On Fri, 21 Sep 2007 01:08:23 +0100, "George Dishman"
wrote:


"Henri Wilson" HW@.... wrote in message
.. .
On Wed, 19 Sep 2007 21:36:10 +0100, "George Dishman"
wrote:
"Henri Wilson" HW@.... wrote in message
...
On Mon, 17 Sep 2007 19:23:54 +0100, "George Dishman"

Multiple images starts at what I called the critical
distance given by d = c^2/a where a is the peak radial
acceleration. Variation of 1.5 mags means a ratio of
4:1 in luminosity which means the speed unification
distance is about 75% of the critical distance. Other
values are

Percentage of
Mag 1:n critical
1 2.5 60.19%
1.5 4.0 74.88%
2 6.3 84.15%
3 15.8 93.69%
4 39.8 97.49%
5 100.0 99.00%
6 251.2 99.60%
7 631.0 99.84%
8 1584.9 99.94%
9 3981.1 99.97%

Isn't it magical how we see variations of up to 9 mags
in some stars yet NEVER see multiple images.

It is true some stars are reported to vary by 7-9 mags. This cannot be
explained solely by c+v effects.

Of course it can Henry, that's the point. Suppose
we take an arbitrary figure of 10 light years
for the speed equalisation distance in the space
surrounding some star. To get 1.5 mag variation
you need a critical distance of 13.35 light years.
To get 9 mag that needs to be 10.0025 light years,
just a 25% increase in the peak orbital
acceleration.

What the hell are you talking about?

Check the numbers, see for yourself. For a given
value of peak acceleration, draw a graph of peak
variation as a function of distance.

What do you think my program does?

Magnitude changes of 2-3 are easy to produce without the appearance of
peaks in
the brightnes curves.

Sure, but there is no law that says stars can't
orbit a little faster or have slightly higher
eccentricity so your problem is explaining why
we _don't_ see such spikes. The answer of course
is that every sysytem we see only shows VDoppler,
never ADoppler.

You get 10/10 for stubborness.

Do the arithemetic and see for yourself.

the computer does it...

Even more
remarkable when you realise the critical distance depends
on inclination while the speed equalisation depends only
on the nature of the ISM, is that the heliosphere?

I have explained the role of inclination.

You miss the point. Suppose a star shows 5 mags
variation to astronomers on Earth and the orbit
has an inclination of 45 degrees. An astronomer
on a nearby star might see an inclination of
44.44 degrees, and he would see a variation of
9 mags because of the slightly higher peak
acceleration. At 44.42 degrees, he sees multiple
images. There is no reason why that shouldn't be
us, you need fairies to carefullt orient all the
inclications in the galaxy so that they have just
the right angle to little old Earth :-)

...and this by a person who boasts about his maths ability....
How embarassing for you....

Check the numbers then, they are correct. You
have never grasped this problem because you
don't do the arithmetic.

.....and I really thought you understood this stuff....
Unification ensures we wont see multiple images....we wont even see magnitude
changes due to BaTh above about .

.. all my numbers above were done in a few
minutes with a spreadsheet, but you couldn't
write that and don't understand the consequences
because you don't practice your maths.

Your numbers are completely wrong...
I think you have confused logs with exponentials...

I think you are forgetting that magnitudes
are logarithmic, or perhaps that ADoppler
goes as 1/(c^2-da). As d approaches c^2/a,
you get a very rapid rise because the
denominator approaches zero.

the situation never arises...


George


Henri Wilson. ASTC,BSc,DSc(T)

www.users.bigpond.com/hewn/index.htm