Cosmic acceleration rediscovered

In article ,
greywolf42 wrote:
Feel free to provide the excerpt that the claimed results (1 in 100,000)
were below the physical resolution of the instruments (1 in 10,000).

The claimed result was 13 microK.
The physical resultion of the instrument was 4 microK.

COBE's detectors don't read in micro K. They read in intensity at a series
of microwave wavelengths.

COBE has three instruments, DIRBE (Diffuse Infrared Background
Experiment), DMR (Differential Microwave Radiomters), and FIRAS (Far
Infrared Absolute Spectrophotometer).




And these two statements weren't in the quote about COBE. Which was:

The claimed result for the quadrupole is in fact in the quote about
COBE. I know, cause I'm the one who posted that quote. Check out the
sentence in all caps.

==================
We have analyzed the first year of data from the Differential
Microwave Radiometers (DMR) on the Cosmic Background Explorer
(COBE). The data show the dipole anisotropy, Galactic emission, and
instrument noise, and detect statistically significant ( 7sigma)
structure that is well-described as scale-invariant fluctuations
with a Gaussian distribution. The major portion of the observed
structure cannot be attributed to known systematic errors in the
instrument, artifacts generated in the data processing or known
Galactic emission. The structure is consistent
with a thermal spectrum at 31, 53, and 90 GHz as expected for cosmic
microwave background anisotropy.

The rms sky variation, smoothed to a total 10ffi FWHM Gaussian, is 30
+- 5 microK for Galactic latitude |b| 20deg data with the dipole
anisotropy
removed. THE RMS COSMIC QUADRUPOLE AMPLITUDE IS 13 +- 4 MICROK. The
angular auto-correlation of the signal in each radiometer channel and
cross-correlation between channels are consistent and give an angular
power-law spectrum with index n = 1.1 +- 0.5, and an
rms-quadrupole-normalized amplitude of 16 +- 4 microK (\Delta T =T ss
6 x 10** -6). These features are in accord with the
Harrison-Zel'dovich (scale-invariant, n = 1) spectrum predicted by
models of inflationary cosmology. The low overall fluctuation
amplitude is consistent with theoretical predictions of the minimal
level gravitational potential variations that would give rise to the
observed present day structure.^L
==================

Your claims (and Bjoern's) are false.

My claim is supported by the quoted text.

In article ,
greywolf42 wrote:
However, I do know that the algorithms used
in WMAP, for example, are false-signal producers.

And you know that how? What algorithms are used in WMAP?

But
since WMAP assumes there *are* signals,

And what is your support for this assertion?

But I specified the problem. The signals claimed are below the physical
resolution of the detector.

What value are you using for the physcial resolution of the detector?
The paper I quoted has the value for the signal above the value of the
resolution.

greywolf42 wrote:
Greg Hennessy wrote in message
...

In article ,
greywolf42 wrote:

Have you even *READ* the COBE papers?
http://aether.lbl.gov/www/projects/c...r_final_apj.ps


You snipped the part where it COBE found variations in the data,
something you claimed it didn't find.

Probably he simply did not understand that your quote addressed
the issue.

Feel free to provide the excerpt that the claimed results (1 in 100,000)
were below the physical resolution of the instruments (1 in 10,000).

The claimed result was 13 microK.
The physical resultion of the instrument was 4 microK.


COBE's detectors don't read in micro K. They read in intensity at a series
of microwave wavelengths.

Which can be translated to a measurement of temperature,
by looking at the definition of surface brightness given
by Greg Hennessy already. Your point?


And these two statements weren't in the quote about COBE. Which was:
==================
We have analyzed the first year of data from the Differential
Microwave Radiometers (DMR) on the Cosmic Background Explorer
(COBE). The data show the dipole anisotropy, Galactic emission, and
instrument noise, and detect statistically significant ( 7sigma)
structure that is well-described as scale-invariant fluctuations
with a Gaussian distribution. The major portion of the observed
structure cannot be attributed to known systematic errors in the
instrument, artifacts generated in the data processing or known
Galactic emission. The structure is consistent
with a thermal spectrum at 31, 53, and 90 GHz as expected for cosmic
microwave background anisotropy.

