| Topic: |
Science > Physics |
| User: |
"Thomas Smid" |
| Date: |
28 Mar 2006 09:16:11 AM |
| Object: |
General Relativity and Gravity |
Hi,
It is generally claimed that gravity in GR is not a force but a
consequence of space curvature which in turn is created by the masses.
Should then not GR be able to describe the motion of masses
self-consistently without use of the gravitational constant G? The
Schwartzschild metric depends on the Schwartzschild radius R=2GM/c^2,
and G obviously has to be determined by (force) measurements.
Also, how can gravity depend on the curvature of space? A curvature can
not be defined locally by just one point, so how would a point mass
'know' about the local curvature? (the usual 'rubber sheet' example
where one mass produces a 'dent' into which the other mass falls is in
my opinion not adequate as the mass follows here merely the tangential
component of the gravity force (obviously it can't follow the force
component normal to the rubber surface as this component is neutralized
by the oppositely acting molecular force the rubber exerts on the
mass)).
Thomas
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| User: "Henry Haapalainen" |
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| Title: Re: General Relativity and Gravity |
28 Mar 2006 03:21:00 PM |
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"Thomas Smid" <thomas.smid@gmail.com> kirjoitti viestissä
news:1143558971.699628.53080@v46g2000cwv.googlegroups.com...
Hi,
It is generally claimed that gravity in GR is not a force but a
consequence of space curvature which in turn is created by the masses.
Should then not GR be able to describe the motion of masses
self-consistently without use of the gravitational constant G? The
Schwartzschild metric depends on the Schwartzschild radius R=2GM/c^2,
and G obviously has to be determined by (force) measurements.
Also, how can gravity depend on the curvature of space? A curvature can
not be defined locally by just one point, so how would a point mass
'know' about the local curvature? (the usual 'rubber sheet' example
where one mass produces a 'dent' into which the other mass falls is in
my opinion not adequate as the mass follows here merely the tangential
component of the gravity force (obviously it can't follow the force
component normal to the rubber surface as this component is neutralized
by the oppositely acting molecular force the rubber exerts on the
mass)).
Thomas
In falling space theory those mistakes of relativity have been corrected. It
also tells, why gravitational constant is needed in some calculations.
Falling space: http://www.wakkanet.fi/~fields/
Henry Haapalainen
.
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| User: "Mike" |
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| Title: Re: General Relativity and Gravity |
28 Mar 2006 04:29:23 PM |
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Henry Haapalainen wrote:
"Thomas Smid" <thomas.smid@gmail.com> kirjoitti viestiss=E4
news:1143558971.699628.53080@v46g2000cwv.googlegroups.com...
Hi,
It is generally claimed that gravity in GR is not a force but a
consequence of space curvature which in turn is created by the masses.
Should then not GR be able to describe the motion of masses
self-consistently without use of the gravitational constant G? The
Schwartzschild metric depends on the Schwartzschild radius R=3D2GM/c^2,
and G obviously has to be determined by (force) measurements.
Also, how can gravity depend on the curvature of space? A curvature can
not be defined locally by just one point, so how would a point mass
'know' about the local curvature? (the usual 'rubber sheet' example
where one mass produces a 'dent' into which the other mass falls is in
my opinion not adequate as the mass follows here merely the tangential
component of the gravity force (obviously it can't follow the force
component normal to the rubber surface as this component is neutralized
by the oppositely acting molecular force the rubber exerts on the
mass)).
Thomas
In falling space theory those mistakes of relativity have been corrected.=
It
also tells, why gravitational constant is needed in some calculations.
Falling space: http://www.wakkanet.fi/~fields/
if you are holding some space and it falls, can you pick it up?
hahahahahahahahahaha
hello crank
Mike
=20
Henry Haapalainen
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| User: "Sam Wormley" |
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| Title: Re: General Relativity and Gravity |
28 Mar 2006 07:00:09 PM |
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Henry Haapalainen wrote:
Falling space: http://www.wakkanet.fi/~fields/
Henry Haapalainen
Henry writes, in http://www.wakkanet.fi/~fields/, "Gravity appears
to be really strange, something inexplicable by theory. This view
has been stated at some time and appears to be well founded. When
an object falls in a gravity field, it seems to be in accelerating
motion. However, this is not so, the acceleration is only apparent".
Perhaps Henry would be interesting in a geometric interpretation of
gravity... Something that would account for "real" acceleration
effects, bending of light, orbital effects, rotational effects and
black holes.
.
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| User: "Henry Haapalainen" |
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| Title: Re: General Relativity and Gravity |
29 Mar 2006 03:29:52 PM |
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"Sam Wormley" <swormley1@mchsi.com> kirjoitti viestissä
news:t_kWf.61776$oL.35742@attbi_s71...
Henry Haapalainen wrote:
Falling space: http://www.wakkanet.fi/~fields/
Henry Haapalainen
Henry writes, in http://www.wakkanet.fi/~fields/, "Gravity appears
to be really strange, something inexplicable by theory. This view
has been stated at some time and appears to be well founded. When
an object falls in a gravity field, it seems to be in accelerating
motion. However, this is not so, the acceleration is only apparent".
Perhaps Henry would be interesting in a geometric interpretation of
gravity... Something that would account for "real" acceleration
effects, bending of light, orbital effects, rotational effects and
black holes.
If somebody else than Sam Wormley had asked that question, I would probably
answer. In the theory all of those have been answered to. You just have to
read.
Henry Haapalainen
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| User: "Sam Wormley" |
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| Title: Re: General Relativity and Gravity |
29 Mar 2006 04:49:47 PM |
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Henry Haapalainen wrote:
"Sam Wormley" <swormley1@mchsi.com> kirjoitti viestissä
news:t_kWf.61776$oL.35742@attbi_s71...
Henry Haapalainen wrote:
Falling space: http://www.wakkanet.fi/~fields/
Henry Haapalainen
Henry writes, in http://www.wakkanet.fi/~fields/, "Gravity appears
to be really strange, something inexplicable by theory. This view
has been stated at some time and appears to be well founded. When
an object falls in a gravity field, it seems to be in accelerating
motion. However, this is not so, the acceleration is only apparent".
Perhaps Henry would be interesting in a geometric interpretation of
gravity... Something that would account for "real" acceleration
effects, bending of light, orbital effects, rotational effects and
black holes.
If somebody else [other] than Sam Wormley had asked that question, I would probably
answer[ed]. In the [my] theory all of those have been answered to[o]. You just have to
read.
Henry Haapalainen
Some sentences in your paragraphs are just plain wrong.... Perhaps it is the
language. Here are some examples:
A1
Gravity appears to be really strange, something inexplicable by theory.
This view has been stated at some time and appears to be well founded.
When an object falls in a gravity field, it seems to be in accelerating
motion. However, this is not so, the acceleration is only apparent. We
who observe it are ourselves in accelerating motion as we stand on the
surface of the Earth, and we experience the acceleration as the surface
of the Earth pushing us upwards.
If we could see events from the
"correct" perspective, we would observe that freely falling objects
move forwards at a constant velocity.
