Why are the 'Fixed Stars' so FIXED?

On Wed, 19 Sep 2007 21:17:09 +0100, "George Dishman"
wrote:


"Henri Wilson" HW@.... wrote in message
.. .
On Mon, 17 Sep 2007 19:33:27 +0100, "George Dishman"

wrote:


"Henri Wilson" HW@.... wrote in message
...


Your model consists of two equations:

v = u+c

and

dv/ds =(c/n - v)/R

Neither has any oscillatory term.


Now write the equation for the wave in a moving organ pipe..whilst it is
producing sound...

Why would I do that? It isn't a solution to either
of the equations so not part of your theory.

You obviously don't understand my photon theory..

You don't have a photon theory, you only talk
about them as bursts of classical waves but
your equations do not produce photons. See
my description of your 'EM' program for an
analysis and then think what you could do to
your equations to make such a pattern self
sustaining during propagation. I don't think
it is possible.

See my reply to Ghost in the other thread.

....oscillations that are not directly related to the
man made electric waves they often make up.

Then you need some equation that links the two,
your model is again lacking.

I don't need an equation. I can imagine the result.

Then you have no theory, you just imagine you
have one.

I admit it isn't complete..

See my response to your new post. If you have any
new equations to add, let me know but the summary
is based on the two you have.

The definition of a "black body" is that it
absorbs all incident radiation. That means
absorptivity = 1 and by Kirschoff's Law that
means emissivity = 1.

Yes well Ok ,

OK, so think twice before saying I am confused next
time, you have been wrong every time.

that's not what I meant to convey. What I meant was "emitters can
have a black body type curve but an emissivity of less than 1".

"Gray atmosphere" I believe is the term in this
field, but Kirchoff's Law still applies and since
the opacity increases with depth you still cannot
see through to deeper layers beyond the photosphere.

But you just asked how long it took light from the sun's core to reach the
surface. If t reaches the surface at all, surely you can 'see' it.

You see the surface from which it was last emitted.
Think of an opaque film with light shining on it.
The film heats up and emits black-body radiation
so a black plastic film with sunlight might glow
at 300K even though the spectrum illuminating it
is over 5000K. You can still feel the warmth but
there will be a slight delay if a shaow passes
over it. Try it with a black plastic bag in a frame
in your back garden.

Not convincing....

While on that subject. I once decided to heat my computer room in winter by
hanging a black plastic curtin over the large north facing window ( I live in
OZ). This worked well, acting as both a re-radiator and a double
insulator...The room warmed very efficiently and was dark enough for me to see
the screen.
There was only one problem...after several weeks, I started to get a sore
throat whenever I sat there for long. I finally woke up to the fact that the
plastic was slowly decaying and giving off what was probably a pretty deadly
gas...It didn't kill me but I wouldn't recommend it to anyone else.

I later rigged up a similar curtain using sheet metal and achieved the same
result....plenty of warmth with little bright sunlight...



Perhaps, I have yet to see any verification test
results for that, but you allow for harmonics and
contributions from a second source which mean the
total is not Keplerian.

My program also incorporates 'egg shaped stars', for instance in tidal
lock.
This is another possible source of an apparent 'overtone'

Right, nothing to do with a Keplerian orbit at all.
Now you understand the point.

...the curve shape is still Keplerian....

Nope, you have extras, it is just arbitrary but
no matter, we know it isn't an orbit anyway.

George




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