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). _______________________________________________ time-nuts mailing list -- [email protected] To unsubscribe, go to https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts and follow the instructions there.
