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Oz
Jan 20, 2012, 1:27:55 PM1/20/12
to
Hi guys, hope you are all well.
>From time to time I consider stuff. One thing, that was never answered
in the various sci.phy.res threads was actually how to view simple GR
spacetime, like, well here on earth for example.
Its important to realise that I am NOT talking about corrections top
Newtonian gravity. Let us consider ourselves, sat in a chair on the
surface of planet earth.
It seems to me that special relativity suggests that space is moving
towards the centre of the earth at g. Once I saw this it seemed an
inevitable consequence, after all a geodesic path (that of a freely
falling body) is to accelerate at g downwards. As far as I can see this
is an inevitable consequence although it has to be said that both I and
the falling body are also travelling through time.
I am not sure (but I suspect) that the gridlike, static co-ordinates of
the Scwharzschild metric are an artefact (or better, property) of the
metric. In particular I note that it assumes a A static spacetime is one
in which all metric components are independent of the time coordinate t.
That's fine, but planets, following ageodesic, move.
Comments?
--
=======================
Oz
Yes, THAT Oz....
robert bristow-johnson
Jan 20, 2012, 2:56:52 PM1/20/12
to
On 1/20/12 1:27 PM, Oz wrote:
>
>
> From time to time I consider stuff. One thing, that was never answered
> in the various sci.phy.res threads was actually how to view simple GR
> spacetime, like, well here on earth for example.
>
> Its important to realise that I am NOT talking about corrections top
> Newtonian gravity. Let us consider ourselves, sat in a chair on the
> surface of planet earth.
>
> It seems to me that special relativity suggests that space is moving
> towards the centre of the earth at g.
i don't think that special relativity speaks to gravitation at all.
>
>
> From time to time I consider stuff. One thing, that was never answered
> in the various sci.phy.res threads was actually how to view simple GR
> spacetime, like, well here on earth for example.
>
> Its important to realise that I am NOT talking about corrections top
> Newtonian gravity. Let us consider ourselves, sat in a chair on the
> surface of planet earth.
>
> It seems to me that special relativity suggests that space is moving
> towards the centre of the earth at g.
i think what general relativity says is that if you put a box around
that chair (and you) sitting on the surface of the planet so that you
cannot see out, that there is no physical experiment you can perform
that will differentiate between that case (sitting on the surface of the
planet) and accelerating in free space "upward" (in the direction toward
above your head) at a rate of g. GR says that, locally, both situations
are the same.
> Once I saw this it seemed an
> inevitable consequence, after all a geodesic path (that of a freely
> falling body) is to accelerate at g downwards.
accelerating at g.
--
r b-j r...@audioimagination.com
"Imagination is more important than knowledge."
Oz
Jan 21, 2012, 12:14:46 PM1/21/12
to
robert bristow-johnson <r...@audioimagination.com> writes
>i think what general relativity says is that if you put a box around
>that chair (and you) sitting on the surface of the planet so that you
>cannot see out, that there is no physical experiment you can perform
>that will differentiate between that case (sitting on the surface of the
>planet) and accelerating in free space "upward" (in the direction toward
>above your head) at a rate of g. GR says that, locally, both situations
>are the same.
Indeed so, but that's not the question. If we take a fleet of free
moving observers (typically clock carrying) we can define an inertial
frame. Actually I think its the only inertial frame, certainly locally.
Then this frame is accelerating towards the centre of the earth, with
respect to an observer at the surface, who is (we agree) on an
accelerating frame. I agree that its easy to determine which is the
accelerating frame.
>> Once I saw this it seemed an
>> inevitable consequence, after all a geodesic path (that of a freely
>> falling body) is to accelerate at g downwards.
>
>the geodesic path *is*, from a Euclidian perspective, downward and
>accelerating at g.
Ok, we are agreed then.
The question then is, where does the spacetime go?
Clearly at the centre of the earth there is no acceleration and the
spacetime is locally static and unchanging. Its certainly not "piling
up" anywhere, it just seems to dissipate 'onto' mass. In fact it seems
to be a valid point of view that mass is continually absorbing
spacetime.
Odd?
Worth some thought experiments?
>On 1/20/12 1:27 PM, Oz wrote:
>>
>> It seems to me that special relativity suggests that space is moving
>> towards the centre of the earth at g.
