Cosmic acceleration rediscovered

George Dishman wrote in message
...

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

The plasma fireworks people disagree -- because space is pre-existing. And
all velocities are *real*.

If the theory includes tired light to explain red shift
then I would look on it as a hybrid of the two.

Fair enough.

There's nothing wrong with that.
Regardless, it doesn't affect what we are trying to do.

OK. One tangent killed.

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.

The current big bang model(s) are not limited to mere systematic increases
in distance. Take a look at any "expanding balloon" or "raisin loaf"
analogy. Space is continually "expanding" (i.e. being created) between
galaxies. The galaxies aren't moving *through* space, so much as being
carried along.

In a PF model, space is pre-existing. And all motion is *through* space.

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.

No problem. I've replied to that post he
http://groups-beta.google.com/group/...b7ff68147cd2aa

I think we can abandon this threadling.

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

Bjoern Feuerbacher wrote in message
...
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?

Quite simply, WMAP does not measure temperature variations directly. WMAP
calculations are completely circular, and are processed until it confirms
prior beliefs.

[snip]

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

George Dishman wrote in message
...

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


{snip abandoned attempt at description}

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.

Delta E. Yes.

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.

Yes.

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.

Yes.

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.

Oops, here is your error.

The aether temperature is not merely a function of the local addition of
energy from starlight degradation. (Temperature is a function of E, not of
dE.) Starlight degradation is a miniscule contribution to the pre-existing
aether energy density. After all, the aether temperature (energy content)
is what drives gravitation. In fact, gravitation is a competing effect --
lowering the local aether temperature a tiny amount, in the vicinity of the
star.

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?

Very close to uniform. Not uniform. Normal gravitation and light energy
are tiny variations on the pre-existing local energy density. (The sun's
surface gravitation is about 1 part in 10^8 of the maximum gravitational
acceleration.)

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

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

snip
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.

Oops, here is your error.

The aether temperature is not merely a function of the local addition of
energy from starlight degradation. (Temperature is a function of E, not
of
dE.) Starlight degradation is a miniscule contribution to the
pre-existing
aether energy density. ...

Ah, the penny drops, thanks. Eddington showed energy
in starlight was equivalent to circa 2.8K but only a
small fraction of that transfers to the aether. 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.

George

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.

"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 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?

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.

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.

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.

snip non-contentious parts

There is no
significant reduction of photon numbers 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.

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

Black body radiators are also perfect absorbers.

-:-

To analyse the above using Ned's test,

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

IMO it is.

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.

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.

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.

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

(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"?

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.

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.

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. Ned's graph is correct for tired light.

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.

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. Now note that in
the Plasma Fireworks model, some of the redshift is due
to motion so the amount of energy loss due to tired light
would be less hence mu would have a smaller value which
we could find if we could separate out the motion part.
If objects were moving apart fast enough, that could
explain all the redshift and hence mu would be zero.

Since the volume of any region of space increases as the
cube of its dimensions, that reduces the photon density
and hence the intensity. If there is no motion 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.

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.

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

George

"George Dishman" wrote in message
...

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

A small clarification, the "And down" is correct but
your "black" should be "red". The shells are really
continuous from zero to infinity as I'm sure you
realised.

George

"g" == greywolf42 writes:

g George Dishman wrote in message
g ...

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.

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

This was a favorite statement of Grote Reber, the first radio
astronomer. I've never quite known what to make of it. Reber
essentially invented the field of radio astronomy, so I'm inclined to
take seriously any of his suggestions. On the other hands, he was
wrong at times. Moreover, he never suggested a physical mechanism by
which the MBR would be produced, and he knew as much about radio
antennas as anybody.

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

--
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No means no, stop rape. | http://patriot.net/%7Ejlazio/
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"g" == greywolf42 writes:

g Joseph Lazio wrote in message
g ...

Looking at Table 4, we see that the error bars on all the
measurements are substantial fractions (...) 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.

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

I suspect that there are lots of people (not astronomers) who spend
time processing data streams to see if signals are present. Indeed, I
suspect that the entire field of signal processing would be a lot more
boring if one could only process data streams for which one knows a
signal to be present.

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.

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. If I estimate the mean and the uncertainty in the mean
from a set of data, find that the mean is consistent with zero, and
conclude that there is no signal present, how is that a problem?
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?

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.

More generally, of course, knowing the uncertainties in the individual
data is one of the great challenges in experimental and observational
sciences. There are ways to estimate the uncertainties (for instance,
calculating the standard deviation of the data), but they make certain
assumptions, and one wants to be careful to check that those
assumptions are valid.

However, it's also one of the reasons you glib dismissals of
experimental results that you don't like rings so hollow among those of
us with experience in data processing.

--
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greywolf42 wrote:
Bjoern Feuerbacher wrote in message
...

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?


Quite simply, WMAP does not measure temperature variations directly. WMAP
calculations are completely circular, and are processed until it confirms
prior beliefs.

Support that claim, please.

You know that you accuse quite a lot of physicists here of
either fraud or incompetence, don't you?


Bye,
Bjoern