Cosmic acceleration rediscovered

"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