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August 23

Planck's law 1901 article and reverse function

At the end of the chapter "...Wien's displacement...", after equation (8), Max Planck gives the formula:
$U=\nu \cdot f_1(\frac{\theta }{\nu })$
Then a new formula:
$\theta =\nu \cdot f_2(\frac{U}{\nu })$
Ok, but the second formula that follows from it is incomprehensible to me:
$\frac{1}{\theta }=\frac{1}{\nu }\cdot f_3(\frac{U}{\nu })$
One has:
$f_3(\frac{U}{\nu })=\frac{1}{f_2(\frac{U}{\nu })}???$
Any idea ?
Malypaet (talk) 12:58, 23 August 2024 (UTC)

So $\theta =\nu \cdot f_2(\frac{U}{\nu })$ and $\frac{1}{\theta }=\frac{1}{\nu }\cdot \frac{1}{f_2(\frac{U}{\nu })}$, it's just taking the reciprocal of both sides. But I don't do physics so I'm probably missing the point.  Card Zero  (talk) 13:38, 23 August 2024 (UTC)

My question is:
on what logic can we write:
$f_3(\frac{U}{\nu })=\frac{1}{f_2(\frac{U}{\nu })}???$
Malypaet (talk) 14:14, 23 August 2024 (UTC)

Still unsure if I'm really helping, but so long as I don't have to know anything about black-body radiation or whatever,

If $a=b\cdot c$ then $\frac{1}{a}=\frac{1}{b}\cdot \frac{1}{c}$, so

If $a=b\cdot f_2(x)$ then $\frac{1}{a}=\frac{1}{b}\cdot \frac{1}{f_2(x)}$, and

If $\frac{1}{a}=\frac{1}{b}\cdot f_3(x)$ then

$f_3(x)=\frac{1}{f_2(x)}.$ But I'm just filling space until somebody comes along who knows what you were getting at.  Card Zero  (talk) 15:08, 23 August 2024 (UTC)

Nobody knows what Malypaet is trying to get at... The answer here, I guess, is simply that $f_3$ is a new name for $1/f_2$, nothing more, nothing less. Planck doesn't know what $f_2$ looks like (all he knows is that its argument is $U/\nu$), and he doesn't know what $f_3$ looks like (all he knows is that, because $f_3=1/f_2$, it is also a function of $U/\nu$). --Wrongfilter (talk) 15:46, 23 August 2024 (UTC)

Written like that, we can admit it. In his combinatorial demonstration we find this analogy of functions between logarithms and exponentials. But he does not write it.
Thank you. Malypaet (talk) 18:59, 23 August 2024 (UTC)