# Where do the Lennard-Jones dimensionless/reduced units come from?

I'm a little bit confused as to where the expressions for the Lennard-Jones Potential Dimensionless units shown in this Wikipedia chart come from. I know they're derived from the L-J force equation but I've been trying to do that for dimensionless time and I can't find an example anywhere.

This is the force I'm using:

$$\vec { f _ { i j } } = \left( \frac { 48 \epsilon } { \sigma ^ { 2 } } \right) \left[ \left( \frac { \sigma } { r _ { i j } } \right) ^ { 14 } - \frac { 1 } { 2 } \left( \frac { \sigma } { r _ { i j } } \right) ^ { 8 } \right] \vec { r } _ { i j }$$

$$t^*=\frac{t}{\sigma}\sqrt{\frac{\epsilon}{m}}$$

But I have no idea how to get from force to reduced time. Can someone nudge me in the right direction?

All of the dimensionless quantities in the chart are derived using simple dimensional analysis. The basic premise is as follows: you are given a length $$\sigma$$, an energy $$\epsilon$$, and a mass $$m$$. Different combinations of these quantities will give additional constants with different units. In order to convert a parameter into its dimensionless form, you divide by the particular combination of these constants which has the same units.

Therefore, to get the reduced time, we must divide the time by a combination of $$\sigma$$, $$\epsilon$$, and $$m$$ that has units of time. Suppose we take a product of arbitrary powers of these three constants: $$\sigma^a \epsilon^b m^c$$. The units of this combination are:

$$\textrm{m}^a\cdot\textrm{J}^b\cdot\textrm{kg}^c=\textrm{kg}^{b+c}\cdot\textrm{m}^{a+2b}\cdot \textrm{s}^{-2b}$$

Finding a combination which has units of seconds is equivalent to solving the equation:

$$\textrm{kg}^{b+c}\cdot\textrm{m}^{a+2b}\cdot\textrm{s}^{-2b}=\textrm{s}^1$$

It should be clear that $$-2b=1$$, so $$b=-1/2$$. We also have that $$b+c=0$$, which means that $$c=1/2$$, and $$a+2b=0$$, which means that $$a=1$$. Plugging these powers back into our original product, we see that we can make a new constant $$\tau$$ which has units of time:

$$\tau=\sigma^1\epsilon^{-1/2} m^{1/2}=\sigma\sqrt{\frac{m}{\epsilon}}=\sqrt{\frac{\sigma^2 m}{\epsilon}}$$

So our reduced time $$t^*$$ is

$$t^*=\frac{t}{\tau}=t\sqrt{\frac{\epsilon}{\sigma^2 m}}$$

This approach is quite general, and can be used in any situation when you need to derive a constant in certain units from a collection of constants with different units. In fact, this procedure is how the Planck length, mass, energy, etc. are defined, where the given collection of constants is $$G$$, $$\hbar$$, and $$c$$.

From your question (and the link) force has the same units as the combination $$\frac{\epsilon}{\sigma^2}\sigma=\frac{\epsilon}{\sigma}$$ since $$r/\sigma$$ is dimensionless. Now, force has units of [kg][m][$$\tau^{-2}$$] we have $$\frac{\epsilon}{\sigma}=\frac{M\sigma}{\tau^2} \quad \Rightarrow \quad \frac{1}{\tau}=\frac{\epsilon}{M\sigma^2}$$ and thus $$t^*=\frac{t}{\tau}=\frac{t}{\sigma}\sqrt{\frac{\epsilon}{M}}\, .$$ with $$M$$ the mass parameter.