W11HW5 - 1 Problem 11.8 (part a only) Substituting k0 = !=c...

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Unformatted text preview: 1 Problem 11.8 (part a only) Substituting k0 = !=c into Jackson's equation 11.29:  H ~k k 0 = k0 ¡ ~   !H ! = k c c  Note that since we're given that the light is either parallel or anti-parallel to the velocity of ~k the uid, ¡ ~ = k . Approximating this to rst-order, we realize that is small while % 1: !H ! % c k c  ! ! ! H % c u ! = c ! (c=n) ! ! H = !n Solving for n(! ): n(! ) = Taylor expanding about ! = ! H : (1) ! !H ! H H ) + ! ! @n n(! ) % n(! 1! @! !=! 0 Substituting in equation (1): where we've let ! H % ! . @n n(! ) % n(! H ) + ( !n) @! !=! % n(!) !n @n(!) @! 1 n(! ) % % 0   1 @n(! ) 1 1 ! n(! ) @!   1 @ 1 + ! n(! ) @! 1 (2) where we've used a rst-order binomial series approximation. 1 2 (! ) n % % 1 @n(! ) 2 1 ! n2 (! ) @!   @n(! ) 1 1 + 2 ! n2 (! ) @!   (3) Now, we use the velocity addition formula, noting that the velocity of the uid is either parallel or anti-parallel to the light: uH + v 1 + uH v=c2 c +v c 1 + nH =H =n 1 + =nH n 1 + =nH   c H) 1 % nH (1 + n nH   c = H 1 + nH H n n c % nH + v nvH2 u= 0 Because nH = n(! H ) and ! % ! H , we can substitute in equations (2) and (3):     c @n(! ) v @n(! ) u% 1 + ! +v 2 1 + 2 ! n(! ) @! n (! ) @! c v! @n(! ) v v! @n(! ) = + +v 2 +2 2 n(! ) n(! ) @! n (! ) n (! ) @!   1 ! @n(! ) ! @n(! ) c +v 1 2 + +2 2 = n(! ) n (! ) n(! ) @! n (! ) @! Dropping the last term (because is small) yields: u%  1 ! @n(! ) c +v 1 2 + n(! ) n (! ) n(! ) @!  recalling that v may be either parallel (in which case v is positive) or anti-parallel (in which case v is negative) to the light. 2 ...
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This note was uploaded on 12/08/2011 for the course PHYSICS 505 taught by Professor Liu during the Winter '08 term at University of Michigan.

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W11HW5 - 1 Problem 11.8 (part a only) Substituting k0 = !=c...

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