[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
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.
[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,
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.
[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-.
OTOH the force of gravity, as measured by 'small' g,
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.Indeed, there is little point, because you go on harping about your r,
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.
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 anyone except a profound idiot would thinkExperiment shows that clocks on the geoid run at constant rates
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.
wrt each other.
Note that the geoid is not a surface of constant r,
nor a surface of constant g,
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.Indeed, there is little point, because you go on harping about your r,
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.
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".Experiment shows that clocks on the geoid run at constant rates
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.
wrt each other. Note that the geoid is not a surface of constant r,
nor a surface of constant g,
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.Indeed, there is little point, because you go on harping about your r,
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.
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".Experiment shows that clocks on the geoid run at constant rates
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.
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)
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.Indeed, there is little point, because you go on harping about your r, and you are ignoring all sound advice by others.
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.
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".Experiment shows that clocks on the geoid run at constant rates
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.
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.So you missed all points, again. I'll simplify.
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)
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,
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.Indeed, there is little point, because you go on harping about your r, and you are ignoring all sound advice by others.
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.
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".Experiment shows that clocks on the geoid run at constant rates
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.
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.So you missed all points, again. I'll simplify.
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)
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.
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.Indeed, there is little point, because you go on harping about your r,
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.
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".Experiment shows that clocks on the geoid run at constant rates
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.
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.So you missed all points, again. I'll simplify.
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)
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 ofToo bad if you didn't say it, for those are the measured values.
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?So you have nothing to say,
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.
beyond agreeing that general relativity gives the right answer,
Jan
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,
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.Indeed, there is little point, because you go on harping about your r,
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.
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".Experiment shows that clocks on the geoid run at constant rates
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.
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.So you missed all points, again. I'll simplify.
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)
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 ofToo bad if you didn't say it, for those are the measured values.
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?So you have nothing to say,
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.
beyond agreeing that general relativity gives the right answer,
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.Indeed, there is little point, because you go on harping about
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.
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".Experiment shows that clocks on the geoid run at constant rates
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.
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.So you missed all points, again. I'll simplify.
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)
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 theToo bad if you didn't say it, for those are the measured values.
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^2So you have nothing to say,
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.
beyond agreeing that general relativity gives the right answer,
Jan
"... a right answer".
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.Indeed, there is little point, because you go on harping about
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.
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".Experiment shows that clocks on the geoid run at constant rates
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.
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.So you missed all points, again. I'll simplify.
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)
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 theToo bad if you didn't say it, for those are the measured values.
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^2So you have nothing to say,
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.
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.
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.Indeed, there is little point, because you go on harping about your r, and you are ignoring all sound advice by others.
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.
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".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,
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.
nor a surface of constant g,
A straw man argument if ever you make.So you missed all points, again. I'll simplify.
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)
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^2So you have nothing to say,
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.
beyond agreeing that general relativity gives the right answer,
Jan
"... a right answer".There is only one,
Jan
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 saidToo bad if you didn't say it, for those are the measured values.
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 knowSo you have nothing to say,
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.
beyond agreeing that general relativity gives the right answer,
Jan
"... a right answer".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.
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.Indeed, there is little point, because you go on harping about your r, and you are ignoring all sound advice by others.
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.
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".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,
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.
nor a surface of constant g,
A straw man argument if ever you make.So you missed all points, again. I'll simplify.
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)
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^2So you have nothing to say,
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.
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,
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.[...]
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.So you have nothing to say,
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.
beyond agreeing that general relativity gives the right answer,
Jan
"... a right answer".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 beNo prolem, the whole idea applies in the Newtonian approximation
that the geodesy is always instantaneously evaluated everywhere.
to GR anyway. (so v^2 << c^2)
Jan
On 11/13/23 3:07 PM, Lou wrote:
a true classical model does not use acceleration to describe theWell, there is no such thing as a "true" model in physics -- physical
force of gravity.
models are valid or invalid, but we humans can never know whether they
are "true".
NM is DIFFERENT
from relativity, and wrong.
Still more nonsense. The force of gravity is minus the gradient of the gravitational potential. They have NEVER been "equal" as you suppose.
On 11/13/23 3:07 PM, Lou wrote:
a true classical model does not use acceleration to describe theWell, there is no such thing as a "true" model in physics -- physical
force of gravity.
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 predictMore nonsense. In Newtonian mechanics, time is universal and NM predicts zero "time dilation" under any and all circumstances. NM is DIFFERENT
the change of resonant frequencies of atoms at different potentials.
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.
