[time-nuts] Q/noise of Earth as an oscillator

Jim Palfreyman jim77742 at gmail.com
Thu Jul 28 18:39:36 EDT 2016

Hi All,

Tom gave me a nudge to look here - I hadn't been following this thread.

For those that don't know, I study pulsars and so the way we measure what
pulsars do could be relevant to this discussion.

First, I have never heard of a Q measure when referencing a pulsar. I think
the key here is that it's not resonating as such. Rotating yes, resonating

Pulsars spin and slow down due to giving off energy (magnetic dipole
radiation). So in the pulsar world we mainly refer to spin frequency (F0)
and frequency derivative (F1). With some of the younger and more "erratic"
pulsars, F2 (and further) can be modelled.

Here's some data on the Vela pulsar (hot off the presses - measured just

F0              11.1867488542579
F1              -1.55859177352837e-11
F2              1.23776878287221e-21

Vela is young and erratic. Millisecond pulsars are outstanding clocks.
Here's the data for J0437-4715 - one of the most stable pulsars we know

F0              173.6879458121843
F1              -1.728361E-15

I'm sure the "Q" of Vela would be pretty decent - but I can tell you now,
as a time-keeper, she's useless.

Jim Palfreyman

On 28 July 2016 at 20:50, Tony Finch <dot at dotat.at> wrote:

> Neville Michie <namichie at gmail.com> wrote:
> > The conical pendulum has a simple form of a weight on a string, instead
> > of oscillating in one plane as a conventional pendulum, it swings around
> > in a circular orbit in the horizontal plane. It has a definite resonant
> > frequency.
> I don't think it does have a resonant frequency, any more than the Earth
> does: the angular velocity of the pendulum is sqrt(g/h) where h is the
> height of the pendulum; give it more energy, it swings higher, so h is
> smaller, so the frequency is higher.
> Tony.
> --
> f.anthony.n.finch  <dot at dotat.at>  http://dotat.at/  -  I xn--zr8h
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