# Page:Über einen die Erzeugung und Verwandlung des Lichtes betreffenden heuristischen Gesichtspunkt.pdf/7

φ can be reduced to a function of only one variable by expressing that the entropy of radiation between reflecting walls is not changed by adiabatic compression. We won't go into that however, but investigate right away how the function φ can be obtained from the radiation law of the black body.

In the case of "black body radiation" ρ is such a function of ν that for a given energy the entropy is a maximum, that is, that

${\displaystyle \delta \int \limits _{0}^{\infty }\phi (\rho ,\nu )d\nu =0,}$

When

${\displaystyle \delta \int \limits _{0}^{\infty }\rho d\nu =0.}$

From this it follows that for any choice of δρ as function of ν

${\displaystyle \int \limits _{0}^{\infty }\left({\frac {\partial \phi }{\partial \rho }}-\lambda \right)\delta \rho d\nu =0,}$

Where λ is independent of ν. Thus ${\displaystyle \partial \phi /\partial \rho }$ is independent of ν

For the temperature increase of dT of a black body radiation of volume v = 1 the following equation is valid:

${\displaystyle dS=\int \limits _{\nu =0}^{\nu =\infty }{\frac {\partial \phi }{\partial \rho }}d\rho d\nu ,}$

or, since ${\displaystyle \partial \phi /\partial \rho }$ is independent of ν:

${\displaystyle dS={\frac {\partial \phi }{\partial \rho }}dE.}$

Since dE is equal to the transferred heat, and the process is reversible we also have:

${\displaystyle dS={\frac {1}{T}}dE.}$

Equating formulas gives:

${\displaystyle {\frac {\partial \phi }{\partial \rho }}={\frac {1}{T}}.}$

This is the black body radiation law. So it's