Unformatted text preview: Chapter 5: Waves 23 3. Ez and Bz are zero everywhere: the Transversal electromagnetic mode (TEM). Than holds: k = and vf = vg , just as if here were no waveguide. Further k I so there exists no cutoff R, frequency. In a rectangular, 3 dimensional resonating cavity with edges a, b and c the possible wave numbers are given n1 n2 n3 , ky = , kz = This results in the possible frequencies f = vk/2 in the cavity: by: kx = a b c f= v 2 n2 n2 n2 y x z + 2 + 2 a2 b c For a cubic cavity, with a = b = c, the possible number of oscillating modes N L for longitudinal waves is given by: 4a3 f 3 NL = 3v 3 Because transversal waves have two possible polarizations holds for them: N T = 2NL . 5.6 Nonlinear wave equations
The Van der Pol equation is given by: dx d2 x 2 + 0 x = 0  0 (1  x2 ) dt2 dt 2 1 x2 can be ignored for very small values of the amplitude. Substitution of x e it gives: =
1 2 2 ). 1 2 0 (i 1 2 0 . The lowestorder instabilities grow as While x is growing, the 2nd term becomes larger 1 and diminishes the growth. Oscillations on a time scale 0 can exist. If x is expanded as x = x (0) + (1) 2 (2) x + x + and this is substituted one obtains, besides periodic, secular terms t. If it is assumed that there exist timescales n , 0 N with n /t = n and if the secular terms are put 0 one obtains: d dt 1 2 dx dt 2 2 + 1 0 x2 2 = 0 (1  x2 ) dx dt 2 This is an energy equation. Energy is conserved if the lefthand side is 0. If x 2 > 1/, the righthand side changes sign and an increase in energy changes into a decrease of energy. This mechanism limits the growth of oscillations. The KortewegDe Vries equation is given by: u u u 3u +  au + b2 3 = 0 t x x x
nonlin dispersive This equation is for example a model for ionacoustic waves in a plasma. For this equation, soliton solutions of the following form exist: d u(x  ct) = cosh2 (e(x  ct)) with c = 1 + 1 ad and e2 = ad/(12b2 ). 3 ...
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 Spring '10
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 Energy, Electromagnet, Frequency, Trigraph

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