Gravity waves, energy and momentum
Edward Green
an example of it.
Gravity waves travel through vacuum, where by definition T = 0 : no
energy or momentum flux.
Gravity waves are thought to be detectable by their effect on matter;
presumably by putting some test masses in relative motion. This
creates local energy.
So where did the energy come from?
Androcles
"Edward Green" <spamsp...@netzero.com> wrote in message
news:c6f9a50c-4732-4a32...@i20g2000prf.googlegroups.com...
Observed gravity waves:
http://www.stormy.ca/tides/
eric gisse
<spamsp...@netzero.com> wrote:
>I"ve heard that "energy conservation fails in GR" and perhaps this is
>an example of it.
It doesn't strictly fail - energy is simply a quantity that has no
single definition, none of which are true in general.
>
>Gravity waves travel through vacuum, where by definition T = 0 : no
>energy or momentum flux.
No. Perturbations of the metric (gravitational radiation) have a
nonzero stress-energy tensor, but it isn't linear in the metric so it
doesn't appear in the linearized theory.
>
>Gravity waves are thought to be detectable by their effect on matter;
>presumably by putting some test masses in relative motion. This
>creates local energy.
>
>So where did the energy come from?
The energy came from the system that emitted the gravitational
radiation in the first place.
The effect is analogous to electromagnetism: gravitational radiation
decreases the time rate of change of the mass quadrupole moment in the
same way electromagnetic radiation decreases the rate of change of the
charge dipole moment.
Igor
The source of the gravity waves. Although locally, it doesn't have to
come from anywhere, since the concept of energy is not covariant.
All we can really say in GR is that energy is conserved globally.
Edward Green
> On Fri, 31 Oct 2008 14:00:40 -0700 (PDT), Edward Green
>
> >I"ve heard that "energy conservation fails in GR" and perhaps this is
> >an example of it.
>
> It doesn't strictly fail - energy is simply a quantity that has no
> single definition, none of which are true in general.
Hmm... is that more or less problematic than the Poynting vector?
> >Gravity waves travel through vacuum, where by definition T = 0 : no
> >energy or momentum flux.
>
> No. Perturbations of the metric (gravitational radiation) have a
> nonzero stress-energy tensor, but it isn't linear in the metric so it
> doesn't appear in the linearized theory.
Now I'm more confused. I that T was the "matter" term in GR. That
presumably includes normal forms of mass, as well as light (and for
completeness neutrinos). But I didn't know it included a contribution
from the gravitational field itself (which I presume we may call
whatever it is that "perturbs" space(time) in the absence of ordinary
mass and fields (whereby "ordinary" fields I mean all those not
gravitational).
Also, I thought the "linearized" theory was a special modification,
and that standard GR was the "unlinearized" form. Is this a yet
_more_ non-linear (and ad-hoc) form you speak of, with additional
terms?
> >Gravity waves are thought to be detectable by their effect on matter;
> >presumably by putting some test masses in relative motion. This
> >creates local energy.
>
> >So where did the energy come from?
>
> The energy came from the system that emitted the gravitational
> radiation in the first place.
Presumably. But I was suggesting there was nothing like a localized
conservation relation -- which, however, you apparently say is an
incorrect surmise, because there can be a non-zero T in even massless,
radiationless, vacuum?
> The effect is analogous to electromagnetism: gravitational radiation
> decreases the time rate of change of the mass quadrupole moment in the
> same way electromagnetic radiation decreases the rate of change of the
> charge dipole moment.
But in EM we have the Poynting vector, which at least gives us some
accounting of local energy flow.
Edward Green
> On Oct 31, 5:00 pm, Edward Green <spamspamsp...@netzero.com> wrote:
<...>
> > So where did the energy come from?
>
> The source of the gravity waves. Although locally, it doesn't have to
> come from anywhere, since the concept of energy is not covariant.
> All we can really say in GR is that energy is conserved globally.
That might be a nice trick if we can't define an unambiguous integrand
to _get_ "global energy"!
Sue...
> an example of it.
Indeed... Noether and Hilbert.
http://www.physics.ucla.edu/~cwp/articles/noether.asg/noether.html
>
> Gravity waves travel through vacuum, where by definition T = 0 : no
> energy or momentum flux.
>
Gravity waves move just as light... through the
free space medium.
http://en.wikipedia.org/wiki/Free_space#Properties_of_free_space
The Origin of Gravity
Authors: C. P. Kouropoulos
http://arxiv.org/abs/physics/0107015
http://arxiv.org/abs/physics/0107015v1
> Gravity waves are thought to be detectable by their effect on matter;
Duh... Pick up an anvil with one hand and a
feather with the other hand. Can you detect
a difference?
> presumably by putting some test masses in relative motion. This
> creates local energy.
>
> So where did the energy come from?
The energy comes from the formation of the matter.
Atoms, protons neutrons...
Sue...
Tom Roberts
> All we can really say in GR is that energy is conserved globally.
Can't even say that, in general.
Tom Roberts
Tom Roberts
> Perturbations of the metric (gravitational radiation) have a
> nonzero stress-energy tensor,
Not true, linear or otherwise. Vacuum in GR means T=0. Period.
Note, however, when a pulse of gravitational radiation interacts with
matter, T(before) != T(after), where T is the energy-momentum tensor of
the matter. So gravitational radiation can TRANSFER energy and momentum,
even though it CONTAINS none itself.
There are various stress-energy pseudo-tensors useful in certain
situations, and gravitational radiation contributes to some of them. But
they are not "the" stress-energy tensor (aka the energy-momentum tensor;
its components in locally-inertial coordinates contain energy density
[T_00], momentum density [T_0i, i=1,2,3], and stress density [T^ij]).
Tom Roberts
Edward Green
> eric gisse wrote:
> > Perturbations of the metric (gravitational radiation) have a
> > nonzero stress-energy tensor,
>
> Not true, linear or otherwise. Vacuum in GR means T=0. Period.
>
> Note, however, when a pulse of gravitational radiation interacts with
> matter, T(before) != T(after), where T is the energy-momentum tensor of
> the matter. So gravitational radiation can TRANSFER energy and momentum,
> even though it CONTAINS none itself.
Hmm... so what does this do to the claims that T is "divergence free",
if in some situations it can apparently pop into existence -- or at
least augment -- without troubling to connect with a local flow? That
sounds like source (and the source of the gravitational radiation
would be a sink) to me.
eric gisse
<tjrobe...@sbcglobal.net> wrote:
>eric gisse wrote:
>> Perturbations of the metric (gravitational radiation) have a
>> nonzero stress-energy tensor,
>
>Not true, linear or otherwise. Vacuum in GR means T=0. Period.
My understanding was that it was only _true_ vacuum to first order,
and that there was a second order contribution that is "pretty much"
vacuum.
>
>Note, however, when a pulse of gravitational radiation interacts with
>matter, T(before) != T(after), where T is the energy-momentum tensor of
>the matter. So gravitational radiation can TRANSFER energy and momentum,
>even though it CONTAINS none itself.
Guess not!
Tom Roberts
> On Nov 1, 11:41 am, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
>> Note, however, when a pulse of gravitational radiation interacts with
>> matter, T(before) != T(after), where T is the energy-momentum tensor of
>> the matter. So gravitational radiation can TRANSFER energy and momentum,
>> even though it CONTAINS none itself.
>
> Hmm... so what does this do to the claims that T is "divergence free",
> if in some situations it can apparently pop into existence -- or at
> least augment -- without troubling to connect with a local flow? That
> sounds like source (and the source of the gravitational radiation
> would be a sink) to me.
Consider the Einstein field equation (projected onto an arbitrary
coordinate basis):
R_ij = T_ij - 2 T g_ij [T = g^kl T_kl]
Now it is clear that the {T_ij} are related only to the Ricci tensor
{R_ij}. Gravitational waves don't affect the Ricci tensor, they exist
wholly in the Weyl tensor. Remember that the Riemann curvature tensor
can be uniquely decomposed into the Ricci and Weyl tensors. So arbitrary
gravitational waves can exist in vacuum (T_ij = 0).
When a gravitational wave intersects some matter with T_ij != 0, the
equation of motion is (as always):
D_i T^ij = 0
Note that the gravitational wave appears here ONLY in the connection
inside the {D_i}. With time-varying components in the connection, it is
necessary for some of the {T^ij} to be time-varying in precisely the
right manner to cancel them. So in a manifold with a gravitational wave
pulse intersecting some matter, the matter can initially be quiescent,
but after the pulse passes it is moving.
Tom Roberts
Ken S. Tucker
> eric gisse wrote:
> > Perturbations of the metric (gravitational radiation) have a
> > nonzero stress-energy tensor,
>
> Not true, linear or otherwise. Vacuum in GR means T=0. Period.
Ok, but when light is deflected by the Sun,
by what we term noneuclidean spacetime curvature,
I would suggest "T" is non-zero and proportional
to the deflecting Mass M, such that the quantity
of curvature G in G=T is proportional to density.
Density depends on the Volume in density = M/V,
and then what's the Volume of a point? (rhetorical),
well it's zero. So now we're obligated to create
a non-zero V.
Regards
Ken S. Tucker
Edward Green
> On Nov 1, 7:41 am, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
>
> > eric gisse wrote:
> > > Perturbations of the metric (gravitational radiation) have a
> > > nonzero stress-energy tensor,
>
> > Not true, linear or otherwise. Vacuum in GR means T=0. Period.
>
> Ok, but when light is deflected by the Sun,
> by what we term noneuclidean spacetime curvature,
> I would suggest "T" is non-zero and proportional
> to the deflecting Mass M, such that the quantity
> of curvature G in G=T is proportional to density.
Apparently, and I don't understand the theory, Ken, this is adequately
explained by T=0 vacuum terms in the vicinity of the sun.
Spacetime may be curved either in the presence of absence of matter,
but that the presence of matter allows richer curvatures, as
additional terms, or degrees of freedom, are activated: these terms
manifest in deviations of the volume of small balls and cones, insofar
as I understand it, from their Euclidean values (implication: in
vacuum, the volume of small balls and cones, whatever else may be
going on, are their Euclidean values).
eric gisse
<dyna...@vianet.on.ca> wrote:
>On Nov 1, 7:41 am, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
>> eric gisse wrote:
>> > Perturbations of the metric (gravitational radiation) have a
>> > nonzero stress-energy tensor,
>>
>> Not true, linear or otherwise. Vacuum in GR means T=0. Period.
>
>Ok, but when light is deflected by the Sun,
>by what we term noneuclidean spacetime curvature,
>I would suggest "T" is non-zero and proportional
>to the deflecting Mass M, such that the quantity
>of curvature G in G=T is proportional to density.
You are wrong and clueless. Stop commenting until you get educated.
eric gisse
<spamsp...@netzero.com> wrote:
>On Nov 1, 6:33 pm, "Ken S. Tucker" <dynam...@vianet.on.ca> wrote:
>> On Nov 1, 7:41 am, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
>>
>> > eric gisse wrote:
>> > > Perturbations of the metric (gravitational radiation) have a
>> > > nonzero stress-energy tensor,
>>
>> > Not true, linear or otherwise. Vacuum in GR means T=0. Period.
>>
>> Ok, but when light is deflected by the Sun,
>> by what we term noneuclidean spacetime curvature,
>> I would suggest "T" is non-zero and proportional
>> to the deflecting Mass M, such that the quantity
>> of curvature G in G=T is proportional to density.
