Cosmic acceleration rediscovered

George Dishman wrote in message
...
References lost due to ISP problem.

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

snip

Above you said that not all the red shift needed to
be due to tired light and suggested this was different
from BB.

Yes.

I agree that gravitational and proper motion
effects are present as well but that applies to BB too
so what did you mean above?

That the BB considers all redshift to be due to doppler due to
expansion -- with miniscule corrections from peculiar
motion dopplers and gravitational shifts. The BB
acknowledges that gravitational redshift, *real* motion
dopplers, and electron scattering do exist. But they are only minor
corrections on the whole.

Laying aside gravitational and other secondary effects
(S-Z etc.), the essential difference is that expansion
deals with the systematic motion while proper motion
is essentially that which departs from the overall trend.

Closer. But many theories have systematic motions that are not due to
cosmic expansion. For example, the "plasma fireworks" model has systematic
doppler redshifts due to *true* motion. The BB theory incorporates an
additional assumption about physics and the expansion of *space* (or the
universe), that is not included in the plasma fireworks model.

Tired light theories presume the *additional* source of "tired light",
from photon energy degradation.

Yep, got that.

But tired light theories therefore consider that
(non-expansion) doppler effects from "peculiar" or systematic motions
will not simply be small corrections to the cosmic expansion signature.

I still don't see the difference. To take a crude example,
the motions of individual galaxies in a cluster relative to
the centre of momentum would be proper motion

The separation rate of the individual galaxies, relative to the CoM will
include a cosmic expansion coefficient in the BB.

while that of
the CoM itself would be mainly expansion but with an
equivalent proper motion relative to neighbouring clusters.
Is that not the same in both views?

No.

Any systematic outward motion that would contribute to the
first order coefficient is simply expansion

Not in the BB. There is spatial (universal) expansion and there is motion
through space expansion.

so you seem to
be suggesting that tired light could include an element of
expansion.

It could include an element of expansion by real motion through pre-existing
space.

Subtracting the tired light part would then give
you still a big bang model but with a much greater age.

It would not be a 'big bang' model. It would be a "real" explosion-type
model (i.e. "plasma fireworks"). And yes, the PF model has "ages of the
universe" of hundreds of billions to trillions of years. Though the meaning
of teh term "age of the univserse" is somewhat different between the two.

To avoid a big bang scenario, you need to explain all the
systematic red shift with something other than motion

But tired light models don't need a big bang scenario.

and
at the moment I'll have to take that as photon energy loss
unless you can identify another contributor.

Even plasma fireworks does not require the big bang. The latter adds
creation of space.

Tired
light theories agree with the BB that gravitational redshift and
electron scattering will be minor corrections on the whole.

Yes, I think those are common.

OK, that will let me illustrate how I think Ned's test
applies. I'll essentially throw a strawman at you and
you can then correct the errors in my understanding of
your model and we will see if that solves the problem.

Just doing the best you can with your view of Ned's site will be fine.
But use a *real* tired light theory. (i.e. Vigiers.) I have no desire
to deal with yet another strawman.

I'll ask you about your view. If that includes elements
of Vigiers then fine but if you want to make the case that
a tired light theory can satisfy the tests, it is for you
to make that case.

These are *your* "tests." It is up to you to support your claim about their
use. Feel free to use the tired light defining equation: dE = - mu E dx.
And feel free to use 1/mu = 4.2GPc (your R, from prior posts). I would
expect that you will perform a calculation to reproduce something similar to
Ned's curves (which are unsupported on Ned's site -- and trivially
incorrect).

Please let me know if this is insufficiently defined for your test.

--
greywolf42
ubi dubium ibi libertas
{remove planet for return e-mail}

Bjoern Feuerbacher wrote in message
...
George Dishman wrote:
Much snipped to try to get back to the physics.

"greywolf42" wrote in message
...

(replying to greywolf here, not to George...)

[snip]

I thought you and Bjoern were being deliberately obtuse,

Well, that's one of the big problems when discussing with you:
that you constantly invent motives for your discussion partners,
instead of actually trying to understand what they are asking
and saying.

And yes, this is indeed an ad hominem. An eye for an eye...