Acceleration is defined at dv/dt
http://scienceworld.wolfram.com/physics/Acceleration.html
A constant velocity (for any observer) ==> acceleration is zero
But objects in free fall are experiencing acceleration, therefore
the velocity is not constant. Velocity is relative, acceleration
is not. Your written words imply that you don't have an understanding
of some pretty fundamental physics.
Gravity is not a force, but
something else. But what is the correct perspective?
A2
The first important theory in the history of research into gravity was
Isaac Newton's theory of gravity. Newton noticed the odd fact that a
feather and a stone fall at the same speed, if air resistance is not
taken into account. A feather and the Moon will also fall at the same
speed. If a feather were in the Moon's place, it would orbit the Earth
as the Moon does now.
You would think that the gravity between two
massive objects would arise from the interaction of their masses, but
this is not the case.
The force of gravity is proportional to their masses in Newton's
Law of gravitation and GTR
http://scienceworld.wolfram.com/physics/Gravity.html
The rest of the paragraph falls apart from here...
Nevertheless, the moon has its own effect, as in
the Earth-Moon system the Earth does not remain "in place", instead the
objects revolve around their common centre of mass. This extra motion
does not properly fit into any equation depicting gravity. But it
exists, and its effect on the movements of objects can be calculated
separately. In the theory of falling space, this motion is separated
from gravity, and its cause is termed the tidal force (old term, but
now in a new wider context). Thus, there are two separate factors in
celestial mechanics: gravity (non-force) and tidal force (force). More
will be said about the tidal force later.
And so does the rest of the "writings" of "publisher" Henry Haapalainen!
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| User: "Henry Haapalainen" |
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| Title: Re: General Relativity and Gravity |
30 Mar 2006 04:15:28 PM |
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"Sam Wormley" <swormley1@mchsi.com> kirjoitti viestissä
news:faEWf.67072$oL.35398@attbi_s71...
Henry Haapalainen wrote:
"Sam Wormley" <swormley1@mchsi.com> kirjoitti viestissä
news:t_kWf.61776$oL.35742@attbi_s71...
Henry Haapalainen wrote:
Falling space: http://www.wakkanet.fi/~fields/
Henry Haapalainen
Henry writes, in http://www.wakkanet.fi/~fields/, "Gravity appears
to be really strange, something inexplicable by theory. This view
has been stated at some time and appears to be well founded. When
an object falls in a gravity field, it seems to be in accelerating
motion. However, this is not so, the acceleration is only apparent".
Perhaps Henry would be interesting in a geometric interpretation of
gravity... Something that would account for "real" acceleration
effects, bending of light, orbital effects, rotational effects and
black holes.
If somebody else [other] than Sam Wormley had asked that question, I
would probably
answer[ed]. In the [my] theory all of those have been answered to[o].
You just have to
read.
Henry Haapalainen
Some sentences in your paragraphs are just plain wrong.... Perhaps it is
the
language. Here are some examples:
A1
Gravity appears to be really strange, something inexplicable by theory.
This view has been stated at some time and appears to be well founded.
When an object falls in a gravity field, it seems to be in accelerating
motion. However, this is not so, the acceleration is only apparent. We
who observe it are ourselves in accelerating motion as we stand on the
surface of the Earth, and we experience the acceleration as the surface
of the Earth pushing us upwards.
If we could see events from the
"correct" perspective, we would observe that freely falling objects
move forwards at a constant velocity.
Acceleration is defined at dv/dt
http://scienceworld.wolfram.com/physics/Acceleration.html
A constant velocity (for any observer) ==> acceleration is zero
But objects in free fall are experiencing acceleration, therefore
the velocity is not constant. Velocity is relative, acceleration
is not. Your written words imply that you don't have an understanding
of some pretty fundamental physics.
Then read this, and tell me what it is that you do not understand:
I. - EXPEDITION (from falling space theory)
Let us make a voyage into four-dimensional space (falling space is the
fourth dimension). In the theory, I say that the events should be seen from
a correct perspective, and the correct perspective is the basic state where
we experience no force acting on us.
You are floating somewhere in space. But then in the vicinity you see a
point that starts to enlarge. You identify it as a planet that seems to
approach you with increasing speed. You can only wait and fear for the
worst. But the planet does not hit you after all, it passes close by. And a
strange thing happens: in passing, it changes its direction. You stay still
all the time, and not even a hair moves on your head. If you had been
sleeping, you would know nothing about it.
Now we change the scenario: two planets approach you from separate
directions. Luckily they miss you again, although they both go pretty close.
Will you float serenely again? Not this time. You get into quite a turmoil,
and that is because of tidal force.
As you float in space, you can move almost at the speed of light, comparing
to something other, or stay still - it will not affect your state, and you
are not aware of it.
Gravity is not a force, but
something else. But what is the correct perspective?
A2
The first important theory in the history of research into gravity was
Isaac Newton's theory of gravity. Newton noticed the odd fact that a
feather and a stone fall at the same speed, if air resistance is not
taken into account. A feather and the Moon will also fall at the same
speed. If a feather were in the Moon's place, it would orbit the Earth
as the Moon does now.
You would think that the gravity between two
massive objects would arise from the interaction of their masses, but
this is not the case.
The force of gravity is proportional to their masses in Newton's
Law of gravitation and GTR
http://scienceworld.wolfram.com/physics/Gravity.html
The rest of the paragraph falls apart from here...
Nevertheless, the moon has its own effect, as in
the Earth-Moon system the Earth does not remain "in place", instead the
objects revolve around their common centre of mass. This extra motion
does not properly fit into any equation depicting gravity. But it
exists, and its effect on the movements of objects can be calculated
separately. In the theory of falling space, this motion is separated
from gravity, and its cause is termed the tidal force (old term, but
now in a new wider context). Thus, there are two separate factors in
celestial mechanics: gravity (non-force) and tidal force (force). More
will be said about the tidal force later.
And so does the rest of the "writings" of "publisher" Henry
Haapalainen!
If you really want to learn falling space you have to change your attitude.
Henry Haapalainen
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| User: "Sam Wormley" |
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| Title: Re: General Relativity and Gravity |
30 Mar 2006 04:30:44 PM |
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Henry Haapalainen wrote:
...read this, and tell me what it is that you do not understand:
I. - EXPEDITION (from falling space theory)
Let us make a voyage into four-dimensional space (falling space is the
fourth dimension). In the theory, I say that the events should be seen from
a correct perspective, and the correct perspective is the basic state where
we experience no force acting on us.
You are floating somewhere in space. But then in the vicinity you see a
point that starts to enlarge. You identify it as a planet that seems to
approach you with increasing speed. You can only wait and fear for the
worst. But the planet does not hit you after all, it passes close by. And a
strange thing happens: in passing, it changes its direction. You stay still
all the time, and not even a hair moves on your head. If you had been
sleeping, you would know nothing about it.
Now we change the scenario: two planets approach you from separate
directions. Luckily they miss you again, although they both go pretty close.
Will you float serenely again? Not this time. You get into quite a turmoil,
and that is because of tidal force.
As you float in space, you can move almost at the speed of light, comparing
to something other, or stay still - it will not affect your state, and you
are not aware of it.
I'm falling around the Sun as we speak... I feel no force on me... but with
appropriate instruments I can measure the tidal force between to points along
a line toward the Sun. Newton and GTR are in agreement with my observations
and measurements.
