Hello Christine, hi Steven,
I thought we might try figuring out the units by making the integrand a
constant.
Suppose our system is two dimensional, 3 furlong wide and 4 furlong high:
(set! geometry-lattice (make lattice (size 3 4)))
(set! resolution 20)
(init-fields)
(integrate-field-function '() (lambda (x) 1))
This gives 11.6094140625, or almost 3 furlong*4 furlong=12 furlong squared.
It suggests that the units of the integral are the product of the units
specified for the cell volume.
Surprisingly, the result for the integral is not quite the same as what one
gets by specifying the volume directly:
(integrate-field-function '() (lambda (x) 1) (volume (center 0 0) (size 3
4)))
This gives 11.91265625 for me.
And
(integrate-field-function '() (lambda (x) 1) (volume (center 0 0) (size 100
100)))
gives 12.
Ummm.... ?
Best,
M.
----- Original Message -----
From: "Christine Kreuzer" <[EMAIL PROTECTED]>
To: <[email protected]>
Sent: Saturday, March 10, 2007 8:25 PM
Subject: [Meep-discuss] integrate-field-functions
Dear Steven, dear meep-users,
I have a question concerning the integrate-field-function.
What are the units of the integration volume if I integrate over the
whole computational cell?
If I want to calculate the mode volume, for example.
I use (integrate-field-function) to get the integral(eps*|E|²) over
the whole computational cell and (max-abs-field-function) to get
max(eps*|E|²). Dividing integral through max gives the modal volume,
but in which units?
In a³?
Thank you very much for any help!
Best regards,
Christine
--
Universität des Saarlandes
AG Prof. Dr. Christoph Becher
Fachrichtung 7.3 (Technische Physik)
Geb. E2.6, Zimmer 2.04
D-66123 Saarbrücken
Phone:+49(0)681 302 3418
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E-mail: [EMAIL PROTECTED]
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