CMBR? Not in the Big Bang Universe.

CMBR? Not in the Big Bang Universe.
-----

For some time I've been trying to understand why the spectral
energy density graph plot of the 2.73 K CMBR, per formula [1]
(2 * pi * f^3) / (c^2 * (exp(h * f / (k * T)) - 1)), is nothing
like a 2.73 K blackbody radiator plot according to formula [2]
(2 * pi * h * c^2) / (b^5 * ((exp((h * f) / (k * T))) - 1))
(b is wavelength)

The graph plot of intensity per frequency unit along a scale of
frequencies can be easily converted for direct comparison with
formula [2] by converting frequency to wavelength with (c / f) and
plotting the curve on the same graph scale as for formula [2].
Whatever shape the curves may follow, 5.35 cycles per cm is the peak
point along the emissive power curve for a 2.73 K radiator according
to formula [1], and that is found to be 1 / 5.35 = .187 cm
wavelength. But this is not so according to formula [2], which
gives the peak wavelength as .106 cm.

It matters not how the numbers are (commonly) juggled, when the
two curves are compared, the asymmetric relationship between the
curve peaks (and the curves as well) is always constant.

The following graphs referred to below were generated using
formulas [1] and [2]. I've made no attempt to sketch them in
ASCII for obvious reasons. The graphs are stored at
http://www.ozemail.com.au/~mkeon/monpol.html
I've also included the text.

Graph 1 demonstrates that the peak of a 2.73 K curve per formula
[2] aligns with the peak of a 4.816 K curve according to formula
[1].

Graph 2 shows the alternative alignment, which is between
a 2.73 K radiator per [1] and 1.55 K radiator per [2].

Adding to [1], a 1.76 * T multiplier for temperature or changing
the base of the exponential function to 1.76, sets the peak of a
2.73 K curve per formula [1] to align with a 2.73 K curve peak per
formula [2], but that would certainly raise a few questions.

The perfect alignment of the 1.55 K curve per [2] and the 2.73 K
curve per [1] is achieved by taking the square root of the emissive
power for each wavelength along the 1.55 K curve, and adding an
appropriate multiplier for the comparison.
Graph 3

The square root inclusion implies that the longer wavelengths have
been stretched by a greater margin than the shorter wavelengths.
But that's not possible. Why would the expansion be locally
asymmetric? Over a wavelength?? A simple multiplier accounts for
the expansion of the entire blackbody curve.

There is no reason whatever why the expanding Big Bang Universe
would shift the peak of the emission curve, **or the curve shape**,
away from that of a natural blackbody radiator.

Dimension around a blackbody radiation detector in the 4000 K
Universe has doubled in all three dimensions when the temperature
of the Universe has fallen to 2000 K, so wavefront areas destined
to reach the detector from the 4000 K era will have reduced to 1/4
when they arrive. If wavelengths could have remained constant the
total radiation energy received would be reduced to 1/4.

The 1/4 energy reduction is further affected because the wavelengths
have of course doubled, thus only half the number of wavelengths are
passing into the detector per time, reducing the total radiation
energy received from **every individual** wavelength to 1/8. And
that's the final result from the expansion. No other energy losses
can possibly be accounted for.

Graph 4 shows the relationship between true 4000 K - 2000 K
blackbody curves and the expanded curve from the 4000 K era. The
radiation energy from each wavelength for the expanded curve is
four times greater than for the real 2000 K blackbody curve.

Multiplying the radiation energy for each wavelength of the proper
2000 K radiator curve by four, shows that the expanded curve aligns
with the shape of a true blackbody curve (raised above the baseline
for obvious reasons). http://www.ozemail.com.au/~mkeon/mon5.gif
http://www.ozemail.com.au/~mkeon/mon6.gif

According to the two formulae, the asymmetry between the true
blackbody and the CMBR curve was present right from the initial
CMBR transmission.

Apart from the CMBR aligning with the wrong curve shape, there's
still the quandary of how to explain the enormous amount of missing
radiation energy that is not removed in the expansion. At the very
first doubling of dimension, that is already four times greater
than would be expected from a true 2000 K radiator
(4000^2 / 2000^2 = 4). By the time the expansion has diminished
the temperature of the Universe to 2.73 K, that additional energy
would rise to 4000^2 / 2.73^2 = 2.147E+6 times greater than for
the proper 2.73 blackbody radiator. Being the focal point of that
much microwave energy, I would expect that I would be well and
truly cooked by now.

The sphere radius around the detector from which the background
radiation was generated when the Universe first became transparent
was expanding away from the detector at the speed of light
(radiation was traveling from everywhere to everywhere at the
speed of light). Regardless of the expansion rate of the Universe,
throughout the expansion, the background source from the 4000 K
realm that arrives at the detector was generated in the 4000 K
environment. Every part of the CMBR was generated in that realm.
The matter content involved in generating the background was thus
increasing at a rate that would exactly counter the decreasing
wavefront areas, from increasingly distant sources, that are
falling on the detector.

The 2D wavefront expanding with dimension and a simple count of wave
numbers arriving at the detector accounts for the entire energy
losses. Nothing else.

The Big Bang Theory fails the CMBR test.
But not so The Zero Origin Concept, which can be found at
http://www.ozemail.com.au/~mkeon/the1-1a.html It paints
a rather ugly Universe compared to the inconsequential Big Bang
Universe.

