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For a calculation of the fluctuations around the expected value of the Casimir force (i.e., variances), see the chapter by Barton in  . For more on the vacuum catastrophe, see [5, 316, 483, 511, 629].Integrate barcode in .netuse visual studio .net crystal bar code encoder toembed bar code in .netProblems Barcode barcode library on .netusing barcode generating for vs .net control to generate, create barcode image in vs .net applications.2.1 Do the Fourier integral (2.66) using (2.67) with the approximation (2.69). This integral is wellsuited for contour integration. Note that for t - t' < 0, the contour must be closed on the lower half of the complex plane not to get any contribution from the semicircle arch when its radius is made infinitely large. The converse holds for t - t' > o. 2.2 Calculate how the pendulum oscillates with the driving force given by (2.71), which is switched on at t = 0 and off at t = T. Use the Green's function method (2.64), with the approximate Green's function (2.70). 2.3 Show that the solution of (2.75) with the driving force F t) = Aexp(d) sinwt is given by (2.79). 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Use this result to sum into a closed-form expression the series solution obtained in Problem 2.5. Show that the closed-form solution is given by(x t) - { (A/2v)~ [J(x + xo) - f(x - Xo + L)] , (A/2v)~ [J(x + xo) - f(x - Xo - L)]for - L < x - Xo < 0, for 0 < x - Xo < L,FIAT LUX!where f(~)~(z) = b]._ ei(w-iE}t cos ([w - i ]~/v) = (W-l, ) sin ([w-u ]L/ v ) . . .and ~(z) stands for the imaginary part of the complex number z [e.g., if z = a + ib,2.7 Take the limit by (2.83).---+ 0 of the result of Problem 2.6 and show that y(x, t) is given ---+ 00,2.8 Go back to the result of Problem 2.6 and take the limit L afterward, the limit ---+ O. Show that this recovers (2.79).and only 2.9 For the cavity shown in Figure 2.7, the boundary conditions (see Appendix B) are given by zI\. E(x, y, l, t) = 0, -z I\. E(x, y,O, t) = 0,y I\. E(x,L,z,t)= 0,-YI\.E(x,O,z,t) =0 xI\.E(L,y,z,t) = 0,-x I\. E(O, y, z, t) =Using the ansatz X = xfx(x, y, z)+ Yfy(x, y, z) + zfz(x, y, z), where fAx, y, z) = N x cos kxx sin kyy sin kzz, fy(x, y, z) = Ny sin kxx cos kyysin kzz, fz(x, y, z) = N z sinkxxsin kyy cos kzz show that these boundary conditions and (A.tO) imply that kx = n x 7r/L, ky = n y7r/ L, and kz = n z7r/l, wheren x, ny, andn z are integers. Now use the divergence condition V X= 0 to show that if anyone of k x , ky, and kz vanishes, there is only one polarization, but otherwise there are two polarizations. See Sec. to.2 of  for a nice pedagogical account of the modes in a box. 2.10 Use the well-known delta function identitykzl,~.L = 7r,