On Sunday, 12 November 2023 at 20:08:58 UTC, J. J. Lodder wrote:
> Lou wrote:
>
> > On Sunday, 12 November 2023 at 15:08:45 UTC, J. J. Lodder wrote:
> > > Lou wrote:
> > >
> > > > On Sunday, 12 November 2023 at 13:30:18 UTC, J. J. Lodder wrote:
> > > > > Lou wrote:
> > > > >
> > > > > > On Saturday, 11 November 2023 at 20:47:35 UTC, J. J. Lodder wrote:
> > > > > > > Lou wrote:
> > > > > > >
> > > > > > > > On Friday, 10 November 2023 at 20:00:54 UTC, J. J. Lodder wrote:
> > > > > > > > > [summary: gravity and clock rates for misled kiddies]
> > > > > > > > >
> > > > > > > > > General relativity predicts that all freely falling clocks
> > > > > > > > > will run at their own inherent rate. [by postulate]
> > > > > > > > > It also predicts that clocks at different places,
> > > > > > > > > and with different velocities will be seen to run at different
> > > > > > > > > rates, -when compared with each other-.
> > > > > > > > > It also predicts that accelerations do not affect clock rates,
> > > > > > > > > so the results can be extended to non-inertial clocks,
> > > > > > > > > such as clocks at rest at different altitudes on Earth.
> > > > > > > > >
> > > > > > > > > According to general relativity all clock effects are purely
> > > > > > > > > kinematic, so derivable from the metric tensor.
> > > > > > > > > Doing the sums for weak fields results in velocity effects
> > > > > > > > > being given by Lorentz factors, and 'gravitational' effects
> > > > > > > > > being given by the variations in Newtonian potential.
> > > > > > > > > So far, so good, and in agreement with experimental results.
> > > > > > > > >
> > > > > > > > > Now there are people such as for example 'Lou' in this forum,
> > > > > > > > > who cannot or will not accept or understand this.
> > > > > > > > > They hold that obverved clock effects must be due to 'gravity'
> > > > > > > > > affecting the workings of the clock, somehow.
> > > > > > > > > In other words, they ascribe the observed clock effects
> > > > > > > > > to physical causes, 'gravity' affecting the workings of clocks,
> > > > > > > > > rather than to intrinsic space-time effects.
> > > > > > > > >
> > > > > > > > > Fortunately it is easy to settle the point by experiment.
> > > > > > > > > GR predicts that all clocks on the rotating geoid on Earth
> > > > > > > > > must run at the same rate, when compared with each other.
> > > > > > > > > Experiment bears this out, to accuracies approaching 10^-15.
> > > > > > > > > This is of immense practical importance,
> > > > > > > > > because it is the basis for realising the SI second.
> > > > > > > > > (on which -all- physical measurement depends nowadays)
> > > > > > > > >
> > > > > > > > > OTOH the force of gravity, as measured by 'small' g,
> > > > > > > > > the acceleration of gravity, varies markedly over the geoid.
> > > > > > > > > (by about 0.5%, between the poles and the equator)
> > > > > > > > >
> > > > > > > > > If (the force of) 'gravity' influenced the rate of the clocks
> > > > > > > > > there should be an effect of geographical latitude
> > > > > > > > > on the rate of clocks.
> > > > > > > > > The idea that 'gravity' affects the rate at which clocks run
> > > > > > > > > is a misconception without basis in observed fact,
> > > > > > > > >
> > > > > > > >
> > > > > > > > A desperately misguided post from JJ.
> > > > > > > > You did not really read any of my posts. If you did...Then you
> > > > > > > > deliberately ignored the fact that I *very* explicitly stated that
> > > > > > > > in a classical model "little g" is acceleration only. Not force.
> > > > > > > > And you ignored the fact that I very clearly stated that force
> > > > > > > > on the atoms at different altitudes in a classical model should be
> > > > > > > > calculated using r.
> > > > > > > Indeed, there is little point, because you go on harping about
> > > > > > > your r, and you are ignoring all sound advice by others.
> > > > > > > You can go on obfuscating because you limit yourself
> > > > > > > to situations with spherical symmetry.
> > > > > > >
> > > > > > > So to see the errors of your ways you should consider situations
> > > > > > > where spherical symmetry does not hold.
> > > > > > > Then the surfaces of constant potential do not coincide
> > > > > > > with surfaces of constant acceleration, or constant r.
> > > > > > > > Not the m/s^2 acceleration of r^2 in "little g".
