"Henri Wilson" HW@.... wrote in message
...
On Tue, 03 Jul 2007 02:31:52 -0700, George Dishman

wrote:

On 3 Jul, 03:22, HW@....(Henri Wilson) wrote:
On Mon, 02 Jul 2007 06:48:25 -0700, George Dishman
wrote:

Of course they are, the same equations apply.
Multiplication is associative so modulating white
light by a sine wave is exactly the same as
modulating a sine wave carrier with a white noise
signal - e.g. voice on AM.

voice is not white noise George.

Aspirates henry, but stop changing the subject, it
should be obvious that they are as easily definable
mathematically as conventional sidebands on an AM
radio.

Not very convincing George.

It's called the "Commutative Law" in maths, a*b = b*a.

Yes, as I said above, both apply and ADoppler is
the extra factor beyond VDoppler that only appears
in ballisitic theory.

ADoppler is usually by far the dominant factor.

Forget the mantra Henry and think. We have proved
there is no ADoppler on pulsars and for contact
binaries where the speeds are the highest and
ADoppler should show up.

Both have a fixed 'sphere' around them. All light leaves at about the same
speed, c wrt the barycentre of the pair.

Look at your Basic program again, you insisted the
spheres were separate and moving in the bottom picture.

If you look at the radius
curves, there is no ADoppler on Cepheids either,
so no, it is not "usually .. dominant", it is
non-existent in every case we have examined.

Where did you get that idea?

Mental block Henry, I've told you dozens of times.

That's OK, you need to match the radius curve for
this example. That will resolve whether there is
any ADoppler effect or if it is only VDoppler.

VDoppler can't produce large magnitude changes like these.

I know, ballistic theory cannot explain Cepheid
curves.

Quite wrong George.

Fact Henry, though apparently your inability to
accept you have been in error prevents you
realising it.

I have explained all that I have tried to match.


Nope, I have ignored that as we agreed some weeks
ago. I am simply differentiating the radius to get
the velocity and then the acceleration.

You aren't using it as an indicator of bunching (or photon density).

Yes I am, I am comparing it against the luminosity
curve which is a direct count of photons in a given
band.

You clearly don't understand the principle involved.

Mental problems again? Remember I had to write the
BaTh equation for speed equalisation for you, I
understand the principles far better than you.

which IS velocity
dependent....h.(c+v)/lambda.
You are completely ignoring the principle factor involved, which is
'number of
photons arriving per second'.

On the contrary, that is the only factor I am
considering at the moment.

You aren't considering the bunching effect due to velocity differences.

Yes I am, that is the ADoppler part which should
be proportional to the top of my three plots, but
the curve shape is completely wrong - that shows
there is no ADoppler.

George, think about this:


A


B


C



1__2___3____4_____5______6____-v,a D


An accelerating source emits pulses of light at equal time intervals at
the
points shown. The speed of each pulse is c wrt its source.
You can easily imagine how the pulses bunch together as they approach the
three
points A, B, C and D.


I think ".. the three points A, B, C and D." have been
messed up by Usenet. Just consider point D, we are only
interested in the radial components and any transverse
component of the speed affects the radial only via
Pythagoras as a second order speed term so changes the
result by less than one part per million.

You can see that the pulse density distribution at A, B, C and D can be
manipulated by curving the path of the source.

Nothing you are saying is relevant to this process.

Think again. Consider points 1 through 6 as the surface
of the star as it expands away from the centre. The
photon rates you will calculate will depend on the speed
and acceleration of the surface, both of which can be
found by differentiating the radius.

That's a good one George....

It's your own Henry, do the sums.

f'=f(c+v)/c

OK, now apply that to the carrier and sideband
frequencies independently. Then inverse transform
the three to get the received waveform. What speed
does the modulation travel at? Show your working ;-)

I don't see the point.
What are you getting at.

Yet again Henry, consideration of sidebands allows
you to calculate the Doppler shift directly from
the speed of modulating pulses hence a speed of
c+v determines the shift.

You don't have a model for individual photons.

