• #### Clock rates don't depend on 'gravity'

From J. J. Lodder@21:1/5 to All on Fri Nov 10 21:00:50 2023
[summary: gravity and clock rates for misled kiddies]

General relativity predicts that all freely falling clocks
will run at their own inherent rate. [by postulate]
It also predicts that clocks at different places,
and with different velocities will be seen to run at different rates,
-when compared with each other-.
It also predicts that accelerations do not affect clock rates,
so the results can be extended to non-inertial clocks,
such as clocks at rest at different altitudes on Earth.

According to general relativity all clock effects are purely kinematic,
so derivable from the metric tensor.
Doing the sums for weak fields results in velocity effects
being given by Lorentz factors, and 'gravitational' effects
being given by the variations in Newtonian potential.
So far, so good, and in agreement with experimental results.

Now there are people such as for example 'Lou' in this forum,
who cannot or will not accept or understand this.
They hold that obverved clock effects must be due to 'gravity'
affecting the workings of the clock, somehow.
In other words, they ascribe the observed clock effects
to physical causes, 'gravity' affecting the workings of clocks,
rather than to intrinsic space-time effects.

Fortunately it is easy to settle the point by experiment.
GR predicts that all clocks on the rotating geoid on Earth
must run at the same rate, when compared with each other.
Experiment bears this out, to accuracies approaching 10^-15.
This is of immense practical importance,
because it is the basis for realising the SI second.
(on which -all- physical measurement depends nowadays)

OTOH the force of gravity, as measured by 'small' g,
the acceleration of gravity, varies markedly over the geoid.
(by about 0.5%, between the poles and the equator)

If (the force of) 'gravity' influenced the rate of the clocks
there should be an effect of geographical latitude
on the rate of clocks.
This is not observed to be the case, so this idea stands falsified.

The idea that 'gravity' affects the rate at which clocks run
is a misconception without basis in observed fact,

Jan

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From mitchrae3323@gmail.com@21:1/5 to J. J. Lodder on Fri Nov 10 20:46:35 2023
On Friday, November 10, 2023 at 12:00:54 PM UTC-8, J. J. Lodder wrote:
[summary: gravity and clock rates for misled kiddies]

General relativity predicts that all freely falling clocks
will run at their own inherent rate. [by postulate]
It also predicts that clocks at different places,
and with different velocities will be seen to run at different rates,
-when compared with each other-.
It also predicts that accelerations do not affect clock rates,
so the results can be extended to non-inertial clocks,
such as clocks at rest at different altitudes on Earth.

According to general relativity all clock effects are purely kinematic,
so derivable from the metric tensor.
Doing the sums for weak fields results in velocity effects
being given by Lorentz factors, and 'gravitational' effects
being given by the variations in Newtonian potential.
So far, so good, and in agreement with experimental results.

Now there are people such as for example 'Lou' in this forum,
who cannot or will not accept or understand this.
They hold that obverved clock effects must be due to 'gravity'
affecting the workings of the clock, somehow.
In other words, they ascribe the observed clock effects
to physical causes, 'gravity' affecting the workings of clocks,
rather than to intrinsic space-time effects.

Fortunately it is easy to settle the point by experiment.
GR predicts that all clocks on the rotating geoid on Earth
must run at the same rate, when compared with each other.
Experiment bears this out, to accuracies approaching 10^-15.
This is of immense practical importance,
because it is the basis for realising the SI second.
(on which -all- physical measurement depends nowadays)

OTOH the force of gravity, as measured by 'small' g,
the acceleration of gravity, varies markedly over the geoid.
(by about 0.5%, between the poles and the equator)

If (the force of) 'gravity' influenced the rate of the clocks
there should be an effect of geographical latitude
on the rate of clocks.
This is not observed to be the case, so this idea stands falsified.

The idea that 'gravity' affects the rate at which clocks run
is a misconception without basis in observed fact,

Jan

There is different rotation that will slow time jan.
Gravity will slow down time. Einstein was right...
on both accounts.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From J. J. Lodder@21:1/5 to mitchr...@gmail.com on Sat Nov 11 10:44:19 2023
mitchr...@gmail.com <mitchrae3323@gmail.com> wrote:

On Friday, November 10, 2023 at 12:00:54?PM UTC-8, J. J. Lodder wrote:
[summary: gravity and clock rates for misled kiddies]

General relativity predicts that all freely falling clocks
will run at their own inherent rate. [by postulate]
It also predicts that clocks at different places,
and with different velocities will be seen to run at different rates,
-when compared with each other-.
It also predicts that accelerations do not affect clock rates,
so the results can be extended to non-inertial clocks,
such as clocks at rest at different altitudes on Earth.

According to general relativity all clock effects are purely kinematic,
so derivable from the metric tensor.
Doing the sums for weak fields results in velocity effects
being given by Lorentz factors, and 'gravitational' effects
being given by the variations in Newtonian potential.
So far, so good, and in agreement with experimental results.

Now there are people such as for example 'Lou' in this forum,
who cannot or will not accept or understand this.
They hold that obverved clock effects must be due to 'gravity'
affecting the workings of the clock, somehow.
In other words, they ascribe the observed clock effects
to physical causes, 'gravity' affecting the workings of clocks,
rather than to intrinsic space-time effects.

Fortunately it is easy to settle the point by experiment.
GR predicts that all clocks on the rotating geoid on Earth
must run at the same rate, when compared with each other.
Experiment bears this out, to accuracies approaching 10^-15.
This is of immense practical importance,
because it is the basis for realising the SI second.
(on which -all- physical measurement depends nowadays)

OTOH the force of gravity, as measured by 'small' g,
the acceleration of gravity, varies markedly over the geoid.
(by about 0.5%, between the poles and the equator)

If (the force of) 'gravity' influenced the rate of the clocks
there should be an effect of geographical latitude
on the rate of clocks.
This is not observed to be the case, so this idea stands falsified.

The idea that 'gravity' affects the rate at which clocks run
is a misconception without basis in observed fact,

Jan

There is different rotation that will slow time jan.
Gravity will slow down time. Einstein was right...
on both accounts.

Jan

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Lou@21:1/5 to J. J. Lodder on Sat Nov 11 06:28:17 2023
On Friday, 10 November 2023 at 20:00:54 UTC, J. J. Lodder wrote:
[summary: gravity and clock rates for misled kiddies]

General relativity predicts that all freely falling clocks
will run at their own inherent rate. [by postulate]
It also predicts that clocks at different places,
and with different velocities will be seen to run at different rates,
-when compared with each other-.
It also predicts that accelerations do not affect clock rates,
so the results can be extended to non-inertial clocks,
such as clocks at rest at different altitudes on Earth.

According to general relativity all clock effects are purely kinematic,
so derivable from the metric tensor.
Doing the sums for weak fields results in velocity effects
being given by Lorentz factors, and 'gravitational' effects
being given by the variations in Newtonian potential.
So far, so good, and in agreement with experimental results.

Now there are people such as for example 'Lou' in this forum,
who cannot or will not accept or understand this.
They hold that obverved clock effects must be due to 'gravity'
affecting the workings of the clock, somehow.
In other words, they ascribe the observed clock effects
to physical causes, 'gravity' affecting the workings of clocks,
rather than to intrinsic space-time effects.

Fortunately it is easy to settle the point by experiment.
GR predicts that all clocks on the rotating geoid on Earth
must run at the same rate, when compared with each other.
Experiment bears this out, to accuracies approaching 10^-15.
This is of immense practical importance,
because it is the basis for realising the SI second.
(on which -all- physical measurement depends nowadays)

OTOH the force of gravity, as measured by 'small' g,
the acceleration of gravity, varies markedly over the geoid.
(by about 0.5%, between the poles and the equator)

If (the force of) 'gravity' influenced the rate of the clocks
there should be an effect of geographical latitude
on the rate of clocks.
This is not observed to be the case, so this idea stands falsified.

The idea that 'gravity' affects the rate at which clocks run
is a misconception without basis in observed fact,

A desperately misguided post from JJ.
You did not really read any of my posts. If you did...Then you
deliberately ignored the fact that I *very* explicitly stated that
in a classical model “little g” is acceleration only. Not force.
And you ignored the fact that I very clearly stated that force
on the atoms at different altitudes in a classical model should be
calculated using r.
Not the m/s^2 acceleration of r^2 in “little g”.
Seeing as everyone except a profound idiot would think
acceleration = force.
And If you actually read my posts rather than thump your bible,
you would realise that I also said that force is what Laplace called gravitational potential. And what Newton referred to as a scalar field.
And what Einstein used to calculate his GR clock rate effects.
(Notice the r of Laplace’s gravitational potential and Newton’s scalar field is also the r used in GR. Not r^2 of little g.)
So if you claim that experiment shows no change of clock
rates at different sea level latitudes. Then you have not
only confirmed the predictions of GR.. You have also confirmed
the predictions of classical theory. Seeing as they both use r to
accurately calculate tick rates at different altitudes.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From J. J. Lodder@21:1/5 to Lou on Sat Nov 11 21:47:31 2023
Lou <noelturntive@live.co.uk> wrote:

On Friday, 10 November 2023 at 20:00:54 UTC, J. J. Lodder wrote:
[summary: gravity and clock rates for misled kiddies]

General relativity predicts that all freely falling clocks
will run at their own inherent rate. [by postulate]
It also predicts that clocks at different places,
and with different velocities will be seen to run at different rates,
-when compared with each other-.
It also predicts that accelerations do not affect clock rates,
so the results can be extended to non-inertial clocks,
such as clocks at rest at different altitudes on Earth.

According to general relativity all clock effects are purely kinematic,
so derivable from the metric tensor.
Doing the sums for weak fields results in velocity effects
being given by Lorentz factors, and 'gravitational' effects
being given by the variations in Newtonian potential.
So far, so good, and in agreement with experimental results.

Now there are people such as for example 'Lou' in this forum,
who cannot or will not accept or understand this.
They hold that obverved clock effects must be due to 'gravity'
affecting the workings of the clock, somehow.
In other words, they ascribe the observed clock effects
to physical causes, 'gravity' affecting the workings of clocks,
rather than to intrinsic space-time effects.

Fortunately it is easy to settle the point by experiment.
GR predicts that all clocks on the rotating geoid on Earth
must run at the same rate, when compared with each other.
Experiment bears this out, to accuracies approaching 10^-15.
This is of immense practical importance,
because it is the basis for realising the SI second.
(on which -all- physical measurement depends nowadays)

OTOH the force of gravity, as measured by 'small' g,
the acceleration of gravity, varies markedly over the geoid.
(by about 0.5%, between the poles and the equator)

If (the force of) 'gravity' influenced the rate of the clocks
there should be an effect of geographical latitude
on the rate of clocks.
This is not observed to be the case, so this idea stands falsified.

The idea that 'gravity' affects the rate at which clocks run
is a misconception without basis in observed fact,

A desperately misguided post from JJ.
You did not really read any of my posts. If you did...Then you
deliberately ignored the fact that I *very* explicitly stated that
in a classical model "little g" is acceleration only. Not force.
And you ignored the fact that I very clearly stated that force
on the atoms at different altitudes in a classical model should be
calculated using r.

Indeed, there is little point, because you go on harping about your r,
and you are ignoring all sound advice by others.
You can go on obfuscating because you limit yourself
to situations with spherical symmetry.

So to see the errors of your ways you should consider situations
where spherical symmetry does not hold.
Then the surfaces of constant potential do not coincide
with surfaces of constant acceleration, or constant r.

Not the m/s^2 acceleration of r^2 in "little g".
Seeing as everyone except a profound idiot would think
acceleration = force.
And If you actually read my posts rather than thump your bible,
you would realise that I also said that force is what Laplace called gravitational potential. And what Newton referred to as a scalar field.
And what Einstein used to calculate his GR clock rate effects.
(Notice the r of Laplace's gravitational potential and Newton's scalar
field is also the r used in GR. Not r^2 of little g.)
So if you claim that experiment shows no change of clock
rates at different sea level latitudes. Then you have not
only confirmed the predictions of GR.. You have also confirmed
the predictions of classical theory. Seeing as they both use r to
accurately calculate tick rates at different altitudes.

Experiment shows that clocks on the geoid run at constant rates
wrt each other. Note that the geoid is not a surface of constant r,
nor a surface of constant g,

Jan

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Maciej Wozniak@21:1/5 to J. J. Lodder on Sat Nov 11 22:41:18 2023
On Friday, 10 November 2023 at 21:00:54 UTC+1, J. J. Lodder wrote:
[summary: gravity and clock rates for misled kiddies]

General relativity predicts that all freely falling clocks
will run at their own inherent rate. [by postulate]
It also predicts that clocks at different places,
and with different velocities will be seen to run at different rates,
-when compared with each other-.

doesn't predict it at all.

OTOH the force of gravity, as measured by 'small' g,

JJ, poor halfbrain, don't you even know that
your idiot guru has refuted the force of gravirty?
The ignorance of relativistic morons keep amazing
me after all these years.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Lou@21:1/5 to J. J. Lodder on Sun Nov 12 03:17:06 2023
On Saturday, 11 November 2023 at 20:47:35 UTC, J. J. Lodder wrote:
Lou wrote:

On Friday, 10 November 2023 at 20:00:54 UTC, J. J. Lodder wrote:
[summary: gravity and clock rates for misled kiddies]

General relativity predicts that all freely falling clocks
will run at their own inherent rate. [by postulate]
It also predicts that clocks at different places,
and with different velocities will be seen to run at different rates, -when compared with each other-.
It also predicts that accelerations do not affect clock rates,
so the results can be extended to non-inertial clocks,
such as clocks at rest at different altitudes on Earth.

According to general relativity all clock effects are purely kinematic, so derivable from the metric tensor.
Doing the sums for weak fields results in velocity effects
being given by Lorentz factors, and 'gravitational' effects
being given by the variations in Newtonian potential.
So far, so good, and in agreement with experimental results.

Now there are people such as for example 'Lou' in this forum,
who cannot or will not accept or understand this.
They hold that obverved clock effects must be due to 'gravity'
affecting the workings of the clock, somehow.
In other words, they ascribe the observed clock effects
to physical causes, 'gravity' affecting the workings of clocks,
rather than to intrinsic space-time effects.

Fortunately it is easy to settle the point by experiment.
GR predicts that all clocks on the rotating geoid on Earth
must run at the same rate, when compared with each other.
Experiment bears this out, to accuracies approaching 10^-15.
This is of immense practical importance,
because it is the basis for realising the SI second.
(on which -all- physical measurement depends nowadays)

OTOH the force of gravity, as measured by 'small' g,
the acceleration of gravity, varies markedly over the geoid.
(by about 0.5%, between the poles and the equator)

If (the force of) 'gravity' influenced the rate of the clocks
there should be an effect of geographical latitude
on the rate of clocks.
This is not observed to be the case, so this idea stands falsified.

The idea that 'gravity' affects the rate at which clocks run
is a misconception without basis in observed fact,

A desperately misguided post from JJ.
You did not really read any of my posts. If you did...Then you deliberately ignored the fact that I *very* explicitly stated that
in a classical model "little g" is acceleration only. Not force.
And you ignored the fact that I very clearly stated that force
on the atoms at different altitudes in a classical model should be calculated using r.
Indeed, there is little point, because you go on harping about your r,
and you are ignoring all sound advice by others.

‘Sound advice’ ? That’s a joke considering your ‘sound advice’ consists of telling me to ignore the facts , and pretend that acceleration=force.

You can go on obfuscating because you limit yourself
to situations with spherical symmetry.

So to see the errors of your ways you should consider situations
where spherical symmetry does not hold.
Then the surfaces of constant potential do not coincide
with surfaces of constant acceleration, or constant r.

You seem to contradict yourself here. Below, dont you suggest
that it doesn’t matter where on the sphere one is,.. calculations
using r of potential always give the same constant tick rate?

Not the m/s^2 acceleration of r^2 in "little g".
Seeing as anyone except a profound idiot would think
acceleration = force.
And If you actually read my posts rather than thump your bible,
you would realise that I also said that force is what Laplace called gravitational potential. And what Newton referred to as a scalar field. And what Einstein used to calculate his GR clock rate effects.
(Notice the r of Laplace's gravitational potential and Newton's scalar field is also the r used in GR. Not r^2 of little g.)
So if you claim that experiment shows no change of clock
rates at different sea level latitudes. Then you have not
only confirmed the predictions of GR.. You have also confirmed
the predictions of classical theory. Seeing as they both use r to accurately calculate tick rates at different altitudes.
Experiment shows that clocks on the geoid run at constant rates
wrt each other.

I haven’t seen this reference before but I assume by “on
the geoid” you mean *all clocks at the earths surface*?

Note that the geoid is not a surface of constant r,
nor a surface of constant g,

Im not familiar with this particular information you cite, but
it makes sense that where both classical and GR calculate
tick rates with r they would show constant rates regardless
of local mass distributions in the earth. Because to calculate
using r, in the various formulae for classical or GR one always
uses *total mass* of the planet.
Not local mass.
So regardless of where on the geoid you are, as long as you
are on the surface...the WHOLE mass of earth is pulling on you.
And the total mass is constant regardless of your location
on the surface of the planet.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Lou@21:1/5 to J. J. Lodder on Sun Nov 12 04:37:31 2023
On Saturday, 11 November 2023 at 20:47:35 UTC, J. J. Lodder wrote:
Lou wrote:

On Friday, 10 November 2023 at 20:00:54 UTC, J. J. Lodder wrote:
[summary: gravity and clock rates for misled kiddies]

General relativity predicts that all freely falling clocks
will run at their own inherent rate. [by postulate]
It also predicts that clocks at different places,
and with different velocities will be seen to run at different rates, -when compared with each other-.
It also predicts that accelerations do not affect clock rates,
so the results can be extended to non-inertial clocks,
such as clocks at rest at different altitudes on Earth.

According to general relativity all clock effects are purely kinematic, so derivable from the metric tensor.
Doing the sums for weak fields results in velocity effects
being given by Lorentz factors, and 'gravitational' effects
being given by the variations in Newtonian potential.
So far, so good, and in agreement with experimental results.

Now there are people such as for example 'Lou' in this forum,
who cannot or will not accept or understand this.
They hold that obverved clock effects must be due to 'gravity'
affecting the workings of the clock, somehow.
In other words, they ascribe the observed clock effects
to physical causes, 'gravity' affecting the workings of clocks,
rather than to intrinsic space-time effects.

Fortunately it is easy to settle the point by experiment.
GR predicts that all clocks on the rotating geoid on Earth
must run at the same rate, when compared with each other.
Experiment bears this out, to accuracies approaching 10^-15.
This is of immense practical importance,
because it is the basis for realising the SI second.
(on which -all- physical measurement depends nowadays)

OTOH the force of gravity, as measured by 'small' g,
the acceleration of gravity, varies markedly over the geoid.
(by about 0.5%, between the poles and the equator)

If (the force of) 'gravity' influenced the rate of the clocks
there should be an effect of geographical latitude
on the rate of clocks.
This is not observed to be the case, so this idea stands falsified.

The idea that 'gravity' affects the rate at which clocks run
is a misconception without basis in observed fact,

A desperately misguided post from JJ.
You did not really read any of my posts. If you did...Then you deliberately ignored the fact that I *very* explicitly stated that
in a classical model "little g" is acceleration only. Not force.
And you ignored the fact that I very clearly stated that force
on the atoms at different altitudes in a classical model should be calculated using r.
Indeed, there is little point, because you go on harping about your r,
and you are ignoring all sound advice by others.
You can go on obfuscating because you limit yourself
to situations with spherical symmetry.

So to see the errors of your ways you should consider situations
where spherical symmetry does not hold.
Then the surfaces of constant potential do not coincide
with surfaces of constant acceleration, or constant r.
Not the m/s^2 acceleration of r^2 in "little g".
Seeing as everyone except a profound idiot would think
acceleration = force.
And If you actually read my posts rather than thump your bible,
you would realise that I also said that force is what Laplace called gravitational potential. And what Newton referred to as a scalar field. And what Einstein used to calculate his GR clock rate effects.
(Notice the r of Laplace's gravitational potential and Newton's scalar field is also the r used in GR. Not r^2 of little g.)
So if you claim that experiment shows no change of clock
rates at different sea level latitudes. Then you have not
only confirmed the predictions of GR.. You have also confirmed
the predictions of classical theory. Seeing as they both use r to accurately calculate tick rates at different altitudes.
Experiment shows that clocks on the geoid run at constant rates
wrt each other. Note that the geoid is not a surface of constant r,
nor a surface of constant g,

A straw man argument if ever you make.
Yes I’ve looked at your ‘geoid’ now and how it varies slightly by about 200m
relative to the reference geoid and how technically the r distance doesn’t exactly follow the geoid surface. That makes sense. Splitting hairs though
on your part to pretend somehow this rules out a classical model
which uses r. I notice you didn’t actually specify why it would.
In fact it doesn’t rule out in any way a classical model
any more than it would rule out GR.
Because in a classical calculation if one needs to assume *exactly* the
*total* mass M of the earth at r, then yes to be *absolutely* accurate
the geoid surface has to be used. Not the actual distance r.
But the same applies to GR. And the fact remains that
generally, the force of gravity in a classical model follows r not r^2.
(And to please the pedant JJ,... with ever so small meter length
fluctuations in the exact distance of r to also be taken into account)

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From J. J. Lodder@21:1/5 to Lou on Sun Nov 12 14:30:13 2023
Lou <noelturntive@live.co.uk> wrote:

On Saturday, 11 November 2023 at 20:47:35 UTC, J. J. Lodder wrote:
Lou wrote:

On Friday, 10 November 2023 at 20:00:54 UTC, J. J. Lodder wrote:
[summary: gravity and clock rates for misled kiddies]

General relativity predicts that all freely falling clocks
will run at their own inherent rate. [by postulate]
It also predicts that clocks at different places,
and with different velocities will be seen to run at different rates, -when compared with each other-.
It also predicts that accelerations do not affect clock rates,
so the results can be extended to non-inertial clocks,
such as clocks at rest at different altitudes on Earth.

According to general relativity all clock effects are purely kinematic, so derivable from the metric tensor.
Doing the sums for weak fields results in velocity effects
being given by Lorentz factors, and 'gravitational' effects
being given by the variations in Newtonian potential.
So far, so good, and in agreement with experimental results.

Now there are people such as for example 'Lou' in this forum,
who cannot or will not accept or understand this.
They hold that obverved clock effects must be due to 'gravity' affecting the workings of the clock, somehow.
In other words, they ascribe the observed clock effects
to physical causes, 'gravity' affecting the workings of clocks,
rather than to intrinsic space-time effects.

Fortunately it is easy to settle the point by experiment.
GR predicts that all clocks on the rotating geoid on Earth
must run at the same rate, when compared with each other.
Experiment bears this out, to accuracies approaching 10^-15.
This is of immense practical importance,
because it is the basis for realising the SI second.
(on which -all- physical measurement depends nowadays)

OTOH the force of gravity, as measured by 'small' g,
the acceleration of gravity, varies markedly over the geoid.
(by about 0.5%, between the poles and the equator)

If (the force of) 'gravity' influenced the rate of the clocks
there should be an effect of geographical latitude
on the rate of clocks.
This is not observed to be the case, so this idea stands falsified.

The idea that 'gravity' affects the rate at which clocks run
is a misconception without basis in observed fact,

A desperately misguided post from JJ.
You did not really read any of my posts. If you did...Then you deliberately ignored the fact that I *very* explicitly stated that
in a classical model "little g" is acceleration only. Not force.
And you ignored the fact that I very clearly stated that force
on the atoms at different altitudes in a classical model should be calculated using r.
Indeed, there is little point, because you go on harping about your r,
and you are ignoring all sound advice by others.
You can go on obfuscating because you limit yourself
to situations with spherical symmetry.

So to see the errors of your ways you should consider situations
where spherical symmetry does not hold.
Then the surfaces of constant potential do not coincide
with surfaces of constant acceleration, or constant r.
Not the m/s^2 acceleration of r^2 in "little g".
Seeing as everyone except a profound idiot would think
acceleration = force.
And If you actually read my posts rather than thump your bible,
you would realise that I also said that force is what Laplace called gravitational potential. And what Newton referred to as a scalar field. And what Einstein used to calculate his GR clock rate effects.
(Notice the r of Laplace's gravitational potential and Newton's scalar field is also the r used in GR. Not r^2 of little g.)
So if you claim that experiment shows no change of clock
rates at different sea level latitudes. Then you have not
only confirmed the predictions of GR.. You have also confirmed
the predictions of classical theory. Seeing as they both use r to accurately calculate tick rates at different altitudes.
Experiment shows that clocks on the geoid run at constant rates
wrt each other. Note that the geoid is not a surface of constant r,
nor a surface of constant g,

A straw man argument if ever you make.
Yes I've looked at your 'geoid' now and how it varies slightly by about 200m relative to the reference geoid and how technically the r distance doesn't exactly follow the geoid surface. That makes sense. Splitting hairs though
on your part to pretend somehow this rules out a classical model
which uses r. I notice you didn't actually specify why it would.
In fact it doesn't rule out in any way a classical model
any more than it would rule out GR.
Because in a classical calculation if one needs to assume *exactly* the *total* mass M of the earth at r, then yes to be *absolutely* accurate
the geoid surface has to be used. Not the actual distance r.
But the same applies to GR. And the fact remains that
generally, the force of gravity in a classical model follows r not r^2.
(And to please the pedant JJ,... with ever so small meter length
fluctuations in the exact distance of r to also be taken into account)

So you missed all points, again. I'll simplify.
The geoid surface is by definition an equipotential surface of the
Newtonian potential.
So it coincides (almost) with the mean sea level.
The geoid is (to a very good approximation) an ellipsoid of revolution.
The small differences between geoid and ellipsoid
(due to slightly irregular mass distributions inside the Earth)
don't matter for what follows.

Now, on the geoid, and at the poles, we have: r < average g > average, potential = constant
On the geoid, at mid-latitudes we have r = average, g = average,
potential = same constant
On the geoid, at the equator, we have r > average, g < average,
potential = still the same constant, by definition of the geoid.

The differences are huge, r = 6357-6378 km, g = 9.863-9.798 m/s2,
compared to clock stabilities of 10^-15.

What is your prediction for the rates of clocks in those three places?
No verbiage, just say faster, slower, or the same,
and if you can by how much,

Jan

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Lou@21:1/5 to J. J. Lodder on Sun Nov 12 06:15:30 2023
On Sunday, 12 November 2023 at 13:30:18 UTC, J. J. Lodder wrote:
Lou wrote:

On Saturday, 11 November 2023 at 20:47:35 UTC, J. J. Lodder wrote:
Lou wrote:

On Friday, 10 November 2023 at 20:00:54 UTC, J. J. Lodder wrote:
[summary: gravity and clock rates for misled kiddies]

General relativity predicts that all freely falling clocks
will run at their own inherent rate. [by postulate]
It also predicts that clocks at different places,
and with different velocities will be seen to run at different rates,
-when compared with each other-.
It also predicts that accelerations do not affect clock rates,
so the results can be extended to non-inertial clocks,
such as clocks at rest at different altitudes on Earth.