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| User: "Henry Haapalainen" |
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| Title: Re: General Relativity and Gravity |
30 Mar 2006 05:13:09 PM |
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"Sam Wormley" <swormley1@mchsi.com> kirjoitti viestissä
news:o_YWf.865380$x96.836802@attbi_s72...
Henry Haapalainen wrote:
...read this, and tell me what it is that you do not understand:
I. - EXPEDITION (from falling space theory)
Let us make a voyage into four-dimensional space (falling space is the
fourth dimension). In the theory, I say that the events should be seen
from
a correct perspective, and the correct perspective is the basic state
where
we experience no force acting on us.
You are floating somewhere in space. But then in the vicinity you see a
point that starts to enlarge. You identify it as a planet that seems to
approach you with increasing speed. You can only wait and fear for the
worst. But the planet does not hit you after all, it passes close by.
And a
strange thing happens: in passing, it changes its direction. You stay
still
all the time, and not even a hair moves on your head. If you had been
sleeping, you would know nothing about it.
Now we change the scenario: two planets approach you from separate
directions. Luckily they miss you again, although they both go pretty
close.
Will you float serenely again? Not this time. You get into quite a
turmoil,
and that is because of tidal force.
As you float in space, you can move almost at the speed of light,
comparing
to something other, or stay still - it will not affect your state, and
you
are not aware of it.
I'm falling around the Sun as we speak... I feel no force on me... but
with
appropriate instruments I can measure the tidal force between to points
along
a line toward the Sun. Newton and GTR are in agreement with my
observations
and measurements.
Newton and GTR can not be in agreement. it is either acceleration or free
fall. (HH)
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| User: "Sam Wormley" |
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| Title: Re: General Relativity and Gravity |
30 Mar 2006 06:47:24 PM |
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Henry Haapalainen wrote:
Newton and GTR can not be in agreement. it is either acceleration or free
fall. (HH)
You are cornfused, Henry Haapalainen!
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| User: "Henry Haapalainen" |
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| Title: Re: General Relativity and Gravity |
31 Mar 2006 03:41:46 PM |
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"Sam Wormley" <swormley1@mchsi.com> kirjoitti viestissä
news:w__Wf.865519$x96.190625@attbi_s72...
Henry Haapalainen wrote:
Newton and GTR can not be in agreement. it is either acceleration or
free
fall. (HH)
You are cornfused, Henry Haapalainen!
Didn't you just tell everybody that you are ignorant? (HH)
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| User: "DaveL Top Poster Extraordinaire" |
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| Title: Re: General Relativity and Gravity |
30 Mar 2006 11:02:40 AM |
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I like this theory. A lot of it makes sense. I think it's not far from the
truth. I have always imagined Einstein's curved space-time as falling
space, or more specifically, aether moving toward all matter.
DaveL
"Henry Haapalainen" <kirppu@kolumbus.fi> wrote in message
news:e0c9e5$9n5$1@phys-news4.kolumbus.fi...
In falling space theory those mistakes of relativity have been corrected.
It
also tells, why gravitational constant is needed in some calculations.
Falling space: http://www.wakkanet.fi/~fields/
Henry Haapalainen
.
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| User: "Henry Haapalainen" |
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| Title: Re: General Relativity and Gravity |
30 Mar 2006 04:58:12 PM |
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"DaveL" <Top Poster Extraordinaire> kirjoitti viestissä
news:8-Odna2Y78eYkrHZRVn-qg@comcast.com...
I like this theory. A lot of it makes sense. I think it's not far from
the
truth. I have always imagined Einstein's curved space-time as falling
space, or more specifically, aether moving toward all matter.
DaveL
Thank you, DaveL. If someone like you wants to ask some questions, I will be
happy to answer. Please, ask anything that need clarification.
Henry Haapalainen
"Henry Haapalainen" <kirppu@kolumbus.fi> wrote in message
news:e0c9e5$9n5$1@phys-news4.kolumbus.fi...
In falling space theory those mistakes of relativity have been
corrected.
It
also tells, why gravitational constant is needed in some calculations.
Falling space: http://www.wakkanet.fi/~fields/
Henry Haapalainen
.
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| User: "T Wake" |
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| Title: Re: General Relativity and Gravity |
30 Mar 2006 11:52:51 AM |
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"DaveL" <Top Poster Extraordinaire> wrote in message
news:8-Odna2Y78eYkrHZRVn-qg@comcast.com...
I like this theory. A lot of it makes sense.
Good for you. Nether of these are requirements for a scientific theory
though.
I think it's not far from the truth.
Well, we can all have our opinions.
I have always imagined Einstein's curved space-time as falling space, or
more specifically, aether moving toward all matter.
Interestingly, what you imagine and reality aren't always the same.
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| User: "Hexenmeister" |
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| Title: Re: General Relativity and Gravity |
30 Mar 2006 03:23:09 PM |
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What does your emotional responses have to do with physics?
"DaveL" <Top Poster Extraordinaire> wrote in message
news:8-Odna2Y78eYkrHZRVn-qg@comcast.com...
|I like this theory.
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| User: "Koobee Wublee" |
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| Title: Re: General Relativity and Gravity |
28 Mar 2006 04:42:24 PM |
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"Thomas Smid" <thomas.smid@gmail.com> wrote in message
news:1143558971.699628.53080@v46g2000cwv.googlegroups.com...
It is generally claimed that gravity in GR is not a force but a
consequence of space curvature which in turn is created by the masses.
This claim is totally false. Consider the segment of spacetime with
Schwarzschild metric.
ds^2 = c^2 g_00 dt^2 - dr^2 / g_00 - r^2 dH^2 +...
It tells us that time is dilated near mass as well as radial
displacement of space is curved.
Now, consider another spacetime.
ds^2 = c^2 g_00 dt^2 - dr^2 - r^2 dH^2 +...
Where
** g_00 = 1 - 2 U
** U = G M / c^2 / r
Here we only have time dilation near mass, and space is not curved in
anyway.
Working out the math, both spacetime give the same Newtonian
gravitational law. All you need is to have gravitational time dilation
to explain gravitation. Curved space is a separate issue. Claiming
curvature of spacetime causing gravitation is not accurate.
Both spacetime predict photon deflection, but Schwarzschild gives twice
the amount. Both fail miserably at predicting Mercury's orbital
anomaly, but adding a 2nd order term (Taylor series expansion) to g_00
of the non-Schwarzschld spacetime (g_00 = 1 - 2 U + ? U^2), you can
find this constant to explain Mercury's orbital anomaly.
Schwarzschild space, of course, predicts antigravity at high speed. It
predicts a photon would increase its speed to infinity while the
non-Schwarzschild spacetime does not.
Should then not GR be able to describe the motion of masses
self-consistently without use of the gravitational constant G? The
Schwartzschild metric depends on the Schwartzschild radius R=2GM/c^2,
and G obviously has to be determined by (force) measurements.
Since curvature in space does not cause gravitation, gravity must be a
force. It is caused by gravitational time dilation.