>
>i don't think that special relativity speaks to gravitation at all.
Oops, correct, I meant GR of course.
>>
>> It seems to me that special relativity suggests that space is moving
>> towards the centre of the earth at g.
>
>i don't think that special relativity speaks to gravitation at all.
>i think what general relativity says is that if you put a box around
>that chair (and you) sitting on the surface of the planet so that you
>cannot see out, that there is no physical experiment you can perform
>that will differentiate between that case (sitting on the surface of the
>planet) and accelerating in free space "upward" (in the direction toward
>above your head) at a rate of g. GR says that, locally, both situations
>are the same.
moving observers (typically clock carrying) we can define an inertial
frame. Actually I think its the only inertial frame, certainly locally.
Then this frame is accelerating towards the centre of the earth, with
respect to an observer at the surface, who is (we agree) on an
accelerating frame. I agree that its easy to determine which is the
accelerating frame.
>> Once I saw this it seemed an
>> inevitable consequence, after all a geodesic path (that of a freely
>> falling body) is to accelerate at g downwards.
>
>the geodesic path *is*, from a Euclidian perspective, downward and
>accelerating at g.
The question then is, where does the spacetime go?
Clearly at the centre of the earth there is no acceleration and the
spacetime is locally static and unchanging. Its certainly not "piling
up" anywhere, it just seems to dissipate 'onto' mass. In fact it seems
to be a valid point of view that mass is continually absorbing
spacetime.
Odd?
Worth some thought experiments?
mpc755
Jan 21, 2012, 12:15:21 PM1/21/12
to
is the pushing back and pressure exerted toward the Earth by the
aether displaced by the Earth.
'Ether and the Theory of Relativity by Albert Einstein'
http://www-groups.dcs.st-and.ac.uk/~history/Extras/Einstein_ether.html
"the state of the [ether] is at every place determined by connections
with the matter and the state of the ether in neighbouring places, ...
disregarding the causes which condition its state."
The state of the aether at every place determined by connections with
the matter and the state of the aether in neighboring places is the
state of displacement of the aether.
What is presently postulated as non-baryonic dark matter is aether.
Aether has mass. Aether physically occupies three dimensional space.
Aether is physically displaced by matter. Aether displaced by matter
pushes back toward the matter.
This is what Voyager found evidence of.
'NASA's Voyager Hits New Region at Solar System Edge'
http://www.nasa.gov/home/hqnews/2011/dec/HQ_11-402_AGU_Voyager.html
"Voyager is showing that what is outside is pushing back. ... Like
cars piling up at a clogged freeway off-ramp, the increased intensity
of the magnetic field shows that inward pressure from interstellar
space is compacting it."
The aether displaced by the solar system is pushing back and exerting
inward pressure toward the solar system. This is gravity.
Displaced aether pushing back toward matter is gravity.
Ken S. Tucker
Jan 21, 2012, 7:39:23 PM1/21/12
to
On Jan 20, 10:27 am, Oz <O...@mailcatch.com> wrote:
> Hi guys, hope you are all well.
>
> >From time to time I consider stuff. One thing, that was never answered
>
> in the various sci.phy.res threads was actually how to view simple GR
> spacetime, like, well here on earth for example.
>
> Its important to realise that I am NOT talking about corrections top
> Newtonian gravity. Let us consider ourselves, sat in a chair on the
> surface of planet earth.
>
> It seems to me that 'general' relativity suggests that space is moving
> Hi guys, hope you are all well.
>
> >From time to time I consider stuff. One thing, that was never answered
>
> in the various sci.phy.res threads was actually how to view simple GR
> spacetime, like, well here on earth for example.
>
> Its important to realise that I am NOT talking about corrections top
> Newtonian gravity. Let us consider ourselves, sat in a chair on the
> surface of planet earth.
>
> towards the centre of the earth at g. Once I saw this it seemed an
> inevitable consequence, after all a geodesic path (that of a freely
> falling body) is to accelerate at g downwards. As far as I can see this
> is an inevitable consequence although it has to be said that both I and
> the falling body are also travelling through time.
>
> I am not sure (but I suspect) that the gridlike, static co-ordinates of
> the Scwharzschild metric are an artefact (or better, property) of the
> metric. In particular I note that it assumes a A static spacetime is one
> in which all metric components are independent of the time coordinate t.