>
>Apparently, and I don't understand the theory, Ken, this is adequately
>explained by T=0 vacuum terms in the vicinity of the sun.
It is. Ken is an untutored imbecile.
Ken S. Tucker
as it relates to your "grok" :-).
Carrying on this post...
Let there be a unit Mass M at the origin.
Let the density "T" be defined at any point
"r" from the origin as proportional to M/r^3,
(Mass/volume).
Let the Curvature "K" be defined as the rate of
change of gravitational acceleration, proportional
to K =d(M/r^2)/dr = M/r^3.
We see K=T is proportional, and is a simplified
derivation of the Einstein Field Equation direct
from Newton's gravity.
In Newton's gravity, the K is a tidal effect.
Regards
Ken S. Tucker
Edward Green
> On Sat, 1 Nov 2008 16:50:30 -0700 (PDT), Edward Green
>
>
>
> <spamspamsp...@netzero.com> wrote:
> >On Nov 1, 6:33 pm, "Ken S. Tucker" <dynam...@vianet.on.ca> wrote:
> >> On Nov 1, 7:41 am, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
>
> >> > eric gisse wrote:
> >> > > Perturbations of the metric (gravitational radiation) have a
> >> > > nonzero stress-energy tensor,
>
> >> > Not true, linear or otherwise. Vacuum in GR means T=0. Period.
>
> >> Ok, but when light is deflected by the Sun,
> >> by what we term noneuclidean spacetime curvature,
> >> I would suggest "T" is non-zero and proportional
> >> to the deflecting Mass M, such that the quantity
> >> of curvature G in G=T is proportional to density.
>
> >Apparently, and I don't understand the theory, Ken, this is adequately
> >explained by T=0 vacuum terms in the vicinity of the sun.
>
He is a polite gentleman. That counts for something, so I won't be
pejoratively labeling him, even when he happens to be mistaken.
eric gisse
Except that isn't an actual definition of curvature.
>
>We see K=T is proportional, and is a simplified
>derivation of the Einstein Field Equation direct
>from Newton's gravity.
Except it is wrong.
>
>In Newton's gravity, the K is a tidal effect.
Tidal forces have a direct derivation through the Riemann tensor. Try
again.
>Regards
>Ken S. Tucker
Ken S. Tucker
On Nov 2, 12:14 am, Edward Green <spamspamsp...@netzero.com> wrote:
> On Nov 1, 8:13 pm, eric gisse <jowr.pi.nos...@gmail.com> wrote:
> > On Sat, 1 Nov 2008 16:50:30 -0700 (PDT), Edward Green
>
> > <spamspamsp...@netzero.com> wrote:
> > >On Nov 1, 6:33 pm, "Ken S. Tucker" <dynam...@vianet.on.ca> wrote:
> > >> On Nov 1, 7:41 am, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
>
> > >> > eric gisse wrote:
> > >> > > Perturbations of the metric (gravitational radiation) have a
> > >> > > nonzero stress-energy tensor,
>
> > >> > Not true, linear or otherwise. Vacuum in GR means T=0. Period.
>
> > >> Ok, but when light is deflected by the Sun,
> > >> by what we term noneuclidean spacetime curvature,
> > >> I would suggest "T" is non-zero and proportional
> > >> to the deflecting Mass M, such that the quantity
> > >> of curvature G in G=T is proportional to density.
>
> > >Apparently, and I don't understand the theory, Ken, this is adequately
> > >explained by T=0 vacuum terms in the vicinity of the sun.
Yes, the G_uv=0 is a covariant representation of
Newtons LAPlacian of POTential, however I'll quote
AE's GR1916 chp 16, "We require the equation
corresponding to Poisson's equation,
LAP(POT) = 4 pi rho density ", which then yields
G_uv = T_uv.
> He is a polite gentleman. That counts for something, so I won't be
> pejoratively labeling him, even when he happens to be mistaken.
LOL. I'm sure Edward you have read the so-called
qualified versions and explanations on the net
and in a few texts, there're fairly well generic,
so there is a need for a fresh insight into the
G_uv=T_uv, even for some easy conversation.
That's why I posted on K=T yesterday directly
from Newton's / Poisson's Theory.
Fun stuff.
Ken S. Tucker
eric gisse
No, he is not. He accuses me of plaigarism and then runs away like a
coward and because he knows it isn't true.
Tom Roberts
> On Nov 1, 6:33 pm, "Ken S. Tucker" <dynam...@vianet.on.ca> wrote:
>> Ok, but when light is deflected by the Sun,
>> by what we term noneuclidean spacetime curvature,
>> I would suggest "T" is non-zero and proportional
>> to the deflecting Mass M, such that the quantity
>> of curvature G in G=T is proportional to density.
You need to be more precise. Yes, T != 0 inside the surface of the sun,
but T = 0 to excellent approximation outside that surface. In
particular, for many measurements of Shapiro delay and light deflection
the solar atmosphere is completely negligible and computing with T = 0
is adequately accurate.
But, of course, the curvature of spacetime is nonzero near but outside
the surface of the sun, due to the nearby matter inside that surface.
Indeed, out well past the orbit of Pluto this is true.
> Apparently, and I don't understand the theory, Ken, this is adequately
> explained by T=0 vacuum terms in the vicinity of the sun.
Yes. Even thoug T is not exactly zero (due to solar corone, solar
atmosphere, and those pesky planets :-), assumint T = 0 in these
computations is sufficiently accurate.
> Spacetime may be curved either in the presence of absence of matter,
> but that the presence of matter allows richer curvatures, as
> additional terms, or degrees of freedom, are activated: these terms
> manifest in deviations of the volume of small balls and cones, insofar
> as I understand it, from their Euclidean values (implication: in
> vacuum, the volume of small balls and cones, whatever else may be
> going on, are their Euclidean values).
Yes. When T = 0, only the Weyl tensor can be nonzero, Ricci must be
zero; when T != 0, both the Weyl and Ricci tensors can be nonzero. Weyl
and Ricci combine to make up the Riemann curvature tensor.
Tom Roberts
Koobee Wublee
> "Ken S. Tucker" wrote:
> >Ok, but when light is deflected by the Sun,
> >by what we term noneuclidean spacetime curvature,
> >I would suggest "T" is non-zero and proportional
> >to the deflecting Mass M, such that the quantity
> >of curvature G in G=T is proportional to density.
>
> You are wrong and clueless. Stop commenting until you get educated.
Why don’t you, a multi-year super-senior, follow your own advice?
Bette yet, why don’t you graduate from junior high school first.
Especially, [g]_ij does not mean the matrix but the elements which are
all single-element scalars to the matrix [g]. <shrug>
Koobee Wublee
> Edward Green wrote:
> >I"ve heard that "energy conservation fails in GR" and perhaps this is
> >an example of it.
>
> It doesn't strictly fail - energy is simply a quantity that has no
> single definition, none of which are true in general.
In another words, you, a multi-year super-senior, has no idea what you
are talking about. <shrug>
> >Gravity waves travel through vacuum, where by definition T = 0 : no
> >energy or momentum flux.
>
> No. Perturbations of the metric (gravitational radiation) have a
> nonzero stress-energy tensor, but it isn't linear in the metric so it
> doesn't appear in the linearized theory.
You absolutely do not know what you are talking about. <shrug>
> >Gravity waves are thought to be detectable by their effect on matter;
> >presumably by putting some test masses in relative motion. This
> >creates local energy.
>
> >So where did the energy come from?
>
> The energy came from the system that emitted the gravitational
> radiation in the first place.
>
> The effect is analogous to electromagnetism: gravitational radiation
> decreases the time rate of change of the mass quadrupole moment in the
> same way electromagnetic radiation decreases the rate of change of the
> charge dipole moment.
You remind me of an elderly component keeper. When taught about the
issues of electrostatic, he started to put everything in electrostatic
bags. When I told him not to waste these electrostatic bags on bolts
and nuts, he refused to listen to reasoning. His mind was very made-
up.
Koobee Wublee
> eric gisse wrote:
> > Perturbations of the metric (gravitational radiation) have a
> > nonzero stress-energy tensor,
>
> Not true, linear or otherwise. Vacuum in GR means T=0. Period.
Gravitational waves have nothing to do with the field equations.
<shrug>
> Note, however, when a pulse of gravitational radiation interacts with
> matter, T(before) != T(after), where T is the energy-momentum tensor of
> the matter. So gravitational radiation can TRANSFER energy and momentum,
> even though it CONTAINS none itself.
Hmmm... This sentence above contains no logical rational. Who are
your trying to kid? The folks like that multi-year super-senior?
> There are various stress-energy pseudo-tensors useful in certain
> situations, and gravitational radiation contributes to some of them. But
> they are not "the" stress-energy tensor (aka the energy-momentum tensor;
> its components in locally-inertial coordinates contain energy density
> [T_00], momentum density [T_0i, i=1,2,3], and stress density [T^ij]).
It sounds like you are very vague in understanding of how the
gravitational waves are derived. <shrug>
Koobee Wublee
> My understanding was that it was only _true_ vacuum to first order,
> and that there was a second order contribution that is "pretty much"
> vacuum.
>
> [...]
>
> Guess not!
You understand jack shit. <shrug>
Koobee Wublee
> Edward Green wrote:
> You need to be more precise. Yes, T != 0 inside the surface of the sun,
> but T = 0 to excellent approximation outside that surface. In
> particular, for many measurements of Shapiro delay and light deflection
> the solar atmosphere is completely negligible and computing with T = 0
> is adequately accurate.
Don’t you do <shrug> anymore?
> But, of course, the curvature of spacetime is nonzero near but outside
> the surface of the sun, due to the nearby matter inside that surface.
> Indeed, out well past the orbit of Pluto this is true.
<shrug>
> > Apparently, and I don't understand the theory, Ken, this is adequately
> > explained by T=0 vacuum terms in the vicinity of the sun.
>
> Yes. Even thoug T is not exactly zero (due to solar corone, solar
> atmosphere, and those pesky planets :-), assumint T = 0 in these
> computations is sufficiently accurate.
Hmmm... With impeccable spelling and grammar, you have just lost
apart. Is this on purpose? Are you making fun of Mr. Tucker? If so,
this is so uncharacteristic of you making fun of some creature far
less developed than you are. Did you just come back from a party or
something?
> > Spacetime may be curved either in the presence of absence of matter,
> > but that the presence of matter allows richer curvatures, as
> > additional terms, or degrees of freedom, are activated: these terms
> > manifest in deviations of the volume of small balls and cones, insofar
> > as I understand it, from their Euclidean values (implication: in
> > vacuum, the volume of small balls and cones, whatever else may be
> > going on, are their Euclidean values).
>
> Yes. When T = 0, only the Weyl tensor can be nonzero, Ricci must be
> zero; when T != 0, both the Weyl and Ricci tensors can be nonzero. Weyl
> and Ricci combine to make up the Riemann curvature tensor.
You have not said anything that is important. <shrug> Weyl tensor is
actually not important in GR. I do not know what makes you throw in
the very irrelevant Weyl tensor from Mr. Greens’ comment above. Are
you still in control of yourself?
Ken S. Tucker
> Edward Green wrote:
> > On Nov 1, 6:33 pm, "Ken S. Tucker" <dynam...@vianet.on.ca> wrote:
> >> Ok, but when light is deflected by the Sun,
> >> by what we term noneuclidean spacetime curvature,
> >> I would suggest "T" is non-zero and proportional
> >> to the deflecting Mass M, such that the quantity
> >> of curvature G in G=T is proportional to density.
