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March 11th, 2019, 04:51 PM   #1
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3 = [f_p][ln(b)][b^(f_p)]/ [b^(f_p) - 1].

On Wednesday, February 8, 2017 at 12:20:05 PM UTC-5, Simon Roberts wrote:
> Result (version 5.4): (all version prior to feb 08 2017 may be wrong)
>
> (14->15) 3 = [f_p][ln(b)][b^(f_p)]/ [b^(f_p) - 1].
>
> where,
>
> (6) b = e^[48 x 10^(-12)/ T](Kelvins*sec).
>
> and,
>
f_p is the frequency of peak intensity
>
> Planck's Law for a given temperature, T.
>
> Proof.
>
> Givens.
>
> (1) B(f,T) = c_0*(f^3/(b^f - 1)) where,
>
> (2) b = e^(h/(k_b*T))
>
> (3) k_b = 1.38 x 10^(-23) Joules / Kelvin.
>
> (4) h = 6.626×10-34 Joule.sec
>
> (5) h/(k_b*T) = [48 x 10^(-12)/ T](Kelvins*sec).
>
> (6) b = e^[48 x 10^(-12)/ T](Kelvins*sec).
>
> (7) ln(b) = [48 x 10^(-12)/ T](Kelvins*sec).
>
> ( c_0 = 2*h / c^2.
>
> (9) c = 299 792 458 m / s.
>
> (10) c_0 = 2(6.626×10-34 Joule.sec/[299.792458x10^(6) m / s]^2.
>
> (10.1) c_0 = .147448 x 10^(-49) joules*sec^3/meter^2
>
> (10.2) c_0 = 14.7448 x 10^(-51) joules*sec^3/meter^2
>
> End Givens.
>
> Main section of proof.
>
> Taking the natural logarithm of B(f),
>
> (11) ln(B(f)) = ln(c_0) + 3ln(f) - ln(b^f - 1),
>
> Taking derivative of (11) w.r.t. f yields,
>
> (12) B'(f)/B(f) = 3/(f_p) - [ln(b)b^(f_p)]/ [b^(f_p) - 1].
>
> Assuming B(f) =/= 0 and f = f_p: B(f_p) is a maxima;
>
> (d/df)B(f) = B'(f_p)= 0.
>
> (12->13) 0 = 3/(f_p) - [ln(b)b^(f_p)]/ [b^(f_p) - 1].
>
> (13 -> 14) 3/(f_p) = [ln(b)b^(f_p)]/ [b^(f_p) - 1].
>
> (14->15) 3 = [f_p][ln(b)][b^(f_p)]/ [b^(f_p) - 1].
>
> where,
>
> (6) b = e^[48 x 10^(-12)/ T](Kelvins*sec).
>
> and,
>
f_p is the frequency of peak intensity of Planck's Law,

for a given temperature T.

Simon C. Robert
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March 17th, 2019, 05:01 PM   #2
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March 17th, 2019, 06:06 PM   #3
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March 17th, 2019, 06:08 PM   #4
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March 17th, 2019, 06:29 PM   #5
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Quote:
Originally Posted by retenshun View Post
On Wednesday, February 8, 2017 at 12:20:05 PM UTC-5, Simon Roberts wrote:
> Result (version 5.4): (all version prior to feb 08 2017 may be wrong)
>
> (14->15) 3 = [f_p][ln(b)][b^(f_p)]/ [b^(f_p) - 1].
>
> where,
>
> (6) b = e^[48 x 10^(-12)/ T](Kelvins*sec).
>
> and,
>
f_p is the frequency of peak intensity
>
> Planck's Law for a given temperature, T.
>
> Proof.
>
> Givens.
>
> (1) B(f,T) = c_0*(f^3/(b^f - 1)) where,
>
> (2) b = e^(h/(k_b*T))
>
> (3) k_b = 1.38 x 10^(-23) Joules / Kelvin.
>
> (4) h = 6.626×10-34 Joule.sec
>
> (5) h/(k_b*T) = [48 x 10^(-12)/ T](Kelvins*sec).
>
> (6) b = e^[48 x 10^(-12)/ T](Kelvins*sec).
>
> (7) ln(b) = [48 x 10^(-12)/ T](Kelvins*sec).
>
> ( c_0 = 2*h / c^2.
>
> (9) c = 299 792 458 m / s.
>
> (10) c_0 = 2(6.626×10-34 Joule.sec/[299.792458x10^(6) m / s]^2.
>
> (10.1) c_0 = .147448 x 10^(-49) joules*sec^3/meter^2
>
> (10.2) c_0 = 14.7448 x 10^(-51) joules*sec^3/meter^2
>
> End Givens.
>
> Main section of proof.
>
> Taking the natural logarithm of B(f),
>
> (11) ln(B(f)) = ln(c_0) + 3ln(f) - ln(b^f - 1),
>
> Taking derivative of (11) w.r.t. f yields,
>
> (12) B'(f)/B(f) = 3/(f_p) - [ln(b)b^(f_p)]/ [b^(f_p) - 1].
>
> Assuming B(f) =/= 0 and f = f_p: B(f_p) is a maxima;
>
> (d/df)B(f) = B'(f_p)= 0.
>
> (12->13) 0 = 3/(f_p) - [ln(b)b^(f_p)]/ [b^(f_p) - 1].
>
> (13 -> 14) 3/(f_p) = [ln(b)b^(f_p)]/ [b^(f_p) - 1].
>
> (14->15) 3 = [f_p][ln(b)][b^(f_p)]/ [b^(f_p) - 1].
>
> where,
>
> (6) b = e^[48 x 10^(-12)/ T](Kelvins*sec).
>
> and,
>
f_p is the frequency of peak intensity of Planck's Law,

for a given temperature T.

Simon C. Robert
What, specifically, is your question?

-Dan
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