However, we see below that you understood all along. You *were* simply
being deliberately obtuse.

and implying that
you didn't understand how luminosity (magnitude) could be considered
distance. I was specifically addressing your statement: "what is
observed is redshift versus magnitude or some other indirect measure of
distance." Magnitude is an indirect measure of distance (since the days
of Wirtz and Hubble). So, we have a curve of redshift vs. distance.

Yes, we can indeed get such a curve from the luminosity-redshift curve
in the reference you gave.

Then your claim that you didn't understand my statement was simply your
being deliberately obtuse.

And the observed curve is exponential.

I don't think so. Where do you get this from?

From the reference I gave. Just like the last two times you asked this in
this thread, Bjoern. Look at the data. Fit an exponential curve to it.
Voila!

At least within error bars.

If you think so: Care to test if an exponential curve, or the curve
predicted by the BBT, using the parameters from WMAP, fits
better to the data, including the error bars?

Why bother? The exponential explanation is valid. Which is the point at
issue. We aren't discussing WMAP at all, because WMAP does not directly
measure the redshift-distance curve.

--
greywolf42
ubi dubium ibi libertas
{remove planet for return e-mail}

George Dishman wrote in message
...

The majority of this reply addresses Ned Wright's page.
I'll deal with a tired light test in my other reply to
try to focus each branch of the thread. I'm hoping this
one can die soon.

Amen.

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

snip

I've already repeatedly shown where you've misunderstood real tired
light theories, and why the test isn't applicable. You insist on
examining only strawman theories. The test isn't applicable,
because Ned's "test" assumes CMBR temperature reduction from
the BB cosmogenesis.

And I've repeatedly pointed out that there is no temperature
change implied or assumed. I'll snip some further comments
based on this misunderstanding.

snip

You've *claimed* that Ned either did not mean to use -- or didn't use -- BB
CMBR cooling for the time it takes for light to travel from z = 0.1. Ned's
site doesn't say that, but let us continue.

I want you to respond to what _I_ am writing.

I am responding to what you are writing. You keep writing about Ned's
page, and Ned's test. Therefore, I must address Ned's page, and Ned's
test.

And you claimed "you need to apply it to a specific theory to find out
whether it can falsify it or not". If you really believe that Ned needs
to change his site to reflect this view, why not say so?

There is nothing wrong with Ned's page for what it tries
to do IMO. YMMV.

But Ned's site does not apply it's claim to "a specific theory." That *IS*
contrary to what you earlier said an application of Ned's test must do.

"Some diehards refuse to face these facts, and continue to push tired light
models of the CMB, but these models do not agree with the observations."

Nope. It can only be applied by assuming that the BB is correct, and
the temperature that Ned arbitrarily selected (solely to impugn the
actual theories) came from the BB.

Wrong, it comes from choosing an arbirtrary figure
of z=0.1 and choosing the temperature to match the
peak wavelength after tired light is applied. The
intensity then follows.

Well, this *is* a different rationale than Ned provided.

Thank you. That's all I meant by listening to my view.

OK. I think Ned provided one rationale, and you are definitely providing
one that is different than the one I believe Ned is providing.

However, this
can't be true for Ned's curves. Because the intensity between Ned's
blue and red curves is unchanged. Photon frequency lowers in a
tired light model, *because* energy is lost. Yet Ned's red curve shows
energy per photon as unchanged after (you claim) tired light shifting
has taken place. It appears that when you include the energy loss, the
red curve once again drops to the black (observed) curve.

Note that the Y scale is in M Jy / sr. 1 Jansky is
10^-26 watts per square meter per Hz. The "per Hz"
factor compensates for this because an interval of
1Hz at the receiver started out as 1.1Hz at the
source.

The "per Hz" does not compensate for this effect. Ned's curve appears to
show a shift *without* energy removal. Note that the "resultant" (red)
curve has the perfect shape to match the observations -- if you merely
divide the intensity values by Ned's arbitrary factor of 1.331.

Let's take a quick look at Ned's support for his curves:
"Assume that the CMB starts out as a T = (1+z)*To = 2.998 K blackbody, which
is the blue curve. Because the photons only lose energy but do not decrease
their density, the resulting red curve is not a blackbody at To = 2.725, but
is instead (1+z)3 = 1.331 times a blackbody."