Also, how can gravity depend on the curvature of space? A curvature can
not be defined locally by just one point, so how would a point mass
'know' about the local curvature? (the usual 'rubber sheet' example
where one mass produces a 'dent' into which the other mass falls is in
my opinion not adequate as the mass follows here merely the tangential
component of the gravity force (obviously it can't follow the force
component normal to the rubber surface as this component is neutralized
by the oppositely acting molecular force the rubber exerts on the
mass)).
When Hilbert was trying to come up with the field equations, I think he
was shooting for the ones that give only gravitational time dilation.
But the Lagangian resulted in Einstein-Hilbert Action was patched
together like that monster Frankenstein, it had embarrassing side
effects. Curvature in space is one of them.
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| User: "Bilge" |
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| Title: Re: General Relativity and Gravity |
29 Mar 2006 02:34:38 AM |
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Koobee Wublee:
Both spacetime predict photon deflection, but Schwarzschild gives twice
the amount. Both fail miserably at predicting Mercury's orbital
anomaly, but adding a 2nd order term (Taylor series expansion) to g_00
of the non-Schwarzschld spacetime (g_00 = 1 - 2 U + ? U^2), you can
find this constant to explain Mercury's orbital anomaly.
The schwarzschild metric gives the correct precession of mercury.
The reason you can't get the right result is that you are determined
to remain ignorant, despite having your errors pointed out to you.
Schwarzschild space, of course, predicts antigravity at high speed. It
predicts a photon would increase its speed to infinity while the
non-Schwarzschild spacetime does not.
Care to prove that?
.
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| User: "Koobee Wublee" |
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| Title: Re: General Relativity and Gravity |
29 Mar 2006 11:17:30 PM |
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"Bilge" <dubious@radioactivex.lebesque-al.net> wrote in message
news:slrne2koq2.5r.dubious@radioactivex.lebesque-al.net...
It is very surprising that you don't have any objection that
gravitation is caused by gravitational time dilation only regardless
whether space is curved or not. So, gravity must be a force indeed not
that mumble jumble about the curvature of spacetime crap.
The schwarzschild metric gives the correct precession of mercury.
The reason you can't get the right result is that you are determined
to remain ignorant, despite having your errors pointed out to you.
We will leave this as another chapter of discussion later.
Schwarzschild space, of course, predicts antigravity at high speed. It
predicts a photon would increase its speed to infinity while the
non-Schwarzschild spacetime does not.
Care to prove that?
In the meantime, since the curvature of spacetime is not what the root
cause of gravity, it is ridiculous to claim any event taking place must
follow the shortest path in spacetime. The Principle of Least should
indeed be what any event taking place follows. However, I assum for
the purpose of this discussion you insist on the shortest segment of
spacetime. So, starting with the following,
ds^2 = c^2 g dt^2 - dr^2 / g - r^2 dH^2
Where
** g = 1 - 2 U
** U = G M /c^2 / r
** Motion confines to equitorial plane.
Dividing both sides by ds^2, the Lagrangian that serves as the density
to a segment of spacetime is
L = c^2 g (dt/ds)^2 - (dr/ds)^2 / g - r^2 (dH/dt)^2 = 1
This Lagrangian above satisfies the traversed spacetime in which this
event takes place. The traversed spacetime is
delta s = integral( ds ) = integral( L ds)
Through Calculus of Variations, we yield the following Euler-Lagrange
Equations.
** eq. [t] ** d(@L/@(dt/ds))/ds = @L@t
** eq. [H] ** d(@L/@(dH/ds))/ds = @L@H
** eq. [r] ** d(@L/@(dr/ds))/ds = @L@r
Expanding equation [t], we have equation [t1] as follows.
d^2t/ds^2 = - (dg/dr) (dr/ds) (dt/ds) / g
Expanding equation r, we have the following equation [r1].
d^2r/ds^2 - g r (dH/ds)^2 = - (g / 2) (dg/dr) (c^2 (dt/ds)^2 -
(dr/ds)^2 / g^2)
As you know, the first term the equation above is equation [r2] as
follows.
d^2r/ds^2 = (d^2r/dt^2) (dt/ds)^2 + (dr/dt) (d^2t/ds^2)
From equation [t1], equation [r2] becomes equation [r3] as follows.
d^2r/ds^2 = (d^2r/dt^2) (dt/ds)^2 - (dg/dr) (dr/dt)^2 (dt/ds)^2
Using equation [r3], equation [r1] multiplied by (ds/dt)^2 becomes
equation [r4] below.
d^2r/dt^2 - g r (dH/dt)^2 = - (g / 2) (dg/dr) (c^2 - 3 (dr/dt)^2 / g^2)
At the following condition, gravitational acceleration becomes
repulsive.
c^2 - 3 (dr/dt)^2 / g^2 < 0
Or
(dr/dt)^2 / c^2 > g^2 / 3
This means a photon would be accelerated to infinite speed. See how
absurd GR really is.
Back to our Mercury's orbit, as you know all planets exhibits the same
type of anomaly regardless of eccentricity. So, if we average
Mercury's orbit to circular, we have the following conditions.
** d^2r/dt^2 = 0
** dr/dt = 0
So, equation [r4] becomes
- g r (dH/dt)^2 = -(g / 2) (dg/dr)
Or
r^2 (dH/dt)^2 = U
It is as Newtonian as you can get to the first order. This indicates
that GR does not predict any anomaly to Mercury's orbit.
To exhibit the same type of observed anomaly, we need to establish
r^2 (dH/dt)^2 = U (1 + 3 U)
The above means on every revolution, the advance is (3 U) in radian.
Therefore, GR fails at this observation. The error was left over from
Besso's copy of Gerber's derivation which Einstein took it as his own.
----- Original Message -----
From: "Koobee Wublee" <koobee.wublee@gmail.com>
Newsgroups: sci.physics,sci.physics.relativity
Sent: Tuesday, March 28, 2006 02:42 PM
Subject: Re: General Relativity and Gravity
"Thomas Smid" <thomas.smid@gmail.com> wrote in message
news:1143558971.699628.53080@v46g2000cwv.googlegroups.com...
It is generally claimed that gravity in GR is not a force but a
consequence of space curvature which in turn is created by the masses.
This claim is totally false. Consider the segment of spacetime with
Schwarzschild metric.
ds^2 = c^2 g_00 dt^2 - dr^2 / g_00 - r^2 dH^2 +...
It tells us that time is dilated near mass as well as radial
displacement of space is curved.
Now, consider another spacetime.
ds^2 = c^2 g_00 dt^2 - dr^2 - r^2 dH^2 +...
Where
** g_00 = 1 - 2 U
** U = G M / c^2 / r
Here we only have time dilation near mass, and space is not curved in
anyway.
Working out the math, both spacetime give the same Newtonian
gravitational law. All you need is to have gravitational time dilation
to explain gravitation. Curved space is a separate issue. Claiming
curvature of spacetime causing gravitation is not accurate.
Both spacetime predict photon deflection, but Schwarzschild gives twice
the amount. Both fail miserably at predicting Mercury's orbital
anomaly, but adding a 2nd order term (Taylor series expansion) to g_00
of the non-Schwarzschld spacetime (g_00 = 1 - 2 U + ? U^2), you can
find this constant to explain Mercury's orbital anomaly.
