On 26 Sep, 11:44, HW@....(clueless Henri Wilson) wrote:
On Wed, 26 Sep 2007 01:28:34 -0700, George Dishman wrote:
On 25 Sep, 22:36, HW@....(Dr. Henri Wilson) wrote:
On Tue, 25 Sep 2007 05:47:38 -0700, George Dishman wrote:
What I said is a statement of what is observed,
any valid theory has to predict that behaviour.

George, admit it. Nobody including you has much of a clue about the true nature
of light.

You obviously don't, QED is a complete theory
of light and gives arguably the most accurate
predictions of any physical theory.

George, QED doesn't tell us anything about the nature of light in transit.

You are back to the old philosophical debate
about whether something exists when it is not
being observed. QED tells you all the _science_.

The only facts we know about light relate to its arrival.
Light spends almost all of its time traveling.

To get interference from two detectors spaced 100 metres apart requires
coherence. If a single photon stretches right across then what you said above
is wrong. A 100 m wide photon will strike more than one atom.
Even YOU should see that.

Of course. A simple telescope mirror has to
have the whole surface accurate to a fraction
of a wavelength because it all affects each
photon, but the photon only ever interacts
with _one_ atom when it finally hits the CCD
or photoelectric plate in a PM tube.

You're the one becoming incoherent now George.

I clearly overestimated your knowledge again, I
thought you knew something about the photoelectric
effect. These are simple statements of fact.

You claim
to be familiar with the photoelectric effect
so you should know that already. The same is
true in the VLTI configuration, each photon
is affected by the paths through both
telescopes and all of both their surfaces, but
the detector still sees them as individual
particles. Those are the facts, what you have
to do is invent a theory (a set of equations)
that predicts that and the resulting interference
pattern.

You aren't talking about interference.

Yes I am, interferometers are regularly used
on astronomical telescopes.

You're talking about a simple image.

An image is formed by interference Henry, Paul
told you how.

much more complex, involving limb darkening and lots
of other stuff but those are the basics.

It sounds as though all you are seeing is a fuzzy image of the star.

There are different ways to configure the detection.
In 2D mode what you get is an image crossed by
interference fringes. In 1D mode you get a graph
which looks like a static version of your EM
illustration, a windowed sine wave.

You've lost me George.

I warned you it was complex, I'll try to
upload some synthesised images so you can
see what I mean but it won't be for a few
days.

Of course not, as I told you before, it is the
superposition of the displaced interference
patterns that carries the information....

I could just as easily carry information about varying light speed.

Don';t change the subject, it is the pattern that
provides the information as I ahve told you before,
not a single photon which is your strawman.

Here we go again...every time I ask George a difficult question he accuses me
of changing the subject...

You didn't ask a question, you made a statement
about a different topic. What other things light
might tell us in addition to what we are discussing
is hardly useful.

Tough, the radius is observed to vary and the
interferometric, photometric and integrated
velocity measurements are all similar.

.they are willusions.

Not according to ballistic theory, it says
they are valid.

bull

A statement of fact Henry.

I hope you are impressed George.

No, ballistic theory is wrong as shown by Sagnac
but I am pleased that you have learnt a little
local astrophysics, that's what this game is all
about ;-)

Sagnac disproves SR. ...

Don't waste your time with stupid word games, it would
be better spent if you learned some schoolboy calculus.

I know more about calculus than you do.

ROFL, you just tried to tell me equation [3]
should be (c+v)/n, obviously you cannot even
differentiate an exponential :-)

What I suggested has nothing to do wtih calculus.

So you don't even recognise a first order
differential equation when you see it!

George, you're problem is you copy maths from websites to try to appear
knowledgable when you really haven't a clue what it all means.

Henry, your problem is that you haven't learned
any maths, equation [3] is trivial schoolboy
stuff and I was writing ones like it when I was
13 years old. I still use equations like that in
my work regularly. You seem to think it is some
fancy PhD stuff but the truth is you will find
it in any decent secondary school textbook.
Looking at some web pages suggests The Aussie
level for this stuff should be around year 8
in secondary school.

For equation [3] we want the speed to change as
the light travels, We could write an equation so
for dv/ds or dv/dt, I choose dv/ds for a couple
of reasons, notably because the mechanism must
involve interactions which can be defined as a
number per distance based on a density.

We know the light will speed up if it is too slow
and slow down if it is too fast so if v is large,
the rate of change is negative while if the speed
is low the rate is positive. That means v enters
with a negative sign. The simplest form is first
order and that will be at worst an adequate
approximation as long as v c, and may be exact,
so the basic form must be:

dv/ds = A * (B - v)

where A and B are positive constants. In case
you don't recognise it, the bit on the left is a
derivative meaning the rate at which v changes
as a function of s, the distance travelled, and
derivatives are one aspect of caclulus.

We know the eventual speed is c/n and at that
speed the rate of change is zero so if the bit
in brackets is to be zero constant when v=c/n,
the value of B must be c/n so:

dv/ds = A * (c/n - v)

The left hand side has units of speed/distance
and (c/n - v) has units of speed so constant A
must have units of inverse distance. It is easier
then to define R = 1 / A and hence:

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

where now R has units of distance. Hey presto
Henry, you have equation [3]. Easy, wasn't it.

I prefer to avoid it where possible.

Hardly surprising given your incompetence, but
then that's physics anyway and all you want to
do is talk philosophy.

George, I suggest you take a real course in physics instead of dreaming you
already have.

I have an honours degree in physics already, and
it's a real one, not like your fake paper. What
_you_ need to do is take the Aussie equivalent of
an English "A Level" in maths, or at least get up
to your 8th grade level so you can understand
equations like the one above without me having to
explain them to you. With a bit of practice you
should be able to write them for yourself.

George