Static Gravitational Field...
Mike_Fontenot
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
Tom Roberts
> 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.
Let me use the usual notation used in these newsgroups, which is
basically TeX expressions without braces or dollar signs. So I'll write
that equation:
g_m0 = 0
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.
Now I must interpret Dirac's words in terms of these more modern
definitions:
* a "static gravitational field" in a given region means the
manifold itself is static in that region.
* a "static coordinate system" makes sense only in a static
region; it uses the timelike Killing vector as its time
coordinate, and applies the other 3 coordinates to the
orthogonal 3-space such that every integral curve of the
Killing vector has constant spatial coordinates throughout
the region of validity of the c.s. (which need not span
the entire static region of the manifold).
So in a region with a "static gravitational field" using "static
coordinates" we have:
d/dx^0 g_uv = 0 [d = partial derivative]
-- this is just another way of stating that d/dx^0 is a Killing vector
(necessarily timelike). We also have:
g_m0 = 0 [m=1,2,3]
-- this is just stating that the spatial coordinates are orthogonal to
the time coordinate.
> 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?
If g_m0 != 0, then the spatial coordinate m is not orthogonal to the
time coordinate, but the above definitions require that all 3 spatial
coordinates be orthogonal to it.
Tom Roberts
Ken S. Tucker
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
Oh No
deal with Killing vectors, and I don't much like them myself, because
what seems to happen at that point is that you substitute technical
definitions and formulae when what is wanted is intuitive understanding.
Also I would have a preference for applying "stationary" and "static" to
the metric, rather than to the manifold, because if pushed I would have
to say that I am not sure what applying it to the manifold means, except
in terms of how it is applied to the metric.
Ultimately though, it comes down to the same thing, and I would
therefore just repeat the key point. He said
So in a region with a "static gravitational field" using "static
coordinates" we have:
d/dx^0 g_uv = 0 [d = partial derivative]
. We also have:
g_m0 = 0 [m=1,2,3]
-- this is just stating that the spatial coordinates are orthogonal to
the time coordinate.
and remark that this is what we have by applying sr locally. Iow when
this is not so, we are moving wrt the coordinate system.
Regards
--
Charles Francis
moderator sci.physics.foundations.
charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and
braces)
Ken S. Tucker
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
Tom Roberts
> we're using dg00/dt=0, as the
> definition of a static g-field is that ok?
In general, no. See my earlier response for the relevant definitions.
With x^0 notated as t, you should use dg_uv/dt=0 for u,v=0,1,2,3.
But in the Newtonian approximation using Newtonian coordinates [#], g_00
is the only component significantly different from Minkowski spacetime,
so the correct formula reduces to yours.
[#] the original poster quoted Dirac's "static coordinate
system"; these are a subset of them.
Tom Roberts
Tom Roberts
> I can't really improve on Tom's answer, but I know that Dirac does not
> deal with Killing vectors, and I don't much like them myself, because
> what seems to happen at that point is that you substitute technical
> definitions and formulae when what is wanted is intuitive understanding.
Hmmm. The original poster indicated he wanted a proof, and technical definitions
are required for that.
The "intuitive understanding" of Killing vectors is that they represent
symmetries of the manifold (with metric) -- sliding everything equally along a
Killing vector changes nothing. So a timelike Killing vector represents time
invariance; an appropriately circular Killing vector represents rotational
invariance, etc.
> Also I would have a preference for applying "stationary" and "static" to
> the metric, rather than to the manifold, because if pushed I would have
> to say that I am not sure what applying it to the manifold means, except
> in terms of how it is applied to the metric.
In physics, the manifold is really a "manifold with metric" and the two cannot
be meaningfully separated. Mathematically, of course, a given manifold can be
given numerous different metrics, and your preference clearly holds.
Physicists usually omit the "with metric"; mathematicians rarely do.
> Ultimately though, it comes down to the same thing, and I would
> therefore just repeat the key point. He said
>> So in a region with a "static gravitational field" using "static
>> coordinates" we have:
>> d/dx^0 g_uv = 0 [d = partial derivative]
>> . We also have:
>> g_m0 = 0 [m=1,2,3]
>> -- this is just stating that the spatial coordinates are orthogonal to
>> the time coordinate.
>
> and remark that this is what we have by applying sr locally.
No. SR always applies locally (to an accuracy that is determined by how large
"locally" is), in ANY manifold of GR. Those equations are a MUCH stronger
statement about the structure of the manifold (with metric).
