>
> http://groups.google.gr/group/sci.physics.relativity/tree/browse_frm/...
>
> This statement of yours as indicated clearly by badd was an outcome of
> loose use of terminology, i.e. calling a segment length a metric. See:
No, it is not based on any loose use of terminology. I did not call a
segment a metric. I have shown mathematically very clearly exactly
what I meant.
> http://groups.google.gr/group/sci.physics.relativity/tree/browse_frm/...
>
> But as it was shown to you by Carlip in:
>
> http://groups.google.gr/group/sci.physics.relativity/tree/browse_frm/...
>
> all you have to do is compute the curvature tensor and if it has
> non-zero components then spacetime is curved. The choice of coordinate
> system is irrelevant.
My point is that your statement above is utterly wrong. The metric
along cannot tell you if there is any curvature in space. You also
need to know the choice of coordinate system to complete the analysis.
Do you ever read the post I applied to you? Apparently, not.
> Therefore, your original question was answered completely and if you
> still insist it is either because you do not understand it or because
> you have something else in mind that you do not state very clearly and
> you will have to be more rigorous and explicit about it.
My originally question has not been answered with a correct answer yet.
carlip...@physics.ucdavis.edu
> Where
You are confusing the *metric* (gI or gJ) with the *components of the
metric* (gI_ij or gJ_ij).
>> As a simple analogy, think of a vector on an ordinary flat plane, pointing
>> along the x axis. To make the arithmetic simple, suppose the vector has
>> length 5. Geometrically, that vector can be represented as an arrow; it
>> exists independent of any coordinate system. If you want to describe its
>> *components* in your coordinate system, you can represent them as a pair
>> (0,5).
Oops. That should have been (5,0).
>> Now rotate you axes 30 degrees, while leaving the vector -- the arrow --
>> unchanged. In your new coordinates, the components of the vector will be
>> (3,4). But the vector hasn't changed; all that has changed is your
>> arbitrary choice of how to describe it.
> It appears to be a good analogy. Unfortunately, in this case, (gI =
> gJ) because in each case it is the same scenario of flat space.
That's irrelevant. The point is that a single vector can have
different components depending on what coordinates you use, but
that doesn't change the fact that you're talking about the same
vector. The same is true for a metric.
> Given the same flat space in 2 dimensions, the following example would be
> better.
> ds^2 = dx^2 + dy^2 = dr^2 + r^2 dtheta^2 = Flat space
Fine. Here's a vector, with an objective geometric meaning:
draw a circle of radius R, and take a vector of length 5 that
is tangent to the circle where it crosses the y axis at y=-R.
In Cartesian coordinates, that vector will again have components
(5,0). In polar coordinates, it will have components (0,5/R)
[no radial component, an angular component that makes its length 5].
But it's still the same vector.
> Now, we have
> ** gI = [1, 0], dqI^2 = [dx dx, dx dy]
> ** gI = [0, 1], dqI^2 = [dy dx, dy dy]
> ** gJ = [1, 0], dqJ^2 = [dr dr, dr dtheta]
> ** gJ = [0, r^2], dqj^2 = [dtheta dr, dtheta dtheta]
> How can you justify (gI = gJ) since obviously (dqI^2 != dqJ^2)?
How can you justify that a vector pointing along the y axis is a
single vector when it has components (5,0) in Cartesian coordinates
and (0,5/R) in polar coordinates? It's the same question
>> In the same way, the metric tensor does not depend on a choice of coordinates.
>> Its *components* in a particular coordinate system depend on which particular
>> coordinate system you choose, but that's not because the metric is different,
>> it's because you've chosen a different way to describe it. That's what tensor
>> analysis is all about.
> Is not gJ (a metric) a function of r while gI is an identity matrix
> which is constant?
The *components* are constant in Cartesian coordinates and not in polar
coordinates. That's because the polar coordinate basis is a function
of r, so of course components in that basis will depend on r.
Steve Carlip
Eric Gisse
Koobee Wublee wrote:
[...]