The rms sky variation, smoothed to a total 10ffi FWHM Gaussian, is 30
+- 5 microK for Galactic latitude |b| 20deg data with the dipole
anisotropy
removed. The rms cosmic quadrupole amplitude is 13 +- 4 microK. The
angular auto-correlation of the signal in each radiometer channel and
cross-correlation between channels are consistent and give an angular
power-law spectrum with index n = 1.1 +- 0.5, and an
rms-quadrupole-normalized amplitude of 16 +- 4 microK (\Delta T =T ss
6 x 10** -6). These features are in accord with the
Harrison-Zel'dovich (scale-invariant, n = 1) spectrum predicted by
models of inflationary cosmology. The low overall fluctuation
amplitude is consistent with theoretical predictions of the minimal
level gravitational potential variations that would give rise to the
observed present day structure.^L
==================

Your claims (and Bjoern's) are false.

Can't you read? Or how did you manage to miss the sentence
containing 13 +- 4 microK at least two times now?

And what about the sentence that there was statistically
significant ( 7 sigma!!!) structure detected? Do you want
to say that these physicists are incapable of doing a proper
statistical analysis of the data, and the 7 sigma are simply
an error by them? Or that they lied? Or what???


Bye,
Bjoern

greywolf42 wrote:
Joseph Lazio wrote in message
...

"g" == greywolf42 writes:

g Joseph Lazio wrote in message
g ...

[snip]

g If you feed such algorithms random noise, the will still provide
g you with an appearance of a signal.

I'm sure some algorithms can fail in that manner. There are many
algorithms in signal processing, though. I can think of some simple
cases where your statement is easily false.


I don't know of any that don't produce false signals. They all are designed
to enchance slight deviations. However, I do know that the algorithms used
in WMAP, for example, are false-signal producers. In discussing with Ted
Bunn, we have found that the algoritms could produce false signals.

Was this in a thread here in sci.astro?


But
since WMAP assumes there *are* signals, then Ted figured it doesn't matter.
Which is called circular logic.

And you simply conveniently ignore that the signals found by
WMAP are nicely consistent with those found by COBE?


[snip]


Indeed, a simple example is estimating the mean and uncertainty in
the mean from a set of data.

g Yes. And a sample of random noise *will* give you a mean and an
g uncertainty in the mean. It doesn't mean that you have a real
g signal.

You don't specify the kind of random noise to which you're referring,
but, yes, random noise can have a mean. Perhaps the most basic is the
normal distribution, which is specified completely by its mean and
variance.

So? Without a better specified problem, your objection is somewhat
meaningless.


But I specified the problem. The signals claimed are below the physical
resolution of the detector.

Which is shown to be false by looking at the quote about
COBE. May I remind you again of the 7 sigma, and of
the 13 +- 4 microK?


[snip]


Conversely, if I estimate the mean and its uncertainty, find that it
is not consistent with zero, and conclude that there is a signal, what
have I done wrong?


That would depend upon how you "found" that the mean and uncertainty were
not "consistent with zero."

Err, what's your problem? A mean is non consistent with zero
if it is larger than 3 standard deviations.


If -- as in the case under discussion -- you
were claiming a result below the resolution of the detector -- then you
would be wrong.

The results were not below the resolution, no matter how often
you claim that.



In the basic case of data with approximately equal uncertainties,
the mean is given by (1/N)*sum{x}, where {x} are the data and N is
the total number of data, and the uncertainty in the mean is given
by s/\sqrt(N), where s is the uncertainty in measuring the
individual values of x.

g But you don't know the "uncertainty" in measuring the individual
g values of a set of data, beforehand. You may know the theoretical
g precision of the apparatus.