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 theSo you have nothing to say,
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.
beyond agreeing that general relativity gives the right answer,
Jan
"... a right answer".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 beNo prolem, the whole idea applies in the Newtonian approximation
that the geodesy is always instantaneously evaluated everywhere.
to GR anyway. (so v^2 << c^2)
Jan
A spherical or elliptical geoid?
On Monday, 13 November 2023 at 22:58:54 UTC, Tom Roberts wrote:
On 11/13/23 3:07 PM, Lou wrote:Exactly....it’s false. Gravity force isn’t acceleration based on
a true classical model does not use acceleration to describe theWell, there is no such thing as a "true" model in physics --
force of gravity.
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).
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.
[... more nonsense]
On Monday, 13 November 2023 at 22:58:54 UTC, Tom Roberts wrote:
On 11/13/23 3:07 PM, Lou wrote:Exactly....it’s false. Gravity force isn’t acceleration based on r^2.
a true classical model does not use acceleration to describe theWell, there is no such thing as a "true" model in physics -- physical
force of gravity.
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).
Thats what I have been trying to get you lot to understand
this whole time.
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 modelNONSENSE! NOBODY thinks that. YOU are confused. The only "classical
acceleration=force.
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.
The only nonsense is your silly claim that I said a classical modelThey do it to make sure a classical model can’t correctly predictMore nonsense. In Newtonian mechanics, time is universal and NM predicts
the change of resonant frequencies of atoms at different potentials.
zero "time dilation" under any and all circumstances. NM is DIFFERENT
from relativity, and wrong.
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.
The force of gravity increases the closer you get to the mass.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.
Proportional to r.
That’s what potential really is.
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?
On 11/14/23 4:06 AM, Lou wrote:
On Monday, 13 November 2023 at 22:58:54 UTC, Tom Roberts wrote:Oh for goodness sake! I forgot how stupid and ignorant you are.
On 11/13/23 3:07 PM, Lou wrote:Exactly....it’s false. Gravity force isn’t acceleration based on
a true classical model does not use acceleration to describe theWell, there is no such thing as a "true" model in physics --
force of gravity.
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).
r^2.
NM is known to be false
precession of the perihelion of mercury, or the "gravitational time dilation" exhibited by GPS satellites.
On 11/14/23 4:06 AM, Lou wrote:
On Monday, 13 November 2023 at 22:58:54 UTC, Tom Roberts wrote:Oh for goodness sake! I forgot how stupid and ignorant you are.
On 11/13/23 3:07 PM, Lou wrote:Exactly....it’s false. Gravity force isn’t acceleration based on
a true classical model does not use acceleration to describe theWell, there is no such thing as a "true" model in physics --
force of gravity.
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).
r^2.
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 thisYou have no hope of doing that, because a) YOU don't understand it, and
whole time.
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,
acceleration = -grad potential.
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.
On 11/14/2023 5:06 AM, Lou wrote:
On Monday, 13 November 2023 at 22:58:54 UTC, Tom Roberts wrote:No, force is the mass * acceleration. Newton's Second Law, you know.
On 11/13/23 3:07 PM, Lou wrote:Exactly....it’s false. Gravity force isn’t acceleration based on r^2.
a true classical model does not use acceleration to describe theWell, there is no such thing as a "true" model in physics -- physical
force of gravity.
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).
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 understandNewton says you're completely wrong.
this whole time.
Einstein realised this and used potential.
Not for the equivalent of gravitational acceleration or force!
Only if you are a liar. And ignore the facts.LaplaceWrong.
realised that gravity force was potential. Newton knew this and
called gravity force a scalar field. And others like Levy also
proposed this
No it doesn’t. You said yourself tick rates are proportionally to r.The ridiculous part of your argument is that although do you admitr^2 does work in Newton's approximation.
r^2 doesn’t work
...you cannot bear to have anyone point outClassical Newtonian gravity uses 1/r^2 for acceleration and force. You
that to make a classical theory work...one must use r of potential.
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.Nope. Newton's Second Law relates force and acceleration. F=ma.
Dont pretend it isn’t.
Only relativists (or idiots) think in a classical modelNONSENSE! NOBODY thinks that. YOU are confused. The only "classical
acceleration=force.
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 tryingYou are (deliberately!) leaving out the mass in Newton's
to pretend in a classical model that m/s^2= force.
It isn’t . ITS CALLED ACCELERATION. Einstein realised this.
force/acceleration relationship. Again, F=ma.
