[time-nuts] Pendulums & Atomic Clocks & Gravity

Brooke Clarke brooke at pacific.net
Sat May 26 17:27:19 EDT 2007


Hi Tom:

I get it.  It's really "g" that matters and not the force (F = m * g).

Have Fun,

Brooke Clarke
http://www.PRC68.com
http://www.precisionclock.com



Tom Van Baak wrote:
>>Then a pendulum calculator at:
>>http://www.ajdesigner.com/phppendulum/simple_pendulum_equation_period.php
>>allows solving any of 8 pendulum related equations.
> 
> 
> Note that this simplistic l/g formula does not characterize the
> period of precision pendulum clocks, as it ignores factors
> such as circular error, and a host of other static and dynamic
> corrections. But it should be good enough for your SWCC.
> 
> 
>>Tom has pointed out that the stability limit on pendulum clocks is in the area
>>of 1E-7 because of the complex effect of the Sun and Moon on the value of "g".
>>http://www.leapsecond.com/hsn2006/ch1.htm
> 
> 
> You'll enjoy ch2 and ch3 too; when they're ready...
> 
> 
>>Gravity also effects atomic clocks, see: http://www.leapsecond.com/great2005/
>>and this puts a limit on what can be done with any atomic clock that's on
>>Earth.  "g" will always have minor fluctuations (noise) due to all sorts of
>>things like the Sun, Moon, planets, asteroids, earthquakes, etc.  It's still a
>>direct "g" effect called red shift like (U2 − U1)/c2, where the Us are
>>gravitational potentials, only smaller by c squared.
> 
> 
> I think 99.999% the Moon and Sun. You can ignore all other
> objects. True, mathematically, any object of any mass has a
> non-zero effect on g, but if you do the tide calculations even
> something as massive and as "close" as Jupiter is so far down
> in the noise (a few millionths the effect of the moon) that we
> don't worry about it for delta g calculations.
> 
> 
>>I expect that in not too many years the official master clocks will no longer
>>be on Earth, but instead in satellites.  There "g" is precisely known to be
>>zero.  Since GPS satellites are excellent for time transfer that's where they
>>will be.  The ensemble will be the full constellation.
> 
> 
> No, g isn't zero at all. Remember g is inverse square to the
> distance from Earth. The other way to think about it is that
> satellites wouldn't be (free-fall orbiting) satellites if g were
> zero, eh?
> 
> 
>>In "Time Too Good to Be True" Kleppner
>>http://www.physicstoday.org/vol-59/iss-3/p10.html
>>says "...a primary standard in space would not overcome the problem of
>>comparing time or frequency at different locations on Earth."
>>
>>I don't understand why that would be the case with a standard in a GPS
>>satellite.  Granted E-18 can not now be done using the current GPS system, but
>>when atomic clocks get into the E-16 or better area and are in satellites, I
>>think the quality of time transfer will keep up.
> 
> 
> I believe the point he is making is that time/frequency on earth
> is dependent on altitude. Altitude is both a function of place and
> of time (e.g., time of day, time of month, as in tides). And since
> altitude is wiggling up and down by fractions of a meter (due to
> earth tides) you have a real time transfer problem down in the
> 18th decimal place.
> 
> By analogy, how do you ultra-precisely measure the elevation
> of a lake when there are winds, waves, or ripples? Now think
> about ripples on the space-time pond.
> 
> If you want more references on tides, let me know.
> Also, related, here's a humbling four-part paper on:
> 
> What Does Height Really Mean?
> http://www.aagsmo.org/resources.htm
> 
> /tvb
> 
> 
> 
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