A general potential energy for a multi-particle system

[etotheipi]

For a system of two or more particles, it is customary to define potential energy functions $V_{ij}$ between pairs of particles, so that the total conservative force (not necessarily total) on any given particle is $$\mathbf{F}_i = \sum_{j\neq i} -\nabla_i V_{ij}$$as a sum over all other particles $j$, evaluated at the position of $i$. Also, $V_{ij} = V_{ij}(\mathbf{r}_i - \mathbf{r}_j)$ depends only on the relative separation of both particles.

I wondered if it were ever possible to find a total potential energy function, which would have the property that if $V = V(\mathbf{r}_1, \mathbf{r}_2, \dots \mathbf{r}_n)$, then that same force $\mathbf{F}_i$ would be $$\mathbf{F}_i = -\nabla_i V$$ for any choice of particle $i$? $\nabla$ is a linear operator so it seems reasonable to suggest $V = V_{12} + V_{13} + \dots + V_{n-1, n}$, but I'm not sure if it is possible to evaluate the derivatives of e.g. the potential energy between particles 3 and 4, at the position of particle 2 (e.g. does $\nabla_2 V_{34}$ make sense?), although this would have to be zero.

Thanks!

 

[troglodyte]

I would say no because of this pairwise description which you have stated above.Maybe some combinatorial arguments are missing to expand the concept to include also such cases.

 

[anuttarasammyak]

From the two equations of OP, It V should be
$$V = \frac{1}{2}\sum_{i} \sum_{j,j \neq i}V_{ij}$$
As for your concern
$$\nabla_iV_{jk}=0$$
for $i \neq j,k$

 

[etotheipi]

From the two equations of OP, It V should be
$$V = \frac{1}{2}\sum_{i} \sum_{j,j \neq i}V_{ij}$$
As for your concern
$$\nabla_iV_{jk}=0$$
for $i \neq j,k$

I agree, this is what I meant by $V = V_{12} + V_{13} + \dots + V_{n-1, n}$.

If I consider a 3 particle system s.t. $V = V_{12} + V_{13} + V_{23}$ then $$-\nabla_{\mathbf{r}_1}V = -\nabla_{\mathbf{r}_1}V_{12} -\nabla_{\mathbf{r}_1}V_{13} -\nabla_{\mathbf{r}_1}V_{23} = \mathbf{F}_{12} + \mathbf{F}_{13} + \mathbf{0} = \mathbf{F}_1$$ So in fact I think it does work that $\mathbf{F}_i = -\nabla_i V$

 

[vanhees71]

For a system of two or more particles, it is customary to define potential energy functions $V_{ij}$ between pairs of particles, so that the total conservative force (not necessarily total) on any given particle is $$\mathbf{F}_i = \sum_{j\neq i} -\nabla_i V_{ij}$$as a sum over all other particles $j$, evaluated at the position of $i$. Also, $V_{ij} = V_{ij}(\mathbf{r}_i - \mathbf{r}_j)$ depends only on the relative separation of both particles.

I wondered if it were ever possible to find a total potential energy function, which would have the property that if $V = V(\mathbf{r}_1, \mathbf{r}_2, \dots \mathbf{r}_n)$, then that same force $\mathbf{F}_i$ would be $$\mathbf{F}_i = -\nabla_i V$$ for any choice of particle $i$? $\nabla$ is a linear operator so it seems reasonable to suggest $V = V_{12} + V_{13} + \dots + V_{n-1, n}$, but I'm not sure if it is possible to evaluate the derivatives of e.g. the potential energy between particles 3 and 4, at the position of particle 2 (e.g. does $\nabla_2 V_{34}$ make sense?), although this would have to be zero.

Thanks!

Indeed you are right. The two-body form is only quite common (e.g., in the astronomical many-body problem). In general nothing fundamental (i.e., symmetry principles of Newtonian mechanics) forbid more complicated $N$-body forces, and indeed that's what has to be taken into account in the nuclear many-body problem. Here are some references:

https://www.physicsforums.com/threa...y-when-carefully-analysed.979739/post-6263544

 

[etotheipi]

@vanhees71 thanks for the reference. There seem to be two distinct ideas here. The first is of a system of N particles interacting via only two body forces s.t. $$V = \frac{1}{2}\sum_{i} \sum_{j,j \neq i}V_{ij}$$with the corresponding relation of $$\mathbf{F}_i = -\nabla_i V = -\sum_{j \neq i} \nabla_i V_{ij}$$and the second idea is one of N particles interacting via an N body force for which $V = V(\mathbf{r}_1, \mathbf{r}_2, \dots, \mathbf{r}_N)$ and cannot be broken down into pairs $V_{ij}$.

It's reasonable perhaps to generalise to a system of $N$ particles interacting via $k$ body forces such that $V = V(\mathbf{r}_1, \mathbf{r}_2, \dots, \mathbf{r}_N) = V_{1,2, \dots, k} + V_{1,3, \dots, k+1} + \dots + V_{N-k+1, N-k+2, \dots, N}$.

 

[etotheipi]

I found another nice explanation here about many body potential energies, such that in a system of interacting particles $$V_{TOT} = \sum_i^N V_1(\vec{r}_i) + \sum_{i<j}^N V_2(\vec{r}_i, \vec{r}_j) + \sum_{i<j<k}^N V_3(\vec{r}_i, \vec{r}_j, \vec{r}_k) + \dots$$ such that $\vec{F}_i = -\nabla_{\vec{r}_i} V_{TOT}$, and $V_2$, $V_3$, etc. arise due to 2/3/etc. particle interactions respectively. Very interesting stuff!

Often any given three particle potential energy is parameterised by $V_3 = V_3(r_{ij}, r_{ik}, \theta_{ijk})$.