Constants

Real units to dimensionless units

Boltzmann constant: kB = 1.3806 10-23 J/K = 8.314 J/mol/K = 1.987 cal/mol/K

The thermal energy at 298K is: kBT = 8.314 J/mol/K 298K = 2478 J/mol = 4.114 ⋅ 10-21 J = 2.478 kJ/mol = 0.59 kcal/mol

van der Waals strength: 0.4 - 4.0 kJ/mol = 0.1 - 4 kcal/mol same order as kT, range 0.3 - 0.6 nm

Coulombic repulsion: the Bjerrum length lB = 0.7 nm in water (dielectric constant, epsilon = 80), which means electrostatic energy becomes equal kBT at about 0.7 nm. In a highly ionic medium, lB would be reduced.

Hydrogen bonding (electrostatic attraction): 1 - 5 kcal/mol -> about 2 - 10 times stronger than kBT, range = 0.3 nm

Reduced time unit

Usually in a simulation the time unit is defined based on the length scale, mass and energy scale:

τ = σ sqrt(m/ε),

where

m = 10-21 kg

ε = 50 kJ/mol = 50e+3/6.023e23 = 8.3e-20 Nm = 8.3e-20 kg m2/s2

σ = 5e-9 m

which give 

τ = 5e-9 m sqrt(1e-21 kg / (8.3e-20 kg m^2/s^2)) = 5e-10 s = 0.5e-9 s = 0.5 ns.

This means that for the given choice of mass, energy and length scales, the time unit is 0.5 nanoseconds. With that time unit, a time step of 0.002τ corresponds to 0.001 ns, or 1 ps.

Energy scale

1 kcal/mol = 0.0433634 eV; then 1 (kcal/mol)/e = 0.0433634 V = 43.3634 mV

1 kBT (at T = 300K) = 0.59 kcal/mol; then 1 kT/e = 0.59 kcal/mol/e = 0.59 * 43.3634 mV = 25.584406 mV

At T = 298K, an interaction energy of 50 kJ/mol corresponds to a reduced energy of ε/kBT = 50 103 J/(6.023  1023) / (4.114  10-21J ) = 20.0, where NA = 6.023 ⋅ 1023 is the Avogadro's number.

Charge reduced unit

The charge reduced unit is defined as: (See http://lammps.sandia.gov/doc/units.html)

where ε is the energy scale.

For q = +3e, σ = 5 nm, if the energy scale is chosen to be the thermal energy ε = kBT at T = 298K then:

q* = \frac{3 ⋅ 1.6 ⋅ 10-19C}{\sqrt{4 π ⋅ 8.85 ⋅ 10-12 C2/J/m ⋅ (5 ⋅ 10-9m) ⋅ 1.3806 ⋅ 10-23 J/K ⋅ 298K}} = \frac{4.8 ⋅ 10-19}{4.783 ⋅ 10-20} = 10.0  

If the energy scale is chosen to be 50 kJ/mol (20 times greater than thermal energy at T = 298K as shown above), then the charge reduced unit would be 10.0/\sqrt{20.0} = 2.23.

If the length scale is chosen to be 0.5 nm, then the charge reduced unit would be: 10.0/\sqrt{0.1} = 31.6

If the length scale is chosen to be 1.0 nm, then the charge reduced unit would be: 10.0/\sqrt{0.2} = 22.4

Electrical field

Plugging in the values of distance unit (σ) and energy unit (ε), we obtain

E* = E (4 π ε0σε)^1/2 σ / ε = E sqrt(4 ⋅ 3.1415926 ⋅ 8.85 ⋅ 10^{-12} C^2/J/m ⋅ 5e-9 m ⋅ 50e3/6.023e+23 Nm) ⋅ 5e-9/ 50e3/6.023e+23 Nm = E (1.29e-8 C/N), 

where perm0 is the vacuum permittivity (=8.85e-12 C^2/J/m) and ε = 50 kJ/mol. Since 1 C/N = 1 m/V (meter/volt), we get the electrical field in real units: E = E*/1.29e-8 V/m. 

