"Dr. Henri Wilson" HW@.... wrote in message
...
On Sun, 20 Apr 2008 22:52:29 +0100, "OG" wrote:


"Dr. Henri Wilson" HW@.... wrote in message
. ..
The planet's reflection could easily be mistaken for emission from a
cooler
star since it spectrum lines would be doppler shifted 180 out of phase
wrt
the
star.


What does 'doppler shifted 180 out of phase' actually mean ?

This is obviously far too hard for you. Stop making a fool of yourself and
go
quietly away.

You have still to explain how doppler shifting can cause phase change.

On Wed, 23 Apr 2008 11:59:18 +0200, "Paul B. Andersen"
wrote:

Dr. Henri Wilson wrote:
On Tue, 22 Apr 2008 20:58:48 +0200, "Paul B. Andersen"
wrote:

That's because I am quoting messages by one:
"A small hot star reflecting off a very large orbiting WCH could
easily result in two different spectra, B and K, shifted 180 out
of phase."

You didn't read it properly.

I quoted you litterally, and I read what you wrote.

You omitted vital words.

I said the light from a small hot star when reflected off a close orbiting
planet could easily be mistaken for light being emitted from a second
star....since the doppler shifts of light from the two would be 180 out of
phase.

The wavelength distribution would also likely be different, probably with the
mode shifted towards the red, making the 'second star' appear cooler.

Not much better.
Light reflected off a planet will never
be mistaken for the light from a star.

Oh?
What would be the main differences?

But is this an attempt to flee your statement about Algol,
namely that the secondary K2 spectrum could easily be a reflection
of the primary B8 spectrum from a large cool dead star or planet?

Do you retract this statement, or do you defend it?

Or is this another case where you didn't say what you said? :-)

B8 is considerably hotter than a K2.
Where is the problem?
Reflected B8 light could easily lose much of the shorter wavelength
contributions.
The radiation curve could conceivably resemble that of a faint K2.


Henri Wilson. ASTC,BSc,DSc(T)
www.users.bigpond.com/hewn/index.htm

.....specialising in teaching physics to engineers and mathematicians....

On Wed, 23 Apr 2008 12:22:58 +0100, "OG" wrote:


"Dr. Henri Wilson" HW@.... wrote in message
.. .
On Sun, 20 Apr 2008 22:52:29 +0100, "OG" wrote:


"Dr. Henri Wilson" HW@.... wrote in message
...
The planet's reflection could easily be mistaken for emission from a
cooler
star since it spectrum lines would be doppler shifted 180 out of phase
wrt
the
star.


What does 'doppler shifted 180 out of phase' actually mean ?

This is obviously far too hard for you. Stop making a fool of yourself and
go
quietly away.

You have still to explain how doppler shifting can cause phase change.

I appreciate it is difficult....and maybe you are also Norwegian.

Here is a linear analogy of a ring gyro:

Two identical oscillators are positioned at different distances from a distant
'detection point', D. They are emitting continuous waves and are initially in
phase.

S1_________________________________D
S2

Since the distance between the oscillators and point D is different,the number
of wavelengths in each path is not the same.
At a particular instant, they are set moving towards D at different speeds,
such that they arrive at D together.

Since they move through different path lengths, the number of wavecrests from
each oscillator arriving at D before the two sources arrive is also different,
causing the observed phase displacement.

Alternatively, the frequency of 'wavecrest arrival' at D from the two
oscillators is not the same. The two rays appear doppler shifted by (c+v)/c and
(c-v)/c. Since they travel for the same time and are initially in phase, they
are out of phase when they meet at the detector.

Get it now?





Henri Wilson. ASTC,BSc,DSc(T)
www.users.bigpond.com/hewn/index.htm

.....specialising in teaching physics to engineers and mathematicians....

Dr. Henri Wilson wrote:
On Wed, 23 Apr 2008 11:59:18 +0200, "Paul B. Andersen"
wrote:

Dr. Henri Wilson wrote:
On Tue, 22 Apr 2008 20:58:48 +0200, "Paul B. Andersen"
wrote:

That's because I am quoting messages by one:
"A small hot star reflecting off a very large orbiting WCH could
easily result in two different spectra, B and K, shifted 180 out
of phase."
You didn't read it properly.
I quoted you litterally, and I read what you wrote.

You omitted vital words.

Henri Wilson never said what he said, eh?

I said the light from a small hot star when reflected off a close orbiting
planet could easily be mistaken for light being emitted from a second
star....since the doppler shifts of light from the two would be 180 out of
phase.

