Applied Electromagnetic Field Theory Chapter 29 -- Electromagnetic Radiation and Infinitesimal Dipol

[@baghdadiabdellatif1581]

Thank you so much. Greetings from north Africa (Algerian desert)

[@elwood.downey]

This is wonderful. I'm an EE who has worked with antennas for many years and this is the first time I've seen an explanation for WHY the darn things radiate - it's the kinks that create the transverse field lines. Thank you so much.

[@veronicanoordzee6440]

Not really. The kink is only a consequence of the charge being accelerated.

[@baghdadiabdellatif1581]

Give him some time ​@@veronicanoordzee6440

[@jacobvandijk6525]

@ 6:38 In fact, dS = r-hat. dA, with r-hat being the dimensionless unit-vector pointing in the radial direction of the Poynting-vector. The dot-product (in eq. 29.1) is between the Poynting-vector S and this unit-vector, resulting in | S | . | r-hat | . cos(a) = | E | . | H | . cos(a), with 'a' being the angle between the E- and the H-vector; and remember that the length of r-hat = 1. The E- and H-vector are orthogonal and thus, sin(a) = 1. The result of a dot-product is a scalar, that we can take out of the integral. Then, in eq. 29.2 the surface-area A = integral of dA.

[@jacobvandijk6525]

@ 13:52 The field lines of a charged particle moving at constant velocity (incl. v = 0) do not propagate or radiate. There is only radiation during the acceleration of a charged particle. Therefore, during the period t (>>> the period of acceleration dt) there can't be radiation all the time. Read this paragraph: @ 15:10. I don't see why the period t can't be equal to dt? The outcome is the same. But if I'm wrong in this, please correct me.

[@ryancusack5909]

Does figure 29.4 ignore the transient behavior of the acceleration? The horizontal axis shows the continuous region only for v*t, the motion after the acceleration occurred. I don't see where the acceleration actually occurred though.