[time-nuts] Pendulums and Atomic Clocks and Gravity :probably more than you want to know...

Bill Beam wbeam at gci.net
Fri Jun 1 16:12:03 EDT 2007


I was afraid someone would say 'Riemann tensor'
The problem with the Riemann tensor is that I don't
think that anyone here can write in down in detail
for this problem (let alone solve it).  I surely can not.

I also don't think that anyone here is ready for the
idea that there is no such thing as gravitational force,
and that in the absence of any other force everything
is in free fall.  World lines and geodesics, oh my!
(inside joke).

The problem of test particles released inside a satellite
in circular orbit is easily solvable using pre-newtonian
physics.  Kepler's laws are completely adaquate.  The 
first and second laws are enough.  The third can be
ignored since, for small changes all things are linear
to first approximation.

For an eureka moment, look up Keplers laws and ask
how the kinetic and potential energy is changed by
displaceing the test particle from the satellite center
of mass.

Next problem:  What is the circulation of the atmosphere
inside the satellite?  (Here, don't forget to consider effect
of differences of orbit inclination.)?  The atmosphere can
be treated as a collection of test particles (to first order).

Finally: be a skeptic, especially in a forum like this one.  They
let almost anyone in here.  Can you verify what is said?
Run the numbers.  Are references cited?  Remember 'cold
fusion'?  Even those guys wore suits and ties, had advanced
degrees and worked at a (formerly) prestigious institution.

Simple is best.

Bill - retired, have time, will travel.

On Fri, 01 Jun 2007 11:36:29 -0400, mike c wrote:

>Dear Dr. Bruce and Bill B and all timenuts in this thread. My two cents 
>about the
>conundrums and how physicists think about atomic clocks in geodesy 
>around the earth.
> For each mass in space, whatever its orbit, if there are no other 
>fields (ex: electric/magnetic or  drag forces) acting on it, then  it 
>sees space and time locally as if it
>were an inertial observer in an intertial
>frame. There is no gravitational field in that frame...technically the 
>metric is exactly that of flat
>space locally and one of the curvature quantities, the so-called Ricci 
>tensor, vanishes identically
>(and not just locally!) The point is, that all that is left of gravity 
>when you are falling is the tides...that is test masses released from 
>rest in your local inertial frame BUT at a distance from you will not 
>remain stationary, but will start to move relative to you -even though 
>are no other forces than gravity present-. Technically this is due to 
>the fact that there
>is another curvature quantity that is non-zero in the space around a 
>massive body, it is the so called Riemann tensor. This is just how GR 
>talks about tidal forces, which is (almost) all that remains of gravity.
>    Now, releasing those masses, atleast for a few orbits the masses 
>will appear to kindof orbit around each other and the mass that defined 
>the original frame. If you are very far from the parent body (and 
>neglecting effects from other bodies outside the system and between the 
>masses themselves) then they will continue to stay close and orbit for a 
>long time. Classically, if the orbits had different major axes they will 
>eventually walk off because - like two precision clocks- they will 
>dephase. As this happens, no SV will be large enough.  If however the 
>masses are released
>so that their major axis is the same, then even in full GR and being 
>close to the parent body,
>they will stay close to one another, 'orbiting' each other as it were, 
>forever. (We are
>neglecting some truly minute effects here) So a very small SV would -in 
>principle- do.
>    Hope that wasn't too long/confusing...this is all in the framework 
>of GR, but, hey,
>maybe there is more to gravity and time and space than that...
>    - Mike
>clock and gravity guy
>time-nuts mailing list
>time-nuts at febo.com

Bill Beam

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