According to general relativity all clock effects are purely kinematic,
so derivable from the metric tensor.
Doing the sums for weak fields results in velocity effects
being given by Lorentz factors, and 'gravitational' effects
being given by the variations in Newtonian potential.
So far, so good, and in agreement with experimental results.

Now there are people such as for example 'Lou' in this forum,
who cannot or will not accept or understand this.
They hold that obverved clock effects must be due to 'gravity' affecting the workings of the clock, somehow.
In other words, they ascribe the observed clock effects
to physical causes, 'gravity' affecting the workings of clocks, rather than to intrinsic space-time effects.

Fortunately it is easy to settle the point by experiment.
GR predicts that all clocks on the rotating geoid on Earth
must run at the same rate, when compared with each other.
Experiment bears this out, to accuracies approaching 10^-15.
This is of immense practical importance,
because it is the basis for realising the SI second.
(on which -all- physical measurement depends nowadays)

OTOH the force of gravity, as measured by 'small' g,
the acceleration of gravity, varies markedly over the geoid.
(by about 0.5%, between the poles and the equator)

If (the force of) 'gravity' influenced the rate of the clocks
there should be an effect of geographical latitude
on the rate of clocks.
This is not observed to be the case, so this idea stands falsified.

The idea that 'gravity' affects the rate at which clocks run
is a misconception without basis in observed fact,

A desperately misguided post from JJ.
You did not really read any of my posts. If you did...Then you deliberately ignored the fact that I *very* explicitly stated that
in a classical model "little g" is acceleration only. Not force.
And you ignored the fact that I very clearly stated that force
on the atoms at different altitudes in a classical model should be calculated using r.
Indeed, there is little point, because you go on harping about your r, and you are ignoring all sound advice by others.
You can go on obfuscating because you limit yourself
to situations with spherical symmetry.

So to see the errors of your ways you should consider situations
where spherical symmetry does not hold.
Then the surfaces of constant potential do not coincide
with surfaces of constant acceleration, or constant r.
Not the m/s^2 acceleration of r^2 in "little g".
Seeing as everyone except a profound idiot would think
acceleration = force.
And If you actually read my posts rather than thump your bible,
you would realise that I also said that force is what Laplace called gravitational potential. And what Newton referred to as a scalar field.
And what Einstein used to calculate his GR clock rate effects.
(Notice the r of Laplace's gravitational potential and Newton's scalar field is also the r used in GR. Not r^2 of little g.)
So if you claim that experiment shows no change of clock
rates at different sea level latitudes. Then you have not
only confirmed the predictions of GR.. You have also confirmed
the predictions of classical theory. Seeing as they both use r to accurately calculate tick rates at different altitudes.
Experiment shows that clocks on the geoid run at constant rates
wrt each other. Note that the geoid is not a surface of constant r,
nor a surface of constant g,

A straw man argument if ever you make.
Yes I've looked at your 'geoid' now and how it varies slightly by about 200m
relative to the reference geoid and how technically the r distance doesn't exactly follow the geoid surface. That makes sense. Splitting hairs though on your part to pretend somehow this rules out a classical model
which uses r. I notice you didn't actually specify why it would.
In fact it doesn't rule out in any way a classical model
any more than it would rule out GR.
Because in a classical calculation if one needs to assume *exactly* the *total* mass M of the earth at r, then yes to be *absolutely* accurate
the geoid surface has to be used. Not the actual distance r.
But the same applies to GR. And the fact remains that
generally, the force of gravity in a classical model follows r not r^2. (And to please the pedant JJ,... with ever so small meter length fluctuations in the exact distance of r to also be taken into account)
So you missed all points, again. I'll simplify.
The geoid surface is by definition an equipotential surface of the
Newtonian potential.
So it coincides (almost) with the mean sea level.
The geoid is (to a very good approximation) an ellipsoid of revolution.
The small differences between geoid and ellipsoid
(due to slightly irregular mass distributions inside the Earth)
don't matter for what follows.

Now, on the geoid, and at the poles, we have: r < average g > average, potential = constant
On the geoid, at mid-latitudes we have r = average, g = average,
potential = same constant
On the geoid, at the equator, we have r > average, g < average,
potential = still the same constant, by definition of the geoid.

The differences are huge, r = 6357-6378 km, g = 9.863-9.798 m/s2,
compared to clock stabilities of 10^-15.

What is your prediction for the rates of clocks in those three places?
No verbiage, just say faster, slower, or the same,
and if you can by how much,

If you tried reading my posts you wouldn’t be pretending I said the force of gravity is 9.863-9.798 m/s2.
That’s r^2 and it’s called acceleration. You don’t seem to know that m/s^2
is acceleration!!! Since when does Force=acceleration?
In all my posts I state very clearly that in a classical model the force of gravity is modelled with GM/r.
And I already responded to your point on geoids that yes if you want to split hairs
the geoid surface varies from r by up to 200 meters. Which is why
very accurate measurements of clock rates will show constant rates at the surface of the geoid only. And not to r. But that’s still consistent with a classical model
as much as with GR.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From J. J. Lodder@21:1/5 to Lou on Sun Nov 12 16:08:40 2023
Lou <noelturntive@live.co.uk> wrote:

On Sunday, 12 November 2023 at 13:30:18 UTC, J. J. Lodder wrote:
Lou wrote:

On Saturday, 11 November 2023 at 20:47:35 UTC, J. J. Lodder wrote:
Lou wrote:

On Friday, 10 November 2023 at 20:00:54 UTC, J. J. Lodder wrote:
[summary: gravity and clock rates for misled kiddies]

General relativity predicts that all freely falling clocks
will run at their own inherent rate. [by postulate]
It also predicts that clocks at different places,
and with different velocities will be seen to run at different rates, -when compared with each other-.
It also predicts that accelerations do not affect clock rates,
so the results can be extended to non-inertial clocks,
such as clocks at rest at different altitudes on Earth.

According to general relativity all clock effects are purely kinematic, so derivable from the metric tensor.
Doing the sums for weak fields results in velocity effects
being given by Lorentz factors, and 'gravitational' effects
being given by the variations in Newtonian potential.
So far, so good, and in agreement with experimental results.

Now there are people such as for example 'Lou' in this forum,
who cannot or will not accept or understand this.
They hold that obverved clock effects must be due to 'gravity' affecting the workings of the clock, somehow.
In other words, they ascribe the observed clock effects
to physical causes, 'gravity' affecting the workings of clocks, rather than to intrinsic space-time effects.

Fortunately it is easy to settle the point by experiment.
GR predicts that all clocks on the rotating geoid on Earth
must run at the same rate, when compared with each other. Experiment bears this out, to accuracies approaching 10^-15.
This is of immense practical importance,
because it is the basis for realising the SI second.
(on which -all- physical measurement depends nowadays)

OTOH the force of gravity, as measured by 'small' g,
the acceleration of gravity, varies markedly over the geoid.
(by about 0.5%, between the poles and the equator)

If (the force of) 'gravity' influenced the rate of the clocks
there should be an effect of geographical latitude
on the rate of clocks.
This is not observed to be the case, so this idea stands falsified.

The idea that 'gravity' affects the rate at which clocks run
is a misconception without basis in observed fact,

A desperately misguided post from JJ.
You did not really read any of my posts. If you did...Then you deliberately ignored the fact that I *very* explicitly stated that
in a classical model "little g" is acceleration only. Not force.
And you ignored the fact that I very clearly stated that force
on the atoms at different altitudes in a classical model should be calculated using r.
Indeed, there is little point, because you go on harping about your r, and you are ignoring all sound advice by others.
You can go on obfuscating because you limit yourself
to situations with spherical symmetry.

So to see the errors of your ways you should consider situations
where spherical symmetry does not hold.
Then the surfaces of constant potential do not coincide
with surfaces of constant acceleration, or constant r.
Not the m/s^2 acceleration of r^2 in "little g".
Seeing as everyone except a profound idiot would think
acceleration = force.
And If you actually read my posts rather than thump your bible,
you would realise that I also said that force is what Laplace
called gravitational potential. And what Newton referred to as a scalar field.
And what Einstein used to calculate his GR clock rate effects. (Notice the r of Laplace's gravitational potential and Newton's scalar
field is also the r used in GR. Not r^2 of little g.)
So if you claim that experiment shows no change of clock
rates at different sea level latitudes. Then you have not
only confirmed the predictions of GR.. You have also confirmed
the predictions of classical theory. Seeing as they both use r to accurately calculate tick rates at different altitudes.
Experiment shows that clocks on the geoid run at constant rates
wrt each other. Note that the geoid is not a surface of constant r,
nor a surface of constant g,

A straw man argument if ever you make.
Yes I've looked at your 'geoid' now and how it varies slightly by
about 200m relative to the reference geoid and how technically the r distance doesn't exactly follow the geoid surface. That makes sense. Splitting hairs though on your part to pretend somehow this rules out
a classical model which uses r. I notice you didn't actually specify
why it would. In fact it doesn't rule out in any way a classical model any more than it would rule out GR.
Because in a classical calculation if one needs to assume *exactly* the *total* mass M of the earth at r, then yes to be *absolutely* accurate the geoid surface has to be used. Not the actual distance r.
But the same applies to GR. And the fact remains that
generally, the force of gravity in a classical model follows r not r^2. (And to please the pedant JJ,... with ever so small meter length fluctuations in the exact distance of r to also be taken into account)
So you missed all points, again. I'll simplify.
The geoid surface is by definition an equipotential surface of the Newtonian potential.
So it coincides (almost) with the mean sea level.
The geoid is (to a very good approximation) an ellipsoid of revolution.
The small differences between geoid and ellipsoid
(due to slightly irregular mass distributions inside the Earth)
don't matter for what follows.

Now, on the geoid, and at the poles, we have: r < average g > average, potential = constant
On the geoid, at mid-latitudes we have r = average, g = average,
potential = same constant
On the geoid, at the equator, we have r > average, g < average,
potential = still the same constant, by definition of the geoid.

The differences are huge, r = 6357-6378 km, g = 9.863-9.798 m/s2,
compared to clock stabilities of 10^-15.

What is your prediction for the rates of clocks in those three places?
No verbiage, just say faster, slower, or the same,
and if you can by how much,

If you tried reading my posts you wouldn't be pretending I said the force of gravity is 9.863-9.798 m/s2.

Too bad if you didn't say it, for those are the measured values.

That's r^2 and it's called acceleration. You don't seem to know that m/s^2
is acceleration!!! Since when does Force=acceleration?
In all my posts I state very clearly that in a classical model the force of gravity is modelled with GM/r.
And I already responded to your point on geoids that yes if you want to
split hairs the geoid surface varies from r by up to 200 meters. Which is
why very accurate measurements of clock rates will show constant rates at
the surface of the geoid only. And not to r. But that's still consistent
with a classical model as much as with GR.

So you have nothing to say,
beyond agreeing that general relativity gives the right answer,

Jan

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Ross Finlayson@21:1/5 to J. J. Lodder on Sun Nov 12 09:34:52 2023
On Sunday, November 12, 2023 at 7:08:45 AM UTC-8, J. J. Lodder wrote:
Lou <noeltu...@live.co.uk> wrote:

On Sunday, 12 November 2023 at 13:30:18 UTC, J. J. Lodder wrote:
Lou wrote:

On Saturday, 11 November 2023 at 20:47:35 UTC, J. J. Lodder wrote:
Lou wrote:

On Friday, 10 November 2023 at 20:00:54 UTC, J. J. Lodder wrote:
[summary: gravity and clock rates for misled kiddies]

General relativity predicts that all freely falling clocks
will run at their own inherent rate. [by postulate]
It also predicts that clocks at different places,
and with different velocities will be seen to run at different rates, -when compared with each other-.
It also predicts that accelerations do not affect clock rates, so the results can be extended to non-inertial clocks,
such as clocks at rest at different altitudes on Earth.

According to general relativity all clock effects are purely kinematic, so derivable from the metric tensor.
Doing the sums for weak fields results in velocity effects
being given by Lorentz factors, and 'gravitational' effects being given by the variations in Newtonian potential.
So far, so good, and in agreement with experimental results.

Now there are people such as for example 'Lou' in this forum, who cannot or will not accept or understand this.
They hold that obverved clock effects must be due to 'gravity' affecting the workings of the clock, somehow.
In other words, they ascribe the observed clock effects
to physical causes, 'gravity' affecting the workings of clocks, rather than to intrinsic space-time effects.

Fortunately it is easy to settle the point by experiment.
GR predicts that all clocks on the rotating geoid on Earth
must run at the same rate, when compared with each other. Experiment bears this out, to accuracies approaching 10^-15. This is of immense practical importance,
because it is the basis for realising the SI second.
(on which -all- physical measurement depends nowadays)

OTOH the force of gravity, as measured by 'small' g,
the acceleration of gravity, varies markedly over the geoid.
(by about 0.5%, between the poles and the equator)

If (the force of) 'gravity' influenced the rate of the clocks there should be an effect of geographical latitude
on the rate of clocks.
This is not observed to be the case, so this idea stands falsified.

The idea that 'gravity' affects the rate at which clocks run
is a misconception without basis in observed fact,

A desperately misguided post from JJ.
You did not really read any of my posts. If you did...Then you deliberately ignored the fact that I *very* explicitly stated that in a classical model "little g" is acceleration only. Not force. And you ignored the fact that I very clearly stated that force
on the atoms at different altitudes in a classical model should be calculated using r.
Indeed, there is little point, because you go on harping about your r,
and you are ignoring all sound advice by others.
You can go on obfuscating because you limit yourself
to situations with spherical symmetry.

So to see the errors of your ways you should consider situations where spherical symmetry does not hold.
Then the surfaces of constant potential do not coincide
with surfaces of constant acceleration, or constant r.
Not the m/s^2 acceleration of r^2 in "little g".
Seeing as everyone except a profound idiot would think acceleration = force.
And If you actually read my posts rather than thump your bible, you would realise that I also said that force is what Laplace called gravitational potential. And what Newton referred to as a scalar field.
And what Einstein used to calculate his GR clock rate effects. (Notice the r of Laplace's gravitational potential and Newton's scalar
field is also the r used in GR. Not r^2 of little g.)
So if you claim that experiment shows no change of clock
rates at different sea level latitudes. Then you have not
only confirmed the predictions of GR.. You have also confirmed
the predictions of classical theory. Seeing as they both use r to accurately calculate tick rates at different altitudes.
Experiment shows that clocks on the geoid run at constant rates
wrt each other. Note that the geoid is not a surface of constant r, nor a surface of constant g,

A straw man argument if ever you make.
Yes I've looked at your 'geoid' now and how it varies slightly by about 200m relative to the reference geoid and how technically the r distance doesn't exactly follow the geoid surface. That makes sense. Splitting hairs though on your part to pretend somehow this rules out a classical model which uses r. I notice you didn't actually specify why it would. In fact it doesn't rule out in any way a classical model any more than it would rule out GR.
Because in a classical calculation if one needs to assume *exactly* the
*total* mass M of the earth at r, then yes to be *absolutely* accurate the geoid surface has to be used. Not the actual distance r.
But the same applies to GR. And the fact remains that
generally, the force of gravity in a classical model follows r not r^2.
(And to please the pedant JJ,... with ever so small meter length fluctuations in the exact distance of r to also be taken into account)
So you missed all points, again. I'll simplify.
The geoid surface is by definition an equipotential surface of the Newtonian potential.
So it coincides (almost) with the mean sea level.
The geoid is (to a very good approximation) an ellipsoid of revolution. The small differences between geoid and ellipsoid
(due to slightly irregular mass distributions inside the Earth)
don't matter for what follows.

Now, on the geoid, and at the poles, we have: r < average g > average, potential = constant
On the geoid, at mid-latitudes we have r = average, g = average, potential = same constant
On the geoid, at the equator, we have r > average, g < average, potential = still the same constant, by definition of the geoid.

The differences are huge, r = 6357-6378 km, g = 9.863-9.798 m/s2, compared to clock stabilities of 10^-15.

What is your prediction for the rates of clocks in those three places? No verbiage, just say faster, slower, or the same,
and if you can by how much,

If you tried reading my posts you wouldn't be pretending I said the force of
gravity is 9.863-9.798 m/s2.
Too bad if you didn't say it, for those are the measured values.
That's r^2 and it's called acceleration. You don't seem to know that m/s^2 is acceleration!!! Since when does Force=acceleration?
In all my posts I state very clearly that in a classical model the force of
gravity is modelled with GM/r.
And I already responded to your point on geoids that yes if you want to split hairs the geoid surface varies from r by up to 200 meters. Which is why very accurate measurements of clock rates will show constant rates at the surface of the geoid only. And not to r. But that's still consistent with a classical model as much as with GR.
So you have nothing to say,
beyond agreeing that general relativity gives the right answer,

Jan

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Maciej Wozniak@21:1/5 to J. J. Lodder on Sun Nov 12 09:49:31 2023
On Sunday, 12 November 2023 at 14:30:18 UTC+1, J. J. Lodder wrote:

What is your prediction for the rates of clocks in those three places?
No verbiage, just say faster, slower, or the same,
and if you can by how much,

How about - it depends what clocks?

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Lou@21:1/5 to J. J. Lodder on Sun Nov 12 10:56:18 2023
On Sunday, 12 November 2023 at 15:08:45 UTC, J. J. Lodder wrote:
Lou wrote:

On Sunday, 12 November 2023 at 13:30:18 UTC, J. J. Lodder wrote:
Lou wrote:

On Saturday, 11 November 2023 at 20:47:35 UTC, J. J. Lodder wrote:
Lou wrote:

On Friday, 10 November 2023 at 20:00:54 UTC, J. J. Lodder wrote:
[summary: gravity and clock rates for misled kiddies]

General relativity predicts that all freely falling clocks
will run at their own inherent rate. [by postulate]
It also predicts that clocks at different places,
and with different velocities will be seen to run at different rates, -when compared with each other-.
It also predicts that accelerations do not affect clock rates, so the results can be extended to non-inertial clocks,
such as clocks at rest at different altitudes on Earth.

According to general relativity all clock effects are purely kinematic, so derivable from the metric tensor.
Doing the sums for weak fields results in velocity effects
being given by Lorentz factors, and 'gravitational' effects being given by the variations in Newtonian potential.
So far, so good, and in agreement with experimental results.

Now there are people such as for example 'Lou' in this forum, who cannot or will not accept or understand this.
They hold that obverved clock effects must be due to 'gravity' affecting the workings of the clock, somehow.
In other words, they ascribe the observed clock effects
to physical causes, 'gravity' affecting the workings of clocks, rather than to intrinsic space-time effects.

Fortunately it is easy to settle the point by experiment.
GR predicts that all clocks on the rotating geoid on Earth
must run at the same rate, when compared with each other. Experiment bears this out, to accuracies approaching 10^-15. This is of immense practical importance,
because it is the basis for realising the SI second.
(on which -all- physical measurement depends nowadays)

OTOH the force of gravity, as measured by 'small' g,
the acceleration of gravity, varies markedly over the geoid.
(by about 0.5%, between the poles and the equator)

If (the force of) 'gravity' influenced the rate of the clocks there should be an effect of geographical latitude
on the rate of clocks.
This is not observed to be the case, so this idea stands falsified.

The idea that 'gravity' affects the rate at which clocks run
is a misconception without basis in observed fact,

A desperately misguided post from JJ.
You did not really read any of my posts. If you did...Then you deliberately ignored the fact that I *very* explicitly stated that in a classical model "little g" is acceleration only. Not force. And you ignored the fact that I very clearly stated that force
on the atoms at different altitudes in a classical model should be calculated using r.
Indeed, there is little point, because you go on harping about your r,
and you are ignoring all sound advice by others.
You can go on obfuscating because you limit yourself
to situations with spherical symmetry.

So to see the errors of your ways you should consider situations where spherical symmetry does not hold.
Then the surfaces of constant potential do not coincide
with surfaces of constant acceleration, or constant r.
Not the m/s^2 acceleration of r^2 in "little g".
Seeing as everyone except a profound idiot would think acceleration = force.
And If you actually read my posts rather than thump your bible, you would realise that I also said that force is what Laplace called gravitational potential. And what Newton referred to as a scalar field.
And what Einstein used to calculate his GR clock rate effects. (Notice the r of Laplace's gravitational potential and Newton's scalar
field is also the r used in GR. Not r^2 of little g.)
So if you claim that experiment shows no change of clock
rates at different sea level latitudes. Then you have not
only confirmed the predictions of GR.. You have also confirmed
the predictions of classical theory. Seeing as they both use r to accurately calculate tick rates at different altitudes.
Experiment shows that clocks on the geoid run at constant rates
wrt each other. Note that the geoid is not a surface of constant r, nor a surface of constant g,

A straw man argument if ever you make.
Yes I've looked at your 'geoid' now and how it varies slightly by about 200m relative to the reference geoid and how technically the r distance doesn't exactly follow the geoid surface. That makes sense. Splitting hairs though on your part to pretend somehow this rules out a classical model which uses r. I notice you didn't actually specify why it would. In fact it doesn't rule out in any way a classical model any more than it would rule out GR.
Because in a classical calculation if one needs to assume *exactly* the
*total* mass M of the earth at r, then yes to be *absolutely* accurate the geoid surface has to be used. Not the actual distance r.
But the same applies to GR. And the fact remains that
generally, the force of gravity in a classical model follows r not r^2.
(And to please the pedant JJ,... with ever so small meter length fluctuations in the exact distance of r to also be taken into account)
So you missed all points, again. I'll simplify.
The geoid surface is by definition an equipotential surface of the Newtonian potential.
So it coincides (almost) with the mean sea level.
The geoid is (to a very good approximation) an ellipsoid of revolution. The small differences between geoid and ellipsoid
(due to slightly irregular mass distributions inside the Earth)
don't matter for what follows.

Now, on the geoid, and at the poles, we have: r < average g > average, potential = constant
On the geoid, at mid-latitudes we have r = average, g = average, potential = same constant
On the geoid, at the equator, we have r > average, g < average, potential = still the same constant, by definition of the geoid.

The differences are huge, r = 6357-6378 km, g = 9.863-9.798 m/s2, compared to clock stabilities of 10^-15.

What is your prediction for the rates of clocks in those three places? No verbiage, just say faster, slower, or the same,
and if you can by how much,

If you tried reading my posts you wouldn't be pretending I said the force of
gravity is 9.863-9.798 m/s2.
Too bad if you didn't say it, for those are the measured values.
That's r^2 and it's called acceleration. You don't seem to know that m/s^2 is acceleration!!! Since when does Force=acceleration?
In all my posts I state very clearly that in a classical model the force of
gravity is modelled with GM/r.
And I already responded to your point on geoids that yes if you want to split hairs the geoid surface varies from r by up to 200 meters. Which is why very accurate measurements of clock rates will show constant rates at the surface of the geoid only. And not to r. But that's still consistent with a classical model as much as with GR.
So you have nothing to say,
beyond agreeing that general relativity gives the right answer,

I suppose there isn’t much more to say to a person such as yourself who thinks that force=acceleration.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From J. J. Lodder@21:1/5 to Ross Finlayson on Sun Nov 12 21:08:53 2023
Ross Finlayson <ross.a.finlayson@gmail.com> wrote:

On Sunday, November 12, 2023 at 7:08:45?AM UTC-8, J. J. Lodder wrote:
Lou <noeltu...@live.co.uk> wrote:

On Sunday, 12 November 2023 at 13:30:18 UTC, J. J. Lodder wrote:
Lou wrote:

On Saturday, 11 November 2023 at 20:47:35 UTC, J. J. Lodder wrote:
Lou wrote:

On Friday, 10 November 2023 at 20:00:54 UTC, J. J. Lodder wrote:
[summary: gravity and clock rates for misled kiddies]

General relativity predicts that all freely falling clocks
will run at their own inherent rate. [by postulate]
It also predicts that clocks at different places,
and with different velocities will be seen to run at different rates, -when compared with each other-.
It also predicts that accelerations do not affect clock rates, so the results can be extended to non-inertial clocks,
such as clocks at rest at different altitudes on Earth.

According to general relativity all clock effects are purely kinematic, so derivable from the metric tensor.
Doing the sums for weak fields results in velocity effects being given by Lorentz factors, and 'gravitational' effects being given by the variations in Newtonian potential.
So far, so good, and in agreement with experimental results.

Now there are people such as for example 'Lou' in this forum, who cannot or will not accept or understand this.
They hold that obverved clock effects must be due to 'gravity' affecting the workings of the clock, somehow.
In other words, they ascribe the observed clock effects
to physical causes, 'gravity' affecting the workings of clocks, rather than to intrinsic space-time effects.

Fortunately it is easy to settle the point by experiment.
GR predicts that all clocks on the rotating geoid on Earth
must run at the same rate, when compared with each other. Experiment bears this out, to accuracies approaching 10^-15. This is of immense practical importance,
because it is the basis for realising the SI second.
(on which -all- physical measurement depends nowadays)

OTOH the force of gravity, as measured by 'small' g,
the acceleration of gravity, varies markedly over the geoid. (by about 0.5%, between the poles and the equator)

If (the force of) 'gravity' influenced the rate of the clocks there should be an effect of geographical latitude
on the rate of clocks.
This is not observed to be the case, so this idea stands falsified.

The idea that 'gravity' affects the rate at which clocks run
is a misconception without basis in observed fact,

A desperately misguided post from JJ.
You did not really read any of my posts. If you did...Then you deliberately ignored the fact that I *very* explicitly stated that
in a classical model "little g" is acceleration only. Not force. And you ignored the fact that I very clearly stated that force
on the atoms at different altitudes in a classical model should be
calculated using r.
Indeed, there is little point, because you go on harping about
You can go on obfuscating because you limit yourself
to situations with spherical symmetry.

