Cosmic acceleration rediscovered

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

My starting assumption was the same as you say above:

"The 'returning' is a local effect (at the point of fractional
return). dE = - const E, at the point of inspection."

Sorry, I did not consider that your starting assumption, because of the
large gap -- and because the next statement does not follow from the
above.

The E in that equation falls as the inverse square

This is your incorrect assumption. The E in the above equation does not
fall as the inverse square. The equation above is the pure linear form.
The only degradation is due to the fractional removal from every photon
(which continues in a straight line). Inverse square is the reduction in
the number of photons per unit area, as a spherical source spreads over
the
surface of larger spherical shells.

hence so does dE.

We are talking past each other so let me try another
way. Consider a area of 1m^2 at a distance of 1000AU
from the Sun. In one second an energy E passes through
that square. That will deposit a small fraction dE
into the aether.

Now consider an area of 1m^2 at a distance of 2000AU.
The energy passing through in one second will be
almost E/4 but very slightly less due to tired light.
Over this distance though the tired light loss will
be negligible compared to the r^-2 loss. The energy
deposited into the aether here will therefore be no
more than dE/4 and in fact slightly less.

If the temperature of the aether at 1000AU is T, that
at 2000AU should be very close to (but slightly less
than) T/sqrt(2) since the power from a black body
radiator is proportional to T^4.

I don't think so. It still seems to me that dE in
the equation you posted will follow an inverse square
from each (point) source of E.

As demonstrated above, you are incorrect. A tired light model will always
be (slightly) below a pure inverse square model.

Yes, that's what I'm saying, though I had ignored the
(slightly) part as it is negligible over short ranges
and it adds to the effect anyway. You are arguing that
the temperature is uniform aren't you?

George

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

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.

OK, this is more about terminology again, I would consider
that a variant of a big-bang model. If the theory includes
tired light to explain red shift then I would look on it
as a hybrid of the two. There's nothing wrong with that.
Regardless, it doesn't affect what we are trying to do.

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.

Big bang describes the idea that at large scales, distances
between objects are systematically increasing. It doesn't
yet go back to creation since the theories break down at
least at the Planck time.

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.

That's fine. I have posted the example using exactly
those postulates. The id is



It may show up as a new thread though I used the same
subject line. Outlook Express is producing reference
lines in excess of 1000 characters, which is the
maximum according to the NNTP protocol, and my new
ISP's server is rejecting them. I had to post a new
message instead of a reply to get round it, sorry.

George

George Dishman wrote in message
...

First I'll draw together a few bits that I think
sum up most of the relevant parts of the discussion
though it's quite possible I'll miss some.

You did miss the fundamental point that the MBR in my favorite theory comes
from the antennae of our measuring devices.

line length reduced in quoted material to minimise
problems later

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

snip

In other areas however, you have said that the
aether is at a uniform temperature.

I've said that we don't expect it to be heated in
the local region of stars. It is roughly uniform.

snip

.... There is no
significant change in temperature with distance
and time.

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

snip

For example, in my favorite theory (a Maxwellian/
LeSage theory), the CMBR is simply an EM hum from
electrons bound to hydrogen.

But that doesn't produce the spectrum we see,

Sure it does. It's a thermal emission.
...
lambda = (1.0E-30) (2.99e+8) / (1.602E-19) = 1.87 E-3 m

Computing the equivalent blackbody temperature spectrum
of this emission gives:

T = .51 / 100 lambda = 2.73 deg K.


"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.

and its dipole moment

The motion of the solar system through the aether.

as well as why the
cosmological red-shift is exponential with distance

Any tired light theory. (In this case, the slight
imperfection in the aether.)


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

snip
Cloud emissions show different spectra
which we can use to determine temperature while the
CMBR is remarkably uniform.

The local region of space (a few tens of parsecs) is
expected to have an aether that is remarkably uniform
in temperature.


"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.

The temperature might
be the same everywhere but the intensity radiated
would be higher in regions with more electrons.

Again, you are thinking cosmogenic.

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

That's why I said I'd work with your numbers. Let's
keep it simple. One does not need to have all the
answers at the beginning.....

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.

Ok, I think we have enough to try this with what
you've said. The aim is partly show how Ned's test
can be applied and also to give a starting point so
that you can try to produce a self-consistent model
that explains the spectrum of the CMBR.

Actually, no. The purpose of these threads was to see if Ned's complaints
against tired light theories were valid. And since tired light theories
don't need to explain the CMBR, Ned's claims (based on cosmogenic CMBR) are
not valid.

If you want
to use Vigier or whatever, feel free but I don't
expect you to say explanation A can produce the
spectrum while explanation B can explain the dipole
if A and B are mutually exclusive. As you said, we
don't need to have all the answers at the beginning
but we should be able to sort them out in the course
of the discussion. So the strawman is this:

-:-

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.