If mankind doesn't stick around, smart enough and long enough
to learn how to bend the rules of the Universe, you, me and the
gatepost are guaranteed an eternal hell that has no limit to how
deep it can go. I wouldn't hold my breath though, it doesn't look
like we'll even make it over the very first little hurdle. A trip
back to the dark ages will fairly well seal our fate.

Isn't it about time for a reality check folk?


--
Max Keon

"Max Keon" skrev i melding
...
CMBR? Not in the Big Bang Universe.
-----

For some time I've been trying to understand why the spectral
energy density graph plot of the 2.73 K CMBR, per formula [1]
(2 * pi * f^3) / (c^2 * (exp(h * f / (k * T)) - 1)), is nothing
like a 2.73 K blackbody radiator plot according to formula [2]
(2 * pi * h * c^2) / (b^5 * ((exp((h * f) / (k * T))) - 1))
(b is wavelength)

And why is that?
I have shown you this before, it is quite simple:
dW/df = (2 *pi *h* f^3) / (c^2 * (exp(h * f / (k * T)) - 1)),
f = c/b, df/db = -c/b^2
dW/db = (dW/df)*(df/db)
dW/db = -(2*pi*h*c^2) / (b^5*((exp((h*c)/ k*T*b))) - 1))

The graph plot of intensity per frequency unit along a scale of
frequencies can be easily converted for direct comparison with
formula [2] by converting frequency to wavelength with (c / f) and
plotting the curve on the same graph scale as for formula [2].

No, you cannot.
If you insert f = c/b in [1], it is still dW/df, which is different
from dW/db.

Whatever shape the curves may follow, 5.35 cycles per cm is the peak
point along the emissive power curve for a 2.73 K radiator according
to formula [1], and that is found to be 1 / 5.35 = .187 cm
wavelength. But this is not so according to formula [2], which
gives the peak wavelength as .106 cm.

dW/db = -c/b^2* dW/df
so it is quite obvious that they don't peak at the same frequency/wavelength.

But I showed you this before, and I am sure you commented it,
so you did see it.

So why this nonsense again?

Paul

Paul B. Andersen wrote:

"Max Keon" skrev i melding
...
CMBR? Not in the Big Bang Universe.
-----

For some time I've been trying to understand why the spectral
energy density graph plot of the 2.73 K CMBR, per formula [1]
(2 * pi * f^3) / (c^2 * (exp(h * f / (k * T)) - 1)), is nothing
like a 2.73 K blackbody radiator plot according to formula [2]
(2 * pi * h * c^2) / (b^5 * ((exp((h * f) / (k * T))) - 1))
(b is wavelength)

And why is that?
I have shown you this before, it is quite simple:
dW/df = (2 *pi *h* f^3) / (c^2 * (exp(h * f / (k * T)) - 1)),
f = c/b, df/db = -c/b^2
dW/db = (dW/df)*(df/db)
dW/db = -(2*pi*h*c^2) / (b^5*((exp((h*c)/ k*T*b))) - 1))

The graph plot of intensity per frequency unit along a scale of
frequencies can be easily converted for direct comparison with
formula [2] by converting frequency to wavelength with (c / f) and
plotting the curve on the same graph scale as for formula [2].

No, you cannot.
If you insert f = c/b in [1], it is still dW/df, which is different
from dW/db.

Whatever shape the curves may follow, 5.35 cycles per cm is the peak
point along the emissive power curve for a 2.73 K radiator according
to formula [1], and that is found to be 1 / 5.35 = .187 cm
wavelength. But this is not so according to formula [2], which
gives the peak wavelength as .106 cm.

dW/db = -c/b^2* dW/df
so it is quite obvious that they don't peak at the same
frequency/wavelength.

I'm trying to picture what you are describing, but it just doesn't
add up. You are saying that the wavelength that emits the greatest
energy quantity from a blackbody radiator is dependent on which
formula is used? That can't possibly be. If a .106 cm wavelength
carries the greatest energy quantity, then it carries the greatest
energy quantity. How can a .187 cm wavelength also claim to carry
the greatest energy quantity, from the same radiator temperature?

I'll try a more hands on approach.
From a graph of the CMBR, plotted according to formula [1] above,
I note that the frequency of oscillation which carries the greatest
energy quantity is roughly 5.3 cycles per cm. I record that
information and, with a simple calculation, I determine that the
wavelength at that frequency is 1 / 5.3 = .188 cm. I can now use
this data for a comparison with the peak of the power curve plotted
for a 2.73 K radiator according to formula [2] above, which peaks
at roughly .11 cm. I then use an appropriate multiplier for spectral
energy density per [1], or the emissive power per [2] to bring
either into an alignment with the other, for a direct comparison.

But no amount of juggling can make the wavelengths attributed to
the two peak power points coincide.

Unfortunately you haven't discovered a way to bend the rules of the
Universe, you've merely shown that the curve shape to which the CMBR
was made to align was based on a flawed formula. And if you genuinely
believe in what you are saying, you have also demonstrated that maths
can befuddle the minds of even the best.