Funny. I seem to recall an exchange between you and Tedd Bunn in
which Tedd was expressing concern about the uncertainty in the data,
whereas you were quite confident in the results.


Odd that you'd recall something like this and not bother to find the
reference.

Odd that you did that yourself above (mention a discussion
with Ted Bunn without providing a reference), but here now
whine when Joseph Lazio does do that.



More generally, of course, knowing the uncertainties in the individual
data is one of the great challenges in experimental and observational
sciences.


That phrase indicates that someone is challenged in this arena, all right.
First, I think you mean "individual datum". Data is plural.

Yes. What's the problem?


And an
individual datum has no statistical uncertainty.

Err, hint: he talked about the uncertainties in the *data*.
Plural.


Groups of measurements of
a single parameter (i.e. data) have a resulting statistical uncertainty.

You are confusing theoretical precision with experimental uncertainty.

To me, it looks as if it is you who is confusing the two.


There are ways to estimate the uncertainties (for instance,
calculating the standard deviation of the data),


That is the *only* way to get statistical uncertainty.

Are you an expert on statistical analysis of data?


but they make certain assumptions,


There are no specific assumptions at all. Except that one is not varying
any independent parameters (i.e. you are measuring the same thing).

Are you an expert on statistical analysis of data?


and one wants to be careful to check that those
assumptions are valid.


And -- in the case under discussion -- "they" are not checking the
assumptions.

How do you know?


[snip]


Bye,
Bjoern

greywolf42 wrote:
Joseph Lazio wrote in message
...


[snip]

I also don't understand how this would explain observations of the
temperature of the CMBR in other galaxies


Since we aren't in other galaxies, there are no such observations. Claims
otherwise are based on circular logic.

Please explain what exactly the circular logic is there.



nor how it would explain the
SZ effect. (The antenna "knows" when we are looking at a cluster of
galaxies and adjusts the resulting signal accordingly?)


Quite simply, the claimed observation "SZ effect" is an artifact of circular
theories and dedicated theorists.

In other words: you simply deny the data.


As noted in recent posts, my understanding of the S-Z effect is that the
inspiration behind the S-Z effect is fine (if there IS as CMBR, then hot
electrons will distort the CMB spectrum toward the blue). The problem
arises in execution. Where excessive zeal and sloppy terminology leads one
to hunt for miniscule reductions in intensity of specific MBR wavelengths.
Literally dozens of experiments were done that "should have been" sufficient
precision -- but all they found was noise. A few more recent experiments
have "removed systematic errors" by computer processing. And claim
resolutions below the physical resolutions of the apparatus.

Yet again your nonsensical claims about being below the
physical resolution.



Bye,
Bjoern

greywolf42 wrote:
George Dishman wrote in message
...

"greywolf42" wrote in message
m...


[snip]


I don't see the difference (unless you mean the specific
temperature). Does this "electron vortex noise" have a
blackbody spectrum or not?


Yes, the spectrum mimics the blackbody shape. No the source is not the
temperature of the electron.

Why is the spectrum that of a blackbody?

[snip]

See my other post. If the hydrogen (or whatever is the
source of the radiatin) isn't moving relative to the
detector then you don't get a Doppler effect and you
don't explain the dipole.


That's relativity, not aether theory. The electrons themselves are
distorted by their motion through the aether. Hence, so is the emission.

So electrons have a finite size and are compressible, or
what?

How exactly is the radiation produced by the electrons,
and why does a compressing of the electrons by the Lorentz
factor lead to a Doppler shift in the emitted radiation?


[snip]


To analyse the above using Ned's test,

Correction: *YOUR* test. This isn't Ned's test.

IMO it is.


We've been through that, and agreed that I will listen to *your* version.

Which is identical to Ned's version, as far as I can see.


[snip]



Looking at Ned's graph, the local (z ~ 0) electron
hum would be measured as the black line other than
being scaled down by the factor k.

Since the k applies to all source densities (distant and local), the "k"
factor here will be a wash.

I'm not familiar with that term, what do you mean?


