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From: Ken S. Tucker on 6 Dec 2009 22:18
On Dec 5, 1:49 pm, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote:
> On Dec 4, 1:15 pm, Mike_Fontenot <mlf...(a)comcast.net> wrote:
>
>
>
> > Dirac (in his GR book) says that the metric for a static gravitational
> > field, referred to a static coordinate system, has g{sub m0} = 0. I.e.,
> > with the metric written as a matrix (with the zero-th row and column
> > representing the time coordinate), the zero-th row and the zero-th
> > column will be all zeros, except for the zero-th element.
>
> > I thought I had proven that, but I recently decided my proof isn't
> > correct. Anyone have a proof of that?
>
> > Also, I'm not certain that I understand the phrase "referred to a static
> > coordinate system". I THINK that just means that, for a static
> > gravitational field, the coordinate system CAN be chosen so that the
> > relative relationships between the three spatial coordinates are the
> > same for ANY value of the time coordinate. Is that correct? And is
> > there a more precise way to say that?
> > Mike Fontenot
>
> To Mike,
> At the point you're at, I'd suggest working out the geodesic
> for a circular orbit, because the distance from a large central
> gravitating Mass (M) and a small test mass (m) is constant.
> I can post a bit of math if you're interested, and see what you
> and the group thinks.
> Regards
> Ken S. Tucker
>
> ======================================= MODERATOR'S COMMENT:
> I don't think this has much to do with Mike's question, but I shall let it pass. CF

We (our science club), could use some advice: By specifying
a 'circular' orbit, relating M and m, we're using dg00/dt=0, as the
definition of a static g-field is that ok?
Thanks
Ken S. Tucker
PS: This summer we (wife and I) hosted star party's at our new
Skyview property, and have a 4" computerized Goto refractor, a
4.5" Newtonian Reflector, a 60 mm refractor, and 7x50 binos.
As temperatures get cold our new study is indoor microscopy,
using "Ken-a-Vision" scopes and remote controlled telescopes.
Ken
From: Igor on 7 Dec 2009 13:14
On Dec 6, 10:18 pm, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote:
> On Dec 5, 1:49 pm, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote:
>
>
>
> > On Dec 4, 1:15 pm, Mike_Fontenot <mlf...(a)comcast.net> wrote:
>
> > > Dirac (in his GR book) says that the metric for a static gravitational
> > > field, referred to a static coordinate system, has g{sub m0} = 0.  I.e.,
> > > with the metric written as a matrix (with the zero-th row and column
> > > representing the time coordinate), the zero-th row and the zero-th
> > > column will be all zeros, except for the zero-th element.
>
> > > I thought I had proven that, but I recently decided my proof isn't
> > > correct.  Anyone have a proof of that?
>
> > > Also, I'm not certain that I understand the phrase "referred to a static
> > > coordinate system".  I THINK that just means that, for a static
> > > gravitational field, the coordinate system CAN be chosen so that the
> > > relative relationships between the three spatial coordinates are the
> > > same for ANY value of the time coordinate.  Is that correct?  And is
> > > there a more precise way to say that?
> > >         Mike Fontenot
>
> > To Mike,
> > At the point you're at, I'd suggest working out the geodesic
> > for a circular orbit, because the distance from a large central
> > gravitating Mass (M) and a small test mass (m) is constant.
> > I can post a bit of math if you're interested, and see what you
> > and the group thinks.
> > Regards
> > Ken S. Tucker
>
> > ======================================= MODERATOR'S COMMENT:
> >  I don't think this has much to do with Mike's question, but I shall let it pass. CF
>
> We (our science club), could use some advice: By specifying
> a 'circular' orbit, relating M and m, we're using dg00/dt=0, as the
> definition of a static g-field is that ok?
> Thanks
> Ken S. Tucker
> PS: This summer we (wife and I) hosted star party's at our new
> Skyview property, and have a 4" computerized Goto refractor, a
> 4.5" Newtonian Reflector, a 60 mm refractor, and 7x50 binos.
> As temperatures get cold our new study is indoor microscopy,
> using "Ken-a-Vision" scopes and remote controlled telescopes.
> Ken

A static metric is defined as one that only depends on the spatial
coordinates.

From: Tom Roberts on 7 Dec 2009 19:43
Igor wrote:
> A static metric is defined as one that only depends on the spatial
> coordinates.

No. It's considerably more complicated than that, because the
coordinates themselves are arbitrary, and in a small enough region of
any manifold with metric one could always come up with an outlandish
coordinate system such that the "metric only depends on the spatial
coordinates".


Here are the relevant definitions, in geometrical terms; all manifolds
are Lorentzian (i.e. 4-d para-compact semi-Riemannian manifolds with
signature 2; they are usually also Hausdorff):

A region of a manifold is said to be "stationary" iff there is a
timelike Killing vector throughout the region.