> That's fine, but planets, following ageodesic, move.
> Comments?
> Oz
> inevitable consequence, after all a geodesic path (that of a freely
> falling body) is to accelerate at g downwards. As far as I can see this
> is an inevitable consequence although it has to be said that both I and
> the falling body are also travelling through time.
>
> I am not sure (but I suspect) that the gridlike, static co-ordinates of
> the Scwharzschild metric are an artefact (or better, property) of the
> metric. In particular I note that it assumes a A static spacetime is one
> in which all metric components are independent of the time coordinate t.
> That's fine, but planets, following ageodesic, move.
> Comments?
> Yes, THAT Oz....
THAT OZ, one of the founding fathers of our illustrious SPF theoretics
lab!
About moving space: IMO is an ok analogy, rather like the rubber
sheet.
The Schwarz. CS's provides for a departure from 'flat' space time
measureable by
light rays and trajectories.
Superficially it should work, but what does one do with the Lorentz
Transform, that in GR is replaced by,
d^2 = g_uv dx^u dx^v, (1)
Specifically, the 'classical' LT is for uniform velocity.
Regards
Ken S. Tucker
Oz
Jan 22, 2012, 11:08:18 AM1/22/12
to
Ken S. Tucker <dyna...@vianet.on.ca> writes
not active there these days either.
>About moving space: IMO is an ok analogy, rather like the rubber
>sheet. The Schwarz. CS's provides for a departure from 'flat' space time
>measureable by
>light rays and trajectories.
>Superficially it should work, but what does one do with the Lorentz
>Transform, that in GR is replaced by,
>
>d^2 = g_uv dx^u dx^v, (1)
>
>Specifically, the 'classical' LT is for uniform velocity.
I'm not absolutely sure what you are saying here. Naively a flat
spacetime is 'static' in the sense that all objects are stationary in
the sense (one at any rate) that a lightbeam echo from one and back
again continuously reports the same delay. For low speeds this can be
replaced by a ruler.
The problem I have, and had even on the spr GR tutorial, is how we
express what we clearly see here on earth, which is an accelerating
frame of reference, as defined by co-moving observers in an inertial
frame. Where do the numbers come from? It should be simple, but nobody,
then and now, can actually point to an expression, isolate a term, and
evaluate it to give g.
Can't be hard, surely?
In the mean time I am forced to the view I offered earlier, which is
that spacetime is funnelling down onto massive particles. In fact this
rate is (to first order) directly proportional to the mass. If this is
so, its quite interesting in my view.
Where is charles when you need him?
Or even Baez, come to that!
>
>THAT OZ, one of the founding fathers of our illustrious SPF theoretics
>lab!
Er, um, yes. Ought to have altered it, it was for another group and I'm
>THAT OZ, one of the founding fathers of our illustrious SPF theoretics
>lab!
not active there these days either.
>About moving space: IMO is an ok analogy, rather like the rubber
>sheet. The Schwarz. CS's provides for a departure from 'flat' space time
>measureable by
>light rays and trajectories.
>Superficially it should work, but what does one do with the Lorentz
>Transform, that in GR is replaced by,
>
>d^2 = g_uv dx^u dx^v, (1)
>
>Specifically, the 'classical' LT is for uniform velocity.
spacetime is 'static' in the sense that all objects are stationary in
the sense (one at any rate) that a lightbeam echo from one and back
again continuously reports the same delay. For low speeds this can be
replaced by a ruler.
The problem I have, and had even on the spr GR tutorial, is how we
express what we clearly see here on earth, which is an accelerating
frame of reference, as defined by co-moving observers in an inertial
frame. Where do the numbers come from? It should be simple, but nobody,
then and now, can actually point to an expression, isolate a term, and
evaluate it to give g.
Can't be hard, surely?
In the mean time I am forced to the view I offered earlier, which is
that spacetime is funnelling down onto massive particles. In fact this
rate is (to first order) directly proportional to the mass. If this is
so, its quite interesting in my view.
Where is charles when you need him?
Or even Baez, come to that!