>
> You need to be more precise. Yes, T != 0 inside the surface of the sun,
> but T = 0 to excellent approximation outside that surface. In
> particular, for many measurements of Shapiro delay and light deflection
> the solar atmosphere is completely negligible and computing with T = 0
> is adequately accurate.
>
> But, of course, the curvature of spacetime is nonzero near but outside
> the surface of the sun, due to the nearby matter inside that surface.
> Indeed, out well past the orbit of Pluto this is true.
Laplace's equation, \/^2 (p) = 0
Poisson's equation, \/^2 (p) = density.
(I'm using \/ as nabla, p for potential)
Laplace's provides the 1/r^2, but the Poisson
supplies the numerator too.
I posted a simple example on Nov.1 in this thread.
Regards
Ken S. Tucker
eric gisse
Just keep telling yourself that.
eric gisse
Actually it does. Look up matrix index notation, and stop imposing
your non-standard notation on a well established field.
So nice to know this latest episode of your documented and admitted
cluelessness has not significantly affected your arrogant stupidity.
eric gisse
<koobee...@gmail.com> wrote:
>On Nov 1, 7:41 am, Tom Roberts wrote:
>> eric gisse wrote:
>
>> > Perturbations of the metric (gravitational radiation) have a
>> > nonzero stress-energy tensor,
>>
>> Not true, linear or otherwise. Vacuum in GR means T=0. Period.
>
>Gravitational waves have nothing to do with the field equations.
><shrug>
Since you have abundantly established that you do not understand the
notation, how the field equations are derived, how the field equations
are used to derive solutions, what covariance means, what the metric
means, what curvature means, how the metric relates to computing area
or volume, what symmetries mean, why the metric is symmetric, why the
connection coeffecients are symmetric, or what Birkhoff's theorem
means, you are unfit to judge.
The derivation of gravitational waves is straightfoward from
perturbation theory. I'd give references but who would I be kidding?
You are stupid.
>
>> Note, however, when a pulse of gravitational radiation interacts with
>> matter, T(before) != T(after), where T is the energy-momentum tensor of
>> the matter. So gravitational radiation can TRANSFER energy and momentum,
>> even though it CONTAINS none itself.
>
>Hmmm... This sentence above contains no logical rational. Who are
>your trying to kid? The folks like that multi-year super-senior?
You are unfit to judge.
>
>> There are various stress-energy pseudo-tensors useful in certain
>> situations, and gravitational radiation contributes to some of them. But
>> they are not "the" stress-energy tensor (aka the energy-momentum tensor;
>> its components in locally-inertial coordinates contain energy density
>> [T_00], momentum density [T_0i, i=1,2,3], and stress density [T^ij]).
>
>It sounds like you are very vague in understanding of how the
>gravitational waves are derived. <shrug>
Again - you are unfit to judge.
>
>
Ken S. Tucker
> On Nov 2, 3:10 pm, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
>
>
>
> > Edward Green wrote:
> > > On Nov 1, 6:33 pm, "Ken S. Tucker" <dynam...@vianet.on.ca> wrote:
> > >> Ok, but when light is deflected by the Sun,
> > >> by what we term noneuclidean spacetime curvature,
> > >> I would suggest "T" is non-zero and proportional
> > >> to the deflecting Mass M, such that the quantity
> > >> of curvature G in G=T is proportional to density.
>
> > You need to be more precise. Yes, T != 0 inside the surface of the sun,
> > but T = 0 to excellent approximation outside that surface. In
> > particular, for many measurements of Shapiro delay and light deflection
> > the solar atmosphere is completely negligible and computing with T = 0
> > is adequately accurate.
>
> > But, of course, the curvature of spacetime is nonzero near but outside
> > the surface of the sun, due to the nearby matter inside that surface.
> > Indeed, out well past the orbit of Pluto this is true.
==============
> Laplace's equation, \/^2 (p) = 0
> Poisson's equation, \/^2 (p) = density.
> (I'm using \/ as nabla, p for potential)
>
> Laplace's provides the 1/r^2, but the Poisson
> supplies the numerator too.
> I posted a simple example on Nov.1 in this thread.
Let me try to explain this a bit better, (correct me
if I'm wrong).
The Laplace Eq. \/^2 (p) = 0 , is a field equation,
with the solution being p=M/r where M can be any
constant number.
IOW's the Laplacian field Eq. does NOT solve M.
OTOH, Poisson's field equation \/^2 (p) = density
will solve M and p , such as, given the density
to be M/r^3 , then Poisson's becomes,
\/^2 (p) = M/r^3
and then p = M/r, and M is defined from the density,
to have a single defined value for M, because
Poisson's field equation is stricter than Laplaces.
In GR theory the Poisson is written as G_uv=T_uv
whereas the Laplacian is G_uv=0, with the relation
being G_uv == \/^2 (p) and density = T_uv.
I should point out I use G_uv=T_uv always because
G_uv=0 implies G_uv is an invariant, which is NOT
the truth, IOW's G_uv=0 is NOT a law of physics.
Regards
Ken S. Tucker
eric gisse
<koobee...@gmail.com> wrote:
>On Oct 31, 2:59 pm, eric gisse < wrote:
>> Edward Green wrote:
>
>> >I"ve heard that "energy conservation fails in GR" and perhaps this is
>> >an example of it.
>>
>> It doesn't strictly fail - energy is simply a quantity that has no
>> single definition, none of which are true in general.
>
>In another words, you, a multi-year super-senior, has no idea what you
>are talking about. <shrug>
You are unfit to judge, and most definitely unfit to criticize since
you couldn't even work through a derivation of the field equations
with your hand held every step of the way.
eric gisse
<dyna...@vianet.on.ca> wrote:
[...]
>I should point out I use G_uv=T_uv always because
>G_uv=0 implies G_uv is an invariant, which is NOT
>the truth, IOW's G_uv=0 is NOT a law of physics.
Learn the difference between 'invariant' and 'covariant'.
>Regards
>Ken S. Tucker
Hayek
> On Sun, 2 Nov 2008 21:16:36 -0800 (PST), Koobee Wublee
> <koobee...@gmail.com> wrote:
>
>> On Nov 1, 7:41 am, Tom Roberts wrote:
>>> eric gisse wrote:
>>>> Perturbations of the metric (gravitational radiation) have a
>>>> nonzero stress-energy tensor,
>>> Not true, linear or otherwise. Vacuum in GR means T=0. Period.
>> Gravitational waves have nothing to do with the field equations.
>> <shrug>
>
> Since you have abundantly established that you do not understand the
> notation, how the field equations are derived, how the field equations
> are used to derive solutions, what covariance means, what the metric
> means, what curvature means, how the metric relates to computing area
> or volume, what symmetries mean, why the metric is symmetric, why the
> connection coeffecients are symmetric, or what Birkhoff's theorem
> means, you are unfit to judge.
We talk about physics, not mathematics.
An electromagnetic wave, and that is about the only waves we know that
propagate in empty space, has nothing do to with a wave. It is a
particle that is allowed to move in an inertialess zone., and this
particle in his zone propagates by inertia.
But suppose we have a gravity wave caused by a pulsating mass.
The mass would have to pulsate by varying its mass, I do not see how
this could be done, but suppose. Thus we have a spherical wave, were the
metric is denser the crest, and were it is less dense, the wave bottom.
Now, in the denser region, the speed of light is slower, but also the
speed at which gravity propagates. And in the less dense region it is
vice versa. So the bottom will overtake the crest and the wave would
selfdestruct and the effects would cancel.
Uwe Hayek.
--
Als ik nu op dit moment geld transfereer [in Belgi隴 naar een
andere rekening staat dat een uur later daar gecrediteerd.
-- Boutros Gali, realiteitsdeskundige.
eric gisse
wrote:
>eric gisse wrote:
>> On Sun, 2 Nov 2008 21:16:36 -0800 (PST), Koobee Wublee
>> <koobee...@gmail.com> wrote:
>>
>>> On Nov 1, 7:41 am, Tom Roberts wrote:
>>>> eric gisse wrote:
>>>>> Perturbations of the metric (gravitational radiation) have a
>>>>> nonzero stress-energy tensor,
>>>> Not true, linear or otherwise. Vacuum in GR means T=0. Period.
>>> Gravitational waves have nothing to do with the field equations.
>>> <shrug>
>>
>> Since you have abundantly established that you do not understand the
>> notation, how the field equations are derived, how the field equations
>> are used to derive solutions, what covariance means, what the metric
>> means, what curvature means, how the metric relates to computing area
>> or volume, what symmetries mean, why the metric is symmetric, why the
>> connection coeffecients are symmetric, or what Birkhoff's theorem
>> means, you are unfit to judge.
>
>We talk about physics, not mathematics.
>
>An electromagnetic wave, and that is about the only waves we know that
>propagate in empty space, has nothing do to with a wave. It is a
>particle that is allowed to move in an inertialess zone., and this
>particle in his zone propagates by inertia.
Here comes the thread hijack about your pet idiocies and stupidities
regarding inertia.
>
>But suppose we have a gravity wave caused by a pulsating mass.
DOA. Spherical pulsations do not generate gravitational radiation -
variations in the quadrupole moment do.
>The mass would have to pulsate by varying its mass, I do not see how
>this could be done, but suppose. Thus we have a spherical wave, were the
>metric is denser the crest, and were it is less dense, the wave bottom.
Untutored idiot. Is PSR 1913+16 varying its' mass? No.
>
>Now, in the denser region, the speed of light is slower, but also the
>speed at which gravity propagates. And in the less dense region it is
>vice versa. So the bottom will overtake the crest and the wave would
>selfdestruct and the effects would cancel.
The speed of light is a constant, and gravitational radiation
propagates at the speed of light.
This same stupidity could be used to 'argue' why light cannot pass
through an object with a non-unity index of refraction.
>
>Uwe Hayek.
Or you could learn the theory you are criticizing so you won't have to
write paragraph after paragraph of nonsense whenever you butt in on
something you obviously don't understand.
Juan R.
> I"ve heard that "energy conservation fails in GR"
Yes, this is sometimes considered its greater problem.
It may be remarked it is a problem *only* with geometric approach to
gravity, not of field theory of gravity.
> and perhaps this is an
> example of it.
Another known example is the cosmological model of expanding universe.
The law of conservation of energy cannot be applied.
> Gravity waves travel through vacuum, where by definition T = 0 : no
> energy or momentum flux.
>
> Gravity waves are thought to be detectable by their effect on matter;
> presumably by putting some test masses in relative motion. This creates
> local energy.
>
> So where did the energy come from?
Nowhere?
Precisely, some author doubt that gravitational waves can be detectable
without any *physical* energy mechanism.
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Juan R.
> Edward Green wrote on Fri, 31 Oct 2008 14:00:40 -0700:
>
>> I"ve heard that "energy conservation fails in GR"
>
> Yes, this is sometimes considered its greater problem.
>
> It may be remarked it is a problem *only* with geometric approach to
> gravity, not of field theory of gravity.
>
>> and perhaps this is an
>> example of it.
>
> Another known example is the cosmological model of expanding universe.
> The law of conservation of energy cannot be applied.
A third example is thermodynamics. Without an energy conservation law,
there is difficulties to correctly define entropy and all the rest of
thermodynamic in the framework of General Relativity.
There is not such problems in a field approach to gravity.
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Hayek
It is not a spherical pulsation, it the varying of mass.