So to see the errors of your ways you should consider situations where spherical symmetry does not hold.
Then the surfaces of constant potential do not coincide
with surfaces of constant acceleration, or constant r.
Not the m/s^2 acceleration of r^2 in "little g".
Seeing as everyone except a profound idiot would think acceleration = force.
And If you actually read my posts rather than thump your bible, you would realise that I also said that force is what Laplace called gravitational potential. And what Newton referred to as a scalar field.
And what Einstein used to calculate his GR clock rate effects. (Notice the r of Laplace's gravitational potential and
Newton's scalar field is also the r used in GR. Not r^2 of
little g.)
So if you claim that experiment shows no change of clock
rates at different sea level latitudes. Then you have not
only confirmed the predictions of GR.. You have also confirmed the predictions of classical theory. Seeing as they both use r to accurately calculate tick rates at different altitudes.
Experiment shows that clocks on the geoid run at constant rates
wrt each other. Note that the geoid is not a surface of constant r, nor a surface of constant g,

A straw man argument if ever you make.
Yes I've looked at your 'geoid' now and how it varies slightly by about 200m relative to the reference geoid and how technically the r distance doesn't exactly follow the geoid surface. That makes sense. Splitting hairs though on your part to pretend somehow this rules out a classical model which uses r. I notice you didn't actually specify why it would. In fact it doesn't rule out in any way a classical model
any more than it would rule out GR.
Because in a classical calculation if one needs to assume
*exactly* the *total* mass M of the earth at r, then yes to be *absolutely* accurate the geoid surface has to be used. Not the actual distance r.
But the same applies to GR. And the fact remains that generally,
the force of gravity in a classical model follows r not r^2.
(And to please the pedant JJ,... with ever so small meter length fluctuations in the exact distance of r to also be taken into account)
So you missed all points, again. I'll simplify.
The geoid surface is by definition an equipotential surface of the Newtonian potential.
So it coincides (almost) with the mean sea level.
The geoid is (to a very good approximation) an ellipsoid of revolution. The small differences between geoid and ellipsoid
(due to slightly irregular mass distributions inside the Earth)
don't matter for what follows.

Now, on the geoid, and at the poles, we have: r < average g > average, potential = constant
On the geoid, at mid-latitudes we have r = average, g = average, potential = same constant
On the geoid, at the equator, we have r > average, g < average, potential = still the same constant, by definition of the geoid.

The differences are huge, r = 6357-6378 km, g = 9.863-9.798 m/s2, compared to clock stabilities of 10^-15.

What is your prediction for the rates of clocks in those three places? No verbiage, just say faster, slower, or the same,
and if you can by how much,

If you tried reading my posts you wouldn't be pretending I said the
force of gravity is 9.863-9.798 m/s2.
Too bad if you didn't say it, for those are the measured values.
That's r^2 and it's called acceleration. You don't seem to know that m/s^2
is acceleration!!! Since when does Force=acceleration?
In all my posts I state very clearly that in a classical model the
force of gravity is modelled with GM/r.
And I already responded to your point on geoids that yes if you want
to split hairs the geoid surface varies from r by up to 200 meters.
Which is why very accurate measurements of clock rates will show
constant rates at the surface of the geoid only. And not to r. But
that's still consistent with a classical model as much as with GR.
So you have nothing to say,
beyond agreeing that general relativity gives the right answer,

Jan

There is only one,

Jan

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From J. J. Lodder@21:1/5 to Lou on Sun Nov 12 21:08:54 2023
Lou <noelturntive@live.co.uk> wrote:

On Sunday, 12 November 2023 at 15:08:45 UTC, J. J. Lodder wrote:
Lou wrote:

On Sunday, 12 November 2023 at 13:30:18 UTC, J. J. Lodder wrote:
Lou wrote:

On Saturday, 11 November 2023 at 20:47:35 UTC, J. J. Lodder wrote:
Lou wrote:

On Friday, 10 November 2023 at 20:00:54 UTC, J. J. Lodder wrote:
[summary: gravity and clock rates for misled kiddies]

General relativity predicts that all freely falling clocks
will run at their own inherent rate. [by postulate]
It also predicts that clocks at different places,
and with different velocities will be seen to run at different rates, -when compared with each other-.
It also predicts that accelerations do not affect clock rates, so the results can be extended to non-inertial clocks,
such as clocks at rest at different altitudes on Earth.

According to general relativity all clock effects are purely kinematic, so derivable from the metric tensor.
Doing the sums for weak fields results in velocity effects being given by Lorentz factors, and 'gravitational' effects being given by the variations in Newtonian potential.
So far, so good, and in agreement with experimental results.

Now there are people such as for example 'Lou' in this forum, who cannot or will not accept or understand this.
They hold that obverved clock effects must be due to 'gravity' affecting the workings of the clock, somehow.
In other words, they ascribe the observed clock effects
to physical causes, 'gravity' affecting the workings of clocks, rather than to intrinsic space-time effects.

Fortunately it is easy to settle the point by experiment.
GR predicts that all clocks on the rotating geoid on Earth
must run at the same rate, when compared with each other. Experiment bears this out, to accuracies approaching 10^-15. This is of immense practical importance,
because it is the basis for realising the SI second.
(on which -all- physical measurement depends nowadays)

OTOH the force of gravity, as measured by 'small' g,
the acceleration of gravity, varies markedly over the geoid. (by about 0.5%, between the poles and the equator)

If (the force of) 'gravity' influenced the rate of the clocks there should be an effect of geographical latitude
on the rate of clocks.
The idea that 'gravity' affects the rate at which clocks run
is a misconception without basis in observed fact,

A desperately misguided post from JJ.
You did not really read any of my posts. If you did...Then you deliberately ignored the fact that I *very* explicitly stated that
in a classical model "little g" is acceleration only. Not force. And you ignored the fact that I very clearly stated that force
on the atoms at different altitudes in a classical model should be
calculated using r.
Indeed, there is little point, because you go on harping about
You can go on obfuscating because you limit yourself
to situations with spherical symmetry.

So to see the errors of your ways you should consider situations where spherical symmetry does not hold.
Then the surfaces of constant potential do not coincide
with surfaces of constant acceleration, or constant r.
Not the m/s^2 acceleration of r^2 in "little g".
Seeing as everyone except a profound idiot would think acceleration = force.
And If you actually read my posts rather than thump your bible, you would realise that I also said that force is what Laplace called gravitational potential. And what Newton referred to as a scalar field.
And what Einstein used to calculate his GR clock rate effects. (Notice the r of Laplace's gravitational potential and
Newton's scalar field is also the r used in GR. Not r^2 of
little g.)
So if you claim that experiment shows no change of clock
rates at different sea level latitudes. Then you have not
only confirmed the predictions of GR.. You have also confirmed the predictions of classical theory. Seeing as they both use r to accurately calculate tick rates at different altitudes.
Experiment shows that clocks on the geoid run at constant rates
wrt each other. Note that the geoid is not a surface of constant r, nor a surface of constant g,

A straw man argument if ever you make.
Yes I've looked at your 'geoid' now and how it varies slightly by about 200m relative to the reference geoid and how technically the r distance doesn't exactly follow the geoid surface. That makes sense. Splitting hairs though on your part to pretend somehow this rules out a classical model which uses r. I notice you didn't actually specify why it would. In fact it doesn't rule out in any way a classical model
any more than it would rule out GR.
Because in a classical calculation if one needs to assume
*exactly* the *total* mass M of the earth at r, then yes to be *absolutely* accurate the geoid surface has to be used. Not the actual distance r.
But the same applies to GR. And the fact remains that generally,
the force of gravity in a classical model follows r not r^2.
(And to please the pedant JJ,... with ever so small meter length fluctuations in the exact distance of r to also be taken into account)
So you missed all points, again. I'll simplify.
The geoid surface is by definition an equipotential surface of the Newtonian potential.
So it coincides (almost) with the mean sea level.
The geoid is (to a very good approximation) an ellipsoid of revolution. The small differences between geoid and ellipsoid
(due to slightly irregular mass distributions inside the Earth)
don't matter for what follows.

Now, on the geoid, and at the poles, we have: r < average g > average, potential = constant
On the geoid, at mid-latitudes we have r = average, g = average, potential = same constant
On the geoid, at the equator, we have r > average, g < average, potential = still the same constant, by definition of the geoid.

The differences are huge, r = 6357-6378 km, g = 9.863-9.798 m/s2, compared to clock stabilities of 10^-15.

What is your prediction for the rates of clocks in those three places? No verbiage, just say faster, slower, or the same,
and if you can by how much,

If you tried reading my posts you wouldn't be pretending I said the
force of gravity is 9.863-9.798 m/s2.
Too bad if you didn't say it, for those are the measured values.
That's r^2 and it's called acceleration. You don't seem to know that m/s^2
is acceleration!!! Since when does Force=acceleration?
And I already responded to your point on geoids that yes if you want to split hairs the geoid surface varies from r by up to 200 meters. Which is why very accurate measurements of clock rates will show constant rates at the surface of the geoid only. And not to r. But that's still consistent with a classical model as much as with GR.
So you have nothing to say,
beyond agreeing that general relativity gives the right answer,

I suppose there isn't much more to say to a person such as yourself who thinks that force=acceleration.

OK, so you give up,

Jan

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Ross Finlayson@21:1/5 to J. J. Lodder on Mon Nov 13 08:51:40 2023
On Sunday, November 12, 2023 at 12:08:57 PM UTC-8, J. J. Lodder wrote:
Ross Finlayson <ross.a.f...@gmail.com> wrote:
On Sunday, November 12, 2023 at 7:08:45?AM UTC-8, J. J. Lodder wrote:
Lou <noeltu...@live.co.uk> wrote:

On Sunday, 12 November 2023 at 13:30:18 UTC, J. J. Lodder wrote:
Lou wrote:

On Saturday, 11 November 2023 at 20:47:35 UTC, J. J. Lodder wrote:
Lou wrote:

On Friday, 10 November 2023 at 20:00:54 UTC, J. J. Lodder wrote:
[summary: gravity and clock rates for misled kiddies]

General relativity predicts that all freely falling clocks will run at their own inherent rate. [by postulate]
It also predicts that clocks at different places,
and with different velocities will be seen to run at different
rates, -when compared with each other-.
It also predicts that accelerations do not affect clock rates,
so the results can be extended to non-inertial clocks,
such as clocks at rest at different altitudes on Earth.

According to general relativity all clock effects are purely kinematic, so derivable from the metric tensor.
Doing the sums for weak fields results in velocity effects being given by Lorentz factors, and 'gravitational' effects being given by the variations in Newtonian potential.
So far, so good, and in agreement with experimental results.

Now there are people such as for example 'Lou' in this forum,
who cannot or will not accept or understand this.
They hold that obverved clock effects must be due to 'gravity'
affecting the workings of the clock, somehow.
In other words, they ascribe the observed clock effects
to physical causes, 'gravity' affecting the workings of clocks,
rather than to intrinsic space-time effects.

Fortunately it is easy to settle the point by experiment.
GR predicts that all clocks on the rotating geoid on Earth must run at the same rate, when compared with each other. Experiment bears this out, to accuracies approaching 10^-15. This is of immense practical importance,
because it is the basis for realising the SI second.
(on which -all- physical measurement depends nowadays)

OTOH the force of gravity, as measured by 'small' g,
the acceleration of gravity, varies markedly over the geoid. (by about 0.5%, between the poles and the equator)

If (the force of) 'gravity' influenced the rate of the clocks
there should be an effect of geographical latitude
on the rate of clocks.
This is not observed to be the case, so this idea stands falsified.

The idea that 'gravity' affects the rate at which clocks run is a misconception without basis in observed fact,

A desperately misguided post from JJ.
You did not really read any of my posts. If you did...Then you deliberately ignored the fact that I *very* explicitly stated that
in a classical model "little g" is acceleration only. Not force.
And you ignored the fact that I very clearly stated that force on the atoms at different altitudes in a classical model should be
calculated using r.
Indeed, there is little point, because you go on harping about your r, and you are ignoring all sound advice by others.
You can go on obfuscating because you limit yourself
to situations with spherical symmetry.

So to see the errors of your ways you should consider situations where spherical symmetry does not hold.
Then the surfaces of constant potential do not coincide
with surfaces of constant acceleration, or constant r.
Not the m/s^2 acceleration of r^2 in "little g".
Seeing as everyone except a profound idiot would think acceleration = force.
And If you actually read my posts rather than thump your bible,
you would realise that I also said that force is what Laplace called gravitational potential. And what Newton referred to as a
scalar field.
And what Einstein used to calculate his GR clock rate effects. (Notice the r of Laplace's gravitational potential and Newton's scalar field is also the r used in GR. Not r^2 of little g.)
So if you claim that experiment shows no change of clock
rates at different sea level latitudes. Then you have not
only confirmed the predictions of GR.. You have also confirmed the predictions of classical theory. Seeing as they both use r to
accurately calculate tick rates at different altitudes.
Experiment shows that clocks on the geoid run at constant rates wrt each other. Note that the geoid is not a surface of constant r,
nor a surface of constant g,

A straw man argument if ever you make.
Yes I've looked at your 'geoid' now and how it varies slightly by about 200m relative to the reference geoid and how technically the r
distance doesn't exactly follow the geoid surface. That makes sense.
Splitting hairs though on your part to pretend somehow this rules out
a classical model which uses r. I notice you didn't actually specify
why it would. In fact it doesn't rule out in any way a classical model
any more than it would rule out GR.
Because in a classical calculation if one needs to assume *exactly* the *total* mass M of the earth at r, then yes to be *absolutely* accurate the geoid surface has to be used. Not the actual distance r.
But the same applies to GR. And the fact remains that generally, the force of gravity in a classical model follows r not r^2.
(And to please the pedant JJ,... with ever so small meter length fluctuations in the exact distance of r to also be taken into account)
So you missed all points, again. I'll simplify.
The geoid surface is by definition an equipotential surface of the Newtonian potential.
So it coincides (almost) with the mean sea level.
The geoid is (to a very good approximation) an ellipsoid of revolution.
The small differences between geoid and ellipsoid
(due to slightly irregular mass distributions inside the Earth) don't matter for what follows.

Now, on the geoid, and at the poles, we have: r < average g > average,
potential = constant
On the geoid, at mid-latitudes we have r = average, g = average, potential = same constant
On the geoid, at the equator, we have r > average, g < average, potential = still the same constant, by definition of the geoid.

The differences are huge, r = 6357-6378 km, g = 9.863-9.798 m/s2, compared to clock stabilities of 10^-15.

What is your prediction for the rates of clocks in those three places?
No verbiage, just say faster, slower, or the same,
and if you can by how much,

If you tried reading my posts you wouldn't be pretending I said the force of gravity is 9.863-9.798 m/s2.
Too bad if you didn't say it, for those are the measured values.
That's r^2 and it's called acceleration. You don't seem to know that m/s^2
is acceleration!!! Since when does Force=acceleration?
In all my posts I state very clearly that in a classical model the force of gravity is modelled with GM/r.
And I already responded to your point on geoids that yes if you want to split hairs the geoid surface varies from r by up to 200 meters. Which is why very accurate measurements of clock rates will show constant rates at the surface of the geoid only. And not to r. But that's still consistent with a classical model as much as with GR.
So you have nothing to say,
beyond agreeing that general relativity gives the right answer,

Jan

There is only one,

Jan

There's only one metric? How about when there are three?
What about when normed-spaces aren't metric spaces and vice-versa?

If boost addition is part of GR, then it has to be setup to be
that the geodesy is always instantaneously evaluated everywhere.

That "really, force is always virtual", or, "there are no truly inelastic collisions", that, "force is a dimensionless quantity just as a matter
of projection a linear, impulse, classical quantity, in a fixed vector field", these are some of the things Einstein considers in his attack on Newton's laws,
as what follows from "Out of my later years", Einstein's opinion about
not just relativity, but also Newton's "classical".

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From J. J. Lodder@21:1/5 to Ross Finlayson on Mon Nov 13 21:34:30 2023
Ross Finlayson <ross.a.finlayson@gmail.com> wrote:

On Sunday, November 12, 2023 at 12:08:57?PM UTC-8, J. J. Lodder wrote:
Ross Finlayson <ross.a.f...@gmail.com> wrote:
On Sunday, November 12, 2023 at 7:08:45?AM UTC-8, J. J. Lodder wrote:
Lou <noeltu...@live.co.uk> wrote:
On Sunday, 12 November 2023 at 13:30:18 UTC, J. J. Lodder wrote:
[-]
Now, on the geoid, and at the poles, we have: r < average g > average, potential = constant
On the geoid, at mid-latitudes we have r = average, g = average, potential = same constant
On the geoid, at the equator, we have r > average, g < average, potential = still the same constant, by definition of the geoid.

The differences are huge, r = 6357-6378 km, g = 9.863-9.798 m/s2, compared to clock stabilities of 10^-15.

What is your prediction for the rates of clocks in those three places? No verbiage, just say faster, slower, or the same, and
if you can by how much,

If you tried reading my posts you wouldn't be pretending I said
the force of gravity is 9.863-9.798 m/s2.
Too bad if you didn't say it, for those are the measured values.
That's r^2 and it's called acceleration. You don't seem to know
that m/s^2 is acceleration!!! Since when does Force=acceleration?
In all my posts I state very clearly that in a classical model the force of gravity is modelled with GM/r.
And I already responded to your point on geoids that yes if you
want to split hairs the geoid surface varies from r by up to 200 meters. Which is why very accurate measurements of clock rates
will show constant rates at the surface of the geoid only. And not
to r. But that's still consistent with a classical model as much
as with GR.
So you have nothing to say,
beyond agreeing that general relativity gives the right answer,

Jan

There is only one,

Jan

There's only one metric? How about when there are three?

One geoid, and one right answer, not one metric.

What about when normed-spaces aren't metric spaces and vice-versa?

If boost addition is part of GR, then it has to be setup to be
that the geodesy is always instantaneously evaluated everywhere.

No prolem, the whole idea applies in the Newtonian approximation
to GR anyway. (so v^2 << c^2)

Jan

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Lou@21:1/5 to J. J. Lodder on Mon Nov 13 13:07:14 2023
On Sunday, 12 November 2023 at 20:08:58 UTC, J. J. Lodder wrote:
Lou wrote:

On Sunday, 12 November 2023 at 15:08:45 UTC, J. J. Lodder wrote:
Lou wrote:

On Sunday, 12 November 2023 at 13:30:18 UTC, J. J. Lodder wrote:
Lou wrote:

On Saturday, 11 November 2023 at 20:47:35 UTC, J. J. Lodder wrote:
Lou wrote:

On Friday, 10 November 2023 at 20:00:54 UTC, J. J. Lodder wrote:
[summary: gravity and clock rates for misled kiddies]

General relativity predicts that all freely falling clocks will run at their own inherent rate. [by postulate]
It also predicts that clocks at different places,
and with different velocities will be seen to run at different
rates, -when compared with each other-.
It also predicts that accelerations do not affect clock rates,
so the results can be extended to non-inertial clocks,
such as clocks at rest at different altitudes on Earth.

According to general relativity all clock effects are purely kinematic, so derivable from the metric tensor.
Doing the sums for weak fields results in velocity effects being given by Lorentz factors, and 'gravitational' effects being given by the variations in Newtonian potential.
So far, so good, and in agreement with experimental results.

Now there are people such as for example 'Lou' in this forum,
who cannot or will not accept or understand this.
They hold that obverved clock effects must be due to 'gravity'
affecting the workings of the clock, somehow.
In other words, they ascribe the observed clock effects
to physical causes, 'gravity' affecting the workings of clocks,
rather than to intrinsic space-time effects.

Fortunately it is easy to settle the point by experiment.
GR predicts that all clocks on the rotating geoid on Earth must run at the same rate, when compared with each other. Experiment bears this out, to accuracies approaching 10^-15. This is of immense practical importance,
because it is the basis for realising the SI second.
(on which -all- physical measurement depends nowadays)

OTOH the force of gravity, as measured by 'small' g,
the acceleration of gravity, varies markedly over the geoid. (by about 0.5%, between the poles and the equator)

If (the force of) 'gravity' influenced the rate of the clocks
there should be an effect of geographical latitude
on the rate of clocks.
The idea that 'gravity' affects the rate at which clocks run is a misconception without basis in observed fact,

A desperately misguided post from JJ.
You did not really read any of my posts. If you did...Then you deliberately ignored the fact that I *very* explicitly stated that
in a classical model "little g" is acceleration only. Not force.
And you ignored the fact that I very clearly stated that force on the atoms at different altitudes in a classical model should be
calculated using r.
Indeed, there is little point, because you go on harping about your r, and you are ignoring all sound advice by others.
You can go on obfuscating because you limit yourself
to situations with spherical symmetry.

So to see the errors of your ways you should consider situations where spherical symmetry does not hold.
Then the surfaces of constant potential do not coincide
with surfaces of constant acceleration, or constant r.
Not the m/s^2 acceleration of r^2 in "little g".
Seeing as everyone except a profound idiot would think acceleration = force.
And If you actually read my posts rather than thump your bible,
you would realise that I also said that force is what Laplace called gravitational potential. And what Newton referred to as a
scalar field.
And what Einstein used to calculate his GR clock rate effects. (Notice the r of Laplace's gravitational potential and Newton's scalar field is also the r used in GR. Not r^2 of little g.)
So if you claim that experiment shows no change of clock
rates at different sea level latitudes. Then you have not
only confirmed the predictions of GR.. You have also confirmed the predictions of classical theory. Seeing as they both use r to
accurately calculate tick rates at different altitudes.
Experiment shows that clocks on the geoid run at constant rates wrt each other. Note that the geoid is not a surface of constant r,
nor a surface of constant g,

A straw man argument if ever you make.
Yes I've looked at your 'geoid' now and how it varies slightly by about 200m relative to the reference geoid and how technically the r
distance doesn't exactly follow the geoid surface. That makes sense.
Splitting hairs though on your part to pretend somehow this rules out
a classical model which uses r. I notice you didn't actually specify
why it would. In fact it doesn't rule out in any way a classical model
any more than it would rule out GR.
Because in a classical calculation if one needs to assume *exactly* the *total* mass M of the earth at r, then yes to be *absolutely* accurate the geoid surface has to be used. Not the actual distance r.
But the same applies to GR. And the fact remains that generally, the force of gravity in a classical model follows r not r^2.
(And to please the pedant JJ,... with ever so small meter length fluctuations in the exact distance of r to also be taken into account)
So you missed all points, again. I'll simplify.
The geoid surface is by definition an equipotential surface of the Newtonian potential.
So it coincides (almost) with the mean sea level.
The geoid is (to a very good approximation) an ellipsoid of revolution.
The small differences between geoid and ellipsoid
(due to slightly irregular mass distributions inside the Earth) don't matter for what follows.

Now, on the geoid, and at the poles, we have: r < average g > average,
potential = constant
On the geoid, at mid-latitudes we have r = average, g = average, potential = same constant
On the geoid, at the equator, we have r > average, g < average, potential = still the same constant, by definition of the geoid.

The differences are huge, r = 6357-6378 km, g = 9.863-9.798 m/s2, compared to clock stabilities of 10^-15.

What is your prediction for the rates of clocks in those three places?
No verbiage, just say faster, slower, or the same,
and if you can by how much,

If you tried reading my posts you wouldn't be pretending I said the force of gravity is 9.863-9.798 m/s2.
Too bad if you didn't say it, for those are the measured values.
That's r^2 and it's called acceleration. You don't seem to know that m/s^2
is acceleration!!! Since when does Force=acceleration?
And I already responded to your point on geoids that yes if you want to
split hairs the geoid surface varies from r by up to 200 meters. Which is
why very accurate measurements of clock rates will show constant rates at
the surface of the geoid only. And not to r. But that's still consistent
with a classical model as much as with GR.
So you have nothing to say,
beyond agreeing that general relativity gives the right answer,

I suppose there isn't much more to say to a person such as yourself who thinks that force=acceleration.
OK, so you give up,

I give up trying to make you understand that a true classical model
does not use acceleration to describe the force of gravity. Only
relativists (or idiots) think in a classical model acceleration=force.

They do it to make sure a classical model can’t correctly predict the
change of resonant frequencies of atoms at different potentials.

Force of gravity was called potential by Laplace. Force of gravity was called Scalar field by Newton and is also used by Relativity to model the effects of gravity at different altitudes.
Which means that for a classical model which also correctly uses potential as force of gravity at different altitudes then the answer to your silly question is...
constant for all 3. As with GR.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Tom Roberts@21:1/5 to Lou on Mon Nov 13 16:58:40 2023
On 11/13/23 3:07 PM, Lou wrote:
a true classical model does not use acceleration to describe the
force of gravity.

Well, there is no such thing as a "true" model in physics -- physical
models are valid or invalid, but we humans can never know whether they
are "true". But we can know when they are false, and the "classical
model" known as Newtonian mechanics (NM) is known to be false (but it is
often useful as an approximation).

As far as the relationship between acceleration and force is concerned,
here in the context of NM, what is used is Newton's second law:
F = m a
where F is the total force on an object, m is its mass, and a is its acceleration. This is implicitly relative to some inertial frame; F, m,
and a are all invariant under Galilean transforms (change of coordinates
to a different inertial frame).

Only relativists (or idiots) think in a classical model
acceleration=force.

NONSENSE! NOBODY thinks that. YOU are confused. The only "classical
model" here is Newtonian mechanics, and in NM the second law applies:
F = m a

They do it to make sure a classical model can’t correctly predict
the change of resonant frequencies of atoms at different potentials.

More nonsense. In Newtonian mechanics, time is universal and NM predicts
zero "time dilation" under any and all circumstances. NM is DIFFERENT
from relativity, and wrong.

Force of gravity was called potential by Laplace.[...]

Still more nonsense. The force of gravity is minus the gradient of the gravitational potential. They have NEVER been "equal" as you suppose.

You REALLY need to learn basic physics. Your GUESSES AND FANTASIES are
wrong.

Tom Roberts

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Ross Finlayson@21:1/5 to J. J. Lodder on Mon Nov 13 16:21:59 2023
On Monday, November 13, 2023 at 12:34:35 PM UTC-8, J. J. Lodder wrote:
Ross Finlayson <ross.a.f...@gmail.com> wrote:
On Sunday, November 12, 2023 at 12:08:57?PM UTC-8, J. J. Lodder wrote:
Ross Finlayson <ross.a.f...@gmail.com> wrote:
On Sunday, November 12, 2023 at 7:08:45?AM UTC-8, J. J. Lodder wrote:
Lou <noeltu...@live.co.uk> wrote:
On Sunday, 12 November 2023 at 13:30:18 UTC, J. J. Lodder wrote:
[-]
Now, on the geoid, and at the poles, we have: r < average g > average, potential = constant
On the geoid, at mid-latitudes we have r = average, g = average, potential = same constant
On the geoid, at the equator, we have r > average, g < average, potential = still the same constant, by definition of the geoid.

The differences are huge, r = 6357-6378 km, g = 9.863-9.798 m/s2,
compared to clock stabilities of 10^-15.

