THE CASIMIR FORCE in .NET

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THE CASIMIR FORCE
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Figure 2.10 The diagram on the left represents the reflection on the perfectly reflecting plate of a plane wave whose electric field is perpendicular to the plane of incidence. If you follow this wave along, you see from the diagram that to satisfy the boundary conditions (Appendix B), the electric field must invert its direction relative to the wave on reflection: Before reflection, the electric field is pointing to the right side of the plane of incidence (for someone facing the direction of propagation); after reflection it points to the left side of the incidence plane. The diagram on the right shows that when the electric field is on the plane of incidence, its projection on the surface of the plate changes direction on reflection (as demanded by the boundary conditions) without the electric field changing direction relative to the wave.
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We take the amplitude of the incident positive-frequency plane-wave component to be [169] (2.176) where aa(k) is the annihilation operator of the mode with wave vector k and polarization a, and (2.177) is the modified vacuum field strength. 41 We obtain the positive-frequency part of the electric field, summing up all the modified plane waves, by varying the direction and polarization of the incident plane wave,
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where the integration is restricted to half space (I.e., positive k z ), because of the perfect plate at z = 0, and the annihilation and creation operators obey the usual continuum commutation relations (2.179)
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41There is an extra factor 2 in comparison to the free-space case because of the perfect plate at z eliminates half the space [35,169,448,495\.
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The label f3 stands either for cav, the field between the plates, or for out, the field outside. The mode functions obtained by our scattering method, also known as the modes of the universe, are
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(2.180)
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e~~~(r,k) = ik vaceiK..r.e1l cosBsinkzz,
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(2.181) (2 . 182)
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,"caY r, , <:' . "11,2 (k) = -Zvvac eiK..r.e II sin BCOS k zZ,
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,"out (k) =tZ 1\ K.vvace iK..r.e .L [. k zz+t'2 r .d B) sinusin (k zZ-U , .. ' <:' ~)] r, sin t.L(B) . ~ .
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(2.183)
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,"out r, . , <:' ~)] B "11,1 (k) =tK.vvace iK..r.e II [. kzZ-t'2 rll (B) SllluSln (k zZ-U COS, sin tll(B) . ~ .
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(2.184)
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,"out r, , <:' ~)]. B "11.2 (k) = -Zvvac eiK..r.e II [ k zZ-t'2 rll (B) SllluCOS (k zZ-u Sill, COS tll(B) . ~
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(2.185)
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where the functions .e (k) = to(B) o - 1 - rno ro(B) exp(i28)
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(2.186)
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describe the resonances of the parallel-plate cavity. Now let us take the limit of perfect reflectivity. In this limit r 0 - t rno and to - t O. Then we can easily see that the field outside the parallel-plate cavity is given by
t.L - 0
tll-O
lim E:ut(r) =
d k vac eiK..r e- io {iZ 1\ ka.L (k) sin(kzz - 8)
+ [ikcosBsin(kzz - 8) - zsinBcos(kzz - 8)]a ll (k)}.
(2.187)
This limit is more subtle for the field between the plates. If you are not careful, you might think that the field will vanish because the transmissivity in the nominator of .eo vanishes. But.e o is not what it seems at first sight. Remember that we mentioned above that it describes the cavity resonances. Notice that 2 2 I.e 12 = [1 - Ito 1 /2 + -lto l2 rno cos(28 + 80 )] Ito l /2 o (1-ltoI2)sin2(28+80)+ltoI4/4'
(2.188)
where we have used that Irol2 + It o l2 = 1 (i.e., the plate does not absorb any radiation) and called 80 the phase of ro [i.e., ro = Irol exp(80 )]. In the limit of