8
$\begingroup$

The internal potential energy of an $N$ particle system is a general function of the coordinates of the particles: $U(r_1,...,r_N)$. In some approximations and expansions - e.g. virial expansion - it is assumed that the potential energy is a sum of pairwise interactions:

$$U(r_1,...,r_N) = \sum_{ij}u(r_{ij})$$

with $r_{ij} = |r_i - r_j|$. But most of the interactions between particles I am aware of are pairwise per se. The only three-particle interaction I know of is a chemical reaction which needs a catalyst.

Which other specific examples of three- or $n$-particle interactions are there?

Addendum: I found this in the Wikipedia article on reaction mechanisms:

In general, reaction steps involving more than three molecular entities do not occur.

$\endgroup$
4
  • 1
    $\begingroup$ Monopole-induced dipole interactions come to mind. For example, one monopole can induce a transient dipole on a second particle, which can then interact with a third monopole. $\endgroup$ Apr 1, 2014 at 13:56
  • 1
    $\begingroup$ @DumpsterDoofus: but if you look at your monopole-dipole system at the level of individual electrons and protons the potential is a simple pairwise sum. It only appears otherwise because at the larger scale you are ignoring the motions of the individual charges. $\endgroup$ Apr 1, 2014 at 14:55
  • $\begingroup$ There could be something like induced dipoles interacting with each other in a collision, where atom A induces a dipole in atoms B and C, and B and C interact with each other as a result. The induced dipole would intrinsically depend on the distance between B-A and C-A, as well as the distance between A-B. $\endgroup$
    – webb
    Apr 1, 2014 at 16:19
  • $\begingroup$ Reaction steps involving more than three molecular entities do not occur because of collision theory proposed limits. $\endgroup$
    – Sensebe
    Apr 24, 2014 at 17:22

6 Answers 6

4
+50
$\begingroup$

The pair interaction is only a convenient approximation. More detailed approximations need multiparticle interactions. The most well-known one is the Axilrod-Teller 3-body potential which gives the first corrections to the $r^{-6}$ pair interaction describing for small distances the van der Waals repulsion of neutral atoms.

Another common way to model multiparticle interactions is the embedded atom model. Here the multiparticle interaction is not separable into k-body terms.

In general, a real system has some N-particle potential, which for theoretical reasons is translation, rotation, and permutation invariant. Theory gives no other restrictions. Thus which specific potential is correct must be decided by experiment (or deduced from a more detailed model). Pair potentials are simply the simplest class of approximations.

$\endgroup$
5
  • $\begingroup$ Even though I do appreciate this specific answer, I do wonder why one always starts learning about interactions as pair interactions - that's the usual way to think about interactions - and only by accident I learn that pair interactions are only a convenient approximation. Approximation of what? And in which more general context? $\endgroup$ Apr 23, 2014 at 22:33
  • $\begingroup$ This is because the physics tradition proceeds (unlike the rtradition of mathematicians, which I prefer) through learning by example. So textbooks just consider the simplest case that is enough to illustrate the typical methods and its difficulties. The complexities come early enough when one is treating real applications rather than teaching examples, since then one often cannot ignore more complex terms. $\endgroup$ Apr 24, 2014 at 8:47
  • $\begingroup$ Approximations of what? Of the true multiparticle potential, which is just an arbitrary translation, rotation, and permutation invariant potential. A real system has some N-particle potential, but which one must be decided by experiment (or deduced from a more detailed model). Pair potentials are simply the simplest class of approximations. $\endgroup$ Apr 24, 2014 at 8:50
  • $\begingroup$ The Axilrod-Teller potential is an effective potential involving three polarizable molecules. Already the words "polarizable" and "molecule" indicate that A-T is not fundamental. The fundamental interactions in the quantum mechanical Hamiltonian are still pair-wise Coulomb potentials. The A-T potential is a very useful approximation to the fundamental pair-wise Coulomb interactions, not the other way around. We learn about pair-wise potentials because they are fundamental. $\endgroup$
    – garyp
    May 5, 2014 at 13:25
  • $\begingroup$ @garyp: all potentials are effective potentials. For example, the Coulomb interactions are obtained by neglecting higher order terms that appear in a more exact QED solution. These corrections are important for larger atoms, e.g., to explain the color of gold or the liquidity of mercury. $\endgroup$ May 5, 2014 at 14:10
3
$\begingroup$

I'm not sure whether you're asking for examples of interactions where three particles interact simultaneously, or for systems where pairwise interactions are not adequate to describe the dynamics. So I'll offer you some of both!

Three-particle systems

You're made of the product of a three-particle interaction: the triple-$\alpha$ process, where three helium nuclei fuse to form carbon. The intermediate state, beryllium-8, has a lifetime around 10-17 seconds. This is long by nuclear physics standards — enough time for light to cross the Be-8 nucleus 108 times — but short enough that it takes roughly a solar mass of hot, electron-degenerate helium before there is enough beryllium for carbon production to take place at any realistic rate. This is why essentially no nuclei heavy than mass 7 were synthesized during the big bang. (And if it weren't for the coincidence that carbon has an excited state right near the energy of a 8Be and 4He together at rest, there might be no carbon in the universe at all.) But this is fundamentally a sequence of pairwise interactions, just taking place on very different timescales.

