The electromagnetic beams emission requires:

• The energy source.
• The non-zero Umov-Poynting vector, which indicates the direction and power of radiation.
SI CGS Simplified $$\tag{1}$$ $\mathbf{\overrightarrow{S}}=\mathbf{\overrightarrow{E}}\times\mathbf{\overrightarrow{H}}$ $\mathbf{\overrightarrow{S}}=\frac{c}{4\pi}\mathbf{\overrightarrow{E}}\times\mathbf{\overrightarrow{H}}$ $\mathbf{\overrightarrow{S}}=\mathbf{\overrightarrow{E}}\times\mathbf{\overrightarrow{H}}c^2$

The volumetric energy density $$w$$, and the flux (power) $$P$$ of radiation with a speed $$v$$, which passes through a unit of area, is: $w=\frac{|\mathbf{\overrightarrow{S}}|}{v}\tag{2}$ $P=|\mathbf{\overrightarrow{S}}|=wv\tag{3}$ The electromagnetic energy is a sum of the electric ("Electric field and Lorentz force", 5) and the magnetic ("Magnetic energy", 2) energies: $w=w_B+w_E=\frac{\mathbf{\overrightarrow{B}}\mathbf{\overrightarrow{H}}}{2}+\frac{\mathbf{\overrightarrow{D}}\mathbf{\overrightarrow{E}}}{2}\tag{4 – SI}$ $w=w_B+w_E=\frac{\mathbf{\overrightarrow{B}}\mathbf{\overrightarrow{H}}}{8\pi}+\frac{\mathbf{\overrightarrow{D}}\mathbf{\overrightarrow{E}}}{8\pi}\tag{4 – CGS}$ $w=w_B+w_E=\frac{\mathbf{\overrightarrow{B}}\mathbf{\overrightarrow{H}}с^2}{2}+\frac{\mathbf{\overrightarrow{D}}\mathbf{\overrightarrow{E}}}{2}\tag{4 – Sim.}$ The electromagnetic beam, unlike the arbitrary aetheric beam, has a maximum possible speed, when all the ámers are moving in the same direction. As a result, the magnetic field vector is always perpendicular to the velocity vector. The electric field exists with respect to the surrounding objects according to the law ("Electric field and Lorentz force", 3): $\mathbf{\overrightarrow{E}}=\mathbf{\overrightarrow{B}}\times\mathbf{\overrightarrow{v}}\tag{5 – SI, Sim.}$ $\mathbf{\overrightarrow{E}}=\mathbf{\overrightarrow{B}}\times\frac{\mathbf{\overrightarrow{v}}}{c}\tag{5 – CGS}$ Since the light has an electric field, it has a magnetic effect ("Magnetic effect of current", 3): $\mathbf{\overrightarrow{H}}=\mathbf{\overrightarrow{v}}\times\mathbf{\overrightarrow{D}}\tag{6 - SI}$ $\mathbf{\overrightarrow{H}}=4\pi\frac{\mathbf{\overrightarrow{v}}}{c}\times\mathbf{\overrightarrow{D}}\tag{6 - CGS}$ $\mathbf{\overrightarrow{H}}=\frac{1}{c^2}\mathbf{\overrightarrow{v}}\times\mathbf{\overrightarrow{D}}\tag{6 - Sim.}$ The beam vectors $$\mathbf{\overrightarrow{D}}(\mathbf{\overrightarrow{E}})$$, $$\mathbf{\overrightarrow{B}}(\mathbf{\overrightarrow{H}})$$ and $$\mathbf{\overrightarrow{v}}$$ are orthogonal, so their modules in the equations (5, 6) can be used: $E=Bv\;\;\;\;\;H=vD\tag{7 - SI}$ $E=B\frac{v}{c}\;\;\;\;\;H=4\pi\frac{v}{c}D\tag{7 - CGS}$ $E=Bv\;\;\;\;\;H=\frac{vD}{c^2}\tag{7 - Sim.}$ The consequence from (7) is the propagation laws:

SI CGS Simplified $$\tag{8}$$ $w=BH=DE$ $EH=DBv^2$ $c=\frac{1}{\sqrt{\varepsilon_0\mu_0}}$ $w=\frac{BH}{4\pi}=\frac{DE}{4\pi}$ $EHc^2=4\pi DBv^2$ $w=BHc^2=DE$ $EHc^2=DBv^2$ $$v$$ is a beam speed; $$c$$ is a speed of light in the vacuum.

When the distance from a source increases, the radiation can lose its fields magnitudes, but it can not lose its speed, because both its momentum and its energy are conserved simultaneously ("Aether model", 7 and 8).