Why are the 'Fixed Stars' so FIXED?

On 8 May 2007 23:58:13 -0700, George Dishman wrote:

On 9 May, 00:41, HW@....(Henri Wilson) wrote:
On 8 May 2007 00:41:49 -0700, George Dishman wrote:
On 8 May, 00:06, HW@....(Henri Wilson) wrote:
On Mon, 7 May 2007 11:17:54 +0100, "George Dishman" wrote:
"Henri Wilson" HW@.... wrote in messagenews:mkqt331o5mk4ifujqvseogifnioaqpd62e@4ax .com...

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

You have been claiming that the speed didn't appear in
the equation and that wavelength couldn't change. One
or the other is wrong. You also claimed the formula
used frequency instead of wavelength but that too isn't
true. Naturally you can replace the wavelength by speed
over frequency but that just reintroduces speed in the
equation.

Desperate again George?

I'm having to teach you basic algebra yet again Henry.

Lambda_i is absolute and all we need.

Lambda_r doesn't enter into this.

Lambda_i isn't enough, if you want to use it you need
to know v and u as well but the grating doesn't measure
them. Remember all we know is the angle phi so you
can turn the second version round to get

Lambda_r = D * sin(phi)

but that's as far as you go. Your first equation isn't
usable because v and u aren't known so in BaTh a
grating doesn't measure Lambda_i, only Lambda_r.

George, Lambda_i is known. It is absolute and universal for a particular
spectral line.

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.

The difference between the measured angle and the expected one is a measure of
c+u/c+v
(Can we assume u is zero?).

No, you can't assume that but even if you did you
don't know v or lamda_i, the whole point of using a
grating is to _measure_ something you don't know.

Don't try to wriggle out with this distraction George.

Conventionally v=0, u=0 and lambda_r = lambda_i
but in BaTh none of those are known. The
measurement of phi tells you lambda_r only.

George, the wavelength of, say, Halpha is absolute and universal.

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.

The equation uses points of equal phase to calculate the angle of the wavefront
of the diffracted beam.

Yes, your basic equations are right but you are left
with two unknowns. Essentially the incident speed
and wavelength are 'conjugate' as you used the term
in relation to pitch and velocity in your simulation so
you don't know either. Going the extra step to express
it in terms of Lambda_r resolves the problem.

You are missing the point.
The BaTh says Lambda_i is absolute for any known spectral line.

No it doesn't, it says Lambda _emitted_ is known but that
isn't the wavelength of the lght incident on the grating.

This is a separate factor but like I said, it doesn't matter anyway. All
velocity changes during flight are accompanied by a corresponding shifts in
absolute wavelength.
The time, 'lambda/'c+v'' remains constant.
So diffraction angle remains the same.

Let's assume that u =0, ie., the reflected light moves at c wrt the GRATING.

The result is as I said: Sin(phi)=D/lambda.(c/(c+v)), for 1st order
diffraction.

However knowing D and phi still leaves two unknowns,
lambda and v, so cannot be solved for either.

No, lambda is known George.

Sorry Henry, you forgot speed equalisation.

forget it ....it doesn't matter.

Speed is included in the equation....so the BaTh explains what is observed.

Lambda_r = D * sin(phi)

George, if you want to measure lambda_r, you will have to put another grating
in the diffracted beam.

No, the equation for a single grating in BaTh, what you
called the "grating equation" tells you Lambda_r, not
Lambda_i.

please read again george.


In the useable form, speed is not included in the equation.

SR does not.

SR gives the same equation but since we know the
speed is c we also have

Lambda_i = Lambda_r

in the frame of the grating.

You've definitely lost it this ime George.
We are talking about the BaTh....not SR....

You said "SR does not.", I just corrected your error.

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

Lambda_i is known.

No, or there would be no point in measuring it.

It isn't measured. It is known and used to calculate relative source speeds.


Don't be stupid, both theories give the same equation.
However, in BaTh a grating cannot measure what you
call the 'absolute wavelength', only the reflected
wavelength. That's a limitation which suggests you
would need other instruments to find v and u.

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.

SR's one infers that the HST gratings would
NOT detect its own orbital movement.

Rubbish, don't try guessing Henry, you don't know
anything about SR so you're not going to get it right.
You know perfectly well that the conventional
grating equation is what I've shown above.

