Why are the 'Fixed Stars' so FIXED?

Another post my ISP dropped:

"Henri Wilson" HW@.... wrote in message
...
On 11 May 2007 07:47:37 -0700, George Dishman
wrote:
On 11 May, 02:21, HW@....(Henri Wilson) wrote:
On 10 May 2007 04:09:06 -0700, George Dishman
wrote:

No it isn't, you are forgetting that it gets changed by speed
equalisation. If the wavelength didn't change, there would
be no Doppler shift whatsoever.

That's another issue, leave it out for now.

We will come back to that below.

It is fundamental, the purpose of a grating is to
measure an unknown wavelength.

No it isn't. It uses the angle of diffraction to measure the velocity of
the
source, knowing the true absolute wavelength.

Rubbish, what is the "absolute wavelength" of a tunable
dye laser?

George, I'm going to cut this thread because it is only repeating what is
said
in the other one.

I'll do the same, we have covered most of the
ground already.

If you want to use a grating to measure ABSOLUTE wavelength, just make
sure the
source is at rest wrt the grating and there will be no problem.

And how do I do that for a distant star Henry, get real.

George the grating hasn't but the observer should know how to recognize
certain
lines..

The only way you can recognise a line id if you know
its wavelength Henry, that's why you use a grating to
measure it.

If you are using that in your grating, even if it HAS changed during
travel,
the assumption that it hasn't will still allow the observed
diffraction angle
to be used to calculate initial c+v.

It is an assumption Henry, the fact remains that what a
grating measures in BaTh is the reflected wavelength.

No George. Its diffraction angles are sensitive to wavelength ...

Exactly, that is what it does, nothing else. What use is
made of that result is nothing to do with the way a grating
works which is what we are discussing.

You DO know the absolute wavelength of certain spectral lines.

No you don't, the wavelength depends on the proper
motion of the source star and the cosmological and
gravitational redshifts.


It still says the same thing and it is still wrong, for a single
grating, the angle of deflection depends on the reflected
wavelength:

You're still bogged down in the constant c model.

ROFL, Henry the whole point of my argument is that the
change of speed on reflection means that lambda_r is
noit the same as lambda_i. If I was using "constant c"
the problem wouldn't arise.

It doesn't matter what u is.

Yes it does, the wavelength gets changed by (c+u)/(c+v).
If I was using "the constant c model" as you suggested
I would NOT need to distinguish between lambda_i and
lambda_r which I have done consistently. Your inaccurate
criticism belies a total failure to understand the true
situation.

The fact remains that my equation explains why hubble should detect its
own
motion whilst yours does NOT.

You still don't get it Henry, my equation is just your
equation with the algebra that was beyond you completed.
What _both_ equations show is that the angle phi depends
_only_ on the reflected wavelength, and in fact if lambda_i
is "absolute" then your equation suggests Hubble should
_not_ detect its own motion _unless_ the grating changes
the wavelength depending on the incident speed. Nothing in
your theory says that happens because you haven't shown
the BaTh equation for reflected speed.

SR does not predict that the HST should detect variations in
diffraction angle
due to its own orbit speeds.

Yes it does, it produces the same equation.

Light sped is not in the SR equation.

The equation is the same for both, N * lambda = D sin(phi).

The difference is that in SR lambda_r = lambda_i so we
don't need to distinguish.

That is only true if v=u.

You were referring to SR, v=u=0

I believe that u = c.

I think you mean u=0, the reflected speed is c+u.

It should for a transmission grating...so a comparison of
diffracted angles produced by the two types should give us a measure of
any
difference between u and v.

In fact, I now have an experiment.
Set up the HST to watch a particular star, monitor the angular change of
Hred
as it orbits, using both a transmission and a reflection grating.

Does the Hubble have both transmission and reflection
gratings fitted Henry? If not, are you going to pay for
the missing one to be installed? You're not living in the
real world.

Compare the
two.
If there's a difference, it will prove that u is not equal to v or c in my
equation as SR is definitely wrong. If there is no difference it wont tell
us
much at all.

SR says v=u and so far there is no evidence to
contradict that.

BaTh can't lose.

Let me give you a simpler experiment, you don't even
need a grating. If v =/= u the angle of reflection
from a mirror will differ from the conventional rule.
Set up an experiment in the lab like Ives and Stillwell
with light from a beam of atoms being reflected off a
mirror at 45 degrees. Then change the beam speed and see
if the light is reflected at a different angle. I think
you will find this is done quite often, for example
perhaps looking at the spectrum of accelerator beams,
but it shouldn't be hard to set up in the lab either.