The wavelength distribution would also likely be different, probably with the
mode shifted towards the red, making the 'second star' appear cooler.
Not much better.
Light reflected off a planet will never
be mistaken for the light from a star.

Oh?
What would be the main differences?

If you measure the visual spectrum from one of the terrestial
planets and try to determine the spectral class as if it was
a stellar spectrum, you will find that it is a G2 spectrum
like the Sun. Of course there would be differences telling you
that the spectrum isn't from a star, but the main point
is that it cannot be mistaken for any other type of spectrum
than G2.

See why below.

But is this an attempt to flee your statement about Algol,
namely that the secondary K2 spectrum could easily be a reflection
of the primary B8 spectrum from a large cool dead star or planet?

Do you retract this statement, or do you defend it?

Or is this another case where you didn't say what you said? :-)

B8 is considerably hotter than a K2.
Where is the problem?
Reflected B8 light could easily lose much of the shorter wavelength
contributions.
The radiation curve could conceivably resemble that of a faint K2.

How do you manage to stay this ignorant after having
discussed these issues for years?

I have told you this numerous times, PLEASE LEARN IT THIS TIME:
The spectral class of a star is determined by the relative
positions and strengths of the absorption lines,
not by where the black body spectrum peaks.
That's why a Doppler shift doesn't affect the determination
of the spectral class.
There is a strong (one to one) correlation between
the spectral class and the temperature of a star,
so when the spectral class is determined, so is the
temperature.

Look at the B0 spectrum he
http://cass.ucsd.edu/public/tutorial...s/O-Gspect.gif
compare it to the K0 spectrum he
http://cass.ucsd.edu/public/tutorial...s/G-Mspect.gif
You can also see the spectra he
http://www.astro.umd.edu/~ssm/ASTR220/OBAFGKM.html

Note that the B spectrum has few absorption lines in
the red end of the spectrum, while the K spectrum
has a lot of absorption lines in the red end.

The spectra are very different, _and have *very* different
sets of absorption lines_. There is no way you can make
a B spectrum look like a K spectrum by "shifting it towards
the red."

If you want to learn about the determination of spectra,
go he
http://www3.gettysburg.edu/~marschal/clea/speclab.html
and download the SpecLab program.
You can then measure the spectra of different stars
with a virtual telescope, and classify them by comparing
them to standard spectra.
Have fun. :-)

Bottom line:
You have to be - if not a moron - extremely ignorant of
the most basic issues of astronomy not to realize that
the statement:
"A small hot star reflecting off a very large orbiting WCH could
easily result in two different spectra, B and K, shifted 180 out
of phase."
is incredible stupid.

But staying ignorant about the issues you talk about
every day for years is a speciality of yours.
Isn't it?


--
Paul

http://home.c2i.net/pb_andersen/

OG wrote:
"Dr. Henri Wilson" HW@.... wrote in message
...
On Sun, 20 Apr 2008 22:52:29 +0100, "OG" wrote:

"Dr. Henri Wilson" HW@.... wrote in message
...
The planet's reflection could easily be mistaken for emission from a
cooler
star since it spectrum lines would be doppler shifted 180 out of phase
wrt
the
star.

What does 'doppler shifted 180 out of phase' actually mean ?
This is obviously far too hard for you. Stop making a fool of yourself and
go
quietly away.

You have still to explain how doppler shifting can cause phase change.

And your are obviously asking what it means in this context:
Henri Wilson:
" .. its spectrum lines would be doppler shifted 180 out of phase
wrt the star."

I see that Henri has tried to answer this question, but he has obviously
forgotten what he is talking about, and is giving an incredible
confused answer. :-)

It is quite simple.
We are talking about the B8 spectrum and the K2 spectrum from
respectively the primary and the secondary component of the Algol binary.
Since the components are orbiting each other in circular orbits,
the radial velocity of the stars will vary sinusoidally, and
the variation will be 180 degrees out of phase.
When one star is approaching, the other is receding, and vice versa.
Since the radial velocity varies, the Doppler shift of
the spectrum will vary sinusoidally. The spectrum is blue shifted
when the star is approaching, red shifted when it is receding.
So when the A8 spectrum is blue shifted, the K2 spectrum is red
shifted, and vice versa.

Loosely said:
"The two spectra are Doppler shifted 180 out of phase."