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

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

The BaTh equation says it will.

An definite victory for hte BaTh wouldn't you say?

Just wrong on every count, you can't even work out
what your own theory says about a grating.

You tried to introduce a red herring and it was promptly caught and eaten.

The BaTh also explains sagnac.

Sagnac doesn't need an "explanation", it is a simple
measurement of OWLS from a moving source and the
result is c which falsifies Ritz's theory. There is a
superficial 'explanation' which I expected you to put
forward a couple of years ago but maybe you have
spotted the problem in it already. Anyway, as it stands
at the moment, you don't have a theory that is
compatible with Sagnac or the Shapiro delay.

I can see I will have to go right through this again.

The question George, now is, "does light reflect from a moving mirror at the
incident angle and speed, wrt the mirror...or does it reflect at c wrt the
mirror and at an angle detemined by the BaTh grating equation?"

Dealt with three years ago, the incident light moves at
c wrt the mirror so the question is moot, the reflected
light also moves at c wrt the mirror whichever model
you adopt and the incident and reflected angles are
equal. The Sagnac experiment doesn't have a grating
in it so I don't know why you even mention that, seems
like you have lost the plot this time Henry.

Specular reflection is a limit case of grating diffraction.



My point is simply that you guessed what the
equation would contain rather than working it out.
When you got round to it, I'm sure it only took a
few minutes but you have now discovered that your
assumptions were inaccurate, speed does not
appear in the final equation, only the reflected
wavelength:

Lambda_r = D * sin(phi)

Can you not get it into you head George, lambda_i is universal and known.

Not according to ballistic theory. You still don't
understand the predictions of your own theory.

It doesn't affect the result if it DOES change during travel. Lambda_i/(c+vi)
is constant.

You also suggested it used the frequency but that
also isn't true because you don't know c+u which
is needed to get frequency from Lambda_r.

Assume u =0....although it might not be....

If it might not be then you can't assume, but even
if you do, you don't know v and you don't know
lambda_i or you wouldn't be trying to measure it
in the first place. The BaTh grating equation is:

Lambda_r = D * sin(phi)

That's all you can say.

I only need Lambda_e/(c+v).
The ratio remains constant during any extinction that takes place.


George



www.users.bigpond.com/hewn/index.htm

Einstein's Relativity - the greatest HOAX since jesus christ's virgin mother.

On 10 May, 01:30, HW@....(Henri Wilson) wrote:
On 8 May 2007 23:58:13 -0700, George Dishman wrote:
On 9 May, 00:41, HW@....(Henri Wilson) wrote:
On 8 May 2007 00:41:49 -0700, George Dishman wrote:
On 8 May, 00:06, HW@....(Henri Wilson) wrote:
On Mon, 7 May 2007 11:17:54 +0100, "George Dishman" wrote:
"Henri Wilson" HW@.... wrote in messagenews:mkqt331o5mk4ifujqvseogifnioaqpd62e@4ax .com...

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

You have been claiming that the speed didn't appear in
the equation and that wavelength couldn't change. One
or the other is wrong. You also claimed the formula
used frequency instead of wavelength but that too isn't
true. Naturally you can replace the wavelength by speed
over frequency but that just reintroduces speed in the
equation.

Desperate again George?

I'm having to teach you basic algebra yet again Henry.

Lambda_i is absolute and all we need.

Lambda_r doesn't enter into this.

Lambda_i isn't enough, if you want to use it you need
to know v and u as well but the grating doesn't measure
them. Remember all we know is the angle phi so you
can turn the second version round to get

Lambda_r = D * sin(phi)

but that's as far as you go. Your first equation isn't
usable because v and u aren't known so in BaTh a
grating doesn't measure Lambda_i, only Lambda_r.

George, Lambda_i is known. It is absolute and universal for a particular
spectral line.

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.

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

The difference between the measured angle and the expected one is a measure of
c+u/c+v
(Can we assume u is zero?).

No, you can't assume that but even if you did you
don't know v or lamda_i, the whole point of using a
grating is to _measure_ something you don't know.

Don't try to wriggle out with this distraction George.

It is not a distraction, a grating is a measuring instrument
and what it measures in BaTh is the _reflected_ wavelength,
nothing else. Other parameters my be inferred using
assumptions but they are not measured.

