Henri Wilson wrote:
On Sun, 26 Aug 2007 23:14:54 +0200, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
Explain the 90 deg phase lag then George.
Not hard at all to explain why the curves are different.
If the radius of the star didn't change, it is obvious
from Planck's blackbody equation that the luminosity
variation due to the changing temperature is much
bigger in visible light (V-band) than it is in IR (K-band 2.2u).
So since the radius changes as well, it is clear that
the luminosity variation due to the changing surface area
will be relatively more important in IR than it is in visible light.
That's why the IR-light curve has it maximum and minimum
at ca. phase 0.4 and 0.9 respectivly, just like the angular
diameter curve.
Compare fig 2 (K-mag) and fig2 and 3.
http://www.arxiv.org/PS_cache/astro-.../0402244v1.pdf

The V light curve will be more dominated by the temperature,
which has its maximum and minimum at phase 1 and 0.65 respectively.
Compare fig 2. (V-mag) in document above to fig 4.7 in document
below:
http://ses.library.usyd.edu.au/bitst....0014whole.pdf
http://tinyurl.com/26q3xh

...and I'm sure you could produce an equally nonsensical theory if the
published figures were entirely diffent.

The 'nonsensical' theory is Planck's black body radiation law.
Which is so well confirmed that not even you will question it.
Or do you? :-)

But since you find the verbal description above unconvincing,
(you didn't understand it, did you?) let's do the calculation properly.

As the primary, measured data, I will use the temperature in fig. 4.3 in:
http://ses.library.usyd.edu.au/bitst....0014whole.pdf
and the radius curve in fig 3 in:
http://www.arxiv.org/PS_cache/astro-.../0402244v1.pdf

The question is: What will the K (2.2u) and V (0.5u) light curves be
according to Planck's blackbody radiation law?

The result is shown in the table below. Here a
Int = the surface radiation intensity relative to the intensity at phase 0.
Lum = the luminosity (intensity*area) relative to the luminosity at phase 0.
Mag = the magnitude relative to the magnitude at phase 0.

The Intensity is calculated from Planck's black body radiation law.
Planck(T,lambda). (Look it up if you don't know it.)
K Int = Planck(T,2.2u)/Planck(5600,2.2u)
V int = Planck(T,0.5u)/Planck(5600,0.5u)

Phase: 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
Temp: 5600 5550 5250 5050 4950 4900 4850 4950 5050 5400
Radius: 2.78 2.88 3.10 3.20 3.20 3.15 3.08 2.95 2.77 2.62
Area: 1.00 1.07 1.24 1.32 1.32 1.28 1.23 1.13 0.99 0.89
K Int: 1.00 0.98 0.89 0.84 0.81 0.79 0.78 0.81 0.84 0.94
K lum: 1.00 1.06 1.11 1.11 1.07 1.02 0.95 0.91 0.83 0.83
K mag: 0.00 -0.06 -0.12 -0.11 -0.07 -0.02 0.05 0.11 0.20 0.20
V Int: 1.00 0.95 0.71 0.57 0.51 0.48 0.45 0.51 0.57 0.83
V lum: 1.00 1.02 0.88 0.76 0.67 0.61 0.55 0.57 0.57 0.73
V mag: 0.00 -0.03 0.14 0.31 0.43 0.53 0.64 0.61 0.62 0.34

Compare K mag and V mag to the curves in fig.1 in
http://www.arxiv.org/PS_cache/astro-.../0402244v1.pdf

The fit is so good that one could think I have cheated.
But I haven't. You can check the calculations yourself,
if you don't believe me.
The fit is good simply because Planck's black body radiation law
is correct, and a Cepheid is what it is known to be - a pulsating star.

Do you still find Planck's blackbody radiation law nonsensical, Henri?

Paul