Now where do *you* get Ned's "original" intensity for the blue curve? I
cannot determine this from your interpretation of Ned's test. For the
intensity would be based on the density of your source gas shell.

Tired light effectively compresses the power
emitted into a narrower band.

Umm, no. Tired light does not compress power bands. This may simply be a
sloppy usage on your part. The relative energy-frequency curve merely
shifts. It does not compress the band, unless you arbitrarily have set a
minimum intensity as defining the edges of the "band."

snip

So are you saying that the CMBR isn't produced within
tens of parsecs? I would assume it was the integral of
contributions over many GPc

That *IS* the BB assumption, again.

No, I mean that it will be a number of times the
characteristic length we worked out for tired light,
i.e. several times 4.2GPc.

That's the BB assumption. Tired light theories do not need nonlocal
sources. In this case, "nonlocal" means "outside of the receiving
apparatus."

unless extinction plays a
significant part. You need to tell me over what range
extinction becomes important if we are to move forward.

I have no need to feed your attempts to throw random objections around.

At the moment, we are discussing Ned's argument. Or your version of it.
Please do so, or drop it.

Fine, but don't complain later that my argument is
flawed because I didn't take extinction into account.

You have my permission to chastise me, if I do.

small snip - moved later

big snip

Let me make it simple,
is the temperature of the aether constant, falling
or rising in your model, or doing something else I
haven't thought of?

As repeatedly noted, it is approximately constant with time. Which is
why I favor it. See the very next statement of my post:

OK, that's fine. (Your next statement didn't address the
time variation of the temperature.)

Actually, it did. The closed cycle part took care of it.

I assume there is a
contribution from EM pasing through, which includes
both ambient starlight and the CMBR itself. However the
effect is not to remove photons as in extinction but just
to reduce their energy creating the tired light effect.

That is my favorite assumption at the moment. However, there is no a
priori need for such an assumption.

There is a need to have a specific model to work
with if you want specific comments. Otherwise all
you can get is hand-waving.

We agree then, that Ned's page is merely hand-waving. For he attempts
to address all tired light theories with his claims. But Ned doesn't
actually address even one single, specific model.

No, it is not handwaving. It is an accurate quantitaive
explanation of a test that can be applied but you need
to know the details of the theory before applying it.

But that's not what Ned does. Ned claims that *all* tired light theories
are disproved by his test.

"Some diehards refuse to face these facts, and continue to push tired light
models of the CMB, but these models do not agree with the observations."

And that is why it is merely hand-waving. Because Ned never addresses a
real tired light theory:

snip

Ned's
source is at 420MPc which I can understand would be
non-local by normal standards, but could be 'local'
by cosmological standards. Where do you want to draw
the line?

snip

3) The local region for the MMBR (measured microwave background
radiation) is the antenna of the device you are using. According to the
matter theory that I favor.

If you are suggesting the instrument designers didn't
consider locally generated thermal noise in the receiver,
that is obvious nonsense. It is a major problem and has
to be designed for, tested and calibrated.

I'm *not* addressing what designers typically consider "thermal noise" of
the receiver. That is merely the noise due to the temperature of the
matter -- which varies with matter temperature. What I am referring to is
the aether electron "hum" that arises from the interaction of the electron
(standing wave) with the aether corpuscles.

Not one CMBR device -- to my knowledge -- has ever attempted an isolation
test. That is, a test to determine whether the signal was actually produced
"within" the antenna -- or whether it had an external source. For example,
I know for a fact that Penzias and Wilson did not do this test. They *did*
cut out the antenna connection. But they did not put their antenna in an
isolation chamber.

I'd be happy to be corrected, if you know of anyone who has done this. (And
this has been discussed on the newsgroups for at least a decade.)

Note that this resolves Ned's little conundrum quite nicely:
"The FIRAS data .... requires that the CMB come from ... distances less than
0.25 Mpc."

Ned's model is at a single temperature. Time
variation of that temperature is of no
relevance to the argument.