Daryl McCullough
Jan 23, 2012, 11:12:09 AM1/23/12
to
On Friday, January 20, 2012 1:27:55 PM UTC-5, Oz wrote:
> I am not sure (but I suspect) that the gridlike, static co-ordinates of
> the Scwharzschild metric are an artefact (or better, property) of the
> metric. In particular I note that it assumes a A static spacetime is one
> in which all metric components are independent of the time coordinate t.
> That's fine, but planets, following a geodesic, move.
Right. So if there are planets circling a star, then the metric is not
> I am not sure (but I suspect) that the gridlike, static co-ordinates of
> the Scwharzschild metric are an artefact (or better, property) of the
> metric. In particular I note that it assumes a A static spacetime is one
> in which all metric components are independent of the time coordinate t.
> That's fine, but planets, following a geodesic, move.
precisely equal to the Schwarzschild metric. What you can do, to get
an approximate answer in that case, is to use an iterative approach:
The first approximation is the Schwarzschild metric due to the
star's mass.
Then you compute the paths of the planets in this metric.
The second approximation to the metric adjusts the Schwarzschild
metric to take into account the gravitational effects of the planets.
Then use the new metric to compute the new paths. Which gives you
a third approximation to the metric.
Hopefully, there is a way to iterate that converges quickly. The
resulting metric will *not* by time-independent, and will not be
spherically symmetric, either.
I'm not sure that that answers your question, because I'm not sure
what your question was.
brad
Jan 23, 2012, 11:11:59 AM1/23/12
to
On Jan 22, 11:08 am, Oz <O...@mailcatch.com> wrote:
> The problem I have, and had even on the spr GR tutorial, is how we
> express what we clearly see here on earth, which is an accelerating
> frame of reference, as defined by co-moving observers in an inertial
> frame. Where do the numbers come from?
It's the time metric that varies with depth in a gravitational field.
> The problem I have, and had even on the spr GR tutorial, is how we
> express what we clearly see here on earth, which is an accelerating
> frame of reference, as defined by co-moving observers in an inertial
> frame. Where do the numbers come from?
So, if you
are sitting in your chair your time metric is constant. I think it's
simplistic
to simply state, "that's an 'accelerating frame of reference'." It's
more
accurate to consider it an inertial frame because the time metric
doesn't change.
>It should be simple, but nobody,
> then and now, can actually point to an expression, isolate a term, and
> evaluate it to give g.
momentum.
When 2 self gravitating bodies interact via stable orbital
relationship their fields
combine to generate a new configuration with their combined
spacetimes.
M MOMENTUM2
O
M this volume can
E can be modeled
N as pressure associated ((A crude diagram!!!))
T with the G fields of
U each body.
M
1
Pressure is associated with the energy of the field and is a source
itself. When the
orbit is stable this pressure is stable.(provided masses stay stable)
The point is that momentum is the source of the Lorentz Transform in
SR, but for stable
orbits it can also be associated with GR which is why GR reduces to SR
in each local reference
frame.
> Can't be hard, surely?
>
> In the mean time I am forced to the view I offered earlier, which is
> that spacetime is funnelling down onto massive particles. In fact this
> rate is (to first order) directly proportional to the mass. If this is
> so, its quite interesting in my view.
illustrate the changing metric.
Aetherists need an aether to generate all physical phenomena. In one
respect their view is valid. Einstein modeled a theory that only
required
a maleable spacetime. This abrogated the need for an aether. But,
Einstein
also recognized that he had no model for how mass and space interact!
He had the maths, but no description of the actual physical
interaction.
This is your original question. It was settled by noting that the
paths were
curved so the safest answer was "curved spacetime". (whatever that
is! )
Brad
Ken S. Tucker
Jan 23, 2012, 11:11:26 AM1/23/12
to
On Jan 22, 8:08 am, Oz <O...@mailcatch.com> wrote:
> Ken S. Tucker <dynam...@vianet.on.ca> writes
language
(math) you're comfortable with.
> In the mean time I am forced to the view I offered earlier, which is
> that spacetime is funnelling down onto massive particles. In fact this
> rate is (to first order) directly proportional to the mass. If this is
> so, its quite interesting in my view.
>
> Where is charles when you need him?
> Or even Baez, come to that!
If there is a specific oddity in their online publications, some of
the
fellows in the group might clarify the meaning.