How could it if the Newtonian force is proportional to 1/r^2 but the perceived frequency difference goes as 1/r?The only nonsense is your silly claim that I said a classical model predicts time dilation. That’s GR.They do it to make sure a classical model can’t correctly predictMore nonsense. In Newtonian mechanics, time is universal and NM predicts >> zero "time dilation" under any and all circumstances. NM is DIFFERENT
the change of resonant frequencies of atoms at different potentials.
from relativity, and wrong.
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 frequenciesBut we also know they don't go as 1/r.
under external force for longer than GR has been around.
Nope. Proportional to 1/r^2.The force of gravity increases the closer you get to the mass. 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.
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 isBecause Einstein wasn't showing the effect of Newtonian forces. He found
OK for Albert to use r of potential to model the effects of gravity?
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.
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.Indeed, there is little point, because you go on harping about your r, and you are ignoring all sound advice by others.
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.
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".Experiment shows that clocks on the geoid run at constant rates
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.
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.So you missed all points, again. I'll simplify.
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)
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,
[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]
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.Indeed, there is little point, because you go on harping about your r, and you are ignoring all sound advice by others.
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.
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".Experiment shows that clocks on the geoid run at constant rates
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.
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.So you missed all points, again. I'll simplify.
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)
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.
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
Those rate difference are (and must) be taken into account
to compute TAI.
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.Indeed, there is little point, because you go on harping about your r,
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.
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".Experiment shows that clocks on the geoid run at constant rates
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.
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.So you missed all points, again. I'll simplify.
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)
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.At last. You are a bit late, considering your grandiose claims.
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.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.
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.Indeed, there is little point, because you go on harping about
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.
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".Experiment shows that clocks on the geoid run at constant rates
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.
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.So you missed all points, again. I'll simplify.
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)
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.At last. You are a bit late, considering your grandiose claims.
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.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.
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.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.
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.
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".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,
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.
nor a surface of constant g,
A straw man argument if ever you make.So you missed all points, again. I'll simplify.
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)
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.At last. You are a bit late, considering your grandiose claims.
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 relativistsStandards laboratory all over the world have been keeping clusters of atomic clocks, and they have been comparing them.
don't like evidence but please... back up your claims with a few citations.
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.You can be as wrong and content as you want to be.
I will be content to use r.
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.I'm quite sure that the people you mention have no need for that,
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.
On Wednesday, 15 November 2023 at 04:59:55 UTC, Volney wrote:
On 11/14/2023 5:06 AM, Lou wrote:So why are you pretending the force of gravity is measured as little g (m/s^2)?
On Monday, 13 November 2023 at 22:58:54 UTC, Tom Roberts wrote:
On 11/13/23 3:07 PM, Lou wrote: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.
a true classical model does not use acceleration to describe theWell, there is no such thing as a "true" model in physics -- physical
force of gravity.
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).
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 understandNewton says you're completely wrong.
this whole time.
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.
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
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 modelNONSENSE! NOBODY thinks that. YOU are confused. The only "classical
acceleration=force.
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.
The only nonsense is your silly claim that I said a classical modelThey do it to make sure a classical model can’t correctly predictMore nonsense. In Newtonian mechanics, time is universal and NM predicts >>>> zero "time dilation" under any and all circumstances. NM is DIFFERENT
the change of resonant frequencies of atoms at different potentials.
from relativity, and wrong.
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.
It’s only dishonest relativists who pretend that force=acceleration in a classical model.
And we have known harmonic oscillators do change frequenciesBut we also know they don't go as 1/r.
under external force for longer than GR has been around.
Really.? That’s odd. Obviously you forgot.
GPS clock rates match those predicted by a classical model
using force of gravity proportional to r.
The force of gravity increases the closer you get to the mass.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.
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!!!
Because Einstein wasn't showing the effect of Newtonian forces. He foundYou 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?
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.
Looks like your idols ...Schwarzschild, Pound Rebka, Einstein etc etc
...all used r!!
No, the GPS clock ticks at 1 second per second.
You lie. Why do you lie?Oh, stupid Mike, whoever is treating physics seriously
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!
On 11/15/2023 4:40 AM, Lou wrote:
On Wednesday, 15 November 2023 at 04:59:55 UTC, Volney wrote:Never. Force, according to Newton's second law, is F=mg since little g
On 11/14/2023 5:06 AM, Lou wrote:So why are you pretending the force of gravity is measured as little g (m/s^2)?
On Monday, 13 November 2023 at 22:58:54 UTC, Tom Roberts wrote:No, force is the mass * acceleration. Newton's Second Law, you know.
