Re: Redshift without expansion


"sean" <jaymoseley@xxxxxxxxxxx> wrote in message
news:1151078619.732634.136850@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

George Dishman wrote:
If I understand Sean's diagram, he is suggesting that at
some point after passing the first duck, every second
wave vanishes for no apparent reason. The waves travel
at constant speed but subsequent ducks observe a lower
frequency. Of course the means the only frequency change
that wouldn't produce sidebands would be a factor of
two, think what would happen to the next duck if every
third wave vanished.
E 000000*00000000000000000000000*000000000
D 00000*00000000000A00000000000*0000000000
C 0000*00000000000*00000000000*00000000000*
B 000*00000*00000*00000*00000*00000*00000*
A 00*00*00*00*00*00*00*00*00*00*00*00*00*
O-source

(I've corrected an error on the 'A' line)

George wrote"...Of course the means the only frequency change
that wouldn't produce sidebands would be a factor of
two, think what would happen to the next duck if every
third wave vanished."

If I can respond to this let me say that its incorrect for you to
say that a frequency change of anything other than a half would
not work with the model. You make this mistake because you havent fully
appreciated the diagram and what it represents. First of all the lines
of * on the paper dont refer to waves or particles. They merely draw
lines
between observed peaks

Peak of what? I understand that to be the time at which
the EM field has its highest value.

and illustrate that the speed is constant and
the
brightness decreases inversely to an increase in distance.

Brightness? There is no indication of how high
each peak might be, only the time when it occurs
for different observer distances from the source.

So far, that
matches known observations. But lets see if you are right and this
rule is broken if I choose a frequency other than 1/2 or a factor of 2.
Take the above example of mine where its frequencies 3-6-12-24.

Ignoring the letter 'A' in the diagram above, the
frequencies are:

E 1/24
D 1/24
C 1/12
B 1/6
A 1/3
O not shown

So at B it is half of the frequency at A, at C it
is half of what it was at B, similarly for D but
you seen to have the diagram wrong at E.

You claim that a third between 12-6 is 8 and it wont produce
the same results. But you are wrong.Here is the pattern
one gets from frequencies at 8 beats per time unit.

Drop every third from the above using just A and B
and you get this:

B 000*00*00000*00*00000*00*00000*00*00000*
A 00*00*00*00*00*00*00*00*00*00*00*00*00*

The pattern at B is no longer regular.

Notice
how the speed is still constant where the frequencies decrease.
And significantly, although its not shown here Ive made samples
with several frequencies overlap on one graph and they all decrease
at the same rate and over the same distances. So for instance every
time one goes from 4-8-16-32 its one distance unit, or if one goes
from 1.5-3-6-12-24-48 its still the same distance unit. So
2-4 is the same distance as 3-6 or 4.5-9)
The rule is that every doubling of frequency no matter what
the original frequency is, it always accompanies a standard
unit of distance, ie a constant speed.

E00000*0000000000000000000000000000000*
D0000*000000000000000*000000000000000*0
C000*0000000*0000000*0000000*0000000*00
B00*000*000*000*000*000*000*000*000*000*0
A0*0*0*0*0*0*0*0*0*0*0*0*0*0*0*0*080*0*0*0*
0-source


That would allow z values of 0, 1, 3, 7, 15
but nothing inbetween.

George


.

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