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Re: Force Field [message #45491]

Wed, 14 September 2005 04:03

panblosky
Messages: 17
Registered: May 2005

Junior Member

>
> Assuming that you are happy with first-order, centered differences, you
> want something like this
>
> grad_x = (SHIFT(phi, -1, 0) - SHIFT(phi, 1, 0))/(2.0*dx)
> grad_y = (SHIFT(phi, 0, -1) - SHIFT(phi, 0, 1))/(2.0*dy)
>
> where phi is the potential It think this handles the periodic boundary
> conditions properly, but check to make sure.
>
> Cheers, Ken Bowman

Thanks Ken, it does work, and really fast!!

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[Message index]

		Re: Force Field By: panblosky on Wed, 14 September 2005 04:03
		Re: Force Field By: Kenneth P. Bowman on Tue, 13 September 2005 19:13
		Re: Force Field By: David Fanning on Tue, 13 September 2005 11:22

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