tvb@LeapSecond.com said:

What is the Q of the Earth?  It might be on one of your web pages, but I
don't remember seeing it.  Google found a few mentions, but I didn't find a
number.

I did find an interesting list of damping mechanisms in a geology book.
Geology-nuts are as nutty as time-nuts.  Many were discussing damping of
seismic waves rather than rotation.

I've seen mention that the rotation rate of the Earth changed by a few
microseconds per day as a result of the 2011 earthquake in Japan.  Does that
show up in any data?  Your recent graph doesn't go back that far and it's got
a full scale of 2000 microseconds so a few is going to be hard to see.

--
These are my opinions.  I hate spam.

42

On Sat, Jul 23, 2016 at 7:59 PM, Hal Murray hmurray@megapathdsl.net wrote:

Hal Murray wrote:

3.1415 * 86400 * 365 * 100 / 0.002 = 5e12, or 5 trillion.  Here's why:

(1)
Among other things, the quality-factor, or Q is a measure of how slowly a free-running oscillator runs down. There are lots of examples of periodic or damped oscillatory motion that have Q -- RC or LC circuit, tuning fork, pendulum, vibrating quartz; yes, even a rotating planet in space.

In the high vacuum of space you'd think the Earth would rotate forever, but the moon affects earth's rotation; it causes "tidal friction". Astronomers have measured the slowing as roughly "2 ms per day per century".

For example, the average length of a day in 1900 was about 86400.000 seconds and the average length of a day in 2000 was about 86400.002 seconds. That's a deceleration, or slight loss in rate, or slight gain in period, of 2 ms/day per century. Since the spinning earth oscillator is slowing down, you can calculate its Q factor.

(2)
The common definition of Q is 2pi times the number of periods it takes for the energy to fall to 37% (1/e). Although correct and simple, it is sometimes awkward to use this definition. I mean, it can take forever to lose 63% of your energy if Q is really high or the period is really long.

Another definition of Q is 2pi over dE/E, the latter term being called the "decrement", which is the relative energy loss per period. For example, if your wine glass oscillator loses 1% of its energy each period, then its Q = 2pi/1%, or 628.

This formula for Q is more convenient because you can do it with data from a single period. For earth the period is 1 day. So what's the decrement, the relative energy loss for one day?

(3)
Rotational energy goes as w^2, where w (omega) is the rotation rate. This means energy drops twice as fast as rotation rate drops.

And how fast does the earth rotation rate drop? That would be 2 ms per day per century, which is 2.3e-8 per century, or 2.3e-10 per year, or 6.3e-13 per day. So the relative loss of energy per period (day) is twice that: 1.3e-12. Plug that into the decrement definition of Q and your calculator gives 5e12.

So the Q of the earth is 5 trillion.

https://www.google.com/#q=3.1415++86400++365+*+100+/+0.002

/tvb

p.s.

This Q is really high, and you might expect a similarly superb Allan deviation. But no, not at all:

http://leapsecond.com/museum/earth/

The reason the ADEV is so poor compared to the Q is that Earth, as a timekeeper, is an unstable mess inside and out. There are all sorts of massive perturbing factors from the liquid core to the swirling atmosphere to the annual redistribution of water and biomass; these all affect the short-term performance.

----- Original Message -----
From: "Hal Murray" hmurray@megapathdsl.net
To: "Tom Van Baak" tvb@leapsecond.com; "Discussion of precise time and frequency measurement" time-nuts@febo.com
Cc: hmurray@megapathdsl.net
Sent: Saturday, July 23, 2016 5:59 PM
Subject: Q/noise of Earth as an oscillator

Hoi Tom,

On Tue, 26 Jul 2016 12:36:37 -0700
"Tom Van Baak" tvb@LeapSecond.com wrote:

I am not sure you can apply this definition of Q onto earth. Q is defined
for harmonic oscillators (or oscillators that can be approximated by an
harmonic oscillator) but the earth isn't oscillating, it's rotating.
While, for time keeping purposes, similar in nature, the physical
description of both are different.

		Attila Kinali

--
Malek's Law:
Any simple idea will be worded in the most complicated way.

On Wed, 27 Jul 2016 00:08:47 +0200, you wrote:

It seems reasonable to me.  The calculation returns a dimensionless
number which does represent the run-down time.  Any reference duration
like hours, months, or years would have returned the same result.

Hal Murray wrote:

Right. The IERS graphs I posted are real measurements of earth rotation. The earthquake / tsunami numbers are theoretical calculations only; numbers far smaller than what can be measured.

A couple of guys at NASA have carefully modeled all of this and the predictions are quite interesting. It's just that the official NASA/JPL press releases, once filtered by the popular press, give the impression that these are worrisome or real measurable effects on the earth's rotation rate or axis tilt. Instead they are order(s) of magnitude below what can be measured, given the level of VLBI resolution or earth rotation noise.