--
Max Keon

"Max Keon" skrev i melding ...
Paul B. Andersen wrote:

"Max Keon" skrev i melding
...
CMBR? Not in the Big Bang Universe.
-----

For some time I've been trying to understand why the spectral
energy density graph plot of the 2.73 K CMBR, per formula [1]
(2 * pi * f^3) / (c^2 * (exp(h * f / (k * T)) - 1)), is nothing
like a 2.73 K blackbody radiator plot according to formula [2]
(2 * pi * h * c^2) / (b^5 * ((exp((h * f) / (k * T))) - 1))
(b is wavelength)

And why is that?
I have shown you this before, it is quite simple:
dW/df = (2 *pi *h* f^3) / (c^2 * (exp(h * f / (k * T)) - 1)),
f = c/b, df/db = -c/b^2
dW/db = (dW/df)*(df/db)
dW/db = -(2*pi*h*c^2) / (b^5*((exp((h*c)/ k*T*b))) - 1))

The graph plot of intensity per frequency unit along a scale of
frequencies can be easily converted for direct comparison with
formula [2] by converting frequency to wavelength with (c / f) and
plotting the curve on the same graph scale as for formula [2].

No, you cannot.
If you insert f = c/b in [1], it is still dW/df, which is different
from dW/db.

Whatever shape the curves may follow, 5.35 cycles per cm is the peak
point along the emissive power curve for a 2.73 K radiator according
to formula [1], and that is found to be 1 / 5.35 = .187 cm
wavelength. But this is not so according to formula [2], which
gives the peak wavelength as .106 cm.

dW/db = -c/b^2* dW/df
so it is quite obvious that they don't peak at the same
frequency/wavelength.

I'm trying to picture what you are describing, but it just doesn't
add up. You are saying that the wavelength that emits the greatest
energy quantity from a blackbody radiator is dependent on which
formula is used? That can't possibly be. If a .106 cm wavelength
carries the greatest energy quantity, then it carries the greatest
energy quantity. How can a .187 cm wavelength also claim to carry
the greatest energy quantity, from the same radiator temperature?

Why is this so hard to get?
dW/df is energy per frequency unit.
That is, it is how much energy there is in the part of the spectrum
that has frequencies between f and f+1.
dW/dt is energy per wavelength unit
That is, it is how much energy there is in the part of the spectrum
that has wavelengths between b and b+1.

Since f = c/b, it means that the bandwidth df = -c/b^2*db
Thus the energy dW/df in the bandwidth 1 Hz,
it is equal to the energy dW/db in the bandwidth c/b^2 metres.
Thus dB/df = (c/b^2)*dB/db

I'll try a more hands on approach.
From a graph of the CMBR, plotted according to formula [1] above,
I note that the frequency of oscillation which carries the greatest
energy quantity is roughly 5.3 cycles per cm. I record that
information and, with a simple calculation, I determine that the
wavelength at that frequency is 1 / 5.3 = .188 cm. I can now use
this data for a comparison with the peak of the power curve plotted
for a 2.73 K radiator according to formula [2] above, which peaks
at roughly .11 cm. I then use an appropriate multiplier for spectral
energy density per [1], or the emissive power per [2] to bring
either into an alignment with the other, for a direct comparison.

But no amount of juggling can make the wavelengths attributed to
the two peak power points coincide.

Unfortunately you haven't discovered a way to bend the rules of the
Universe, you've merely shown that the curve shape to which the CMBR
was made to align was based on a flawed formula. And if you genuinely
believe in what you are saying, you have also demonstrated that maths
can befuddle the minds of even the best.

Oh, my dear.
You really are a crank, aren't you? :-)
This is a case of simple ignorance of elementary math and physics.
If you don't believe me, why don't you look up "black body radiation" in
an elementary physics textbook and learn it in stead of all this nonsense?
In most books you will find both spectra side by side, and how
you derive the one from the other.
Or you can see:
http://hyperphysics.phy-astr.gsu.edu/hbase/mod6.html

Why you bother to make so much fuss about something which
can be cleared up by half an hour reading, beats me.

Paul

Max Keon wrote in message ...
Paul B. Andersen wrote:

"Max Keon" skrev i melding
...
CMBR? Not in the Big Bang Universe.
-----

For some time I've been trying to understand why the spectral
energy density graph plot of the 2.73 K CMBR, per formula [1]
(2 * pi * f^3) / (c^2 * (exp(h * f / (k * T)) - 1)), is nothing
like a 2.73 K blackbody radiator plot according to formula [2]
(2 * pi * h * c^2) / (b^5 * ((exp((h * f) / (k * T))) - 1))
(b is wavelength)

And why is that?
I have shown you this before, it is quite simple:
dW/df = (2 *pi *h* f^3) / (c^2 * (exp(h * f / (k * T)) - 1)),
f = c/b, df/db = -c/b^2
dW/db = (dW/df)*(df/db)
dW/db = -(2*pi*h*c^2) / (b^5*((exp((h*c)/ k*T*b))) - 1))

The graph plot of intensity per frequency unit along a scale of
frequencies can be easily converted for direct comparison with
formula [2] by converting frequency to wavelength with (c / f) and
plotting the curve on the same graph scale as for formula [2].

No, you cannot.
If you insert f = c/b in [1], it is still dW/df, which is different
from dW/db.

Whatever shape the curves may follow, 5.35 cycles per cm is the peak
point along the emissive power curve for a 2.73 K radiator according
to formula [1], and that is found to be 1 / 5.35 = .187 cm
wavelength. But this is not so according to formula [2], which
gives the peak wavelength as .106 cm.

dW/db = -c/b^2* dW/df
so it is quite obvious that they don't peak at the same
frequency/wavelength.