Schwarzschild space, of course, predicts antigravity at high speed. It
predicts a photon would increase its speed to infinity while the
non-Schwarzschild spacetime does not.
Should then not GR be able to describe the motion of masses
self-consistently without use of the gravitational constant G? The
Schwartzschild metric depends on the Schwartzschild radius R=2GM/c^2,
and G obviously has to be determined by (force) measurements.
Since curvature in space does not cause gravitation, gravity must be a
force. It is caused by gravitational time dilation.
Also, how can gravity depend on the curvature of space? A curvature can
not be defined locally by just one point, so how would a point mass
'know' about the local curvature? (the usual 'rubber sheet' example
where one mass produces a 'dent' into which the other mass falls is in
my opinion not adequate as the mass follows here merely the tangential
component of the gravity force (obviously it can't follow the force
component normal to the rubber surface as this component is neutralized
by the oppositely acting molecular force the rubber exerts on the
mass)).
When Hilbert was trying to come up with the field equations, I think he
was shooting for the ones that give only gravitational time dilation.
But the Lagangian resulted in Einstein-Hilbert Action was patched
together like that monster Frankenstein, it had embarrassing side
effects. Curvature in space is one of them.
.
|
|
|
|
| User: "Koobee Wublee" |
|
| Title: Re: General Relativity and Gravity |
29 Mar 2006 11:17:32 PM |
|
|
"Bilge" <dubious@radioactivex.lebesque-al.net> wrote in message
news:slrne2koq2.5r.dubious@radioactivex.lebesque-al.net...
It is very surprising that you don't have any objection that
gravitation is caused by gravitational time dilation only regardless
whether space is curved or not. So, gravity must be a force indeed not
that mumble jumble about the curvature of spacetime crap.
The schwarzschild metric gives the correct precession of mercury.
The reason you can't get the right result is that you are determined
to remain ignorant, despite having your errors pointed out to you.
We will leave this as another chapter of discussion later.
Schwarzschild space, of course, predicts antigravity at high speed. It
predicts a photon would increase its speed to infinity while the
non-Schwarzschild spacetime does not.
Care to prove that?
In the meantime, since the curvature of spacetime is not what the root
cause of gravity, it is ridiculous to claim any event taking place must
follow the shortest path in spacetime. The Principle of Least should
indeed be what any event taking place follows. However, I assum for
the purpose of this discussion you insist on the shortest segment of
spacetime. So, starting with the following,
ds^2 = c^2 g dt^2 - dr^2 / g - r^2 dH^2
Where
** g = 1 - 2 U
** U = G M /c^2 / r
** Motion confines to equitorial plane.
Dividing both sides by ds^2, the Lagrangian that serves as the density
to a segment of spacetime is
L = c^2 g (dt/ds)^2 - (dr/ds)^2 / g - r^2 (dH/dt)^2 = 1
This Lagrangian above satisfies the traversed spacetime in which this
event takes place. The traversed spacetime is
delta s = integral( ds ) = integral( L ds)
Through Calculus of Variations, we yield the following Euler-Lagrange
Equations.
** eq. [t] ** d(@L/@(dt/ds))/ds = @L@t
** eq. [H] ** d(@L/@(dH/ds))/ds = @L@H
** eq. [r] ** d(@L/@(dr/ds))/ds = @L@r
Expanding equation [t], we have equation [t1] as follows.
d^2t/ds^2 = - (dg/dr) (dr/ds) (dt/ds) / g
Expanding equation r, we have the following equation [r1].
d^2r/ds^2 - g r (dH/ds)^2 = - (g / 2) (dg/dr) (c^2 (dt/ds)^2 -
(dr/ds)^2 / g^2)
As you know, the first term the equation above is equation [r2] as
follows.
d^2r/ds^2 = (d^2r/dt^2) (dt/ds)^2 + (dr/dt) (d^2t/ds^2)
From equation [t1], equation [r2] becomes equation [r3] as follows.
d^2r/ds^2 = (d^2r/dt^2) (dt/ds)^2 - (dg/dr) (dr/dt)^2 (dt/ds)^2
Using equation [r3], equation [r1] multiplied by (ds/dt)^2 becomes
equation [r4] below.
d^2r/dt^2 - g r (dH/dt)^2 = - (g / 2) (dg/dr) (c^2 - 3 (dr/dt)^2 / g^2)
At the following condition, gravitational acceleration becomes
repulsive.
c^2 - 3 (dr/dt)^2 / g^2 < 0
Or
(dr/dt)^2 / c^2 > g^2 / 3
This means a photon would be accelerated to infinite speed. See how
absurd GR really is.
Back to our Mercury's orbit, as you know all planets exhibits the same
type of anomaly regardless of eccentricity. So, if we average
Mercury's orbit to circular, we have the following conditions.
** d^2r/dt^2 = 0
** dr/dt = 0
So, equation [r4] becomes
- g r (dH/dt)^2 = -(g / 2) (dg/dr)
Or
r^2 (dH/dt)^2 = U
It is as Newtonian as you can get to the first order. This indicates
that GR does not predict any anomaly to Mercury's orbit.
To exhibit the same type of observed anomaly, we need to establish
r^2 (dH/dt)^2 = U (1 + 3 U)
The above means on every revolution, the advance is (3 U) in radian.
Therefore, GR fails at this observation. The error was left over from
Besso's copy of Gerber's derivation which Einstein took it as his own.
----- Original Message -----
From: "Koobee Wublee" <koobee.wublee@gmail.com>
Newsgroups: sci.physics,sci.physics.relativity
Sent: Tuesday, March 28, 2006 02:42 PM
Subject: Re: General Relativity and Gravity
"Thomas Smid" <thomas.smid@gmail.com> wrote in message
news:1143558971.699628.53080@v46g2000cwv.googlegroups.com...
It is generally claimed that gravity in GR is not a force but a
consequence of space curvature which in turn is created by the masses.
This claim is totally false. Consider the segment of spacetime with
Schwarzschild metric.
ds^2 = c^2 g_00 dt^2 - dr^2 / g_00 - r^2 dH^2 +...
It tells us that time is dilated near mass as well as radial
displacement of space is curved.
Now, consider another spacetime.
ds^2 = c^2 g_00 dt^2 - dr^2 - r^2 dH^2 +...
Where
** g_00 = 1 - 2 U
** U = G M / c^2 / r
Here we only have time dilation near mass, and space is not curved in
anyway.
Working out the math, both spacetime give the same Newtonian
gravitational law. All you need is to have gravitational time dilation
to explain gravitation. Curved space is a separate issue. Claiming
curvature of spacetime causing gravitation is not accurate.
Both spacetime predict photon deflection, but Schwarzschild gives twice
the amount. Both fail miserably at predicting Mercury's orbital
anomaly, but adding a 2nd order term (Taylor series expansion) to g_00
of the non-Schwarzschld spacetime (g_00 = 1 - 2 U + ? U^2), you can
find this constant to explain Mercury's orbital anomaly.
Schwarzschild space, of course, predicts antigravity at high speed. It
predicts a photon would increase its speed to infinity while the
non-Schwarzschild spacetime does not.
