2

For a system of two interacting particles 1, 2 we get from the conservation of momentum $$ \dot{\bf{p_1}} + \dot{\bf{p_2}} = 0$$ and from conservation of angular momentum $$ \bf{r_1} \times \dot{\bf{p_1}} + \bf{r_2} \times \dot{\bf{p_2}} = 0$$ so $$ \bf{F_1} = -\bf{F_2} $$ and $$ \bf{r_{12}} \times \bf{F_1} = 0 $$ where $ \bf{r_{12}} = \bf{r_1} - \bf{r_2} $ is the separation between particles. So basically only central forces $\bf{F}=\bf{F}\left(|\bf{r_{12}}|\right)$ are allowed.

But for a system of three and more interacting particles we don't have such restrictions. E.g. for 3 particles $$ \bf{F_3} = -\bf{F_1}-\bf{F_2}$$ and $$ \bf{r_{13}} \times \bf{F_1} + \bf{r_{23}} \times \bf{F_2} = 0 $$ (two more likewise expressions for (21,31) and (12,32)). And this is, I believe, as far as one can get in general case.

We can expand $$ \bf{F_1} = \bf{F_{1,2}} + \bf{F_{1,3}} + \bf{R_1} $$ where $\bf{F_{1,2}}$ and $\bf{F_{1,3}}$ are for two-body interactions, but no one forbids the existence of 3-body force $\bf{R_1}\left(\bf{r_{12}}, \bf{r_{23}}, \bf{r_{31}}\right)$.

Are there any known examples or I am missing something?

xaxa
  • 1,582
  • 3
    While not known in the fundamental forces, three body forces do arise in certain effective approximations. For example, the presence of a molecule will polarize two neighbouring molecules, altering their interaction with each other, resulting in an effective three body interaction. This plays a role in scattering and cluster formation in dense gases and liquids for example. Another example is an effective three body force between nucleons, arising in the same sort of way, which is important in nuclear structure calculations. So it's basically short range interactions in dense systems. – Michael Oct 21 '13 at 15:52
  • If I understand what's the question (maybe you could clarify), I think it's a bit strange to assume the conservation of linear/angular momentum only for a system of $N$ points and to refuse to assume it for a system of $2$ points. Isn't it? – pppqqq Oct 21 '13 at 15:53
  • 1
    Related: http://physics.stackexchange.com/a/13657/2451 – Qmechanic Oct 21 '13 at 15:57
  • If the forces act over a non-instantaneous time, the time derivative should be over the cross product of position and momentum, not just momentum. – Rick Sep 29 '16 at 12:08

0 Answers0