The purpose of a grating is to make a measurement, that
is what the 'grating equation' does for you. The thing that
is measured is the reflected wavelength. Note that is
based on you diagram and I have a minor reservation
about it but you need to learn the basic principle before
we look in more detail.

George, gratings are used to determine wavelengths from sources at rest
wrt the
grating.

Gratings deflect light by an angle that depends on the
wavelength, period. Whether the source is moving or not
is not relevant.

The angle will be different, as you know. But moving the grating rather
than
the source ensures that nothing happens to the absolute 'wavelength'.

We are discussing the generic "grating equation" and
what I said remains true.

They don't give the same equation.

Yes they do, both give N * lambda = D * sin(phi)

The BaTh adds a .c/(c+v)...which is what is required.

No, the angle phi only tells you lambda_r, you have no
measure of v.

Study my diagram again George.

You seem to be incapable of doing basic algebra so
here is a corrected diagram that illustrates how a
grating works without requiring the extra step:

http://www.georgedishman.f2s.com/Hen...ic_grating.gif

I told YOU that equation.

No you didn't. You gave a more complex version which
I had to simplify for you:


"George Dishman" wrote in message
...

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

see: www.users.bigpond.com/hewn/bathgrating.jpg

Well done Henry. So your equation is

lambda_i * (c+u)
sin(phi) = ---------------
D * (c+v)

where lambda_i is the wavelength of the _incident_
light.

The wavelength of the reflected light, lambda_r, is
given by

lambda_r c+u
-------- = ---
lambda_i c+v

So your equation can also be written

lambda_r
sin(phi) = --------
D



The points you are missing is that Lambda_i is absolute and known...

No it isn't. We are talking about the generic equation for
a grating when it its used as a measuring instrument. So
far we have proved only

lambda_r = D * sin(phi) / N

because in the instrument itself only D is known and phi
can be mesured.

and that my
equation will allow hubble to detect its own orbital motion with a grating
whilst yours will not.

Sorry Henry wrong on both points. If lambda_i is "absolute"
and v=u as a truly ballistic theory suggests then it couldn't
detect it because lambda_r = lambda_i which is unaffected by
Hubble's speed.

On the other hand in SR, the wavelength isn't absolute and
changes due to Doppler so SR gets it right.

Wavelength of light is intrinsic and cannot change just because a
grating is
moved somewhere, George.

That is your religion Henry, not reality.

The observed is never affected by the observer george.

Doppler shift Henry.

That doesn't affect the observed George.

It does in SR, and in reality.

It only affects the measuring
technique. If you measure a 0.7 um line to be 0.71 um, YOU have made an
error
of judgement somewhere and must compensate for it. In the case of
starlight of
course, you simply allow for the relative movement betwen you and the
star.

SR says gratings are purely wavelength sensitive, George.
SR must be wrong.

Nope, the wavelength changes in reality.

It certainly might change after hitting the grating. So might the
velocity...by
the same ratio. The frequency remains constant.

Nope, in reality frequency depends on observer motion
too Henry, Ives and Stilwell proved the second order
part of SR Doppler.

I said the frequency AFTER reflection must be the same as that BEFORE.
I think that is right.

No, you said the frequency remains constant which is
wrong. The incident and reflected frequencies are the
same but they vary with the velocity of the observer.

Angle phi depends only on lambda_r.

It depends entirely on Lambda_i and c+v/c+u

And that combination is lambda_r.

http://www.georgedishman.f2s.com/Hen...ic_grating.gif

I already told YOU that George.

No, you said the grating equation was:

lambda_i * (c+u)
sin(phi) = ---------------
D * (c+v)

I had to do the algebra to show you that becomes:

lambda_r
sin(phi) = --------
D

Don't try to take the credit now Henry, it's taken
you a week to realise I was right all along.

Lambda_r = D * sin(phi)

That's all you can say.

I only need Lambda_e/(c+v).

You don't know v, angle phi depends _only_ on lambda_r
so that is what is measured, all else is conjecture.

You haven't studied the diagram George.

Since you apparently cannot understand the algebra
required by your own diagram, see the corrected one.

You keep repeating thngs I already know.

Then why do you keep arguing against them? I think
you only post for the fun of arguing.

George