which becomes absorbed or transformed into work.
Altogether, the energy quantity
∫
0
π
/
2
2
π
i
0
D
B
3
sin
ϕ
d
ϕ
1
−
σ
2
sin
2
ϕ
(
B
−
2
i
0
i
+
B
+
2
i
0
i
′
)
=
M
{\displaystyle \int _{0}^{\pi /2}{\frac {2\pi i_{0}D}{{\mathfrak {B}}^{3}}}{\frac {\sin \phi \ d\phi }{\sqrt {1-\sigma ^{2}\sin ^{2}\phi }}}\left({\mathfrak {B}}_{-}^{2}{\frac {i_{0}}{i}}+{\mathfrak {B}}_{+}^{2}{\frac {i_{0}}{i'}}\right)=M}
is absorbed, while the energy quantity
∫
0
π
/
2
2
π
i
0
D
B
3
sin
ϕ
d
ϕ
1
−
σ
2
sin
2
ϕ
(
B
−
2
p
1
c
i
−
B
+
2
p
2
c
i
′
)
=
N
{\displaystyle \int _{0}^{\pi /2}{\frac {2\pi i_{0}D}{{\mathfrak {B}}^{3}}}{\frac {\sin \phi \ d\phi }{\sqrt {1-\sigma ^{2}\sin ^{2}\phi }}}\left({\mathfrak {B}}_{-}^{2}{\frac {p_{1}c}{i}}-{\mathfrak {B}}_{+}^{2}{\frac {p_{2}c}{i'}}\right)=N}
is transformed into work. One can convince oneself, that exactly the same expressions are valid, when for example the black surface
B
{\displaystyle B}
is replaced by a mirror. Namely, the radiation coming from
A
{\displaystyle A}
to
B
{\displaystyle B}
is not absorbed in
B
{\displaystyle B}
, but is partly reflected, i.e. , from the totality of the energy incident in
B
{\displaystyle B}
, the fraction
i
′
i
{\displaystyle {\tfrac {i'}{i}}}
is reflected; it comes back to
A
{\displaystyle A}
, and the fraction
i
0
i
′
{\displaystyle {\frac {i_{0}}{i'}}}
is absorbed there, so that eventually the fraction
i
′
i
⋅
i
0
i
′
=
i
0
i
{\displaystyle {\tfrac {i'}{i}}\cdot {\tfrac {i_{0}}{i'}}={\tfrac {i_{0}}{i}}}
of the radiation propagating at the beginning in the direction from
A
{\displaystyle A}
to
B
{\displaystyle B}
, is absorbed again. Thus
M
{\displaystyle M}
and therefore also
N
{\displaystyle N}
remain unchanged.
Now, if one uses equations (4), (5), (7) and (8), then
M
=
2
π
i
0
D
B
4
∫
0
π
/
2
sin
ϕ
d
ϕ
(
B
−
3
+
B
+
3
)
{\displaystyle M={\frac {2\pi i_{0}D}{{\mathfrak {B}}^{4}}}\int _{0}^{\pi /2}\sin \phi \ d\phi ({\mathfrak {B}}_{-}^{3}+{\mathfrak {B}}_{+}^{3})}
and
N
=
2
π
i
0
D
B
3
∫
0
π
/
2
sin
ϕ
d
ϕ
1
−
σ
2
sin
2
ϕ
{\displaystyle N={\frac {2\pi i_{0}D}{{\mathfrak {B}}^{3}}}\int _{0}^{\pi /2}{\frac {\sin \phi \ d\phi }{\sqrt {1-\sigma ^{2}\sin ^{2}\phi }}}}
⋅
[
σ
2
sin
2
ϕ
(
B
−
2
+
B
+
2
)
+
σ
cos
ϕ
1
−
σ
2
sin
2
ϕ
(
B
−
2
−
B
+
2
)
]
{\displaystyle \cdot \left[\sigma ^{2}\sin ^{2}\phi ({\mathfrak {B}}_{-}^{2}+{\mathfrak {B}}_{+}^{2})+\sigma \cos \phi {\sqrt {1-\sigma ^{2}\sin ^{2}\phi }}({\mathfrak {B}}_{-}^{2}-{\mathfrak {B}}_{+}^{2})\right]}