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U V  W
U total 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
C
A
Direction of the
transformation.
B
C
(uncharged)
B
A, B “common” particles, C “unique” particle which is going to be uncharged.
q q
q q 
V0  k  A C  B C 
rBC 
 rAC
V1  0
q q
q q
q q 
q q
U 0  k  A B  A C  B C   V0  k A B
rAC
rBC 
rAB
 rAB
U1  V1  k
q A qB
rAB
q q
q q 
V    1   V0  V1  1   V0  1   k  A C  B C 
rBC 
 rAC

  q q   qB qC    q A qB 
q q
q q  q q 
q q
 where q    1   q
 
U    V    k A B  k  1    A C  B C   A B   k   A C

C
C
rAB
rBC  rAB 
rBC  rAB 
 rAC

  rAC
So the simulated systems in different
 values differ in charge of the particle C, which

q q 
V    k A B 
rAB  V

U
   
   V1  V0  V0  k  q AqC  qB qC




rBC
 rAC



decreases as
 increases.
During the simulation in given lambda value we just
calculate this quantity where qC is the original charge
of particle C.
V U
U total el. pot. energy
Initial system (total el. pot. energy V0)
Final system (total el. pot. energy V1)
A
C
B
Direction of the
transformation.
A
C
(uncharged)
B
A, B “common” particles, C “unique” particle which is going to be uncharged.
q q
q q
q q 
V0  k  A B  A C  B C 
rAC
rBC 
 rAB
V1  k
q A qB
rAB
q q
q q
 q q   qB qC   
q q
q q 
q q 
q q
q q
q q

V    1   V0  V1  1   k  A B  A C  B C   k A B  k A B  1   k  A C  B C   k A B  k  A C

r
r
r
r
r
r
r
r
r
r
AC
BC 
AB
AB
BC 
AB
AC
BC
 AB
 AC


where
qC    1   qC
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.


V
  V1  V0  k  q AqC  qB qC 

rBC 
 rAC
During the simulation in given lambda value we just
calculate this quantity where qC is the original charge
of particle C.