What is your prediction for the rates of clocks in those three places? No verbiage, just say faster, slower, or the same, and if you can by how much,

If you tried reading my posts you wouldn't be pretending I said the force of gravity is 9.863-9.798 m/s2.
Too bad if you didn't say it, for those are the measured values.
That's r^2 and it's called acceleration. You don't seem to know that m/s^2 is acceleration!!! Since when does Force=acceleration? In all my posts I state very clearly that in a classical model the force of gravity is modelled with GM/r.
And I already responded to your point on geoids that yes if you want to split hairs the geoid surface varies from r by up to 200 meters. Which is why very accurate measurements of clock rates will show constant rates at the surface of the geoid only. And not to r. But that's still consistent with a classical model as much as with GR.
So you have nothing to say,
beyond agreeing that general relativity gives the right answer,

Jan

There is only one,

Jan

There's only one metric? How about when there are three?
One geoid, and one right answer, not one metric.
What about when normed-spaces aren't metric spaces and vice-versa?

If boost addition is part of GR, then it has to be setup to be
that the geodesy is always instantaneously evaluated everywhere.
No prolem, the whole idea applies in the Newtonian approximation
to GR anyway. (so v^2 << c^2)

Jan

A spherical or elliptical geoid?

When it comes to "algebraic geometry, I rather prefer Lefschetz and Picard, and maybe Kodaira,
"algebraic GEOMETRY", to for example Bourbaki or Groethendieck, "ALGEBRAIC geometry".

Then about the geoid, it seemed a good idea to read Kepler's treatise, it's quite good,
through a lens of about a few hundred years. (Re "spherical or elliptical or spherical, ...".)

About the higher-order terms and multipole moment and so on, here in terms of metrization
as it were, such terms live in the "highly-non-linear" but just as well "circuitous", representing
all the dimensioned quantities that result in a standing projection, "dimensionless quantities",
with respect to all higher-order moments, and about here particularly "any acceleration
is all the higher orders of acceleration", and "there are no perfectly inelastic collisions".
That "force is virtual" because for example "it's the potential fields that are real, the
classical is just the sum of all the potentials", makes for both its success in the linear,
and, getting into the mechanics separating linear and rotational, and kinetics and kinematics,
about why energy comes in many kinds, and, momentum these days is in some place,
"pseudo-momentum", with respect to what gets "conserved".

These days Hamiltonians are more often Hamilton-Jacobi, and
usually enough "Lagrangians", if not to be throwing shade on
good old bra-ket and boost addition.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Maciej Wozniak@21:1/5 to Tom Roberts on Tue Nov 14 00:30:11 2023
On Monday, 13 November 2023 at 23:58:54 UTC+1, Tom Roberts wrote:
On 11/13/23 3:07 PM, Lou wrote:
a true classical model does not use acceleration to describe the
force of gravity.
Well, there is no such thing as a "true" model in physics -- physical
models are valid or invalid, but we humans can never know whether they
are "true".

While there is, indeed, nothing "true" in your moronic
mumble - you have no clue what is this "truth" you're

NM is DIFFERENT
from relativity, and wrong.

Oppositely - your Shit is different from NM and wrong.
The mumble of your idiot guru wasn't even consistent.

Still more nonsense. The force of gravity is minus the gradient of the gravitational potential. They have NEVER been "equal" as you suppose.

The force of gravity, as your idiot guru has
announced - doesn't exist. You should really
learn the basics of your moronic religion.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Lou@21:1/5 to Tom Roberts on Tue Nov 14 02:06:23 2023
On Monday, 13 November 2023 at 22:58:54 UTC, Tom Roberts wrote:
On 11/13/23 3:07 PM, Lou wrote:
a true classical model does not use acceleration to describe the
force of gravity.
Well, there is no such thing as a "true" model in physics -- physical
models are valid or invalid, but we humans can never know whether they
are "true". But we can know when they are false, and the "classical
model" known as Newtonian mechanics (NM) is known to be false (but it is often useful as an approximation).

Exactly....it’s false. Gravity force isn’t acceleration based on r^2.
Thats what I have been trying to get you lot to understand
this whole time. Einstein realised this and used potential. Laplace
realised that gravity force was potential. Newton knew this and
called gravity force a scalar field. And others like Levy also
proposed this
The ridiculous part of your argument is that although do you admit
r^2 doesn’t work ...you cannot bear to have anyone point out
that to make a classical theory work...one must use r of potential.
Despite being in the hypocritical position of accepting that
GR theory can ditch r^2…and use r of potential.

As far as the relationship between acceleration and force is concerned,
here in the context of NM, what is used is Newton's second law:
F = m a
where F is the total force on an object, m is its mass, and a is its acceleration. This is implicitly relative to some inertial frame; F, m,
and a are all invariant under Galilean transforms (change of coordinates
to a different inertial frame).

Word salad. R^2 isn’t F=ma. It is a ridiculous F=a. You know it is.
Dont pretend it isn’t.

Only relativists (or idiots) think in a classical model acceleration=force.
NONSENSE! NOBODY thinks that. YOU are confused. The only "classical
model" here is Newtonian mechanics, and in NM the second law applies:
F = m a

No...YOU think acceleration=force. Not me.,You just spent your whole post trying
to pretend in a classical model that m/s^2= force.
It isn’t . ITS CALLED ACCELERATION. Einstein realised this.

They do it to make sure a classical model can’t correctly predict
the change of resonant frequencies of atoms at different potentials.
More nonsense. In Newtonian mechanics, time is universal and NM predicts zero "time dilation" under any and all circumstances. NM is DIFFERENT
from relativity, and wrong.

The only nonsense is your silly claim that I said a classical model
predicts time dilation. That’s GR.
A classical model says no time dilation is occuring. What you
are seeing is a harmonic oscillator (c-133) changing its resonant
frequency due to an external force of gravity.
And we have known harmonic oscillators do change frequencies
under external force for longer than GR has been around.

Force of gravity was called potential by Laplace.[...]

Still more nonsense. The force of gravity is minus the gradient of the gravitational potential. They have NEVER been "equal" as you suppose.

The force of gravity increases the closer you get to the mass.
Proportional to r. That’s what potential really is.
You don’t understand basic physics. The mathematical gimmick
you refer to is based on the flawed interpretation of the observation
that it needs more energy to fly higher.
Of course an idiot would think that if it takes more energy to go
from surface to 3r than from surface to 2r then they would in their
stupidity think that the force of gravity also increases proportional to r.

You REALLY need to learn basic physics. Your GUESSES AND FANTASIES are wrong.

If I’m wrong to model the effects of gravity with r...then why is
OK for Albert to use r of potential to model the effects of gravity?
You don’t even realise what a hypocrite you are.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From J. J. Lodder@21:1/5 to Ross Finlayson on Tue Nov 14 12:50:49 2023
Ross Finlayson <ross.a.finlayson@gmail.com> wrote:

On Monday, November 13, 2023 at 12:34:35?PM UTC-8, J. J. Lodder wrote:
Ross Finlayson <ross.a.f...@gmail.com> wrote:
On Sunday, November 12, 2023 at 12:08:57?PM UTC-8, J. J. Lodder wrote:
Ross Finlayson <ross.a.f...@gmail.com> wrote:
On Sunday, November 12, 2023 at 7:08:45?AM UTC-8, J. J. Lodder wrote:
Lou <noeltu...@live.co.uk> wrote:
On Sunday, 12 November 2023 at 13:30:18 UTC, J. J. Lodder wrote:
[-]
Now, on the geoid, and at the poles, we have: r < average g > average, potential = constant
On the geoid, at mid-latitudes we have r = average, g = average,
potential = same constant
On the geoid, at the equator, we have r > average, g < average, potential = still the same constant, by definition of the geoid.

The differences are huge, r = 6357-6378 km,
g = 9.863-9.798 m/s2, compared to clock stabilities of 10^-15.

What is your prediction for the rates of clocks in those three places? No verbiage, just say faster, slower, or the same, and if you can by how much,

If you tried reading my posts you wouldn't be pretending I said the force of gravity is 9.863-9.798 m/s2.
Too bad if you didn't say it, for those are the measured values.
That's r^2 and it's called acceleration. You don't seem to know that m/s^2 is acceleration!!! Since when does Force=acceleration? In all my posts I state very clearly that in a classical model the
force of gravity is modelled with GM/r.
And I already responded to your point on geoids that yes if you want to split hairs the geoid surface varies from r by up to 200 meters. Which is why very accurate measurements of clock rates will show constant rates at the surface of the geoid only. And not
to r. But that's still consistent with a classical model as much as with GR.
So you have nothing to say,
beyond agreeing that general relativity gives the right answer,

Jan

There is only one,

Jan

There's only one metric? How about when there are three?
One geoid, and one right answer, not one metric.
What about when normed-spaces aren't metric spaces and vice-versa?

If boost addition is part of GR, then it has to be setup to be
that the geodesy is always instantaneously evaluated everywhere.
No prolem, the whole idea applies in the Newtonian approximation
to GR anyway. (so v^2 << c^2)

Jan

A spherical or elliptical geoid?

Whatever it is. As so often with those things,
the reality is being changed by definition.
Instead of that pesky 'mean sea level',
which is hard to determine accurately, and unstable,
the geoid is being redefined as that surface on which an atomic clock
runs at TT, so in practice at TAI time.
The name for it is the 'chronometric geoid'.
The actual shape is quite irregular.

Jan

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Tom Roberts@21:1/5 to Lou on Tue Nov 14 23:00:03 2023
On 11/14/23 4:06 AM, Lou wrote:
On Monday, 13 November 2023 at 22:58:54 UTC, Tom Roberts wrote:
On 11/13/23 3:07 PM, Lou wrote:
a true classical model does not use acceleration to describe the
force of gravity.
Well, there is no such thing as a "true" model in physics --
physical models are valid or invalid, but we humans can never know
whether they are "true". But we can know when they are false, and
the "classical model" known as Newtonian mechanics (NM) is known to
be false (but it is often useful as an approximation).

Exactly....it’s false. Gravity force isn’t acceleration based on
r^2.

Oh for goodness sake! I forgot how stupid and ignorant you are.

NM is known to be false, by a few parts per trillion, such as the
precession of the perihelion of mercury, or the "gravitational time
dilation" exhibited by GPS satellites. Confusing 1/r and 1/r^2 would
involve factors billions or trillions of times larger. GR, of course,
fixes these errors in NM.

Thats what I have been trying to get you lot to understand this
whole time.

You have no hope of doing that, because a) YOU don't understand it, and
b) it is WRONG.

Einstein realised this and used potential.

Not for gravitational force, but rather for gravitational potential --
DUH! He then used approximation techniques to show that in the Newtonian approximation to GR the relevant component of the metric tensor involved
the Newtonian gravitational potential.

Laplace realised that gravity force was potential.

Nope. He was not STUPID AND IGNORANT like you.

Again, in NM for gravity: force != potential;
rather, force = m * -grad potential, and since F = m a,

In GR neither gravitational force nor gravitational potential appears in
the theory. The relative acceleration between small objects due to
gravity is expressed by the Raychaudhuri equation.

[... more nonsense]

Tom Roberts

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Volney@21:1/5 to Lou on Tue Nov 14 23:59:49 2023
On 11/14/2023 5:06 AM, Lou wrote:
On Monday, 13 November 2023 at 22:58:54 UTC, Tom Roberts wrote:
On 11/13/23 3:07 PM, Lou wrote:
a true classical model does not use acceleration to describe the
force of gravity.
Well, there is no such thing as a "true" model in physics -- physical
models are valid or invalid, but we humans can never know whether they
are "true". But we can know when they are false, and the "classical
model" known as Newtonian mechanics (NM) is known to be false (but it is
often useful as an approximation).

Exactly....it’s false. Gravity force isn’t acceleration based on r^2.

No, force is the mass * acceleration. Newton's Second Law, you know.

The acceleration due to gravity, according to Newton, is GM/r^2.
The force on a mass m due to gravity, according to Newton, is GMm/r^2.

Thats what I have been trying to get you lot to understand
this whole time.

Newton says you're completely wrong.

Einstein realised this and used potential.

Not for the equivalent of gravitational acceleration or force!

Laplace
realised that gravity force was potential. Newton knew this and
called gravity force a scalar field. And others like Levy also
proposed this

Wrong.

The ridiculous part of your argument is that although do you admit
r^2 doesn’t work

r^2 does work in Newton's approximation.

...you cannot bear to have anyone point out
that to make a classical theory work...one must use r of potential.

Classical Newtonian gravity uses 1/r^2 for acceleration and force. You
have been claiming classical gravity uses 1/r and it is COMPLETELY WRONG!

Despite being in the hypocritical position of accepting that
GR theory can ditch r^2…and use r of potential.

As far as the relationship between acceleration and force is concerned,
here in the context of NM, what is used is Newton's second law:
F = m a
where F is the total force on an object, m is its mass, and a is its
acceleration. This is implicitly relative to some inertial frame; F, m,
and a are all invariant under Galilean transforms (change of coordinates
to a different inertial frame).

Word salad. R^2 isn’t F=ma. It is a ridiculous F=a. You know it is.
Dont pretend it isn’t.

Nope. Newton's Second Law relates force and acceleration. F=ma.

Only relativists (or idiots) think in a classical model
acceleration=force.
NONSENSE! NOBODY thinks that. YOU are confused. The only "classical
model" here is Newtonian mechanics, and in NM the second law applies:
F = m a

No...YOU think acceleration=force. Not me.,You just spent your whole post trying
to pretend in a classical model that m/s^2= force.
It isn’t . ITS CALLED ACCELERATION. Einstein realised this.

You are (deliberately!) leaving out the mass in Newton's
force/acceleration relationship. Again, F=ma.

They do it to make sure a classical model can’t correctly predict
the change of resonant frequencies of atoms at different potentials.
More nonsense. In Newtonian mechanics, time is universal and NM predicts
zero "time dilation" under any and all circumstances. NM is DIFFERENT
from relativity, and wrong.

The only nonsense is your silly claim that I said a classical model
predicts time dilation. That’s GR.
A classical model says no time dilation is occuring. What you
are seeing is a harmonic oscillator (c-133) changing its resonant
frequency due to an external force of gravity.

How could it if the Newtonian force is proportional to 1/r^2 but the
perceived frequency difference goes as 1/r?

And we have known harmonic oscillators do change frequencies
under external force for longer than GR has been around.

But we also know they don't go as 1/r.

Force of gravity was called potential by Laplace.[...]

Still more nonsense. The force of gravity is minus the gradient of the
gravitational potential. They have NEVER been "equal" as you suppose.

The force of gravity increases the closer you get to the mass.
Proportional to r.

Nope. Proportional to 1/r^2.

That’s what potential really is.

Nope. Potential is proportionate to 1/r. It doesn't even have the units
of force or acceleration.

You REALLY need to learn basic physics. Your GUESSES AND FANTASIES are
wrong.

If I’m wrong to model the effects of gravity with r...then why is
OK for Albert to use r of potential to model the effects of gravity?

Because Einstein wasn't showing the effect of Newtonian forces. He found
a new relationship that simplifies to the potential. It has nothing to
do with Newtonian force or acceleration at all, and it is not due to any
"the change of resonant frequencies of atoms" word salad.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Maciej Wozniak@21:1/5 to Tom Roberts on Tue Nov 14 23:56:33 2023
On Wednesday, 15 November 2023 at 06:00:17 UTC+1, Tom Roberts wrote:
On 11/14/23 4:06 AM, Lou wrote:
On Monday, 13 November 2023 at 22:58:54 UTC, Tom Roberts wrote:
On 11/13/23 3:07 PM, Lou wrote:
a true classical model does not use acceleration to describe the
force of gravity.
Well, there is no such thing as a "true" model in physics --
physical models are valid or invalid, but we humans can never know
whether they are "true". But we can know when they are false, and
the "classical model" known as Newtonian mechanics (NM) is known to
be false (but it is often useful as an approximation).

Exactly....it’s false. Gravity force isn’t acceleration based on
r^2.
Oh for goodness sake! I forgot how stupid and ignorant you are.

NM is known to be false

Samely as Earth was known to be flat.
You're an ignorant, fanatic idiot and your
"knowing" is worthless.

, by a few parts per trillion, such as the
precession of the perihelion of mercury, or the "gravitational time dilation" exhibited by GPS satellites.

A lie. Of course. Time is what clocks indicate, isn't it?
There is no time dilation idiocy for GPS satellites.
Common sense was warning your idiot guru.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Lou@21:1/5 to Tom Roberts on Wed Nov 15 01:53:47 2023
On Wednesday, 15 November 2023 at 05:00:17 UTC, Tom Roberts wrote:
On 11/14/23 4:06 AM, Lou wrote:
On Monday, 13 November 2023 at 22:58:54 UTC, Tom Roberts wrote:
On 11/13/23 3:07 PM, Lou wrote:
a true classical model does not use acceleration to describe the
force of gravity.
Well, there is no such thing as a "true" model in physics --
physical models are valid or invalid, but we humans can never know
whether they are "true". But we can know when they are false, and
the "classical model" known as Newtonian mechanics (NM) is known to
be false (but it is often useful as an approximation).

Exactly....it’s false. Gravity force isn’t acceleration based on
r^2.
Oh for goodness sake! I forgot how stupid and ignorant you are.

Said the guy who thinks force=acceleration
Even idiots aren’t that stupid.

NM is known to be false, by a few parts per trillion, such as the
precession of the perihelion of mercury, or the "gravitational time dilation" exhibited by GPS satellites. Confusing 1/r and 1/r^2 would
involve factors billions or trillions of times larger. GR, of course,
fixes these errors in NM.

NM got lots of things wrong. He falsely put all the mass at the center of the volume of the mass. But,.. If one spreads the mass out across the suns volume or the galaxy disc then the classical predictions of preccession and
galaxy rotation curves can be made consistent with the observations

Thats what I have been trying to get you lot to understand this
whole time.
You have no hope of doing that, because a) YOU don't understand it, and
b) it is WRONG.
Einstein realised this and used potential.
Not for gravitational force, but rather for gravitational potential --

He certainly conned you. He called the force of gravity... potential,
Used r instead...and got the correct results.
Sad part is he just copied Newton’s scalar field.
Newton himself knew the force of gravity was proportional to r.
He never meant the acceleration of little g to be interpreted by idiots as
a force.

DUH! He then used approximation techniques to show that in the Newtonian approximation to GR the relevant component of the metric tensor involved
the Newtonian gravitational potential.
Laplace realised that gravity force was potential.
Nope. He was not STUPID AND IGNORANT like you.

Said the guy who thinks force=acceleration.

Again, in NM for gravity: force != potential;
rather, force = m * -grad potential, and since F = m a,

In GR neither gravitational force nor gravitational potential appears in
the theory. The relative acceleration between small objects due to
gravity is expressed by the Raychaudhuri equation.

Exactly. Finally you woke up and smelt the coffee.
GR does not use little g’s r^2, as I have been trying to tell you guys.
And he does use the r of potential. As I’ve been telling you lot.
Except to con his followers, he refers to potential using other
names like “ Raychaudhuri equation.”
But if it walks like a duck, quacks like a duck and looks like a duck
Then it’s a potential duck.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Lou@21:1/5 to Volney on Wed Nov 15 01:40:50 2023
On Wednesday, 15 November 2023 at 04:59:55 UTC, Volney wrote:
On 11/14/2023 5:06 AM, Lou wrote:
On Monday, 13 November 2023 at 22:58:54 UTC, Tom Roberts wrote:
On 11/13/23 3:07 PM, Lou wrote:
a true classical model does not use acceleration to describe the
force of gravity.
Well, there is no such thing as a "true" model in physics -- physical
models are valid or invalid, but we humans can never know whether they
are "true". But we can know when they are false, and the "classical
model" known as Newtonian mechanics (NM) is known to be false (but it is >> often useful as an approximation).

Exactly....it’s false. Gravity force isn’t acceleration based on r^2.
No, force is the mass * acceleration. Newton's Second Law, you know.

So why are you pretending the force of gravity is measured as little g (m/s^2)? Since when does force=acceleration?

The acceleration due to gravity, according to Newton, is GM/r^2.
The force on a mass m due to gravity, according to Newton, is GMm/r^2.
Thats what I have been trying to get you lot to understand
this whole time.
Newton says you're completely wrong.

No. Newton said the force of gravity was proportional to r.
He called it a scalar field just to confuse idiots.

Einstein realised this and used potential.
Not for the equivalent of gravitational acceleration or force!

Does GR use Gm/r^2 anywhere in its calculations?
No
Does GR use r?
Yes.
Does GR use r to model tick rates at different altitudes?
Yes
Looks like if it’s OK for GR to ignore little g and use r...
It’s OK for a classical model to do the same.
Anyways Newton did say force can be calculated with just r.
He called it a scalar field.

Laplace
realised that gravity force was potential. Newton knew this and
called gravity force a scalar field. And others like Levy also
proposed this
Wrong.
Only if you are a liar. And ignore the facts.

The ridiculous part of your argument is that although do you admit
r^2 doesn’t work
r^2 does work in Newton's approximation.
No it doesn’t. You said yourself tick rates are proportionally to r.
...you cannot bear to have anyone point out
that to make a classical theory work...one must use r of potential.
Classical Newtonian gravity uses 1/r^2 for acceleration and force. You
have been claiming classical gravity uses 1/r and it is COMPLETELY WRONG!

Said the relativist who knows that to correctly calculate the effects
of the force of gravity..one must use r

Despite being in the hypocritical position of accepting that
GR theory can ditch r^2…and use r of potential.

As far as the relationship between acceleration and force is concerned, >> here in the context of NM, what is used is Newton's second law:
F = m a
where F is the total force on an object, m is its mass, and a is its
acceleration. This is implicitly relative to some inertial frame; F, m, >> and a are all invariant under Galilean transforms (change of coordinates >> to a different inertial frame).

Word salad. R^2 isn’t F=ma. It is a ridiculous F=a. You know it is.
Dont pretend it isn’t.
Nope. Newton's Second Law relates force and acceleration. F=ma.

Yes. And notice force does not equal acceleration.
As you are trying to pretend a classical model says.

Only relativists (or idiots) think in a classical model
acceleration=force.
NONSENSE! NOBODY thinks that. YOU are confused. The only "classical
model" here is Newtonian mechanics, and in NM the second law applies:
F = m a

No...YOU think acceleration=force. Not me.,You just spent your whole post trying
to pretend in a classical model that m/s^2= force.
It isn’t . ITS CALLED ACCELERATION. Einstein realised this.
You are (deliberately!) leaving out the mass in Newton's
force/acceleration relationship. Again, F=ma.

Im not. You are. How many times have you said the force of gravity
equals just acceleration. Millions of times.

They do it to make sure a classical model can’t correctly predict
the change of resonant frequencies of atoms at different potentials.
More nonsense. In Newtonian mechanics, time is universal and NM predicts >> zero "time dilation" under any and all circumstances. NM is DIFFERENT
from relativity, and wrong.

The only nonsense is your silly claim that I said a classical model predicts time dilation. That’s GR.
A classical model says no time dilation is occuring. What you
are seeing is a harmonic oscillator (c-133) changing its resonant frequency due to an external force of gravity.
How could it if the Newtonian force is proportional to 1/r^2 but the perceived frequency difference goes as 1/r?

You have a very short attention span. I just told you even Newton
said the force of gravity was r. He called it a scalar field
It’s only dishonest relativists who pretend that force=acceleration in a classical model.

And we have known harmonic oscillators do change frequencies
under external force for longer than GR has been around.
But we also know they don't go as 1/r.

Really.? That’s odd. Obviously you forgot.
GPS clock rates match those predicted by a classical model
using force of gravity proportional to r.

Force of gravity was called potential by Laplace.[...]

Still more nonsense. The force of gravity is minus the gradient of the
gravitational potential. They have NEVER been "equal" as you suppose.

The force of gravity increases the closer you get to the mass. Proportional to r.
Nope. Proportional to 1/r^2.

No Volney. Force does not equal acceleration.

That’s what potential really is.
Nope. Potential is proportionate to 1/r. It doesn't even have the units
of force or acceleration.

Only an idiot would think that if it takes more energy to lift an
object to a higher altitude that must mean that the force of gravity
increases with altitude!!!

You REALLY need to learn basic physics. Your GUESSES AND FANTASIES are
wrong.

If I’m wrong to model the effects of gravity with r...then why is
OK for Albert to use r of potential to model the effects of gravity?
Because Einstein wasn't showing the effect of Newtonian forces. He found
a new relationship that simplifies to the potential. It has nothing to
do with Newtonian force or acceleration at all, and it is not due to any "the change of resonant frequencies of atoms" word salad.

In other words Einstein used r to model the effects of the force of gravity. And to cover it up..he changed the name of “force of gravity” to potential.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Lou@21:1/5 to J. J. Lodder on Wed Nov 15 04:22:21 2023
On Sunday, 12 November 2023 at 13:30:18 UTC, J. J. Lodder wrote:
Lou <volney..@live.co.uk> wrote:

On Saturday, 11 November 2023 at 20:47:35 UTC, J. J. Lodder wrote:
Lou wrote:

On Friday, 10 November 2023 at 20:00:54 UTC, J. J. Lodder wrote:
[summary: gravity and clock rates for misled kiddies]

General relativity predicts that all freely falling clocks
will run at their own inherent rate. [by postulate]
It also predicts that clocks at different places,
and with different velocities will be seen to run at different rates,
-when compared with each other-.
It also predicts that accelerations do not affect clock rates,
so the results can be extended to non-inertial clocks,
such as clocks at rest at different altitudes on Earth.

According to general relativity all clock effects are purely kinematic,
so derivable from the metric tensor.
Doing the sums for weak fields results in velocity effects
being given by Lorentz factors, and 'gravitational' effects
being given by the variations in Newtonian potential.
So far, so good, and in agreement with experimental results.

Now there are people such as for example 'Lou' in this forum,
who cannot or will not accept or understand this.
They hold that obverved clock effects must be due to 'gravity' affecting the workings of the clock, somehow.
In other words, they ascribe the observed clock effects
to physical causes, 'gravity' affecting the workings of clocks, rather than to intrinsic space-time effects.

Fortunately it is easy to settle the point by experiment.
GR predicts that all clocks on the rotating geoid on Earth
must run at the same rate, when compared with each other.
Experiment bears this out, to accuracies approaching 10^-15.
This is of immense practical importance,
because it is the basis for realising the SI second.
(on which -all- physical measurement depends nowadays)

OTOH the force of gravity, as measured by 'small' g,
the acceleration of gravity, varies markedly over the geoid.
(by about 0.5%, between the poles and the equator)

If (the force of) 'gravity' influenced the rate of the clocks
there should be an effect of geographical latitude
on the rate of clocks.
This is not observed to be the case, so this idea stands falsified.

