direction of the transformation

2
Direction of the transformatio n. A B C A B C (uncharged) Initial system (total el. pot. energy U 0 ) A, B “common” particles, C “unique” particle which is going to be uncharged. Final system (total el. pot. energy U 1 ) AB B A BC C B AC C A AB B A r q q k V r q q r q q r q q k U 0 0 AB B A r q q k V U 1 1 BC C B AC C A r q q r q q k V 0 0 1 V BC C B AC C A r q q r q q k V V V V 1 1 1 0 1 0 AB B A BC C B AC C A AB B A BC C B AC C A AB B A r q q r q q r q q k r q q r q q r q q k r q q k V U 1 BC C B AC C A AB B A r q q r q q k V V V V r q q k V U 0 0 1 W V U U total el. pot. energy , V is just the potential of common-unique pairs, W is just the potential of common-common pairs. where C C q q 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 q C is the original charge of particle C.

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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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Page 1: Direction of the                             transformation

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

qq

r

qq

r

qqkU

00

AB

BA

r

qqkVU 11

BC

CB

AC

CA

r

qq

r

qqkV0 01 V

BC

CB

AC

CA

r

qq

r

qqkVVVV 111 010

AB

BA

BC

CB

AC

CA

AB

BA

BC

CB

AC

CA

AB

BA

r

qq

r

qq

r

qqk

r

qq

r

qq

r

qqk

r

qqkVU

1

BC

CB

AC

CAAB

BA

r

qq

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.

Page 2: Direction of the                             transformation

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

qq

r

qq

r

qqkV0

BC

CB

AC

CA

AB

BA

BC

CB

AC

CA

AB

BA

AB

BA

BC

CB

AC

CA

AB

BA

r

qq

r

qqk

r

qqk

r

qq

r

qqk

r

qqk

r

qqk

r

qq

r

qq

r

qqkVVV

111 10

UV U total el. pot. energy

AB

BA

r

qqkV 1

BC

CB

AC

CA

r

qq

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.