Force on the $i$th body due to the $j$th body, and three-body interaction

Question

I am currently studying Classical Mechanics, 5th edition, by Kibble and Berkshire. Chapter 1.2 Newton's Laws says the following:

If we denote the force on the ith body due to the jth body by $\mathbf{F}_{ij}$, then

$$\mathbf{F}_i = \mathbf{F}_{i1} + \mathbf{F}_{i2} + \dots + \mathbf{F}_{iN} = \sum_{j = 1}^N \mathbf{F}_{ij}, \tag{1.2}$$

where of course $\mathbf{F}_{ii} = \mathbf{0}$, since there is no force on the $i$th body due to itself. Note that since the sum on the right side of (1.2) is a vector sum, this equation incorporates the 'parallelogram law' of composition of forces. The two-body forces $\mathbf{F}_{ij}$ must satisfy Newton's third law, which asserts that 'action' and 'reaction' are equal and opposite,

$$\mathbf{F}_{ji} = - \mathbf{F}_{ij}. \tag{1.3}$$

Moreover, $\mathbf{F}_{ij}$ is a function of the positions and velocities (and internal structure) of the $i$th and $j$th bodies, but is unaffected by the presence of other bodies. (It can be argued that this is an unnecessarily restrictive assumption. It would be perfectly possible to include also, say, three-body forces, which depend on the positions and velocities of three particles simultaneously. However, within the realm of validity of classical mechanics, no such forces are known, an deter inclusion would be an inessential complication.)

I found this last part confusing:

Moreover, $\mathbf{F}_{ij}$ is a function fo the positions and velocities (and internal structure) of the $i$th and $j$th bodies, but is unaffected by the presence of other bodies. (It can be argued that this is an unnecessarily restrictive assumption. It would be perfectly possible to include also, say, three-body forces, which depend on the positions and velocities of three particles simultaneously. However, within the realm of validity of classical mechanics, no such forces are known, an deter inclusion would be an inessential complication.)

I don't understand how it is reasonable to assume that, if three bodies are in close proximity, then only two 'affect' each other, and the other doesn't. This is clearly not an accurate generalization of the physical world.

I would greatly appreciate it if people would please take the time to clarify this.

Answer 1

The statement is that the force $F_{ij}$ depends on the position and velocities of objects $i$ and $j$, that is, $r_i,r_j,\dot r_i$ and $\dot r_j$. They then go on to say that in principle, one could remove this restriction, and have the force between objects $i$ and $j$ also depend on a third object $k$, e.g. $F_{ij} = F_{ij}(r_i,r_j,\dot r_i,\dot r_j,r_k,\dot r_k)$.

Note this is distinct from having three objects, and considering the force between $i$ and $j$, $j$ and $k$ and so on. Rather, you have a "three-body force" that does not arise when two objects are in the system, but rather when three objects are present and the force on one depends simultaneously on the two other objects.$^\dagger$

Obviously we do not encounter this in classical mechanics, which is why it is not considered in the book. They are careful to point out this notion doesn't arise in classical mechanics, as it has been posited in other areas of physics like particle physics, e.g. a three-nucleon force.

---

$\dagger$ To further clarify, consider Newton's law modified so that masses $m_i$ and $m_j$ experience no force, but when a third mass enters the picture, mass $m_i$ experiences a force dependent on $m_j$ and $m_k$.

Answer 2

You are actually having a little confusion about the forces involved in a three body system. You are wrong in your understanding that it has been 'assumed' that ''if three bodies are in close proximity, only two 'affect' each other, and the other doesn't.'' This is not right.

What the author says is that there is no such force in the realm of classical mechanics which is a quantity that simultaneously depends on the positions and velocities of three particles. If there would be so, you could then write such a force in the form $\textbf{F}_{ijk}=\textbf{F}_{ijk}(\textbf{r}_i,\textbf{r}_j,\textbf{r}_k,\textbf{v}_i,\textbf{v}_j,\textbf{v}_k)$ , [three indices for three particles]. This would be the force on the $i^{th}$ particle due to the $j^{th}$ and $k^{th}$ particles and would be some equation involving all those variables. It would then be a so called 'three-body force'.

