My Math Forum 3 = [f_p][ln(b)][b^(f_p)]/ [b^(f_p) - 1].

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 March 11th, 2019, 04:51 PM #1 Newbie   Joined: Feb 2019 From: Watertown NY USA Posts: 5 Thanks: 0 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
 March 17th, 2019, 05:01 PM #2 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,801 Thanks: 636 Math Focus: Yet to find out. ?? Thanks from Greens
March 17th, 2019, 06:06 PM   #3
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 Originally Posted by Joppy ??
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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 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

-Dan

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