But ok , it is not even possible.
>> The mass would have to pulsate by varying its mass, I do not see how
>> this could be done, but suppose. Thus we have a spherical wave, were the
>> metric is denser the crest, and were it is less dense, the wave bottom.
>
> Untutored idiot. Is PSR 1913+16 varying its' mass? No.
No, but if you were situated nearby you wuld have mass coming close and
dissappearing again, this indeed a fluctuating gravitational field, i
will refrain from using the word inertial field, since you seem allergic
to the EP.
>
>> Now, in the denser region, the speed of light is slower, but also the
>> speed at which gravity propagates. And in the less dense region it is
>> vice versa. So the bottom will overtake the crest and the wave would
>> selfdestruct and the effects would cancel.
>
> The speed of light is a constant, and gravitational radiation
> propagates at the speed of light.
This is what you get from picking at the efe's without knowing what your
doing. Again, proof that no ones understands GR, from the equations.
Mass creates the gravitational field right ? And were it is stronger
clocks run slower ? Well Einstein says, there is an inseparable
connection between the speed of light and time. And time is what you
read on a clock. So were the field is stronger c is lower. From the outside.
> This same stupidity could be used to 'argue' why light cannot pass
> through an object with a non-unity index of refraction.
>
> Or you could learn the theory you are criticizing so you won't have to
> write paragraph after paragraph of nonsense whenever you butt in on
> something you obviously don't understand.
Work on your own stupidity before accusing someone else.
Start reading Clifford Will's "was Einstein right" and leave the picking
at equations for math class.
You clearly can't understand GR from the equations, but no-one could.
Even Einstein lost his inertia in them.
Uwe Hayek.
Tom Roberts
>> Laplace's provides the 1/r^2, but the Poisson
>> supplies the numerator too.
One does not arbitrarily get to "choose" among these equations, one must
apply the one that is appropriate to what you are trying to do --
whichever one provides an accurate model.
If you are trying to solve Newtonian gravitation, using its F=ma and
F=GmM/r^2 via a potential, then you MUST use Poisson's equation,
Laplace's equation is completely irrelevant (except for the fact that
Poisson's equation reduces to it in regions where the mass density is zero).
> Let me try to explain this a bit better, (correct me
> if I'm wrong).
Good parenthetical, as in several places you are wrong.
> The Laplace Eq. \/^2 (p) = 0 , is a field equation,
> with the solution being p=M/r where M can be any
> constant number.
Yes it is a field equation; no you did not give the general solution.
> IOW's the Laplacian field Eq. does NOT solve M.
This is true, but not because of the (wrong) reason you gave.
> OTOH, Poisson's field equation \/^2 (p) = density
> will solve M and p , such as, given the density
> to be M/r^3 , then Poisson's becomes,
> \/^2 (p) = M/r^3
Yes, but irrelevant, as no sensible object has such a density.
A general solution to Poisson's equation is well known, and comes via
the Green's function approach:
\phi(r) = \integral d^3r' \rho(r')/|r-r'|
where both r and r' are 3-vector positions, \phi(r) is
the potential [your (p)], and \rho(r') is the mass
density at r'.
As Poisson's equation is linear, other solutions can be added at will,
so one needs boundary conditions to select among the infinite number of
solutions. Note that boundary conditions have also implicitly been
applied to select the above Green's function.
> Poisson's field equation is stricter than Laplaces.
Not "stricter", but rather CORRECT FOR THIS PROBLEM.
Note that all the above is Newtonian gravitation, and is completely
irrelevant in GR.
> In GR theory the Poisson is written as G_uv=T_uv
> whereas the Laplacian is G_uv=0, with the relation
> being G_uv == \/^2 (p) and density = T_uv.
This is all wrong. If one ignores sensible usages of words and
introduces PUNs on just about every word you use, than one can claim a
loose analogy as you claim, but that's all (i.e. your "relation" is a
rather loose analogy, not an equality as you write).
> I should point out I use G_uv=T_uv [...]
You use this construct a lot: "I use ...". That's rather silly, implying
that you are in control of which equations are valid. You aren't, and as
I said above the choice of which equation to use is dictated by Nature
and the models/theories of physics, not by Ken S. Tucker.
> always because
> G_uv=0 implies G_uv is an invariant, which is NOT
> the truth,
Hmmm. If G_uv=0 then the {G_uv} ARE invariant. But only in that
particular case. That is, the components of the zero tensor is
invariant. G itself, of course, is always invariant (it is a tensor).
> IOW's G_uv=0 is NOT a law of physics.
This is true. The related law of physics is G_uv = T_uv, known as the
Einstein field equations. These 16 equations are covariant, and when
written in terms of the tensors it is a single invariant equation, even
though the individual components of the tensors are neither covariant
nor invariant -- this is a general property of tensor equations.
Tom Roberts
Ken S. Tucker
On Nov 3, 7:45 am, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
> Ken S. Tucker wrote:
> >> Laplace's provides the 1/r^2, but the Poisson
> >> supplies the numerator too.
>
> One does not arbitrarily get to "choose" among these equations, one must
> apply the one that is appropriate to what you are trying to do --
> whichever one provides an accurate model.
Yes, and AE choose Poisson's.
> If you are trying to solve Newtonian gravitation, using its F=ma and
> F=GmM/r^2 via a potential, then you MUST use Poisson's equation,
> Laplace's equation is completely irrelevant (except for the fact that
> Poisson's equation reduces to it in regions where the mass density is zero).
>
> > Let me try to explain this a bit better, (correct me
> > if I'm wrong).
>
> Good parenthetical, as in several places you are wrong.
Well I've *proved* the math, and posted it,
however I'll let you decide if that is valid
for you. It's a free country, you can go your
own way!
> > The Laplace Eq. \/^2 (p) = 0 , is a field equation,
> > with the solution being p=M/r where M can be any
> > constant number.
>
> Yes it is a field equation; no you did not give the general solution.
>
> > IOW's the Laplacian field Eq. does NOT solve M.
>
> This is true, but not because of the (wrong) reason you gave.
In the spec'd single body problem it's ok.
> > OTOH, Poisson's field equation \/^2 (p) = density
> > will solve M and p , such as, given the density
> > to be M/r^3 , then Poisson's becomes,
> > \/^2 (p) = M/r^3
>
> Yes, but irrelevant, as no sensible object has such a density.
Then you need to learn what density is.
> A general solution to Poisson's equation is well known, and comes via
> the Green's function approach:
>
> \phi(r) = \integral d^3r' \rho(r')/|r-r'|
> where both r and r' are 3-vector positions, \phi(r) is
> the potential [your (p)], and \rho(r') is the mass
> density at r'.
Of course, but we don't need a general solution,
we're in applications to a single body.
> As Poisson's equation is linear, other solutions can be added at will,
> so one needs boundary conditions to select among the infinite number of
> solutions. Note that boundary conditions have also implicitly been
> applied to select the above Green's function.
>
> > Poisson's field equation is stricter than Laplaces.
>
> Not "stricter", but rather CORRECT FOR THIS PROBLEM.
>
> Note that all the above is Newtonian gravitation, and is completely
> irrelevant in GR.
Wrong, see GR1916 chp16.
> > In GR theory the Poisson is written as G_uv=T_uv
> > whereas the Laplacian is G_uv=0, with the relation
> > being G_uv == \/^2 (p) and density = T_uv.
>
> This is all wrong. If one ignores sensible usages of words and
> introduces PUNs on just about every word you use, than one can claim a
> loose analogy as you claim, but that's all (i.e. your "relation" is a
> rather loose analogy, not an equality as you write).
Tom it's ok as an abstract.
> > I should point out I use G_uv=T_uv [...]
>
> You use this construct a lot: "I use ...". That's rather silly, implying
> that you are in control of which equations are valid. You aren't, and as
> I said above the choice of which equation to use is dictated by Nature
> and the models/theories of physics, not by Ken S. Tucker.
Tom, Ken S. Tucker recommends you understand
Guv=Tuv. Much of the low class mathematical
tensor gobly-gook (including your's) imparts
NO good understanding. I chose to use Newton
as a springboard into the EFE's.
> > always because
> > G_uv=0 implies G_uv is an invariant, which is NOT
> > the truth,
>
> Hmmm. If G_uv=0 then the {G_uv} ARE invariant. But only in that
> particular case.
There is NO such thing as case invariancy.
>That is, the components of the zero tensor is
> invariant. G itself, of course, is always invariant (it is a tensor).
>
> > IOW's G_uv=0 is NOT a law of physics.
>
> This is true. The related law of physics is G_uv = T_uv, known as the
> Einstein field equations. These 16 equations are covariant, and when
> written in terms of the tensors it is a single invariant equation, even
> though the individual components of the tensors are neither covariant
> nor invariant -- this is a general property of tensor equations.
Of course, did you type that because you like
hearing yourself type :-) ?
> Tom Roberts
Regards
Ken S. Tucker
stric...@gmail.com
> Gravity waves...
Are you speaking tlhIngan Hol ?
carlip...@physics.ucdavis.edu
> On Nov 3, 7:45 am, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
> > Ken S. Tucker wrote:
[...]
> > > OTOH, Poisson's field equation \/^2 (p) = density
> > > will solve M and p , such as, given the density
> > > to be M/r^3 , then Poisson's becomes,
> > > \/^2 (p) = M/r^3
> >
> > Yes, but irrelevant, as no sensible object has such a density.
> Then you need to learn what density is.
No, Tom is absolutely right here. The definition of mass density is that
if you integrate it over any finite region, you obtain the mass in that
region.
Your expression implies that the "density" is nonzero in completely
empty space. If you apply your "density" to, say, the Sun, you will find
that you are claiming the "mass" is mostly outside the surface of the
Sun. In fact, with your expression, the total "mass" of the Sun is infinite
-- just integrate your "density" over all space.
Further, if you do the math, you will find that the solution to your
equation
\/^2 (p) = M/r^3
is
p = -M/r (1+ ln(r/r_0) )
where ln is the natural log and r_0 is an integration constant. This
gives a gravitational field vastly different from the Newtonian one,
and clearly ruled out by observation.
Steve Carlip
Androcles
<stric...@gmail.com> wrote in message
news:e760b91f-a507-467c...@f37g2000pri.googlegroups.com...
On Oct 31, 4:00 pm, Edward Green <spamspamsp...@netzero.com> wrote:
> Gravity waves...
Are you speaking tlhIngan Hol ?
======================================
Here's a gravity wave, right under his nose:
http://tinyurl.com/5sav8h
carlip...@physics.ucdavis.edu
> eric gisse wrote:
> > Perturbations of the metric (gravitational radiation) have a
> > nonzero stress-energy tensor,
> Not true, linear or otherwise. Vacuum in GR means T=0. Period.
> Note, however, when a pulse of gravitational radiation interacts with
> matter, T(before) != T(after), where T is the energy-momentum tensor
> of the matter. So gravitational radiation can TRANSFER energy and
> momentum,
Right. This analysis was first suggested by Feynman at the 1957 Chapel
Hill Conference. He proposed a thought experiment in which a bead on
a stick has a gravitational wave pass by; the wave will move the bead
back and forth, and the resulting friction will heat the stick.
> even though it CONTAINS none itself.
