The CHARMM22 Force Field

The form of the potential energy function we will use in this exercise is taken directly from CHARMM22 and given by the following equation [1]:

$$V = \sum_{\mathrm{bonds}}{k_b \left(b-b_0\right)^2} + \sum_{\mathrm{a... ...sum_{\mathrm{dihedrals}}{k_\phi \left[1+\cos\left(n\phi - \delta\right)\right]}$$

$$+ \sum_{\mathrm{impropers}}{k_\omega \left(\omega-\omega_0\right)^2} + \sum_{\mathrm{Urey-Bradley}}{k_u \left(u-u_0\right)^2}$$

$$+ \sum_{\mathrm{nonbonded}}{\epsilon \left[\left(\frac{R_{\mathrm... ...{R_{\mathrm{min}_{ij}}}{r_{ij}}\right)^6\right]+\frac{q_iq_j}{\epsilon r_{ij}}}$$ (1)

The first term in the energy function accounts for the bond stretches where $k_b$ is the bond force constant and $b-b_0$ is the distance from equilibrium that the atom has moved. The second term in the equation accounts for the bond angles where $k_\theta$ is the angle force constant and $\theta-\theta_0$ is the angle from equilibrium between 3 bonded atoms. The third term is for the dihedrals (a.k.a. torsion angles) where $k_\phi$ is the dihedral force constant, $n$ is the multiplicity of the function, $\phi$ is the dihedral angle and $\delta$ is the phase shift. The fourth term accounts for the impropers (i.e. out of plane bending), where $k_\omega$ is the force constant and $\omega-\omega_0$ is the out of plane angle. The Urey-Bradley component (cross-term accounting for angle bending using 1,3 nonbonded interactions) comprises the fifth term, where $k_U$ is the respective force constant and $U$ is the distance between the 1,3 atoms in the harmonic potential. Nonbonded interactions between pairs of atoms $(i,j)$ are represented by the last two terms. By definition, the nonbonded forces are only applied to atom pairs separated by at least three bonds. The VDW energy is calculated with a standard 12-6 Lennard-Jones potential and the electrostatic energy with a Coulombic potential.