>
> > Please, go LEARN THE BASICS. You are very confused.
>
> You need to learn the basics of matrices.
Amazing.
I wish I could display that kind of confidence even when a half dozen
people are telling me how wrong I am with explicit examples.
Koobee Wublee
> Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > ds^2 = gI_ij dqI^i dqI^j = gJ_ij dqJ^i dqJ^j
> > Where
> > ** ds = Actual segment of real space
> > ** gI = Metric associated with observer I's choice of coordinate, qI
> > ** gJ = Metric associated with observer J's choice of coordinate, qJ
> > ** i, j = 1, 2, 3 that represent the 3 spatial dimensions
> > Since ds^2 is invariant, that is very audacious to say (gI = gJ) when
> > (qI != qJ).You are confusing the *metric* (gI or gJ) with the *components of the
> metric* (gI_ij or gJ_ij).
>
> [...]
>
> > ds^2 = dx^2 + dy^2 = dr^2 + r^2 dtheta^2 = Flat space
>
> Fine. Here's a vector, with an objective geometric meaning:
> draw a circle of radius R, and take a vector of length 5 that
> is tangent to the circle where it crosses the y axis at y=-R.
> In Cartesian coordinates, that vector will again have components
> (5,0). In polar coordinates, it will have components (0,5/R)
> [no radial component, an angular component that makes its length 5].
> But it's still the same vector.
<dx, dy> in rectangular coordinate is the same as
<(x dx + y dy) / sqrt(x^2 + y^2), (x dy - y dx) / sqrt(x^2 + y^)>
in spherically symmetric polar coordinate. In your example,
(x = 0, y = -R) and (dx = 5, dy = 0), the tangent vector in polar
coordinate is indeed <0, 5 / R>. However, this is a very bad
example because it is already defined as
The actual vector = the metric * the coordinate vector.
Given a choice of coordinate vector, the metric is such coordinate
dependent that the overall result is universally agreed upon by
whatever the coordinate system one wishes to choose.
> > Now, we have
> > ** gI = [1, 0], dqI^2 = [dx dx, dx dy]
> > ** gI = [0, 1], dqI^2 = [dy dx, dy dy]
> > ** gJ = [1, 0], dqJ^2 = [dr dr, dr dtheta]
> > ** gJ = [0, r^2], dqj^2 = [dtheta dr, dtheta dtheta]
> > How can you justify (gI = gJ) since obviously (dqI^2 != dqJ^2)?
>
> How can you justify that a vector pointing along the y axis is a
> single vector when it has components (5,0) in Cartesian coordinates
> and (0,5/R) in polar coordinates? It's the same question
Again, your example is bad. A better way to write your vector is
[1, 0] [dx] = [cos(theta), - r sin(theta)] [dr]
[0, 1] [dy] = [sin(theta), r cos(theta)] [dtheta]
Where your metrics are
** [1, 0] and [cos(theta), - r sin(theta)]
** [0, 1] and [sin(theta), r cos(theta)]
And your coordinate vectors are
** [dx] and [dr]
** [dy] and [dtheta]
Since the metric [1, 0] is not the same as
Since the metric [0, 1] is not the same as
the metric [cos(theta), - r sin(theta)], the coordinate
the metric [sin(theta), r cos(theta)], the coordinate
vectors [dx] and [dr] have to be different to give
vectors [dy] and [dtheta] have to be different to give
the same overall tangent vector.
> >> In the same way, the metric tensor does not depend on a choice of coordinates.
> >> Its *components* in a particular coordinate system depend on which particular
> >> coordinate system you choose, but that's not because the metric is different,
> >> it's because you've chosen a different way to describe it. That's what tensor
> >> analysis is all about.
> > Is not gJ (a metric) a function of r while gI is an identity matrix
> > which is constant?
>
> The *components* are constant in Cartesian coordinates and not in polar
> coordinates. That's because the polar coordinate basis is a function
> of r, so of course components in that basis will depend on r.
This is so obvious. Another way to state the case is that the metric
is dependent on the choice of coordinate system.
