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.

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.
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.

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.

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.

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.


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.

Lambda_i is known.

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

The lesson Henry, is to work out the equation before you
start telling people what it contains.

The BaTh wins again.

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)

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.

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.

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.

The BaTh wins yet again.

Your obsession is getting the better of you, try to
calm down. For the grating (as for the MMX), both
theories give the same result and for Ives and
Stilwell, Sagnac and the Shapiro delay BaTh fails.

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.

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.

George