The idea that 'gravity' affects the rate at which clocks run
is a misconception without basis in observed fact,

A desperately misguided post from JJ.
You did not really read any of my posts. If you did...Then you deliberately ignored the fact that I *very* explicitly stated that
in a classical model "little g" is acceleration only. Not force.
And you ignored the fact that I very clearly stated that force
on the atoms at different altitudes in a classical model should be calculated using r.
Indeed, there is little point, because you go on harping about your r, and you are ignoring all sound advice by others.
You can go on obfuscating because you limit yourself
to situations with spherical symmetry.

So to see the errors of your ways you should consider situations
where spherical symmetry does not hold.
Then the surfaces of constant potential do not coincide
with surfaces of constant acceleration, or constant r.
Not the m/s^2 acceleration of r^2 in "little g".
Seeing as everyone except a profound idiot would think
acceleration = force.
And If you actually read my posts rather than thump your bible,
you would realise that I also said that force is what Laplace called gravitational potential. And what Newton referred to as a scalar field.
And what Einstein used to calculate his GR clock rate effects.
(Notice the r of Laplace's gravitational potential and Newton's scalar field is also the r used in GR. Not r^2 of little g.)
So if you claim that experiment shows no change of clock
rates at different sea level latitudes. Then you have not
only confirmed the predictions of GR.. You have also confirmed
the predictions of classical theory. Seeing as they both use r to accurately calculate tick rates at different altitudes.
Experiment shows that clocks on the geoid run at constant rates
wrt each other. Note that the geoid is not a surface of constant r,
nor a surface of constant g,

A straw man argument if ever you make.
Yes I've looked at your 'geoid' now and how it varies slightly by about 200m
relative to the reference geoid and how technically the r distance doesn't exactly follow the geoid surface. That makes sense. Splitting hairs though on your part to pretend somehow this rules out a classical model
which uses r. I notice you didn't actually specify why it would.
In fact it doesn't rule out in any way a classical model
any more than it would rule out GR.
Because in a classical calculation if one needs to assume *exactly* the *total* mass M of the earth at r, then yes to be *absolutely* accurate
the geoid surface has to be used. Not the actual distance r.
But the same applies to GR. And the fact remains that
generally, the force of gravity in a classical model follows r not r^2. (And to please the pedant JJ,... with ever so small meter length fluctuations in the exact distance of r to also be taken into account)
So you missed all points, again. I'll simplify.
The geoid surface is by definition an equipotential surface of the
Newtonian potential.
So it coincides (almost) with the mean sea level.
The geoid is (to a very good approximation) an ellipsoid of revolution.
The small differences between geoid and ellipsoid
(due to slightly irregular mass distributions inside the Earth)
don't matter for what follows.

Now, on the geoid, and at the poles, we have: r < average g > average, potential = constant
On the geoid, at mid-latitudes we have r = average, g = average,
potential = same constant
On the geoid, at the equator, we have r > average, g < average,
potential = still the same constant, by definition of the geoid.

The differences are huge, r = 6357-6378 km, g = 9.863-9.798 m/s2,
compared to clock stabilities of 10^-15.

What is your prediction for the rates of clocks in those three places?
No verbiage, just say faster, slower, or the same,
and if you can by how much,

I’m wondering if this claim of yours is another piece of BS.
I’ve looked up measuring clock rates at the geoid and so far
it seems this actually hasn’t been done yet. It’s only considered
possible with the latest tech.
So I know relativists don’t like evidence but please...
back up your claims with a few citations.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Maciej Wozniak@21:1/5 to J. J. Lodder on Wed Nov 15 05:47:49 2023
On Friday, 10 November 2023 at 21:00:54 UTC+1, J. J. Lodder wrote:
[summary: gravity and clock rates for misled kiddies]

General relativity predicts that all freely falling clocks
will run at their own inherent rate. [by postulate]

What to expect from a mumbling inconsistently idiot.
Has he ever heard of pendulum clocks?

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From J. J. Lodder@21:1/5 to Lou on Wed Nov 15 22:01:50 2023
Lou <noelturntive@live.co.uk> wrote:

On Sunday, 12 November 2023 at 13:30:18 UTC, J. J. Lodder wrote:
Lou <volney..@live.co.uk> wrote:

On Saturday, 11 November 2023 at 20:47:35 UTC, J. J. Lodder wrote:
Lou wrote:

On Friday, 10 November 2023 at 20:00:54 UTC, J. J. Lodder wrote:
[summary: gravity and clock rates for misled kiddies]

General relativity predicts that all freely falling clocks
will run at their own inherent rate. [by postulate]
It also predicts that clocks at different places,
and with different velocities will be seen to run at different rates, -when compared with each other-.
It also predicts that accelerations do not affect clock rates,
so the results can be extended to non-inertial clocks,
such as clocks at rest at different altitudes on Earth.

According to general relativity all clock effects are purely kinematic, so derivable from the metric tensor.
Doing the sums for weak fields results in velocity effects
being given by Lorentz factors, and 'gravitational' effects
being given by the variations in Newtonian potential.
So far, so good, and in agreement with experimental results.

Now there are people such as for example 'Lou' in this forum,
who cannot or will not accept or understand this.
They hold that obverved clock effects must be due to 'gravity' affecting the workings of the clock, somehow.
In other words, they ascribe the observed clock effects
to physical causes, 'gravity' affecting the workings of clocks, rather than to intrinsic space-time effects.

Fortunately it is easy to settle the point by experiment.
GR predicts that all clocks on the rotating geoid on Earth
must run at the same rate, when compared with each other. Experiment bears this out, to accuracies approaching 10^-15.
This is of immense practical importance,
because it is the basis for realising the SI second.
(on which -all- physical measurement depends nowadays)

OTOH the force of gravity, as measured by 'small' g,
the acceleration of gravity, varies markedly over the geoid.
(by about 0.5%, between the poles and the equator)

If (the force of) 'gravity' influenced the rate of the clocks
there should be an effect of geographical latitude
on the rate of clocks.
This is not observed to be the case, so this idea stands falsified.

The idea that 'gravity' affects the rate at which clocks run
is a misconception without basis in observed fact,

A desperately misguided post from JJ.
You did not really read any of my posts. If you did...Then you deliberately ignored the fact that I *very* explicitly stated that
in a classical model "little g" is acceleration only. Not force.
And you ignored the fact that I very clearly stated that force
on the atoms at different altitudes in a classical model should be calculated using r.
Indeed, there is little point, because you go on harping about your r, and you are ignoring all sound advice by others.
You can go on obfuscating because you limit yourself
to situations with spherical symmetry.

So to see the errors of your ways you should consider situations
where spherical symmetry does not hold.
Then the surfaces of constant potential do not coincide
with surfaces of constant acceleration, or constant r.
Not the m/s^2 acceleration of r^2 in "little g".
Seeing as everyone except a profound idiot would think
acceleration = force.
And If you actually read my posts rather than thump your bible,
you would realise that I also said that force is what Laplace
called gravitational potential. And what Newton referred to as a scalar field.
And what Einstein used to calculate his GR clock rate effects. (Notice the r of Laplace's gravitational potential and Newton's scalar field is also the r used in GR. Not r^2 of little g.)
So if you claim that experiment shows no change of clock
rates at different sea level latitudes. Then you have not
only confirmed the predictions of GR.. You have also confirmed
the predictions of classical theory. Seeing as they both use r to accurately calculate tick rates at different altitudes.
Experiment shows that clocks on the geoid run at constant rates
wrt each other. Note that the geoid is not a surface of constant r,
nor a surface of constant g,

A straw man argument if ever you make.
Yes I've looked at your 'geoid' now and how it varies slightly by
about 200m relative to the reference geoid and how technically the r distance doesn't exactly follow the geoid surface. That makes sense. Splitting hairs though on your part to pretend somehow this rules out
a classical model which uses r. I notice you didn't actually specify
why it would. In fact it doesn't rule out in any way a classical model any more than it would rule out GR.
Because in a classical calculation if one needs to assume *exactly* the *total* mass M of the earth at r, then yes to be *absolutely* accurate the geoid surface has to be used. Not the actual distance r.
But the same applies to GR. And the fact remains that
generally, the force of gravity in a classical model follows r not r^2. (And to please the pedant JJ,... with ever so small meter length fluctuations in the exact distance of r to also be taken into account)
So you missed all points, again. I'll simplify.
The geoid surface is by definition an equipotential surface of the Newtonian potential.
So it coincides (almost) with the mean sea level.
The geoid is (to a very good approximation) an ellipsoid of revolution.
The small differences between geoid and ellipsoid
(due to slightly irregular mass distributions inside the Earth)
don't matter for what follows.

Now, on the geoid, and at the poles, we have: r < average g > average, potential = constant
On the geoid, at mid-latitudes we have r = average, g = average,
potential = same constant
On the geoid, at the equator, we have r > average, g < average,
potential = still the same constant, by definition of the geoid.

The differences are huge, r = 6357-6378 km, g = 9.863-9.798 m/s2,
compared to clock stabilities of 10^-15.

What is your prediction for the rates of clocks in those three places?
No verbiage, just say faster, slower, or the same,
and if you can by how much,

I'm wondering if this claim of yours is another piece of BS.
I've looked up measuring clock rates at the geoid and so far
it seems this actually hasn't been done yet.

At last. You are a bit late, considering your grandiose claims.

It's only considered possible with the latest tech. So I know relativists don't like evidence but please... back up your claims with a few
citations.

Standards laboratory all over the world have been keeping clusters of
atomic clocks, and they have been comparing them.
Over fifty years ago they noticed systematic differences in clock rates
between standards laboratories, with those situated at higher altitudes
running faster. (as seen by clocks near sea level)
Accurate comparisons over long times established that those rate
differences are proportional to the difference in Newtonian potential
between their locations.
(as predicted by general relativity, in the Newtonian limit)

Those rate difference are (and must) be taken into account
to compute TAI. All this has been going on for decades,
to stabilities of order 10^-15,
and all accurate timekeeping depends on it.

Note that your rantings about r versus r^2 do not come into this at all.
They cannot do so, because r varies to order 10^-3 over the geoid,
many orders of magnitude more than observed clock rates.

You would do better to learn some real physics,

Jan

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Maciej Wozniak@21:1/5 to J. J. Lodder on Wed Nov 15 21:49:53 2023
On Wednesday, 15 November 2023 at 22:01:54 UTC+1, J. J. Lodder wrote:

Accurate comparisons over long times established that those rate
differences are proportional to the difference in Newtonian potential
between their locations.
(as predicted by general relativity

A lie, of course, your fellow idiot Tom could explain
you that The Shit is predicting the same rate everywhere..

, in the Newtonian limit)

Those rate difference are (and must) be taken into account
to compute TAI.

Sure, not even you are stupid enough to treat this
"time dilation mumble" seriously.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Lou@21:1/5 to J. J. Lodder on Thu Nov 16 03:11:30 2023
On Wednesday, 15 November 2023 at 21:01:54 UTC, J. J. Lodder wrote:
Lou <jjlodder @..uk> wrote:

On Sunday, 12 November 2023 at 13:30:18 UTC, J. J. Lodder wrote:
Lou <@JJ.comwrote:

On Saturday, 11 November 2023 at 20:47:35 UTC, J. J. Lodder wrote:
Lou wrote:

On Friday, 10 November 2023 at 20:00:54 UTC, J. J. Lodder wrote:
[summary: gravity and clock rates for misled kiddies]

General relativity predicts that all freely falling clocks
will run at their own inherent rate. [by postulate]
It also predicts that clocks at different places,
and with different velocities will be seen to run at different rates, -when compared with each other-.
It also predicts that accelerations do not affect clock rates, so the results can be extended to non-inertial clocks,
such as clocks at rest at different altitudes on Earth.

According to general relativity all clock effects are purely kinematic, so derivable from the metric tensor.
Doing the sums for weak fields results in velocity effects
being given by Lorentz factors, and 'gravitational' effects being given by the variations in Newtonian potential.
So far, so good, and in agreement with experimental results.

Now there are people such as for example 'Lou' in this forum, who cannot or will not accept or understand this.
They hold that obverved clock effects must be due to 'gravity' affecting the workings of the clock, somehow.
In other words, they ascribe the observed clock effects
to physical causes, 'gravity' affecting the workings of clocks, rather than to intrinsic space-time effects.

Fortunately it is easy to settle the point by experiment.
GR predicts that all clocks on the rotating geoid on Earth
must run at the same rate, when compared with each other. Experiment bears this out, to accuracies approaching 10^-15. This is of immense practical importance,
because it is the basis for realising the SI second.
(on which -all- physical measurement depends nowadays)

OTOH the force of gravity, as measured by 'small' g,
the acceleration of gravity, varies markedly over the geoid.
(by about 0.5%, between the poles and the equator)

If (the force of) 'gravity' influenced the rate of the clocks there should be an effect of geographical latitude
on the rate of clocks.
This is not observed to be the case, so this idea stands falsified.

The idea that 'gravity' affects the rate at which clocks run
is a misconception without basis in observed fact,

A desperately misguided post from JJ.
You did not really read any of my posts. If you did...Then you deliberately ignored the fact that I *very* explicitly stated that in a classical model "little g" is acceleration only. Not force. And you ignored the fact that I very clearly stated that force
on the atoms at different altitudes in a classical model should be calculated using r.
Indeed, there is little point, because you go on harping about your r,
and you are ignoring all sound advice by others.
You can go on obfuscating because you limit yourself
to situations with spherical symmetry.

So to see the errors of your ways you should consider situations where spherical symmetry does not hold.
Then the surfaces of constant potential do not coincide
with surfaces of constant acceleration, or constant r.
Not the m/s^2 acceleration of r^2 in "little g".
Seeing as everyone except a profound idiot would think acceleration = force.
And If you actually read my posts rather than thump your bible, you would realise that I also said that force is what Laplace called gravitational potential. And what Newton referred to as a scalar field.
And what Einstein used to calculate his GR clock rate effects. (Notice the r of Laplace's gravitational potential and Newton's scalar field is also the r used in GR. Not r^2 of little g.)
So if you claim that experiment shows no change of clock
rates at different sea level latitudes. Then you have not
only confirmed the predictions of GR.. You have also confirmed
the predictions of classical theory. Seeing as they both use r to accurately calculate tick rates at different altitudes.
Experiment shows that clocks on the geoid run at constant rates
wrt each other. Note that the geoid is not a surface of constant r, nor a surface of constant g,

A straw man argument if ever you make.
Yes I've looked at your 'geoid' now and how it varies slightly by about 200m relative to the reference geoid and how technically the r distance doesn't exactly follow the geoid surface. That makes sense. Splitting hairs though on your part to pretend somehow this rules out a classical model which uses r. I notice you didn't actually specify why it would. In fact it doesn't rule out in any way a classical model any more than it would rule out GR.
Because in a classical calculation if one needs to assume *exactly* the
*total* mass M of the earth at r, then yes to be *absolutely* accurate the geoid surface has to be used. Not the actual distance r.
But the same applies to GR. And the fact remains that
generally, the force of gravity in a classical model follows r not r^2.
(And to please the pedant JJ,... with ever so small meter length fluctuations in the exact distance of r to also be taken into account)
So you missed all points, again. I'll simplify.
The geoid surface is by definition an equipotential surface of the Newtonian potential.
So it coincides (almost) with the mean sea level.
The geoid is (to a very good approximation) an ellipsoid of revolution. The small differences between geoid and ellipsoid
(due to slightly irregular mass distributions inside the Earth)
don't matter for what follows.

Now, on the geoid, and at the poles, we have: r < average g > average, potential = constant
On the geoid, at mid-latitudes we have r = average, g = average, potential = same constant
On the geoid, at the equator, we have r > average, g < average, potential = still the same constant, by definition of the geoid.

The differences are huge, r = 6357-6378 km, g = 9.863-9.798 m/s2, compared to clock stabilities of 10^-15.

What is your prediction for the rates of clocks in those three places? No verbiage, just say faster, slower, or the same,
and if you can by how much,

I'm wondering if this claim of yours is another piece of BS.
I've looked up measuring clock rates at the geoid and so far
it seems this actually hasn't been done yet.
At last. You are a bit late, considering your grandiose claims.
It's only considered possible with the latest tech. So I know relativists don't like evidence but please... back up your claims with a few citations.
Standards laboratory all over the world have been keeping clusters of
atomic clocks, and they have been comparing them.
Over fifty years ago they noticed systematic differences in clock rates between standards laboratories, with those situated at higher altitudes running faster. (as seen by clocks near sea level)
Accurate comparisons over long times established that those rate
differences are proportional to the difference in Newtonian potential between their locations.
(as predicted by general relativity, in the Newtonian limit)

Those rate difference are (and must) be taken into account
to compute TAI. All this has been going on for decades,
to stabilities of order 10^-15,
and all accurate timekeeping depends on it.

Interesting trivia thanks. But as for classical theory and as with most relativists,...when discussing clock rates vs altitude.
I will be content to use r.

Note that your rantings about r versus r^2 do not come into this at all. They cannot do so, because r varies to order 10^-3 over the geoid,
many orders of magnitude more than observed clock rates.

Irrelevent pedantry.
Your reference points out it wasn’t until the late 70s that the physics community
*including relativists* realised r wasn’t following the density distribution of the earths gravitational mass by a very small variation of about +- 100m error
from the assumed radius 6371000m of r.
Looks like on your rational...all predictions by GR was BS before 1977.
And, even then, since 1977 any relativist (including Paul and Volney) who uses r but doesn’t mention geoid isn’t a physicist according to JJ.
Oh dear Shock Horror!!
Looks like your idols ...Schwarzschild, Pound Rebka, Einstein etc etc
...all used r!!
I suppose you think they all need to learn some basic physics.
If only they had studied under JJ.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From J. J. Lodder@21:1/5 to Lou on Thu Nov 16 17:46:59 2023
Lou <noelturntive@live.co.uk> wrote:

On Wednesday, 15 November 2023 at 21:01:54 UTC, J. J. Lodder wrote:
Lou <jjlodder @..uk> wrote:

On Sunday, 12 November 2023 at 13:30:18 UTC, J. J. Lodder wrote:
Lou <@JJ.comwrote:

On Saturday, 11 November 2023 at 20:47:35 UTC, J. J. Lodder wrote:
Lou wrote:

On Friday, 10 November 2023 at 20:00:54 UTC, J. J. Lodder wrote:
[summary: gravity and clock rates for misled kiddies]

General relativity predicts that all freely falling clocks
will run at their own inherent rate. [by postulate]
It also predicts that clocks at different places,
and with different velocities will be seen to run at different rates, -when compared with each other-.
It also predicts that accelerations do not affect clock rates, so the results can be extended to non-inertial clocks,
such as clocks at rest at different altitudes on Earth.

According to general relativity all clock effects are purely kinematic, so derivable from the metric tensor.
Doing the sums for weak fields results in velocity effects being given by Lorentz factors, and 'gravitational' effects being given by the variations in Newtonian potential.
So far, so good, and in agreement with experimental results.

Now there are people such as for example 'Lou' in this forum, who cannot or will not accept or understand this.
They hold that obverved clock effects must be due to 'gravity' affecting the workings of the clock, somehow.
In other words, they ascribe the observed clock effects
to physical causes, 'gravity' affecting the workings of clocks, rather than to intrinsic space-time effects.

Fortunately it is easy to settle the point by experiment.
GR predicts that all clocks on the rotating geoid on Earth
must run at the same rate, when compared with each other. Experiment bears this out, to accuracies approaching 10^-15. This is of immense practical importance,
because it is the basis for realising the SI second.
(on which -all- physical measurement depends nowadays)

OTOH the force of gravity, as measured by 'small' g,
the acceleration of gravity, varies markedly over the geoid. (by about 0.5%, between the poles and the equator)

If (the force of) 'gravity' influenced the rate of the clocks there should be an effect of geographical latitude
on the rate of clocks.
This is not observed to be the case, so this idea stands falsified.

The idea that 'gravity' affects the rate at which clocks run
is a misconception without basis in observed fact,

A desperately misguided post from JJ.
You did not really read any of my posts. If you did...Then you deliberately ignored the fact that I *very* explicitly stated that
in a classical model "little g" is acceleration only. Not force. And you ignored the fact that I very clearly stated that force
on the atoms at different altitudes in a classical model should be
calculated using r.
Indeed, there is little point, because you go on harping about
your r, and you are ignoring all sound advice by others. You can
go on obfuscating because you limit yourself to situations with spherical symmetry.

So to see the errors of your ways you should consider situations where spherical symmetry does not hold.
Then the surfaces of constant potential do not coincide
with surfaces of constant acceleration, or constant r.
Not the m/s^2 acceleration of r^2 in "little g".
Seeing as everyone except a profound idiot would think acceleration = force.
And If you actually read my posts rather than thump your bible, you would realise that I also said that force is what Laplace called gravitational potential. And what Newton referred to as a scalar field.
And what Einstein used to calculate his GR clock rate effects. (Notice the r of Laplace's gravitational potential and Newton's scalar field is also the r used in GR. Not r^2 of little g.)
So if you claim that experiment shows no change of clock
rates at different sea level latitudes. Then you have not
only confirmed the predictions of GR.. You have also confirmed the predictions of classical theory. Seeing as they both use r to accurately calculate tick rates at different altitudes.
Experiment shows that clocks on the geoid run at constant rates
wrt each other. Note that the geoid is not a surface of constant r, nor a surface of constant g,

A straw man argument if ever you make.
Yes I've looked at your 'geoid' now and how it varies slightly by about 200m relative to the reference geoid and how technically the r distance doesn't exactly follow the geoid surface. That makes sense. Splitting hairs though on your part to pretend somehow this rules out a classical model which uses r. I notice you didn't actually specify why it would. In fact it doesn't rule out in any way a classical model
any more than it would rule out GR.
Because in a classical calculation if one needs to assume
*exactly* the *total* mass M of the earth at r, then yes to be *absolutely* accurate the geoid surface has to be used. Not the actual distance r.
But the same applies to GR. And the fact remains that generally,
the force of gravity in a classical model follows r not r^2. (And
to please the pedant JJ,... with ever so small meter length fluctuations in the exact distance of r to also be taken into account)
So you missed all points, again. I'll simplify.
The geoid surface is by definition an equipotential surface of the Newtonian potential.
So it coincides (almost) with the mean sea level.
The geoid is (to a very good approximation) an ellipsoid of revolution. The small differences between geoid and ellipsoid
(due to slightly irregular mass distributions inside the Earth)
don't matter for what follows.

Now, on the geoid, and at the poles, we have: r < average g > average, potential = constant
On the geoid, at mid-latitudes we have r = average, g = average, potential = same constant
On the geoid, at the equator, we have r > average, g < average, potential = still the same constant, by definition of the geoid.

The differences are huge, r = 6357-6378 km, g = 9.863-9.798 m/s2, compared to clock stabilities of 10^-15.

What is your prediction for the rates of clocks in those three places? No verbiage, just say faster, slower, or the same,
and if you can by how much,

I'm wondering if this claim of yours is another piece of BS.
I've looked up measuring clock rates at the geoid and so far
it seems this actually hasn't been done yet.
At last. You are a bit late, considering your grandiose claims.
It's only considered possible with the latest tech. So I know relativists don't like evidence but please... back up your claims with a few citations.
Standards laboratory all over the world have been keeping clusters of atomic clocks, and they have been comparing them.
Over fifty years ago they noticed systematic differences in clock rates between standards laboratories, with those situated at higher altitudes running faster. (as seen by clocks near sea level)
Accurate comparisons over long times established that those rate differences are proportional to the difference in Newtonian potential between their locations.
(as predicted by general relativity, in the Newtonian limit)

Those rate difference are (and must) be taken into account
to compute TAI. All this has been going on for decades,
to stabilities of order 10^-15,
and all accurate timekeeping depends on it.

Interesting trivia thanks. But as for classical theory and as with most relativists,...when discussing clock rates vs altitude.
I will be content to use r.

You can be as wrong and content as you want to be.
Don't expect people to listen to you though.

Note that your rantings about r versus r^2 do not come into this at all. They cannot do so, because r varies to order 10^-3 over the geoid,
many orders of magnitude more than observed clock rates.

Irrelevent pedantry.

Yes, merely pointing out that you are off by ten orders of magnitude.

Your reference points out it wasn't until the late 70s that the physics community *including relativists* realised r wasn't following the density distribution of the earths gravitational mass by a very small variation of about +- 100m error from the assumed radius 6371000m of r.
Looks like on your rational...all predictions by GR was BS before 1977.
And, even then, since 1977 any relativist (including Paul and Volney) who uses r but doesn't mention geoid isn't a physicist according to JJ.
Oh dear Shock Horror!!
Looks like your idols ...Schwarzschild, Pound Rebka, Einstein etc etc
...all used r!!
I suppose you think they all need to learn some basic physics.
If only they had studied under JJ.

I'm quite sure that the people you mention have no need for that,

Jan

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Lou@21:1/5 to J. J. Lodder on Thu Nov 16 09:37:12 2023
On Thursday, 16 November 2023 at 16:47:03 UTC, J. J. Lodder wrote:
Lou <albertco.uk> wrote:

On Wednesday, 15 November 2023 at 21:01:54 UTC, J. J. Lodder wrote:
Lou <jjlodder @..uk> wrote:

On Sunday, 12 November 2023 at 13:30:18 UTC, J. J. Lodder wrote:
Lou <@JJ.comwrote:

On Saturday, 11 November 2023 at 20:47:35 UTC, J. J. Lodder wrote:
Lou wrote:

On Friday, 10 November 2023 at 20:00:54 UTC, J. J. Lodder wrote:
[summary: gravity and clock rates for misled kiddies]

General relativity predicts that all freely falling clocks will run at their own inherent rate. [by postulate]
It also predicts that clocks at different places,
and with different velocities will be seen to run at different
rates, -when compared with each other-.
It also predicts that accelerations do not affect clock rates,
so the results can be extended to non-inertial clocks,
such as clocks at rest at different altitudes on Earth.

According to general relativity all clock effects are purely kinematic, so derivable from the metric tensor.
Doing the sums for weak fields results in velocity effects being given by Lorentz factors, and 'gravitational' effects being given by the variations in Newtonian potential.
So far, so good, and in agreement with experimental results.

Now there are people such as for example 'Lou' in this forum,
who cannot or will not accept or understand this.
They hold that obverved clock effects must be due to 'gravity'
affecting the workings of the clock, somehow.
In other words, they ascribe the observed clock effects
to physical causes, 'gravity' affecting the workings of clocks,
rather than to intrinsic space-time effects.

Fortunately it is easy to settle the point by experiment.
GR predicts that all clocks on the rotating geoid on Earth must run at the same rate, when compared with each other. Experiment bears this out, to accuracies approaching 10^-15. This is of immense practical importance,
because it is the basis for realising the SI second.
(on which -all- physical measurement depends nowadays)

OTOH the force of gravity, as measured by 'small' g,
the acceleration of gravity, varies markedly over the geoid. (by about 0.5%, between the poles and the equator)

If (the force of) 'gravity' influenced the rate of the clocks
there should be an effect of geographical latitude
on the rate of clocks.
This is not observed to be the case, so this idea stands falsified.