> Iow when
> this is not so, we are moving wrt the coordinate system.
I'm not sure to whom "we" refers. Nor can I see how to fix it -- if the first
equation above is not so, then the time coordinate is not a Killing vector, and
if the second is not so then the coordinates are not "static" (in Dirac's sense,
which I updated to modern terminology earlier). But I see no "motion wrt the
coordinate system" here (which would be expressed as dx^i/dx^0 != 0 for i=1,2,3
with the {x^i} representing "our" position at time x^0).
Tom Roberts
Ken S. Tucker
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_geode...
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
Mike_Fontenot
>
> So in a region with a "static gravitational field" using "static
> coordinates" we have:
> . We also have:
> g_m0 = 0 [m=1,2,3]
> -- this is just stating that the spatial coordinates are orthogonal to
> the time coordinate.
>
I DO understand that in 3D spatial coordinates, that a diagonal metric
implies (and is implied by) orthogonality. But I haven't been able to
wrap my mind around what orthogonality really means for the time
coordinate and any of the three spatial coordinates in 4D
spacetime...I've got to think about that some more.
The direction I was heading was this: that for a static gravitational
field, and a choice of a static coodinate system, that the "spatial
contribution" to (ds)^2 is independent of dt. I.e., for dt = dx^0 = 0,
and some arbitrary but fixed choices of dx^m, we can compute
ds^2 = (sum) dx^m * dx^m == D_0,
(where the index m denotes index values of 1,2, and 3, but not 0, versus
greek indices which denote all four coordinates). D_0 is the "spatial
contribution" to ds^2 when there is NO time displacement (and, in that
case, D_0 is of course the TOTAL ds^2, because there is no time
contribution).
The static requirement (in addition to the obvious requirement that the
time derivatives of all metric elements are zero) would be that the
above "spatial contribution for dt = 0", D_0, will ALWAYS be the spatial
contribution to ds^2, for ANY choice of dt.
So, the requirement for static field and static coordinates would be
that for the original choice of nonzero x^m, and with ANY choice of x^0,
we would require that
ds^2 = dx^0 * dx^0 + D_0
ds^2 = dx^0 * dx^0 + (sum) dx^m * dx^m.
But since it is ALWAYS true (for any field, and for any choice of
coordinate system) that
ds^2 = (sum) dx^alpha * dx^alpha,
then the requirement is clearly equivalent to the requirement that
g_0m = g_m0 = 0.
This seems intuitively reasonable to me, but I don't know if it was what
Dirac had in mind or not.
Mike Fontenot
Tom Roberts
> I DO understand that in 3D spatial coordinates, that a diagonal metric
> implies (and is implied by) orthogonality. But I haven't been able to
> wrap my mind around what orthogonality really means for the time
> coordinate and any of the three spatial coordinates in 4D
> spacetime...I've got to think about that some more.
When g_m0 = 0 for m=1,2,3, it obviously means the spatial coordinates are
orthogonal to the time coordinate. That implies that the integral curves of
d/dx^0 have constant spatial coordinates -- a clock at rest in these coordinates
displays the time coordinate if g_00 = -1 (possibly with constant offset).
Relative to a locally inertial frame, coordinates with some g_m0 != 0 are
accelerated relative to the inertial frame. Note the converse is not true.
> The direction I was heading was this: [...]
What you write is correct, but is merely re-writing the original equation with
different symbols (such as your D_0).
For a static coordinate system in a static region of spacetime, the set of
points with a given value of the time coordinate forms a 3-d spatial submanifold
such that no point is in the past or future of any other point (Hawking and
Ellis call this an achronal surface).
Tom Roberts
Mike_Fontenot
ANOTHER CORRECTION, third attempt: (The "^" in ds^2 got lost in
processing my previous correction...sorry for all the false starts and
wasted bandwidth).
_____________________________________________________
Oh No wrote:
>
> So in a region with a "static gravitational field" using "static
> coordinates" we have:
[...]
> . We also have:
> g_m0 = 0 [m=1,2,3]
> -- this is just stating that the spatial coordinates are orthogonal to
> the time coordinate.
>
I DO understand that in 3D spatial coordinates, that a diagonal metric
implies (and is implied by) orthogonality. But I haven't been able to
wrap my mind around what orthogonality really means for the time
coordinate and any of the three spatial coordinates in 4D
spacetime...I've got to think about that some more.