Here are three recent press releases:

2005 "NASA Details Earthquake Effects on the Earth"
http://www.jpl.nasa.gov/news/news.php?feature=716

2010 "Chilean Quake May Have Shortened Earth Days"
http://www.nasa.gov/topics/earth/features/earth-20100301.html

2011 "Japan Quake May Have Shortened Earth Days, Moved Axis"
http://www.nasa.gov/topics/earth/features/japanquake/earth20110314.html

Google for Richard Gross and Benjamin Fong Chao for more information.

/tvb

----- Original Message -----
From: "Hal Murray" hmurray@megapathdsl.net
To: "Tom Van Baak" tvb@leapsecond.com; "Discussion of precise time and frequency measurement" time-nuts@febo.com
Cc: hmurray@megapathdsl.net
Sent: Saturday, July 23, 2016 5:59 PM
Subject: Q/noise of Earth as an oscillator

Around mid 1978 a scientific magazine had an item on the shift of axis
of the earth due to earthquakes.  Sorry but searching has not found the
item. My, not to be  relied on memory, thinks that it was from the
French Bureau International de l'Heure in Science or Scientific American

Cheers,
Will

On 07/27/2016 10:45 AM, Tom Van Baak wrote:

On Wed, Jul 27, 2016 at 8:08 AM, Attila Kinali attila@kinali.ch wrote:

"I am not sure you can apply this definition of Q onto earth."

It  doesn't make sense to me either.

If you mark a point on the surface of a sphere then you can observe
that point as the sphere
rotates and count rotations to make a clock. If you think of just a
circle, then a point on it viewed in a rectilinear coordinate system
executes simple harmonic motion so the motion of that point looks like
an oscillator, so that much is OK.

But unlike the LCR circuit, the pendulum and quartz crystal, the
sphere's rotational motion does not have a
resonant frequency. Another way of characterizing the Q of an
oscillator, the relative width of the resonance, makes
no sense in this context.

It seems to me that the model of the earth as an oscillator is
misapplied and that the 'Q' is not a meaningful number.
I think the confusion arises here because of a conflation of a
rotation of the sphere (which marks out a time interval) with an
oscillation. Both can be used to define an energy lost per unit time
but the former doesn't have anything to do with the properties of an
oscillator.

Something else that indicates that the model is suspect is that the
apparently high 'Q' implies a stability which the earth does not have,
as Tom observes. Viewed another way, this suggests that the model is
inappropriate because it leads to an incorrect conclusion.

Time for bed. I'll probably lie awake thinking about this now :-)

Cheers
Michael

On Wed, Jul 27, 2016 at 8:08 AM, Attila Kinali attila@kinali.ch wrote:

On 7/27/16 5:43 AM, Michael Wouters wrote:

There's also the thing that "things that resonate" typically have energy
transferring back and forth between modes or components: E field and H
field for an antenna; kinetic vs potential energy for pendulums and
weight/spring; charge and current (C & L, really E field/H field again).

Spinning earth is more of an "rotational inertia and loss" thing, with
zero frequency, just the exponential decay term.

If you think of a single measurand in any of these scenarios you have at
the core some sort of exp(-kt)cos(omegat+phi) and we're relating Q to
the coefficient k.

Hi Michael,

I sympathize with both your and Attila's comments and would like to dig deeper for the truth on this.

Clearly both the earth and a pendulum (and many other periodic systems) exhibit a decay of energy, when you remove the periodic restoring force. And if you take the classic definition Q = 2 pi * total energy / energy lost per cycle then it would seem earth has a Q factor.

In fact, if you use real energy numbers you get:

• total rotational energy of earth is 2.14e29 J
• energy lost per cycle (day) is 2.7e17 J
• so Q = 2pi * 2.14e29 / 2.7e17 = 5e12, the same 5 trillion as my earlier calculation.

But your point about resonance is a good one and it has always intrigued me. Is this one difference between a pendulum and the earth as timekeepers?

On the other hand, if you swept the earth with an external powerful frequency in the range well below to well above 1.16e-5 Hz (1/86164 s) would you not see a resonance peak right at the center? Given the mass of the planet and its pre-existing rotational energy, it seems like there is a "resonance", a preference to remain at its current frequency. Plus it has a slow decay due to internal friction. This sounds like any other timing system with Q to me.

Or imagine a planet the same size as earth made from a Mylar balloon. Much less mass. Give it the same rotational speed. Much easier to increase or decrease its energy by applying external force. Far lower Q than earth, yes?

It might also be useful at this point, to:

read the history Q and its definition:
http://www.collinsaudio.com/Prosound_Workshop/The_story_of_Q.pdf

and read the patent in which Q first appeared:
http://leapsecond.com/pages/Q/1927-US1628983.pdf

or view the first paragraph in which Q appeared:
http://leapsecond.com/pages/Q/1927-Q-patent-600x300.gif

/tvb

----- Original Message -----
From: "Michael Wouters" michaeljwouters@gmail.com
To: "Discussion of precise time and frequency measurement" time-nuts@febo.com; attila@kinali.ch
Sent: Wednesday, July 27, 2016 5:43 AM
Subject: Re: [time-nuts] Q/noise of Earth as an oscillator