LARMOR'S RELATIONSHIP

1. Classical Larmor's formula

The following figure gives the expressions of the
electric and magnetic fields in the terms of
retarded potentials(Liénard-Wichert potentials)
for a charged and accelerated particle. 





We have:
B = E/c
The pointing vector is S = E x B/ μ0 ( Power/area)
Recall: ε0μ0c2 = 1.

E is perpendicular to B, 
Then S = E B / μ0 = E2/cμ0 = cε0E2
Using the expression of E:






We have dP= /dA = S(θ) ( P the power is the ratio of energy on time =
magnitude of the Poynting power times area)

Then : P = ∫ S(θ) dA
Where dA = r2dr = 2π r2 sin θ dθ
P = 2π ∫ S(θ) r2 sin θ dθ (from 0 to 2π)
	= 2π(e2a2/16π2ε0c3) ∫ sin3 θ dθ

∫ sin3 θ dθ = ∫ sin2θ(- dcosθ) dθ = 
∫ (cos2θ - 1)dcosθ = (1/3)(cos3θ - cosθ) (from 0 t0 π) = 4/3

Finaly, the Larmor's relationship takes the following expression:





That is the power emitted by a charged particle (electron)
accelerated at a, via the electromagnetic wave (E,B). 

2. Relativistic case

We can demonstrate that in the relativistic case;
The expression of the electric field E takes the form:






The maximum for S(θ) is given by:
dS(θ) / dθ = 0
Which gives:
θMAX = arcos ( (1/3β) (1 + 15 β2)1/2 - 1))
Where β = v/c and v is the particle's velocity.
Recall that in the relativistic case, v is related to β then to γ
and γ = E( total energy) / m0 c2 
(m0 electron rest mass)

In the classical case: β << 1 : θMAX = π/2
In the very relativistic case (v  ≈ c) : θMAX = 0