In a time-reversed world, there is recent evidence that iron-45, on the proton drip line, decays to chromium-43 by emitting two protons at once, rather than emitting the two protons sequentially or a bound diproton which decays later.

Of course, most weak decays have three particles in the final state (the daughter nucleus, the beta particle, and the neutrino); however the high-energy theory of the weak force describes this as a sequence of two two-body decays.

Three-body forces

Usually we describe what's happening inside a nucleus in terms of protons and neutrons. However, protons and neutrons are just the lowest-energy bound states of quantum chromodynamics; at very short distance scales, nucleons are made of many constituents, which interact strongly with each other where the nucleons overlap. In precision modeling of nuclei, one finds no pairwise interactions which describe all of the light nuclei. However there are several successful nuclear potentials which contain two- and three-nucleon forces.

You can compare the three-body force in nuclear physics to tidal forces arising in gravitation. The earth's tidal bulges arise because people with the moon overhead fall towards the moon faster than the rest of the earth does, while people with the moon underfoot feel the earth falling towards the moon away from them. This is a straightforward effect of the purely two-body gravitational force, which has major impacts on the long-term evolution of two-body systems (for instance, the moon's motion away from the earth as the pair become tidally locked, or the destruction of satellites within the Roche limit), and which vanishes if you use the otherwise-useful approximation that the objects involved are point masses. You could say in this case that gravitational tides are a "three-body" force that arises because there internal degrees of freedom to the earth-moon system that you'd prefer to suppress in your model.

The nuclear three-body force is the same sort of thing — except that unlike the tides, we don't actually have a quantitative theory for what those internal degrees of freedom are. I suspect that this is also what's happening in the interatomic potentials mentioned by Arnold. A complicated object like a nucleon or an atom may simply be a little different when it has close neighbors than when it's alone, and the sum of a two- and three- (and higher-)body interactions is a compact was to account for the difference.

$\endgroup$
4
  • $\begingroup$ The Three-Nucleon Force Is Not Made by Nature says those three-body interactions aren't real, but simply a theoretical artifact. $\endgroup$
    – juanrga
    Apr 13, 2019 at 12:09
  • $\begingroup$ @juanrga From 1984. Our understanding of the nuclear environment has advanced a lot since then, and the consensus is still that we make more accurate predictions if we include three-nucleon forces. The analogy to gravitational tides (which arise from purely two-body forces in many-body systems) is valuable. $\endgroup$
    – rob
    Apr 13, 2019 at 13:43
  • $\begingroup$ Yes, but the article doesn't say that they don't give better predictions, the article says that those forces aren't fundamental. $\endgroup$
    – juanrga
    Apr 15, 2019 at 17:47
  • $\begingroup$ @juanrga I think that "what's the relationship between this reference and the contemporary understanding of three body forces in nuclear physics?" would make a good follow-up question. I'm not really equipped to quickly give you an overview of the current literature, and if I were, a comment thread wouldn't be the place to do it. $\endgroup$
    – rob
    Apr 15, 2019 at 23:12
2
$\begingroup$

You start your question with a fundamental statement from mechanics, and end with an example from a very complex system. Newtonian mechanics fundamentally involves only pair-wise forces at a microscopic level. When you build up a more complex system, like that of a catalyzed chemical reaction, you hide the very complicated microscopic phenomena by introducing higher levels of abstraction. For example, we don't discuss the individual collisions of a gas against a piston, instead we discuss pressure.

I seem to recall that someone (Mach?) has shown that interactions that require three objects (triple-wise potentials?) break Newtonian mechanics. That's just a hazy memory though. Furthermore, a three-body potential that is not the sum of pair-wise potentials would break superposition of forces. In any event, restricting attention to (microscopic) pair-wise potentials seems to work in all cases I'm aware of. @ArnoldNeumaier in his answer suggests that N-wise potentials work. I'd like to be corrected if that is true.

$\endgroup$
1
  • $\begingroup$ On the level of nuclei and eelectrons, the pairwise Coulomb interaction is an excellent approximation (though not perfect). On the level of atoms, triple interactions of Axilrod-Teller type are already necessary for high quality models. Both levels are microscopic. $\endgroup$ Apr 25, 2014 at 7:56
1
$\begingroup$

The only three-particle interaction I know of is a chemical reaction which needs a catalyst.
Which other specific examples of three- or n-particle interactions are there?

The most famous example is probably $\mathrm{H_2 + I + I \to 2HI}$

Max Bodenstein proposed in 1894 this termolecular mechanism. It is confirmed that this occurs. See this article by James Anderson.

$\endgroup$
0
$\begingroup$

Generally speaking, this is a property obtained from the linearity of the problem allowing for a superposition principle, which will breakdown for nonlinear problems. E.g., particle hydrodynamic interactions can be described as above in a Stokes flow, but cannot in a Navier-Stokes flow or viscoelastic flow.

$\endgroup$
1
  • $\begingroup$ Thanks, Joce. But how can those interactions be - positively - described? (You just say how the can not be described.) $\endgroup$ Apr 18, 2014 at 8:59
0
$\begingroup$

According to an "expert" many-particle interactions e.g. in water have to do with

“[...] the cooperative effects of the H bonds. [...]it is because of these effects that the “dipole” moment of a molecule in condensed phase is greatly enhanced compared to that of a molecule in gas phase [...]”

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.