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U total el. p ot. e nergy , V is just the potential of common-unique pairs, W is just the potential of common-common pairs. Initial system (total el. p ot. e nergy U 0 ). Final system (total el. p ot. e nergy U 1 ). A. A. C. - PowerPoint PPT Presentation
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Direction of the transformation.
A
B
CA
B
C (uncharged)
Initial system (total el. pot. energy U0)
A, B “common” particles, C “unique” particle which is going to be uncharged.
Final system (total el. pot. energy U1)
AB
BA
BC
CB
AC
CA
AB
BA
r
qqkV
r
r
r
qqkU
00
AB
BA
r
qqkVU 11
BC
CB
AC
CA
r
r
qqkV0 01 V
BC
CB
AC
CA
r
r
qqkVVVV 111 010
AB
BA
BC
CB
AC
CA
AB
BA
BC
CB
AC
CA
AB
BA
r
r
r
qqk
r
r
r
qqk
r
qqkVU
1
BC
CB
AC
CAAB
BA
r
r
qqkVVV
Vrqq
kVU
001
WVU U total el. pot. energy , V is just the potential of common-unique pairs, W is just the potential of common-common pairs.
where CC qq 1
So the simulated systems in different values differ in charge of the particle C, which decreases as increases.
During the simulation in given lambda value we just calculate this quantity where qC is the original charge of particle C.
Direction of the transformation.
A
B
CA
B
C (uncharged)
Initial system (total el. pot. energy V0)
A, B “common” particles, C “unique” particle which is going to be uncharged.
Final system (total el. pot. energy V1)
BC
CB
AC
CA
AB
BA
r
r
r
qqkV0
BC
CB
AC
CA
AB
BA
BC
CB
AC
CA
AB
BA
AB
BA
BC
CB
AC
CA
AB
BA
r
r
qqk
r
qqk
r
r
qqk
r
qqk
r
qqk
r
r
r
qqkVVV
111 10
UV U total el. pot. energy
AB
BA
r
qqkV 1
BC
CB
AC
CA
r
r
qqkVV
V01
During the simulation in given lambda value we just calculate this quantity where qC is the original charge of particle C.
CC qq 1where So the simulated systems in different values differ in charge of the particle C, which decreases as increases.
If this interpretation is OK, why we need 2 simultaneous sander threads for MD run with given lambda value if the simulated systems differ just in charge of particle C ? So just normal (one sander thread) MD should be OK for each lambda value simply just using actual charge value for C particle.