Ned's model is at a temperature of 2.998 K at emission (time #1) and a
temperature of 2.73 at reception (time #2).

section snipped from above

At a temperature of 2.998 K, instead of the 2.73 K measured "here".
That *IS* two different temperatures at two different locations
(420 MPc apart).

... It's as if you were
additionally assuming that the temperature
was 2.725K now as well as 2.998K at the time
of emission.

That's what Ned's graph is based upon --

This is where you are making your mistake. The page is
based on one fact, that the temperature measured here
and now is 2.725K. As an example, he then considers the
postulate that it was emitted at 2.998K at a distance
of 400MPc and the peak has been moved by tired light
to coincide with the peak frequency of a black body of
2.725K. There is no suggestion that the actual source
is at 2.725K at any time.

That is *your* description. Unfortuantely, Ned doesn't provide any such
description. What he provides is just this:

"The expanding balloon analogy for cosmological models can be used to show
this. ... Note that the galaxies (yellow blobs) do not grow, but the
distance between galaxies grows, and that the photons move and shift from
blue to red as the Universe expands, and the photon density goes down. But
in the tired light model, illustrated below, the density does not go down.
Assume that the CMB starts out as a T = (1+z)*To = 2.998 K blackbody...."

AFAICT, Ned *IS* specifically addressing the expanding balloon analogy of
the BB. And Ned *IS* therefore specifically addressing the fact that the
"temperature of space" is cooling.

Your approach would need no reliance upon the BB balloon analogy at all.

aside from being an obviously
incorrect calculation.

See above, the units are not what you thought.

Thanks, but Ned's calculation is merely a method of producing an arbitrary
factor.

snip

ONLY big bang theory requires the
temperature to change with time.

No change with time is implied, in the example it
is a source at constant temperature of 2.998K at
z=0.1

The time used by Ned is the time it takes for light to travel from
z=0.1. That's where he got the temperature (the BB).

Wrong, he got it from by back-calculating from
the current observed temperature, and you already
said that.

Then why did you say "wrong?"

Because you said "The time used by Ned" suggesting
there is a time variation involved. There isn't,
only the shift of peak frequency due to tired light
acting over a distance.

And there is a time difference between emission and absorbtion. Enough for
the light to obtain a redshift of 0.1 in the BB model. That is my
understanding of Ned's analysis. It is not *your* analysis.

If the distant source is still at that temperature,
in the future we will continue to see
the CMBR at the same temperature as at present.

This is a spurious argument. For we can only measure the present.

I hope you now follow the above comment. Your model has
the temperature as constant so the measured spectrum
would also be constant in time.

*NED's* model is the BB balloon universe. The temperature varies with time.

*YOUR* model does not.

No, it only assumes the temperature was 2.998K at the
time and location of emission.

Which is a slowly falling temperature. You aren't dumb. Tell me,
why do you feel the need to make excuses for Ned's slime site?

Because you are making an incorrect assumption that
means you fail to understand the test. I intend to
apply the same test but to your distributed source
which is at uniform temperature so you need to open
your mind a bit and take the blinkers off.

To what "distributed source" are you referring? I think you are again
looking at nonlocal sources.

I thought you said the CMBR was emitted by electrons
being excited by the aether,

Yes. I have attempted to clarify the situation for you. *ALL* electrons are
effected by the aether. Including the antenna of the Penzias and Wilson
device, and all other such devices. No experimental system that I am aware
of, has ever tested to eliminate this possibility. P&Z never made this
test. After P&Z, everyone simply assumed it was "cosmic" in origin (i.e.
outside the mechanism) -- and never made this test.

though now you seem to be
claiming that the people who designed COBE, WMAP and
all the other instruments that have measured the CMBR
can't tell the difference between the CMBR and thermal
noise in the instrument. If that's your best alternative
to the big bang, I don't think much of it!

Your strawman argument is noted. And ignored.

snip

He doesn't, you need to re-appraise the page.

Not based on Ned's writings. *You* have come up with a different
rationalization than Ned states. However, your reasoning does not
support Ned's graph. (Or vice versa.)

snip

That's what I intend to disprove, but you have to get
rid of your preconceptions and start thinking about
the physics instead.