Conventionally v=0, u=0 and lambda_r = lambda_i
but in BaTh none of those are known. The
measurement of phi tells you lambda_r only.

George, the wavelength of, say, Halpha is absolute and universal.

You forget the Doppler shift due to proper motion of the
source. The grating has no idea the light is hydrogen
alpha, the fact is that the angle of deflection depends
_only_ on the reflected wavelength.

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.

snip repetition

....
George, if you want to measure lambda_r, you will have to put another grating
in the diffracted beam.

No, the equation for a single grating in BaTh, what you
called the "grating equation" tells you Lambda_r, not
Lambda_i.

please read again george.

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

Lambda_r = D * sin(phi) / N

Anything beyond that requires assumptions and calculations.

In the useable form, speed is not included in the equation.

SR does not.

SR gives the same equation but since we know the
speed is c we also have

Lambda_i = Lambda_r

in the frame of the grating.

You've definitely lost it this ime George.
We are talking about the BaTh....not SR....

You said "SR does not.", I just corrected your error.

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.

Lambda_i is known.

No, or there would be no point in measuring it.

It isn't measured. It is known and used to calculate relative source speeds.

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.

Don't be stupid, both theories give the same equation.
However, in BaTh a grating cannot measure what you
call the 'absolute wavelength', only the reflected
wavelength. That's a limitation which suggests you
would need other instruments to find v and u.

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.

SR's one infers that the HST gratings would
NOT detect its own orbital movement.

Rubbish, don't try guessing Henry, you don't know
anything about SR so you're not going to get it right.
You know perfectly well that the conventional
grating equation is what I've shown above.

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

That is your religion Henry, not reality.

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

Nope, the wavelength changes in reality.

The BaTh equation says it will.

An definite victory for hte BaTh wouldn't you say?

Just wrong on every count, you can't even work out
what your own theory says about a grating.

You tried to introduce a red herring and it was promptly caught and eaten.

Hook line and sinker.

....
Dealt with three years ago, the incident light moves at
c wrt the mirror so the question is moot, the reflected
light also moves at c wrt the mirror whichever model
you adopt and the incident and reflected angles are
equal. The Sagnac experiment doesn't have a grating
in it so I don't know why you even mention that, seems
like you have lost the plot this time Henry.

Specular reflection is a limit case of grating diffraction.

OK.

My point is simply that you guessed what the
equation would contain rather than working it out.
When you got round to it, I'm sure it only took a
few minutes but you have now discovered that your
assumptions were inaccurate, speed does not
appear in the final equation, only the reflected
wavelength:

Lambda_r = D * sin(phi)

Can you not get it into you head George, lambda_i is universal and known.

Not according to ballistic theory. You still don't
understand the predictions of your own theory.

It doesn't affect the result if it DOES change during travel. Lambda_i/(c+vi)
is constant.

Angle phi depends only on lambda_r.

You also suggested it used the frequency but that
also isn't true because you don't know c+u which
is needed to get frequency from Lambda_r.

Assume u =0....although it might not be....

If it might not be then you can't assume, but even
if you do, you don't know v and you don't know
lambda_i or you wouldn't be trying to measure it
in the first place. The BaTh grating equation is:

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.

George

On 10 May 2007 04:09:06 -0700, George Dishman wrote:

On 10 May, 01:30, HW@....(Henri Wilson) wrote:
On 8 May 2007 23:58:13 -0700, George Dishman wrote:
On 9 May, 00:41, HW@....(Henri Wilson) wrote:

You have been claiming that the speed didn't appear in
the equation and that wavelength couldn't change. One
or the other is wrong. You also claimed the formula
used frequency instead of wavelength but that too isn't
true. Naturally you can replace the wavelength by speed
over frequency but that just reintroduces speed in the
equation.

Desperate again George?

I'm having to teach you basic algebra yet again Henry.

Lambda_i is absolute and all we need.

Lambda_r doesn't enter into this.

Lambda_i isn't enough, if you want to use it you need
to know v and u as well but the grating doesn't measure
them. Remember all we know is the angle phi so you
can turn the second version round to get

Lambda_r = D * sin(phi)

but that's as far as you go. Your first equation isn't
usable because v and u aren't known so in BaTh a
grating doesn't measure Lambda_i, only Lambda_r.