However, in the realm of classical mechanics, all forces are 'two-body' forces which means that it is of the form $\textbf{F}_{ij}=\textbf{F}_{ij}(\textbf{r}_i,\textbf{r}_j,\textbf{v}_i,\textbf{v}_j)$ , [two indices for two particles]. Now to clarify this, let us take for simplicity a system of three particles. Then the different forces involved in the system will be $\textbf{F}_{12}$,$\textbf{F}_{21}$, $\textbf{F}_{13}$, $\textbf{F}_{31}$, $\textbf{F}_{23}$ and $\textbf{F}_{32}$ . Hence, the force on say, the 1st particle will be the combined force on it due to the second and the third particles, i.e., $\textbf{F}_{1j}=\textbf{F}_{12}+\textbf{F}_{13}$ . So, here the effect of all the forces on the first particle has been taken into account. None of them has been neglected as you were thinking by mistake. Only, the way of taking all of them into account is the vector addition of the mutual 'two-body' forces and not a single 'three-body force'. This applies for more than three particles as well. Hope you have understood.

Answer 3

There are already two good answers but I have the feeling that something is still missing. In fact, consider a system of more than 2 particles. Formally, the force on the particle $1$ is

$$\textbf{F}_{\text{1 due to 2..N}}=\textbf{f}(\textbf{r}_1|\textbf{r}_2,..,\textbf{r}_N) \, .$$

In general, this is a non-linear function of the position $\textbf{r}_1$ given the positions $\textbf{r}_2,..,\textbf{r}_N$. Apart from the positions, you may also want to consider the velocities or other degrees of freedom (spin maybe) of the particles, but this is not fundamental to discuss the point.

Therefore, $\textbf{F}_{\text{1 due to 2..N}}$ looks like a $N$-body force, but typically $\textbf f$ can be decomposed as

$$ \textbf{f}(\textbf{r}_1|\textbf{r}_2,..,\textbf{r}_N) = \sum_{j=2...N} \textbf{q}(\textbf{r}_1|\textbf{r}_j) $$

If you can do this, then $\textbf f$ is not a "genuine" $N$-body force, but rather it is just the total force due to the sum of 2-body interactions.

You are forced to consider 3-body forces if you discover that the above decomposition is impossible (or if you discover that you cannot reproduce the observed dynamics of your system bu considering only two-body interactions). If this is the case, your physical system may be well described by considering a more general decomposition:

$$ \textbf{f}(\textbf{r}_1|\textbf{r}_2,..,\textbf{r}_N) = \sum_{j=2...N} \textbf{q}(\textbf{r}_1|\textbf{r}_j) + \sum_{i<j=3...N} \textbf{h}(\textbf{r}_1|\textbf{r}_i,\textbf{r}_j) $$

In this case the $ \textbf{h}$ are your "genuine three body forces", i.e. it is impossible to split $ \textbf{h}$ as $ \textbf{h}(1|ab)=\textbf{m}(1|a)+\textbf{m}(1|b)$, where $\textbf{m}$ is a two-body interaction (possibly different from $\textbf{q}$).

Now, the question why terms like $\textbf{h}(1|ab)$ are theoretically possible but "not observed" (if someone knows examples that do not involve QCD or quantum stuff please comment) observed in the "macroscopic" world is interesting and I do not have a complete answer for that. However, my feeling is that it has something to do with the kind of interactions that mediate the "action at a distance". For example, in celestial mechanics the gravitational potential is a linear superposition of the single-particle potentials: this ensures the two-body nature of the gravitational attraction. The same for Coulomb forces. Since General Relativity is non-linear, I suspect that there could be some post-Newtonian correction to the motion of a system of "planets" that gives rise to tree-body forces (but maybe those corrections are so small that are never considered).