This depends on what you mean by energy and momentum. Gravitational
radiation has no stress-energy tensor, and more generally no *local*
energy or momentum density. But a gravitational wave does carry Bondi
energy, which is invariantly defined, but is a total energy, not obtainable
from a local density. One can also define "quasilocal" energy. This is a
covariant but nonlocal object, but under many circumstances -- including
the description of weak, short-wavelenth gravitational waves in a slowly
varying background -- it can be approximated by a local density.
Steve Carlip
eric gisse
wrote:
[snip]
>>> The mass would have to pulsate by varying its mass, I do not see how
>>> this could be done, but suppose. Thus we have a spherical wave, were the
>>> metric is denser the crest, and were it is less dense, the wave bottom.
>>
>> Untutored idiot. Is PSR 1913+16 varying its' mass? No.
>No, but if you were situated nearby you wuld have mass coming close and
>dissappearing again, this indeed a fluctuating gravitational field, i
>will refrain from using the word inertial field, since you seem allergic
>to the EP.
I'm allergic to the invented terminology of delusional, untutored
idiots.
>>
>>> Now, in the denser region, the speed of light is slower, but also the
>>> speed at which gravity propagates. And in the less dense region it is
>>> vice versa. So the bottom will overtake the crest and the wave would
>>> selfdestruct and the effects would cancel.
>>
>> The speed of light is a constant, and gravitational radiation
>> propagates at the speed of light.
>
>This is what you get from picking at the efe's without knowing what your
>doing. Again, proof that no ones understands GR, from the equations.
See? Delusional, untutored, and definitely an idiot.
http://arxiv.org/abs/gr-qc/9812067
>
>Mass creates the gravitational field right ? And were it is stronger
>clocks run slower ? Well Einstein says, there is an inseparable
>connection between the speed of light and time. And time is what you
>read on a clock. So were the field is stronger c is lower. From the outside.
The speed of light remains constant, albiet only locally. Get an
education.
>
>> This same stupidity could be used to 'argue' why light cannot pass
>> through an object with a non-unity index of refraction.
No comment?
>
>
>
>>
>> Or you could learn the theory you are criticizing so you won't have to
>> write paragraph after paragraph of nonsense whenever you butt in on
>> something you obviously don't understand.
>
>Work on your own stupidity before accusing someone else.
>Start reading Clifford Will's "was Einstein right" and leave the picking
>at equations for math class.
Oh look the untutored idiot once again dismisses the usage of
mathematics in physics.
>
>You clearly can't understand GR from the equations, but no-one could.
Typical delusional crank belief that just because the /crank/ can't
understand, /nobody/ can understand.
eric gisse
<juanR...@canonicalscience.com> wrote:
>Edward Green wrote on Fri, 31 Oct 2008 14:00:40 -0700:
>
>> I"ve heard that "energy conservation fails in GR"
>
>Yes, this is sometimes considered its greater problem.
By you, and you alone.
>
>It may be remarked it is a problem *only* with geometric approach to
>gravity, not of field theory of gravity.
Here we go with the incessant pushing of irrelevant and pointless
alternative theories that you don't even understand.
>
>> and perhaps this is an
>> example of it.
>
>Another known example is the cosmological model of expanding universe.
>The law of conservation of energy cannot be applied.
Sure it can. There exists a time-like Killing vector for the entire
set of FRW cosmologies.
>
>> Gravity waves travel through vacuum, where by definition T = 0 : no
>> energy or momentum flux.
>>
>> Gravity waves are thought to be detectable by their effect on matter;
>> presumably by putting some test masses in relative motion. This creates
>> local energy.
>>
>> So where did the energy come from?
>
>Nowhere?
>
>Precisely, some author doubt that gravitational waves can be detectable
>without any *physical* energy mechanism.
...and some authors doubt that special relativity is even a
mathematical theory. Clearly a higher standard must be applied.
eric gisse
<juanR...@canonicalscience.com> wrote:
>"Juan R." González-Álvarez wrote on Mon, 03 Nov 2008 14:18:15 +0100:
>
>> Edward Green wrote on Fri, 31 Oct 2008 14:00:40 -0700:
>>
>>> I"ve heard that "energy conservation fails in GR"
>>
>> Yes, this is sometimes considered its greater problem.
>>
>> It may be remarked it is a problem *only* with geometric approach to
>> gravity, not of field theory of gravity.
>>
>>> and perhaps this is an
>>> example of it.
>>
>> Another known example is the cosmological model of expanding universe.
>> The law of conservation of energy cannot be applied.
>
>A third example is thermodynamics. Without an energy conservation law,
>there is difficulties to correctly define entropy and all the rest of
>thermodynamic in the framework of General Relativity.
More ignorant spew! Energy conservation is explicitly encoded by
virtue of the stress-energy tensor being divergence free, and is
sufficient for doing thermodynamics on curved space-time. See chapter
22 of MTW.
>
>There is not such problems in a field approach to gravity.
Do you get paid per shill, or what the fuck? You can never - ever -
not bring up alternative approaches no matter how off-topic or
pointless they are. Plus with your rampant misunderstandings about the
primary theory, how can you /possibly/ be qualified to objectively
evaluate alternative theories?
Ken S. Tucker
On Nov 3, 12:53 pm, carlip-nos...@physics.ucdavis.edu wrote:
> Ken S. Tucker <dynam...@vianet.on.ca> wrote:
>
> > Hi Tom
> > On Nov 3, 7:45 am, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
> > > Ken S. Tucker wrote:
>
> [...]
>
> > > > OTOH, Poisson's field equation \/^2 (p) = density
> > > > will solve M and p , such as, given the density
> > > > to be M/r^3 , then Poisson's becomes,
> > > > \/^2 (p) = M/r^3
>
> > > Yes, but irrelevant, as no sensible object has such a density.
> > Then you need to learn what density is.
>
> No, Tom is absolutely right here.
"absolutely right", is a dictum, I'm listening.
> The definition of mass density is that
> if you integrate it over any finite region, you obtain the mass in that
> region.
Ok, that's one way of doing it.
> Your expression implies that the "density" is nonzero in completely
> empty space. If you apply your "density" to, say, the Sun, you will find
> that you are claiming the "mass" is mostly outside the surface of the
> Sun. In fact, with your expression, the total "mass" of the Sun is infinite
> -- just integrate your "density" over all space.
Well integrated over all space is the integration
of probability, so naturally you'll get the unit
mass "M".
I simply pointed out...
+++++
Let there be a unit Mass M at the origin.
Let the density "T" be defined at any point
"r" from the origin as proportional to M/r^3,
(Mass/volume).
Let the Curvature "K" be defined as the rate of
change of gravitational acceleration, proportional
to K =d(M/r^2)/dr = M/r^3.
We see K=T is proportional, and is a simplified
derivation of the Einstein Field Equation direct
from Newton's gravity.
In Newton's gravity, the K is a tidal effect.
----------
> Further, if you do the math, you will find that the solution to your
> equation
> \/^2 (p) = M/r^3
> is
> p = -M/r (1+ ln(r/r_0) )
> where ln is the natural log and r_0 is an integration constant. This
> gives a gravitational field vastly different from the Newtonian one,
> and clearly ruled out by observation.
This is math, we have one dimension relating
the Mass "M" at origin to a point displaced by
a radius "r", so I double diff on "r" like,
d^2 (M/r)/dr^2 = 2M/r^3
works for me.
> Steve Carlip
Regards
Ken S. Tucker
carlip...@physics.ucdavis.edu
> On Nov 3, 12:53 pm, carlip-nos...@physics.ucdavis.edu wrote:
> > Ken S. Tucker <dynam...@vianet.on.ca> wrote:
[...]
> > Your expression implies that the "density" is nonzero in completely
> > empty space.
I note that you didn't respond to this.
> > If you apply your "density" to, say, the Sun, you will find
> > that you are claiming the "mass" is mostly outside the surface of the
> > Sun. In fact, with your expression, the total "mass" of the Sun is infinite
> > -- just integrate your "density" over all space.
> Well integrated over all space is the integration
> of probability, so naturally you'll get the unit
> mass "M".
No. The expression you wrote down has an infinite integral. There's
no way to normalize it.
[...]
> > Further, if you do the math, you will find that the solution to your
> > equation
> > \/^2 (p) = M/r^3
> > is
> > p = -M/r (1+ ln(r/r_0) )
> > where ln is the natural log and r_0 is an integration constant. This
> > gives a gravitational field vastly different from the Newtonian one,
> > and clearly ruled out by observation.
> This is math, we have one dimension relating
> the Mass "M" at origin to a point displaced by
> a radius "r", so I double diff on "r" like,
> d^2 (M/r)/dr^2 = 2M/r^3
> works for me.
It works for you because you don't understand spherical coordinates. The
part of the Laplacian involving r derivatives isn't d^2/dr^2, it's
1/r^2 d/dr (r^2 d/dr)
You can find this in any introductory book on mathematical physics. It's
elementary vector calculus.
Steve Carlip
Edward Green
> Tom Roberts <tjroberts...@sbcglobal.net> wrote:
> > eric gisse wrote:
> > > Perturbations of the metric (gravitational radiation) have a
> > > nonzero stress-energy tensor,
> > Not true, linear or otherwise. Vacuum in GR means T=0. Period.
> > Note, however, when a pulse of gravitational radiation interacts with
> > matter, T(before) != T(after), where T is the energy-momentum tensor
> > of the matter. So gravitational radiation can TRANSFER energy and
> > momentum,
>
> Right. This analysis was first suggested by Feynman at the 1957 Chapel
> Hill Conference. He proposed a thought experiment in which a bead on
> a stick has a gravitational wave pass by; the wave will move the bead
> back and forth, and the resulting friction will heat the stick.
I suppose it is a thing known with certainty that the beads will slide
back and forth on the stick, vs. the stick and beads together
deforming in a way which only has global significance, but no local
consequences?
Koobee Wublee
> Koobee Wublee wrote:
> >Gravitational waves have nothing to do with the field equations.
> ><shrug>
>
> Since you have abundantly established that you do not understand the
> notation, how the field equations are derived, how the field equations
> are used to derive solutions, what covariance means, what the metric
> means, what curvature means, how the metric relates to computing area
> or volume, what symmetries mean, why the metric is symmetric, why the
> connection coeffecients are symmetric, or what Birkhoff's theorem
> means, you are unfit to judge.
Well, let me say that again.
Gravitational waves have nothing to do with the field equations.
<shrug>
> The derivation of gravitational waves is straightfoward from
> perturbation theory. I'd give references but who would I be kidding?
> You are stupid.
Bullshit! There is no way to establish gravitational waves from the
field equations. <shrug>
eric gisse
<koobee...@gmail.com> wrote:
>On Nov 2, 11:06 pm, eric gisse wrote:
>> Koobee Wublee wrote:
>
>> >Gravitational waves have nothing to do with the field equations.
>> ><shrug>
>>
>> Since you have abundantly established that you do not understand the
>> notation, how the field equations are derived, how the field equations
>> are used to derive solutions, what covariance means, what the metric
>> means, what curvature means, how the metric relates to computing area
>> or volume, what symmetries mean, why the metric is symmetric, why the
>> connection coeffecients are symmetric, or what Birkhoff's theorem
>> means, you are unfit to judge.
>
>Well, let me say that again.
>
>Gravitational waves have nothing to do with the field equations.
><shrug>
Last time you uttered this nonsense, I asked you to show the error in
the derivation of gravitational radiation from the linearized theory.
You couldn't.
http://groups.google.com/group/sci.physics/msg/cf988b5deba2f957?dmode=source
In fact, when I presented you with a document containing the deivation
you ignored it and then went into a huge off-topic rant about the GR
Lagrangian. Which you *just* proved you know exactly fuck-all about.
