<Dd> H φ = i I δ l k 4 π r e − i k r sin ⁡ (θ) E θ = i ζ 0 I δ l k 4 π r e − i k r sin ⁡ (θ) (\ displaystyle (\ begin (aligned) H_ (\ phi) & = i (\ frac (I \, \ delta \ ell \, k) (4 \ pi \, r)) e ^ (- i \, k \, r) \, \ sin (\ theta) \ \ E_ (\ theta) & = i (\ frac (\ zeta _ (0) \, I \, \ delta \ ell \, k) (4 \ pi \, r)) e ^ (- i \, k \, r) \, \ sin (\ theta) \ end (aligned))). </Dd> <P> The far field pattern is thus seen to consist of a transverse electromagnetic (TEM) wave, with electric and magnetic fields at right angles to each other and at right angles to the direction of propagation (the direction of r, as we assumed the source to be at the origin). The electric polarization, in the θ direction, is coplanar with the source current (in the Z direction), while the magnetic field is at right angles to that, in the φ direction . It can be seen from these equations, and also in the animation, that the fields at these distances are exactly in phase . Both fields fall according to ​ ⁄, with the power thus falling according to ​ ⁄ as dictated by the inverse square law . </P> <P> If one knows the far field radiation pattern due to a given antenna current, then it is possible to compute the radiation resistance directly . For the above fields due to the Hertzian dipole, we can compute the power flux according to the Poynting vector, resulting in a power (as averaged over one cycle) of: </P> <Dl> <Dd> ⟨ S ⟩ = 1 2 R e (E × H ∗). (\ displaystyle \ langle \ mathbf (S) \ rangle = (\ frac (1) (2)) \ mathrm (Re) \ left (\ mathbf (E) \ times \ mathbf (H) ^ (*) \ right).) </Dd> </Dl>

Derivation of directivity of half wave dipole antenna