The idea that 'gravity' affects the rate at which clocks run is a misconception without basis in observed fact,

A desperately misguided post from JJ.
You did not really read any of my posts. If you did...Then you deliberately ignored the fact that I *very* explicitly stated that
in a classical model "little g" is acceleration only. Not force.
And you ignored the fact that I very clearly stated that force on the atoms at different altitudes in a classical model should be
calculated using r.
Indeed, there is little point, because you go on harping about your r, and you are ignoring all sound advice by others. You can go on obfuscating because you limit yourself to situations with spherical symmetry.

So to see the errors of your ways you should consider situations where spherical symmetry does not hold.
Then the surfaces of constant potential do not coincide
with surfaces of constant acceleration, or constant r.
Not the m/s^2 acceleration of r^2 in "little g".
Seeing as everyone except a profound idiot would think acceleration = force.
And If you actually read my posts rather than thump your bible,
you would realise that I also said that force is what Laplace called gravitational potential. And what Newton referred to as a
scalar field.
And what Einstein used to calculate his GR clock rate effects. (Notice the r of Laplace's gravitational potential and Newton's
scalar field is also the r used in GR. Not r^2 of little g.) So if you claim that experiment shows no change of clock
rates at different sea level latitudes. Then you have not
only confirmed the predictions of GR.. You have also confirmed the predictions of classical theory. Seeing as they both use r to
accurately calculate tick rates at different altitudes.
Experiment shows that clocks on the geoid run at constant rates wrt each other. Note that the geoid is not a surface of constant r,
nor a surface of constant g,

A straw man argument if ever you make.
Yes I've looked at your 'geoid' now and how it varies slightly by about 200m relative to the reference geoid and how technically the r
distance doesn't exactly follow the geoid surface. That makes sense.
Splitting hairs though on your part to pretend somehow this rules out
a classical model which uses r. I notice you didn't actually specify
why it would. In fact it doesn't rule out in any way a classical model
any more than it would rule out GR.
Because in a classical calculation if one needs to assume *exactly* the *total* mass M of the earth at r, then yes to be *absolutely* accurate the geoid surface has to be used. Not the actual distance r.
But the same applies to GR. And the fact remains that generally, the force of gravity in a classical model follows r not r^2. (And to please the pedant JJ,... with ever so small meter length fluctuations in the exact distance of r to also be taken into account)
So you missed all points, again. I'll simplify.
The geoid surface is by definition an equipotential surface of the Newtonian potential.
So it coincides (almost) with the mean sea level.
The geoid is (to a very good approximation) an ellipsoid of revolution.
The small differences between geoid and ellipsoid
(due to slightly irregular mass distributions inside the Earth) don't matter for what follows.

Now, on the geoid, and at the poles, we have: r < average g > average,
potential = constant
On the geoid, at mid-latitudes we have r = average, g = average, potential = same constant
On the geoid, at the equator, we have r > average, g < average, potential = still the same constant, by definition of the geoid.

The differences are huge, r = 6357-6378 km, g = 9.863-9.798 m/s2, compared to clock stabilities of 10^-15.

What is your prediction for the rates of clocks in those three places?
No verbiage, just say faster, slower, or the same,
and if you can by how much,

I'm wondering if this claim of yours is another piece of BS.
I've looked up measuring clock rates at the geoid and so far
it seems this actually hasn't been done yet.
At last. You are a bit late, considering your grandiose claims.
It's only considered possible with the latest tech. So I know relativists
don't like evidence but please... back up your claims with a few citations.
Standards laboratory all over the world have been keeping clusters of atomic clocks, and they have been comparing them.
Over fifty years ago they noticed systematic differences in clock rates between standards laboratories, with those situated at higher altitudes running faster. (as seen by clocks near sea level)
Accurate comparisons over long times established that those rate differences are proportional to the difference in Newtonian potential between their locations.
(as predicted by general relativity, in the Newtonian limit)

Those rate difference are (and must) be taken into account
to compute TAI. All this has been going on for decades,
to stabilities of order 10^-15,
and all accurate timekeeping depends on it.

Interesting trivia thanks. But as for classical theory and as with most relativists,...when discussing clock rates vs altitude.
I will be content to use r.
You can be as wrong and content as you want to be.
Don't expect people to listen to you though.
Note that your rantings about r versus r^2 do not come into this at all. They cannot do so, because r varies to order 10^-3 over the geoid,
many orders of magnitude more than observed clock rates.

Irrelevent pedantry.
Yes, merely pointing out that you are off by ten orders of magnitude.

Oh Deary me! Better tell all those relativists and Newtonian and GR reference pages...
They can’t use r anymore. It varies by up to ...oh my god ! Up to 200 meters every 637100 meters!
This day will go down in history. Before Lodder & After Lodder.

Your reference points out it wasn't until the late 70s that the physics community *including relativists* realised r wasn't following the density distribution of the earths gravitational mass by a very small variation of about +- 100m error from the assumed radius 6371000m of r.
Looks like on your rational...all predictions by GR was BS before 1977. And, even then, since 1977 any relativist (including Paul and Volney) who uses r but doesn't mention geoid isn't a physicist according to JJ.
Oh dear Shock Horror!!
Looks like your idols ...Schwarzschild, Pound Rebka, Einstein etc etc ...all used r!!
I suppose you think they all need to learn some basic physics.
If only they had studied under JJ.
I'm quite sure that the people you mention have no need for that,

Your right, they died years before you showed them up for
incorrectly using r in their formulas.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Volney@21:1/5 to Lou on Thu Nov 16 19:56:30 2023
On 11/15/2023 4:40 AM, Lou wrote:
On Wednesday, 15 November 2023 at 04:59:55 UTC, Volney wrote:
On 11/14/2023 5:06 AM, Lou wrote:
On Monday, 13 November 2023 at 22:58:54 UTC, Tom Roberts wrote:
On 11/13/23 3:07 PM, Lou wrote:
a true classical model does not use acceleration to describe the
force of gravity.
Well, there is no such thing as a "true" model in physics -- physical
models are valid or invalid, but we humans can never know whether they >>>> are "true". But we can know when they are false, and the "classical
model" known as Newtonian mechanics (NM) is known to be false (but it is >>>> often useful as an approximation).

Exactly....it’s false. Gravity force isn’t acceleration based on r^2. >> No, force is the mass * acceleration. Newton's Second Law, you know.

So why are you pretending the force of gravity is measured as little g (m/s^2)?
Since when does force=acceleration?

Never. Force, according to Newton's second law, is F=mg since little g
is an acceleration (9.81 m/s^2 at sea level)

The acceleration due to gravity, according to Newton, is GM/r^2.
The force on a mass m due to gravity, according to Newton, is GMm/r^2.
Thats what I have been trying to get you lot to understand
this whole time.
Newton says you're completely wrong.

No. Newton said the force of gravity was proportional to r.
He called it a scalar field just to confuse idiots.

Lie.

Einstein realised this and used potential.
Not for the equivalent of gravitational acceleration or force!

Another lie. GR math is completely different from classical gravity. It
does come up with the same answers as classical gravity which is an approximation.

Does GR use Gm/r^2 anywhere in its calculations?
No

Because that's classical gravity.

Does GR use r?
Yes.

As an approximation.

Does GR use r to model tick rates at different altitudes?
Yes

Misleading. Perceived tick rates is not force.

Looks like if it’s OK for GR to ignore little g and use r...

Little g is from classical physics. GR states gravity isn't even a force
but approximating it as one is GMm/r^2 and for earth's surface gives the acceleration little g.

It’s OK for a classical model to do the same.
Anyways Newton did say force can be calculated with just r.

You lie.

The ridiculous part of your argument is that although do you admit
r^2 doesn’t work

r^2 does work in Newton's approximation.

No it doesn’t. You said yourself tick rates are proportionally to r.

The PERCEIVED tick rate due to blueshift, for weak field gravity, is proportional to 1/r. Classical force goes as GMm/r^2.

...you cannot bear to have anyone point out
that to make a classical theory work...one must use r of potential.

Classical Newtonian gravity uses 1/r^2 for acceleration and force. You
have been claiming classical gravity uses 1/r and it is COMPLETELY WRONG!

Said the relativist who knows that to correctly calculate the effects
of the force of gravity..one must use r

Completely wrong. GM/r does not even have units of acceleration or
force. It has units of meters^2/seconds^2 which doesn't even have a
secondary unit name.

As far as the relationship between acceleration and force is concerned, >>>> here in the context of NM, what is used is Newton's second law:
F = m a
where F is the total force on an object, m is its mass, and a is its
acceleration. This is implicitly relative to some inertial frame; F, m, >>>> and a are all invariant under Galilean transforms (change of coordinates >>>> to a different inertial frame).

Word salad. R^2 isn’t F=ma. It is a ridiculous F=a. You know it is.
Dont pretend it isn’t.

Nope. Newton's Second Law relates force and acceleration. F=ma.

Yes. And notice force does not equal acceleration.
As you are trying to pretend a classical model says.

Nope. F=ma is not F=a.

Only relativists (or idiots) think in a classical model
acceleration=force.
NONSENSE! NOBODY thinks that. YOU are confused. The only "classical
model" here is Newtonian mechanics, and in NM the second law applies:
F = m a

No...YOU think acceleration=force. Not me.,You just spent your whole post trying
to pretend in a classical model that m/s^2= force.
It isn’t . ITS CALLED ACCELERATION. Einstein realised this.

You are (deliberately!) leaving out the mass in Newton's
force/acceleration relationship. Again, F=ma.

Im not. You are. How many times have you said the force of gravity
equals just acceleration. Millions of times.

No, never. The formula for classical force is GMm/r^2. I never said
force is acceleration, classical force is mass * acceleration. Again,
you are deliberately leaving out the mass and accusing me of what you did.

They do it to make sure a classical model can’t correctly predict
the change of resonant frequencies of atoms at different potentials.
More nonsense. In Newtonian mechanics, time is universal and NM predicts >>>> zero "time dilation" under any and all circumstances. NM is DIFFERENT
from relativity, and wrong.

The only nonsense is your silly claim that I said a classical model
predicts time dilation. That’s GR.
A classical model says no time dilation is occuring. What you
are seeing is a harmonic oscillator (c-133) changing its resonant
frequency due to an external force of gravity.

How could it if the Newtonian force is proportional to 1/r^2 but the
perceived frequency difference goes as 1/r?

You have a very short attention span. I just told you even Newton
said the force of gravity was r.

No, he did not. You are lying when you say that. Newton explicitly
stated the force formula is GMm/r^2. Inversely proportional to r^2, not proportional to r.

He called it a scalar field
It’s only dishonest relativists who pretend that force=acceleration in a classical model.

And we have known harmonic oscillators do change frequencies
under external force for longer than GR has been around.
But we also know they don't go as 1/r.

Really.? That’s odd. Obviously you forgot.
GPS clock rates match those predicted by a classical model
using force of gravity proportional to r.

No, the GPS clock ticks at 1 second per second. The signal is
blueshifted when received on earth so the timebase used for the
transmission frequencies runs a bit slow to compensate. And, of course, classical force is proportional to 1/r^2. If you disagree, argue with
Newton.

Force of gravity was called potential by Laplace.[...]

Still more nonsense. The force of gravity is minus the gradient of the >>>> gravitational potential. They have NEVER been "equal" as you suppose.

The force of gravity increases the closer you get to the mass.
Proportional to r.

Nope. Proportional to 1/r^2.

No Volney. Force does not equal acceleration.

That's why I said "proportional to 1/r^2. The amount of proportion is
the mass m. The classical Newtonian formula is GMm/r^2.

That’s what potential really is.

Nope. Potential is proportionate to 1/r. It doesn't even have the units
of force or acceleration.

Only an idiot would think that if it takes more energy to lift an
object to a higher altitude that must mean that the force of gravity increases with altitude!!!

The force doesn't. The potential is different.

You REALLY need to learn basic physics. Your GUESSES AND FANTASIES are >>>> wrong.

If I’m wrong to model the effects of gravity with r...then why is
OK for Albert to use r of potential to model the effects of gravity?
Because Einstein wasn't showing the effect of Newtonian forces. He found
a new relationship that simplifies to the potential. It has nothing to
do with Newtonian force or acceleration at all, and it is not due to any
"the change of resonant frequencies of atoms" word salad.

In other words Einstein used r to model the effects of the force of gravity. And to cover it up..he changed the name of “force of gravity” to potential.

You lie. Why do you lie?

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Volney@21:1/5 to Lou on Fri Nov 17 00:56:24 2023
On 11/16/2023 6:11 AM, Lou wrote:

Looks like your idols ...Schwarzschild, Pound Rebka, Einstein etc etc
...all used r!!

And none of them EVER said that any force or acceleration was
proportional to 1/r. Too bad for you!

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Maciej Wozniak@21:1/5 to Volney on Thu Nov 16 23:09:40 2023
On Friday, 17 November 2023 at 01:56:37 UTC+1, Volney wrote:

No, the GPS clock ticks at 1 second per second.

Sutre it does. And it doesn't tick at 1(socend of a
relativistic idiot) per second.

You lie. Why do you lie?
Oh, stupid Mike, whoever is treating physics seriously
must end impudently lying.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Lou@21:1/5 to Volney on Fri Nov 17 01:01:45 2023
On Friday, 17 November 2023 at 05:56:29 UTC, Volney wrote:
On 11/16/2023 6:11 AM, Lou wrote:

Looks like your idols ...Schwarzschild, Pound Rebka, Einstein etc etc ...all used r!!
And none of them EVER said that any force or acceleration was
proportional to 1/r. Too bad for you!

You don’t seem to realise your own con.
Just because they called force of gravity by some other name in GR...
Doesn’t mean it’s not force of gravity.
Here I’ll do it. In classical physics I’ll call gravity force jujube. And jujube
uses r to model the strength of jujube at different altitudes.
There see! Call the force of gravity: jujube or metric or potential or gravitational time dilation or gravity well and you can pretend it’s not
the force of gravity
Fortunately only idiots will fall for that bit of snake oil.
Or maybe I should say...Unfortunately.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Lou@21:1/5 to Volney on Fri Nov 17 01:14:45 2023
On Friday, 17 November 2023 at 00:56:37 UTC, Volney wrote:
On 11/15/2023 4:40 AM, Lou wrote:
On Wednesday, 15 November 2023 at 04:59:55 UTC, Volney wrote:
On 11/14/2023 5:06 AM, Lou wrote:
On Monday, 13 November 2023 at 22:58:54 UTC, Tom Roberts wrote:
On 11/13/23 3:07 PM, Lou wrote:
a true classical model does not use acceleration to describe the
force of gravity.
Well, there is no such thing as a "true" model in physics -- physical >>>> models are valid or invalid, but we humans can never know whether they >>>> are "true". But we can know when they are false, and the "classical >>>> model" known as Newtonian mechanics (NM) is known to be false (but it is
often useful as an approximation).

Exactly....it’s false. Gravity force isn’t acceleration based on r^2.
No, force is the mass * acceleration. Newton's Second Law, you know.

So why are you pretending the force of gravity is measured as little g (m/s^2)?
Since when does force=acceleration?
Never. Force, according to Newton's second law, is F=mg since little g
is an acceleration (9.81 m/s^2 at sea level)

The acceleration due to gravity, according to Newton, is GM/r^2.
The force on a mass m due to gravity, according to Newton, is GMm/r^2. >>> Thats what I have been trying to get you lot to understand
this whole time.
Newton says you're completely wrong.

No. Newton said the force of gravity was proportional to r.
He called it a scalar field just to confuse idiots.
Lie.

Einstein realised this and used potential.
Not for the equivalent of gravitational acceleration or force!
Another lie. GR math is completely different from classical gravity. It
does come up with the same answers as classical gravity which is an approximation.

Does GR use Gm/r^2 anywhere in its calculations?
No
Because that's classical gravity.
Does GR use r?
Yes.
As an approximation.
Does GR use r to model tick rates at different altitudes?
Yes
Misleading. Perceived tick rates is not force.
Looks like if it’s OK for GR to ignore little g and use r...
Little g is from classical physics. GR states gravity isn't even a force
but approximating it as one is GMm/r^2 and for earth's surface gives the acceleration little g.
It’s OK for a classical model to do the same.
Anyways Newton did say force can be calculated with just r.
You lie.

The ridiculous part of your argument is that although do you admit
r^2 doesn’t work

r^2 does work in Newton's approximation.

No it doesn’t. You said yourself tick rates are proportionally to r.
The PERCEIVED tick rate due to blueshift, for weak field gravity, is proportional to 1/r. Classical force goes as GMm/r^2.
...you cannot bear to have anyone point out
that to make a classical theory work...one must use r of potential.

Classical Newtonian gravity uses 1/r^2 for acceleration and force. You
have been claiming classical gravity uses 1/r and it is COMPLETELY WRONG!

Said the relativist who knows that to correctly calculate the effects
of the force of gravity..one must use r
Completely wrong. GM/r does not even have units of acceleration or
force. It has units of meters^2/seconds^2 which doesn't even have a secondary unit name.

As far as the relationship between acceleration and force is concerned, >>>> here in the context of NM, what is used is Newton's second law:
F = m a
where F is the total force on an object, m is its mass, and a is its >>>> acceleration. This is implicitly relative to some inertial frame; F, m, >>>> and a are all invariant under Galilean transforms (change of coordinates
to a different inertial frame).

Word salad. R^2 isn’t F=ma. It is a ridiculous F=a. You know it is. >>> Dont pretend it isn’t.

Nope. Newton's Second Law relates force and acceleration. F=ma.

Yes. And notice force does not equal acceleration.
As you are trying to pretend a classical model says.
Nope. F=ma is not F=a.

Only relativists (or idiots) think in a classical model
acceleration=force.
NONSENSE! NOBODY thinks that. YOU are confused. The only "classical >>>> model" here is Newtonian mechanics, and in NM the second law applies: >>>> F = m a

No...YOU think acceleration=force. Not me.,You just spent your whole post trying
to pretend in a classical model that m/s^2= force.
It isn’t . ITS CALLED ACCELERATION. Einstein realised this.

You are (deliberately!) leaving out the mass in Newton's
force/acceleration relationship. Again, F=ma.

Im not. You are. How many times have you said the force of gravity
equals just acceleration. Millions of times.
No, never. The formula for classical force is GMm/r^2. I never said
force is acceleration, classical force is mass * acceleration. Again,
you are deliberately leaving out the mass and accusing me of what you did.

They do it to make sure a classical model can’t correctly predict >>>>> the change of resonant frequencies of atoms at different potentials. >>>> More nonsense. In Newtonian mechanics, time is universal and NM predicts
zero "time dilation" under any and all circumstances. NM is DIFFERENT >>>> from relativity, and wrong.

The only nonsense is your silly claim that I said a classical model
predicts time dilation. That’s GR.
A classical model says no time dilation is occuring. What you
are seeing is a harmonic oscillator (c-133) changing its resonant
frequency due to an external force of gravity.

How could it if the Newtonian force is proportional to 1/r^2 but the
perceived frequency difference goes as 1/r?

You have a very short attention span. I just told you even Newton
said the force of gravity was r.
No, he did not. You are lying when you say that. Newton explicitly
stated the force formula is GMm/r^2. Inversely proportional to r^2, not proportional to r.

He called it a scalar field
It’s only dishonest relativists who pretend that force=acceleration in a classical model.

And we have known harmonic oscillators do change frequencies
under external force for longer than GR has been around.
But we also know they don't go as 1/r.

Really.? That’s odd. Obviously you forgot.
GPS clock rates match those predicted by a classical model
using force of gravity proportional to r.
No, the GPS clock ticks at 1 second per second. The signal is
blueshifted when received on earth so the timebase used for the
transmission frequencies runs a bit slow to compensate. And, of course, classical force is proportional to 1/r^2. If you disagree, argue with Newton.

Under GR clocks are considered to all tick at 1s per second and the
change in tick rates proportional to r is caused by the force of gravity (Called potential in GR) red or blueshifting the imaginary photon.
But under a classical model harmonic oscillators natural frequencies
resonate at different rates at different altitudes proportional to r
due to external force (of gravity).
And you cannot prove that this is a false claim.
Anymore than I can prove the claim of GR is false.
Because the same observations confirm predictions by both
theories.
there is a force of gravity proportional to r. Newton called it scalar
field and many others including Laplace called it potential.And we
already have a precedence for harmonic oscillators natural f changing
due to external force going way back well before Albert.
But there is no precedence or previous observations confirming
time dilation or gravitational time dilation.

Force of gravity was called potential by Laplace.[...]

Still more nonsense. The force of gravity is minus the gradient of the >>>> gravitational potential. They have NEVER been "equal" as you suppose. >>>>
The force of gravity increases the closer you get to the mass.
Proportional to r.

Nope. Proportional to 1/r^2.

No Volney. Force does not equal acceleration.
That's why I said "proportional to 1/r^2. The amount of proportion is
the mass m. The classical Newtonian formula is GMm/r^2.

That’s what potential really is.

Nope. Potential is proportionate to 1/r. It doesn't even have the units >> of force or acceleration.

Only an idiot would think that if it takes more energy to lift an
object to a higher altitude that must mean that the force of gravity increases with altitude!!!
The force doesn't. The potential is different.

You REALLY need to learn basic physics. Your GUESSES AND FANTASIES are >>>> wrong.

If I’m wrong to model the effects of gravity with r...then why is
OK for Albert to use r of potential to model the effects of gravity?
Because Einstein wasn't showing the effect of Newtonian forces. He found >> a new relationship that simplifies to the potential. It has nothing to
do with Newtonian force or acceleration at all, and it is not due to any >> "the change of resonant frequencies of atoms" word salad.

In other words Einstein used r to model the effects of the force of gravity.
And to cover it up..he changed the name of “force of gravity” to potential.
You lie. Why do you lie?

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Tom Roberts@21:1/5 to Lou on Fri Nov 17 11:44:54 2023
On 11/17/23 3:01 AM, Lou wrote:
On Friday, 17 November 2023 at 05:56:29 UTC, Volney wrote:
On 11/16/2023 6:11 AM, Lou wrote:

Looks like your idols ...Schwarzschild, Pound Rebka, Einstein
etc etc ...all used r!!
And none of them EVER said that any force or acceleration was
proportional to 1/r. Too bad for you!

You don’t seem to realise your own con.

No "con". It's just that YOU DO NOT UNDERSTAND THE DIFFERENCE BETWEEN GRAVITATIONAL FORCE AND GRAVITATIONAL POTENTIAL. The error is YOURS.

Just because they called force of gravity by some other name in GR... Doesn’t mean it’s not force of gravity.

There is no force of gravity in GR. The closest thing to it is certain components of the geometrical connection (which must be multiplied by
the mass of the object to compute "gravitational force" on the object),
small objects only, weak fields only, speeds << c only.

Here I’ll do it. In classical physics I’ll call gravity force
jujube. And jujube uses r to model the strength of jujube at
different altitudes.

This is BLATANTLY WRONG. In classical physics, gravitational force is proportional to 1/r^2. You REALLY need to learn basic physics.

Fortunately only idiots will fall for that bit of snake oil.

Yes, YOU are trying to foist snake oil on your readers. STOP IT!

Tom Roberts

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Volney@21:1/5 to Lou on Sat Nov 18 02:12:02 2023
On 11/17/2023 4:01 AM, Lou wrote:
On Friday, 17 November 2023 at 05:56:29 UTC, Volney wrote:
On 11/16/2023 6:11 AM, Lou wrote:

Looks like your idols ...Schwarzschild, Pound Rebka, Einstein etc etc
...all used r!!
And none of them EVER said that any force or acceleration was
proportional to 1/r. Too bad for you!

You don’t seem to realise your own con.
Just because they called force of gravity by some other name in GR... Doesn’t mean it’s not force of gravity.
Here I’ll do it. In classical physics I’ll call gravity force jujube. And jujube
uses r to model the strength of jujube at different altitudes.

Those last two sentences contradict each other. Either jujube is a
force, or it varies proportionately to r. You can't have both, because classical force is GMm/r^2. But I'll play along and say jujube varies as r.

There see! Call the force of gravity: jujube or metric or potential or gravitational time dilation or gravity well and you can pretend it’s not the force of gravity

Since jujube and potential vary according to 1/r, they won't have units
of force, so neither one can be force.

Fortunately only idiots will fall for that bit of snake oil.

So why are you trying to use that snake oil as some sort of example?

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Lou@21:1/5 to Volney on Sat Nov 18 04:20:52 2023
On Saturday, 18 November 2023 at 07:12:06 UTC, Volney wrote:
On 11/17/2023 4:01 AM, Lou wrote:
On Friday, 17 November 2023 at 05:56:29 UTC, Volney wrote:
On 11/16/2023 6:11 AM, Lou wrote:

Looks like your idols ...Schwarzschild, Pound Rebka, Einstein etc etc >>> ...all used r!!
And none of them EVER said that any force or acceleration was
proportional to 1/r. Too bad for you!

You don’t seem to realise your own con.
Just because they called force of gravity by some other name in GR... Doesn’t mean it’s not force of gravity.
Here I’ll do it. In classical physics I’ll call gravity force jujube. And jujube
uses r to model the strength of jujube at different altitudes.
Those last two sentences contradict each other. Either jujube is a
force, or it varies proportionately to r. You can't have both, because classical force is GMm/r^2. But I'll play along and say jujube varies as r.
There see! Call the force of gravity: jujube or metric or potential or gravitational time dilation or gravity well and you can pretend it’s not the force of gravity
Since jujube and potential vary according to 1/r, they won't have units
of force, so neither one can be force.

So you think Force is defined by m/s^2?
I thought m/s^2 refers to acceleration.

Fortunately only idiots will fall for that bit of snake oil.
So why are you trying to use that snake oil as some sort of example?

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Volney@21:1/5 to Lou on Sat Nov 18 16:26:05 2023
On 11/18/2023 7:20 AM, Lou wrote:
On Saturday, 18 November 2023 at 07:12:06 UTC, Volney wrote:
On 11/17/2023 4:01 AM, Lou wrote:
On Friday, 17 November 2023 at 05:56:29 UTC, Volney wrote:
On 11/16/2023 6:11 AM, Lou wrote:

Looks like your idols ...Schwarzschild, Pound Rebka, Einstein etc etc >>>>> ...all used r!!
And none of them EVER said that any force or acceleration was
proportional to 1/r. Too bad for you!

