time@radio.sent.com said:

It's energy loss in both cases.

Is there a term other than Q that is used to describe the rate of energy loss
for things that aren't oscillators?

--
These are my opinions.  I hate spam.

On 7/31/16 5:19 PM, Hal Murray wrote:

cooling (as in hot things)
discharge (as in capacitors and batteries)
leakage (as in pressure vessels)
loss

An irreversible process would be a better description versus energy loss.
Like joule heating (resistance, friction).

On Sunday, 31 July 2016, Hal Murray hmurray@megapathdsl.net wrote:

Hi

ummm…. Q is the general term of rate of energy loss and we just happen to apply
it to oscillators in a very elegant fashion….

Bob

Hal Murray wrote:

I've seen "energy dissipation" or "energy decrement".

Before Q was used to describe line width of atomic and optical clocks,
before Q was used to describe performance of simple harmonic oscillators and quartz crystals,
before or while Q was used to describe the ratio of reactance to resistance in a coil,
horologists that worked with precision pendulum clocks needed a way to characterize the amount of energy lost per period.

This was very important because the energy lost each swing (due to all forms of friction) had to be replaced in order to keep the pendulum oscillator running. And the timekeeping performance seemed to be related to how small or how consistent this energy was. I'm sure "decrement" was used before then, but I'm not well-read enough in science history to have any idea. Still...

What I do know is this abstract from a 1938 paper [1], written by an early time nut:

THE DISSIPATION OF ENERGY BY A PENDULUM SWINGING IN AIR, By E. C. ATKINSON

ABSTRACT. The decrement of a pendulum falls slowly with the amplitude: hence the
need for determinations based on small changes of angle. The resulting errors of observation
lead to erratic values but not to systematic error. The result of measurements with a
seconds pendulum enclosed in a case is shown by a smoothed curve, the departure from
observed times being expressed by smoothing fractions, and a smoothing figure is a
measure of this departure for the whole or part of the experiment. >From the decrement
the rate of loss of energy is calculated. This 7 kg. pendulum with amplitude 53' dissipates
a Board of Trade Unit (which serves a 70 w. lamp for 14 hours) in rather over 100,000
years. Experiments with different pendulums are described by which the component
losses due to suspension, rod, and bob are found. Suspension springs made from thin
strip clamped in chaps dissipate large and variable amounts of energy compared with
springs made from thick strip ground thin in the middle. The variable losses are associated
with variable rates of the pendulum. The cylindrical case adds considerably to the air
resistance. The measured loss due to a gravity impulse lever is little in excess of the
computed loss from collision with the pendulum: for a seconds pendulum 1/2000 part of
the free pendulum loss.

This article is also a favorite of mine because it talks about a "root mean square of a series of such fractions may be called the smoothing figure for the series, and its smallness is an indication of the confidence which may be placed in the result." -- a 1938 precursor of the two-sample variance, or Allan deviation.

/tvb

Hal:

Jim:

Scott:

Notice that these are all energy losses over time; gradual processes with perhaps an exponential time constant, but without cycles or periods. We know not to apply Q in these scenarios.

But when you have an oscillator, or a resonator, or (as I suggest) a "rotator", it seems to make sense to use Q to describe the normalized rate of decay. So three keys to Q: you need energy; you need energy loss; you need cycles over which that loss repeatedly occurs.

We use units of time (for example, SI seconds) when we describe a rate. But here's why Q is unitless -- you normalize the energy (using E / dE) and you also normalize the time (by cycle). No Joules. No seconds. So having period is fundamental to Q. It's this unitless character of Q (in both energy and time) that makes it portable from one branch of science to another. And if you measure in radians you can even get rid of the 2*pi factor ;-)

Without controversy, lots of articles define Q as 2*pi times {total energy} / {energy lost per cycle}. To me, a slowly decaying spinning Earth meets the three criteria. It appears to follow both the letter and the spirit of Q.

Bob:

Oh, no. Now we have both quality factor and elegance factor!

/tvb

Hi Tom,

You said: "you need energy; you need energy loss; you need cycles over
which that loss repeatedly occurs."

With regard to the earth, where is the first one? Sure it was there at the
start when the solar system formed, but where is it now?

Jim

On 1 August 2016 at 12:16, Tom Van Baak tvb@leapsecond.com wrote:

I still claim that there is no natural frequency associated with the
rotation of a body. The periodic nature of the rotating body motion is
confusing you. The choice of a coordinate system is what is confusing.

As I pointed out, what's the difference between an inertial body moving
in a straight line and a rotating body? Let's say you take a body moving
in free space with a small energy loss due to interactions with thin
interstellar gas. You measure it on a ruler with marks every 628 meters
(to use my example of a point on the Earth 100 meters from the pole as a
comparison). That's the same as making an astronomical measurement on
the Earth at a distant star which is overhead every 24 hours. In both
cases the point moves 628 meters every 24 hours (measured with an atomic
clock). We generate a tick when the Earth has rotated to the same
relative position and when the body moving in a straight line reaches
the next 628 meter mark. Both objects generate a tick every 24 hours,
but the velocity is each case (angular or linear) is unconstrained by
any periodic physical processes.

What makes the rotating body (Earth) suitable for study as a harmonic
oscillator with Q in this case? There is no energy transfer during each
rotation. Should we establish a Q value for the body in free space
straight line motion?  Both bodies have mass, inertia, a nearly constant
velocity (linear and angular), and a slight loss. Each generates a tick
every 24 hours (using the atomic clock as a reference). If we unwrap the
polar coordinates and view the Earth rotation angle as increasing
monotonically (or make marks every 628 meters on the scale measuring the
free space body with straight line travel) they are identical.