Should then not GR be able to describe the motion of masses
self-consistently without use of the gravitational constant G? The
Schwartzschild metric depends on the Schwartzschild radius R=2GM/c^2,
and G obviously has to be determined by (force) measurements.
Since curvature in space does not cause gravitation, gravity must be a
force. It is caused by gravitational time dilation.
Also, how can gravity depend on the curvature of space? A curvature can
not be defined locally by just one point, so how would a point mass
'know' about the local curvature? (the usual 'rubber sheet' example
where one mass produces a 'dent' into which the other mass falls is in
my opinion not adequate as the mass follows here merely the tangential
component of the gravity force (obviously it can't follow the force
component normal to the rubber surface as this component is neutralized
by the oppositely acting molecular force the rubber exerts on the
mass)).
When Hilbert was trying to come up with the field equations, I think he
was shooting for the ones that give only gravitational time dilation.
But the Lagangian resulted in Einstein-Hilbert Action was patched
together like that monster Frankenstein, it had embarrassing side
effects. Curvature in space is one of them.
.
|
|
|
|
| User: "Dirk Van de moortel" |
|
| Title: Re: General Relativity and Gravity |
29 Mar 2006 03:23:05 AM |
|
|
"Bilge" <dubious@radioactivex.lebesque-al.net> wrote in message news:slrne2koq2.5r.dubious@radioactivex.lebesque-al.net...
Koobee Wublee:
Both spacetime predict photon deflection, but Schwarzschild gives twice
the amount. Both fail miserably at predicting Mercury's orbital
anomaly, but adding a 2nd order term (Taylor series expansion) to g_00
of the non-Schwarzschld spacetime (g_00 = 1 - 2 U + ? U^2), you can
find this constant to explain Mercury's orbital anomaly.
The schwarzschild metric gives the correct precession of mercury.
The reason you can't get the right result is that you are determined
to remain ignorant, despite having your errors pointed out to you.
Schwarzschild space, of course, predicts antigravity at high speed. It
predicts a photon would increase its speed to infinity while the
non-Schwarzschild spacetime does not.
Care to prove that?
It's easy, when one starts with:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/NewLagrangian.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/NewPotential.html
;-)
Dirk Vdm
.
|
|
|
| User: "Andreas Most" |
|
| Title: Re: General Relativity and Gravity |
29 Mar 2006 03:49:57 AM |
|
|
Dirk Van de moortel wrote:
"Bilge" <dubious@radioactivex.lebesque-al.net> wrote in message news:slrne2koq2.5r.dubious@radioactivex.lebesque-al.net...
Koobee Wublee:
Schwarzschild space, of course, predicts antigravity at high speed. It
predicts a photon would increase its speed to infinity while the
non-Schwarzschild spacetime does not.
Care to prove that?
It's easy, when one starts with:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/NewLagrangian.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/NewPotential.html
;-)
Dirk Vdm
You may also want to look at
gr-qc/0505099
(also gr-qc/0505098 )
There was some discussion about that recently here
http://groups.google.co.uk/group/sci.physics/browse_thread/thread/e19c135451aebfdc/30c9f36c98a5a64d?lnk=st&q=antigravity+beams&rnum=5#30c9f36c98a5a64d
Andreas.
.
|
|
|
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|
| User: "Mike" |
|
| Title: Re: General Relativity and Gravity |
29 Mar 2006 08:46:00 AM |
|
|
Koobee Wublee wrote:
"Thomas Smid" <thomas.smid@gmail.com> wrote in message
news:1143558971.699628.53080@v46g2000cwv.googlegroups.com...
It is generally claimed that gravity in GR is not a force but a
consequence of space curvature which in turn is created by the masses.
This claim is totally false. Consider the segment of spacetime with
Schwarzschild metric.
ds^2 = c^2 g_00 dt^2 - dr^2 / g_00 - r^2 dH^2 +...
It tells us that time is dilated near mass as well as radial
displacement of space is curved.
Now, consider another spacetime.
ds^2 = c^2 g_00 dt^2 - dr^2 - r^2 dH^2 +...
Where
** g_00 = 1 - 2 U
** U = G M / c^2 / r
Here we only have time dilation near mass, and space is not curved in
anyway.
Working out the math, both spacetime give the same Newtonian
gravitational law. All you need is to have gravitational time dilation
to explain gravitation. Curved space is a separate issue. Claiming
curvature of spacetime causing gravitation is not accurate.
Nobody claims "curvature of spacetime causing gravitation" except
people like you who misunderstand the notion of causality. There are
several notions of causality other than a cause being antecedant to its
effect. You narrow view of causality is contradicted by simple
experiments everyone can do at home.
Both spacetime predict photon deflection, but Schwarzschild gives twice
the amount. Both fail miserably at predicting Mercury's orbital
anomaly, but adding a 2nd order term (Taylor series expansion) to g_00
of the non-Schwarzschld spacetime (g_00 = 1 - 2 U + ? U^2), you can
find this constant to explain Mercury's orbital anomaly.
Schwarzschild space, of course, predicts antigravity at high speed. It
predicts a photon would increase its speed to infinity while the
non-Schwarzschild spacetime does not.
Should then not GR be able to describe the motion of masses
self-consistently without use of the gravitational constant G? The
Schwartzschild metric depends on the Schwartzschild radius R=2GM/c^2,
and G obviously has to be determined by (force) measurements.
Since curvature in space does not cause gravitation, gravity must be a
force. It is caused by gravitational time dilation.
What kind of logic is this? have you ever had an introductory course in
first order predicate logic or even categorical logic to know about the
fallacies of "slippery slope", 'faulty causality" and "non sequitur",
which you managed in an increadible way to combine in one and only
conditional?
Also, how can gravity depend on the curvature of space? A curvature can
not be defined locally by just one point, so how would a point mass
'know' about the local curvature? (the usual 'rubber sheet' example
where one mass produces a 'dent' into which the other mass falls is in
my opinion not adequate as the mass follows here merely the tangential
component of the gravity force (obviously it can't follow the force
component normal to the rubber surface as this component is neutralized
by the oppositely acting molecular force the rubber exerts on the
mass)).
When Hilbert was trying to come up with the field equations, I think he
was shooting for the ones that give only gravitational time dilation.
But the Lagangian resulted in Einstein-Hilbert Action was patched
together like that monster Frankenstein, it had embarrassing side
effects. Curvature in space is one of them.
God knows what you are talking about. You insist amongst other things
that gravity must be a force. I must inform you that other than
impressed forces, all other forces are intellectual constructions and
not actual entities. You are left only with impressed forces. maybe you
are the person who once insisted gravity is the effect of ultra light
minute particles hitting mass at FTL speeds. I think I recall that.
You should understand by now why you are wrong. For one, besides the
obvious rebuttal of this ridiculus hypothesis an infinite universe in
size for the concept to work. failure.
Mike
.
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|
|
| User: "Bill Hobba" |
|
| Title: Re: General Relativity and Gravity |
28 Mar 2006 05:33:01 PM |
|
|
"Thomas Smid" <thomas.smid@gmail.com> wrote in message
news:1143558971.699628.53080@v46g2000cwv.googlegroups.com...
Hi,
It is generally claimed that gravity in GR is not a force but a
consequence of space curvature
Wrong - space-time curvature.
which in turn is created by the masses.