If two observers sit at points with different curvature in space, the
same vector would be observed very differently but still the same
vector. This general case is the central theme of my point. This is
where my argument stands out.
JanPB
>
> My point is that your statement above is utterly wrong. The metric
> along cannot tell you if there is any curvature in space.
The metric alone determines the curvature tensor and therefore it alone
tells you if there is any curvature in space.
In practical computations one uses coordinate systems to get the
_components_ of the curvature tensor. This is just a computational
device, not a requirement.
--
Jan Bielawski
JanPB
> On Nov 3, 9:02 am, "JanPB" <film...@gmail.com> wrote:
> > Koobee Wublee wrote:
>
> > > gI_ij dqI^i dqI^j = gJ_ij dqJ^i dqJ^j = ds^2 = Scalar
> >
> > > No, you are absolutely confused. (gI_ij dqI^i dqI^j) nor (gJ_ij dqJ^i
> > > dqJ^j) are tensors. They are scalars.
> >
> > No, they are tensors. I know what you are trying to get at but you are
> > using a thoroughly antiquated notation from the time calculus was not
> > an airt tight science yet. Here is the breakdown of the symbol
> > gI_ij dqI^i dqI^j:
> >
> > gI_ij => a function (one for each pair of i,j with i<=j),
> > dqI^i => a covector (aka. 1-form, aka. (1,0)-tensor), namely the
> > exterior derivative of the coordinate function qI^i (one for each i),
> > dqI^i dqI^j => a symmetric tensor product of two such tensors (one
> > such term for each pair i,j with i<=j), where "symmetric tensor
> > product" means "1/2 (dqI^i X dqI^j + dqI^j X dqI^i), where "X" denotes
> > the standard tensor product. Each such term is thus a (2,0)-tensor,
> > ...hence gI_ij dqI^i dqI^j is a (2,0)-tensor (being a linear
> > combination of same).
>
> As I have already pointed out, you don't even understand what a dot
> product is. The above example should be correctly written as follows.
>
> ds^2 = gI_ij dqI^i dqI^j = SUM[i = 1 to 3](SUM[j = 1 to 3](gI_ij dqI_i
> dqJ_j)) = A number
You are using the antiquated "meaning" of the symbol ds^2 according to
which the formula above represents a "dot product of infinitesimal
vectors", and so "ds" itself is an "infinitesimal number". Nobody uses
this "definition" in the actual work anymore as it's in fact
meaningless (there are no such things as "infinitesimal numbers", let
alone "dot product of infinitesimal numbers"). Any attempt at making
this definition correct by means of limiting processes to represent
"infinitesimals" inevitably leads to the concept of *tensor* operating
on *tangent vectors*. The RHS is such a tensor, and hence so is the
LHS.
This expression becomes an honest dot product _only_ when you evaluate
it on a pair of tangent vectors and it is then equal to the dot product
of those two particular vectors and is then obviously a different
number for different vectors (none of these numbers being necessarily
"small", let alone "infinitesimal").
The reason for maintaining the notation "ds^2" instead of the plain "g"
to denote this tensor (both notations are common) is that the element
of length of a curve is usually denoted by "ds" (it is a 1-density, not
a number, again) and such element of length can be thought of loosely
as a "square root" of the tensor ds^2 *pulled back* (restricted) to the
curve:
(pulled-back-ds^2) = ||v||^2 dt^2
...where v is the curve's velocity vector (this is what the pullback
gives you) and dt is the usual coordinate 1-form on the curve. Hence:
ds = "square root of(||v||^2 dt^2)" = ||v|| dt
...which is the standard formula found even in basic vector calculus.
> A number is a scalar.
>
> > > Still, the metrics are gI_ij and gJ_ij WHICH ARE TENSORS and are the
> > > victims of the errors created by tensor analyses.
> >
> > No, these are the components of these tensors. They change with the
> > coordinates.
>
> I meant gI and gJ are tensors. They should change with choice of
> coordinate system.
Tensors never change with coordinate systems.
--
Jan Bielawski