The classic ad hominem. If you intend to disprove it, feel free.

It was not intended as an insult of any form, above
you say my view differs from Ned's and that suggests
you have considered my words separately which is all
I wanted.

I do consider your model separate from Ned's description.

within the posted error bars." In
short, only theories with changing temperature of space (specifically
those that use the BB rate-of-change) have a problem with Ned's false
assertions in the first place. There is no reason to expect a change
of temperature ... except in the BB.

I think you will eventally start arguing the opposite
of that, but see what comes.

Nonsubstantive, ad hominem.

Again that was not an insult, just an expectation
of where the physics will lead us.

The above is pure ad hominem.

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.

That's fine. That means I don't need to worry about
including what-if's to cover it. It'll make the
discussion easier.

--
greywolf42
ubi dubium ibi libertas
{remove planet for return e-mail}

greywolf42 wrote:
Bjoern Feuerbacher wrote in message
...

George Dishman wrote:

Much snipped to try to get back to the physics.

"greywolf42" wrote in message
. com...

(replying to greywolf here, not to George...)

[snip]


I thought you and Bjoern were being deliberately obtuse,

Well, that's one of the big problems when discussing with you:
that you constantly invent motives for your discussion partners,
instead of actually trying to understand what they are asking
and saying.

And yes, this is indeed an ad hominem. An eye for an eye...


However, we see below that you understood all along.

I guessed vaguely that you *maybe* mean that. Do you suggest that I base
my answers to you on what I *guess* what you meant, instead of
*asking* you what you meant?


You *were* simply being deliberately obtuse.

I was asking for clarification. What on earth is your problem
with that?


and implying that
you didn't understand how luminosity (magnitude) could be considered
distance. I was specifically addressing your statement: "what is
observed is redshift versus magnitude or some other indirect measure of
distance." Magnitude is an indirect measure of distance (since the days
of Wirtz and Hubble). So, we have a curve of redshift vs. distance.

Yes, we can indeed get such a curve from the luminosity-redshift curve
in the reference you gave.


Then your claim that you didn't understand my statement was simply your
being deliberately obtuse.

Where, exactly, did I claim that I did not understand your
statement?


And the observed curve is exponential.

I don't think so. Where do you get this from?


From the reference I gave. Just like the last two times you asked this in
this thread, Bjoern. Look at the data. Fit an exponential curve to it.
Voila!

I *have* looked at the data (repeatedly now), and I *still* don't see
why on earth you think that an exponential curve fits that data.

Care to demonstrate that? Will you finally stop using mere rhetoric
and come up with something quantitative?



At least within error bars.

If you think so: Care to test if an exponential curve, or the curve
predicted by the BBT, using the parameters from WMAP, fits
better to the data, including the error bars?


Why bother?

So you don't think it is necessary to back up your claims?


The exponential explanation is valid.

That's so far an unsupported assertion. Take the data, do a
chi-squared fit, and demonstrate that an exponential curve
indeed fits the data.


Which is the point at issue.

The point at issue is that you keep making the unsupported assertion
that an exponential curve is detected.


We aren't discussing WMAP at all, because WMAP does not directly
measure the redshift-distance curve.

I did not claim that it does. Read again what I actually wrote,
please.


Bye,
Bjoern

In article ,
greywolf42 wrote:

Looking at Table 4, we see that the error bars on all the
measurements are substantial fractions (and often many times the
value of) the claimed "measured" value. Which clearly demonstrates
my point about noise processing.

No, it doesn't. A basic aspect of signal processing is dealing with
and extracting signals from data streams for which the signal-to-noise
ratio is less than unity. Indeed, a simple example is estimating the
mean and uncertainty in the mean from a set of data. 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. Even in
the case for which x ~ s, if N is large enough it is possible to make
fairly precise measurements.

--
Lt. Lazio, HTML police | e-mail:
No means no, stop rape. | http://patriot.net/%7Ejlazio/
sci.astro FAQ at http://sciastro.astronomy.net/sci.astro.html

In article ,
Bjoern Feuerbacher wrote:
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.

It has become to me he did not understand the S-Z effect.

Temperature is measured by the *overall* spectrum ... not by an arbitrary
slice of it!


