Silicon refractive index The dispersion producing the best fit to measurements at ambient room temperature (293 K) [D.F. Edwards, Silicon (Si). In: E.D. Palik, Editor, Handbook of Optical Constants of Solids, Academic Press Inc. (1985) pp. 547-569, ISBN 0-12-544420-6] with the following modified Sellmeier expression ( ǫ = 11 . 7, λ 1 = 1 . 1 µ m): B λ 2 n 2 ( λ ) = ǫ + A 1 λ 2 + λ 2 − λ 2 1 12.1 12.0 n 2 or Ε ( Λ ) 11.9 11.8 11.7 2 4 6 8 10 12 Wavelength, Λ 1/5
Approximations Large value of ǫ ≈ 12 makes reasonably good a model of perfectly conducting metal. We assume a metal bar of radius a and length l located at a distance d from the beam orbit. If the bunch length σ z ≫ d , then the field on the bar is a slow function of time, and one can use electrostatic approximation to solve the fields. 2/5
Calculations ζ is coordinate measured along the bar. The potential generated at point ζ of the bar at time t is � d 2 + ζ 2 φ B ( t , ζ ) = 2 λ B ( t ) ln where λ B is the charge per unit length of the bunch Q e − t 2 c 2 / 2 σ 2 λ B ( t ) = √ z 2 πσ z This potential should be compensated by the image charge on the bar. Λ ( t , ζ ) is the charge per unit length of the bar. In the limit a ≪ l , d , the potential generated by the image charge on the surface of the bar is � ∞ φ im ( t , ζ ) ≈ 2 Λ ( ζ ) ln 2 l a − d ζ ( Λ ′ ( ζ + ξ ) − Λ ′ ( ζ − ξ )) ln ( ξ/ l ) 0 The sum of two potentials does not depend on ζ φ im ( t , ζ ) + φ B ( t , ζ ) = φ 0 ( t ) 3/5
Solution To solve the equation, neglect the integral. This introduces a relative error of order of 1 / ln ( 2 l / a ) . √ d 2 + ζ 2 Λ ( ζ ) = φ 0 − 2 λ B ( t ) ln 2 ln ( 2 d / a ) The constant φ 0 is found from the condition that the total charge on the � bar is zero: Λ ( ζ ) d ζ = 0. The electric field of the bar kicks the beam. Further calculations will be numerical and require more specificity. I need better knowledge about d , a , l . 4/5
Geometry of the experiment 5/5
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