Page:Zur Theorie der Strahlung bewegter Koerper.djvu/15

This becomes by differentiation of equation (3):

$={\frac {2\pi i_{0}}{\mathfrak {B}}}\sin \phi \ d\phi {\frac {\left({\sqrt {1-\sigma ^{2}\sin ^{2}\phi }}-\sigma \ \cos \phi \right)^{2}}{\sqrt {1-\sigma ^{2}\sin ^{2}\phi }}}=$ $={\frac {2\pi i_{0}}{\mathfrak {B}}}{\frac {{\mathfrak {B}}_{-}^{2}}{{\mathfrak {B}}^{2}{\sqrt {1-\sigma ^{2}\sin ^{2}\phi }}}}\sin \phi \ d\phi$

Since this energy is located in a cylinder of height $D$, then the energy quantity falling upon $B$ in this direction is:

${\frac {2\pi i_{0}D}{{\mathfrak {B}}^{3}}}\cdot {\frac {{\mathfrak {B}}_{-}^{2}}{\sqrt {1-\sigma ^{2}\sin ^{2}\phi }}}\sin \phi \ d\phi$

Previously (§ 1) we saw, that from the incident radiation $2\pi \ i\ cos\phi \ sin\phi \ d\phi$, the fraction $2\pi \ i_{0}\ \cos \phi \sin \phi \ d\phi$ is absorbed, while the fraction $2\pi \ p_{1}\ \cos \phi \sin \phi d\phi$ is transformed into work. Thus from the totality of the incident energy quantity, the fraction ${\tfrac {i_{0}}{i}}$ is absorbed and the fraction ${\tfrac {p_{1}c}{i}}$ is transformed into work.

Thus from the just incident energy quantity, the fraction

${\frac {2\pi i_{0}D}{{\mathfrak {B}}^{3}}}\cdot {\frac {{\mathfrak {B}}_{-}^{2}}{\sqrt {1-\sigma ^{2}\sin ^{2}\phi }}}\sin \phi \ d\phi \cdot {\frac {i_{0}}{i}}$

is absorbed; the fraction

${\frac {2\pi i_{0}D}{{\mathfrak {B}}^{3}}}\cdot {\frac {{\mathfrak {B}}_{-}^{2}}{\sqrt {1-\sigma ^{2}\sin ^{2}\phi }}}\sin \phi \ d\phi \cdot {\frac {p_{1}c}{i}}$

is transformed into work.

In a similar way one recognizes, that the energy quantity

${\frac {2\pi i_{0}D}{{\mathfrak {B}}^{3}}}\cdot {\frac {{\mathfrak {B}}_{+}^{2}}{\sqrt {1-\sigma ^{2}\sin ^{2}\phi }}}\sin \phi \ d\phi$

falls upon $A$, and that we have to multiply this energy quantity with ${\frac {i_{0}}{i}}$ and $-{\frac {p_{2}c}{i}}$ respectively, to obtain the fraction of them