Van der Waals radius (let's denote it as $r_w$) of an atom, according to Wikipedia, is the radius of an imaginary hard sphere representing the distance of closest approach for another atom, and, to my understanding, quantifies a thermodynamical property of atomic gases.
While van der Waals length (denoted as $R_{vdw}$), according to Eq.(30) in arXiv0812.1496 (page 10) is defined by inter-atomic potential, which is purely quantum-mechanical.
They should be different as one can find that, $r_w=1.82 Å$ (see here) for Lithium (average of isotopes?), while $R_{vdw}=16.5 Å$ for Lithium-6, according to table I in arXiv0812.1496.
So, what's the difference? And basically how to calculate both? I guess that $r_w$ partially depends on $R_{vdw}$.

  • $\begingroup$ Minor comment to the post (v1): In the future please link to abstract pages rather than pdf files, e.g., arxiv.org/abs/0812.1496 $\endgroup$
    – Qmechanic
    Oct 30, 2017 at 18:38
  • $\begingroup$ thx@Qmechanic♦ and I've modified the first position, but may I ask why? Isn't it much more direct and convenient to access the original file? $\endgroup$
    – Mathieu
    Oct 31, 2017 at 0:29
  • $\begingroup$ Not everyone has access to fast internet. $\endgroup$
    – Qmechanic
    Oct 31, 2017 at 0:34

1 Answer 1


In the van der Waals equation of state for a real gas ($k$ is Boltzmann's constant),


$b$ is the van der Waals volume,

$$b = \frac{4}{3}\pi r_w^3.$$

On the other hand, we can give ourselves a microscopic model of the gas, postulating that the interaction between the molecules or atoms is strongly repulsive at short distance and weakly attractive at long distance, behaving as $1/r^6$ in the latter case. This power comes from the fact that atoms or molecules are neutral and therefore that only dipole-dipole interaction can occur. From there, it is possible to derive the van der Waals equation and to relate the coefficient of $1/r^6$ to $b$, either heuristically by working directly with the van der Waals equation [1], or more rigorously by first expanding the van der Waals equation in powers of $N/V$ (virial expansion). Amusingly, I have just revised an answer of mine performing this computation. This is purely within classical mechanics and it only requires to postulate an interaction potential between the gas molecules or atoms. For example, the Sutherland potential

$$\phi(r) = \begin{cases} +\infty,& r < r^*,\\ -k\epsilon\left(\dfrac{r^*}{r}\right)^6, & r > r^*, \end{cases}$$

results in

$$b = \frac{2\pi r^{*3}}{3}.$$

That is to say,


$r^*$ and $r_w$ are therefore quite similar and it makes sense from the formula of the potential where $r^*$ shall be of the order of an atom or molecule size.

Let's now move to the article you quote. It is concerned with cold atoms which are governed by quantum mechanics, as shown by the interaction potential the authors postulate:


The physics is therefore completely different from that of the derivation of the van der Waals equation of state, which does not apply at all for such cold atoms anyway. However, we could throw caution to the wind and just equate the coefficients of $1/r^6$!! In [1], it is shown that by defining

$$\begin{align} R_{vdw}&=\frac{1}{2}\left(\frac{2\mu C_6}{\hbar^2}\right)^\frac{1}{4},\\ E_{vdw}&=\frac{\hbar^2}{2\mu}\frac{1}{R_{vdw}^2}, \end{align}$$

the potential between cold atoms can be rewritten as

$$\frac{V_l(R)}{E_{vdw}} = -\frac{16}{r^6}+\frac{l(l+1)}{r^2},$$

where $r=R/R_{vdw}$.

So we end up with

$$k\epsilon r^{*6} = 16R_{vdw}^6 E_{vdw},$$


$$r^* = \left(16\frac{E_{vdw}/k}{\epsilon}\right)^\frac{1}{6}R_{vdw}.$$

So for Lithium, using the values you quoted, we end up with

$$16\frac{E_{vdw}/k}{\epsilon} = 1.8\times 10^{-6}.$$

This sounds very suspicious to have 6 orders of magnitude between the strength of the two potentials. Unfortunately, all the elements considered in your article have boiling points of the order of 1000 ${}^o$C and van der Waals equation coefficients are therefore hard to come by. But to me, it clearly looks like we are in totally different regimes and that there is no direct relation between $r^*$ and $R_{vdw}$. Unless I made a mistake in the equations…

[1] Terrell L. Hill, Derivation of the complete van der Waals' equation from statistical mechanics., Journal of Chemical Education 25 (1948), 347

[2] Kevin M. Jones, Eite Tiesinga, Paul D. Lett, and Paul S. Julienne, Ultracold photoassociation spectroscopy: long-range molecules and atomic scattering, Rev. Mod. Phys. 78 (2006), 483--535


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