Lou <noelturntive@live.co.uk> wrote:

> On Sunday, 12 November 2023 at 13:30:18 UTC, J. J. Lodder wrote:
> > Lou wrote:
> >
> > > On Saturday, 11 November 2023 at 20:47:35 UTC, J. J. Lodder wrote:
> > > > Lou wrote:
> > > >
> > > > > On Friday, 10 November 2023 at 20:00:54 UTC, J. J. Lodder wrote:
> > > > > > [summary: gravity and clock rates for misled kiddies]
> > > > > >
> > > > > > General relativity predicts that all freely falling clocks
> > > > > > will run at their own inherent rate. [by postulate]
> > > > > > It also predicts that clocks at different places,
> > > > > > and with different velocities will be seen to run at different
> > > > > > rates, -when compared with each other-.
> > > > > > It also predicts that accelerations do not affect clock rates,
> > > > > > so the results can be extended to non-inertial clocks,
> > > > > > such as clocks at rest at different altitudes on Earth.
> > > > > >
> > > > > > According to general relativity all clock effects are purely
> > > > > > kinematic, so derivable from the metric tensor.
> > > > > > Doing the sums for weak fields results in velocity effects
> > > > > > being given by Lorentz factors, and 'gravitational' effects
> > > > > > being given by the variations in Newtonian potential.
> > > > > > So far, so good, and in agreement with experimental results.
> > > > > >
> > > > > > Now there are people such as for example 'Lou' in this forum,
> > > > > > who cannot or will not accept or understand this.
> > > > > > They hold that obverved clock effects must be due to 'gravity'
> > > > > > affecting the workings of the clock, somehow.
> > > > > > In other words, they ascribe the observed clock effects
> > > > > > to physical causes, 'gravity' affecting the workings of clocks,
> > > > > > rather than to intrinsic space-time effects.
> > > > > >
> > > > > > Fortunately it is easy to settle the point by experiment.
> > > > > > GR predicts that all clocks on the rotating geoid on Earth
> > > > > > must run at the same rate, when compared with each other.
> > > > > > Experiment bears this out, to accuracies approaching 10^-15.
> > > > > > This is of immense practical importance,
> > > > > > because it is the basis for realising the SI second.
> > > > > > (on which -all- physical measurement depends nowadays)
> > > > > >
> > > > > > OTOH the force of gravity, as measured by 'small' g,
> > > > > > the acceleration of gravity, varies markedly over the geoid.
> > > > > > (by about 0.5%, between the poles and the equator)
> > > > > >
> > > > > > If (the force of) 'gravity' influenced the rate of the clocks
> > > > > > there should be an effect of geographical latitude
> > > > > > on the rate of clocks.
> > > > > > This is not observed to be the case, so this idea stands falsified.
> > > > > >
> > > > > > The idea that 'gravity' affects the rate at which clocks run
> > > > > > is a misconception without basis in observed fact,
> > > > > >
> > > > >
> > > > > A desperately misguided post from JJ.
> > > > > You did not really read any of my posts. If you did...Then you
> > > > > deliberately ignored the fact that I *very* explicitly stated that
> > > > > in a classical model "little g" is acceleration only. Not force.
> > > > > And you ignored the fact that I very clearly stated that force
> > > > > on the atoms at different altitudes in a classical model should be
> > > > > calculated using r.
> > > > Indeed, there is little point, because you go on harping about your r,
> > > > and you are ignoring all sound advice by others.
> > > > You can go on obfuscating because you limit yourself
> > > > to situations with spherical symmetry.
> > > >
> > > > So to see the errors of your ways you should consider situations
> > > > where spherical symmetry does not hold.
> > > > Then the surfaces of constant potential do not coincide
> > > > with surfaces of constant acceleration, or constant r.
> > > > > Not the m/s^2 acceleration of r^2 in "little g".
> > > > > Seeing as everyone except a profound idiot would think
> > > > > acceleration = force.
> > > > > And If you actually read my posts rather than thump your bible,
> > > > > you would realise that I also said that force is what Laplace
> > > > > called gravitational potential. And what Newton referred to as a
> > > > > scalar field.
> > > > > And what Einstein used to calculate his GR clock rate effects.
> > > > > (Notice the r of Laplace's gravitational potential and Newton's scalar
> > > > > field is also the r used in GR. Not r^2 of little g.)
> > > > > So if you claim that experiment shows no change of clock
> > > > > rates at different sea level latitudes. Then you have not
> > > > > only confirmed the predictions of GR.. You have also confirmed
> > > > > the predictions of classical theory. Seeing as they both use r to
> > > > > accurately calculate tick rates at different altitudes.
> > > > Experiment shows that clocks on the geoid run at constant rates
> > > > wrt each other. Note that the geoid is not a surface of constant r,
> > > > nor a surface of constant g,
> > > >
> > > A straw man argument if ever you make.
> > > Yes I've looked at your 'geoid' now and how it varies slightly by
> > > about 200m relative to the reference geoid and how technically the r
> > > distance doesn't exactly follow the geoid surface. That makes sense.
> > > Splitting hairs though on your part to pretend somehow this rules out
> > > a classical model which uses r. I notice you didn't actually specify
> > > why it would. In fact it doesn't rule out in any way a classical model
> > > any more than it would rule out GR.
> > > Because in a classical calculation if one needs to assume *exactly* the
> > > *total* mass M of the earth at r, then yes to be *absolutely* accurate
> > > the geoid surface has to be used. Not the actual distance r.
> > > But the same applies to GR. And the fact remains that
> > > generally, the force of gravity in a classical model follows r not r^2.
> > > (And to please the pedant JJ,... with ever so small meter length
> > > fluctuations in the exact distance of r to also be taken into account)
> > So you missed all points, again. I'll simplify.
> > The geoid surface is by definition an equipotential surface of the
> > Newtonian potential.
> > So it coincides (almost) with the mean sea level.
> > The geoid is (to a very good approximation) an ellipsoid of revolution.
> > The small differences between geoid and ellipsoid
> > (due to slightly irregular mass distributions inside the Earth)
> > don't matter for what follows.
> >
> > Now, on the geoid, and at the poles, we have: r < average g > average,
> > potential = constant
> > On the geoid, at mid-latitudes we have r = average, g = average,
> > potential = same constant
> > On the geoid, at the equator, we have r > average, g < average,
> > potential = still the same constant, by definition of the geoid.
> >
> > The differences are huge, r = 6357-6378 km, g = 9.863-9.798 m/s2,
> > compared to clock stabilities of 10^-15.
> >
> > What is your prediction for the rates of clocks in those three places?
> > No verbiage, just say faster, slower, or the same,
> > and if you can by how much,
> >
>
> If you tried reading my posts you wouldn't be pretending I said the force of
> gravity is 9.863-9.798 m/s2.

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

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

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

Jan