The direction I was heading was this: that for a static gravitational
field, and a choice of a static coodinate system, that the "spatial
contribution" to (ds)^2 is independent of dt. I.e., for dt = dx0 = 0,
and some arbitrary but fixed choices of dx^j, we can compute
ds^2 = (sum) g_mn * dx^m * dx^n == D_0,
(where the indices m and n denote index values of 1,2, and 3, but not 0,
versus greek indices which denote all four coordinates). D_0 is the
"spatial contribution" to ds^2 when there is NO time displacement (and,
in that case, D_0 is of course the TOTAL ds^2, because there is no time
contribution).
The static requirement (in addition to the obvious requirement that the
time derivatives of all metric elements are zero) would be that the
above "spatial contribution for dt = 0", D_0, will ALWAYS be the spatial
contribution to ds^2, for ANY choice of dt.
So, the requirement for static field and static coordinates would be
that for the original choice of nonzero x^j, and with ANY choice of x0,
we would require that
ds^2 = g_00 * dx0 * dx0 + D_0
ds^2 = g_00 * dx0 * dx0 + (sum) g_mn * dx^m * dx^n.
But since it is ALWAYS true (for any field, and for any choice of
coordinate system) that
ds^2 = (sum) g_alpha_beta * dx^alpha * dx^beta,
Mike_Fontenot
Sorry, I just realized that I omitted ALL of the metric elements in my
submission of a few minutes ago! Here is the corrected version:
________________________________________________
Oh No wrote:
>
> So in a region with a "static gravitational field" using "static
> coordinates" we have:
[...]
> . We also have:
> g_m0 = 0 [m=1,2,3]
> -- this is just stating that the spatial coordinates are orthogonal to
> the time coordinate.
>
I DO understand that in 3D spatial coordinates, that a diagonal metric
implies (and is implied by) orthogonality. But I haven't been able to
wrap my mind around what orthogonality really means for the time
coordinate and any of the three spatial coordinates in 4D
spacetime...I've got to think about that some more.
The direction I was heading was this: that for a static gravitational
field, and a choice of a static coodinate system, that the "spatial
contribution" to (ds)2 is independent of dt. I.e., for dt = dx0 = 0,
and some arbitrary but fixed choices of dx^j, we can compute
ds2 = (sum) g_mn * dx^m * dx^n == D_0,
(where the indices m and n denote index values of 1,2, and 3, but not 0,
versus greek indices which denote all four coordinates). D_0 is the
"spatial contribution" to ds2 when there is NO time displacement (and,
in that case, D_0 is of course the TOTAL ds2, because there is no time
contribution).
The static requirement (in addition to the obvious requirement that the
time derivatives of all metric elements are zero) would be that the
above "spatial contribution for dt = 0", D_0, will ALWAYS be the spatial
contribution to ds2, for ANY choice of dt.
So, the requirement for static field and static coordinates would be
that for the original choice of nonzero x^j, and with ANY choice of x0,
we would require that
ds2 = g_00 * dx0 * dx0 + D_0
ds2 = g_00 * dx0 * dx0 + (sum) g_mn * dx^m * dx^n.
But since it is ALWAYS true (for any field, and for any choice of
coordinate system) that
ds2 = (sum) g_alpha_beta * dx^alpha * dx^beta,
Tom Roberts
> 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.
Those are NOT instances of a static gravitational field (as being discussed in
this thread). Those are merely instances of locations at which the metric is
constant. The key notion you missed is that a static g-field is static
throughout a REGION of the manifold, not just at a single point, or on a path or
surface.
Need I point out that a region of a manifold necessarily has
the same dimensionality as the manifold itself? Your orbit is
only 1-D, and your surface is only 3-D, but the manifold is of
course 4-D.
Exercise for Ken: explain why your examples are only 1-D and 3-D.
[Ken, this is a VERY simple exercise, intended to permit you
to assess your own understanding. Or lack thereof.]
There is no such thing as a static g-field with two masses, unless they are
rigidly connected (in which case they act as a single object; beware of the
impossibility of truly rigid connections in relativity). "Sitting in a chair"
qualifies, and a universe consisting of just a planet and you sitting motionless
(forever) on a chair could be a static manifold.
Exercise for Ken: I said COULD be -- what other conditions are
required for that to be a static manifold?
[Ken, this is a VERY simple exercise, intended to permit you
to assess your own understanding. Or lack thereof.]
> We can employ the geodesic as ref'd here, [...]
