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Unformatted text preview: U LJ ( r ) = 4 r 12 r 6 . (29) I There are (at least) two ways how to drive this potential to zero: I Shift and tilt the potential to get U ( r ) and U ( r ) continuous at r c . I Use a third order polynomial for r [ r c , r c + r ] . Notes I Shiftandtilt requires playing around with the actual potential equation: U ( r ) = U LJ ( r )  ( rr c ) U LJ ( r c ) U LJ ( r c ) . (30) I Now U ( r c ) = 0 as is U ( r c ) = 0. I However, since we are changing the equation, the tting must be redone with the new equation. U ( r ) r shiftandtilt U LJ P ( r ) I Or, as said, we can solve constants for the polynomial P ( r ) = ar 3 + br 3 + cr + d for r [ r c , r c + r ] with conditions: P ( r c ) = U LJ ( r c ) P ( r c ) = U LJ ( r c ) P ( r c + r c ) = P ( r c + r c ) = (31) Notes I Note that even smoothlooking potential functions can have problematic forces (even if continuous). Notes...
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This note was uploaded on 02/14/2012 for the course CSE 6590 taught by Professor Kotakoski during the Winter '12 term at York University.
 Winter '12
 Kotakoski

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