Temperature is only defined for a system in equilibrium. The photons
are not in equlibrium after being inverse compton scattered, hence
don't follow the black body curve.

Why does it make sense then to talk about the change in
temperature???

Because you make observations at a finite band, with a certain
bandwidth. Each observations measures a surface brightness, which
gives an equivalent temperature. When in the rayleigh taylor region,
the change in surface brightness is a decrement, meaning the
background behind the cluster is of lower brightness when seen in that
band. If observed by a telescope at a frequency in the Wein region,
the surface brightness would be increased.

For any specific wavelength, a
temperature can be calculated from the observed specific intensity via
B = 2 nu^2 kT/c^2.


http://scienceworld.wolfram.com/phys...mperature.html

Only valid in the Rayleigh-Jeans region of the Planck curve.

No, that is a definition, valid every where. The brightness
temperature defined in the equation does not necessarily match a
physical temperature. Often in astronomy we observe non-thermal
affects, and we still assign a brightness temperater to them.

Since the photons do not have a black body spectrum anymore,
why is it valid to use that formula?

Because the formula is not dependant on the photons being black body.

And confused temperature decrements with intensity decrements.


For a black body they are related.

But after the scattering, there *is* no black body spectrum
anymore!

There still exists a relationship between the surface brightness
temperature and the intensity.

Apparently greywolf, who is, as well all know, an expert on
radio astronomy and error analysis...

Sometimes his arrogance leaves me really speachless!

Yea, like when he tells me I'm using astronomical language
incorrectly.

In article ,
Greg Hennessy wrote:
It has become to me he did not understand the S-Z effect.

It has become clear to me he did not understand the S-Z effect.

Note to self: Don't post before coffee.

Greg Hennessy wrote:
In article ,
Bjoern Feuerbacher wrote:

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.


It has become [clear] to me he did not understand the S-Z effect.

He does not understand quite a lot about cosmology...



Temperature is measured by the *overall* spectrum ... not by an arbitrary
slice of it!


Temperature is only defined for a system in equilibrium. The photons
are not in equlibrium after being inverse compton scattered, hence
don't follow the black body curve.

Why does it make sense then to talk about the change in
temperature???


Because you make observations at a finite band, with a certain
bandwidth.

Clear.


Each observations measures a surface brightness, which
gives an equivalent temperature.

"equivalent" in what sense?

That a black body of this temperature would have the same
surface brightness?


When in the rayleigh taylor region,
the change in surface brightness is a decrement, meaning the
background behind the cluster is of lower brightness when seen in that
band. If observed by a telescope at a frequency in the Wein region,
the surface brightness would be increased.

Yes, I got that from your last posts already. ;-)


For any specific wavelength, a
temperature can be calculated from the observed specific intensity via
B = 2 nu^2 kT/c^2.


http://scienceworld.wolfram.com/phys...mperature.html

Only valid in the Rayleigh-Jeans region of the Planck curve.


No, that is a definition, valid every where.

One defines an "(equivalent) brightness temperature" in this way?
I vaguely remember having read that already once in an astronomy book.
But the web page you cited does not say that.


The brightness
temperature defined in the equation does not necessarily match a
physical temperature. Often in astronomy we observe non-thermal
affects, and we still assign a brightness temperater to them.

Thanks for the explanation. I did not know (or had forgotten)
the term "brightness temperature".


[snip some repetitions, based on this misunderstanding]


Bye,
Bjoern

In article ,
Bjoern Feuerbacher wrote:
Each observations measures a surface brightness, which
gives an equivalent temperature.

"equivalent" in what sense?

Well, if you observe a thermal source in the rayleigh taylor region,
the brightness temperature will equal the physical temperature.

For something like the S-Z effect, where wat you are looking at is
close to, but not exactly, a black body, the surface brightness you
measure from the telescope is to first order proportional to the
temperature you would assign.

In a true BB curve, the and you compare to objects at different
temperatures, the flux at a given frequency of the hotter object is
always greater than the cooler object. The spectrum of the photons
after being inverse compton scattered by the SZ effect are in some
places higher flux than the unscattered BB curve, in some places
lower. This is because the number of photons in a BB curve is
proportional to T**3, but while the inverse compton effect adds
energy, it does not add photons.

