# What is the link between the meaning of excluded volume and its mathematical derivation?

When we examine real polymer chains we have to consider the interactions between single monomers. Therefore we consider a Lennard-Jones-like potential for bringing two monomers together and calculate the probability of the distance r between the monomers with a Boltzmann-distribution.

Then we define the so-called "Mayer f-function" as the difference between the Boltzmann factor and the case of no interaction or infinite distance:

$$f(r) = exp[-U(r)/k_BT]-1$$

The derivation can be visualized with the following graphs: Now we define the excluded volume v as the negative of the integral of the Mayer f-function over all space:

$$v = -\int f(r)dr = \int 1-exp[-U(r)/k_BT]dr$$

This excluded volume is now interpreted and used as the volume occupied by one monomer, which can therefore not be occupied by anything else (Excluded volume - Wikipedia).

How can I derive this interpretation from the mathematical calculation? I have my difficulties with getting a clear picture of the physical meaning of the excluded volume by just looking at the above derivation.

This interpretation is only strictly true in the special case of hard particles, that is when the interaction potential is infinite if the particles overlap, and zero if they do not. In that case, it is easy to see that the Mayer function is $$-1$$ for overlap, and $$0$$ for non-overlap. The negative sign is taken into account explicitly in the definition of $$v$$, so the integral over all space just becomes an integral of $$1$$ over the region where the particles overlap, in other words an excluded volume.

In the more general case, the result still has the units of volume, albeit a temperature-dependent "effective" excluded volume.

• Okay, I understand it in the case of hard particles then. But the graph of the Mayer function above, which I have copied from the lecture notes takes other values than -1 and 0 too. Is that for soft particles then? In this case the integral is also over the region where the particles don't overlap but are very close to each other. Does this volume just get subtracted from the overlapping region? May 13 '19 at 15:00
• The contribution from the attractive term in the potential, which gives rise to the positive bulge in $f(r)$, does indeed get subtracted from the excluded volume arising from the overlap region. It is less helpful (in my opinion) to think of this as a "volume", although since $f$ is dimensionless, it gives something with the units of volume when integrated. Incidentally, to be clear, your integral should be over 3D space or, equivalently, include a $4\pi r^2$ factor.
– user197851
May 14 '19 at 3:04
• I'm still confused. When we calculate the free energy of the interaction, we use the excluded volume to calculate the probability for a contact between the two particles: $$F_{Int} = kT \cdot v \cdot \frac{N}{R^3} \cdot N$$ where $v\cdot\frac{N}{R^3}$ is that probability. I'm still not sure what I should think of this excluded volume v. May 15 '19 at 16:30
• I think I've addressed the point in your question. Now, you've introduced new quantities that were not part of your question, without defining them, and asked something new, so I can't offer much help there. I can say that the non-ideal part of $F$ can be rigorously expanded in powers of the density: the virial expansion. You can look it up. The coefficient of the leading, pairwise, term involves an integral over $f(r)$. Presumably this is it. It is not necessary to interpret it literally as an excluded volume. Maybe you need to ask a new question, giving more details of the derivation.
– user197851
May 16 '19 at 4:14
• In the case of hard particles, the excluded volume would just equal the particle volume itself. It seems like the attractive term of the Lennard-Jones potential allows the particles to push in each other. Can this be explained by quantum mechanics? Jun 23 '19 at 16:08