Download 18324-CHARMM_Units

I have another question regarding the CHARMM force field. FYI, I’m using a Cartesian
coordinate system with all positions in angstroms, velocity in angstroms/second, and
accelerations in angstroms/second2. My masses are in Daltons and charges in electron
volts. Please see what I did to address energy evaluation and acceleration derivation with
the bond interaction and angle interaction, and if you can, please help me finish the
dihedral acceleration. I really appreciate your help.
Bond Interaction
EB  kb (r  ro )2
Energy evaluation
kcal
.
mol  ang 2
If all positions are in angstroms then when the distance formula is applied to obtain
kcal
(r  ro ) it is in angstroms and the energy is in
.
mol
With the bond interaction the bond force constant, kb , is stored in units of
EB  kb (r  ro )2
EB


kcal
kcal
2
 mol  ang 2  ang   mol
Dynamics
First the bond constant needs to be converted from kcal to joules by multiplication of
kg  m 2
4184 J/kcal. We then equate joules to
. From here we convert kg to g by
s2
multiplication of 103 and m2 to ang2 by multiplication of 1020. ang2 cancels with ang2
and g/mol can also be written as Daltons. The resulting new bond constant unit is
Daltons/s2.
kb

kcal
mol  ang 2
kg
mol  s 2
joules
kg  m 2
 4184 joules 
x




kcal
mol  ang 2
mol  ang 2  s 2
 10 20 ang 
kg  ang 2
x

s 2  mol  ang 2
 m 
 10 3 g 
g
daltons
x


2
mol  s
s2
 kg 
Given the new unit for the bond constant we take the derivate of the bond energy,
Eb  2kb (r  ro ) and we plug in the difference in distance, (r  ro ) , in angstroms. This
daltons  ang
yields a force in
. This force is then divided by the mass of the system in
s2
ang
Daltons, yielding an acceleration in 2 .
s
Angle Interaction
E  k (  o )2
Energy Evaluation – The energy evaluation for angles is the same as bonds, just use
everything as is except the theta must be in radians.
Dynamics –
kcal
. We follow the rubric from bonding and go from
mol  radian 2
kcal to joules, joules to kg-m2/s2, m2 to ang2 and simplify kg/mol to Daltons.
The angle constant is in
kcal
mol  radian 2
joules
kg  m 2
 4184 joules 
x




kcal
mol  radian 2
s 2  mol  radian 2
daltons  m 2
 2
s  radian 2
 10 3 g 
x
 kg 
2
 1010 ang 
daltons  ang 2
x

s 2  radian 2
 m 
For clarity I separated the semi-final unit into two parts:
 daltons  ang   ang 
 2


s  radian   radian 
The first term, when multiplied by the difference in angle (which has units of radians)
and then divided by the mass (in Daltons), yields an acceleration in ang/s2. However, I
 ang 
still have a unit of 
left over.
 radian 
I thought that this unit may have something to do with torque. Consider a fixed angle,
arbitrary in size, and points along the two vectors forming the angle. As we move farther
along the vectors, if the torque of the angle is fixed then, given that   rxF , if r gets
incrementally bigger, that F gets incrementally smaller.
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
Therefore, we impose the following multiplier. The definition of a radian is the angle
where the arc length is equal to the radius. In otherwords for a given system, where atom
“i” is at the origin and atom “j” is on one of the vectors, 1 radian  dist(ij) in angstroms.
1rad
Therefore we can take our constant and impose a miltiplier of
making the total
dist(ij)
acceleration,
ai 
2k (  o )(4184x10 23 )
, where j is the central atom in the angle.
mass(i)xdist(ij)
Dihedrals
The dihedral energy is given as E  k  k cos(n ) . Now if the energy is assumed to
kcal
kcal
as the other terms are than the dihedral force constant must also be in
.
mol
mol
kg  m 2
Similarly as before, the force constant unit goes from kcal to joules to
to
mol  s 2
be in
daltons  ang 2
. Now the derivative in this case is slightly more difficult but in every
s2
case we end up with some trig function, call it f ( ) , multiplied by the constant k .
If we assume the trig function has no units, as it mustn’t, then at present we have an extra
“angstrom” unit in our acceleration. The only thing I can think is that the acceleration on
the i-th atom in the diheral i-j-k-l is somehow scaled by distance i-j again. I kind of feel
like I’m just arbitrarily throwing in multipliers here though.