On 11/13/23 3:07 PM, Lou wrote:Exactly....it’s false. Gravity force isn’t acceleration based on r^2.
a true classical model does not use acceleration to describe theWell, 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
force of gravity.
often useful as an approximation).
Since when does force=acceleration?
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.Lie.
He called it a scalar field just to confuse idiots.
Another lie. GR math is completely different from classical gravity. ItEinstein realised this and used potential.
Not for the equivalent of gravitational acceleration or force!
does come up with the same answers as classical gravity which is an approximation.
Does GR use Gm/r^2 anywhere in its calculations?Because that's classical gravity.
No
Does GR use r?As an approximation.
Yes.
Does GR use r to model tick rates at different altitudes?Misleading. Perceived tick rates is not force.
Yes
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.You lie.
Anyways Newton did say force can be calculated with just r.
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 effectsCompletely wrong. GM/r does not even have units of acceleration or
of the force of gravity..one must use r
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.Nope. F=ma is not F=a.
As you are trying to pretend a classical model says.
Only relativists (or idiots) think in a classical modelNONSENSE! NOBODY thinks that. YOU are confused. The only "classical >>>> model" here is Newtonian mechanics, and in NM the second law applies: >>>> F = m a
acceleration=force.
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 gravityNo, never. The formula for classical force is GMm/r^2. I never said
equals just acceleration. Millions of times.
force is acceleration, classical force is mass * acceleration. Again,
you are deliberately leaving out the mass and accusing me of what you did.
The only nonsense is your silly claim that I said a classical modelThey 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 predictszero "time dilation" under any and all circumstances. NM is DIFFERENT >>>> from relativity, and wrong.
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 NewtonNo, he did not. You are lying when you say that. Newton explicitly
said the force of gravity was r.
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 frequenciesBut we also know they don't go as 1/r.
under external force for longer than GR has been around.
Really.? That’s odd. Obviously you forgot.No, the GPS clock ticks at 1 second per second. The signal is
GPS clock rates match those predicted by a classical model
using force of gravity proportional to r.
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.
The force of gravity increases the closer you get to the mass.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. >>>>
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 anThe force doesn't. The potential is different.
object to a higher altitude that must mean that the force of gravity increases with altitude!!!
Because Einstein wasn't showing the effect of Newtonian forces. He found >> a new relationship that simplifies to the potential. It has nothing toYou 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?
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.You lie. Why do you lie?
And to cover it up..he changed the name of “force of gravity” to potential.
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, EinsteinAnd none of them EVER said that any force or acceleration was
etc etc ...all used r!!
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.
Fortunately only idiots will fall for that bit of snake oil.
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 etcAnd none of them EVER said that any force or acceleration was
...all used r!!
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.
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.Those last two sentences contradict each other. Either jujube is a
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.
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 gravitySince 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?
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:Those last two sentences contradict each other. Either jujube is a
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.
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 gravitySince 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.
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 theWell, 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
force of gravity.
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.
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:Oh for goodness sake! I forgot how stupid and ignorant you are.
On 11/13/23 3:07 PM, Lou wrote:Exactly....it’s false. Gravity force isn’t acceleration based on r^2.
a true classical model does not use acceleration to describe theWell, there is no such thing as a "true" model in physics --
force of gravity.
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).
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
He certainly conned you. He called the force of gravity... potential,Thats what I have been trying to get you lot to understand thisYou have no hope of doing that, because a) YOU don't understand it, and
whole time.
b) it is WRONG.
Einstein realised this and used potential.Not for gravitational force, but rather for gravitational 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,
acceleration = -grad potential.
In GR neither gravitational force nor gravitational potential appears in the theory. The relative acceleration between small objects due toExactly. Finally you woke up and smelt the coffee.
gravity is expressed by the Raychaudhuri equation.
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.
On Friday, 17 November 2023 at 05:56:29 UTC, Volney wrote:
On 11/16/2023 6:11 AM, Lou wrote:
You don’t seem to realise your own con.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!
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.
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:Oh for goodness sake! I forgot how stupid and ignorant you are.
On 11/13/23 3:07 PM, Lou wrote:Exactly....it’s false. Gravity force isn’t acceleration based on r^2.
a true classical model does not use acceleration to describe theWell, there is no such thing as a "true" model in physics --
force of gravity.
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).
Said the guy who thinks force=accelerationNonsense.
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
He certainly conned you. He called the force of gravity... potential,Thats what I have been trying to get you lot to understand thisYou have no hope of doing that, because a) YOU don't understand it, and b) it is WRONG.
whole time.