"k" is *your* term, above.

Err, he *obviously* asked for the meaning of your term "wash",
not for the meaning of "k" - see the next question directly
below!

*Who* is deliberately obtuse here?


(Because you have to determine the source
density from what you measured.)

The factor is present in the equations but may be
able to be determined empirically, is that what
you meant by "a wash"?


Close enough.

I'm also not familiar with the term "a wash". Probably
you meant it in the sense of definition 4 at
www.m-w.org?



[snip]


No, each shell would produce a different red curve
depending on the source temperature, the distance
(hence z), the k factor for that shell and the
integral of k for all shells closer to us which
will partially hide more distant shells.


Why are you now assuming that the temperatures are different at each source?

He is not necessarily assuming that above. Why do you think
so? He merely takes the *possibility* into account that this
is so.


For the peak of each curve to match the black curve,
each shell must be at a temperature of (1+z)T so
the farther back into the past you look, the higher
the temperature.


But we don't need each individual curve to match the black curve. If you
are arguing against your uniform-external-source model, you need to show
that the integrated signature doesn't match the shape of the received curve.

Well, how likely is it that the integrated signature would
give a blackbody curve again if the individual curves
do not match a blackbody spectrum? That would indeed be an
incredible coincidence.


(Of course you'd also have to justify your Earth-centered universal
temperature distribution.)

What would work better against that strawman is an integration over constant
density and constant temperature sources.

Which is exactly what George suggested originally, IIRC.



[snip]


I hope my other post cleared that up. The peaks
are equal if you allow for the energy loss due to
graphing against frequency. They are not equal if
you don't.


A substantive physics test will not rely upon the type of graphing used.

I get the faint suspicion that you did not understand the
argument.


[snip]


Actually, the observed spectrum looks like it will match a blackbody --
because you have tried to work within a "per nu" function. While the
strawman requires decay in the photon -- not wavelength.

No, tired light reduces the energy by (1+z) while
the Stefan-Boltzmann Law increases the total power
by (1+z)^4 leaving the discrepancy of (1+z)^3. That
is the point of Ned's page.


Based on what assumption of source density? You keep ignoring this
question.

He obviously uses the simplest possible assumption (constant
density) here.


[snip]


Since the volume of any region of space increases as the
cube of its dimensions, that reduces the photon density
and hence the intensity.


You've fallen into the BB assumptions again. Photon density remains constant
per unit volume as you arbitrarily expand the volume of space you are
considering in your "region". Unless you expand space (which is a BB
assumption). PF does not expand space. TL does not expand space.

And George was also not talking about expansion of space above.
Your point?



[snip]


then there
will be an error in intensity of (1+z)^3 while if the
motion causes expansion by a scaling of exactly 1+z then
the result exactly matches a black body. Intermediate
amounts of motion would give an intermediate intensity
factor.


Total non-sequiteur. That's the BB model again.

No, it isn't. Why on earth do you think so?



[snip]


I don't currently know of any such theories.

Which is what I said a long time ago, I'm prepared to
consider tired light theories but I don't know of any
that can explain the dipole and the spectrum of the CMBR.


TIRED
LIGHT
THEORIES
HAVE
NO
NEED
TO
EXPLAIN
BIG BANG
ASSUMPTIONS!

Neither the dipole nor the spectrum of the CMBR are
"big bang assumptions". They are data. Why do tired light
theorists think that they have to explain only one set
of data which the BBT addresses (the red shift), but can
ignore others?


Cosmic origin of the MBR is a BB assumption. It does not exist in any tired
light theory.

George did not say it does.



[snip]



You have been
claiming that you approach this as a "test", that can only be applied to a
specific theory at a time. Now, you slip back into several universal claims
that "tired light" is disproved.

I have not seen George saying in any of these posts that
"tired light" is disproved by this.


[snip]

Bye,
Bjoern

In article ,
Bjoern Feuerbacher wrote:
COBE's detectors don't read in micro K. They read in intensity at a series
of microwave wavelengths.