--
Paul

http://home.c2i.net/pb_andersen/

Dr. Henri Wilson wrote:
On Wed, 23 Apr 2008 12:22:58 +0100, "OG" wrote:

"Dr. Henri Wilson" HW@.... wrote in message
...
On Sun, 20 Apr 2008 22:52:29 +0100, "OG" wrote:

"Dr. Henri Wilson" HW@.... wrote in message
...
The planet's reflection could easily be mistaken for emission from a
cooler
star since it spectrum lines would be doppler shifted 180 out of phase
wrt
the
star.

What does 'doppler shifted 180 out of phase' actually mean ?
This is obviously far too hard for you. Stop making a fool of yourself and
go
quietly away.
You have still to explain how doppler shifting can cause phase change.

I appreciate it is difficult....and maybe you are also Norwegian.

Well, let's hope he is not from down under,
because being upside down seems to lead to confusion:

Here is a linear analogy of a ring gyro:

Two identical oscillators are positioned at different distances from a distant
'detection point', D. They are emitting continuous waves and are initially in
phase.

S1_________________________________D
S2

Since the distance between the oscillators and point D is different,the number
of wavelengths in each path is not the same.
At a particular instant, they are set moving towards D at different speeds,
such that they arrive at D together.

Since they move through different path lengths, the number of wavecrests from
each oscillator arriving at D before the two sources arrive is also different,
causing the observed phase displacement.

Alternatively, the frequency of 'wavecrest arrival' at D from the two
oscillators is not the same. The two rays appear doppler shifted by (c+v)/c and
(c-v)/c. Since they travel for the same time and are initially in phase, they
are out of phase when they meet at the detector.

Get it now?

Try standing on your head.
It must be better, because it can't be worse.

--
Paul

http://home.c2i.net/pb_andersen/

"Dr. Henri Wilson" HW@.... wrote in message
...
On Wed, 23 Apr 2008 12:22:58 +0100, "OG" wrote:


"Dr. Henri Wilson" HW@.... wrote in message
. ..
On Sun, 20 Apr 2008 22:52:29 +0100, "OG"
wrote:


"Dr. Henri Wilson" HW@.... wrote in message
m...
The planet's reflection could easily be mistaken for emission from a
cooler
star since it spectrum lines would be doppler shifted 180 out of phase
wrt
the
star.


What does 'doppler shifted 180 out of phase' actually mean ?

This is obviously far too hard for you. Stop making a fool of yourself
and
go
quietly away.

You have still to explain how doppler shifting can cause phase change.

I appreciate it is difficult....and maybe you are also Norwegian.
Here is a linear analogy of a ring gyro:

Two identical oscillators are positioned at different distances from a
distant
'detection point', D. They are emitting continuous waves and are initially
in
phase.

S1_________________________________D
S2

Since the distance between the oscillators and point D is different,the
number
of wavelengths in each path is not the same.
At a particular instant, they are set moving towards D at different
speeds,
such that they arrive at D together.

When you say '*they* arrive at D together' - what is the *they* that you are
referring to? 'S1 & S2 themselves' or the 'waves from S1 and S2'

Since they move through different path lengths, the number of wavecrests
from
each oscillator arriving at D before the two sources arrive is also
different,
causing the observed phase displacement.

The difference in path length isn't constant, does this mean that the phase
displacement is also not constant?

Alternatively, the frequency of 'wavecrest arrival' at D from the two
oscillators is not the same. The two rays appear doppler shifted by
(c+v)/c and
(c-v)/c. Since they travel for the same time and are initially in phase,
they
are out of phase when they meet at the detector.

Where does v come from?

On Thu, 24 Apr 2008 14:25:17 +0200, "Paul B. Andersen"
wrote:

OG wrote:
"Dr. Henri Wilson" HW@.... wrote in message
...
On Sun, 20 Apr 2008 22:52:29 +0100, "OG" wrote:

"Dr. Henri Wilson" HW@.... wrote in message
...
The planet's reflection could easily be mistaken for emission from a
cooler
star since it spectrum lines would be doppler shifted 180 out of phase
wrt
the
star.

What does 'doppler shifted 180 out of phase' actually mean ?
This is obviously far too hard for you. Stop making a fool of yourself and
go
quietly away.

You have still to explain how doppler shifting can cause phase change.

And your are obviously asking what it means in this context:
Henri Wilson:
" .. its spectrum lines would be doppler shifted 180 out of phase
wrt the star."

Are you drunk? We are discussing Sagnac.

