Re: This is What Einstein Actually Did.

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

Henri Wilson wrote:
On Wed, 26 Jul 2006 14:33:25 +0200, "Paul B. Andersen"
<paul.b.andersen@xxxxxxxxxxxxxxxx> wrote:

I hadn't thought about it.
Yes, you have.
I have shown you the calculation below before,
and you "explained" the phenomenon thus:

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.

Here is how conventional theory explains it:

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.
...which I assume is what is observed....then somebody worked backwards and
arrived at the starting point of your argument, ie. that "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
"
You are correct to some extent.
The parameters of the Algol binary are found by
analysis of the observations in _visible light_
whether you use conventional theory or BaT.

The above are the parameters found by conventional
theory from observations in visible light.

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? :-)

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? :-)

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!
When the above parameters were found, nobody
had observed Algol in 10 um infra-red.

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.

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.
In other words, conventional theory predicts exactly
what is observed in 10um IR.

The BaT fails miserably, because it predicts that
the light-curves should be the same in visible light
and 10um IR.

Not necessarily.
The whole model of Algol is different under the BaTh.

... and nobody knows what the BaT predicts. :-)

This is quite funny really. You have based all your reasoning on the assumption that the figures you used
are the right ones when in fact they have been derived using identical
reasoning to your own... only in reverse.

In other words your whole theory and argument is circular.
In other words, you are fleeing again because the BaT can
give no explanation for the phenomenon.

I haven't even thought seriously about it.

Yes, you have.
But I note with interest that you are fleeing
your previous "explanation". :-)

> I'm too busy matching other light
curves with BaTh predictions.

Busy inventing fantasy parameters? :-)

The essence is, Henri:
The observed light curves from Algol in visible light and
in 10 um infra-red are very different.
The BaT predicts that they should be equal.
Conventional theory predicts exactly what is observed.

The conclusion in obvious.

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. :-)

Your arguments are circular.

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

Frustrated, Henri? :-)

Paul
.

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