I'm trying to picture what you are describing, but it just doesn't
add up. You are saying that the wavelength that emits the greatest
energy quantity from a blackbody radiator is dependent on which
formula is used? That can't possibly be. If a .106 cm wavelength
carries the greatest energy quantity, then it carries the greatest
energy quantity. How can a .187 cm wavelength also claim to carry
the greatest energy quantity, from the same radiator temperature?

0.106 cm is 1060 micron (um). It's easier to explain this with
bigger numbers:

The energy is distributed over the spectrum. Suppose you draw
the dW/db graph in units of power per micron. The 'y' value of
the graph at 1060um is the power between 1060um and 1061um
while at 1870um it is the power between 1870um and 1871um.

Those wavelengths convert as follows

um GHz
1060 282.823
1061 282.557 difference 0.267GHz

1870 160.317
1871 160.231 difference 0.086GHz

The amount of power in a 1um wide band at 1060um is spread
over 0.267GHz but the power a 1um wide band at 1870um is
spread over only 0.086GHz.

Now suppose the two 'y' values on the dW/db graph were equal.
When you express that as say the energy per GHz that makes the
'y' value for a band from 160GHz to 161GHz 3.11 times larger
than that a band from 282GHz to 283GHz.

I'll try a more hands on approach.
From a graph of the CMBR, plotted according to formula [1] above,
I note that the frequency of oscillation which carries the greatest
energy quantity is roughly 5.3 cycles per cm. I record that
information and, with a simple calculation, I determine that the
wavelength at that frequency is 1 / 5.3 = .188 cm. I can now use
this data for a comparison with the peak of the power curve plotted
for a 2.73 K radiator according to formula [2] above, which peaks
at roughly .11 cm. I then use an appropriate multiplier for spectral
energy density per [1], or the emissive power per [2] to bring
either into an alignment with the other, for a direct comparison.

But no amount of juggling can make the wavelengths attributed to
the two peak power points coincide.

Unfortunately you haven't discovered a way to bend the rules of the
Universe, you've merely shown that the curve shape to which the CMBR
was made to align was based on a flawed formula. And if you genuinely
believe in what you are saying, you have also demonstrated that maths
can befuddle the minds of even the best.

Your flaw was to treat the values as discrete frequencies
rather than as bands with a finite width. Bands with the
same width in cm have different widths in GHz.

George.

"Paul B. Andersen" wrote in message
...

"Max Keon" skrev i melding
...
Paul B. Andersen wrote:

"Max Keon" skrev i melding
...
CMBR? Not in the Big Bang Universe.
-----

For some time I've been trying to understand why the spectral
energy density graph plot of the 2.73 K CMBR, per formula [1]
(2 * pi * f^3) / (c^2 * (exp(h * f / (k * T)) - 1)), is nothing
like a 2.73 K blackbody radiator plot according to formula [2]
(2 * pi * h * c^2) / (b^5 * ((exp((h * f) / (k * T))) - 1))
(b is wavelength)

And why is that?
I have shown you this before, it is quite simple:
dW/df = (2 *pi *h* f^3) / (c^2 * (exp(h * f / (k * T)) - 1)),
f = c/b, df/db = -c/b^2
dW/db = (dW/df)*(df/db)
dW/db = -(2*pi*h*c^2) / (b^5*((exp((h*c)/ k*T*b))) - 1))

The graph plot of intensity per frequency unit along a scale of
frequencies can be easily converted for direct comparison with
formula [2] by converting frequency to wavelength with (c / f) and
plotting the curve on the same graph scale as for formula [2].

No, you cannot.
If you insert f = c/b in [1], it is still dW/df, which is different
from dW/db.

Whatever shape the curves may follow, 5.35 cycles per cm is the peak
point along the emissive power curve for a 2.73 K radiator according
to formula [1], and that is found to be 1 / 5.35 = .187 cm
wavelength. But this is not so according to formula [2], which
gives the peak wavelength as .106 cm.

dW/db = -c/b^2* dW/df
so it is quite obvious that they don't peak at the same
frequency/wavelength.

I'm trying to picture what you are describing, but it just doesn't
add up. You are saying that the wavelength that emits the greatest
energy quantity from a blackbody radiator is dependent on which
formula is used? That can't possibly be. If a .106 cm wavelength
carries the greatest energy quantity, then it carries the greatest
energy quantity. How can a .187 cm wavelength also claim to carry
the greatest energy quantity, from the same radiator temperature?

Why is this so hard to get?
dW/df is energy per frequency unit.
That is, it is how much energy there is in the part of the spectrum
that has frequencies between f and f+1.
dW/dt is energy per wavelength unit
That is, it is how much energy there is in the part of the spectrum
that has wavelengths between b and b+1.

Since f = c/b, it means that the bandwidth df = -c/b^2*db
Thus the energy dW/df in the bandwidth 1 Hz,
it is equal to the energy dW/db in the bandwidth c/b^2 metres.
Thus dB/df = (c/b^2)*dB/db

I'll try a more hands on approach.
From a graph of the CMBR, plotted according to formula [1] above,
I note that the frequency of oscillation which carries the greatest
energy quantity is roughly 5.3 cycles per cm. I record that
information and, with a simple calculation, I determine that the
wavelength at that frequency is 1 / 5.3 = .188 cm. I can now use
this data for a comparison with the peak of the power curve plotted
for a 2.73 K radiator according to formula [2] above, which peaks
at roughly .11 cm. I then use an appropriate multiplier for spectral
energy density per [1], or the emissive power per [2] to bring
either into an alignment with the other, for a direct comparison.

