CMBR? Not in the Big Bang Universe.

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

"Max Keon" skrev i melding
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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?

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