Perhaps a pointer to the problem, GR isn't necessarily complete.
Regards
Ken S. Tucker
> Ken S. Tucker <dynam...@vianet.on.ca> writes
>
>
>
> >THAT OZ, one of the founding fathers of our illustrious SPF theoretics
> >lab!
>
> Er, um, yes. Ought to have altered it, it was for another group and I'm
> not active there these days either.
>
> >About moving space: IMO is an ok analogy, rather like the rubber
> >sheet. The Schwarz. CS's provides for a departure from 'flat' space time
> >measureable by
> >light rays and trajectories.
> >Superficially it should work, but what does one do with the Lorentz
> >Transform, that in GR is replaced by,
>
> >d^2 = g_uv dx^u dx^v, (1)
>
> >Specifically, the 'classical' LT is for uniform velocity.
>
> I'm not absolutely sure what you are saying here. Naively a flat
> spacetime is 'static' in the sense that all objects are stationary in
> the sense (one at any rate) that a lightbeam echo from one and back
> again continuously reports the same delay. For low speeds this can be
> replaced by a ruler.
>
> The problem I have, and had even on the spr GR tutorial, is how we
> express what we clearly see here on earth, which is an accelerating
> frame of reference, as defined by co-moving observers in an inertial
> frame. Where do the numbers come from? It should be simple, but nobody,
> then and now, can actually point to an expression, isolate a term, and
> evaluate it to give g.
>
> Can't be hard, surely?
Yes, Classical GR is fairly straightforward, it depends on the
>
>
> >THAT OZ, one of the founding fathers of our illustrious SPF theoretics
> >lab!
>
> Er, um, yes. Ought to have altered it, it was for another group and I'm
> not active there these days either.
>
> >About moving space: IMO is an ok analogy, rather like the rubber
> >sheet. The Schwarz. CS's provides for a departure from 'flat' space time
> >measureable by
> >light rays and trajectories.
> >Superficially it should work, but what does one do with the Lorentz
> >Transform, that in GR is replaced by,
>
> >d^2 = g_uv dx^u dx^v, (1)
>
> >Specifically, the 'classical' LT is for uniform velocity.
>
> I'm not absolutely sure what you are saying here. Naively a flat
> spacetime is 'static' in the sense that all objects are stationary in
> the sense (one at any rate) that a lightbeam echo from one and back
> again continuously reports the same delay. For low speeds this can be
> replaced by a ruler.
>
> The problem I have, and had even on the spr GR tutorial, is how we
> express what we clearly see here on earth, which is an accelerating
> frame of reference, as defined by co-moving observers in an inertial
> frame. Where do the numbers come from? It should be simple, but nobody,
> then and now, can actually point to an expression, isolate a term, and
> evaluate it to give g.
>
> Can't be hard, surely?
language
(math) you're comfortable with.
> In the mean time I am forced to the view I offered earlier, which is
> that spacetime is funnelling down onto massive particles. In fact this
> rate is (to first order) directly proportional to the mass. If this is
> so, its quite interesting in my view.
>
> Where is charles when you need him?
> Or even Baez, come to that!
the
fellows in the group might clarify the meaning.
Perhaps a pointer to the problem, GR isn't necessarily complete.
Regards
Ken S. Tucker
Tom Roberts
Jan 24, 2012, 2:07:49 AM1/24/12
to
On 1/23/12 1/23/12 10:12 AM, Daryl McCullough wrote:
> So if there are planets circling a star, then the metric is not
> precisely equal to the Schwarzschild metric. What you can do, to get
> an approximate answer in that case, is to use an iterative approach:
> The first approximation is the Schwarzschild metric due to the
> star's mass.
> Then you compute the paths of the planets in this metric.
> The second approximation to the metric adjusts the Schwarzschild
> metric to take into account the gravitational effects of the planets.
> Then use the new metric to compute the new paths. Which gives you
> a third approximation to the metric.
> Hopefully, there is a way to iterate that converges quickly. The
> resulting metric will *not* by time-independent, and will not be
> spherically symmetric, either.
Yes. In the solar system this converges rapidly: for most purposes only the
> So if there are planets circling a star, then the metric is not
> precisely equal to the Schwarzschild metric. What you can do, to get
> an approximate answer in that case, is to use an iterative approach:
> The first approximation is the Schwarzschild metric due to the
> star's mass.