George, Lambda_i is known. It is absolute and universal for a particular
spectral line.

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.

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.



The difference between the measured angle and the expected one is a measure of
c+u/c+v
(Can we assume u is zero?).

No, you can't assume that but even if you did you
don't know v or lamda_i, the whole point of using a
grating is to _measure_ something you don't know.

Don't try to wriggle out with this distraction George.

It is not a distraction, a grating is a measuring instrument
and what it measures in BaTh is the _reflected_ wavelength,
nothing else. Other parameters my be inferred using
assumptions but they are not measured.

explained in the other message.

Conventionally v=0, u=0 and lambda_r = lambda_i
but in BaTh none of those are known. The
measurement of phi tells you lambda_r only.

George, the wavelength of, say, Halpha is absolute and universal.

You forget the Doppler shift due to proper motion of the
source. The grating has no idea the light is hydrogen
alpha, the fact is that the angle of deflection depends
_only_ on the reflected wavelength.

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

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 but its main
function (we're talking astronomy remember) is to determine source speed.

Certainly it CAN be and has widely been used to measure wavelength if the
source is at rest wrt the grating.

snip repetition

...
George, if you want to measure lambda_r, you will have to put another grating
in the diffracted beam.

No, the equation for a single grating in BaTh, what you
called the "grating equation" tells you Lambda_r, not
Lambda_i.

please read again george.

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.

Lambda_r = D * sin(phi) / N

Anything beyond that requires assumptions and calculations.

George, I'm talking BaTh not fairyland physics.



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.

Lambda_i is known.

No, or there would be no point in measuring it.

It isn't measured. It is known and used to calculate relative source speeds.

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.
This whole discussion is about the relatively moving situation.
So don't try to distract me with irrelevancies....

Don't be stupid, both theories give the same equation.
However, in BaTh a grating cannot measure what you
call the 'absolute wavelength', only the reflected
wavelength. That's a limitation which suggests you
would need other instruments to find v and u.

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.

SR's one infers that the HST gratings would
NOT detect its own orbital movement.

Rubbish, don't try guessing Henry, you don't know
anything about SR so you're not going to get it right.
You know perfectly well that the conventional
grating equation is what I've shown above.

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.

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.


Dealt with three years ago, the incident light moves at
c wrt the mirror so the question is moot, the reflected
light also moves at c wrt the mirror whichever model
you adopt and the incident and reflected angles are
equal. The Sagnac experiment doesn't have a grating
in it so I don't know why you even mention that, seems
like you have lost the plot this time Henry.

Specular reflection is a limit case of grating diffraction.

OK.

My point is simply that you guessed what the
equation would contain rather than working it out.
When you got round to it, I'm sure it only took a
few minutes but you have now discovered that your
assumptions were inaccurate, speed does not
appear in the final equation, only the reflected
wavelength:

Lambda_r = D * sin(phi)

Can you not get it into you head George, lambda_i is universal and known.

Not according to ballistic theory. You still don't
understand the predictions of your own theory.

It doesn't affect the result if it DOES change during travel. Lambda_i/(c+vi)
is constant.

Angle phi depends only on lambda_r.

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

You also suggested it used the frequency but that
also isn't true because you don't know c+u which
is needed to get frequency from Lambda_r.

Assume u =0....although it might not be....

If it might not be then you can't assume, but even
if you do, you don't know v and you don't know
lambda_i or you wouldn't be trying to measure it
in the first place. The BaTh grating equation is:

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.


George



www.users.bigpond.com/hewn/index.htm

Einstein's Relativity - the greatest HOAX since jesus christ's virgin mother.

On 11 May, 02:21, HW@....(Henri Wilson) wrote:
On 10 May 2007 04:09:06 -0700, George Dishman wrote:
On 10 May, 01:30, HW@....(Henri Wilson) wrote:
On 8 May 2007 23:58:13 -0700, George Dishman wrote:
On 9 May, 00:41, HW@....(Henri Wilson) wrote:
....
I'm having to teach you basic algebra yet again Henry.

Lambda_i is absolute and all we need.

Lambda_r doesn't enter into this.

