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# week7 - 1 Monday October 31 Relativistic Charged Particles...

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1 Monday, October 31: Relativistic Charged Particles As I was saying, before the midterm exam intervened, in an inertial frame of reference K there exists an electric field ~ E and a magnetic field ~ B at a space- time location ( ~ r, t ). Another inertial frame of reference K 0 is moving relative to the first at a constant velocity ~v = ~ βc . In the K 0 frame of reference, the spacetime location ( ~ r, t ) transforms to ( ~ r 0 , t 0 ) by the Lorentz transformation. In the K 0 frame of reference, ~ E 0 k = ~ E k (1) and ~ B 0 k = ~ B k , (2) where ~ E k and ~ B k are the components of the E and B fields parallel to the relative velocity vector ~ β . The transformations for the components perpen- dicular to β are more complicated, and thus more interesting. ~ E 0 = γ ( ~ E + ~ β × ~ B ) (3) and ~ B 0 = γ ( ~ B - ~ β × ~ E ) , (4) where γ is the usually relativistic “gamma factor”, γ (1 - β 2 ) - 1 / 2 . We can, of course, also the transformations from the primed frame K 0 to the unprimed frame K , by switching primed and unprimed quantities and changing the sign of ~ β : ~ E = γ ( ~ E 0 - ~ β × ~ B 0 ) (5) and ~ B = γ ( ~ B 0 + ~ β × ~ E 0 ) . (6) If there exists a single charged particle that moves with a constant velocity (as seen from an inertial frame of reference), transforming its electromagnetic field between one inertial frame and another is a fairly straightforward task, outlined in section 4.6 of the textbook. One interesting application is a charged particle which is moving past you (an inertial observer) on a straight line at a constant speed. The relative speed of you and the charged particle is v = βc . The distance of closest approach between you and the particle 1

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(the impact parameter ) is b . Choose a coordinate system so that you, in the K frame of reference, are at ~ r = b ˆ e y , and the particle is moving along the x axis, with ~ r = vt ˆ e x . (Thus, I am choosing the moment t = 0 to be the moment of closest approach between you and the charged particle.) In the K 0 frame of reference, in which the particle is at rest, the ~ B 0 field is zero, and the ~ E 0 field is a simple inverse square relation: ~ E 0 = q~ r 0 ( r 0 ) 3 . (7) In the K frame of reference, in which you are at rest, the ~ B field is no longer zero and the vecE field is no longer isotropic around the particle. In the limit of highly relativistic motion ( β 1, γ 1), the nonzero components of the electromagnetic field are E x - qγct ( γ 2 c 2 t 2 + b 2 ) 3 / 2 (8) E y qγb ( γ 2 c 2 t 2 + b 2 ) 3 / 2 (9) B z E y . (10) The E field is maximized when t = 0, and the point charge is at its closest approach. At this instant, E y B z γ q b 2 , (11) larger by a factor of γ than the value of E y you would measure in the non- relativistic case. On the other hand, E y only has a high value when γc | t | ¿ b , or | t | ¿ 1 γ b c . (12) In the more leisurely non-relativistic case, E y has a high value when v | t | ¿ b , or | t | ¿ b v 1 β b c . (13) In the highly relativistic case, γ 1, and the encounter time is brief. In the non-relativistic case, β ¿ 1, and the encounter time is long. A highly relativistic charged particle moving past you will cause a much higher electric 2
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week7 - 1 Monday October 31 Relativistic Charged Particles...

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