Einstein realised this and used potential.Not for gravitational force, but rather for gravitational 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.It never was by anyone except you who invented the whole fabrication.
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.No, Tom never said that (it contradicts even high school physics).
You can't even read.
Again, in NM for gravity: force != potential;
rather, force = m * -grad potential, and since F = m a,
acceleration = -grad potential.
There is no potential in GR.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 duckAgain, there is no such thing as "gravitational potential" in GR.
Then it’s a potential duck.
Gravity is described by a metric which is a rank-2 tensor, it's
NOT a scalar (which is what potential is).
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 theWell, 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).
force of gravity.
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 understandYou should have known already in high school that acceleration is not force.
this whole time.
Einstein realised this and used potential.This is gobbledygook.
LaplaceMore gobbledygook. You have no idea what you are talking about.
realised that gravity force was potential.
Newton knew this and
called gravity force a scalar field.
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:Those last two sentences contradict each other. Either jujube is a
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.
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 notSince jujube and potential vary according to 1/r, they won't have units >> of force, so neither one can be force.
the force of gravity
So you think Force is defined by m/s^2?YOU were the one who called "jujube" a force, not me! I'll repeat:
I thought m/s^2 refers to acceleration.
Newton's Second Law is force = mass * acceleration.
(and why do you insist something with units of m^2/s^2 is a force?)
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:
What other name?You don’t seem to realise your own con.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!
Just because they called force of gravity by some other name in GR...
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 jujubeNewtonian gravitational potential and the spacetime metric of GR
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
are different things, both physically and mathematically.
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:Those last two sentences contradict each other. Either jujube is a
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.
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 notSince jujube and potential vary according to 1/r, they won't have units >>>> of force, so neither one can be force.
the force of gravity
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?
😂🤣
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:Those last two sentences contradict each other. Either jujube is a
On 11/16/2023 6:11 AM, Lou wrote:
Looks like your idols ...Schwarzschild, Pound Rebka, Einstein etc etcAnd none of them EVER said that any force or acceleration was
...all used r!!
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.
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 notSince jujube and potential vary according to 1/r, they won't have units >>>> of force, so neither one can be force.
the force of gravity
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 IYou contradict yourself. Gravity is modeled inversely proportional to
meant the force of gravity is modelled with r.
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. ButYou can use any name you want, but if it's modeled as potential, it is
could and is called by any different names including jujube if you wished
not a force. Full stop.
GR calls it curved spacetime etc. Newton called it scalar field.None of these are force.
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,And once again, you counter Newton, who states classical force =
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.
GMm/r^2. Why do you counter Newton's classical force?
M/s^2 is acceleration. Force is not measured in m/s^2Nor 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 asNo as I keep reminding you Newton's second law is F=ma, so force of
you try to pretend when you say little g, which uses m/s^2, is the force of gravity.
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.I'm the one who keeps reminding you of Newton's second law, force = mass
You think acceleration is force?
😂🤣
* acceleration, but you like snipping that.
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:Those last two sentences contradict each other. Either jujube is a >>>> force, or it varies proportionately to r. You can't have both, because
On 11/16/2023 6:11 AM, Lou wrote:
Looks like your idols ...Schwarzschild, Pound Rebka, Einstein etc etcAnd none of them EVER said that any force or acceleration was >>>>>> proportional to 1/r. Too bad for you!
...all used r!!
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.
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 orSince jujube and potential vary according to 1/r, they won't have units
gravitational time dilation or gravity well and you can pretend it’s not
the force of gravity
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 IYou contradict yourself. Gravity is modeled inversely proportional to
meant the force of gravity is modelled with r.
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. ButYou can use any name you want, but if it's modeled as potential, it is
could and is called by any different names including jujube if you wished
not a force. Full stop.
GR calls it curved spacetime etc. Newton called it scalar field.None of these are force.
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,And once again, you counter Newton, who states classical force =
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.
GMm/r^2. Why do you counter Newton's classical force?
M/s^2 is acceleration. Force is not measured in m/s^2Nor 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 calledI'm the one who keeps reminding you of Newton's second law, force = mass
acceleration.
You think acceleration is force?
😂🤣
* 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.
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?
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.
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.
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.
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.
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.
The geoid is defined a set of points at the same potential?
All at the same distance from the center?
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.
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.
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.
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.
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.
?Eh, zero potential is out at infinity.
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.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
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.
?Eh, zero potential is out at infinity.
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.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.
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.
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?
Is there an Einstein gravitational potential, is that well defined?
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.
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,
JanThe rate of a clock is dependent on its state of absolution motion.
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.
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?
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