"Henri Wilson" HW@.... wrote in message
...
On Tue, 03 Jul 2007 01:59:47 -0700, George Dishman

wrote:

On 3 Jul, 03:28, HW@....(Henri Wilson) wrote:
On Mon, 02 Jul 2007 00:22:11 -0700, George Dishman
wrote:
On 2 Jul, 01:27, HW@....(Henri Wilson) wrote:
On Sun, 1 Jul 2007 13:13:50 +0100, "George Dishman"
wrote:

There is a single value in the Excel sheet I used
to produce it that defines the width of the raised
cosine as a fraction of the period. It would be
simple to increase that a bit but it won't change
the basic shapes which are produced so I don't see
any point in spending the time to do that, use your
imagination. Barring the rounded edges, the
acceleration is rectangular, the velocity is a
sawtooth and the radius is two quadratics.

Your acceleration curve has an exaggerated length of time during
which
acceleration is constant....it isn't dead constant either.

Indeed, but I always said it was merely illustrative
and the key point here is that real _luminosity_
curves show a steady decline over a signidficant
part of the cycle, they don't show the near constant
values of the acceleration curve.

Ah, I see your problem.

You aren't taking account of the sequential emission delays between the
'pulses'. That is fundamental to the bunching calculations.

You are assuming they are all emitted at the same instant.

Where did you get that daft idea? In fact there
is a small "error" in that I am assuming the
time of arrival is similarly spcaed to the time
of departure but that is correct if the effect
is VDoppler only.

What the hell are you talking about?

I am simply pointing that I have _obviously_ taken
account of the aspect that you said I had missed.
If there weren't "sequential emission delays between
the 'pulses'" then the _transmitted_ frequency would
be infinite.

It is only correct if one assumes Einstein's second postulate.

Garbage, zero time between pulses gives infinite
frequency regardless of speed. I guess you have
some point you are trying to make but you need
try to sort out what you are saying if it is to
be understandable.

For the ADoppler case, it would
have the effect of distorting the X scale of the
plot but not the Y scale, so it would only change
the mark:space ratio of the rectangular curve
which is arbitrary in the illustrative curve
anyway.

George, you seem to have completely lost it.
Racing cars could never overtake each other if you had your way.

Consider two pulses of light emitted from an orbiting source. The second
is
emitted a short time after the first but is moving faster. What happens
George?
The second eventually catches the first, of course.

Of course, but also if they are both emitted while
the source is moving towards the observer, the pair
arrive earlier at their destination than pulses
emitted at the same time but from a source at rest
wrt the barycentre. That is the only effect I have
omitted since the pulse widthis arbitrary anyway
and it is only an illustration, not a simulation.

It so happens that the shape of some elliptical orbits in particular is
such
that pulses emitted at regular interval from 'concave' sections bunch
together
whilst those emitted from the convex, move apart. There are sections from
which
light emitted sequentially over a certain time interval will arrive at an
observer over a much shorter time interval. An observer will see this as
large
brightness increase.

Yes, and that is what I have been discussing all along.

But then the radius would be the second integral
of the luminosity curve, not the first integral,
and that doesn't match observation.

Observation is about photon density, not individual photon energy.

Yes, and the photon density measured as the luminosity
is what I am comparing. It matches the sawtooth velocity
curve far better than the rectangular acceleration.

Let's be clear Henry, according to BaTh, the photon
arrival rate should be the product of the emission
rate, the VDoppler and the ADoppler factors. VDoppler
depends on the velocity curve only but ADoppler
depends on the acceleration and also the speed
equalisation distance. Comparing the luminosity with
the derivatives of the actual radius curve shows that
the speed equalisation must be very short leaving
VDoppler as the dominant effect.

Now include the emission time delay.

It is already included.

No you haven't.

The frequency is not infinite, it is included.

Get it now, George?

Nothing you have said is at odds with my point, but
you seem to be grasping it a bit better now. You
still have some way to go though.

You might understand now.

The only thing I don't understand is why you
are having so much difficulty following what
is essentially a very simple argument.

I agree it is fairly simple.
Why can't you understand it.

Obviously I do, or the transmitted frequency would
be infinite. Where do you get the bizarre idea that
I have not taken it into account?

To summarise - the time between pulses (or wavecrest
emissions) is non-zero. During that time the source
moves some distance towards or away from the observer.
That leads to the VDoppler term, (1+v/c). In addition
the speed for one pulse may differ from the previous
which leads to the ADoppler term, 1/(1-Ra/c^2). The
distance moved and the difference in speed also affect
the time of arrival of each wavecrest which slightly
distorts the resulting curves. So what do you think
I have missed?

George