[time-nuts] Q/noise of Earth as an oscillator
Gabs Ricalde
gsricalde at gmail.com
Sun Jul 31 21:44:51 EDT 2016
On Sat, Jul 30, 2016 at 1:19 AM, Tom Van Baak wrote:
> The remaining question in this thread is if earth Q measurement has actual meaning, that is, if the concept of Q is valid for a slowly decaying rotating object, as it is for a slowly decaying simple harmonic oscillator. And that's were get into the history and definition(s) and applicability of Q to non harmonic oscillators, such as coils, capacitors, atomic clocks, planets, pulsars, etc.
>
> /tvb
>
On Mon, Aug 1, 2016 at 6:50 AM, Bill Byrom wrote:
> My final argument is that the rotation frequency of the Earth is
> affected by tidal friction, but the amplitude of the motion of that
> point 100 meters from the axis is unaffected. The amplitude of a
> harmonic oscillator is directly affected by friction or other losses,
> but the effect on resonant frequency is tiny. So loss effects frequency
> in one situation and amplitude in the other. How can Q relate to both
> situations?
> --
> Bill Byrom N5BB
>
I'll try to answer both. Apologies for mistakes or shortcuts, I will
correct or provide details in future posts if needed.
If we consider this definition:
Q = 2*pi * energy stored / energy lost per cycle,
we can write it as
Q = 2*pi * E / (-dE/d(cycle))
= E / (-dE/d(theta))
=> E = E0*exp(-theta/Q).
The energy decays exponentially if the frequency is constant. The
impulse response of an RLC circuit is an exponentially decaying sinusoid
with a fixed frequency, so Q naturally describes the decay of an RLC
circuit and other resonant systems that behave similarly.
Suppose we want to use Q for a rotating object that gradually slows
down. When a rotating object loses energy, the period increases. From
the definition, the energy of a rotating object with constant Q decays
exponentially when we are counting cycles. But each cycle gets stretched
over time, so the energy decay is slower than exponential over time.
Omitting the derivation, a rotating object with constant Q behaves like
this:
E0 = energy at t = 0
I = moment of inertia
k1 = sqrt(E0/I)
k2 = E0*I
theta = 2*Q*log(k1*t/(Q*sqrt(2)) + 1) rad
omega = k1*2*Q / (k1*t + Q*sqrt(2)) rad/s (1)
energy = k2*Q^2 / (E0/2*t^2 + sqrt(2*k2)*Q*t + I*Q^2)
------------------------------------------------------------------------
Consider the Earth with an LOD increase of 2 ms/century (6.338e-13 s/s).
Suppose LOD = 0 at t = 0. Then
k3 = 6.338e-13/(86400 s)
omega0 = 72921151.467064e-12 rad/s
omega = omega0 * (1 - t*k3) rad/s. (2)
( from http://hpiers.obspm.fr/eop-pc/earthor/ut1lod/UT1.html )
The frequency linearly decreases over time, which is different from the
angular frequency of a rotating object with constant Q (1). Again
omitting the derivation, the changing Q of the Earth over time is
Q = omega0 * (1 - t*k3)^2 / (2*k3). (3)
The Q is around 4.97e12 and decreases very slowly to 4.74e12 after 100
million years, assuming a constant 2 ms/century for that duration.
------------------------------------------------------------------------
If the period linearly increases over time, the angular frequency is
omega = 2*pi / (T0 + k4*t) rad/s (4)
where T0 is the period at t = 0 and k4 is the change in period per
second. This looks like the angular frequency with constant Q (1). Let's
make (4) look like (1):
omega = k1*2*(pi/k4) / (k1*t + T0*k1/k4)
so we set Q = pi/k4 and set k4 equal to the change in the period of (2)
at t = 0. Then we get
Q = omega0 / (2*k3) (5)
which is the same as (3) when t = 0. If we approximate omega0 =
2*pi/86400 (1 cycle = 1 solar day), (5) is the same as tvb's formula
pi * (86400 * 365 * 100 / 0.002).
------------------------------------------------------------------------
To recap:
1. An oscillation with fixed frequency and exponential decay has constant Q.
2. A rotating object with linearly decreasing frequency (2) has
decreasing Q (3).
3. A rotating object with linearly increasing period (4) has constant Q (5).
4. Over short time scales or when Q is very large, (3) is almost equal to (5).
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