sci.physics.relativity

"Henri Wilson" wrote in message
news:4k7pc3tja00ehj6vd1409608cuficcrr98@ ...
> On Tue, 21 Aug 2007 23:44:10 +0100, "George Dishman"
> wrote:
>>"Henri Wilson" wrote in message
>>news:l5ghc39j97u7ukbffe2m32i5kat1cd20g5@ ...
...
>>> The surfeace DOES NOT varty sinusoidally. It varies 'elliptically' and
>>> with
>>> 'hysteresis'. Different layers move slightly out of phase with each
>>> other.
>>
>>There is only one visible layer - Kirchoff's Law.
>
> Actually, there is another solution to this.
> On average, light from each band is emitted at different phases of the
> expansion cycle...but becasue of the different radial speeds,

Henry, the surfaces cannot be more than a few km
apart yet the move together by more than 10^6 km.
The difference in speeds is at least 7 or 8 orders
of magnitude less that the speed of the layer as a
whole and about 12 orders of magnitude less than
the speed of light. You need to stop hand-waving
and do some very basic calculations.

> they arrive
> seemingly IN phase.
> So what I'm suggesting is that the light is emitted from the fairly thin
> surface layer at slightly different times rather than from different
> layers (at
> the same instant)

Do the sums Henry, for toy numbers suppose the
separation varies by +/- 10km peak over 35 days
and the layer as a whole moves by 1,600,000 km
in that time. What is the peak difference in
speed? What is the difference between time of
emission? What is the speed difference between
the layers?

>>>>VDoppler from ADoppler. It's really simple you know,
>>>>if you write a program for a sine wave motion,
>>>>velocity and acceleration are in quadrature so the
>>>>radius curve is shifted by 90 degrees depending on
>>>>VDoppler or ADoppler. Your existing program should
>>>>be able to do that easily, just add a plot for true
>>>>radius.
>>>
>>> You don't understand. Similaly shaped and phased velocity and brightness
>>> curves
>>> can result from V or A doppler.
>>
>>Sure, but there is a 90 degree phase difference
>>which allows us to distinguish them. Fit the
>>velocity curve using VDoppler and the radius
>>curve matches, fit it with ADoppler and it is
>>90 degrees out. You _cannot_ use the luminosity
>>curve because ...
>
> NO THERE IS NOT A 90 degree phase shift.
> If ADoppler dominates, the apparent brightness maximum is due to the
> bunching
> of light emitted at the region of MAXIMUM ACCELERATION....or thereabouts.

Henry, do the sums. In K band, L Car varies by
about mag (by eye from the graph) or a factor
of . The angular diameter varies from about
mas to mas from the following

/38ydvu

That is a factor of and means the disc area
varies by , so if the luminosity varies by
then the brightness varies by / = .

Incidentally, that should illustrate why using
the terms "luminosity" and "brightness" correctly
matters.

So at most ADoppler accounts for % while the
change in diameter alone accounts for 26% and
then you still have the temperature variation to
consider so ADoppler certainly doesn't dominate
and all the evidence is that it doesn't exist.

> BUT because individual photons are also ADOPPLER 'compressed',

Oh whoopeee - yet another ad hoc bodge, what the
heck is 'ADOPPLER compression'?

> the maximum
> APPARENT doppler shift is IN PHASE with the observed brightness maximum.
>
>>> The BIG difference is that only ADoppler can produce large brightness
>>> variations
>>
>>Wrong, the _ratio_ of different bands is independent
>>of both VDoppler and ADoppler factors and tells us the
>>temperature variation, and the temperature variation
>>_alone_ is enough to explain the brightness variation.
>>That together with the radius change, is enough to
>>explain the luminosity curve.
>
> The two bands are emited at slightly different instants... when the radial
> velocities and accelerations are also slightly different. We see them
> arriving
> in phase.

Do the maths Henry, the differences are utterly
negligible.

George