>
>> The derivation of gravitational waves is straightfoward from
>> perturbation theory. I'd give references but who would I be kidding?
>> You are stupid.
>
>Bullshit! There is no way to establish gravitational waves from the
>field equations. <shrug>
Since you don't understand how to derive the field equations, much
less apply perturbation theory to them, I find this a pretty amusing
claim.
You have no idea how gravitational radiation is derived. You make
nonsense appeals to Noether's theorem - another thing you do not
understand - when you are unable to objectively evaluate the
perturbation calculations.
Juan R.
> On Mon, 3 Nov 2008 14:22:44 +0100 (CET), "Juan R." Gonz�lez-�lvarez
> <juanR...@canonicalscience.com> wrote:
>
>>"Juan R." Gonz�lez-�lvarez wrote on Mon, 03 Nov 2008 14:18:15 +0100:
>>
>>> Edward Green wrote on Fri, 31 Oct 2008 14:00:40 -0700:
>>>
>>>> I"ve heard that "energy conservation fails in GR"
>>>
>>> Yes, this is sometimes considered its greater problem.
>>>
>>> It may be remarked it is a problem *only* with geometric approach to
>>> gravity, not of field theory of gravity.
>>>
>>>> and perhaps this is an
>>>> example of it.
>>>
>>> Another known example is the cosmological model of expanding universe.
>>> The law of conservation of energy cannot be applied.
>>
>>A third example is thermodynamics. Without an energy conservation law,
>>there is difficulties to correctly define entropy and all the rest of
>>thermodynamic in the framework of General Relativity.
(snip nonsense from MTW)
>>There is not such problems in a field approach to gravity.
> Do you get paid per shill, or what the fuck? You can never - ever - not
> bring up alternative approaches no matter how off-topic or pointless
> they are. Plus with your rampant misunderstandings about the primary
> theory, how can you /possibly/ be qualified to objectively evaluate
> alternative theories?
Do you mean like when you claimed misunderstandings about the primary
theory of special relativity?
http://canonicalscience.blogspot.com/2008/08/some-samples-of-usenet-fauna-
ii.html
We still laugh :-)
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Juan R.
> On Mon, 3 Nov 2008 14:18:15 +0100 (CET), "Juan R." Gonz�lez-�lvarez
> <juanR...@canonicalscience.com> wrote:
>
>>Edward Green wrote on Fri, 31 Oct 2008 14:00:40 -0700:
>>
>>> I"ve heard that "energy conservation fails in GR"
>>
>>Yes, this is sometimes considered its greater problem.
(snip ignorance)
>
>>It may be remarked it is a problem *only* with geometric approach to
>>gravity, not of field theory of gravity.
(snip ad hominem)
>
>>> and perhaps this is an
>>> example of it.
>>
>>Another known example is the cosmological model of expanding universe.
>>The law of conservation of energy cannot be applied.
(snip more ignorance)
>>Precisely, some author doubt that gravitational waves can be detectable
>>without any *physical* energy mechanism.
(snip off-topic)
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
stric...@gmail.com
1) LIGO
2) Gravity Probe B
I mean, for the absence of gravity waves...
Wait, lest anybody argue with me, these experiments were DESIGNED to
detect g-waves. That they failed within that DESIGN, notwithstanding
the excuses and the equipment malfunction, is evidence for the non-
existence of g-waves.
stric...@gmail.com
> <strich.9...@gmail.com> wrote in message
>
> > Gravity waves...
>
> Are you speaking tlhIngan Hol ?
>
> ======================================
> Here's a gravity wave, right under his nose:
> http://tinyurl.com/5sav8h
The buggers must think a booger is a compactified g-wave :)
Ken S. Tucker
Pls be careful about deleting ref material.
On Nov 3, 5:23 pm, carlip-nos...@physics.ucdavis.edu wrote:
> Ken S. Tucker <dynam...@vianet.on.ca> wrote:
>
> > Hi Steve, it's good to see you posting again.
> > On Nov 3, 12:53 pm, carlip-nos...@physics.ucdavis.edu wrote:
> > > Ken S. Tucker <dynam...@vianet.on.ca> wrote:
>
> [...]
>
> > > Your expression implies that the "density" is nonzero in completely
> > > empty space.
>
> I note that you didn't respond to this.
Well it's because I did NOT imply that in,
density=M/r^3. You'll need to define "empty
space", with particular attention given to
the volume of the "space" you refer to.
I used the volume 4/3 pi r^3 but suppressed
the constant for simplicity.
> > > If you apply your "density" to, say, the Sun, you will find
> > > that you are claiming the "mass" is mostly outside the surface of the
> > > Sun. In fact, with your expression, the total "mass" of the Sun is infinite
> > > -- just integrate your "density" over all space.
> > Well integrated over all space is the integration
> > of probability, so naturally you'll get the unit
> > mass "M".
>
> No. The expression you wrote down has an infinite integral. There's
> no way to normalize it.
Well, it looks straight forward to me.
M = (M/r^3) r^3
or
let the density at any "r" be the Konstant M/r^3
and integrate as,
r
$ K dr^3 = K r^3 = M
0
> > > Further, if you do the math, you will find that the solution to your
> > > equation
> > > \/^2 (p) = M/r^3
> > > is
> > > p = -M/r (1+ ln(r/r_0) )
> > > where ln is the natural log and r_0 is an integration constant. This
> > > gives a gravitational field vastly different from the Newtonian one,
> > > and clearly ruled out by observation.
> > This is math, we have one dimension relating
> > the Mass "M" at origin to a point displaced by
> > a radius "r", so I double diff on "r" like,
> > d^2 (M/r)/dr^2 = 2M/r^3
> > works for me.
>
> It works for you because you don't understand spherical coordinates. The
> part of the Laplacian involving r derivatives isn't d^2/dr^2, it's
> 1/r^2 d/dr (r^2 d/dr)
>
> You can find this in any introductory book on mathematical physics. It's
> elementary vector calculus.
Ok Steve, thanks for the pointers, I'll do
a review and also study p = -M/r (1+ ln(r/r_0) )
I wonder what happens when r_0 = oo.
eric gisse
<juanR...@canonicalscience.com> wrote:
>eric gisse wrote on Mon, 03 Nov 2008 14:35:57 -0900:
>
>> On Mon, 3 Nov 2008 14:22:44 +0100 (CET), "Juan R." González-Álvarez
>> <juanR...@canonicalscience.com> wrote:
>>
>>>"Juan R." González-Álvarez wrote on Mon, 03 Nov 2008 14:18:15 +0100:
>>>
>>>> Edward Green wrote on Fri, 31 Oct 2008 14:00:40 -0700:
>>>>
>>>>> I"ve heard that "energy conservation fails in GR"
>>>>
>>>> Yes, this is sometimes considered its greater problem.
>>>>
>>>> It may be remarked it is a problem *only* with geometric approach to
>>>> gravity, not of field theory of gravity.
>>>>
>>>>> and perhaps this is an
>>>>> example of it.
>>>>
>>>> Another known example is the cosmological model of expanding universe.
>>>> The law of conservation of energy cannot be applied.
>>>
>>>A third example is thermodynamics. Without an energy conservation law,
>>>there is difficulties to correctly define entropy and all the rest of
>>>thermodynamic in the framework of General Relativity.
>
>(snip nonsense from MTW)
It is only 'nonsense' because it directly contradicts your claim.
>
>>>There is not such problems in a field approach to gravity.
>
>> Do you get paid per shill, or what the fuck? You can never - ever - not
>> bring up alternative approaches no matter how off-topic or pointless
>> they are. Plus with your rampant misunderstandings about the primary
>> theory, how can you /possibly/ be qualified to objectively evaluate
>> alternative theories?
>
>Do you mean like when you claimed misunderstandings about the primary
>theory of special relativity?
>
>http://canonicalscience.blogspot.com/2008/08/some-samples-of-usenet-fauna-
>ii.html
>
>We still laugh :-)
Whose 'we' ?
Parse your logs; how many unique visitors a week do you get to that
page? One? Two?
eric gisse
<juanR...@canonicalscience.com> wrote:
>eric gisse wrote on Mon, 03 Nov 2008 14:28:11 -0900:
>
>> On Mon, 3 Nov 2008 14:18:15 +0100 (CET), "Juan R." González-Álvarez
>> <juanR...@canonicalscience.com> wrote:
>>
>>>Edward Green wrote on Fri, 31 Oct 2008 14:00:40 -0700:
>>>
>>>> I"ve heard that "energy conservation fails in GR"
>>>
>>>Yes, this is sometimes considered its greater problem.
>
>(snip ignorance)
>
>>
>>>It may be remarked it is a problem *only* with geometric approach to
>>>gravity, not of field theory of gravity.
>
>(snip ad hominem)
>
>>
>>>> and perhaps this is an
>>>> example of it.
>>>
>>>Another known example is the cosmological model of expanding universe.
>>>The law of conservation of energy cannot be applied.
>
>(snip more ignorance)
"Sure it can. There exists a time-like Killing vector for the entire
set of FRW cosmologies."
Since that is the correct answer and you think it is ignorance,
clearly you are upset at your own ignorance. Why even respond if you
are just going to be a little bitch about it?
eric gisse
<dyna...@vianet.on.ca> wrote:
[...]
>> > > Further, if you do the math, you will find that the solution to your
>> > > equation
>> > > \/^2 (p) = M/r^3
>> > > is
>> > > p = -M/r (1+ ln(r/r_0) )
>> > > where ln is the natural log and r_0 is an integration constant. This
>> > > gives a gravitational field vastly different from the Newtonian one,
>> > > and clearly ruled out by observation.
>> > This is math, we have one dimension relating
>> > the Mass "M" at origin to a point displaced by
>> > a radius "r", so I double diff on "r" like,
>> > d^2 (M/r)/dr^2 = 2M/r^3
>> > works for me.
>>
>> It works for you because you don't understand spherical coordinates. The
>> part of the Laplacian involving r derivatives isn't d^2/dr^2, it's
>> 1/r^2 d/dr (r^2 d/dr)
>>
>> You can find this in any introductory book on mathematical physics. It's
>> elementary vector calculus.
>
>Ok Steve, thanks for the pointers, I'll do
>a review and also study p = -M/r (1+ ln(r/r_0) )
>I wonder what happens when r_0 = oo.
I think you need to study differential equations instead so you can
understand why integration constants are never equal to infinity.
Tom Roberts
> On Nov 3, 4:07 pm, carlip-nos...@physics.ucdavis.edu wrote:
>> This analysis was first suggested by Feynman at the 1957 Chapel
>> Hill Conference. He proposed a thought experiment in which a bead on
>> a stick has a gravitational wave pass by; the wave will move the bead
>> back and forth, and the resulting friction will heat the stick.
>
> I suppose it is a thing known with certainty that the beads will slide
> back and forth on the stick, vs. the stick and beads together
> deforming in a way which only has global significance, but no local
> consequences?
No. It is quite possible that the bead is stuck to the stick, so both
stick and bead deform when the gravitational wave passes by. No problem,
because this performs work on the bead and stick, generating heat.
Remember that atoms in a solid will arrange to be a definite PROPER
distance apart, with the value determined by their atomic properties. A
gravitational wave cannot change their atomic properties, but it can
change the metric of spacetime between the atoms, affecting their
spacing. So when the wave is at a peak along the line between a given
pair of atoms, those atoms will not be at their correct separation, and
they will feel an atomic (EM) force that pushes them in the direction to
make their spacing correct. That force will vary with the wave, and will
do work on them, heating the solid. Of course this is VERY small....