But no amount of juggling can make the wavelengths attributed to
the two peak power points coincide.

Unfortunately you haven't discovered a way to bend the rules of the
Universe, you've merely shown that the curve shape to which the CMBR
was made to align was based on a flawed formula. And if you genuinely
believe in what you are saying, you have also demonstrated that maths
can befuddle the minds of even the best.

Oh, my dear.
You really are a crank, aren't you? :-)
This is a case of simple ignorance of elementary math and physics.
If you don't believe me, why don't you look up "black body radiation" in
an elementary physics textbook and learn it in stead of all this nonsense?
In most books you will find both spectra side by side, and how
you derive the one from the other.
Or you can see:
http://hyperphysics.phy-astr.gsu.edu/hbase/mod6.html

Why you bother to make so much fuss about something which
can be cleared up by half an hour reading, beats me.

Paul



People with preconceived ideas generally don't want to look up equations in
some text book written by some "conspiricist"!

Dave

Paul B. Andersen wrote:

"Max Keon" skrev i melding ...
Paul B. Andersen wrote:

"Max Keon" skrev i melding
...
CMBR? Not in the Big Bang Universe.
-----

For some time I've been trying to understand why the spectral
energy density graph plot of the 2.73 K CMBR, per formula [1]
(2 * pi * f^3) / (c^2 * (exp(h * f / (k * T)) - 1)), is nothing
like a 2.73 K blackbody radiator plot according to formula [2]
(2 * pi * h * c^2) / (b^5 * ((exp((h * f) / (k * T))) - 1))
(b is wavelength)

And why is that?
I have shown you this before, it is quite simple:
dW/df = (2 *pi *h* f^3) / (c^2 * (exp(h * f / (k * T)) - 1)),
f = c/b, df/db = -c/b^2
dW/db = (dW/df)*(df/db)
dW/db = -(2*pi*h*c^2) / (b^5*((exp((h*c)/ k*T*b))) - 1))

The graph plot of intensity per frequency unit along a scale of
frequencies can be easily converted for direct comparison with
formula [2] by converting frequency to wavelength with (c / f) and
plotting the curve on the same graph scale as for formula [2].

No, you cannot.
If you insert f = c/b in [1], it is still dW/df, which is different
from dW/db.

Whatever shape the curves may follow, 5.35 cycles per cm is the peak
point along the emissive power curve for a 2.73 K radiator according
to formula [1], and that is found to be 1 / 5.35 = .187 cm
wavelength. But this is not so according to formula [2], which
gives the peak wavelength as .106 cm.

dW/db = -c/b^2* dW/df
so it is quite obvious that they don't peak at the same
frequency/wavelength.

I'm trying to picture what you are describing, but it just doesn't
add up. You are saying that the wavelength that emits the greatest
energy quantity from a blackbody radiator is dependent on which
formula is used? That can't possibly be. If a .106 cm wavelength
carries the greatest energy quantity, then it carries the greatest
energy quantity. How can a .187 cm wavelength also claim to carry
the greatest energy quantity, from the same radiator temperature?

Why is this so hard to get?
dW/df is energy per frequency unit.
That is, it is how much energy there is in the part of the spectrum
that has frequencies between f and f+1.
dW/dt is energy per wavelength unit
That is, it is how much energy there is in the part of the spectrum
that has wavelengths between b and b+1.

Since f = c/b, it means that the bandwidth df = -c/b^2*db
Thus the energy dW/df in the bandwidth 1 Hz,
it is equal to the energy dW/db in the bandwidth c/b^2 metres.
Thus dB/df = (c/b^2)*dB/db

I'll try a more hands on approach.
From a graph of the CMBR, plotted according to formula [1] above,
I note that the frequency of oscillation which carries the greatest
energy quantity is roughly 5.3 cycles per cm. I record that
information and, with a simple calculation, I determine that the
wavelength at that frequency is 1 / 5.3 = .188 cm. I can now use
this data for a comparison with the peak of the power curve plotted
for a 2.73 K radiator according to formula [2] above, which peaks
at roughly .11 cm. I then use an appropriate multiplier for spectral
energy density per [1], or the emissive power per [2] to bring
either into an alignment with the other, for a direct comparison.

But no amount of juggling can make the wavelengths attributed to
the two peak power points coincide.

Unfortunately you haven't discovered a way to bend the rules of the
Universe, you've merely shown that the curve shape to which the CMBR
was made to align was based on a flawed formula. And if you genuinely
believe in what you are saying, you have also demonstrated that maths
can befuddle the minds of even the best.

Oh, my dear.
You really are a crank, aren't you? :-)
This is a case of simple ignorance of elementary math and physics.
If you don't believe me, why don't you look up "black body radiation" in
an elementary physics textbook and learn it in stead of all this nonsense?
In most books you will find both spectra side by side, and how
you derive the one from the other.
Or you can see:
http://hyperphysics.phy-astr.gsu.edu/hbase/mod6.html

Why you bother to make so much fuss about something which
can be cleared up by half an hour reading, beats me.

I "make so much fuss" because the problem for me had not yet been
satisfactorily resolved. I had not imagined for a second that the
CMBR graph plot would follow a curve shape that was so obviously
different to that plotted according to wavelength, without
justification? Do you really believe that I think you folk are
complete idiots?