Should then not GR be able to describe the motion of masses
self-consistently without use of the gravitational constant G?
It sure can - G can take on any value even 1.
The
Schwartzschild metric depends on the Schwartzschild radius R=2GM/c^2,
and G obviously has to be determined by (force) measurements.
Also, how can gravity depend on the curvature of space? A curvature can
not be defined locally by just one point,
That should be at one point. Yes it can - just as the gradient of a curve
can.
Learn the basics then repost.
Bill
so how would a point mass
'know' about the local curvature? (the usual 'rubber sheet' example
where one mass produces a 'dent' into which the other mass falls is in
my opinion not adequate as the mass follows here merely the tangential
component of the gravity force (obviously it can't follow the force
component normal to the rubber surface as this component is neutralized
by the oppositely acting molecular force the rubber exerts on the
mass)).
Thomas
.
|
|
|
| User: "Thomas Smid" |
|
| Title: Re: General Relativity and Gravity |
29 Mar 2006 03:46:22 AM |
|
|
Bill Hobba wrote:
"Thomas Smid" <thomas.smid@gmail.com> wrote in message
news:1143558971.699628.53080@v46g2000cwv.googlegroups.com...
Hi,
It is generally claimed that gravity in GR is not a force but a
consequence of space curvature
Wrong - space-time curvature.
which in turn is created by the masses.
Should then not GR be able to describe the motion of masses
self-consistently without use of the gravitational constant G?
It sure can - G can take on any value even 1.
Then please demonstrate how you obtain the numerical value for the
gravitational acceleration of the earth in this way.
The
Schwartzschild metric depends on the Schwartzschild radius R=2GM/c^2,
and G obviously has to be determined by (force) measurements.
Also, how can gravity depend on the curvature of space? A curvature can
not be defined locally by just one point,
That should be at one point. Yes it can - just as the gradient of a curve
can.
One can not define a gradient by using just one function value:
the derivative of a function f at point x is defined as the limit of (
f(x+dx)-f(x) )/dx with dx being infinitesimal but different from 0,
that is a point mass at location x would have to 'know' the value of
f(x+dx) which would contradict locality (for corresponding problems in
the time domain it would also violate causality).
Thomas
.
|
|
|
| User: "Bill Hobba" |
|
| Title: Re: General Relativity and Gravity |
29 Mar 2006 04:53:52 PM |
|
|
"Thomas Smid" <thomas.smid@gmail.com> wrote in message
news:1143625582.098764.162850@t31g2000cwb.googlegroups.com...
Bill Hobba wrote:
"Thomas Smid" <thomas.smid@gmail.com> wrote in message
news:1143558971.699628.53080@v46g2000cwv.googlegroups.com...
Hi,
It is generally claimed that gravity in GR is not a force but a
consequence of space curvature
Wrong - space-time curvature.
which in turn is created by the masses.
Should then not GR be able to describe the motion of masses
self-consistently without use of the gravitational constant G?
It sure can - G can take on any value even 1.
Then please demonstrate how you obtain the numerical value for the
gravitational acceleration of the earth in this way.
Meausre it. BTW it has nothing to do with G.
The
Schwartzschild metric depends on the Schwartzschild radius R=2GM/c^2,
and G obviously has to be determined by (force) measurements.
Also, how can gravity depend on the curvature of space? A curvature can
not be defined locally by just one point,
That should be at one point. Yes it can - just as the gradient of a
curve
can.
One can not define a gradient by using just one function value:
the derivative of a function f at point x is defined as the limit of (
f(x+dx)-f(x) )/dx with dx being infinitesimal but different from 0,
that is a point mass at location x would have to 'know' the value of
f(x+dx) which would contradict locality (for corresponding problems in
the time domain it would also violate causality).
Notice my correction - at one point. Manifolds or curves by their
definition are never defined by one point.
Bill
Thomas
.
|
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| User: "Dirk Van de moortel" |
|
| Title: Re: General Relativity and Gravity |
29 Mar 2006 04:46:18 AM |
|
|
"Thomas Smid" <thomas.smid@gmail.com> wrote in message news:1143625582.098764.162850@t31g2000cwb.googlegroups.com...
Bill Hobba wrote:
"Thomas Smid" <thomas.smid@gmail.com> wrote in message
news:1143558971.699628.53080@v46g2000cwv.googlegroups.com...
Hi,
It is generally claimed that gravity in GR is not a force but a
consequence of space curvature
Wrong - space-time curvature.
which in turn is created by the masses.
Should then not GR be able to describe the motion of masses
self-consistently without use of the gravitational constant G?
It sure can - G can take on any value even 1.
Then please demonstrate how you obtain the numerical value for the
gravitational acceleration of the earth in this way.
The
Schwartzschild metric depends on the Schwartzschild radius R=2GM/c^2,
and G obviously has to be determined by (force) measurements.
Also, how can gravity depend on the curvature of space? A curvature can
not be defined locally by just one point,
That should be at one point. Yes it can - just as the gradient of a curve
can.
One can not define a gradient by using just one function value:
What would an imbecile like you know about gradients?
http://groups.google.co.uk/group/sci.physics.relativity/msg/3f9a9fa2b8d3b5ec
Dvdm:
| > 1C) A bomb explodes at a distance D and at time T according
| > to me - but, again, you can use *your* favourite definitions
| > and measurements of distance, speed and time.
| > You freely choose some value for v of the bomb, but let's
| > say you pick some v > 0, to signify that the bomb is going in
| > some positive x-direction.
| > So we know that the equation of motion of the bomb is
| > x - D = v (t-T)
| > where
| > t = time in my frame
| > v = velocity of bomb
| > x = distance of bomb to me at time t
| > D = given distance of explosion event
| > T = given time of explosion event
| > in other words: at time t, the bomb is at distance
| > x = D + v (t-T),
| > so indeed, at the given time T, the bomb is at the given
| > distance D.
| >
| >
| > 2C) If I want the bomb to explode at the given distance D
| > at the given time T, then my outgoing signal must be present
| > there and then, so the equation of motion of the outgoing
| > signal is
| > x - D = c (t-T)
| > where
| > t = time in my frame
| > c = light speed (direction positive x-axis)
| > x = distance of outgoing signal at time t
| > D = given distance of explosion event
| > T = given time of explosion event
| > in other words, at time t, the outgoing signal is at
| > distance
| > x = D + c (t-T),
|
Smid:
| I am not quite with you here: from your equations obtained in 1C
| (x=D+v(t-T)) and 2C (x=D+c(t-T)) one obtains v=c.
Dirk Vdm
.
|
|
|
| User: "Thomas Smid" |
|
| Title: Re: General Relativity and Gravity |
29 Mar 2006 05:28:43 AM |
|
|
Dirk Van de moortel wrote:
"Thomas Smid" <thomas.smid@gmail.com> wrote in message news:1143625582.098764.162850@t31g2000cwb.googlegroups.com...
One can not define a gradient by using just one function value:
the derivative of a function f at point x is defined as the limit of
( f(x+dx)-f(x) )/dx with dx being infinitesimal but different from 0,
that is a point mass at location x would have to 'know' the value of
f(x+dx) which would contradict locality (for corresponding problems in
the time domain it would also violate causality).