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.

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.

This "electron hum" is produced everywhere roughly
uniformly as the electron density does not affect
the emitted intensity.

The electron hum is produced mainly within the antennae of the MBR
detectors.

Each photon of this emitted radiation loses energy
at a rate given by dE/E = dl/L due to a slight
imperfection in the aether. The value of L is
approximately 4.2GPc give or take a factor of 2.
This is independent of frequency producing an
exponential reduction of energy with distance.

True. Though irrelevant.

While the defect reduces the energy of individual
photons, it does not absorb photons.

It's easier to think of it this way, though we don't actually need photons.

There is no
significant reduction of photon numbers due to
grey dust or other possible causes of extinction.

Wherever did you get this one? Extinction is a totally separate concern.
(As it is in all astrophysics.)

The energy lost is transferred to the aether

The organized wave energy within the aether is converted to random motions
of the aether particles.

and
replaces the energy used locally to emit the CMBR
photons, hence energy is locally conserved.

Energy is conserved, even if my favorite version is not correct. The
correct term is "closed system."


-:-

To analyse the above using Ned's test,

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

Oh, and your aren't addressing my model -- which is an artifact of the
matter of which our detectors are constructed. But, go ahead with your
"distant-source" origin analysis. If valid, it can be applied to other (as
yet unknown) theories of the origin of the (C)MBR.

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.

However,
depending on the "cross section" of an electron,
it may be only a fraction k of the amount that
would be emitted by a solid (opaque) surface. This
factor k is adjustable.

The total radiation we receive is then the sum of
the photons from all the shells, however each
photon will be measured at a frequency

and energy

which has
been reduced from that at which it was transmitted
by the tired light effect.

http://www.astro.ucla.edu/~wright/tiredlit.gif

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

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.

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'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.

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.

I don't currently know of any such theories.

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

Joseph Lazio wrote in message
...
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.

That's if you know that you have a signal. Because you sent one.

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

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

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

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.

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

Even in
the case for which x ~ s, if N is large enough it is possible to make
fairly precise measurements.

One can deceive oneself that one is getting precise measurements. For
example, the Hipparcos gravitational bending of light team claimed an
observational precision that was four orders of magnitude below the physical
resolution of their apparatus. COBE analyis didn't find anything until they
were one order of magnitude below the physical resolution of their
apparatus.

Tom Roberts would call this "overaveraging."

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

Greg Hennessy wrote:
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.

Thanks again for the explanations.


Bye,
Bjoern

greywolf42 wrote:

[snip]

COBE analyis didn't find anything until they
were one order of magnitude below the physical resolution of their
apparatus.

If you suggest that the variations in temperature detected by
COBE are simply noise and not a real signal, then how do you
explain that WMAP found the same variations?



[snip]

Bye,
Bjoern

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


[snip]


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


Big bang describes the idea that at large scales, distances
between objects are systematically increasing. It doesn't
yet go back to creation since the theories break down at
least at the Planck time.

I think he did not mean the initial creation of space here,
but wanted to say that the expansion of the universe required
a continuous "creation" of "new" space.


[snip]

Bye,
Bjoern

greywolf42 wrote:
George Dishman wrote in message
...


First I'll draw together a few bits that I think
sum up most of the relevant parts of the discussion
though it's quite possible I'll miss some.


You did miss the fundamental point that the MBR in my favorite theory comes
from the antennae of our measuring devices.

Then why are there variations in the signal depending on the
direction of the antennae? If you claim that these variations
are simply statistical fluctuations, noise, please explain why
the same variations are seen by COBE and WMAP.

And why is *only* the "electron hum" occuring directly
in the antenna detected, and not the emission from all the
other "electron hums" around the antenna, going on into
arbitrary distance?


[snip]


Ok, I think we have enough to try this with what
you've said. The aim is partly show how Ned's test
can be applied and also to give a starting point so
that you can try to produce a self-consistent model
that explains the spectrum of the CMBR.


Actually, no. The purpose of these threads was to see if Ned's complaints
against tired light theories were valid. And since tired light theories
don't need to explain the CMBR,

Why not?


Ned's claims (based on cosmogenic CMBR) are not valid.

What, exactly, do you mean with "cosmogenic"?


[snip]



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.

What is the difference between the two statements?


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.

Do I understand you correctly here? The "electron hum" of
the electrons in the antenna produces microwave radiation,
this radiation moves a very short way in the aether, so that
it is still inside the antenna, and is then detected by the antenna?


[snip]


This "electron hum" is produced everywhere roughly
uniformly as the electron density does not affect
the emitted intensity.


The electron hum is produced mainly within the antennae of the MBR
detectors.

