[time-nuts] Q/noise of Earth as an oscillator
Bill Byrom
time at radio.sent.com
Sun Jul 31 23:27:06 EDT 2016
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:
> Hal:
>> Is there a term other than Q that is used to describe the rate of
>> energy loss
>> for things that aren't oscillators?
>
> Jim:
>> cooling (as in hot things)
>> discharge (as in capacitors and batteries)
>> leakage (as in pressure vessels)
>> loss
>
> Scott:
>> An irreversible process would be a better description versus
>> energy loss.
>> Like joule heating (resistance, friction).
>
> 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:
>> ummmâ€¦. Q is the general term of rate of energy loss and we just
>> happen to apply
>> it to oscillators in a very elegant fashionâ€¦.
>
> Oh, no. Now we have both quality factor and elegance factor!
>
> /tvb
> _________________________________________________
> time-nuts mailing list -- time-nuts at febo.com
> To unsubscribe, go to
> https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
> and follow the instructions there.
More information about the time-nuts
mailing list