Of course such a geodesic is only a 1-D path (in a 4-D manifold). So you could
"employ" it, but it cannot give any understanding about "static g-field",
because it does not define a REGION of the manifold. Of course if you had
instead discussed a congruence of geodesics, that COULD be used to do so....
Exercise for Ken: explain why a geodesic is 1-D.
[Ken, this is a VERY simple exercise, intended to permit you
to assess your own understanding. Or lack thereof.]
Advanced exercise: explain why a congruence of geodesics can
span a 4-D region of the manifold. Hint1: it need not do so.
Hint2: first explain what "congruence" means in this context.
More advanced exercise: explain how a 4-D congruence of
geodesics can be used to determine whether or not the
region it spans is static or not. Hint: I have an idea,
but have not proven it does this....
Tom Roberts
Ken S. Tucker
> Ken S. Tucker wrote:
> > 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.
>
> Those are NOT instances of a static gravitational field (as being discussed in
> this thread). Those are merely instances of locations at which the metric is
> constant. The key notion you missed is that a static g-field is static
> throughout a REGION of the manifold, not just at a single point, or on a path or
> surface.
I'm suggesting obtaining a deeper understanding of
the problem from a study of geodesics.
OTOH, Einstein provides the metrics for a static
field in GR1916, Eq.(70), (available online), but
that should be well known to a GR student.
...
Ken S. Tucker
Mike_Fontenot
> What you write is correct, but is merely re-writing the original
> equation with different symbols (such as your D_0).
What I was trying to do was get from the original assumption of "a
static gravitational field and a static coordinate system" to some
requirement on the metric that I could immediately intuitively see as
following from the "static" assumption. Maybe this would describe what
I was trying to do better:
First, let's explicitly show the arguments of ds^2, with the time
displacement and the spatial displacements shown separately, by writing
ds^2 = ds^2(dt, dx^m).
Then we can denote the specific value of ds^2 for some specific (but
arbitrary) choices of those arguments by writing
ds^2 = ds^2(dt = a, dx^m = b^m) == ds^2(a, b^m).
Now, let D_m denote the way ds^2 CHANGES when the spatial differential
displacements dx^m (initially all zero) are changed to some arbitrary,
but fixed values dx^m = b^m:
D_m(t) = ds^2(dt, b^m) - ds^2(dt, 0),
where D_m is shown as a function of dt, because the value of dt hasn't
yet been specified.
For a general field, the value of D_m(t) will generally be different for
different values of dt. But, it seems intuitively reasonable to me
that, for a static field, and a static coordinate system, that D_m
SHOULD just be some number, the same for ANY value of dt. (Remember
that D_m is defined for some PARTICULAR, fixed choices for dx^m).
So, we can get the value of D_m by setting dt = 0, and calculating
D_m(0) = ds^2(0, b^m) - ds^2(0, 0)
D_m(0) = ds^2(0, b^m)
D_m(0) = (sum) g_mn * b^m * b^n.
Now, note that
ds^2(dt, 0) = g_00 * dt * dt,
So
ds^2(dt, b^m) = D_m + ds^2(dt, 0)
ds^2(dt, b^m) = D_m + g_00 * dt * dt
ds^2(dt, b^m) = (sum) g_mn * b^m * b^n + g_00 * dt * dt.
But that latter equation then implies that
g_0m = g_m0 = 0,
since the general expression for ds^2(dt, b^m) is
ds^2(dx^0, dx^m) = (sum) g_alpha_beta * dx^alpha * dx^beta,
with dx^0 = dt and dx^m = b^m.
So, what seemed to me to be an intuitively reasonable requirement when
the field and coordinate system are static DOES indeed lead to Dirac's
constraint on the g_0m elements of the metric, and that's what I wanted
to accomplish.
Mike Fontenot
Ken S. Tucker
> On Dec 7, 9:12 am, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
>
> > Ken S. Tucker wrote:
> > > we're using dg00/dt=0, as the
> > > definition of a static g-field is that ok?
>
> > In general, no. See my earlier response for the relevant definitions.
> > With x^0 notated as t, you should use dg_uv/dt=0 for u,v=0,1,2,3.
>
> > But in the Newtonian approximation using Newtonian coordinates [#], g_00
> > is the only component significantly different from Minkowski spacetime,
> > so the correct formula reduces to yours.
>
> > [#] the original poster quoted Dirac's "static coordinate
> > system"; these are a subset of them.
