Re: This is What Einstein Actually Did.

Henri Wilson wrote:
On 1 Aug 2006 19:57:06 -0700, "Jerry" <Cephalobus_alienus@xxxxxxxxxxx> wrote:

Henri Wilson wrote:
On Tue, 01 Aug 2006 13:44:05 +0200, "Paul B. Andersen"
<paul.b.andersen@xxxxxxxxxxxxxxxx> wrote:
Hilarious, no? :-)
No.

So what _does_ the BaT predict the light curve of Algol should look like?
Don't know? :-)
You have seen the curves.
Go to
http://www.astro.psu.edu/~mrichards/research/journey.html
Fit the 1920 Angstrom curve
What are your calculated orbital parameters?

This is very interesting. It's a three body curve. Two stars are in the same
eccentric orbit around a larger body. One is 180 degrees behind the other.

See: www.users.bigpond.com/hewn/threebodyalgol.jpg

Add the two curves and what do you get? My program doesn't handle three bodies
yet but I will include that when I get around to it.

So what does the BaT predict the light curve of Algol should look like?
Don't know? :-)
You have seen the curves...
Go to
http://www.astro.psu.edu/~mrichards/research/journey.html
Fit the 5500 Angstrom curve
What are your calculated orbital parameters?

Yes. Different layers of a star have different radial velocities and therefore
will exhibit bigger magnitude changes at the same observer distance.

But you cannot match the light curve of Algol? :-)
You have seen the curves.
Go to
http://www.astro.psu.edu/~mrichards/research/journey.html
Fit the 1.2 micron curve
What are your calculated orbital parameters?

Like I said, different layers of a star have different radial velocities and
therefore will exhibit bigger magnitude changes at the same observer distance.

I am sure the astrophysicists will be glad to learn that
they can observe the interior of the Sun in infrared. :-)

Observation of the secondary minimum at 10um can be found in:
http://tinyurl.com/mywm8
Go to
http://tinyurl.com/mywm8
Fit the 10 micron curve.
What are your calculated orbital parameters?

Don't worry about the parameters.

Why are your calculated orbital parameters all different?

Unification.

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.
Naturally you will come up with such accurate answers when the assumptions made
about the two stars was entirely based on the answers themselves. Once again
you have fallen victim to circular logic.

Show us the spectral dtata Paul...
Prior to 1978, Algol was considered a "single-lined" spectroscopic
eclipsing binary. With improved instrumentation, spectral lines
from the secondary are now resolvable. The Doppler shifts of the
primary and secondary components are simultaneously observable.
http://adsabs.harvard.edu/cgi-bin/nph-bib_query?bibcode=1978ApJ...222L.119T

Algol is a binary star, Henri.

It has three bodies.

Right.
But Algol C is orbiting the inner binary with period 1.86 years.


Plus, the X-ray spectrum of the secondary has been observed
by Chandra, and the measured X-ray Doppler shifts are totally
consistent with the Doppler shifts measured in visible light.

There are lots of double-lined eclipsing spectroscopic binaries,
by the way.

Do a google on
double-lined eclipsing spectroscopic binary

Give up Jerry. c+v is here to stay.

Thanks for giving me a few new ideas, by the way.

Oh, my dear Henri.
You are really something. :-)

Algol is a three body system for some wavelengths
and a two body system in other wavelengths.
The reason is unification.

Keep it up, Henri.
Your new ideas are always entertaining.

Paul
.

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