No, that is a definition, valid every where.

One defines an "(equivalent) brightness temperature" in this way?
I vaguely remember having read that already once in an astronomy book.
But the web page you cited does not say that.

Yes. For example:http://www.astro.cf.ac.uk/observatory/radioback.html

It is often confuses grad students who think that the definition of
brightness temperature is only valid in the RT limit, but it is valid
every where.

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

much snipped again to get to the physics, some parts
moved

However, this
can't be true for Ned's curves. Because the intensity between Ned's
blue and red curves is unchanged. Photon frequency lowers in a
tired light model, *because* energy is lost. Yet Ned's red curve shows
energy per photon as unchanged after (you claim) tired light shifting
has taken place. It appears that when you include the energy loss, the
red curve once again drops to the black (observed) curve.

Note that the Y scale is in M Jy / sr. 1 Jansky is
10^-26 watts per square meter per Hz. The "per Hz"
factor compensates for this because an interval of
1Hz at the receiver started out as 1.1Hz at the
source.

The "per Hz" does not compensate for this effect. Ned's curve appears to
show a shift *without* energy removal. ...

part from later in your replay brought forward

Tired light effectively compresses the power
emitted into a narrower band.

Umm, no. Tired light does not compress power bands. This may simply be a
sloppy usage on your part. The relative energy-frequency curve merely
shifts. It does not compress the band, unless you arbitrarily have set a
minimum intensity as defining the edges of the "band."

Think of the graph as a histogram with bins 1Hz wide,
that's what sets the edges. The photons that go into a
bin from f to f+1 on the red curve come from frequencies
in the range 1.1*f to 1.1*(f+1) on the blue curve. That's
10% more photons than are in the corresponding bin on the
blue curve which is from (1.1*f) to (1.1*f)+1. Of course
each photon carries 10% less energy so the two factors
cancel and the intensity is the same.

Take a more extree example, consider a source received
with z=2. Suppose you measure the intensity at 1MHz. That
is a measure of the power in the band from 1,000,000Hz to
1,000,001Hz. However when the photons were emitted they
had frequencies from 2,000,000Hz to 2,000,002Hz When put
into the histogram though, that range would be split into
two bins, the first from 2,000,000Hz to 2,000,001Hz and
the other from 2,000,001Hz to 2,000,002Hz. There are
therefore twice as many photons in the bin for the received
intensity as in each of the bins for the emitted intensity,
but since each received photon has half the energy when it
was transmitted, the total power in each 1Hz band is the
same. Do you follow yet, it's not my best explanation.

... Note that the "resultant" (red)
curve has the perfect shape to match the observations -- if you merely
divide the intensity values by Ned's arbitrary factor of 1.331.

That's right, so if the source was say rocks at that
temperature and the rocks only covered 75.13% of the
sky then the intensity would be just right. Similarly if
the rocks were at 3.27K and z=0.2 then they would need
to cover exactly 57.87% of the sky to match observation.

I think that might be one of the possible coincidences
that Ned mentions, that to explain the observation, the
amount of sky covered would have to be exactly (1+z)^-3

Let's take a quick look at Ned's support for his curves:
"Assume that the CMB starts out as a T = (1+z)*To = 2.998 K blackbody,
which
is the blue curve. Because the photons only lose energy but do not
decrease
their density, the resulting red curve is not a blackbody at To = 2.725,
but
is instead (1+z)3 = 1.331 times a blackbody."

Now where do *you* get Ned's "original" intensity for the blue curve? I
cannot determine this from your interpretation of Ned's test. For the
intensity would be based on the density of your source gas shell.

I haven't checked the values but I assume it is simply
Planck's Law, in other words it covers the whole sky.

snip
... Tired light theories do not need nonlocal
sources. In this case, "nonlocal" means "outside of the receiving
apparatus."

snip

3) The local region for the MMBR (measured microwave background
radiation) is the antenna of the device you are using. According to
the
matter theory that I favor.

If you are suggesting the instrument designers didn't
consider locally generated thermal noise in the receiver,
that is obvious nonsense. It is a major problem and has
to be designed for, tested and calibrated.