Which can be translated to a measurement of temperature,
by looking at the definition of surface brightness given
by Greg Hennessy already. Your point?

The Differential Microwave Radiometers actually measure the difference
in temperature from two points, about 60 degrees apart. There are six
independant DMR's, measuring three independant bands.

"greywolf42" wrote in message
. ..
George Dishman wrote in message
...

snip
Eddington showed energy
in starlight was equivalent to circa 2.8K but only a
small fraction of that transfers to the aether.

Actually, all of it would transfer back to the aether -- eventually. Only
a
fraction from any one source would transfer at any given dV.

Yes, that's what I meant. You should be getting
used to my 'economical' style by now ;-)

Also,
even if the aether temperature is much higher than
2.8K, your previous comments would get round this if
the transfer of energy to electrons is slow compared
to the re-radiation as thermal energy by the electrons.

The total power transferred just needs to match that
radiated at 2.8K.

You have the general idea.

Now I'm going to add one more bit of confusion. Thermal radiation from
non-fusing sources in thermal equilibrium, such as planetary bodies and
cooled collapsed objects will eventually be radiating just as much power
as
they absorb from gravitation. And that thermal radiation will eventually
be
re-absorbed into the aether. Which is the part of my model that I believe
you now understand.

But since we were mentioning starlight, we have to touch on the issue of
fusion inside stars. Fusion energy is a temporary additional source. So
the amount of heat leaving a fusing star will actually be above that
received from the gravitational force. Until it burns out and cools down.
The original source of the fusion energy would come from the orignal
formation of protons (and anti-protons) out of the original aether. And
the
origin of the aether (and universe) is as yet undescribed by the model.

OK, I'll treat that as a side issue too. There may
be some problems with the details but as long as any
additional heating of bodies is at a low level then it
probably doesn't affect our conversaton significantly.

George

"greywolf42" wrote in message
. ..
George Dishman wrote in message
...

"greywolf42" wrote in message
. ..
George Dishman wrote in message
...

most quotes snipped, one moved later

"greywolf42" wrote
George Dishman wrote:
snip
If he can come up with one that explains the spectrum
of the CMBR

Electron vortex noise from the aether. A local effect
due to electrons bound in hydrogen gas.

your comment on CMBR relevance snipped but
addressed at the end

-:-

Postulates:

The CMBR is produced by "Electron vortex noise
from the aether, a local effect due to electrons
bound in hydrogen gas."

This is not a "postulate" of tired light theory. It is not even a
postulate of my current favorite theory. It is an unavoidable
consequence of the aether-matter model that I favor.

It is just what you said, quoted above.

The quote above never mentions "postulates". It is a conclusion, not an
assumption (postulate).

_I_ am stating these as postulates based on my understanding
of what you have said previously with the noted exception
of the motion of the source that we discuss elsewhere.

The electrons are producing a blackbody spectrum
at an equivalent temperature of roughly 2.73K.

Slight correction: the electron imedance noise gives a signature that
is
equivalent to 2.73 (or 2.81) K.

I don't see the difference (unless you mean the specific
temperature). Does this "electron vortex noise" have a
blackbody spectrum or not?

Yes, the spectrum mimics the blackbody shape. No the source is not the
temperature of the electron.

I wasn't claiming it was, note the word "equivalent".

The solar system is moving through this hydrogen

No such assumption is needed. Under any version. The EM waves
(photons) emitted are based in the aether. Thus, it is motion
through the aether that is important. Not motion relative to
hydrogen.

and as a result there is a Doppler effect which
produces the cosmic dipole moment.

The MBR dipole moment comes from the motion of the detectors through
the
aether. The MBR moves within the local aether fluid.

See my other post. If the hydrogen (or whatever is the
source of the radiatin) isn't moving relative to the
detector then you don't get a Doppler effect and you
don't explain the dipole.

That's relativity, not aether theory. The electrons themselves are
distorted by their motion through the aether. Hence, so is the emission.