Yes, this is backwards from the usual thinking about EM
forces -- usually moving a charge a little bit changes
the force on it because it moves relative to other charges;
here no such motion is involved, but the distance between
charges changes due to the gravitational wave.
Exercise for the reader: compare the atomic relaxation times
of solids to the frequencies of gravitational radiation that
LIGO can detect.
Tom Roberts
Edward Green
> Edward Green wrote:
> > On Nov 3, 4:07 pm, carlip-nos...@physics.ucdavis.edu wrote:
> >> This analysis was first suggested by Feynman at the 1957 Chapel
> >> Hill Conference. He proposed a thought experiment in which a bead on
> >> a stick has a gravitational wave pass by; the wave will move the bead
> >> back and forth, and the resulting friction will heat the stick.
>
> > I suppose it is a thing known with certainty that the beads will slide
> > back and forth on the stick, vs. the stick and beads together
> > deforming in a way which only has global significance, but no local
> > consequences?
>
> No. It is quite possible that the bead is stuck to the stick, so both
> stick and bead deform when the gravitational wave passes by. No problem,
> because this performs work on the bead and stick, generating heat.
Well, I think you know what I meant! Clearly if the bead sticks, we
can still do work on the stick and bead together. I was thinking of
something more like the bead and stick both deforming as if they were
in a fun house mirror, without any local work -- or heating.
Possibly this makes no sense. Forget I said anything.
Koobee Wublee
> Koobee Wublee wrote:
> >Gravitational waves have nothing to do with the field equations.
> ><shrug>
>
> Last time you uttered this nonsense, I asked you to show the error in
> the derivation of gravitational radiation from the linearized theory.
Yes, I did. You just did not understand. You still cannot. <shrug>
http://groups.google.com/group/sci.physics/msg/cf988b5deba2f957?dmode=source
> In fact, when I presented you with a document containing the deivation
> you ignored it and then went into a huge off-topic rant about the GR
> Lagrangian. Which you *just* proved you know exactly fuck-all about.
What is that all about again?
> >Bullshit! There is no way to establish gravitational waves from the
> >field equations. <shrug>
>
> Since you don't understand how to derive the field equations, much
> less apply perturbation theory to them, I find this a pretty amusing
> claim.
Hmmm... These accusations are always useless. They always reflect
your very own incompetence. <shrug>
> [more whining crap snipped]
eric gisse
That's ok, we both know that you are an idiot and that the
protestations to the contrary are merely a formality required to save
face.
Juan R.
> It is only 'nonsense' because it directly contradicts your claim.
You arelady proved before you cannot differentiate nonsense from right
stuff.
http://canonicalscience.blogspot.com/2008/08/some-samples-of-usenet-fauna-
ii.html
>>We still laugh :-)
>
> Whose 'we' ?
Look around you :-)
> Parse your logs; how many unique visitors a week do you get to that
> page? One? Two?
Why are criying me about number of visitors?
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Juan R.
No exactly. They look for *GR* predicted waves, and there is not evidence
for that kind of waves.
If gravity is not geometrical (as some of us think) then they were
looking at wrong place and still there is room to find graviton waves.
Indeed, it seems that spin-0 graviton waves can explain certain
discrepancies for binary pulsars. People just would look for.
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Juan R.
> On Tue, 4 Nov 2008 14:37:02 +0100 (CET), "Juan R." González-Álvarez
> <juanR...@canonicalscience.com> wrote:
>
>>eric gisse wrote on Mon, 03 Nov 2008 14:28:11 -0900:
>>
>>> On Mon, 3 Nov 2008 14:18:15 +0100 (CET), "Juan R." González-Álvarez
>>> <juanR...@canonicalscience.com> wrote:
>>>
>>>>Edward Green wrote on Fri, 31 Oct 2008 14:00:40 -0700:
>>>>
>>>>> I"ve heard that "energy conservation fails in GR"
>>>>
>>>>Yes, this is sometimes considered its greater problem.
>>
>>(snip ignorance)
>>
>>
>>>>It may be remarked it is a problem *only* with geometric approach to
>>>>gravity, not of field theory of gravity.
>>
>>(snip ad hominem)
>>
>>
>>>>> and perhaps this is an
>>>>> example of it.
>>>>
>>>>Another known example is the cosmological model of expanding universe.
>>>>The law of conservation of energy cannot be applied.
>>
>>(snip more ignorance)
(snip more ignorance)
(snip insult)
>>>>Precisely, some author doubt that gravitational waves can be
>>>>detectable without any *physical* energy mechanism.
>>
>>(snip off-topic)
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
eric gisse
<juanR...@canonicalscience.com> wrote:
>eric gisse wrote on Tue, 04 Nov 2008 13:12:12 -0900:
>
>> It is only 'nonsense' because it directly contradicts your claim.
>
>You arelady proved before you cannot differentiate nonsense from right
>stuff.
The overall track record continues to be better than yours.
Why don't you explain why the presentation in MTW is wrong?
>
>http://canonicalscience.blogspot.com/2008/08/some-samples-of-usenet-fauna-
>ii.html
>
>>>We still laugh :-)
>>
>> Whose 'we' ?
>
>Look around you :-)
>
>> Parse your logs; how many unique visitors a week do you get to that
>> page? One? Two?
>
>Why are criying me about number of visitors?
If it proves your point, you could rub it in my face. Since it
doesn't, either you don't know how or it isn't enough to gloat over.
eric gisse
<juanR...@canonicalscience.com> wrote:
>strich.9991 wrote on Tue, 04 Nov 2008 08:04:09 -0800:
>
>> Here are the solid evidence for gravitational waves:
>>
>> 1) LIGO
>> 2) Gravity Probe B
I'd say you are stupid, but I'd have repeated myself.
LIGO has nulled to date - exclusion of a fair amount of models and
reduction in the relevant parameter space. Not enough to seriously
impinge on the theory.
GPB saw what was predicted, though LAGEOS I/II did it sufficiently
well and for a lot less money. But had nothing to do with
gravitational waves.
>>
>> I mean, for the absence of gravity waves...
>>
>> Wait, lest anybody argue with me, these experiments were DESIGNED to
>> detect g-waves. That they failed within that DESIGN, notwithstanding
>> the excuses and the equipment malfunction, is evidence for the non-
>> existence of g-waves.
GPB wasn't designed to detect gravitational radiation. 'tard.
>
>No exactly. They look for *GR* predicted waves, and there is not evidence
>for that kind of waves.
Are you so desparate to find someone who agrees with your stance that
you'll agree with obvious nonsense? GPB wasn't a gravitational wave
search, and you are clearly wrong about there being no evidence.
Making nonsense claims about my knowledge when you agree with obvious
stupidity is rather amusing.
We have at least two decaying binary systems [PSR 1913+16, PSR
J0737-3039] that work to spec. Explain the exactly predicted orbital
decay and match against your claim that there is no evidence.
Claiming there would be no direct evidence would be more honest and
correct, but you have to take that extra step.
>
>If gravity is not geometrical (as some of us think) then they were
>looking at wrong place and still there is room to find graviton waves.
Do you have any evidence to support these non-geometrical theories of
gravity that would justify mentioning them in every post you make to
this newsgroup? Any at all?
No, LIGO and associated detectors nulling so far does /not/ count. Not
enough, at least.
>
>Indeed, it seems that spin-0 graviton waves can explain certain
>discrepancies for binary pulsars. People just would look for.
Bwuh? A spin-0 wave corresponds to monopole radiation. That's silly.
Do you perhaps mean...spin-2? As both spin-0 and spin-1 radiation
[monopole, dipole] is /clearly/ ruled out by observation, I have no
idea what you are talking about.
But since you seem to believe it is true, let's see the literature
reference supporting your assertion.
Of course I'm sure I'll be asking for it again when you blithely snip
everything I write under the guise of non-sequitur arguments.
Juan R.
> On Wed, 5 Nov 2008 10:59:21 +0100 (CET), "Juan R." González-Álvarez
> <juanR...@canonicalscience.com> wrote:
>
>>eric gisse wrote on Tue, 04 Nov 2008 13:12:12 -0900:
>>
>>> It is only 'nonsense' because it directly contradicts your claim.
>>
>>You arelady proved before you cannot differentiate nonsense from right
>>stuff.
>
> The overall track record continues to be better than yours.
Remember you already said you are smarter
(\blockquote
Yea but I'm smarter than both of you. By such a wide margin it boggles
the mind.
)
http://canonicalscience.blogspot.com/2008/08/some-samples-of-usenet-
fauna-ii.html
And it was broadly showed you were *very* smart :-)
> Why don't you explain why the presentation in MTW is wrong?
Because there is not reason to waste time trying to explain something to
one crackpot as you.
Look for "do not argue against proud non-specialists" on
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
And example of Peter and me wasting several days to explain to you some
*elementary* facts of special relativity and mechanics is reported in
http://canonicalscience.blogspot.com/2008/08/some-samples-of-usenet-
fauna-ii.html
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Juan R.
(snip)
stric...@gmail.com
> On Tue, 4 Nov 2008 22:04:49 -0800 (PST), Koobee Wublee
>
>
>
>
>
> <koobee.wub...@gmail.com> wrote:
> >On Nov 4, 12:21 am, eric gisse wrote:
> >> Koobee Wublee wrote:
>
> >> >Gravitational waves have nothing to do with the field equations.
> >> ><shrug>
>
> >> Last time you uttered this nonsense, I asked you to show the error in
> >> the derivation of gravitational radiation from the linearized theory.
>
> >Yes, I did. You just did not understand. You still cannot. <shrug>
>
>
> >> In fact, when I presented you with a document containing the deivation
> >> you ignored it and then went into a huge off-topic rant about the GR
> >> Lagrangian. Which you *just* proved you know exactly fuck-all about.
>
> >What is that all about again?
>
> >> >Bullshit! There is no way to establish gravitational waves from the
> >> >field equations. <shrug>
>
> >> Since you don't understand how to derive the field equations, much
> >> less apply perturbation theory to them, I find this a pretty amusing
> >> claim.
>
> >Hmmm... These accusations are always useless. They always reflect
> >your very own incompetence. <shrug>
>
> >> [more whining crap snipped]
>
> That's ok, we both know that you are an idiot and that the
> protestations to the contrary are merely a formality required to save
Like when you cannot come up with your source, or when you did not
understand standard deviations?
Tom Roberts
> They look for *GR* predicted waves,
Of course, because GR is the current best theory of gravity. Articles
about gravitational waves mean those of GR, unless qualified with some
other theory (which in my experience is rare).
> and there is not evidence
> for that kind of waves.
No DIRECT evidence. But the binary pulsars give indirect evidence of
their existence. The fact that LIGO et al have not detected any signal
is only mildly troubling, as the current estimates for the rate of
detectable signals is about 1 for their elapsed observational time --
that certainly is no sensible refutation of GR.
> If gravity is not geometrical (as some of us think) then they were
> looking at wrong place and still there is room to find graviton waves.
Perhaps. What would be "the right place" to look? Can a search there be
combined with a search for the gravitational waves of GR? How sould
current detectors be modified to become sensitive to these "other"
gravitational waves?
> Indeed, it seems that spin-0 graviton waves can explain certain
> discrepancies for binary pulsars. People just would look for.
