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

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