Electromagnetic radiation can be expanded as a series of terms, with the lowest-order term called the dipole term. To derive the strength of dipole radiation, start with the Maxwell equations in cgs,




where E is the electric field, is the charge density, B is the magnetic field, c is the speed of light, and J is the charge density. From (4),

where differentiation with respect to time and space have been performed in the opposite order. Plugging in (3) yields

Expanding the vector derivative and plugging in (1)



Similarly,

(cf. telegraphy equations). These are wave equations, but they cannot be solved easily. Therefore, define the electric potential and magnetic vector potential by


Equation (1) becomes


And equation (3) becomes




Equations (4) and (7) are still intractable, however, so switch to a gauge which will simplify the form of the equations. The Coulomb gauge requires that

so (16) and (?) simplify to


Now make the simplifying assumption that

This will be true if the electric field is not changing too rapidly. Equation (18) simplifies to

This is the electromagnetic Helmholtz equation. It can be solved by finding its Green's function.Define the Fourier components of A and J such that



We are looking for harmonic solutions to the homogeneous equation, so set

Then the homogeneous equation is


Solving,




since the negative solution is unphysical. This is the Green's function. If k = 0, we must obtain the Green's function corresponding to

so

Integrating over all such solutions


Now, make series expansions of








The solution to (34) in the limit is therefore


This can be simplified by noting that



The solution is



The first term represents electromagnetic radiation, the second electromagnetic induction, and the final an electrostatic field.

The linearly polarized power is then


The equation can also be derived by letting A and p be complex in the electromagnetic Helmholtz equation,

When we obtain the dipole approximation. The magnetic field is then



When , we are in the radiation zone and the electric field is, using the Maxwell equations and





For we are in the radiation zone and

giving a flux of





and the linearly polarized power radiated is


Solving the integral

so the final solution is


Another method follows that of Thomson (Tessman 1967), using the above figure (Bekefi and Barrett 1987, p. 256). A particle starting at position O is accelerated a short interval to point The particle then continues in a straight line at a velocity , and arrives at point at a time An observer at A, a distance from O and angle from the direction of motion would see the radial electric field of the stationary particle; an observer at B, a radial distance r = ct and angle from would see a radial electric field centered on Assume OA and to be parallel. By similar triangles,

where is the retarded time

Then

where it is assumed that

(equivalent to the dipole approximation).



The power flux is then

where is the permeability of free space and the identity

has been used, with the permittivity of free space. The total power is


[
A final method of derivation uses simple vectors.

where is the retarded acceleration.


Using the BAC-CAB rule,


Since is in the direction,



Since


it follows that



[
This is known as the Larmor power.
For two parallel dipoles,