eV/Å units

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 \[1~\text{fs}\].