eV/Å units

Units

For the EAM potential you need to switch from working with Lennard-Jones units to working with real physical units. We outline here briefly how unit conversion works. Our system of units requires units for length \([l]\), time \([t]\), and mass \([m]\). All other units are compound units, for example velocities \([v]=[l]/[t]\), accelerations \([a]=[l]/[t]^2\), forces \([f]=[m][l]/[t]^2\) and energies \([E]=[m][l]^2/[t]^2\). Note that the unit of force can also be expressed through energy and length, \([f]=[E]/[l]\) and it is common to use this denomination in molecular calculations.

eV/Å

Energy, lengths and forces

In molecular calculations we typically fix the units of energy \([E]\) and length \([l]\). This is because we know what typical energy and length scales look like. Molecular calculations use different sets of unit system. We will here use \([E]=\text{eV}\) and \([l]=\text{Å}\), because \(1~\text{eV}\) is roughly the energy stored in a metallic or covalent bond and \(1~\text{Å}\) is roughly the distance between atoms. Note that \(k_B T\) at room temperature \(T=300~\text{K}\) is \(k_B T\approx 25~\text{meV}\).

Other energy units that are used in molecular calculations are \(\text{kJ/mol}\) or \(\text{kcal/mol}\). Please forget immediately that something like \(\text{kcal/mol}\) even exists. To convert from \(\text{eV}\) to SI units, we simply insert the respective units. Remember \(e=1.6\cdot 10^{-19}~\text{C}\), then \(1~\text{eV}\approx 1.6\cdot 10^{-19} C J/C=1.6\cdot 10^{-19} J\).

Forces in molecular calculations are typically reported in units of \(\text{eV}/\text{Å}\), which is the natural force unit for this system of units. Note that in SI units this becomes \(1~\text{eV}/\text{Å}\approx 1.6\cdot 10^{-19}/10^{10}~\text{N}=1.6~\text{nN}\).

Fixing the mass unit

We have now fixed energy and length units. This means we can decide on a third unit, which then fixes our system of units. Let this third unit be the mass, which we set to \([m]=\text{g/mol}\). (This units is useful because atomic masses are tabulated in this unit in the periodic table.) We can then rewrite \([E]=[m][l]^2/[t]^2\) to \([t]=[l]\sqrt{[m]/[E]}\). Inserting our favorite system of units yields \([t]=Å \sqrt{\text{g}/(\text{eV}~\text{mol})}\). Remember that \(\text{mol}=6\cdot 10^{23}\) (it is just a number!), hence \([t]=10^{-10}~\text{m} \sqrt{10^{-3}~\text{kg}/(1.6\cdot 10^{-19} J \times 6\cdot 10^{23})}=1.018\cdot 10^{-14}~\text{s}=10.18~\text{fs}\). This means that if you use \(\text{g/mol}\) for your unit of mass, a unit time step in your calculation has the actual length of \(10.18~\text{fs}\).

Fixing the time unit

Since the time step is something we actively adjust as a user, it is often convenient to fix the time step rather than the mass. A good choice is \([t]=1~\text{fs}\) since this is around the time step that you need to discretize the equations of motion. Again from the units of the energy we obtain the mass units as \([m]=[E][t]^2/[l]^2\). Inserting our system of units yields \([m]=1.6\cdot 10^{-19}~\text{J} \times 10^{-30}~\text{s}^2 / (10^{-20}~\text{m})=1.6\cdot 10^{-29}~\text{kg}\). This is a rather unwieldy number that we want to express in terms of \(\text{g}/\text{mol}\). This gives \([m]=1.6\cdot 10^{-29}~\text{kg} \times \text{mol}/\text{mol} = 1.6\cdot 10^{-29} \times 10^{3}\times 6\cdot 10^{23} ~\text{g}/\text{mol}=0.009649~\text{g}/\text{mol}\). If we have masses in \(\text{g}/\text{mol}\), we need to divide those by \(0.009649\) (multiply them by a factor of \(103.6\)) to get a time step of \(1~\text{fs}\).