Lambda_i isn't enough, if you want to use it you need
to know v and u as well but the grating doesn't measure
them. Remember all we know is the angle phi so you
can turn the second version round to get

Lambda_r = D * sin(phi)

but that's as far as you go. Your first equation isn't
usable because v and u aren't known so in BaTh a
grating doesn't measure Lambda_i, only Lambda_r.

George, Lambda_i is known. It is absolute and universal for a particular
spectral line.

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.

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?

The difference between the measured angle and the expected one is a measure of
c+u/c+v
(Can we assume u is zero?).

No, you can't assume that but even if you did you
don't know v or lamda_i, the whole point of using a
grating is to _measure_ something you don't know.

Don't try to wriggle out with this distraction George.

It is not a distraction, a grating is a measuring instrument
and what it measures in BaTh is the _reflected_ wavelength,
nothing else. Other parameters my be inferred using
assumptions but they are not measured.

explained in the other message.

Conventionally v=0, u=0 and lambda_r = lambda_i
but in BaTh none of those are known. The
measurement of phi tells you lambda_r only.

George, the wavelength of, say, Halpha is absolute and universal.

You forget the Doppler shift due to proper motion of the
source. The grating has no idea the light is hydrogen
alpha, the fact is that the angle of deflection depends
_only_ on the reflected wavelength.

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.

....
George, if you want to measure lambda_r, you will have to put another grating
in the diffracted beam.

No, the equation for a single grating in BaTh, what you
called the "grating equation" tells you Lambda_r, not
Lambda_i.

please read again george.

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.

Lambda_r = D * sin(phi) / N

Anything beyond that requires assumptions and calculations.

George, I'm talking BaTh not fairyland physics.

Same thing.

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.

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.

This whole discussion is about the relatively moving situation.
So don't try to distract me with irrelevancies....

Then don't raise them.

Don't be stupid, both theories give the same equation.
However, in BaTh a grating cannot measure what you
call the 'absolute wavelength', only the reflected
wavelength. That's a limitation which suggests you
would need other instruments to find v and u.

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

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.

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.

....
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

....
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.

George

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.

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.
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.



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.


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

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.
I believe that u = c. 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. 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.

BaTh can't lose.


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'.


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.
The points you are missing is that Lambda_i is absolute and known...and that my
equation will allow hubble to detect its own orbital motion with a grating
whilst yours will not.

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 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.

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.

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.


George



www.users.bigpond.com/hewn/index.htm

Einstein's Relativity - the greatest HOAX since jesus christ's virgin mother.

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

On 14 May 2007 10:30:33 -0700, George Dishman wrote:

Another post my ISP dropped:

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


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, don't act dumb. I know you aren't.
You measure it in the lab...with the source and grating at rest.
Knowing the absolute wavelength of a particular spectral line, you can then use
a grating to measure the relative velocity of a star.




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.

Maybe you aren't acting.....

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.

YOUR equation is the simplified , constant c, one.
MINE is the general equation.

The fact that the HST detects its own orbital motion shows I'm correct and you
are wrong.

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.

I have just told you how to measure it.
Since lambda_r can be found from phi and since lambda_i of a known spectral
line is absolute and universal, c+u/c+v can also be determined from teh
difference between phi and what it SHOULD be if v=u.

In fact we now have a way of checking the constancy of OWLS, albeit not a very
sensitive one. All we need is MY formula and light from a star with high
doppler shift.


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.

yes.

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.

Somebody else might like a Noble .... no good to me.....

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.

Nobody has done it so there's NO evidence either way.

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.

THere possibly IS an experiment of that nature that might work.
Unfortunately I don't have a lab at the moment.


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.

It does..but it isn't the general equation.


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


I wrote that for somebody else then.

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.

George, lambda_i is also known.

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.

This is correct...which is the main reason I reckon u must = 0....or it
certainly cannot = v.

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

Moving the grating cannot affect the light.


Doppler shift Henry.

That doesn't affect the observed George.

It does in SR, and in reality.

but SR is wrong.




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.

That's what I inferred. In steady INERTIAL state, the same number of
'wavecrests' must pass any point in the beam in unit time....whether the
'mirror' is moving wrt the source or not.



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.

It's no big deal, taking credit for theleeding obvious....

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.

look who's talking....


George




www.users.bigpond.com/hewn/index.htm

Einstein's Relativity - the greatest HOAX since jesus christ's virgin mother.