Electrical potential 

where ε is the energy scale. if the energy scale is chosen to be the thermal energy ε = kBT at T = 298K, distance unit (σ) 1 nm, we obtain

V* = V (4 π ε0σε)^1/2 σ / ε = E sqrt(4 ⋅ 3.1415926 ⋅ 8.85 ⋅ 10^{-12} C^2/J/m ⋅ 1e-9 m ⋅ 1.3806e-23 J/K ⋅ 298 K) /(1.3806e-23 J/K ⋅ 298K) = V (5.2 C/J)

where perm0 is the vacuum permittivity (=8.85e-12 C^2/J/m). Since 1 Volt = 1 J/C , we get the electrical field in real units: V = V* / 5.2 (V).

NOTE: The electrical field and potential are derived from the charge, energy and distance units.

Dipole moment reduced unit

The dipole moment reduced unit is defined as (See http://lammps.sandia.gov/doc/units.html)

where

μ is the real dipole moment in Debye (D) unit: 1D = 3.335 10-30C m, ε0 = 8.85  10-12F/m is the vacuum permittivity, σ is the length unit and  ε is the energy unit. 1 F/m = C2/J/m. 

For μ = 100D, σ = 5 nm, ε = kBT at T = 298K,

μ* = \frac{100 ⋅ 3.335 ⋅ 10-30Cm }{\sqrt{4 π ⋅ 8.85 ⋅ 10-12 C2/J/m (5 ⋅ 10-9)3 ⋅m3 1.3806 ⋅ 10-23 J/K ⋅ 298K}}

= \frac{100 ⋅ 3.335 ⋅ 10-30Cm }{2.392 ⋅ 10-28} = 1.394

If the length scale is chosen to be 0.5 nm, then the dipole moment reduced unit would be: 1.394/\sqrt{0.1^3} = 44.0

If the length scale is chosen to be 1.0 nm, then the dipole moment reduced unit would be: 1.394/\sqrt{0.2^3} = 16.0

Mass

'''Nanoparticles''' 

The density of some inorganic materials is in the range of 5-20g/cm3. For example, 

*CdSe: 5.816 g/cm3 

*CdTe: 5.85 g/cm3

*PbS: 7.6 g/cm3

*TiO2: 4.23 g/cm3

*Au: 19.3 g/cm3

*Ag: 10.49 g/cm3

A 5 nm sized nanoparticle will have a mass of roughly <math>10 ⋅ 10^{-3} kg/cm^3 ⋅ (5 ⋅ 10-7cm)^3 ⋅ \pi/6 = 6.5 ⋅ 10^{-22}kg

The molecular weight of protein Cytochrome C (3-4 nm in size) is 12,000 daltons (Da) (1 Da = 1.66 ⋅ 10^{-27} kg), meaning that its weight in kilograms is: 12,000 ⋅ 1.66 ⋅ 10-27 kg = 2 μ 10-23 kg

Concentration

1 M = 1 mol/l = 6.023+e23 / (1e-3 m^3) = 6.023+e23 / (1e-3 1.0e+27 nm^3) = 0.6023 1/nm3

1 1/nm3 = 1.66 M

To convert number density into concentration and vice versa, we need to specify a length unit. For example, if the length unit is 1 nm, then number density rho = 1.0 1/sigma3 corresponds to 0.6 M.

For physiological salt concentration of 0.1 M (100 mM), that is, 0.06 1/nm3, with the unit length of \sigma = 3.0 Angstroms = 0.3 nm, the corresponding dimensionless number density is \rho = 0.06 1/(1/0.3 sigma)^3 = 0.0016 1/\sigma^3.

If sigma = 5 Angstroms, \rho = 0.06 /(1/0.5 sigma)^3 = 0.0075 1/\sigma^3.

For water, density of 1000 kg/m^3 -> 55555 mol/m^3 = (55.6e+3) (6.023e+23) / 1e+30 Angstrom^3 = (0.333e+6) e+23 / e+30 = 0.03333 1/A^3