As you say, I could have cleared this up some time ago with half an
hour of reading. But why couldn't this have been cleared up on a
newsgroup for all to see, a long time ago? To me, your clarification
prior to your last reply fell a long way short of identifying the
problem. It's an elementary problem, but you didn't explain it in
an elementary fashion. But then, why should you?

Of the two options, "ask for clarification" or "make outlandish
comments", I chose the latter because option (1) would probably
have resulted in the same type of derogatory comments, but would
not necessarily achieve anything more than I already had. Option (2)
offsets the them/me credibility balance in favor of the them, which
opens the door for the same comments to be thrown, but it also opens
the door for better clarification.

Thanks for the excellent link.

Anyway, have you had any thoughts on what was obviously part 2 of
the original post? i.e. the enormous shortfall in the spectral
energy density of the microwave background?
---------

And thanks George Dishman, for the analogy (always there to pick
up those who fall by the wayside).


--
Max Keon

"Max Keon" skrev i melding ...
Paul B. Andersen wrote:

"Max Keon" skrev i melding ...
Paul B. Andersen wrote:

"Max Keon" skrev i melding
...
CMBR? Not in the Big Bang Universe.
-----

For some time I've been trying to understand why the spectral
energy density graph plot of the 2.73 K CMBR, per formula [1]
(2 * pi * f^3) / (c^2 * (exp(h * f / (k * T)) - 1)), is nothing
like a 2.73 K blackbody radiator plot according to formula [2]
(2 * pi * h * c^2) / (b^5 * ((exp((h * f) / (k * T))) - 1))
(b is wavelength)

And why is that?
I have shown you this before, it is quite simple:
dW/df = (2 *pi *h* f^3) / (c^2 * (exp(h * f / (k * T)) - 1)),
f = c/b, df/db = -c/b^2
dW/db = (dW/df)*(df/db)
dW/db = -(2*pi*h*c^2) / (b^5*((exp((h*c)/ k*T*b))) - 1))

The graph plot of intensity per frequency unit along a scale of
frequencies can be easily converted for direct comparison with
formula [2] by converting frequency to wavelength with (c / f) and
plotting the curve on the same graph scale as for formula [2].

No, you cannot.
If you insert f = c/b in [1], it is still dW/df, which is different
from dW/db.

Whatever shape the curves may follow, 5.35 cycles per cm is the peak
point along the emissive power curve for a 2.73 K radiator according
to formula [1], and that is found to be 1 / 5.35 = .187 cm
wavelength. But this is not so according to formula [2], which
gives the peak wavelength as .106 cm.

dW/db = -c/b^2* dW/df
so it is quite obvious that they don't peak at the same
frequency/wavelength.

I'm trying to picture what you are describing, but it just doesn't
add up. You are saying that the wavelength that emits the greatest
energy quantity from a blackbody radiator is dependent on which
formula is used? That can't possibly be. If a .106 cm wavelength
carries the greatest energy quantity, then it carries the greatest
energy quantity. How can a .187 cm wavelength also claim to carry
the greatest energy quantity, from the same radiator temperature?

Why is this so hard to get?
dW/df is energy per frequency unit.
That is, it is how much energy there is in the part of the spectrum
that has frequencies between f and f+1.
dW/dt is energy per wavelength unit
That is, it is how much energy there is in the part of the spectrum
that has wavelengths between b and b+1.

Since f = c/b, it means that the bandwidth df = -c/b^2*db
Thus the energy dW/df in the bandwidth 1 Hz,
it is equal to the energy dW/db in the bandwidth c/b^2 metres.
Thus dB/df = (c/b^2)*dB/db

I'll try a more hands on approach.
From a graph of the CMBR, plotted according to formula [1] above,
I note that the frequency of oscillation which carries the greatest
energy quantity is roughly 5.3 cycles per cm. I record that
information and, with a simple calculation, I determine that the
wavelength at that frequency is 1 / 5.3 = .188 cm. I can now use
this data for a comparison with the peak of the power curve plotted
for a 2.73 K radiator according to formula [2] above, which peaks
at roughly .11 cm. I then use an appropriate multiplier for spectral
energy density per [1], or the emissive power per [2] to bring
either into an alignment with the other, for a direct comparison.

But no amount of juggling can make the wavelengths attributed to
the two peak power points coincide.

Unfortunately you haven't discovered a way to bend the rules of the
Universe, you've merely shown that the curve shape to which the CMBR
was made to align was based on a flawed formula. And if you genuinely
believe in what you are saying, you have also demonstrated that maths
can befuddle the minds of even the best.

Oh, my dear.
You really are a crank, aren't you? :-)
This is a case of simple ignorance of elementary math and physics.
If you don't believe me, why don't you look up "black body radiation" in
an elementary physics textbook and learn it in stead of all this nonsense?
In most books you will find both spectra side by side, and how
you derive the one from the other.
Or you can see:
http://hyperphysics.phy-astr.gsu.edu/hbase/mod6.html

Why you bother to make so much fuss about something which
can be cleared up by half an hour reading, beats me.

I "make so much fuss" because the problem for me had not yet been
satisfactorily resolved. I had not imagined for a second that the
CMBR graph plot would follow a curve shape that was so obviously
different to that plotted according to wavelength, without
justification? Do you really believe that I think you folk are
complete idiots?

Yes.

Max Keon wrote in June 2002
| Since all inflation based theories
| have been soundly demolished by the CMBR's non compliance with an
| appropriate black body curve, and by the necessarily enormous
| magnitude of that background radiation, what are the implications
| for GR?

