Direction of the transformation
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Direction of the transformation. PowerPoint PPT Presentation


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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.

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Direction of the transformation.

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Direction of the transformation

Utotal el. pot. energy , V is just the potential of common-unique pairs, W is just the potential of common-common pairs.

Initial system (total el. pot. energy U0)

Final system (total el. pot. energy U1)

A

A

C

Direction of the transformation.

(uncharged)

C

B

B

A, B “common” particles, C “unique” particle which is going to be uncharged.

where

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 qCis the original charge of particle C.


Direction of the transformation

Utotal el. pot. energy

Initial system (total el. pot. energy V0)

Final system (total el. pot. energy V1)

A

A

C

Direction of the transformation.

(uncharged)

C

B

B

A, B “common” particles, C “unique” particle which is going to be uncharged.

So the simulated systems in different values differ in charge of the particle C, which decreases as increases.

where

If this interpretation is OK, why we need 2 simultaneous sander threadsfor 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.

During the simulation in given lambda value we just calculate this quantity where qCis the original charge of particle C.


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