$$ A'^2 = A^2 \frac {1 - \frac {v}{c}} {1 + \frac {v}{c}} $$
From these equations, it appears that for an observer, which moves with
the velocity c towards the source of light, the source should appear
infinitely intense.
§ 8. Transformation of the Energy of the Rays of Light. Theory of the
Radiation-pressure on a perfect mirror.
Since A²/8π is equal to the energy of light per unit volume, we have to
regard A²/8π as the energy of light in the moving system. A′²/A² would
therefore denote the ratio between the energies of a definite
light-complex “measured when moving” and “measured when stationary,” the
volumes of the light-complex measured in K and _k_ being equal. Yet this
is not the case. If _l_, _m_, _n_ are the direction-cosines of the
wave-normal of light in the stationary system, then no energy passes
through the surface elements of the spherical surface
(_x_ - _clt_)² + (_y_ - _cmt_)² + (_z_ - _cnt_)² = R²,
which expands with the velocity of light. We can therefore say, that
this surface always encloses the same light-complex. Let us now consider
the quantity of energy, which this surface encloses, when regarded from
the system _k_, _i.e._, the energy of the light-complex relative to the
system _k_.
Regarded from the moving system, the spherical surface becomes an
ellipsoidal surface, having, at the time τ = 0, the equation:—
$$ (\beta \xi - l \beta \frac {v}{c} \xi)^2 + (\eta - m \beta \frac
{v}{c} \xi)^2 + (\zeta - n \beta \frac {v}{c} \xi)^2 = R^2 $$
If S = volume of the sphere, S′ = volume of this ellipsoid, then a
simple calculation shows that:
$$ \frac {S'}{S} = \frac {\beta}{\sqrt{1 - \frac {v}{c} \cos \Phi}} $$
If E denotes the quantity of light energy measured in the stationary
system, E′ the quantity measured in the moving system, which are
enclosed by the surfaces mentioned above, then
$$ \frac {E'}{E} = \frac {\frac {A'^2}{8\pi} S'}{\frac {A^2}{8\pi}S} =
\frac {1 - \frac {v}{c} \cos \Phi}{\sqrt{1 - \frac {v^2}{c^2}}} $$