I don't need a model, I just need to know that
they deflect by the same angle as the classical
wave on hitting a grating which is proved by
the photomultiplier experiment. Again this is
something I have pointed out dozens of times.
When are you going to stop trying to change the
subject and address the proof?

Again, that is true only if the speed equalisation
distance is large so that it travels a long way
through the ISM at variable speed. The same is true
without the sphere if the light so again it still
appears redundant.

NO NO NO!!!!!. You are quite wrong there George.
The sphere effectively becomes the source. If it moves with the star,
then the
original c+v relationship with Earth holds.

Exactly, so what is the difference from saying it
leaves the star at c+v?

But George, if TWO stars are in close orbit, the common sphere remains
vurtually at rest wrt both.

You spent a long time telling me there were two spheres and
each moved with its parent star. You lost the plot somewhere.

No I haven't. The spheres aren't rigid steel balls.... they behave more
like a
gas and their effect probably drops off with an inverse square law.

Go back and read your posts again.

Their
contributions are additive so two equally sized orbiting stars will end up
with
an almost steady sphere with a couple of small circulating bumps.. ALL
light
leaves that sphere at about c wrt the sphere and NOT at c wrt each star.

There is very little if any 'c+v'. The light from
both leaves at c/n...but is wavelength shifted dring the unification
process.

Surely you can see this.

Yep, but the same is true if the light changes to speed
c at surface of the heliopause as it moves into the ISM
so what does the sphere do?

I just explained.

What you said still doesn't have any effect. Perhaps
you should calculate the result before guessing any more.

You are right, I can't see why you think it makes any
difference.

see above. It makes a difference in cases like contact binaries.

You said the sphere moved with the star.

For a single star or a well separated pair, that is true. The sphere makes
little or no difference to the speed of light leaving the system.

Have a look at your Basic animation, the bottom
diagram.

... what I said was that
we know there are no errors in the derivation
of the predicted angle from the respective
theories and they give different answers.

Ther is an optical lens effect anyway.

That's what we are talking about, effectively
Newton predicts twice the focal length for a
given mass.

I meant an atmospheric lens as well as the gravitational one.

That doesn't come into the maths of GR or Newton,
they give different predictions.

I say much of the bending is optical rather than gravitational.

Nope, it's less than a millionth of the gravitational
bend (from memory, Craig Markwardt posted the details
about a year or more ago in reply to Sean). You can
easily separate the effects since the optical is
frequency dependent while gravitational is not.

I say that my
'spheres' also bend light.


I repeat, I suggest you go and look ;-)

I have
.http://www.users.bigpond.com/hewn/stupidjerry.jpg

is typical

It doesn't show a radius curve.

Why should it?

You have lost the plot Henry, the conversation is about
comparing the first and second derivatives of the radius
to the luminosity to find out whether ballistic theory
says it is VDoppler or ADoppler.



It shows typically observed cepheid brightness curves.


You are missing the whole point.
You are becoming quite clueless George.

Since you can't even work out what was being
calculated, it is you who needs the clue - see
above.

George, I'll let the computer calculatebthe bunching.

Let me give you the clue again - we were talking about
differentiating the radius to get the radial velocity.
You have wandered off on all sorts of tangents.

No, YOU have....

Exactly, so stop trying to change the subject.

because you are starting to realise that I'm right.

ROFL, Henry ballistic theory gets _every_ prediction
wrong unless it the source is at rest. You claimed
that Cepheid luminosity variation was produced by
ADoppler so I have pointed out that is not true
because the curve matches the shape of the velocity,
not the acceleration of the surface. As you of course
realise, it is also three orders of magnitude too small
but that's another matter, you said you were just
matching the shapes and on that basis alone, only
VDoppler gives a match.

Again, You are becoming quite clueless George.

Simple statement of fact Henry.

Even Max Keon thinks you are losing it.

Have a look at Max's first attempts at writing
equations, he though he needed two, one for negative
numbers and another for positive. Both took the
square root of a square. He didn't know the
associative, distributive and commutative laws until
I pointed him at K12 pages. His maths is way behind
yours!

It could still be pretty good then.