I'm *not* addressing what designers typically consider "thermal noise" of
the receiver. That is merely the noise due to the temperature of the
matter -- which varies with matter temperature. What I am referring to is
the aether electron "hum" that arises from the interaction of the electron
(standing wave) with the aether corpuscles.

Not one CMBR device -- to my knowledge -- has ever attempted an isolation
test. That is, a test to determine whether the signal was actually
produced
"within" the antenna -- or whether it had an external source. For
example,
I know for a fact that Penzias and Wilson did not do this test. They
*did*
cut out the antenna connection. But they did not put their antenna in an
isolation chamber.

I'd be happy to be corrected, if you know of anyone who has done this.
(And
this has been discussed on the newsgroups for at least a decade.)

Well I can't imagine they would launch any craft without
testing the equipment and you can't do that except in
a screened room. However, you would have to contact the
team that built the equipment to get confirmation, it's
the sort of thing everyone does without comment, just
normal engineering, so it is unlikely you will find
anything published about it. Perhaps it is simpler just
to point out that since COBE and WMAP produce the same
map of the sky, the data cannot be an artefact of the
detector. Obviously the resolution of WMAP is far better
but just calculate the correlation of the two.

I thought you said the CMBR was emitted by electrons
being excited by the aether,

Yes. I have attempted to clarify the situation for you. *ALL* electrons
are
effected by the aether.

As you said, the dipole indicates that the source of
the radiation is moving past the detector. If this is
"electron hum" from electrons bound in hydrogen as you
said, then that hydrogen must be flowing past the craft
at about 400km/s and the hydrogen is moving in the same
direction regardless of which way the craft is pointed.

Including the antenna of the Penzias and Wilson
device, and all other such devices. No experimental system that I am
aware
of, has ever tested to eliminate this possibility. P&Z never made this
test. After P&Z, everyone simply assumed it was "cosmic" in origin (i.e.
outside the mechanism) -- and never made this test.

How do you think they would calibrate an instrument
that was picking up every microwave oven and mobile
phone for miles around? All such testing and
calibration has to be done in an electrically quiet
environment and that means a screened room. However,
what would be the point? If this noise is generated
by hydrogen in matter anyway, it would still fill any
EMC chanber as well.

... The page is
based on one fact, that the temperature measured here
and now is 2.725K. As an example, he then considers the
postulate that it was emitted at 2.998K at a distance
of 400MPc and the peak has been moved by tired light
to coincide with the peak frequency of a black body of
2.725K. There is no suggestion that the actual source
is at 2.725K at any time.

That is *your* description. Unfortuantely, Ned doesn't provide any such
description. What he provides is just this:

"The expanding balloon analogy for cosmological models can be used to show
this. ... Note that the galaxies (yellow blobs) do not grow, but the
distance between galaxies grows, and that the photons move and shift from
blue to red as the Universe expands, and the photon density goes down. But
in the tired light model, illustrated below, the density does not go down.
Assume that the CMB starts out as a T = (1+z)*To = 2.998 K blackbody...."

AFAICT, Ned *IS* specifically addressing the expanding balloon analogy of
the BB. And Ned *IS* therefore specifically addressing the fact that the
"temperature of space" is cooling.

Your approach would need no reliance upon the BB balloon analogy at all.

snip

*NED's* model is the BB balloon universe. The temperature varies with
time.

*YOUR* model does not.

The top two circles show the balloon analogy for the big
bang. The balloon on the right is bigger and the effect
of expansion is both to redden the light (the blue squiggly
lines become red ones) but also to increase the volume of
space as the cube of the expansion. The fact that the same
number of photons are skulling around in a bigger volume
si what reduces the density.

In contrast, the next pair of circles represents a steady-
state model so the two circles are the same size - no
expansion. Tired light still reddens the light but the
density of squiggly lines remains the same. That's why
there's a difference of (1+z)^3 between the two models.
Note that in this pair, there is _no_ expansion of the
balloon, that's the whole point. Only by keeping the size
the same can he illustrate that the volume doesn't change
hence the photon density doesn't change, only the individual
energies.

George