You don’t seem to realise your own con.
Just because they called force of gravity by some other name in GR...
Doesn’t mean it’s not force of gravity.
Here I’ll do it. In classical physics I’ll call gravity force jujube. And jujube
uses r to model the strength of jujube at different altitudes.
Those last two sentences contradict each other. Either jujube is a
force, or it varies proportionately to r. You can't have both, because
classical force is GMm/r^2. But I'll play along and say jujube varies as r. >>> There see! Call the force of gravity: jujube or metric or potential or
gravitational time dilation or gravity well and you can pretend it’s not >>> the force of gravity
Since jujube and potential vary according to 1/r, they won't have units
of force, so neither one can be force.

So you think Force is defined by m/s^2?
I thought m/s^2 refers to acceleration.

YOU were the one who called "jujube" a force, not me! I'll repeat:
Newton's Second Law is force = mass * acceleration.

(and why do you insist something with units of m^2/s^2 is a force?)

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From JanPB@21:1/5 to Lou on Sat Nov 18 20:42:42 2023
On Tuesday, November 14, 2023 at 2:06:26 AM UTC-8, Lou wrote:
On Monday, 13 November 2023 at 22:58:54 UTC, Tom Roberts wrote:
On 11/13/23 3:07 PM, Lou wrote:
a true classical model does not use acceleration to describe the
force of gravity.
Well, there is no such thing as a "true" model in physics -- physical models are valid or invalid, but we humans can never know whether they
are "true". But we can know when they are false, and the "classical
model" known as Newtonian mechanics (NM) is known to be false (but it is often useful as an approximation).

Exactly....it’s false. Gravity force isn’t acceleration based on r^2.

Obviously force is not acceleration. What's your point?

Thats what I have been trying to get you lot to understand
this whole time.

You should have known already in high school that acceleration is not force.

Einstein realised this and used potential.

This is gobbledygook.

Laplace
realised that gravity force was potential.

More gobbledygook. You have no idea what you are talking about.

Newton knew this and
called gravity force a scalar field.

Irrelevant. Gobbledygook.

The rest of your post is another pyramid of nonsense, not worth further debate.

--
Jan

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From JanPB@21:1/5 to Lou on Sat Nov 18 20:50:52 2023
On Wednesday, November 15, 2023 at 1:53:49 AM UTC-8, Lou wrote:
On Wednesday, 15 November 2023 at 05:00:17 UTC, Tom Roberts wrote:
On 11/14/23 4:06 AM, Lou wrote:
On Monday, 13 November 2023 at 22:58:54 UTC, Tom Roberts wrote:
On 11/13/23 3:07 PM, Lou wrote:
a true classical model does not use acceleration to describe the
force of gravity.
Well, there is no such thing as a "true" model in physics --
physical models are valid or invalid, but we humans can never know
whether they are "true". But we can know when they are false, and
the "classical model" known as Newtonian mechanics (NM) is known to
be false (but it is often useful as an approximation).

Exactly....it’s false. Gravity force isn’t acceleration based on r^2.
Oh for goodness sake! I forgot how stupid and ignorant you are.

Said the guy who thinks force=acceleration
Even idiots aren’t that stupid.
NM is known to be false, by a few parts per trillion, such as the precession of the perihelion of mercury, or the "gravitational time dilation" exhibited by GPS satellites. Confusing 1/r and 1/r^2 would involve factors billions or trillions of times larger. GR, of course, fixes these errors in NM.
NM got lots of things wrong. He falsely put all the mass at the center of the
volume of the mass. But,.. If one spreads the mass out across the suns volume
or the galaxy disc then the classical predictions of preccession and
galaxy rotation curves can be made consistent with the observations
Thats what I have been trying to get you lot to understand this
whole time.
You have no hope of doing that, because a) YOU don't understand it, and
b) it is WRONG.
Einstein realised this and used potential.
Not for gravitational force, but rather for gravitational potential --
He certainly conned you. He called the force of gravity... potential,
Used r instead...and got the correct results.
Sad part is he just copied Newton’s scalar field.

Nonsense.

Newton himself knew the force of gravity was proportional to r.
He never meant the acceleration of little g to be interpreted by idiots as
a force.

It never was by anyone except you who invented the whole fabrication.

DUH! He then used approximation techniques to show that in the Newtonian approximation to GR the relevant component of the metric tensor involved the Newtonian gravitational potential.
Laplace realised that gravity force was potential.
Nope. He was not STUPID AND IGNORANT like you.

Said the guy who thinks force=acceleration.

No, Tom never said that (it contradicts even high school physics).

Again, in NM for gravity: force != potential;
rather, force = m * -grad potential, and since F = m a,

In GR neither gravitational force nor gravitational potential appears in the theory. The relative acceleration between small objects due to
gravity is expressed by the Raychaudhuri equation.
Exactly. Finally you woke up and smelt the coffee.
GR does not use little g’s r^2, as I have been trying to tell you guys. And he does use the r of potential. As I’ve been telling you lot.
Except to con his followers, he refers to potential using other
names like “ Raychaudhuri equation.”

There is no potential in GR.

But if it walks like a duck, quacks like a duck and looks like a duck
Then it’s a potential duck.

Again, there is no such thing as "gravitational potential" in GR.
Gravity is described by a metric which is a rank-2 tensor, it's
NOT a scalar (which is what potential is).

Pick a different hobby.

--
Jan

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From JanPB@21:1/5 to Lou on Sat Nov 18 20:54:08 2023
On Friday, November 17, 2023 at 1:01:49 AM UTC-8, Lou wrote:
On Friday, 17 November 2023 at 05:56:29 UTC, Volney wrote:
On 11/16/2023 6:11 AM, Lou wrote:

Looks like your idols ...Schwarzschild, Pound Rebka, Einstein etc etc ...all used r!!
And none of them EVER said that any force or acceleration was
proportional to 1/r. Too bad for you!
You don’t seem to realise your own con.
Just because they called force of gravity by some other name in GR...

What other name?

Doesn’t mean it’s not force of gravity.

Meaningless and irrelevant.

Here I’ll do it.

Yeah, sure. Let me decimate you:

In classical physics I’ll call gravity force jujube. And jujube
uses r to model the strength of jujube at different altitudes.
There see! Call the force of gravity: jujube or metric or potential or gravitational time dilation or gravity well and you can pretend it’s not the force of gravity

Newtonian gravitational potential and the spacetime metric of GR
are different things, both physically and mathematically.

Fortunately only idiots will fall for that bit of snake oil.
Or maybe I should say...Unfortunately.

You are just yet another arrogant ignoramus. Change your hobby,
this physics thing doesn't work for you.

--
Jan

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Lou@21:1/5 to JanPB on Sun Nov 19 06:12:17 2023
On Sunday, 19 November 2023 at 04:50:54 UTC, JanPB wrote:
On Wednesday, November 15, 2023 at 1:53:49 AM UTC-8, Lou wrote:
On Wednesday, 15 November 2023 at 05:00:17 UTC, Tom Roberts wrote:
On 11/14/23 4:06 AM, Lou wrote:
On Monday, 13 November 2023 at 22:58:54 UTC, Tom Roberts wrote:
On 11/13/23 3:07 PM, Lou wrote:
a true classical model does not use acceleration to describe the
force of gravity.
Well, there is no such thing as a "true" model in physics --
physical models are valid or invalid, but we humans can never know
whether they are "true". But we can know when they are false, and
the "classical model" known as Newtonian mechanics (NM) is known to >> be false (but it is often useful as an approximation).

Exactly....it’s false. Gravity force isn’t acceleration based on r^2.
Oh for goodness sake! I forgot how stupid and ignorant you are.

Said the guy who thinks force=acceleration
Even idiots aren’t that stupid.
NM is known to be false, by a few parts per trillion, such as the precession of the perihelion of mercury, or the "gravitational time dilation" exhibited by GPS satellites. Confusing 1/r and 1/r^2 would involve factors billions or trillions of times larger. GR, of course, fixes these errors in NM.
NM got lots of things wrong. He falsely put all the mass at the center of the
volume of the mass. But,.. If one spreads the mass out across the suns volume
or the galaxy disc then the classical predictions of preccession and galaxy rotation curves can be made consistent with the observations
Thats what I have been trying to get you lot to understand this
whole time.
You have no hope of doing that, because a) YOU don't understand it, and b) it is WRONG.
Einstein realised this and used potential.
Not for gravitational force, but rather for gravitational potential --
He certainly conned you. He called the force of gravity... potential,
Used r instead...and got the correct results.
Sad part is he just copied Newton’s scalar field.
Nonsense.
Newton himself knew the force of gravity was proportional to r.
He never meant the acceleration of little g to be interpreted by idiots as a force.
It never was by anyone except you who invented the whole fabrication.
DUH! He then used approximation techniques to show that in the Newtonian approximation to GR the relevant component of the metric tensor involved the Newtonian gravitational potential.
Laplace realised that gravity force was potential.
Nope. He was not STUPID AND IGNORANT like you.

Said the guy who thinks force=acceleration.
No, Tom never said that (it contradicts even high school physics).
Again, in NM for gravity: force != potential;
rather, force = m * -grad potential, and since F = m a,

In GR neither gravitational force nor gravitational potential appears in the theory. The relative acceleration between small objects due to gravity is expressed by the Raychaudhuri equation.
Exactly. Finally you woke up and smelt the coffee.
GR does not use little g’s r^2, as I have been trying to tell you guys. And he does use the r of potential. As I’ve been telling you lot.
Except to con his followers, he refers to potential using other
names like “ Raychaudhuri equation.”
There is no potential in GR.
But if it walks like a duck, quacks like a duck and looks like a duck
Then it’s a potential duck.
Again, there is no such thing as "gravitational potential" in GR.
Gravity is described by a metric which is a rank-2 tensor, it's
NOT a scalar (which is what potential is).

There is no potential in GR...because potential is called something
else. Curved spacetime metric etc.
All albert did was take potential...and give it a fancy new name.
Still the force of gravity. Only an idiot wouldn’t realise.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Lou@21:1/5 to JanPB on Sun Nov 19 06:14:30 2023
On Sunday, 19 November 2023 at 04:42:44 UTC, JanPB wrote:
On Tuesday, November 14, 2023 at 2:06:26 AM UTC-8, Lou wrote:
On Monday, 13 November 2023 at 22:58:54 UTC, Tom Roberts wrote:
On 11/13/23 3:07 PM, Lou wrote:
a true classical model does not use acceleration to describe the
force of gravity.
Well, there is no such thing as a "true" model in physics -- physical models are valid or invalid, but we humans can never know whether they are "true". But we can know when they are false, and the "classical model" known as Newtonian mechanics (NM) is known to be false (but it is often useful as an approximation).

Exactly....it’s false. Gravity force isn’t acceleration based on r^2.
Obviously force is not acceleration. What's your point?

Better tell Tom and Volney. They are desperate to con everyone into thinking Force =acceleration.

Thats what I have been trying to get you lot to understand
this whole time.
You should have known already in high school that acceleration is not force.
Einstein realised this and used potential.
This is gobbledygook.
Laplace
realised that gravity force was potential.
More gobbledygook. You have no idea what you are talking about.
Newton knew this and
called gravity force a scalar field.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Lou@21:1/5 to Volney on Sun Nov 19 06:08:57 2023
On Saturday, 18 November 2023 at 21:26:08 UTC, Volney wrote:
On 11/18/2023 7:20 AM, Lou wrote:
On Saturday, 18 November 2023 at 07:12:06 UTC, Volney wrote:
On 11/17/2023 4:01 AM, Lou wrote:
On Friday, 17 November 2023 at 05:56:29 UTC, Volney wrote:
On 11/16/2023 6:11 AM, Lou wrote:

Looks like your idols ...Schwarzschild, Pound Rebka, Einstein etc etc >>>>> ...all used r!!
And none of them EVER said that any force or acceleration was
proportional to 1/r. Too bad for you!

You don’t seem to realise your own con.
Just because they called force of gravity by some other name in GR... >>> Doesn’t mean it’s not force of gravity.
Here I’ll do it. In classical physics I’ll call gravity force jujube. And jujube
uses r to model the strength of jujube at different altitudes.
Those last two sentences contradict each other. Either jujube is a
force, or it varies proportionately to r. You can't have both, because
classical force is GMm/r^2. But I'll play along and say jujube varies as r.
There see! Call the force of gravity: jujube or metric or potential or >>> gravitational time dilation or gravity well and you can pretend it’s not
the force of gravity
Since jujube and potential vary according to 1/r, they won't have units >> of force, so neither one can be force.

So you think Force is defined by m/s^2?
I thought m/s^2 refers to acceleration.
YOU were the one who called "jujube" a force, not me! I'll repeat:

You conveniently forgot...When I said jujube meant force I
meant the force of gravity is modelled with r. As in potential. But
could and is called by any different names including jujube if you wished

GR calls it curved spacetime etc. Newton called it scalar field.
Laplace called it gravitational potential. And I sarcastically said you
could call it whatever you want including jujube.
But it doesn’t matter. Because whether or not it’s called gravitational potential,
jujube,curvature of spacetime, gravity well, time dilation, scalar field etc....
it’s always still...the force of gravity.
And it’s always proportional to r. Not r^2
In GR or in classical theory.

M/s^2 is acceleration. Force is not measured in m/s^2

Newton's Second Law is force = mass * acceleration.

(and why do you insist something with units of m^2/s^2 is a force?)

You’re nuts.
YOU are insisting that force is measured in units of m/s^2! Not me.
I’m trying to tell you that force ISNT measured in units of m/s^2 as
you try to pretend when you say little g, which uses m/s^2, is the force
of gravity.
How could it be?
Little g cannot be force because it’s measured in m/s^2. That’s called acceleration.
You think acceleration is force?
😂🤣

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Lou@21:1/5 to JanPB on Sun Nov 19 06:25:32 2023
On Sunday, 19 November 2023 at 04:54:10 UTC, JanPB wrote:
On Friday, November 17, 2023 at 1:01:49 AM UTC-8, Lou wrote:
On Friday, 17 November 2023 at 05:56:29 UTC, Volney wrote:
On 11/16/2023 6:11 AM, Lou wrote:

Looks like your idols ...Schwarzschild, Pound Rebka, Einstein etc etc ...all used r!!
And none of them EVER said that any force or acceleration was proportional to 1/r. Too bad for you!
You don’t seem to realise your own con.
Just because they called force of gravity by some other name in GR...
What other name?
Doesn’t mean it’s not force of gravity.
Meaningless and irrelevant.
Here I’ll do it.
Yeah, sure. Let me decimate you:
In classical physics I’ll call gravity force jujube. And jujube
uses r to model the strength of jujube at different altitudes.
There see! Call the force of gravity: jujube or metric or potential or gravitational time dilation or gravity well and you can pretend it’s not the force of gravity
Newtonian gravitational potential and the spacetime metric of GR
are different things, both physically and mathematically.

Yes Jan. whatever you say.
I drop a rock to the ground. Something pulls it down.
Albert said this pull was because of his “spacetime metric”
and was modelled with r
Classical physics calls this pull “gravitational potential.” and models
it with r
Ahh! So Jan thinks that because Albert called gravitational
potential “spacetime metric” it must be a completely seperate *physically* from gravitational potential.😂🤣
Tell me. In your nutty universe when you drop a rock
is it pulled to the ground by 2 seperate effects ? Potential and metric?
Or are they both just different names for the same thing and
you are just an idiot?

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Volney@21:1/5 to Lou on Sun Nov 19 11:02:12 2023
On 11/19/2023 9:08 AM, Lou wrote:
On Saturday, 18 November 2023 at 21:26:08 UTC, Volney wrote:
On 11/18/2023 7:20 AM, Lou wrote:
On Saturday, 18 November 2023 at 07:12:06 UTC, Volney wrote:
On 11/17/2023 4:01 AM, Lou wrote:
On Friday, 17 November 2023 at 05:56:29 UTC, Volney wrote:
On 11/16/2023 6:11 AM, Lou wrote:

Looks like your idols ...Schwarzschild, Pound Rebka, Einstein etc etc >>>>>>> ...all used r!!
And none of them EVER said that any force or acceleration was
proportional to 1/r. Too bad for you!

You don’t seem to realise your own con.
Just because they called force of gravity by some other name in GR... >>>>> Doesn’t mean it’s not force of gravity.
Here I’ll do it. In classical physics I’ll call gravity force jujube. And jujube
uses r to model the strength of jujube at different altitudes.
Those last two sentences contradict each other. Either jujube is a
force, or it varies proportionately to r. You can't have both, because >>>> classical force is GMm/r^2. But I'll play along and say jujube varies as r.
There see! Call the force of gravity: jujube or metric or potential or >>>>> gravitational time dilation or gravity well and you can pretend it’s not
the force of gravity
Since jujube and potential vary according to 1/r, they won't have units >>>> of force, so neither one can be force.

So you think Force is defined by m/s^2?
I thought m/s^2 refers to acceleration.

YOU were the one who called "jujube" a force, not me! I'll repeat:

You conveniently forgot...When I said jujube meant force I
meant the force of gravity is modelled with r.

You contradict yourself. Gravity is modeled inversely proportional to
r^2, not r. Make up your mind; jujube is potential (GMm/r) or force
(GMm/r^2). It can't be both.

As in potential. But
could and is called by any different names including jujube if you wished

You can use any name you want, but if it's modeled as potential, it is
not a force. Full stop.

GR calls it curved spacetime etc. Newton called it scalar field.
Laplace called it gravitational potential. And I sarcastically said you
could call it whatever you want including jujube.

None of these are force.

But it doesn’t matter. Because whether or not it’s called gravitational potential,
jujube,curvature of spacetime, gravity well, time dilation, scalar field etc....
it’s always still...the force of gravity.
And it’s always proportional to r. Not r^2
In GR or in classical theory.

And once again, you counter Newton, who states classical force =
GMm/r^2. Why do you counter Newton's classical force?

M/s^2 is acceleration. Force is not measured in m/s^2

Nor is force m^2/s^2. But again, Newton's second law is F=ma.

Newton's Second Law is force = mass * acceleration.

(and why do you insist something with units of m^2/s^2 is a force?)

You’re nuts.

I'm not the one claiming force is m^2/s^2!

YOU are insisting that force is measured in units of m/s^2! Not me.

No I keep telling you F=ma, classical force is kg*m/s^2, force = GMm/r^2
and you frequently snip that.

I’m trying to tell you that force ISNT measured in units of m/s^2 as
you try to pretend when you say little g, which uses m/s^2, is the force
of gravity.

No as I keep reminding you Newton's second law is F=ma, so force of
gravity is F=mg.

How could it be?

It isn't. You are simply very confused.

Little g cannot be force because it’s measured in m/s^2. That’s called acceleration.
You think acceleration is force?
😂🤣

I'm the one who keeps reminding you of Newton's second law, force = mass
* acceleration, but you like snipping that.

And you are the one who thinks potential is force. The units are wrong. Potential has units of m^2/s^2 which isn't force.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Lou@21:1/5 to Volney on Sun Nov 19 08:58:00 2023
On Sunday, 19 November 2023 at 16:02:18 UTC, Volney wrote:
On 11/19/2023 9:08 AM, Lou wrote:
On Saturday, 18 November 2023 at 21:26:08 UTC, Volney wrote:
On 11/18/2023 7:20 AM, Lou wrote:
On Saturday, 18 November 2023 at 07:12:06 UTC, Volney wrote:
On 11/17/2023 4:01 AM, Lou wrote:
On Friday, 17 November 2023 at 05:56:29 UTC, Volney wrote:
On 11/16/2023 6:11 AM, Lou wrote:

Looks like your idols ...Schwarzschild, Pound Rebka, Einstein etc etc
...all used r!!
And none of them EVER said that any force or acceleration was
proportional to 1/r. Too bad for you!

You don’t seem to realise your own con.
Just because they called force of gravity by some other name in GR... >>>>> Doesn’t mean it’s not force of gravity.
Here I’ll do it. In classical physics I’ll call gravity force jujube. And jujube
uses r to model the strength of jujube at different altitudes.
Those last two sentences contradict each other. Either jujube is a
force, or it varies proportionately to r. You can't have both, because >>>> classical force is GMm/r^2. But I'll play along and say jujube varies as r.
There see! Call the force of gravity: jujube or metric or potential or >>>>> gravitational time dilation or gravity well and you can pretend it’s not
the force of gravity
Since jujube and potential vary according to 1/r, they won't have units >>>> of force, so neither one can be force.

So you think Force is defined by m/s^2?
I thought m/s^2 refers to acceleration.

YOU were the one who called "jujube" a force, not me! I'll repeat:

You conveniently forgot...When I said jujube meant force I
meant the force of gravity is modelled with r.
You contradict yourself. Gravity is modeled inversely proportional to
r^2, not r. Make up your mind; jujube is potential (GMm/r) or force (GMm/r^2). It can't be both.
As in potential. But
could and is called by any different names including jujube if you wished
You can use any name you want, but if it's modeled as potential, it is
not a force. Full stop.

GR calls it curved spacetime etc. Newton called it scalar field.
Laplace called it gravitational potential. And I sarcastically said you could call it whatever you want including jujube.
None of these are force.
But it doesn’t matter. Because whether or not it’s called gravitational potential,
jujube,curvature of spacetime, gravity well, time dilation, scalar field etc....
it’s always still...the force of gravity.
And it’s always proportional to r. Not r^2
In GR or in classical theory.
And once again, you counter Newton, who states classical force =
GMm/r^2. Why do you counter Newton's classical force?

M/s^2 is acceleration. Force is not measured in m/s^2
Nor is force m^2/s^2. But again, Newton's second law is F=ma.

Newton's Second Law is force = mass * acceleration.

(and why do you insist something with units of m^2/s^2 is a force?)

You’re nuts.
I'm not the one claiming force is m^2/s^2!
YOU are insisting that force is measured in units of m/s^2! Not me.
No I keep telling you F=ma, classical force is kg*m/s^2, force = GMm/r^2
and you frequently snip that.
I’m trying to tell you that force ISNT measured in units of m/s^2 as
you try to pretend when you say little g, which uses m/s^2, is the force of gravity.
No as I keep reminding you Newton's second law is F=ma, so force of
gravity is F=mg.

How could it be?

It isn't. You are simply very confused.
Little g cannot be force because it’s measured in m/s^2. That’s called acceleration.
You think acceleration is force?
😂🤣
I'm the one who keeps reminding you of Newton's second law, force = mass
* acceleration, but you like snipping that.

Another lie from Baloney.
I deliberately left your f=ma in your post above. Just to show everyone how when
you are asked to proved your claim that force is measured in m/s^2
you wriggle out of it, change the subject and say f=ma.
But f=ma isn’t what’s being discussed.
It’s the m/s^2 of little g that you incorrectly insist is force.
So answer the question...is force measured in m/s^2 as you
claim the force of gravity is ?
Oops Volney, I bet you can’t answer that one either so you better change the subject
AGAIN for about the millionth time.

Anyways you also forget the whole reason why this discussion started.
You claimed the force of gravity could not be modelled using r in a classical model. Despite the fact that you were unable to refute the fact that the classical
model predicts the area of the gravity shadow falls off proportional with r.
So to get out of admitting you can’t refute this you changed the subject and pretended that force is measured in m/s^2.
A ridiculous claim which you know is ridiculous.
So when you were asked to prove that force is measured in m/s^2 you changed
the subject again grabbed at straws and said that F=ma.
And when it was pointed out to you that F=ma isn’t measured in m/s^2 and
you knew this is true, you lied again, and changed the subject *again* and claimed that I was saying that force is measured in units
of m^2/s^2. Which I wasn’t.
Another lie and another subject change for desperate Volney
All done to get you off the hook for not being able to refute the fact
that a classical shadow gravity model has the force of gravity proportional
to r.
Amazing pack of lies, especially considering you know that Albert assumed
the force of gravity falls off with r and not Newton’s r^2.
Just because Albert called a duck a flying spacetime metric goblin..doesn’t mean
it isn’t a duck anymore.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Paul Alsing@21:1/5 to Lou on Sun Nov 19 09:57:45 2023
On Sunday, November 19, 2023 at 8:58:03 AM UTC-8, Lou wrote:
On Sunday, 19 November 2023 at 16:02:18 UTC, Volney wrote:
On 11/19/2023 9:08 AM, Lou wrote:
On Saturday, 18 November 2023 at 21:26:08 UTC, Volney wrote:
On 11/18/2023 7:20 AM, Lou wrote:
On Saturday, 18 November 2023 at 07:12:06 UTC, Volney wrote:
On 11/17/2023 4:01 AM, Lou wrote:
On Friday, 17 November 2023 at 05:56:29 UTC, Volney wrote:
On 11/16/2023 6:11 AM, Lou wrote:

Looks like your idols ...Schwarzschild, Pound Rebka, Einstein etc etc
...all used r!!
And none of them EVER said that any force or acceleration was >>>>>> proportional to 1/r. Too bad for you!

You don’t seem to realise your own con.
Just because they called force of gravity by some other name in GR...
Doesn’t mean it’s not force of gravity.
Here I’ll do it. In classical physics I’ll call gravity force jujube. And jujube
uses r to model the strength of jujube at different altitudes.
Those last two sentences contradict each other. Either jujube is a >>>> force, or it varies proportionately to r. You can't have both, because
classical force is GMm/r^2. But I'll play along and say jujube varies as r.
There see! Call the force of gravity: jujube or metric or potential or
gravitational time dilation or gravity well and you can pretend it’s not
the force of gravity
Since jujube and potential vary according to 1/r, they won't have units
of force, so neither one can be force.

So you think Force is defined by m/s^2?
I thought m/s^2 refers to acceleration.

YOU were the one who called "jujube" a force, not me! I'll repeat:

You conveniently forgot...When I said jujube meant force I
meant the force of gravity is modelled with r.
You contradict yourself. Gravity is modeled inversely proportional to
r^2, not r. Make up your mind; jujube is potential (GMm/r) or force (GMm/r^2). It can't be both.
As in potential. But
could and is called by any different names including jujube if you wished
You can use any name you want, but if it's modeled as potential, it is
not a force. Full stop.

GR calls it curved spacetime etc. Newton called it scalar field.
Laplace called it gravitational potential. And I sarcastically said you could call it whatever you want including jujube.
None of these are force.
But it doesn’t matter. Because whether or not it’s called gravitational potential,
jujube,curvature of spacetime, gravity well, time dilation, scalar field etc....
it’s always still...the force of gravity.
And it’s always proportional to r. Not r^2
In GR or in classical theory.
And once again, you counter Newton, who states classical force =
GMm/r^2. Why do you counter Newton's classical force?

M/s^2 is acceleration. Force is not measured in m/s^2
Nor is force m^2/s^2. But again, Newton's second law is F=ma.

Newton's Second Law is force = mass * acceleration.

(and why do you insist something with units of m^2/s^2 is a force?)

