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Additional resources for Answers to selected problems from Jackson's Classical electrodynamics
TE or TM waves propagate along the cylinder. 60) where γ 2 = µ ω 2 − k 2 and ψ is either Ez for TM waves or Bz for TE waves. Solving this equation in cylindrical coordinates (and recognizing that there is no z-dependence), we get solutions ψ = eımφ Jm (γρ)eı(kz−wt) . 61) We already discarded the other solution to the Bessel equation, because it has a singularity at the origin. Lets apply the rest of the boundary conditions for the appropriate wave. For TM waves we know that ψ|s = ψ|ρ=R = 0. 63) where xmn is the n-th root of the m-th order Bessel function.
31) which states u +v u⊥ u = and u⊥ = . 7) 1 + v·u γ 1 + v·u c2 c2 The acceleration parallel to the relative velocity between the reference frames is a = 1 + v·u du − u + v cv2 du du c2 = . 6. THE ROCKET SHIP Use dx1 = u1 c dx0 49 (Jackson page 531) and a = du /dt to conclude that 1+ a a = γ 1+ v2 c2 vu 1− a = 3 1+ c2 v2 c2 3/2 . 9) c2 The acceleration perpendicular to the direction of propagation is u d γ 1+⊥v·u ( c2 ) du⊥ a⊥ = . 10) After computing the differential of the numerator with the chain rule, we obtain du⊥ γ 1 + v·u c2 du⊥ = − v·du c2 v·u + c2 u⊥ γ 1 2.
In other words, what is t? 15) 1 . 16) 1 − β2 The twin in the rocket ship feels a force but does not move in its reference frame: x = 0, so v c β= and γ= dt = γdt . 17) I wrote this in differentials because γ is function of velocity and the velocity of the space ship as seen from earth increases with time. We’ll find this velocity via the acceleration a. We know a = g and that a=a x ˆ= 1+ gt = 1− v 1− v2 c2 3 vu x ˆ=g 1− v2 c2 3/2 x ˆ⇔ c2 x ˆ⇔ −3/2 v2 c2 3/2 3/2 dv v2 =g 1− 2 dt c gdt = v2 c2 1− a dv ⇔ .