If you consider not knowing the level expected of a 12
year old "good" for an adult. The derivatives of the
radius I am discussing are a couple of years ahead of
that and you don't seem to be able to understand them
even after I drew the plots for you.

So find one that gets eclipsed at the fundamental.
Statistically there must be many.

I would like to find one.

Exactly, without them you have a problem to explain.

I don't have any problems.

You do, there are thousands of Cepheids known so if they
were binaries, many should be eclipsing but none are.

I can simulate eccentricity and yaw angle to within
a few percent.

Not without a companion.

Yes you do, you have nothing to compare against the
radius.

The radial velocity of the surface of a star that goes 'huff puff' is
very
similar to that of one in elliptical orbit.

You think? So add the curve to your software
and let's see it.

That's how it works already.

So add the curve and let's see what it predicts,
what's your problem?

Except for one thing George.
VDoppler variations are minute. ADoppler can easily produce
variations up to
mag 4.

However, differentiating the radius twice is nothing
like the luminosity curve, it only matches the velocity.

George, your own suggested method of calculating photon bunching matches
just
about any brightness curve.

Don't try to change the subject Henry.

I thought you would be impressed. You actually achieved something. You
method
is faster than mine even if it is considerably harder to program. It
produces
exactly the same results.

Flattery won't work either. I repeat:

However, differentiating the radius twice is nothing
like the luminosity curve, it only matches the velocity.
However, differentiating the radius twice is nothing
like the luminosity curve, it only matches the velocity.

Got it yet, or won't your mental state allow you to
respond to that point?

You are referring to the TRUE radius variation. It is the OBSERVED
variation
that matters.

I am referring to the radius measured by means of the
angle subtended by the star so I don't see what
distinction you are drawing.

OK, some people have claimed to have seen cepheids actually pulsating.

For goodness sake Henry, what do you think we have been
talking about for the last several weeks ????? The
ESO page is exactly that measurement.

There might be stars that actually do that and it might indeed be
possible to
see them....but I would be very suspicious..

Welcome to the conversation.

I don't think anyone has actuallyseen the radius pulsating.

Yes they have:

http://tinyurl.com/239mw6

The red dots with vertical error bars are the measured
angular diameter in milli arc seconds. Multiply by the
distance to get radius and differentiate to get surface
velocity. Alternatively integrate the measured velocity
curve obtained spectroscopically and you get the
background smooth curve.

That is what we have been talking about for a couple of
weeks now so if you have finally grasped the plot, maybe
you can say something sensible about it this time.

Go ahead then, add the radial distance curve, match
it to the radius curve for L Car and "let your program
provide the answers".

Again, all you are considering is the h.c/lambda energy
effect.....not the
'photon density' one.

Nope, you are lost entirely. I am discussing the
measured luminosity and CCD detectors are photon
counters, not sensitive to the energy. None of
this discussion has been related to photon energy
at any point.

Just let the computer do the sums and produce the curves George.
My program is corrrect.

ROFL, I have pointed out the error in it many, many
times.

There is no error. Even your own suggested method produces the right
answers.

Sure, but you only use that method on one curve when it
applies to both, your "measured velocity" curve is wrong.

Incidentally, I have now included the effects of tidal bulges and have
found
that their effects are very similar to those of a first overtone. I have
matched some brightness curves very closely.

Worthless until you match the radius at the same time,
or find an eclipser so you can match the phase, As you
admitted, you can just as easily match a curve with
VDoppler as with ADoppler

George, for most 'cepheids' the radius is constant.

Nope, the ESO chart for L Car is how a typical
Cepheid behaves.

Note also, if you differentiate the radius, you get
the speed of the surface relative to the barycentre
of the star. Subtract that from the observed velocity
curve and you get the velocity of the barycentre and
guess what, since they are the same, there is no
observed motion of the barycentre at the luminosity
period.

Most cepheids are merely stars in orbit around something dark. Many are
egg
shaped, due to tidal effects.

If you want anyone to believe a simple binary system
has been mistaken for a variable-radius Cepheid then
you will need to show a discrepancy between the
derivative of the angular radius and the spectroscopic
velocity. There is none for L Car and no reason to
think it isn't entirely typical.

George