> Then you compute the paths of the planets in this metric.
> The second approximation to the metric adjusts the Schwarzschild
> metric to take into account the gravitational effects of the planets.
> Then use the new metric to compute the new paths. Which gives you
> a third approximation to the metric.
> Hopefully, there is a way to iterate that converges quickly. The
> resulting metric will *not* by time-independent, and will not be
> spherically symmetric, either.
first approximation is needed. I doubt there is any experiment or observation
that needs more than the second approximation.
Note, however, that this procedure is only appropriate in regions of weak
gravitation -- that is needed because it relies on gravitation being
(approximately) linear. It also requires that the planets be "small" compared to
the star (in our solar system, planets/sun ~ 0.001), so they approximately
follow geodesics. Near the earth, the dimensionless parameter characterizing the
"strength of gravity" is about 1E-6, which is indeed small. Don't expect this to
work near a neutron star or a black hole.
Tom Roberts
Tom Roberts
Jan 24, 2012, 2:09:39 AM1/24/12
to
On 1/20/12 1/20/12 12:27 PM, Oz wrote:
> Let us consider ourselves, sat in a chair on the
> surface of planet earth.
>
> It seems to me that [general] relativity suggests that space is moving
> Let us consider ourselves, sat in a chair on the
> surface of planet earth.
>
> towards the centre of the earth at g.
That is an outrageously naive analogy that is tantamount to being wrong. It is
by no means original with you, but is still utterly inappropriate for GR. Yes,
for tiny objects that are moving slowly relative to the earth surface, it is
true that they accelerate downward at g. But for tiny but fast-moving objects
and for light in vacuum this is not true. This is basically because GR is not
linear, and because velocities add via the Lorentz composition of velocities,
not (3-)vector addition.
It is also wrong in that the notion of space "moving" cannot be
well defined. OBJECTS move, not space. Space is a manifold, and
the concept "motion" simply does not apply.
Objects must be "tiny" so their effects on the geometry of the manifold can be
neglected.
> Once I saw this it seemed an
> inevitable consequence, after all a geodesic path (that of a freely
> falling body) is to accelerate at g downwards.
earth with less acceleration in an earth-surface frame.
Just think about it: if an object initially at rest in your frame moves with
velocity v after a time t, then an object initially moving vertically with
velocity u in your frame cannot move with velocity u+v after time t, because the
Lorentz composition of velocities is not linear. So you cannot really assign a
"universal acceleration" to all objects.
Note that this requires exceedingly good measuring accuracy,
far better than achievable in practice. But you weren't talking
about experimental measurements.
> I am not sure (but I suspect) that the gridlike, static co-ordinates of
> the Scwharzschild metric are an artefact (or better, property) of the
> metric.
over what region they are valid, but coordinates are completely arbitrary --
after all, there are a half-dozen different coordinates commonly used on the
Schwarzschild manifold.
> In particular I note that it assumes a A static spacetime is one
> in which all metric components are independent of the time coordinate t.
when the time coordinate is also a Killing vector of the manifold.
A stationary region of a manifold is one which admits a timelike
Killing vector. A static region of a manifold is stationary and
it is possible to define 3-d space orthogonal to the timelike
killing vector.
> That's fine, but planets, following a geodesic, move.
on the geometry of the manifold can be neglected.
> The question then is, where does the spacetime go?
> Clearly at the centre of the earth there is no acceleration and the
> spacetime is locally static and unchanging. Its certainly not "piling
> up" anywhere, it just seems to dissipate 'onto' mass. In fact it seems
> to be a valid point of view that mass is continually absorbing
> spacetime.
> Naively a flat
> spacetime is 'static' in the sense that all objects are stationary
anyway, which excludes all known objects.
Remember that spacetime is a MODEL, and we inevitably use approximations in
constructing our models. So the APPROXIMATION of a flat spacetime can be useful
in situations where the approximation is good enough for the purpose.
That is, at base, why we can use SR to analyze particle experiments
-- the approximation involved is vastly better than our measuring
accuracy.
brad said: > It's the time metric that varies with depth in a gravitational field.
No. It is the TIME COMPONENT of the metric, in suitable coordinates, that varies
with depth. Such things are tricky in GR, because coordinates are so fungible.
