Re: This is What Einstein Actually Did.

Henri Wilson wrote:
On Mon, 31 Jul 2006 15:41:19 +0200, "Paul B. Andersen"
<paul.b.andersen@xxxxxxxxxxxxxxxx> wrote:

Henri Wilson wrote:
On Fri, 28 Jul 2006 10:17:42 +0200, "Paul B. Andersen"
<paul.b.andersen@xxxxxxxxxxxxxxxx> wrote:

Henri Wilson wrote:
| You have completely overlooked the common situation in which the main star is
| wobbling around an internal barycentre. In that case, IR should have smaller
| radial speeds than visible. According to my model, that would usually cause
| smaller brightness variation in IR than visible.

Hilarious, no? :-)
Don't quote me out of context.

You know that I was refering to a situation where the large central star was
cool and the outside one small and hot.
Henri Wilson wrote October 20. 2005
|Paul B.Andersen wrote:
| >The BaT predicts no difference in the visible light
| >curve and the 10um light curve and thus is proven wrong.
|
| As usual, you are talking nonsense.
|
| You have completely overlooked the common situation in which the main star is
| wobbling around an internal barycentre. In that case, IR should have smaller
| radial speeds than visible. According to my model, that would usually cause
| smaller brightness variation in IR than visible.

We were discussing Algol.

As a general rule, different layers of a star will have quite different radial
velocities, particularly when the barycentre of the system is below its
surface. That could have profound effects on the observed brightness curves at
different wavelengths.

Exactly!
Hilarious, no? :-)


The parameters found by BaT from observations in
visible light are very different.
(And you have to use the light-curve only and
ignore the spectrum, but that's another issue.)
Paul, it matters not one iota to me whether or not Algol is eclipsing. he fat
that its major axis is roughly aligned with the LOS to Earth means that its
BaTh predicted brightness curve is merely superimposed on any eclipse effect.
And since the observed light curve is as predicted by
conventional theory if it is eclipsing, the BaT predicted
light curve must be flat? :-)

Since no decent spectral information is available for Algol, I would prefer not
to say anything concrete at this stage.

So what _does_ the BaT predict the light curve of Algol should look like?
Don't know? :-)

However, Androcles assures us that its orbit plane is almost normal to the LOS
so I cannot see how it is eclipsing at all.
Confused, Henri? :-)
Hard to say what the BaT predicts, isn't it? :-)

The BaTh predicts that many stars will appear to be eclipsing when in fact they
are not. It also predicts that many stars will appear to flare up occasionally,
supposedly due to gravitatioal lensing. The BaTh predicts exactly the same
'lensing' curve for which Einstein is given sole credit. It isn't lensing at
all.

So what does the BaT predict the light curve of Algol should look like?
Don't know? :-)

So what does conventional theory predict we should
observe at 10 um infra-red?
The obvious.
What does the BaTh predict? Not obvious. The unification length is
probablywavelength dependent.
So it's impossible to say what the BaT predicts?
In other words, it predicts nothing.

keep raving Paul. I will continue to match just about any star curve using c+v.

But you cannot match the light curve of Algol? :-)

It is!
The model constructed according to conventional theory is completely wrong.
Quite.
Since conventional theory predicts exactly what is observed,
it must obviously be completely wrong. :-)

Conventional theory comes up with quite ridiculous and unproveable theories as
to what appears to happen and why.

I note with interest that you have snipped my calculation
of what conventional theory predicts.

You will find it again below.
Please specify what you find "quite ridiculous" about it.
Is Planck's radiation law ridiculous?
Is the idea that stars can eclipse each other ridiculous?

You claim that this calculation
We have two stars.
Algol A: temperature Ta = 12000K, radius Ra = 2.88 solar radii
Algol B: temperature Tb = 4880K, radius Rb = 3.54 solar radii

Their relative brightness at the wavelength lambda will be:
Ba/Bb = (Ra/Rb)^2* W(lambda,Ta)/W(lambda,Tb)
where W(lambda,T) is Planck's radiation law.
Now we have:
(Ra/Rb)^2 = 0.66
W(lambda,Ta)/W(lambda,Tb) =
(exp(C/(lambda*Tb))-1)/(exp(C/(lambda*Ta))-1)
where C = 0.00144 m degree

In the visible spectrum lambda = 0.5 um.
W(0,5um,Ta)/W(0,5um,Tb) = 40

So their relative visual brightness will be:
Ba/Bb = 26.
That is A is 26 times brigter than B.
The binary is 27 times brighter than B.

If we assume that the eclipses are 100%,
we get the following brightnesses (B as unit):
No eclipse = 27
B eclipses A: 1 (primary dip)
A eclipses B: 26 (secondary dip)

The deepness of the minima in magnitudes will be:
Primary dip: 2.5*log(27) = 3.58 magnitudes
Secondary dip: 2.5*log(27/26) = 0.04 magnitudes.

We see that the deepness of the primary minimum fits
quite well with what is observed.
But the secondary minimum is hardly observable at all
in the visible spectrum!


Let us calculate what the deepness of the minima would
be in the infra-red, lambda = 10um.
We use the same method as above:

Ba/Bb = (Ra/Rb)^2* W(10um,Ta)/W(10m,Tb) = 1.8

No eclipse = 2.8
B eclipses A: 1 (primary dip)
A eclipses B: 1.8 (secondary dip)

The deepness of the minima in magnitudes will be:
Primary: 2.5*log(2.8) = 1.12 magnitudes
Secondary: 2.5*log(2.8/1.8) = 0.48 magnitudes.

Observation of the secondary minimum at 10um can be found in:
http://tinyurl.com/mywm8

The observed deepness of the secondary minimum is ca. 0.35.
A little less deep than what I calculated it should be.
However, since B is larger than A, the eclipse will not be 100%,
and the minimum _should_ be less deep.

Your arguments are circular.
Quite.
This is obvious from the very convincing arguments you gave,
isn't it? :-)

Frustrated, Henri? :-)

Matching observed brightness curves with the BaTh is anything BUT frustrating,
Paul. It is most satisfying.

My program: www.users.bigpond.com/hewn/variables.exe has been again
streamlined. Even a Norwegian should now be able to use it.

If you do, you will find an interesting relationship between star period and
light speed unification length.

Of course you will find a relationship between star period
and your "light speed unification length".

If we assume the orbit to be roughly circular,
the BaT predicts the following variation in magnitude:

delta_m = 2.5*log((1 + 2*pi*d*v/(p*c2))/(1 - 2*pi*d*v/(p*c2)))
where:
d = distance [m]
v = amplitude of radial velocity [m/s]
p = period [s]

So all you have to do is to pick a d which give
the desired delta_m.
The shorter the period, the shorter d you must pick.

What conclusion can we draw from that relationship, Henri? :-)

Paul
.

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