>
> > Tom Roberts
>
> 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.
>
>
> 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
A correction on indices above,
dU^r/ds = - {r,ab} U^a U^b = 0
(AFAIK) in the two circumstances of the static g-field
I suggested above.
From my study, I think ambiguity does exist when dealing with
'nonorthogonal dynamic spacetime metrics' such as g_i0, wherein
the g_i0 depend upon relativity as defined herein,
http://physics.trak4.com/modern-spacetime.pdf
that intents to extending the ISU c=rt decision to modernize
spacetime, that appears superficially consistent according to,
The evolution of tensor calculus was straightforward and was
applied multidimensionally to static geometry, specifically as
a result of map survey's on a spherical Earth.
Einstein and crew later - mainly successfully - based on
Minkowski's spacetime, fused time into the metric, changing
a static math to one that is dynamic, due to time inclusion.
The move to a dynamic spacetime by AE et al, is (IMO) not
well documented, so I posted some online articles on MST,
based on the 1987 ISU definition of the meter.
It goes deeper though, when rotations are considered, that
rotations (as a current in a loop) reverses the antisymmetric
of the g12 = -g21 in the field.
Regards
Ken S. Tucker
Ken S. Tucker
pardon my ramblings...
On Dec 13, 11:22 pm, "Ken S. Tucker" <dynam...@vianet.on.ca> wrote:
> On Dec 9, 1:30 pm, "Ken S. Tucker" <dynam...@vianet.on.ca> wrote:
...
> that intents to extending the ISU c=r/t decision to modernize
> spacetime, that appears superficially consistent according to,
>
> http://physics.trak4.com/
>
> The evolution of tensor calculus was straightforward and was
> applied multidimensionally to static geometry, specifically as
> a result of map survey's on a spherical Earth.
> Einstein and crew later - mainly successfully - based on
> Minkowski's spacetime, fused time into the metric, changing
> a static math to one that is dynamic, due to time inclusion.
>
> The move to a dynamic spacetime by AE et al, is (IMO) not
> well documented, so I posted some online articles on MST,
> based on the 1987 ISU definition of the meter.
>
> It goes deeper though, when rotations are considered, that
> rotations (as a current in a loop) reverses the antisymmetric
> of the g12 = -g21 in the field.
I think most guys who have studied tensors calculus do
certainly understand it as it was developed for static
geometries, even in multidimensions, the math seems
straightforward to me, and the usual common text book
basically provides us with 3D static spatial.
I'd like to suggest I find a deparature from a static tensor
calculus when time is infused to form spacetime, and then
the symbolic logic of tensors is applied directly therein.
We may demark the math into static and dynamic tensor
analysis, with the latter being the infusion of time as
defined by relativity, via the inter-relation of the the speed
of light, the second, and the meter.
GR (and I'll add QT) has forced a mathematical condition
onto the usual static tensor math such as we are unable
to instantly know any finite part of the surface, but instead
only know it by finite "c" translation of information and in
physics, the relation of effects.
For those reasons, I think we should set forth a definite
"dynamic" tensor analysis.
Regards
Ken S. Tucker
glird
> A region of a manifold is said to be "stationary" iff there is a timelike Killing vector throughout the region.
>
That rules out the LTE and SR. In them, the time co-ordinate t is
orthogonal to the perpendicular axes thus co-incident with the
direction of motion per esynched system.
glird
Tom Roberts
> On Dec 4 2009, 5:31 pm, Tom Roberts wrote:
>> A region of a manifold is said to be "stationary" iff there is a timelike
>> Killing vector throughout the region.
>> 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.
>
> That rules out the LTE and SR.
I'm not sure what you mean by "LTE" [#][%], but the Minkowski manifold of SR is
most definitely static. It is just about the only Lorentzian manifold with an
infinite number of timelike Killing vectors, and there is a 3-d spatial
submanifold orthogonal to each and every one of them throughout the entire manifold.
Just take any inertial frame -- its time coordinate is a
(timelike) Killing vector, and its 3-space is orthogonal
to the time coordinate. These statements are quite clear
from the fact that in such a frame the metric components
are diag(-1,1,1,1) -- the fact that none of them depend on
the time coordinate shows that the time coordinate is a
(timelike) Killing vector; the fact that all off-diagonal
elements are zero shows that the 3 spatial coordinates are
orthogonal to the time coordinate.
[#] If you meant LET, the Lorentz Ether Theory, its only
manifold is that of 3-space, and the Semi-Riemannian
concept of a TIMELIKE Killing vector does not apply.