As you say, I could have cleared this up some time ago with half an
hour of reading. But why couldn't this have been cleared up on a
newsgroup for all to see, a long time ago?

I did. In June 2002
http://www.google.com/groups?q=g:thl...lly.uninett.no

And considering that you in your reply wrote:
| I am gratefully enlightened by your reply, as perhaps are others who
| are following this thread. This is how I learn what goes on in your
| world.
| It seems that I am once again in your debt.

I thought you had got it.

To me, your clarification
prior to your last reply fell a long way short of identifying the
problem. It's an elementary problem, but you didn't explain it in
an elementary fashion. But then, why should you?

A couple of lines simple math is as elementary as I can make it.

Of the two options, "ask for clarification" or "make outlandish
comments", I chose the latter because option (1) would probably
have resulted in the same type of derogatory comments, but would
not necessarily achieve anything more than I already had. Option (2)
offsets the them/me credibility balance in favor of the them, which
opens the door for the same comments to be thrown, but it also opens
the door for better clarification.

Thanks for the excellent link.

Anyway, have you had any thoughts on what was obviously part 2 of
the original post? i.e. the enormous shortfall in the spectral
energy density of the microwave background?

I did,
http://www.google.com/groups?hl=no&l...lly.uninett.no
You wouldn't listen.

Paul

Paul B. Andersen wrote:

"Max Keon" skrev i melding ...
Paul B. Andersen wrote:
"Max Keon" skrev i melding
...
Paul B. Andersen wrote:
"Max Keon" skrev i melding
...
CMBR? Not in the Big Bang Universe.
-----

For some time I've been trying to understand why the spectral
energy density graph plot of the 2.73 K CMBR, per formula [1]
(2 * pi * f^3) / (c^2 * (exp(h * f / (k * T)) - 1)), is nothing
like a 2.73 K blackbody radiator plot according to formula [2]
(2 * pi * h * c^2) / (b^5 * ((exp((h * f) / (k * T))) - 1))
(b is wavelength)

And why is that?
I have shown you this before, it is quite simple:
dW/df = (2 *pi *h* f^3) / (c^2 * (exp(h * f / (k * T)) - 1)),
f = c/b, df/db = -c/b^2
dW/db = (dW/df)*(df/db)
dW/db = -(2*pi*h*c^2) / (b^5*((exp((h*c)/ k*T*b))) - 1))
-------
-------

This is a case of simple ignorance of elementary math and physics.
If you don't believe me, why don't you look up "black body radiation" in
an elementary physics textbook and learn it in stead of all this nonsense?
In most books you will find both spectra side by side, and how
you derive the one from the other.
Or you can see:
http://hyperphysics.phy-astr.gsu.edu/hbase/mod6.html

Why you bother to make so much fuss about something which
can be cleared up by half an hour reading, beats me.

I "make so much fuss" because the problem for me had not yet been
satisfactorily resolved. I had not imagined for a second that the
CMBR graph plot would follow a curve shape that was so obviously
different to that plotted according to wavelength, without
justification? Do you really believe that I think you folk are
complete idiots?

Yes.

Max Keon wrote in June 2002
| Since all inflation based theories
| have been soundly demolished by the CMBR's non compliance with an
| appropriate black body curve, and by the necessarily enormous
| magnitude of that background radiation, what are the implications
| for GR?

That certainly seems to be the way I would do it. I don't think I
was calling anyone an idiot though.

As you say, I could have cleared this up some time ago with half an
hour of reading. But why couldn't this have been cleared up on a
newsgroup for all to see, a long time ago?

I did. In June 2002
http://www.google.com/groups?q=g:thl...lly.uninett.no

Here's the relevant snipped portion from that post:

-"Max Keon" wrote in message
...
- Paul B. Andersen wrote:
-
- "Max Keon" wrote in message
...
- (Qbasic compatible as written)
-
r=((6000/b)^2/((1+(1872000/(b*k)))+((6000/b)/(.0024*k))^3)^2)^2*.0002*k
- Emissive power (r) per (b) wave length for a (k) temperature
radiator.
-
- So what's this?
- It obviously isn't Planck's formula which is:
- r = ((2*pi*c^2*h)/b^5)/(exp(h*c/(b*k*T))-1)
- where h is Plank's constant, b is wavelength as in your formula,
- but k is the Bolzmann constant and T is the temperature.
- Your formula is probably an approximation.

[[[[
11-11-03 Note: That formula was of course an approximation. It was
handy because I could set it up so that the curve would peak
wherever I wanted it to.

In the original post of the current series, my choosing to use
Paul's erroneous equation further down this extract wasn't just a
coincidence. I didn't expect any feedback though because removing
the "h" from "2*pi*h*f^3" makes not one scrap of difference to the
plotted curve shape. Any feedback may have opened up a Pandora's
box.