As noted in another reply, it applies to both since
both derive the measured effects (length contraction,
time dilation, mass anisotropy, etc.) from the Lorentz
Transforms. It surprised me that two quite different
philosophies should produce identical predictions but
that's the way it is.

Sorry, it's not possible to explain the dipole in
that case. I'm not pushing the strawman, feel free
to change your suggested source in any way you think
can explain the dipole.

You are incorrect. However, I must apologize for overly simplifying my
responses -- to the point where I was apparently not clear.

That's ok, you only mentioned motion which is entirely
reasonable and Lorentz factors cannot change the
frequency measured by a real detector. Of course it will
change the "absolute" frequency by the time dilation
factor but that is unmeasureable because the aether
affects the detector too.

due to
grey dust or other possible causes of extinction.

Wherever did you get this one?

I missed a quote, it was in one of your recent
posts. I'll try to find it if you like.

You have a short memory. That was you on 12/27: "Extinction is discussed
by
Perlmutter ss 'grey dust'."

Nope, found it:


"greywolf42" wrote in message
...

Sigh. Extinction due to grey dust is not contained in any tired light
theory. Only in strawmen. Hence I frankly don't care to go out on
another
tangent.


Extinction is a totally separate concern.
(As it is in all astrophysics.)

Black body radiators are also perfect absorbers.

Did you have a relevant point to make?

No, it was a side issue that you might like to
consider separately from this discussion. A
degree of extinction will be a consequence of
your model. Photons will return some energy to
the aether producing the tired light effect as
you say but also some photons will be absorbed
by electrons. The cross section could be very
small though so we can continue to ignore this.

snip
we split
the universe around the solar system into thin,
concentric spherical shells or thickness dR at
radius R. The surface area of each shell can be
thought of as composed of many small cells of
volume dV and the number of such cells is
R^2*dR/dV. Ignoring tired light energy loss, the
amount of radiation we receive from each cell is
proportional to R^-2 (inverse square law) and
proportional to dV hence the total rate of photons
from each shell is independent of R.

Assuming that radiation source density is constant in all dV,
throughout
the universe. If we limit the region of analysis to a thin shell at
distance R_0, then you assume that the radiation source density is
constant throughout the shell.

I would have thought that too but you said:

"greywolf42" wrote in message
...
George Dishman wrote in message
...
snip
The
extra factor to be taken into account in this case
would be the electron density.

Nope. Electron density wouldn't change anything.

Again, I'm not pushing the strawman, correct me if
if that is in error.

Ned's theory requires an electron density for cosmic origin. Your theory
requires an electron density for emission at intermediate location. Mine
does not ... because it is a signal internal to the antenna.

OK, as explained, the dipole (and the correlation of
the anisotropy between missions) rule out an internal
source so I guess we need to consider the source
density again. It's covered by the 'k' factor I
mentioned so I'll leave it for the moment unless you
want to go into it more. BTW, I'm not assuming k is
constant with time but it is the same for a given
shell.

snip
Since the k applies to all source densities (distant and local), the
"k"
factor here will be a wash.

I'm not familiar with that term, what do you mean?

"k" is *your* term, above.

See below:

(Because you have to determine the source
density from what you measured.)

The factor is present in the equations but may be
able to be determined empirically, is that what
you meant by "a wash"?

Close enough.

Ok, I've just never heard it called that before.

However, to that
we must add contributions from greater distances
since there is no appreciable extinction.

First think of a series of shells at z=0.1, z=0.2,
etc.. Each would produce a curve similar to the
black line with the same peak intensity but with
the peak frequency moved to the left.

And down. Each shell would result in the same (black) curve.

No, each shell would produce a different red curve
depending on the source temperature, the distance
(hence z), the k factor for that shell and the
integral of k for all shells closer to us which
will partially hide more distant shells.

Why are you now assuming that the temperatures are different
at each source?

The above is only a statement of fact, the curve would
depend on the source temperature.