> > > > > > > > Seeing as everyone except a profound idiot would think
> > > > > > > > acceleration = force.
> > > > > > > > And If you actually read my posts rather than thump your bible,
> > > > > > > > you would realise that I also said that force is what Laplace
> > > > > > > > called gravitational potential. And what Newton referred to as a
> > > > > > > > scalar field.
> > > > > > > > And what Einstein used to calculate his GR clock rate effects.
> > > > > > > > (Notice the r of Laplace's gravitational potential and
> > > > > > > > Newton's scalar field is also the r used in GR. Not r^2 of
> > > > > > > > little g.)
> > > > > > > > So if you claim that experiment shows no change of clock
> > > > > > > > rates at different sea level latitudes. Then you have not
> > > > > > > > only confirmed the predictions of GR.. You have also confirmed
> > > > > > > > the predictions of classical theory. Seeing as they both use r to
> > > > > > > > accurately calculate tick rates at different altitudes.
> > > > > > > Experiment shows that clocks on the geoid run at constant rates
> > > > > > > wrt each other. Note that the geoid is not a surface of constant r,
> > > > > > > nor a surface of constant g,
> > > > > > >
> > > > > > A straw man argument if ever you make.
> > > > > > Yes I've looked at your 'geoid' now and how it varies slightly by
> > > > > > about 200m relative to the reference geoid and how technically the r
> > > > > > distance doesn't exactly follow the geoid surface. That makes sense.
> > > > > > Splitting hairs though on your part to pretend somehow this rules out
> > > > > > a classical model which uses r. I notice you didn't actually specify
> > > > > > why it would. In fact it doesn't rule out in any way a classical model
> > > > > > any more than it would rule out GR.
> > > > > > Because in a classical calculation if one needs to assume
> > > > > > *exactly* the *total* mass M of the earth at r, then yes to be
> > > > > > *absolutely* accurate the geoid surface has to be used. Not the
> > > > > > actual distance r.
> > > > > > But the same applies to GR. And the fact remains that generally,
> > > > > > the force of gravity in a classical model follows r not r^2.
> > > > > > (And to please the pedant JJ,... with ever so small meter length
> > > > > > fluctuations in the exact distance of r to also be taken into account)
> > > > > So you missed all points, again. I'll simplify.
> > > > > The geoid surface is by definition an equipotential surface of the
> > > > > Newtonian potential.
> > > > > So it coincides (almost) with the mean sea level.
> > > > > The geoid is (to a very good approximation) an ellipsoid of revolution.
> > > > > The small differences between geoid and ellipsoid
> > > > > (due to slightly irregular mass distributions inside the Earth)
> > > > > don't matter for what follows.
> > > > >
> > > > > Now, on the geoid, and at the poles, we have: r < average g > average,
> > > > > potential = constant
> > > > > On the geoid, at mid-latitudes we have r = average, g = average,
> > > > > potential = same constant
> > > > > On the geoid, at the equator, we have r > average, g < average,
> > > > > potential = still the same constant, by definition of the geoid..
> > > > >
> > > > > The differences are huge, r = 6357-6378 km, g = 9.863-9.798 m/s2,
> > > > > compared to clock stabilities of 10^-15.
> > > > >
> > > > > What is your prediction for the rates of clocks in those three places?
> > > > > No verbiage, just say faster, slower, or the same,
> > > > > and if you can by how much,
> > > > >
> > > >
> > > > If you tried reading my posts you wouldn't be pretending I said the
> > > > force of gravity is 9.863-9.798 m/s2.
> > > Too bad if you didn't say it, for those are the measured values.
> > > > That's r^2 and it's called acceleration. You don't seem to know that m/s^2
> > > > is acceleration!!! Since when does Force=acceleration?
> > > > And I already responded to your point on geoids that yes if you want to
> > > > split hairs the geoid surface varies from r by up to 200 meters. Which is
> > > > why very accurate measurements of clock rates will show constant rates at
> > > > the surface of the geoid only. And not to r. But that's still consistent
> > > > with a classical model as much as with GR.
> > > So you have nothing to say,
> > > beyond agreeing that general relativity gives the right answer,
> > >
> > I suppose there isn't much more to say to a person such as yourself who
> > thinks that force=acceleration.
> OK, so you give up,
>
I give up trying to make you understand that a true classical model
does not use acceleration to describe the force of gravity. Only
relativists (or idiots) think in a classical model acceleration=force.

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

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