Incidentally, I've been using a copy of the rest of this extract
as a mousepad for the past month or so. It was taken from my own
files of course.
]]]]

- According to both formulas, the peak of the emission curve for a
- 2.73K radiator is around 1062300nm.
-
-Close enough.
-The Planck equation for the intensity per wavelength spectrum:
-dI/db = ((2*pi*c^2*h)/b^5)/(exp(h*c/(b*k*T))-1) (b is wavelength)
-peaks at the wavelength (in cm) b_m = 0.2897/T
-So at 2.73K the peak is at 0.10612 cm.
-
- But the true 2.73 radiator curve
- peaks at close to 1900000nm. And on top of all that, the curve shape
- is nothing like any mathematically plotted curve.
-
-I know why you think so.
-You are obviously referring to the CMBR spectrum measured by COBE
-as it usually is published. If you look at it carefully, and note
-the units on the axes, you will see that it shows:
- intensity per frequency unit versus frequency.
-
-This is a different form of the Planck formula:
- dI/df = 2*pi*f^3/(c^2*(exp(h*f/(k*T))-1)), f is frequency
-
-since f = c/b and thus df/db = - c/b^2
-we have:
-dI/db = (dI/df)*(df/db) = (c/b^2)*dI/df
-I leave to you to see that this give the form written above.
-
-But the important point is:
-dI/df and dI/db peak at different frequencies/wavelengths!
-And of course dI/df has not the same shape as dI/db!
-
-The CMBR spectrum matches the Planck formula for 2.73K exactly.
-It peaks where it should, and has the shape it should.
-
-If you don't believe that, plot it with a QBASIC program
-using the dI/df equation above.


And considering that you in your reply wrote:
| I am gratefully enlightened by your reply, as perhaps are others who
| are following this thread. This is how I learn what goes on in your
| world.
| It seems that I am once again in your debt.

Many years ago I came across a formula for plotting spectral energy
density per de/df = 8*pi*h*f^3/(c^3*(exp(h*f/(k*T))-1)) I had no
reason to question its validity. But it never made any sense because
c^3 should have read c^2, as you had then just shown me. Using that
formula, I could plot a curve, but it was not a plot of the CMBR
curve as it is usually displayed. I was, and still am, grateful for
that enlightenment.

I thought you had got it.

Perhaps you did Paul. That was probably the right time to ask for
clarification, but I missed that boat.

To me, your clarification
prior to your last reply fell a long way short of identifying the
problem. It's an elementary problem, but you didn't explain it in
an elementary fashion. But then, why should you?

A couple of lines simple math is as elementary as I can make it.

So we are in agreement then?
-------
Continues.
-------


--
Max Keon

Paul B. Andersen wrote:

"Max Keon" skrev i melding ...
Thanks for the excellent link.

Anyway, have you had any thoughts on what was obviously part 2 of
the original post? i.e. the enormous shortfall in the spectral
energy density of the microwave background?

I did,
http://www.google.com/groups?hl=no&l...lly.uninett.no
You wouldn't listen.

No, I wouldn't listen. But I did have a few details muddled at the
time, which resulted in a far greater energy discrepancy than should
have been.

This may be better.
The equally spaced dots in the diagram represent uniformly
distributed matter throughout the Universe at the time when it first
became transparent. The temperature of every dot mass was 4000
degrees Kelvin. The box shape at the center is a global blackbody
radiation detector. Each "o" is one light second from the detector.
One second after transparency occurred the transmissions from each
"o" will enter the detector in unison. Another second later, the
transmissions from each "x" will enter the detector. In the global
picture, there are four times as many dot transmissions arriving
from the "x" radius than there were from the "o" radius. But the
divergence angle from the "o" radius that finds the detector, finds
only 1/4 of the detector when applied from the "x" radius. Thus the
power received reduces to 1/4 multiplied by 4 times the number of
dot sources = 1. So nothing changes. The same conditions apply for
every CMBR source radius.

If there was no expansion, we would each currently be right in the
middle of a 4000 K blackbody radiator.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . .x. . x . .x. . . . . . . . . . . . . .
. . . . . . . . . . . x . . . . . . . . . x . . . . . . . . . . .
. . . . . . . . . .x. . . . . . . . . . . . .x. . . . . . . . . .
. . . . . . . . . . . . . . . . o . . . . . . . . . . . . . . . .
. . . . . . . . x . . . . o . . . . . o . . . . x . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . x . . . .o. . . . . . . . .o. . . . x . . . . . . .
. . . . . . . . . . . . . . . .___. . . . . . . . . . . . . . . .
. . . . . . . x . . . o . . . l l . . . o . . . x . . . . . . .
. . . . . . . . . . . . . . . l___l . . . . . . . . . . . . . . .
. . . . . . . x . . . .o. . . . . . . . .o. . . . x . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . x . . . . o . . . . . o . . . . x . . . . . . . .
. . . . . . . . . . . . . . . . o . . . . . . . . . . . . . . . .
. . . . . . . . . .x. . . . . . . . . . . . .x. . . . . . . . . .
. . . . . . . . . . . x . . . . . . . . . x . . . . . . . . . . .
. . . . . . . . . . . . . .x. . x . .x. . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Consider that the spatial distance in all directions between each
dot has doubled during the time slot between the "o" and "x"
transmission arrival times at the detector. That part of the
divergence angle from each "x" that now falls on the detector will
have reduced to 1/4 * 1/4 = 1/16, so the final global result is a
1/4 power reduction. But the number of wavelengths entering the
detector will have halved, resulting in 1/2 * 1/4 = 1/8 of the
original power.

But a doubled wavelength carries only half the power surely? No,
it damn well doesn't. This is not a wavelength generated in a 2000 K
blackbody radiator, it's a wavelength generated in a 4000 K radiator
that has been stretched to twice its original length. There's a big
difference. And it's an energy difference that **must** be accounted
for if it has gone missing.

But I think you people already know this.


--
Max Keon