> So, if you
> are sitting in your chair your time metric is constant.
sun, planets, and all other massive objects moving relative to you, then the
metric components do not depend on your time coordinate.
> I think it's
> simplistic
> to simply state, "that's an 'accelerating frame of reference'." It's
> more
> accurate to consider it an inertial frame because the time metric
> doesn't change.
not follow uniform straight lines (they are accelerated by the gravity of the
earth). If you actually LOOK at the metric components in those coordinates, they
are not diag(1,-1,-1,-1), which is required for a locally inertial frame in
relativity. If you construct coordinates near your chair with those metric
components, you'll find the coordinate system is accelerating downward at g
(assuming you and your chair were initially at rest in them).
Tom Roberts
Ken S. Tucker
Jan 25, 2012, 10:21:45 AM1/25/12
to
On Jan 23, 11:09 pm, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
> On 1/20/12 1/20/12 12:27 PM, Oz wrote:
>
> > Let us consider ourselves, sat in a chair on the
> > surface of planet earth.
>
> > It seems to me that [general] relativity suggests that space is moving
> > towards the centre of the earth at g.
>
> That is an outrageously naive analogy that is tantamount to being wrong. It is
> by no means original with you, but is still utterly inappropriate for GR. Yes,
> for tiny objects that are moving slowly relative to the earth surface, it is
> true that they accelerate downward at g. But for tiny but fast-moving objects
> and for light in vacuum this is not true. This is basically because GR is not
> linear, and because velocities add via the Lorentz composition of velocities,
> not (3-)vector addition.
I think OZ may replace 2Φ by escape velocity V thus rendering the
> On 1/20/12 1/20/12 12:27 PM, Oz wrote:
>
> > Let us consider ourselves, sat in a chair on the
> > surface of planet earth.
>
> > It seems to me that [general] relativity suggests that space is moving
> > towards the centre of the earth at g.
>
> That is an outrageously naive analogy that is tantamount to being wrong. It is
> by no means original with you, but is still utterly inappropriate for GR. Yes,
> for tiny objects that are moving slowly relative to the earth surface, it is
> true that they accelerate downward at g. But for tiny but fast-moving objects
> and for light in vacuum this is not true. This is basically because GR is not
> linear, and because velocities add via the Lorentz composition of velocities,
> not (3-)vector addition.
gravitional effect on spacetime to be,
dr' = dr / √ (1 - V2 / c2)
dt' = dt √ (1 - V2 / c2)
g_00 ≡ (1 - V2 /c2) , g_11 = 1/g_00
If that's what OZ means by "velocity of space" then it works ok.
It's a good introduction to simple GR effects.
Regards
Ken S. Tucker
-------------------------
ΔΘΛΞΠΣΦΨΩαβγδεζηθκλμνξπρφψ∂∆∑√∫≈≠≡⌠⌡⌡
Tom
Jan 25, 2012, 10:22:01 AM1/25/12
to
On Jan 20, 1:27 pm, Oz <O...@mailcatch.com> wrote:
> Hi guys, hope you are all well.
>
> >From time to time I consider stuff. One thing, that was never answered
>
> in the various sci.phy.res threads was actually how to view simple GR
> spacetime, like, well here on earth for example.
>
> Its important to realise that I am NOT talking about corrections top
> Newtonian gravity. Let us consider ourselves, sat in a chair on the
> surface of planet earth.
>
> It seems to me that special relativity suggests that space is moving
> towards the centre of the earth at g.
You mean general relativity. Even if one wants to think of
>
> >From time to time I consider stuff. One thing, that was never answered
>
> in the various sci.phy.res threads was actually how to view simple GR
> spacetime, like, well here on earth for example.
>
> Its important to realise that I am NOT talking about corrections top
> Newtonian gravity. Let us consider ourselves, sat in a chair on the
> surface of planet earth.
>
> It seems to me that special relativity suggests that space is moving
> towards the centre of the earth at g.
acceleration this way, the action is time symmetric with net
zero acceleration.
General relativity is a classical field theory with no
privileged reference frame, so it doesn't make physical
sense to assign an absolute rest frame -- such as Newtonian
center of mass -- to the interaction of matter and spacetime.
Tom
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