There are several theories which include spin-0 (scalar) fields, both
classical and semi-classical. At present theories like this (e.g.
Brans-Dicke) are experimentally indistinguishable from GR (if not, they
are almost surely refuted). That's why GR remains our best theory of
gravitation.
Tom Roberts
stric...@gmail.com
>
> No DIRECT evidence.
Case closed.
eric gisse
<juanR...@canonicalscience.com> wrote:
>eric gisse wrote on Wed, 05 Nov 2008 01:39:02 -0900:
>
>> On Wed, 5 Nov 2008 10:59:21 +0100 (CET), "Juan R." González-Álvarez
>> <juanR...@canonicalscience.com> wrote:
>>
>>>eric gisse wrote on Tue, 04 Nov 2008 13:12:12 -0900:
>>>
>>>> It is only 'nonsense' because it directly contradicts your claim.
>>>
>>>You arelady proved before you cannot differentiate nonsense from right
>>>stuff.
>>
>> The overall track record continues to be better than yours.
>
>Remember you already said you are smarter
>
>(\blockquote
> Yea but I'm smarter than both of you. By such a wide margin it boggles
> the mind.
>)
>
>http://canonicalscience.blogspot.com/2008/08/some-samples-of-usenet-
>fauna-ii.html
>
>And it was broadly showed you were *very* smart :-)
>
>> Why don't you explain why the presentation in MTW is wrong?
>
>Because there is not reason to waste time trying to explain something to
>one crackpot as you.
You can't, so you hide behind insults and idiocy. Par for the course.
[snip]
eric gisse
<juanR...@canonicalscience.com> wrote:
>eric gisse wrote on Wed, 05 Nov 2008 01:48:48 -0900:
>
>(snip)
>
>http://canonicalscience.blogspot.com/2008/08/some-samples-of-usenet-fauna-
>ii.html
Taking bets on how long Juan R. will use this page to deflect
criticisms of his stupidity, no matter how legitimate.
stric...@gmail.com
> On Thu, 6 Nov 2008 14:24:19 +0100 (CET), "Juan R." González-Álvarez
>
> >eric gisse wrote on Wed, 05 Nov 2008 01:48:48 -0900:
>
> >(snip)
>
> >ii.html
>
> Taking bets on how long Juan R. will use this page to deflect
> criticisms of his stupidity, no matter how legitimate.
I'll bet $10000 you are dishonest and stupid. Where do I collect?
Dono
<juanREM...@canonicalscience.com> wrote:
> strich.9991 wrote on Tue, 04 Nov 2008 08:04:09 -0800:
>
> > Here are the solid evidence for gravitational waves:
>
> > 1) LIGO
> > 2) Gravity Probe B
>
> > I mean, for the absence of gravity waves...
>
> > Wait, lest anybody argue with me, these experiments were DESIGNED to
> > detect g-waves. That they failed within that DESIGN, notwithstanding
> > the excuses and the equipment malfunction, is evidence for the non-
> > existence of g-waves.
>
> No exactly. They look for *GR* predicted waves, and there is not evidence
> for that kind of waves.
>
No, Juanshito
Contrary to your demented ideas, GP-B is not looking for "GR waves",
"predicted" or otherwise. GP-B is not looking for GR waves AT ALL.
stric...@gmail.com
> ... GP-B is not looking for "GR waves"...
Exactly. It cannot look for what does not exist.
Dono
No, cretin.
Ligo is designed for detecting GR waves.
GP-B is designed for detecting something else, old fart.
stric...@gmail.com
Which also does not exist, you stinking greenhorn
Dono
>
>
> > No, cretin.
> > Ligo is designed for detecting GR waves.
> > GP-B is designed for detecting something else, old fart.
>
> Which also does not exist, you stinking greenhorn
You don't know what it is, shithead :-)
Strich.9
I know what it is not, fecal brain :]
Androcles
"Strich.9" <stric...@gmail.com> wrote in message
news:28666e63-6906-46ff...@h23g2000prf.googlegroups.com...
=================================
"GR waves" ---- HAHAHAHAHAHAHAHA...
Ask the cretin what GR waves are.
Strich.9
> "Strich.9" <strich.9...@gmail.com> wrote in message
They spout enough stupidity as it is :)
Juan R.
>> and there is not evidence
>> for that kind of waves.
>
> No DIRECT evidence. But the binary pulsars give indirect evidence of
> their existence. The fact that LIGO et al have not detected any signal
> is only mildly troubling, as the current estimates for the rate of
> detectable signals is about 1 for their elapsed observational time --
> that certainly is no sensible refutation of GR.
In science, "indirect evidence" is never a substitute for a direct
confirmation. This is *why* more direct tests have been driven, and as you
recognize without success.
Indirect tests based in binary pulsars are of little help. Non-
geometrical theories of gravity also explain binary pulsars at least with
the same precision than GR, and the result is also interpreted in terms of
waves, but of a different kind.
>> If gravity is not geometrical (as some of us think) then they were
>> looking at wrong place and still there is room to find graviton waves.
>
> Perhaps. What would be "the right place" to look? Can a search there be
> combined with a search for the gravitational waves of GR? How sould
> current detectors be modified to become sensitive to these "other"
> gravitational waves?
Field theoreticians already suggested where to look.
>> Indeed, it seems that spin-0 graviton waves can explain certain
>> discrepancies for binary pulsars. People just would look for.
>
> There are several theories which include spin-0 (scalar) fields, both
> classical and semi-classical. At present theories like this (e.g.
> Brans-Dicke) are experimentally indistinguishable from GR (if not, they
> are almost surely refuted). That's why GR remains our best theory of
> gravitation.
First, you mix field with metric theories (in rigor BD belongs to the
latter kind).
Second, field theories with spin-0 waves are distinguisable from GR
(which has not spin-0 waves).
Third, I already explained about some authors claim the most accurate
observations of PSR1913+16 revealed the existence of about 1% excess in
energy losses can be interpreted as spin-0 scalar gravitational
radiation. I know of no mechanism to explain that using GR, and you?
This has been debated in recent cosmology conference. Look at proceedings
http://www.canonicalscience.org/en/publicationzone/
canonicalsciencetoday/20080516.html
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Juan R.
> On Nov 5, 2:03 am, "JuanShito R." González-Álvarez
> <juanREM...@canonicalscience.com> wrote:
>> strich.9991 wrote on Tue, 04 Nov 2008 08:04:09 -0800:
>>
>> > Here are the solid evidence for gravitational waves:
>>
>> > 1) LIGO
>> > 2) Gravity Probe B
>>
>> > I mean, for the absence of gravity waves...
>>
>> > Wait, lest anybody argue with me, these experiments were DESIGNED to
>> > detect g-waves. That they failed within that DESIGN, notwithstanding
>> > the excuses and the equipment malfunction, is evidence for the non-
>> > existence of g-waves.
>>
>> No exactly. They look for *GR* predicted waves, and there is not
>> evidence for that kind of waves.
>>
(snip insults and lie)
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Dono
<juanShito...@canonicalscience.com> wrote:
>
> >> > Here are the solid evidence for gravitational waves:
>
> >> > 1) LIGO
> >> > 2) Gravity Probe B
>
> >> > I mean, for the absence of gravity waves...
>
> >> > Wait, lest anybody argue with me, these experiments were DESIGNED to
> >> > detect g-waves. That they failed within that DESIGN, notwithstanding
> >> > the excuses and the equipment malfunction, is evidence for the non-
> >> > existence of g-waves.
>
> >> No exactly. They look for *GR* predicted waves, and there is not
> >> evidence for that kind of waves.
>
Juanshito,
You demonstrated your ignorance, GP-B is not looking for GR waves.
carlip...@physics.ucdavis.edu
[...]
> > No. The expression you wrote down has an infinite integral. There's
> > no way to normalize it.
> Well, it looks straight forward to me.
> M = (M/r^3) r^3
No. Integration is not multiplication.
> or
> let the density at any "r" be the Konstant M/r^3
> and integrate as,
> r
> $ K dr^3 = K r^3 = M
> 0
No. This is, again, an elementary mistake in math.
For a function that depends only on r, the integration measure
is 4\pi r^2 dr. If you have a density \rho = K/r^3, then the total
mass is
\int (K/r^3) 4\pi r^2 dr = 4\pi K \int dr/r = 4\pi K ln r.
If the range of r goes up to infinity, this is infinite.
There's nothing wrong with not understanding spherical coordinates,
or with not knowing how to do an integral. But you shouldn't pretend
that you do understand.
Steve Carlip
Ken S. Tucker
with a wee bit of ascii math...
On Nov 6, 3:05 pm, carlip-nos...@physics.ucdavis.edu wrote:
> Ken S. Tucker <dynam...@vianet.on.ca> wrote:
>
> [...]
>
> > > No. The expression you wrote down has an infinite integral. There's
> > > no way to normalize it.
> > Well, it looks straight forward to me.
> > M = (M/r^3) r^3
>
> No. Integration is not multiplication.
Sometimes it works out the same,
Volume = Length x Width x Height.
> > or
> > let the density at any "r" be the Konstant M/r^3
> > and integrate as,
> > r
> > $ K dr^3 = K r^3 = M
> > 0
>
> No. This is, again, an elementary mistake in math.
Well let's have a look.
Let K be a constant measured by an observer
at a constant distance R from the origin at
which there is a point Mass M then, definite
triple integration produces,
RRR
$$$ K*dr^3 = K*R^3 (Ken1)
000
> For a function that depends only on r, the integration measure
> is 4\pi r^2 dr. If you have a density \rho = K/r^3, then the total
> mass is
>
> \int (K/r^3) 4\pi r^2 dr = 4\pi K \int dr/r = 4\pi K ln r.
That's correct mathematically , however you are
summing (integrating) for ALL observers for all
"r" using an indefinite integral.
What our problem is: solve curvature and density
at a point, and using Eq.(Ken1), with K=M/R^3,
for those values of curvature and density Eq.(Ken1)
works perfectly.
> If the range of r goes up to infinity, this is infinite.
Of course, all you've done mathematically
is to take an infinite number of observers,
each measuring the Mass M, along "r" and then
multiply by M*(oo) = (oo) to get the total mass.
> There's nothing wrong with not understanding spherical coordinates,
> or with not knowing how to do an integral. But you shouldn't pretend that you do understand.
Right and be careful with integration too :-).
I think Eq.(Ken1) will get us a foundation.
> Steve Carlip
Regards
Ken S. Tucker
Juan R.
Dono
Juan R.
(snip)
Dono ignorance of most elementary aspects of both special relativity and
mechanics, his usual lies and insults, his cobards attempts to delete his
usenet messages for hiding his mistakes, and the history of how tried to
falsify Groups ratings voting for himself, before being caught by
administrators, are summarized in
http://canonicalscience.blogspot.com/2008/08/some-samples-of-usenet-fauna-
iii-nasty.html
See also:
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
for general guidelines to beat nasty trolls.
Juan R.
Dono
<juanREM...@canonicalscience.com> wrote:
> (snip)
>
> http://canonicalscience.blogspot.com/2008/08/some-samples-of-usenet-f...
Juanshito,
You are putting a lot of effort on your website that no one reads.
Every time you post here you get embarassed by various members.
Why don't you go and hide under a rock, Juanshito?
PS: Have you googled your name lately?
Dono
Juan R.
Dono
Strich.9
BULL SHITZ
It was made to look at NOTHING.