Why? There are lots of electrons in the universe. Why should
the antennae see mainly the hum coming from the electrons
within themselves?


Each photon of this emitted radiation loses energy
at a rate given by dE/E = dl/L due to a slight
imperfection in the aether. The value of L is
approximately 4.2GPc give or take a factor of 2.
This is independent of frequency producing an
exponential reduction of energy with distance.


True. Though irrelevant.

Why? The radiation is there. Do you claim that it is
*not* detected by the antennae? That that signal is much
weaker than the one produced by the electrons within
the antennae, or what?


[snip]


There is no
significant reduction of photon numbers due to
grey dust or other possible causes of extinction.


Wherever did you get this one? Extinction is a totally separate concern.
(As it is in all astrophysics.)

He did get this from you yourself, after asking you repeatedly
what effect extinction does have.


The energy lost is transferred to the aether


The organized wave energy within the aether is converted to random motions
of the aether particles.

What is "organized wave energy" (what is "organized energy"
in general?), and how does this converting work?


[snip]


To analyse the above using Ned's test,


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

It is.


Oh, and your aren't addressing my model -- which is an artifact of the
matter of which our detectors are constructed.

Err, in another post, you said explicitly that the effect you
propose is *not* an artifact of the detector. Do you know want
to quibble that there is a difference between "the detector" and
"the matter of which our detectors are constructed"?


But, go ahead with your
"distant-source" origin analysis. If valid, it can be applied to other (as
yet unknown) theories of the origin of the (C)MBR.

Do you want to say that there is no such theory around yet?
Do you want to say that *all* theories other than the BBT say
that the CMBR comes from within the detector?

If you did not mean that, please clarify what you meant.


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.

Err, didn't you say that the radiation is independent of the
electron density?



[snip]


which has
been reduced from that at which it was transmitted
by the tired light effect.

http://www.astro.ucla.edu/~wright/tiredlit.gif

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

Err, didn't you say that the radiation is independent of the
electron density?



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.

Do you mean that the peak intensity would move down? Why?


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

Why?

If the peak frequency is moved to the left (to which you
agreed above), the curves obviously are not the same anymore
- they are shifted wrt each other!


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.

Why on earth do you think he assumed that? A frequency shift is
automatically a change in 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.

That is absolutely irrelevant for the argument, as far as I can see.


This changes the results of the integration.

Well, please tell us what the result would be, in your
opinion. Show your work.


[snip]



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.

Why not?


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.
I don't currently know of any such theories.

Didn't Tom Flandern propose something like that? Or do I misremember?


Bye,
Bjoern

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


[snip]


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


Big bang describes the idea that at large scales, distances
between objects are systematically increasing. It doesn't
yet go back to creation since the theories break down at
least at the Planck time.

I think he did not mean the initial creation of space here,
but wanted to say that the expansion of the universe required
a continuous "creation" of "new" space.

[snip]

I think we are all on the same page on this one. (Thank goodness.)

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

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


First I'll draw together a few bits that I think
sum up most of the relevant parts of the discussion
though it's quite possible I'll miss some.


You did miss the fundamental point that the MBR in my favorite theory
comes
from the antennae of our measuring devices.

Then why are there variations in the signal depending on the
direction of the antennae? If you claim that these variations
are simply statistical fluctuations, noise, please explain why
the same variations are seen by COBE and WMAP.

COBE doesn't see *any* variations. At least the detector doesn't. The
claimed variations (1 part in 100,000) are all artifacts of computer
processing of noise below the physical resolution of the COBE instruments
(which are 1 part in 10,000). WMAP data doesn't actually measure these
variations at all.

And why is *only* the "electron hum" occuring directly
in the antenna detected, and not the emission from all the
other "electron hums" around the antenna, going on into
arbitrary distance?

It *is* going on everywhere. But the density in antennas is so much greater
than the densities out in space that it dominates any other signals.

[snip]


Ok, I think we have enough to try this with what
you've said. The aim is partly show how Ned's test
can be applied and also to give a starting point so
that you can try to produce a self-consistent model
that explains the spectrum of the CMBR.


Actually, no. The purpose of these threads was to see if Ned's
complaints
against tired light theories were valid. And since tired light theories
don't need to explain the CMBR,

Why not?

Becuase the MBR is not part of the theory, of course. The fact that the BB
assumes that the MBR is "relic radiation" does not mean that other theories
have to explain it.

Ned's claims (based on cosmogenic CMBR) are not valid.

What, exactly, do you mean with "cosmogenic"?

Created by the origin of the cosmos. No specific material objects required.

Bjoern, I've already gone through all of these questions with George. Read
the prior responses.

[snip]

And I have no need to explain to you all of the rest of theory and
observation beyond what George and I were discussing.

{snip the rest}

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