Remember that my original article restricted everything I
said to the Lorentzian manifolds of GR. LET is experimentally
indistinguishable from SR, but is a different theory.
[%] If you meant Lorentz transform equations, they're
irrelevant as they merely relate inertial coordinates, and
my statements are all purely geometrical, independent of any
coordinate choice. Those equations are part and parcel of SR.
> In them, the time co-ordinate t is orthogonal
> to the perpendicular axes thus co-incident with the direction of motion per
> esynched system.
You clearly do not understand the actual meanings of the words I used. There is
no relevant "direction of motion" [@], and Einstein clock synchronization is
used only to construct the coordinates of an inertial frame; it is of no
geometrical significance.
[@] Such a direction of motion only applies BETWEEN inertial
frames; but my statements were all geometrical in nature,
completely independent of any coordinate system (or frame).
Tom Roberts
glird
TR: I'm not sure what you mean by "LTE" [#][%],
but the Minkowski manifold of SR is most definitely
static. It is just about the only Lorentzian manifold
with an infinite number of timelike Killing vectors,
and there is a 3-d spatial submanifold orthogonal
to each and every one of them throughout the
entire manifold.
[snip]
� Remember that my original article restricted
everything I said to the Lorentzian manifolds
of GR.
>
glird: In them, the time co-ordinate t is orthogonal
to the perpendicular axes thus is co-incident with
the direction of motion per esynched system.
>
TR: You clearly do not understand the actual meanings
of the words I used. >
Do you? If so, please define "inertial frames" and
give us an example.
TR: �Such a direction of motion only applies BETWEEN inertial
frames; but my statements were all geometrical in nature,
completely independent of any coordinate system (or frame). >
While you are at it, give us an example of a
"geometrical" statement that is unrelated to
a frame of reference. [Note that to normal people
a 4 by 4 matrix is a mathematical abstraction,
NOT a geometrical object.]
glird
glird
wrote:
>< When g_m0 = 0 for m=1,2,3, it obviously means the spatial coordinates are orthogonal to the time coordinate. >
It is obviously impossible for three mutually perpendicular lines to
all be perpendicular to a fourth line.
>< That implies that the integral curves of d/dx^0 have constant spatial coordinates -- a clock at rest in these coordinates displays the time coordinate if g_00 = -1 (possibly with constant offset).>
In Relativity, time is the indications of the hands of a clock
REGARDLESSegardless
> Relative to a locally inertial frame, coordinates with some g_m0 != 0 are
> accelerated relative to the inertial frame. Note the converse is not true.
>
> > The direction I was heading was this: [...]
>
> What you write is correct, but is merely re-writing the original equation with
> different symbols (such as your D_0).
>
> For a static coordinate system in a static region of spacetime, the set of
> points with a given value of the time coordinate forms a 3-d spatial submanifold
> such that no point is in the past or future of any other point (Hawking and
> Ellis call this an achronal surface).
>
> Tom Roberts
======================================= MODERATOR'S COMMENT:
Please trim (delete) uncommented quoted text. -fd
Tom Roberts
Any statement about vectors or tensors is unrelated to any coordinates,
including frames of reference. This includes such statements as:
* the curvature of this manifold is zero everywhere
* this object follows a geodesic path in the manifold
* that object does not follow a geodesic path in the manifold
* ... zillions more ...
> [Note that to normal people
> a 4 by 4 matrix is a mathematical abstraction,
> NOT a geometrical object.]
Hmmm. ALL OF SCIENCE is abstractions about the world; in physics these are
primarily mathematical ones.
Nowhere did I mention any "4 by 4 matrix"; note that tensors are NOT matrices.
The phrase "geometrical object" is part of the TECHNICAL vocabulary and refers
to a quantity related only to the geometry of the manifold, unrelated to any
coordinate system. So vectors and tensors are geometrical objects.
[I'm through jumping through hoops to help you understand
very basic stuff like this. You need to STUDY.]
Tom Roberts
Tom Roberts
> On Dec 9 2009, 8:58 pm, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
>> < When g_m0 = 0 for m=1,2,3, it obviously means the spatial coordinates are
>> orthogonal to the time coordinate. >
>
> It is obviously impossible for three mutually perpendicular lines to all be
> perpendicular to a fourth line.
Not true, in a FOUR dimensional manifold, such as in relativity.
Tom Roberts