You’re nuts.
I'm not the one claiming force is m^2/s^2!
YOU are insisting that force is measured in units of m/s^2! Not me.
No I keep telling you F=ma, classical force is kg*m/s^2, force = GMm/r^2 and you frequently snip that.
I’m trying to tell you that force ISNT measured in units of m/s^2 as you try to pretend when you say little g, which uses m/s^2, is the force of gravity.
No as I keep reminding you Newton's second law is F=ma, so force of gravity is F=mg.

How could it be?

It isn't. You are simply very confused.
Little g cannot be force because it’s measured in m/s^2. That’s called
acceleration.
You think acceleration is force?
😂🤣
I'm the one who keeps reminding you of Newton's second law, force = mass
* acceleration, but you like snipping that.

Another lie from Baloney.
I deliberately left your f=ma in your post above. Just to show everyone how when
you are asked to proved your claim that force is measured in m/s^2
you wriggle out of it, change the subject and say f=ma.
But f=ma isn’t what’s being discussed.
It’s the m/s^2 of little g that you incorrectly insist is force.
So answer the question...is force measured in m/s^2 as you
claim the force of gravity is ?
Oops Volney, I bet you can’t answer that one either so you better change the subject
AGAIN for about the millionth time.

Anyways you also forget the whole reason why this discussion started.
You claimed the force of gravity could not be modelled using r in a classical
model. Despite the fact that you were unable to refute the fact that the classical
model predicts the area of the gravity shadow falls off proportional with r. So to get out of admitting you can’t refute this you changed the subject and
pretended that force is measured in m/s^2.
A ridiculous claim which you know is ridiculous.
So when you were asked to prove that force is measured in m/s^2 you changed the subject again grabbed at straws and said that F=ma.
And when it was pointed out to you that F=ma isn’t measured in m/s^2 and you knew this is true, you lied again, and changed the subject *again* and claimed that I was saying that force is measured in units
of m^2/s^2. Which I wasn’t.
Another lie and another subject change for desperate Volney
All done to get you off the hook for not being able to refute the fact
that a classical shadow gravity model has the force of gravity proportional to r.
Amazing pack of lies, especially considering you know that Albert assumed the force of gravity falls off with r and not Newton’s r^2.
Just because Albert called a duck a flying spacetime metric goblin..doesn’t mean
it isn’t a duck anymore.

When it is claimed that f=ma, you don't seem to understand that the units are therefore expressed as kilograms-m/s^2, otherwise known as a Newton of force.

Just read pretty much any beginning physics book for verification!

Perhaps you have a learning disabiliy?

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From carl eto@21:1/5 to All on Sun Nov 19 10:18:38 2023
Einstein's general relativity (1917) is based on Maxwell's equations that are derived using Faraday's induction effect and an ether that does not exist. Plus, Faraday's induction effect is not gravitational. Your ineptitude has resulted in the US debt of
$33T. The deficit in 2023 was$1.7B and the payment on the debt interest was $600B. Next year the projected interest will be$1.2T. In 2028, four years from now, the debt will be $40T and in 2032 the debt will be$60T where the deficit will be 50% of the
budget.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From RichD@21:1/5 to Lou on Sat Nov 25 13:31:42 2023
On November 19, Lou wrote:
I drop a rock to the ground. Something pulls it down.
Albert said this pull was because of his “spacetime metric”
and was modelled with r
Classical physics calls this pull “gravitational potential.” and models it with r
Ahh! So Jan thinks that because Albert called gravitational
potential “spacetime metric” it must be a completely seperate *physically*
from gravitational potential.
Tell me. In your nutty universe when you drop a rock
is it pulled to the ground by 2 seperate effects? Potential and metric?
Or are they both just different names for the same thing and
you are just an idiot?

Given an infinite plane, with mass density W kg/m²

What is the gravitational field at a height h, above the plane?

--
Rich

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From RichD@21:1/5 to J. J. Lodder on Sat Nov 25 13:48:12 2023
On November 10, J. J. Lodder wrote:
General relativity predicts that all freely falling clocks
will run at their own inherent rate.
It also predicts that clocks at different places,
and with different velocities will be seen to run at different rates,
-when compared with each other-.
It also predicts that accelerations do not affect clock rates,
so the results can be extended to non-inertial clocks,
such as clocks at rest at different altitudes on Earth.

According to general relativity all clock effects are purely kinematic,
so derivable from the metric tensor.
Doing the sums for weak fields results in velocity effects
being given by Lorentz factors, and 'gravitational' effects
being given by the variations in Newtonian potential.
So far, so good, and in agreement with experimental results.

They hold that obverved clock effects must be due to 'gravity'
affecting the workings of the clock, somehow.
In other words, they ascribe the observed clock effects
to physical causes, 'gravity' affecting the workings of clocks,
rather than to intrinsic space-time effects.

Fortunately it is easy to settle the point by experiment.
GR predicts that all clocks on the rotating geoid on Earth
must run at the same rate, when compared with each other.
Experiment bears this out, to accuracies approaching 10^-15.
OTOH the force of gravity, as measured by 'small' g,
the acceleration of gravity, varies markedly over the geoid.
(by about 0.5%, between the poles and the equator)

If (the force of) 'gravity' influenced the rate of the clocks
there should be an effect of geographical latitude
on the rate of clocks.
This is not observed to be the case, so this idea stands falsified.

Is the center of the geoid defined at the center of mass of the
planet, or at the geometric center? Center of mass, I presume.

This question becomes pertinent when you talk about the effect
of latitude.

--
Rich

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From J. J. Lodder@21:1/5 to RichD on Sun Nov 26 10:33:10 2023
RichD <r_delaney2001@yahoo.com> wrote:

On November 10, J. J. Lodder wrote:
General relativity predicts that all freely falling clocks
will run at their own inherent rate.
It also predicts that clocks at different places,
and with different velocities will be seen to run at different rates,
-when compared with each other-.
It also predicts that accelerations do not affect clock rates,
so the results can be extended to non-inertial clocks,
such as clocks at rest at different altitudes on Earth.

According to general relativity all clock effects are purely kinematic,
so derivable from the metric tensor.
Doing the sums for weak fields results in velocity effects
being given by Lorentz factors, and 'gravitational' effects
being given by the variations in Newtonian potential.
So far, so good, and in agreement with experimental results.

They hold that obverved clock effects must be due to 'gravity'
affecting the workings of the clock, somehow.
In other words, they ascribe the observed clock effects
to physical causes, 'gravity' affecting the workings of clocks,
rather than to intrinsic space-time effects.

Fortunately it is easy to settle the point by experiment.
GR predicts that all clocks on the rotating geoid on Earth
must run at the same rate, when compared with each other.
Experiment bears this out, to accuracies approaching 10^-15.
OTOH the force of gravity, as measured by 'small' g,
the acceleration of gravity, varies markedly over the geoid.
(by about 0.5%, between the poles and the equator)

If (the force of) 'gravity' influenced the rate of the clocks
there should be an effect of geographical latitude
on the rate of clocks.
This is not observed to be the case, so this idea stands falsified.

Is the center of the geoid defined at the center of mass of the
planet, or at the geometric center? Center of mass, I presume.

There is nothing to 'define'. The geoid is irregularly shaped,
and it doesn't have an obvious geometric centre. (unlike an ellipsoid)
Otoh there is an obvious physical centre: the point at which the
Newtonian potential is lowest.
It is the point at which the acceleration of gravity is exactly zero.

This question becomes pertinent when you talk about the effect
of latitude.

Depends on what you want 'latitude' to mean.
(geometrical or physical)
But those tiny differences between geoid and ellipsoid
are the next thing to worry about.
For the present it suffices to point out that our 'Lou'
is completely and hopelessly and incurably wrong,

Jan

PS For the kiddies: What a GPS unit does for you is to calculate
your position in 3-D space and time using a suitable coordinate system.
Next it projects your position onto the conventional WGS84 ellipsoid
to give you WGS84 latitude, longitude, and altitude.
(because that is what your printed map shows)
Finally it uses a rather crude built in mathematical geoid model
to correct the WGS84 height to geoid height.
It wouldn't do for the thing to tell you consistently
that you are ten meters under water, or floating in the air,
when you are actually standing on a beach.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From RichD@21:1/5 to J. J. Lodder on Sun Nov 26 12:49:04 2023
On November 26, J. J. Lodder wrote:
rate.
It also predicts that accelerations do not affect clock rates,
so the results can be extended to non-inertial clocks,
such as clocks at rest at different altitudes on Earth.
According to general relativity all clock effects are purely kinematic,
so derivable from the metric tensor.
Doing the sums for weak fields results in velocity effects
being given by Lorentz factors, and 'gravitational' effects
being given by the variations in Newtonian potential.

They hold that obverved clock effects must be due to 'gravity'
affecting the workings of the clock, somehow.
In other words, they ascribe the observed clock effects
to physical causes, 'gravity' affecting the workings of clocks,
rather than to intrinsic space-time effects.

Fortunately it is easy to settle the point by experiment.
GR predicts that all clocks on the rotating geoid on Earth
must run at the same rate, when compared with each other.
Experiment bears this out, to accuracies approaching 10^-15.
OTOH the force of gravity, as measured by 'small' g,
the acceleration of gravity, varies markedly over the geoid.
(by about 0.5%, between the poles and the equator)
If (the force of) 'gravity' influenced the rate of the clocks
there should be an effect of geographical latitude
on the rate of clocks.
This is not observed to be the case, so this idea stands falsified.

Is the center of the geoid defined at the center of mass of the
planet, or at the geometric center? Center of mass, I presume.

There is nothing to 'define'. The geoid is irregularly shaped,
and it doesn't have an obvious geometric centre. (unlike an ellipsoid)
Otoh there is an obvious physical centre: the point at which the
Newtonian potential is lowest.

That would be the center of mass.

The geoid is defined a set of points at the same potential?
All at the same distance from the center?

This question becomes pertinent when you talk about the effect
of latitude.

Depends on what you want 'latitude' to mean.
(geometrical or physical)
But those tiny differences between geoid and ellipsoid
are the next thing to worry about.

I don't know the difference.

--
Rich

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Tom Roberts@21:1/5 to RichD on Sun Nov 26 18:30:02 2023
On 11/26/23 2:49 PM, RichD wrote:
On November 26, J. J. Lodder wrote:
there is an obvious physical centre: the point at which the
Newtonian potential is lowest.

That would be the center of mass.

Not quite. If the earth were a perfect ellipsoid with constant density,
then the center of mass would correspond to the geometric center. But it obviously is not a perfect ellipsoid, and clearly does not have constant density.

Note the center of mass of a collection of masses is not necessarily the
point with the lowest Newtonian gravitational potential.

The geoid is defined a set of points at the same potential?

To a physicist, the geoid is the locus of all points on earth that have
the same metric (considering just the earth). For all practical purposes
this is the same as having the same Newtonian gravitational potential.

All at the same distance from the center?

No. Because the earth is not a perfect ellipsoid, and does not have
constant density.

Tom Roberts

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From J. J. Lodder@21:1/5 to RichD on Mon Nov 27 12:54:57 2023
RichD <r_delaney2001@yahoo.com> wrote:

On November 26, J. J. Lodder wrote:
rate.
It also predicts that accelerations do not affect clock rates,
so the results can be extended to non-inertial clocks,
such as clocks at rest at different altitudes on Earth.
According to general relativity all clock effects are purely kinematic, >>> so derivable from the metric tensor.
Doing the sums for weak fields results in velocity effects
being given by Lorentz factors, and 'gravitational' effects
being given by the variations in Newtonian potential.

They hold that obverved clock effects must be due to 'gravity'
affecting the workings of the clock, somehow.
In other words, they ascribe the observed clock effects
to physical causes, 'gravity' affecting the workings of clocks,
rather than to intrinsic space-time effects.

Fortunately it is easy to settle the point by experiment.
GR predicts that all clocks on the rotating geoid on Earth
must run at the same rate, when compared with each other.
Experiment bears this out, to accuracies approaching 10^-15.
OTOH the force of gravity, as measured by 'small' g,
the acceleration of gravity, varies markedly over the geoid.
(by about 0.5%, between the poles and the equator)
If (the force of) 'gravity' influenced the rate of the clocks
there should be an effect of geographical latitude
on the rate of clocks.
This is not observed to be the case, so this idea stands falsified.

Is the center of the geoid defined at the center of mass of the
planet, or at the geometric center? Center of mass, I presume.

There is nothing to 'define'. The geoid is irregularly shaped,
and it doesn't have an obvious geometric centre. (unlike an ellipsoid)
Otoh there is an obvious physical centre: the point at which the
Newtonian potential is lowest.

That would be the center of mass.

No, but close.

The geoid is defined a set of points at the same potential?

Yes, in the Newtonian approximation.
(if the Newtonian approximation isn't valid
there is no such thing as a well-defined potential)

All at the same distance from the center?

Certainy not. The geoid is almost an ellipsoid.
(hence the choice of the metric founding fathers to measure a meridian)

This question becomes pertinent when you talk about the effect
of latitude.

Depends on what you want 'latitude' to mean.
(geometrical or physical)
But those tiny differences between geoid and ellipsoid
are the next thing to worry about.

I don't know the difference.

Latitude is conventional. It is a human construct.
To define it you can do two things.
1) Geometrical. Approximate the geoid by a 'best fitting' ellipsoid,
for example WGS84. (but there are dozens of other datums)
Find the lines of constant latitude by differentiating the ellipsoid.
2) Physical. Get a local level surface, from a water surface, or pool of mercury. (or use a plumb line to find the vertical)
Get a telescope, and measure the elevation of the celestial pole
wrt to the local level, and call that the latitude.
The results would coincide if the Earth were a perfect ellipsoid.

Jan

--
"They are merely conventional signs!" (The Crew)

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From RichD@21:1/5 to Tom Roberts on Mon Nov 27 12:58:50 2023
On November 26, Tom Roberts wrote:
there is an obvious physical centre: the point at which the
Newtonian potential is lowest.

That would be the center of mass.

Note the center of mass of a collection of masses is not necessarily the point with the lowest Newtonian gravitational potential.

?
Isn't that a contradiction, if the center of mass doesn't coincide
with zero potential?

The geoid is defined a set of points at the same potential?

To a physicist, the geoid is the locus of all points on earth that have
the same metric (considering just the earth). For all practical purposes
this is the same as having the same Newtonian gravitational potential.

How is the metric measured? How does one determine empirically if
two separated points share the same metric? That is, without measuring
any clock rate, which is the subject under discussion.

In Newtonian mechanics, one simply measures the gravitational gradient,
easy enough.

--
Rich

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From J. J. Lodder@21:1/5 to RichD on Tue Nov 28 23:09:17 2023
RichD <r_delaney2001@yahoo.com> wrote:

On November 26, Tom Roberts wrote:
there is an obvious physical centre: the point at which the
Newtonian potential is lowest.

That would be the center of mass.

Note the center of mass of a collection of masses is not necessarily the point with the lowest Newtonian gravitational potential.

?
Isn't that a contradiction, if the center of mass doesn't coincide
with zero potential?

Eh, zero potential is out at infinity.

The geoid is defined a set of points at the same potential?

To a physicist, the geoid is the locus of all points on earth that have
the same metric (considering just the earth). For all practical purposes this is the same as having the same Newtonian gravitational potential.

How is the metric measured? How does one determine empirically if
two separated points share the same metric? That is, without measuring
any clock rate, which is the subject under discussion.

In Newtonian mechanics, one simply measures the gravitational gradient,
easy enough.

Yes, 'simple', 'easy enough'. Now go and do it, accurately.
(say to one cm, between BIPM and NIST)

We'll soon reach the point where the potential difference between
distant points can only be measured accurately by comparing clock rates.
At present acurate frequency transport (to 10^-18) is not possible yet,
over long distances, but that is actively being worked on.
Distances of many hundreds of kilometers have been reached already,

Jan

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Ross Finlayson@21:1/5 to J. J. Lodder on Tue Nov 28 17:45:18 2023
On Tuesday, November 28, 2023 at 2:09:21 PM UTC-8, J. J. Lodder wrote:
RichD <r_dela...@yahoo.com> wrote:

On November 26, Tom Roberts wrote:
there is an obvious physical centre: the point at which the
Newtonian potential is lowest.

That would be the center of mass.

Note the center of mass of a collection of masses is not necessarily the point with the lowest Newtonian gravitational potential.

?
Isn't that a contradiction, if the center of mass doesn't coincide
with zero potential?
Eh, zero potential is out at infinity.
The geoid is defined a set of points at the same potential?

To a physicist, the geoid is the locus of all points on earth that have the same metric (considering just the earth). For all practical purposes this is the same as having the same Newtonian gravitational potential.

How is the metric measured? How does one determine empirically if
two separated points share the same metric? That is, without measuring
any clock rate, which is the subject under discussion.

In Newtonian mechanics, one simply measures the gravitational gradient, easy enough.
Yes, 'simple', 'easy enough'. Now go and do it, accurately.
(say to one cm, between BIPM and NIST)

We'll soon reach the point where the potential difference between
distant points can only be measured accurately by comparing clock rates.
At present acurate frequency transport (to 10^-18) is not possible yet,
over long distances, but that is actively being worked on.
Distances of many hundreds of kilometers have been reached already,

Jan

Channels Maupertuis, ....

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Maciej Wozniak@21:1/5 to J. J. Lodder on Tue Nov 28 23:03:29 2023
On Tuesday, 28 November 2023 at 23:09:21 UTC+1, J. J. Lodder wrote:
RichD <r_dela...@yahoo.com> wrote:

On November 26, Tom Roberts wrote:
there is an obvious physical centre: the point at which the
Newtonian potential is lowest.

That would be the center of mass.

Note the center of mass of a collection of masses is not necessarily the point with the lowest Newtonian gravitational potential.

?
Isn't that a contradiction, if the center of mass doesn't coincide
with zero potential?
Eh, zero potential is out at infinity.
The geoid is defined a set of points at the same potential?

To a physicist, the geoid is the locus of all points on earth that have the same metric (considering just the earth). For all practical purposes this is the same as having the same Newtonian gravitational potential.

How is the metric measured? How does one determine empirically if
two separated points share the same metric? That is, without measuring
any clock rate, which is the subject under discussion.

In Newtonian mechanics, one simply measures the gravitational gradient, easy enough.
Yes, 'simple', 'easy enough'. Now go and do it, accurately.
(say to one cm, between BIPM and NIST)

We'll soon reach the point where the potential difference between
distant points can only be measured accurately by comparing clock rates.

Your fellow idiot Tom could explain you, that according
to your moronic religion clocks run always at the same rate.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From J. J. Lodder@21:1/5 to RichD on Sat Dec 2 23:28:05 2023
RichD <r_delaney2001@yahoo.com> wrote:

On November 28, J. J. Lodder wrote:
there is an obvious physical centre: the point at which the
Newtonian potential is lowest.

That would be the center of mass.

Note the center of mass of a collection of masses is not necessarily the >>> point with the lowest Newtonian gravitational potential.

Isn't that a contradiction, if the center of mass doesn't coincide
with zero potential?

Eh, zero potential is out at infinity.

That's an arbitrary number, no objective significance.
Potential refers to energy. Place a test mass at a point, release,
watch it fly.

The point where it remains motionless is the lowest
potential.

The geoid is defined a set of points at the same potential?

To a physicist, the geoid is the locus of all points on earth that have >>> the same metric (considering just the earth). For all practical purposes >>> this is the same as having the same Newtonian gravitational potential.

How is the metric measured? How does one determine empirically if
two separated points share the same metric? That is, without measuring
any clock rate, which is the subject under discussion.

In Newtonian mechanics, one simply measures the gravitational gradient,
easy enough.

Yes, 'simple', 'easy enough'. Now go and do it, accurately.
(say to one cm, between BIPM and NIST)

Observe the test mass acceleration, and use the Lorentz
momentum formula. Calculate the g force, assuming Newton's
model. The Newtonian potential follows directly, if you know
the distribution of mass.

That is not the claim.
It is that relative rates of clocks at different points
depend on the potential difference between those points.

The claim is that in general relativity, clock rate depends only on potential, not on local g. We want to verify this. How to measure potential, locally, independently of g?

Easily verified, on the geoid clocks run at the same rate,
while the local g on the geoid varies a lot,

Jan

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Tom Roberts@21:1/5 to RichD on Sun Dec 3 09:11:47 2023
On 12/1/23 5:14 PM, RichD wrote:
Is there an Einstein gravitational potential, is that well defined?

Not really.

Remember that the Newtonian gravitational force is minus the gradient of
the Newtonian gravitational potential. The quantities closest to that in
GR are the geometrical connection acting as "gravitational force" -- it
is related to the metric components by derivatives, so one can consider
the metric components as the analog to the Newtonian gravitational
potential. Note that the analogy is not very close, and this is
explicitly coordinate dependent, which is counter to the fundamentals of GR.

In a region wit weak fields and velocities << c, GR reduces
approximately to Newtonian gravitation, with the approximation being exceedingly good.

Tom Roberts

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From J. J. Lodder@21:1/5 to Tom Roberts on Sun Dec 3 22:01:20 2023
Tom Roberts <tjoberts137@sbcglobal.net> wrote:

On 12/1/23 5:14 PM, RichD wrote:
Is there an Einstein gravitational potential, is that well defined?

Not really.

Remember that the Newtonian gravitational force is minus the gradient of
the Newtonian gravitational potential. The quantities closest to that in
GR are the geometrical connection acting as "gravitational force" -- it
is related to the metric components by derivatives, so one can consider
the metric components as the analog to the Newtonian gravitational
potential. Note that the analogy is not very close, and this is
explicitly coordinate dependent, which is counter to the fundamentals of GR.

In a region wit weak fields and velocities << c, GR reduces
approximately to Newtonian gravitation, with the approximation being exceedingly good.

Indeed. In particular:
\Phi ~= - 6 10^7 m^2/s^2 (Kiddies: note that ist is negative)
\Delta \Phi ~= 10^4 m^2/s^2 (NIST to BIPM for example, less negative)
c^2 ~~ 10^17 m^2/s^2
\Delta \nu / \nu ~= 10^-13 (NIST faster than BIPM)

Small, but easily observable with today's cesium fountain clocks,
even more so when you average ~ 500 of them to obtain TAI,

Jan

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From J. J. Lodder@21:1/5 to Ken Seto on Thu Dec 7 15:34:06 2023
Ken Seto <setoken47@gmail.com> wrote:

On Friday, November 10, 2023 at 3:00:54?PM UTC-5, J. J. Lodder wrote:
[summary: gravity and clock rates for misled kiddies]

General relativity predicts that all freely falling clocks
will run at their own inherent rate. [by postulate]
It also predicts that clocks at different places,
and with different velocities will be seen to run at different rates,
-when compared with each other-.
It also predicts that accelerations do not affect clock rates,
so the results can be extended to non-inertial clocks,
such as clocks at rest at different altitudes on Earth.

According to general relativity all clock effects are purely kinematic,
so derivable from the metric tensor.
Doing the sums for weak fields results in velocity effects
being given by Lorentz factors, and 'gravitational' effects
being given by the variations in Newtonian potential.
So far, so good, and in agreement with experimental results.

Now there are people such as for example 'Lou' in this forum,
who cannot or will not accept or understand this.
They hold that obverved clock effects must be due to 'gravity'
affecting the workings of the clock, somehow.
In other words, they ascribe the observed clock effects
to physical causes, 'gravity' affecting the workings of clocks,
rather than to intrinsic space-time effects.

Fortunately it is easy to settle the point by experiment.
GR predicts that all clocks on the rotating geoid on Earth
must run at the same rate, when compared with each other.
Experiment bears this out, to accuracies approaching 10^-15.
This is of immense practical importance,
because it is the basis for realising the SI second.
(on which -all- physical measurement depends nowadays)

OTOH the force of gravity, as measured by 'small' g,
the acceleration of gravity, varies markedly over the geoid.
(by about 0.5%, between the poles and the equator)

If (the force of) 'gravity' influenced the rate of the clocks
there should be an effect of geographical latitude
on the rate of clocks.
This is not observed to be the case, so this idea stands falsified.

The idea that 'gravity' affects the rate at which clocks run
is a misconception without basis in observed fact,

Jan
The rate of a clock is dependent on its state of absolution motion.

Absolute motion with respect to what?

The observer assumes that he is in a state of zero absolute motion. Therefore his clock will accumulate clock second at a fastest rate in a gravity environment.
Therefore the observed clock will accumulate clock second at a slower rate.

Jan

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From J. J. Lodder@21:1/5 to RichD on Sun Dec 10 10:48:19 2023
RichD <r_delaney2001@yahoo.com> wrote:

On December 2, J. J. Lodder wrote:
The geoid is defined a set of points at the same potential?

To a physicist, the geoid is the locus of all points on earth that have
the same metric (considering just the earth). For all practical
thpurposes is is the same as having the same Newtonian
thgravitational potential.

How is the metric measured? How does one determine empirically if
two separated points share the same metric? That is, without measuring >>>> any clock rate, which is the subject under discussion.

Observe the test mass acceleration, and use the Lorentz
momentum formula. Calculate the g force, assuming Newton's
model. The Newtonian potential follows directly, if you know
the distribution of mass.

That is not the claim.
It is that relative rates of clocks at different points
depend on the potential difference between those points.

The claim is that in general relativity, clock rate depends only on
potential, not on local g. We want to verify this. How to measure
potential, locally, independently of g?

Easily verified, on the geoid clocks run at the same rate,
while the local g on the geoid varies a lot,

step 1: Identify two distant points on the same geoid surface.
step 2: Observe their clock rates.
step 3: Are they identical?

How to perform step 1, without recourse to clocks (which would
be circular), and ignoring local g, which supposedly varies relative
to the potential?

By the process called levelling.
This is what geometers have been doing for hundreds of years.
Find the local level, or plumb line, and use your theodolite
to find the relative altitude of neighbouring points.
In its simplest form, take a flexible tube, fill with water,
and find points at the same potential.
(yes, I know it is more complicated than that in practice,
if you want to go beyond building a level house)
For points far away use the mean sea level as the reference.
(with the necessary corrections)
To find vertical potential differences calculate \int dz g(z),
going straight up.
There are limits to what can be achieved this way.
In practice a few cm is often the best that can be done.
(which is still good enough, given the limitations of cesium clocks)

In the near future we will have to pass on to 'chronometric levelling',
using clocks to establish level. And no, nothing circular about it.
The two mwthods will still have to agree with each other,
to the realisable accuracy of course.

And yes, clock rates are identical on the (rotating) geoid,
to the accuracy to which they can be measured.
This is a basic fact on which all of physics, and much of modern
technology, depends. Whe couldn't have accurate TAI, and TUC without it.

It is backed up by over 50 years of experience with hundreds atomic
clocks in standards laboratories all over the world.

Jan

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)