@ 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.
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.
@ 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.
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.
Thank you so much. Greetings from north Africa (Algerian desert)
@ 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.
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.
Not really. The kink is only a consequence of the charge being accelerated.
Give him some time @@veronicanoordzee6440
@ 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.
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.