Examples: Minkowski spacetime (infinite number of timelike
killing vectors); the region outside the horizon of Kerr
spacetime (one timelike Killing vector).

A region of a manifold is said to be "static" iff it is stationary and
at every point in the region there exists a neighborhood of the point in
question with a 3-d spatial submanifold orthogonal to the timelike
Killing vector at every point within the neighborhood.

Examples: Minkowski spacetime; the region outside the
horizon of Schwarzschild spacetime.

NOTE: some authors apply "stationary" and "static" to the metric, some
to the manifold itself; I do the latter (because in physics we only
consider a "manifold with metric" and the last two words are
often/usually omitted).


Tom Roberts


From: xxein on 9 Dec 2009 00:38
On Dec 7, 7:43 pm, Tom Roberts <tjroberts...(a)sbcglobal.net> wrote:
> Igor wrote:
> > A static metric is defined as one that only depends on the spatial
> > coordinates.
>
> No. It's considerably more complicated than that, because the
> coordinates themselves are arbitrary, and in a small enough region of
> any manifold with metric one could always come up with an outlandish
> coordinate system such that the "metric only depends on the spatial
> coordinates".
>
> Here are the relevant definitions, in geometrical terms; all manifolds
> are Lorentzian (i.e. 4-d para-compact semi-Riemannian manifolds with
> signature 2; they are usually also Hausdorff):
>
> A region of a manifold is said to be "stationary" iff there is a
> timelike Killing vector throughout the region.
>
>      Examples: Minkowski spacetime (infinite number of timelike
>      killing vectors); the region outside the horizon of Kerr
>      spacetime (one timelike Killing vector).
>
> A region of a manifold is said to be "static" iff it is stationary and
> at every point in the region there exists a neighborhood of the point in
> question with a 3-d spatial submanifold orthogonal to the timelike
> Killing vector at every point within the neighborhood.
>
>      Examples: Minkowski spacetime; the region outside the
>      horizon of Schwarzschild spacetime.
>
> NOTE: some authors apply "stationary" and "static" to the metric, some
> to the manifold itself; I do the latter (because in physics we only
> consider a "manifold with metric" and the last two words are
> often/usually omitted).
>
> Tom Roberts

xxein: Tom? Take your own test. What is the difference between the
physic and your definition of physics to describe it?
From: Ken S. Tucker on 9 Dec 2009 15:56
On Dec 8, 9:38 pm, xxein <xx...(a)comcast.net> wrote:
> On Dec 7, 7:43 pm, Tom Roberts <tjroberts...(a)sbcglobal.net> wrote:
>
>
>
> > Igor wrote:
> > > A static metric is defined as one that only depends on the spatial
> > > coordinates.
>
> > No. It's considerably more complicated than that, because the
> > coordinates themselves are arbitrary, and in a small enough region of
> > any manifold with metric one could always come up with an outlandish
> > coordinate system such that the "metric only depends on the spatial
> > coordinates".
>
> > Here are the relevant definitions, in geometrical terms; all manifolds
> > are Lorentzian (i.e. 4-d para-compact semi-Riemannian manifolds with
> > signature 2; they are usually also Hausdorff):
>
> > A region of a manifold is said to be "stationary" iff there is a
> > timelike Killing vector throughout the region.
>
> > Examples: Minkowski spacetime (infinite number of timelike
> > killing vectors); the region outside the horizon of Kerr
> > spacetime (one timelike Killing vector).
>
> > A region of a manifold is said to be "static" iff it is stationary and
> > at every point in the region there exists a neighborhood of the point in
> > question with a 3-d spatial submanifold orthogonal to the timelike
> > Killing vector at every point within the neighborhood.
>
> > Examples: Minkowski spacetime; the region outside the
> > horizon of Schwarzschild spacetime.
>
> > NOTE: some authors apply "stationary" and "static" to the metric, some
> > to the manifold itself; I do the latter (because in physics we only
> > consider a "manifold with metric" and the last two words are
> > often/usually omitted).
>
> > Tom Roberts
>
> xxein: Tom? Take your own test. What is the difference between the
> physic and your definition of physics to describe it?

So far I have no way of generalizing a "static g-field".
At 1st strike, I find two physical instances of a "static"
g-field, where the g-potential relating two bodies with masses
M and m, remains constant,
1) Circular orbit.
2) m on the surface of M, such as we (m) sit in a chair.

We can employ the geodesic as ref'd here,
http://en.wikipedia.org/wiki/Solving_the_geodesic_equations#The_geodesic_equation

I'll write (in easy ascii) as,

dU^a/ds = - {a,bc} U^a U^b , using the U^a = dx^a/ds.

then I'll *specialize* the CS to a polar with x^1 being
our radius "r", and then set the condition,
(I'm using r as an index to indicate a CS specialized),

dU^r /ds = 0

for a static field to be true in (1) and (2) above.
If that's ok, I'll display the RHS work, using,

{a,bc} U^a U^b = 0

Regards
Ken S. Tucker
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