# What does the “total energy” of a molecule mean for the Chem3D MM2 minimized energy calculation?

I was fiddling around with Chem3D and the MM2 calculation for finding a molecules minimized energy.

Benzene was assigned a total energy of -0.5741 kcal/mol while biphenyl was assigned a total energy of 4.5397 kcal/mol.

What is the "total energy" relative to? In other words, why does benzene have a negative total energy while biphenyl has a positive total energy? What is the 0 energy state that they are being compared to?

• Typically the force field energies are parameterized to attempt to reproduce heats of formation. Otherwise, the answers below are pretty good. – Geoff Hutchison Aug 18 '16 at 19:21

## General definition of energy

Usually, the total energy of a system is subdivided into kinetic energies of all particles and potential energies of the interactions between them, where the later terms are most often limited to the strongest Coulomb interactions only. In atomic units we get then:

$$E = - \sum\limits_{\alpha=1}^{ν} \frac{1}{2 m_{\alpha}} \nabla_{\alpha}^{2} - \sum\limits_{i=1}^{n} \frac{1}{2} \nabla_{i}^{2} - \sum\limits_{\alpha=1}^{ν} \sum\limits_{i=1}^{n} \frac{Z_{\alpha}}{r_{\alpha i}} + \sum\limits_{\alpha=1}^{ν} \sum\limits_{\beta > \alpha} \frac{Z_{\alpha} Z_{\beta}}{r_{\alpha \beta}} + \sum\limits_{i=1}^{n} \sum\limits_{j > i}^{n} \frac{1}{r_{ij}} \, ,$$
where $m_{\alpha}$ is the rest mass of nucleus $\alpha$ and $Z_{\alpha}$ is its atomic number, $r_{\alpha i} = |\vec{r}_{\alpha} - \vec{r}_{i}|$ is the distance between nucleus $\alpha$ and electron $i$, $r_{\alpha \beta} = |\vec{r}_{\alpha} - \vec{r}_{\beta}|$ is the distance between two nuclei $\alpha$ and $β$, and $r_{ij} = |\vec{r}_{i} - \vec{r}_{j}|$ is the distance between two electrons $i$ and $j$.

Now, the zero of energy corresponds to the case when all the terms are zero. Kinetic energy is zero when velocity is zero, while the Coulomb interaction energy is zero when particles are infinitely separated from each other. Thus, the zero of energy corresponds to the case when all the particles are infinitely far away from each other and not moving. In accordance with that choice of the zero of energy, the energy of any stable molecular system in the modeled environment (usually, an isolated molecule in the gas phase) has to be negative.

## Molecular mechanics definition of energy

Molecular mechanics (MM) operates with a different kind of energy. To start from, usually, it is just the potential energy. Secondly, in MM we think about a molecule in terms of atoms and bonds (rather than nuclei and electrons) and the potential energy is divided into contributions from bond stretching, angle bending, torsional rotations around single bonds, non-bonded interactions, etc. $$E = E_{\mathrm{stretch}} + E_{\mathrm{bend}} + E_{\mathrm{torsion}} + E_{\mathrm{nb}} + \dotsc \, ,$$ where the first term, for instance, is typically proportional to the sum of the squares of extensions from the equilibrium length for all the bonds: $$E_{\mathrm{stretch}} = \sum\limits_{\mathrm{bonds}} k (r - r_{\mathrm{e}})^2 \, .$$ Now, we could take the energy corresponding to the equilibrium length $r_{\mathrm{e}}$ as the zero of energy and proceed similarly for the other bonded terms, which would mean that (ignoring usually small non-bonded interactions) the MM potential energy is zero when the molecular system is completely "unstrained" (i.e. when all geometrical parameters have their "natural" equilibrium values). However, different molecules are bonded differently, and thus, the zero of energy would, in general, be defined differently. Each potential energy will thus be computed relative to a different zero of energy which makes the comparison of energies for any two different molecules meaningless.

• Perhaps the steric effects from the methyl groups as well as the repulsive effects from a higher electron density conjugated system contribute to the positive total energy? I'm sure nothing is wrong with the set up and even a simple MM2 model should be capable of approximating a positive or negative energy. – Nova Aug 12 '16 at 21:20
• The Chem3D manual confirms your definition of total energy as sum of kinetic energy and potential energy. However, the “Total Energy” reported by MM2 in Chem3D is actually the total steric energy, which is the sum of the individual terms for bond stretching, bond angle bending, torsion, van der Waals interactions, and electrostatic interactions. (I can reproduce the results given in the question with default values for MM2.) – Loong Aug 12 '16 at 21:28
• The assumption of zeros at infinite separation are false for bonded potentials in force fields (which MM2 is). Take a closer look at the parametrization of force fields. Usually bonded potentials are zero at equilibrium. Hence you will get positive energies when these potentials are coupled while adding electrostatics can go either way. – Deathbreath Aug 18 '16 at 13:47

The total energy is indeed the sum of kinetic and potential energies. In force fields such as MM2, the potentials are split into bonded and non-bonded terms. The bonded terms are generally such that they are zero at their respective global minimum. For instance, a harmonic bond potential has the form $k_{ij}(x_{ij} - x_{ij}^{(0)})^2$. When you combine enough bonds, it is no longer generally possible to find a geometry that all bonds are at respective $x_{ij}^{(0)}$. Since the potential is non-negative, this means the energy has to be non-zero and positive.

The nonbonded terms generally do not follow this definition, but rather tend to zero at infinity, e.g., the Coulomb term or a Lennard-Jones vdW potential. These may furthermore have any sign so you may witness negative or positive energies.

In your specific case of benzene and biphenyl, you need to look at the individual contributions to discern why biphenyl is positive in energy.

• Why do you think that "it is no longer generally possible .. that all bonds are at [the minimum]"? While I agree that in complicated systems, angle, dihedral, non-bonding terms, etc. might cause problems - it's certainly possible to have large systems where all bonds are at the equilibrium bond length. – Geoff Hutchison Aug 18 '16 at 19:24
• I am using "general" in the same sense as two lines in general position aren't parallel and I am talking about the minima of the constituting terms, so while a large system may be at the bond minima of the constituting harmonic terms, it generally won't be. Consider three bodies with 3 bonds and one of those bonds' minimum energy separation exceeds the sum of the other two bonds' minimum energy separations. Add angles and non-bonded terms and the "problem" is exacerbated. – Deathbreath Sep 12 '16 at 13:43