For the peak of each curve to match the black curve,
each shell must be at a temperature of (1+z)T so
the farther back into the past you look, the higher
the temperature.

But we don't need each individual curve to match the black curve. If you
are arguing against your uniform-external-source model, you need to show
that the integrated signature doesn't match the shape of the received
curve.
(Of course you'd also have to justify your Earth-centered universal
temperature distribution.)

I suppose it could be Earth centered and constant in
time at each location but that's not what I meant. I
was considering that the temperature is the same
everywhere at any time from which the conclusion would
be that it was higher in the past.

What would work better against that strawman is an integration over
constant
density and constant temperature sources.

The next paragraph covered that:

The total
would then be the sum of an infinite series of
such curves. It should be clear that essentially
the total observed curve becomes something like a
straight line to the left (lower frequency) of the
locally generated peak. Of course the series of
discrete shells is an approximation as the source
is continuous so to find the real prediction let
dR tend to zero and integrate instead of summing.
The overall intensity of the curve can be adjusted
by changing k but the intensity will always be too
high at frequencies below the peak for a blackbody.
In fact I don't think you will get a peak at all.

That is because you have assumed (incorrectly, I believe) that the
frequency shifts, without losing energy. The problem is (I think)
that you have tried to do your integration in the "per nu"
expression. Energy is lost per photon. Not per unit frequency.
This changes the results of the integration.

I hope my other post cleared that up. The peaks
are equal if you allow for the energy loss due to
graphing against frequency. They are not equal if
you don't.

A substantive physics test will not rely upon the type of graphing used.

I am saying that you need to reconsider the above
paragraph because you previously dismissed it for
a reason that you now know to be incorrect due to
your misreading of what was being graphed.

Ned's graph is correct for tired light.

Here you violate your own rule that Ned's test / *your* test can only be
used against a single theory at a time. Yet here you throw out another
blanket claim that "tired light" is disproved.

I've spent too much time putting this together
and I don't want to spend more time doing the
integration, I think I've said enough so you can
if you wish.

I think we both see the intent of your test.

The point is that, with the stated
strawman, the observed spectrum will not match a
blackbody.

Actually, the observed spectrum looks like it will match a blackbody --
because you have tried to work within a "per nu" function. While the
strawman requires decay in the photon -- not wavelength.

No, tired light reduces the energy by (1+z) while
the Stefan-Boltzmann Law increases the total power
by (1+z)^4 leaving the discrepancy of (1+z)^3. That
is the point of Ned's page.

Based on what assumption of source density? You keep ignoring this
question.

Ned's graph is for 100%, in this discusson I have defined
the coverage as 'k' for any shell. That can be generalised
further as k(t) assuming the source is homogeneous and
isotropic.

So the question is can you change the
strawman, or indeed discard it entirely and
replace it with a real tired light theory, and
show that you can then match the observed
spectrum while still explaining the dipole?

Again, tired light theories have no need to explain the MBR spectrum.
The question before you, is whether your disproof of the strawman is
valid. Then you can keep it in your pocket for use if anyone ever
proffers a combined MBR / tired light theory that presumes that
matter within space is uniform, and gives rise to the MBR constantly
throughout the universe.

Yes and no. Think back to how we got the value of 0.024%
per MPc for mu in the tired light theory. It comes from
the observed redshift versus distance.

Of starlight, yes.

Now note that in
the Plasma Fireworks model, some of the redshift is due
to motion

Yes.

so the amount of energy loss due to tired light
would be less

Yes.

hence mu would have a smaller value

Yes.

which
we could find if we could separate out the motion part.

Which we can't, of course without other theoretical calculations.

If objects were moving apart fast enough, that could
explain all the redshift and hence mu would be zero.

Yes. You can have a PF model without tired light. And you can have a
tired
light model without PF. Redshift-distance alone cannot determine which is
real.

Since the volume of any region of space increases as the
cube of its dimensions, that reduces the photon density
and hence the intensity.

