Hi, Giovanni
gamma in the material definiton corresponds to γ/2π in the formula.
Anduo Hu
At 2010-12-14 01:00:01，meep-discuss-requ...@ab-initio.mit.edu wrote:
>Send meep-discuss mailing list submissions to
> meep-discuss@ab-initio.mit.edu
>
>To subscribe or unsubscribe via the World Wide Web, visit
> http://ab-initio.mit.edu/cgi-bin/mailman/listinfo/meep-discuss
>or, via email, send a message with subject or body 'help' to
> meep-discuss-requ...@ab-initio.mit.edu
>
>You can reach the person managing the list at
> meep-discuss-ow...@ab-initio.mit.edu
>
>When replying, please edit your Subject line so it is more specific
>than "Re: Contents of meep-discuss digest..."
>
>
>Today's Topics:
>
> 1. Re: Verification of material (Au) dispersion (gpipc)
>
>
>----------------------------------------------------------------------
>
>Message: 1
>Date: Sun, 12 Dec 2010 21:50:44 +0100
>From: gpipc <gp...@cup.uni-muenchen.de>
>Subject: Re: [Meep-discuss] Verification of material (Au) dispersion
>To: <meep-discuss@ab-initio.mit.edu>
>Message-ID: <7117cf6421d58bce8459da9f26de6...@cup.uni-muenchen.de>
>Content-Type: text/plain; charset=UTF-8
>
>On Fri, 3 Dec 2010 16:42:11 +0100, "Marell, M.J.H." <m.j.h.mar...@tue.nl>
>wrote:
>> If I?m not mistaking , this only works for dispersive media of which the
>> real part is positive, because for negative epsilon you just get
>> exponential decay
>> There is way to verify your material:
>> http://www.mail-archive.com/meep-discuss@ab-initio.mit.edu/msg03527.html
>>
>> Best regards,
>>
>> Milan
>>
>>
>
>
>Thanks for the nice example - I also have found the example in the
>tutorial a bit confusing: as far as I understand it now, the material
>definition given at the beginning of
>http://ab-initio.mit.edu/wiki/index.php/Meep_Tutorial/Material_dispersion
>
>(set! default-material
> (make dielectric (epsilon 2.25)
> (E-polarizations
> (make polarizability
> (omega 1.1) (gamma 1e-5) (sigma 0.5))
> (make polarizability
> (omega 0.5) (gamma 0.1) (sigma 2e-5))
> )))
>
>
>should correspond to a dispersion function
>
>epsilon = 2.25 + (1.1^2)*0.5/(1.1^2 - f^2 - i*f*1e-5) +
>(0.5^2)*2*1e-5/(.5^2 - f^2 - i*f*0.1)
>
>or in Latex notation
>
>
>
>\varepsilon \left( f \right) =
>
> 2.25 + \frac{{1.1^2 \cdot 0.5}}{{1.1^2 - f^2 - if \cdot 10^{ - 5} }} +
>\frac{{0.5^2 \cdot 2 \cdot 10^{ - 5} }}{{0.5^2 - f^2 - if \cdot 0.1}}
>
>
>
>and not to the formula with the divisions by 2*pi as in the webpage.
>Excuse me if I repeat something that has already been said but the point
>seems rather important to me. Am I missing something simple or is the
>formula for epsilon(f) that appears in the tutorial wrong?
>
>
>Giovanni
>
>--
>================================================
>Giovanni Piredda
>Postdoc - AK Hartschuh
>
>Phone: ++49 - (0) 89/2180-77601
>Fax.: ++49 ? (0) 89/2180-77188
>Room: E2.062
>----------------------------------------
>Message sent by Cup Webmail (Roundcube)
>
>
>
>
>------------------------------
>
>_______________________________________________
>meep-discuss mailing list
>meep-discuss@ab-initio.mit.edu
>http://ab-initio.mit.edu/cgi-bin/mailman/listinfo/meep-discuss
>
>End of meep-discuss Digest, Vol 58, Issue 10
>********************************************
_______________________________________________
meep-discuss mailing list
meep-discuss@ab-initio.mit.edu
http://ab-initio.mit.edu/cgi-bin/mailman/listinfo/meep-discuss