The geometry of the rotating body (Earth) is fooling you into thinking
it's a periodic oscillator. Just because the position is similar after
24 hours doesn't mean anything, since there is no energy storage and
transfer during each rotation. The Earth is reasonably symmetric (for
this discussion), and it has no field which matters for this discussion
which is rotating. It's just matter moving in a constrained circular
fashion due to the geometry and constraints of a rigid body. Change the
coordinate scale to linear (628 meters for each rotation at 100 meters
from the axis) and compare it to the free space object moving in a
straight line. What's the difference?

--
Bill Byrom N5BB

On Sun, Jul 31, 2016, at 09:16 PM, Tom Van Baak wrote:

At the risk of boring everyone, here is an "Alice and Bob" thought
experiment concerning linear and circular movement:

Condition A: Let's say that Alice is in her spacecraft in inertial
straight line travel through nearly free space containing a thin gas
which creates a slight friction to movement. She is traveling at 628
meters per 24 hours adjacent to a distance scale with markers every 628
meters on the space highway. Passing a marker results in a tick. So she
can check her atomic clock every 24 hours at the tick. The thin gas
causes a slight reduction of her velocity (about a 2 millisecond
increase in tick interval every century).

Question: Does the movement of Alice exhibit Q? If so, how can we compute the value? If there a resonant frequency?

Condition B: Bob is in his spacecraft, which is identical to Alice's
model. A rod 200 meters long connects Alice's craft and Bob's craft. Bob
is moving at 628 meters per 24 hours in the opposite direction of Alice,
so that the midpoint of the rod is fixed and unmoving with regards to
the distant stars. Alice will return to the same point in space every 24
hours and pass a marker, generating a tick so she can check her atomic
clock once a day at the tick. Her linear motion has been constrained to
circular motion due to the stiff rod. The same thin gas is present,
resulting in a 2 millisecond increase in tick interval every century.

Question: Now does the movement of Alice exhibit Q? If so, how can we compute the value? If there a resonant frequency?

Consider that in either Condition A or Condition B Alice (and Bob) can
increase or decrease their velocity to receive timing ticks at a
faster or slower rate. But there is no tendency for the velocity to
return to the one tick per day rate (628 meters per 24 hours), as
there would be with a harmonic oscillator. Due to Newton's First Law,
the velocity remains constant unless slowed by friction or affected by
external forces.

We could have started with a rod which was in the limit very long, so
the motion of Alice was only slightly diverted from a straight line
during a 24 hour interval. We could still measure the motion in 628
meter distance intervals on a circular or angular scale. Note that
nothing is fundamentally different if the rod is the exact length which
causes one rotation every 24 hours. The distance traveled and the
inertia resisting change in velocity is the same if the motion is linear
or circular, isn't it?

Bill Byrom N5BB

On Sun, Jul 31, 2016, at 10:27 PM, Bill Byrom wrote:

Hi Jim.

By first one, do you mean where does the initial energy come from?

For a pendulum clock, you supply energy with a lift or a push. For a lift to the side, E = mgh, where h is the height above the base. For a push from center, E = 1/2 mv^2. Either way, it takes all the potential or kinetic E you provide and starts making time from there.

For a rotating clock, you just give it a twist. In this case, E = 1/2 Iw^2, where I is the moment of inertia and w (omega) is angular velocity. For earth the total E is 2.1e29 J. That's the energy number you want, yes?

I don't have data on where the initial swirl of solar system mass came from, or how much of that rotational energy went into our planet and its pesky moon, or Who or what gave that initial twist. The Q is pretty high so I assume you could work backwards, but I leave that to astronomers and cosmologists. I believe the 2 ms/day / century estimate we use is one such measurement.

For more on earth rotation rate, UTC and leap seconds see https://www.ucolick.org/~sla/leapsecs/dutc.html

Surely in the literature there is a pile of information or speculation regarding all the rotational energy in the universe. It seems a common theme everywhere you look; maybe it was as much Big Twist as Big Bang? Perhaps in your Pulsar research you've run across some papers you could share. Off-list is ok, unless you think it has general time-nuts appeal. We're running the risk of spinning off-topic already.

Thanks,
/tvb

----- Original Message -----
From: "Jim Palfreyman" jim77742@gmail.com
To: "Discussion of precise time and frequency measurement" time-nuts@febo.com
Sent: Sunday, July 31, 2016 7:34 PM
Subject: Re: [time-nuts] Q/noise of Earth as an oscillator

Hi Tom,

You said: "you need energy; you need energy loss; you need cycles over
which that loss repeatedly occurs."

With regard to the earth, where is the first one? Sure it was there at the
start when the solar system formed, but where is it now?

Jim

On 1 August 2016 at 12:16, Tom Van Baak tvb@leapsecond.com wrote:

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----- Original Message -----
From: "Jim Palfreyman" jim77742@gmail.com
To: "Discussion of precise time and frequency measurement" time-nuts@febo.com
Sent: Sunday, July 31, 2016 7:34 PM
Subject: Re: [time-nuts] Q/noise of Earth as an oscillator

Hi Tom,

You said: "you need energy; you need energy loss; you need cycles over
which that loss repeatedly occurs."

With regard to the earth, where is the first one? Sure it was there at the
start when the solar system formed, but where is it now?

Jim

On 1 August 2016 at 12:16, Tom Van Baak tvb@leapsecond.com wrote:

time-nuts mailing list -- time-nuts@febo.com
To unsubscribe, go to https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
and follow the instructions there.