Why are the 'Fixed Stars' so FIXED?

"Henri Wilson" HW@.... wrote in message
...
On Wed, 22 Aug 2007 18:39:17 +0100, "George Dishman"
wrote:
"Henri Wilson" HW@.... wrote in message
. ..
On Mon, 20 Aug 2007 03:39:59 +0000 (UTC), bz
wrote:
HW@....(Henri Wilson) wrote in
m:
On Sun, 19 Aug 2007 14:44:51 +0000 (UTC), bz
wrote:
HW@....(Henri Wilson) wrote in
news:q7sec355kreu68a8jpo9qtadisje8qur8t@4ax. com:
...
You haven't a clue either.

the ADoppler shift is slightly different for the layers where the two
bands
dominate (on average)

Maybe somebody needs to give Henry a clue. The layer
is few hundred km thick. The figure for the Sun is
400km and the temperatures are comparable so I'll
use that as an example. The light we see contains
contributions from all depths through that but there
is little light emitted from the top and most of what
is emitted at the bottom is absorbed as the higher
material is opaque. Somewhere in the middle, the
contribution per metre of depth peaks. Now suppose
the location of that surface of peak contribution
varies with frequency by 10% of the depth between
two widely separated filter bands, that's just 40km
difference between the surfaces. Of course as the
layer rises and falls, if those surfaces stay 40km
apart, there is no difference in their motion, but
suppose it varied from 30km to 50km due to some phase
lag between them. You now have a difference in motion
of 10km.

For L Car the radius of the layer varies by about
1.6 million km over the cycle so that's a difference
in motion of 6 parts per million.

According to Henry, this is supposed to explain why
the K band surface brightness varies by 5.3% while in
the V band it is over 50%.

The K band varies by about 35% and the V band by nearly 90%.

Well done Henry, you are actually doing physics now.

the most interesting feature is that the K max lags the V max by about 90.
Can you expain that George?

Easily, brightness is luminosity per unit area. Why do
you think I have been correcting you on that for weeks?

The radius varies by about 12% giving an area variation
of 26%. We would expect the K band brightness variation
of 33% to be mostly due to the change of area and if
you compare the K curve with figure 2 you will see the
maxima are around phase 0.4 with the minimum at 0.9
and the minimum is a sharper turn-round than the max.
In fact the curves are very similar.

For the V band the situation is reversed, the 26% area
change is less than the 90% luminosity change so the V
band curve should be dominated by the temperature curve.
Unfortunately that isn't shown in this paper but if you
look at typical temperature curves you will see they are
similar to luminosity in general.

That corresponds to what we expect from the Planck curve
as the peak of the radiation curve is much higher than
the K band (and of course filter widths matter too).

To check the match you need to combine both radius and
temperature effects. Turning it round, divide the
luminosity by the brightness curve derived from the
temperature curve for the band and you get the area.
The square root of that gives the photometric radius
shown in figures 2 and 3 and as you can see it matches
the integrated velocity and ESO's interferometric angular
radius extremely well.

Your modelling using ballistic theory should be trying
to match the _difference_ between the curves (in terms
of brightness of course), but there is no difference
above the noise.

George