You've fallen into the BB assumptions again. Photon density remains
constant
per unit volume as you arbitrarily expand the volume of space you are
considering in your "region". Unless you expand space (which is a BB
assumption). PF does not expand space. TL does not expand space.

If there is no motion

If there is no motion, then there is no PF model. Why are you throwing
around these self-contradictory arguments? Have you now abandoned
discussion of the pure PF model and gone back to pure tired light?

Interesting, I need to think about that. PF effectively
means the source density is falling but does that
compensate the measured intensity? I'll need to think
a bit more about that.

then there
will be an error in intensity of (1+z)^3 while if the
motion causes expansion by a scaling of exactly 1+z then
the result exactly matches a black body. Intermediate
amounts of motion would give an intermediate intensity
factor.

Total non-sequiteur. That's the BB model again.

Assuming you are right above then it is much simpler. PF
fails by (1+z)^3 as well. As I said before and you said you
understood, the Wein Displacement Law means that the source
temperature T is proportional to (1+z) to get the peak at
the right frequency, the Stephan-Boltzmann Law means the
intensity is raised by T^4 and tired light then reduces the
intensity by (1+z). The remaining ratio is (1+z)^3 which
can be cancelled by expansion. I had thought the expansion
in PF would do but maybe not.

In other words, the amount by which the observed intensity
deviates from that of a black body is an indirect measure
of the value of mu, and if there is no difference then
mu=0, and that means light doesn't tire.

Mu is based on starlight ... not the CMBR.

Once they have been emitted, tired light applies to
_all_ photons.

I don't currently know of any such theories.

Which is what I said a long time ago, I'm prepared to
consider tired light theories but I don't know of any
that can explain the dipole and the spectrum of the CMBR.

TIRED
LIGHT
THEORIES
HAVE
NO
NEED
TO
EXPLAIN
BIG BANG
ASSUMPTIONS!

Tired light doesn't have to explain the workings of
stars, but as you said above "Mu is based on starlight"

Tired light must affect the photons observed as the
CMBR so the CMBR can be used to test tired light. By
considering the spectrum and intensity, that is what
we are doing. Tired light doesn't have to explain
what causes the CMBR but it does have explain why it
doesn't flatten the low-frequency end of the spectrum.
I think you can only do that for specific theories
regarding the source hence I do not claim it applies
to all.

Cosmic origin of the MBR is a BB assumption. It does not exist in any
tired
light theory.

BB does explain them and tired light could, in theory,
also occur in a BB universe,

That is a combination that we have not specifically discussed. But what
you
are attempting to do is arbitrarily force TL into a BB cosmos. If you
have
one, you don't *need* the other.

but the expansion scales as
(1+z)^3 which means that mu has the empirical value of 0
to the limit of the resolution of our measurements.

Total non-sequiteur.


You have been
claiming that you approach this as a "test", that can only be applied to a
specific theory at a time.

I am only applying it to this strawman which, I think,
is similar to your ideas other than having the hydrogen
outside the detctor as required by the dipole. If you
now think it would apply to many others too that's fine,
but I am not claiming this test applies to all tired
light theories.

Now, you slip back into several universal claims
that "tired light" is disproved.

You may make that claim, I do not.

George

"Bjoern Feuerbacher" wrote in message
...
greywolf42 wrote:
George Dishman wrote in message
...

snip

I'm also not familiar with the term "a wash". Probably
you meant it in the sense of definition 4 at
www.m-w.org?

wash (noun)

4 a : worthless especially liquid waste : REFUSE
b : an insipid beverage
c : vapid writing or speech

It still dosn't any much sense to me.

No, each shell would produce a different red curve
depending on the source temperature, the distance
(hence z), the k factor for that shell and the
integral of k for all shells closer to us which
will partially hide more distant shells.


Why are you now assuming that the temperatures are different at each
source?

He is not necessarily assuming that above. Why do you think
so? He merely takes the *possibility* into account that this
is so.

Thanks, at least what I said was clear to you.

George