What would an imbecile like you know about gradients?
Apparently more than you as otherwise you would given a proper response
to my argument.
Thomas
.
|
|
|
| User: "Dirk Van de moortel" |
|
| Title: Re: General Relativity and Gravity |
29 Mar 2006 05:58:01 AM |
|
|
"Thomas Smid" <thomas.smid@gmail.com> wrote in message news:1143631723.506197.257850@g10g2000cwb.googlegroups.com...
Dirk Van de moortel wrote:
"Thomas Smid" <thomas.smid@gmail.com> wrote in message news:1143625582.098764.162850@t31g2000cwb.googlegroups.com...
One can not define a gradient by using just one function value:
the derivative of a function f at point x is defined as the limit of
( f(x+dx)-f(x) )/dx with dx being infinitesimal but different from 0,
that is a point mass at location x would have to 'know' the value of
f(x+dx) which would contradict locality (for corresponding problems in
the time domain it would also violate causality).
What would an imbecile like you know about gradients?
Apparently more than you as otherwise you would given a proper response
to my argument.
What could an obnoxious imbecile like you possibly do with a
proper response to his 'argument'?
http://groups.google.co.uk/group/sci.physics.relativity/msg/c692727b3dbba929
Daryl McCullough:
| > Wait a minute. Are you now in agreement that you
| > were wrong about your *original* reason for saying
| > that Einstein's derivation is inconsistent? You
| > seem to be moving onto a completely different
| > argument.
| >
| > We've discovered algebraic mistakes that you've
| > made, and algebraic mistakes that I've made, but
| > we have not yet found an algebraic mistake that
| > Einstein made.
|
Smid:
| I have never claimed that Einstein made an algebraic
| mistake but that his equations are mathematically
| inconsistent as he worked from the equations
| x-ct=0
| x+ct=0
| which is inconsistent unless everything is identically
| zero. If I had to review a paper containing these
| equations, I would reject it out of hand (whether it
| was written by Einstein or anybody else).
Dirk Vdm
.
|
|
|
| User: "Thomas Smid" |
|
| Title: Re: General Relativity and Gravity |
29 Mar 2006 06:04:26 AM |
|
|
Dirk Van de moortel wrote:
"Thomas Smid" <thomas.smid@gmail.com> wrote in message news:1143631723.506197.257850@g10g2000cwb.googlegroups.com...
Dirk Van de moortel wrote:
"Thomas Smid" <thomas.smid@gmail.com> wrote in message news:1143625582.098764.162850@t31g2000cwb.googlegroups.com...
One can not define a gradient by using just one function value:
the derivative of a function f at point x is defined as the limit of
( f(x+dx)-f(x) )/dx with dx being infinitesimal but different from 0,
that is a point mass at location x would have to 'know' the value of
f(x+dx) which would contradict locality (for corresponding problems in
the time domain it would also violate causality).
What would an imbecile like you know about gradients?
Apparently more than you as otherwise you would have given a proper response
to my argument.
What could an obnoxious imbecile like you possibly do with a
proper response to his 'argument'?
Let this be my problem.
So what is your response to my argument:
One can not define a gradient by using just one function value:
the derivative of a function f at point x is defined as the limit of
( f(x+dx)-f(x) )/dx with dx being infinitesimal but different from 0,
f(x+dx) which would contradict locality (for corresponding problems in
the time domain it would also violate causality).
?
Thomas
.
|
|
|
|
| User: "Thomas Smid" |
|
| Title: Re: General Relativity and Gravity |
29 Mar 2006 06:09:22 AM |
|
|
Dirk Van de moortel wrote:
"Thomas Smid" <thomas.smid@gmail.com> wrote in message news:1143631723.506197.257850@g10g2000cwb.googlegroups.com...
Dirk Van de moortel wrote:
"Thomas Smid" <thomas.smid@gmail.com> wrote in message news:1143625582.098764.162850@t31g2000cwb.googlegroups.com...
One can not define a gradient by using just one function value:
the derivative of a function f at point x is defined as the limit of
( f(x+dx)-f(x) )/dx with dx being infinitesimal but different from 0,
that is a point mass at location x would have to 'know' the value of
f(x+dx) which would contradict locality (for corresponding problems in
the time domain it would also violate causality).
What would an imbecile like you know about gradients?
Apparently more than you as otherwise you would have given a proper response
to my argument.
What could an obnoxious imbecile like you possibly do with a
proper response to his 'argument'?
Let this be my problem.
So what is your response to my argument:
One can not define a gradient by using just one function value:
the derivative of a function f at point x is defined as the limit of
( f(x+dx)-f(x) )/dx with dx being infinitesimal but different from 0,
that is a point mass at location x would have to 'know' the value of
f(x+dx) which would contradict locality (for corresponding problems in
the time domain it would also violate causality).
?
Thomas
.
|
|
|
| User: "Dirk Van de moortel" |
|
| Title: Re: General Relativity and Gravity |
29 Mar 2006 06:44:31 AM |
|
|
"Thomas Smid" <thomas.smid@gmail.com> wrote in message news:1143634162.537172.94730@i39g2000cwa.googlegroups.com...
Dirk Van de moortel wrote:
"Thomas Smid" <thomas.smid@gmail.com> wrote in message news:1143631723.506197.257850@g10g2000cwb.googlegroups.com...
Dirk Van de moortel wrote:
"Thomas Smid" <thomas.smid@gmail.com> wrote in message news:1143625582.098764.162850@t31g2000cwb.googlegroups.com...
One can not define a gradient by using just one function value:
the derivative of a function f at point x is defined as the limit of
( f(x+dx)-f(x) )/dx with dx being infinitesimal but different from 0,
that is a point mass at location x would have to 'know' the value of
f(x+dx) which would contradict locality (for corresponding problems in
the time domain it would also violate causality).
What would an imbecile like you know about gradients?
Apparently more than you as otherwise you would have given a proper response
to my argument.
What could an obnoxious imbecile like you possibly do with a
proper response to his 'argument'?
Let this be my problem.
So what is your response to my argument:
What could a retard like you possible do with responses?
Something like this?
http://groups.google.co.uk/group/sci.physics.relativity/msg/35e09534e108fa99
Smid:
| > > The derivation is not straightforward at all. In fact it
| > > is hopelessly flawed and makes me cringe.
| > > Apart from the issue addressed by me in the quote, the
| > > initial equation x=ut results for instance in dx/dt=u,
| > > whereas further down it reads dx/dt=c and equivalently
| > > dx'/dt'=-u and dx'/dt'=c.
| >
DVdm
| > That is because the equation
| > x = u t
| > is the equation of motion of the origin of S' as seen in the
| > S-frame, which therefore has velocity dx/dt = u according
| > to the S-frame, whereas the equation of motion of the
| > light signal according to the S-frame is
| > x = c t
| > which of course has speed dx/dt = c according to the
| > S-frame.
|
Smid:
| I know that, but this means that the variable x is defined
| twice here
| 1) it is the coordinate of the origin of S',
| 2) it is the coordinate of the light signal, and taking the
| two equations above one gets u=c, which is obviously nonsense.
Dirk Vdm
.
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