Derivation of EM Fields of a Charged Particle using Liénard-Wiechert Potentials - Prof. Na, Study notes of Electromagnetism and Electromagnetic Fields Theory

Download Derivation of EM Fields of a Charged Particle using Liénard-Wiechert Potentials - Prof. Na and more Study notes Electromagnetism and Electromagnetic Fields Theory in PDF only on Docsity! April 9, 2008 Notes for Lecture #31 Two formulations of electromagnetic fields produced by a charged particle moving at constant velocity In Chapter 11 of Jackson (page 559 – Eqs. 11.151-2 and Fig. 11.8), we derived the electric and magnetic field of a particle having charge q moving at velocity v along the x̂1 axis. The results are for the fields at the point r = bx̂2 are: E(x1, x2, x3, t) = E(0, b, 0, t) = q −vγtx̂1 + γbx̂2 (b2 + (vγt)2)3/2 (1) and B(x1, x2, x3, t) = B(0, b, 0, t) = q γβbx̂3 (b2 + (vγt)2)3/2 (2) for the electric and magnetic fields respectively. The denominators of these expressions are easily interpreted as the distance of the particle from the field point, as measured in the particle’s own reference frame. On the other hand, we can consider the same physical problem from the point of view of Liénard-Wiechert potentials: Consider the electric field produced by a point charge q moving on a trajectory described by r0(t) with ρ(r, t) ≡ qδ3(r − r0(t)). Assume that v0(t) ≡ ∂r0(t)/∂t and ∂2r0(t)/∂t2 = 0. Show that the electric field can be written in the form: E(r, t) = q 4π²0 (1− v20/c2)(R− v0R/c) (R− v0 ·R/c)3 −→ Gaussian units q (1− v20/c2)(R− v0R/c) (R− v0 ·R/c)3 , (3) where R ≡ |R(tr)|, R(tr) ≡ r− r0(tr), and where all quantities which depend on time on the right hand side of the equation are evaluated at the retarded time tr ≡ t−R(tr)/c. In the same notation, the magnetic field is given by B = R× E R . (4) If we evaluate this result for the same case as above (Fig. 11.8 of Jackson), v0 ≡ vx̂1, and R(tr) = −vtrx̂1 + bx̂2. In order to relate this result to Eqs. 1 and 2 above, we need to express tr in terms of the known quantities. Noting that R(tr) = c(t− tr) = √ (vtr)2 + b2, (5) we find that tr must be a solution to the quadratic equation: t2r − 2γ2ttr + γ2t2 − γ2b2/c2 = 0 (6)