francisco:
>i was thinking of something along the same line:
>
[quoted text clipped - 6 lines]
>the length of the path traveled by light in vacuum in 1 / 299,792,458 of a
>second.
Right. But since the second itself was originally defined in
terms of distance anyway, it shouldnt seem strange. Before the
second was defined in terms of an atomic transition, it was
defined as 1/31,556,925.9747 of the year 1900, which is just
the distance the earth traversed as it went around the sun that
year.
>currently the second is defined as the duration of 9,192,631,770 vibrations
>of a (specified) radiation emitted by a (specified) isotope of the cesium
>atom.
>vibration or a periodic oscillation of the radiation emitted by the isotope
>is a periodic movement to and fro between two points. let the length of a
>single movement to and fro between two points of the radiation emitted by
>the isotope = L.
>
>does this mean that 9,192,631,770L = length = 1s?
No. The `L' you want is the wavelength. There is nothing oscillating
back and forth in the way you are picturing the to-and-fro motion.
>how fast is one oscillation?
I'm not sure what you mean, but c = \lambda\nu, where \lambda is the
wavelength. The way the correct frequency is found so that the definition
isnt circular, is by tuning the system to peak the resonance. The
resonance depends only on the atomic structure, so the peak will always
occir at the same frequency. Since the definition of a second is completely
arbitrary, the number of cycles, 9,192,631,770, was chosen to coincide as
closely as possible with the older definition.
>what is the length of L?
It's the wavelength of the microwave signal from the cesium.
>if the speed of one oscillation = c, does that mean that 9,192,631,770L =
>299,792,458m?
If the speed of light is `c', then that relation is correct.
c = \lambda\nu
The speed of an oscillation doen't really make sense. I get a wavelength
of 3.26 cm, which is in the microwave regime. It would also be related
to the approximate dimensions of the resonant cavity.