I see that Henri has tried to answer this question, but he has obviously
forgotten what he is talking about, and is giving an incredible
confused answer. :-)

It is quite simple.
We are talking about the B8 spectrum and the K2 spectrum from
respectively the primary and the secondary component of the Algol binary.
Since the components are orbiting each other in circular orbits,
the radial velocity of the stars will vary sinusoidally, and
the variation will be 180 degrees out of phase.
When one star is approaching, the other is receding, and vice versa.
Since the radial velocity varies, the Doppler shift of
the spectrum will vary sinusoidally. The spectrum is blue shifted
when the star is approaching, red shifted when it is receding.
So when the A8 spectrum is blue shifted, the K2 spectrum is red
shifted, and vice versa.

Loosely said:
"The two spectra are Doppler shifted 180 out of phase."

So now you are agreeing with me. Do you also apologise for your stupidity?

In the absence of a third object, the doppler shifts are exactly 180 out of
phase.



Henri Wilson. ASTC,BSc,DSc(T)
www.users.bigpond.com/hewn/index.htm

.....specialising in teaching physics to engineers and mathematicians....

On Thu, 24 Apr 2008 22:43:31 +0100, "OG" wrote:


"Dr. Henri Wilson" HW@.... wrote in message
.. .
On Wed, 23 Apr 2008 12:22:58 +0100, "OG" wrote:


I appreciate it is difficult....and maybe you are also Norwegian.
Here is a linear analogy of a ring gyro:

Two identical oscillators are positioned at different distances from a
distant
'detection point', D. They are emitting continuous waves and are initially
in
phase.

S1_________________________________D
S2

Since the distance between the oscillators and point D is different,the
number
of wavelengths in each path is not the same.
At a particular instant, they are set moving towards D at different
speeds,
such that they arrive at D together.

When you say '*they* arrive at D together' - what is the *they* that you are
referring to? 'S1 & S2 themselves' or the 'waves from S1 and S2'

....a particular infinitesimal element that leaves the source and is split into
the two, Each half goes into a different ray. The two leave together and arrive
at the detector together. One travels further than the other.

The source actually emits two traveling waves in opposite directions..
In the source frame, the wavelength (distance between point of equal phase) in
both rays is the same. Therefore, since the path lengths are different, there
are more whole wavelengths in one ray than the other.

Since they move through different path lengths, the number of wavecrests
from
each oscillator arriving at D before the two sources arrive is also
different,
causing the observed phase displacement.

The difference in path length isn't constant, does this mean that the phase
displacement is also not constant?

At zero rotation speed, there is no fringe displacement.
At any other rotation speed, there is a finite and constant displacement.

Fringe MOVEMENT occurs only during a change in rotation speed.

Alternatively, the frequency of 'wavecrest arrival' at D from the two
oscillators is not the same. The two rays appear doppler shifted by
(c+v)/c and
(c-v)/c. Since they travel for the same time and are initially in phase,
they
are out of phase when they meet at the detector.

Where does v come from?

'v' is the peripheral speed of the ring wrt the nonrotating frame.

Hence the rays move at c+v and c-v wrt that frame.


Henri Wilson. ASTC,BSc,DSc(T)
www.users.bigpond.com/hewn/index.htm

.....specialising in teaching physics to engineers and mathematicians....

"Dr. Henri Wilson" HW@.... wrote in message
...
On Thu, 24 Apr 2008 22:43:31 +0100, "OG" wrote:


"Dr. Henri Wilson" HW@.... wrote in message
. ..
On Wed, 23 Apr 2008 12:22:58 +0100, "OG"
wrote:


I appreciate it is difficult....and maybe you are also Norwegian.
Here is a linear analogy of a ring gyro:

Two identical oscillators are positioned at different distances from a
distant
'detection point', D. They are emitting continuous waves and are
initially
in
phase.

S1_________________________________D
S2

Since the distance between the oscillators and point D is different,the
number
of wavelengths in each path is not the same.
At a particular instant, they are set moving towards D at different
speeds,
such that they arrive at D together.

When you say '*they* arrive at D together' - what is the *they* that you
are
referring to? 'S1 & S2 themselves' or the 'waves from S1 and S2'

...a particular infinitesimal element that leaves the source and is split
into
the two, Each half goes into a different ray. The two leave together and
arrive
at the detector together. One travels further than the other.

I was talking about YOUR analogy. Please stick to the terms of YOUR
explanation

S1, S2 and D
Now - please explain how this helps explain the term - 'doppler shifted 180
out of phase'. If you can.