For more on the delta function, see Appendix F and the references mentioned there. in .NET

Insert ECC200 in .NET For more on the delta function, see Appendix F and the references mentioned there.
For more on the delta function, see Appendix F and the references mentioned there.
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Milonni's book [451] offers a quite comprehensive pedagogical account of the Casimir effect, including a discussion of the curious model that Casimir proposed for the electron. Most of the important literature on the Casimir force is listed in [377], with key advances discussed in detail in the recent review [66]. For a recent measurement of the Casimir force, see [97]. For a calculation of the fluctuations around the expected value of the Casimir force (i.e., variances), see the chapter by Barton in [49] . For more on the vacuum catastrophe, see [5, 316, 483, 511, 629].
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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). Start from equation (2.77) and note that its solution is simply
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Then substitute in (2.76) and do the integral by contour integration. 2.4 Redo Problem 2.3 with the delta function in (2.75) replaced by the Lorentzian function
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Show that y( x) remains finite at x = Xo even in the limit of A --+ 0+, in which the Lorentzian Ll(x - xo) turns into a delta function (F.7) and we recover (2.79). 2.5 Solve (2.81) and substitute the solution back in (2.80) to obtain y x, t) for the finite string. 2.6 Work out the Fourier series for cosku with -L < x < L. 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,
~(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 [537] for a nice pedagogical account of the modes in a box. 2.10 Use the well-known delta function identity
kzl,~.L = 7r,