# Double counting potentials in $N$-Body Problem?

I suspect an error in my “Classical Dynamics” lecture notes but my lecturer doesn’t agree with me so I need your help!

We assume that in the $$n$$-body problem, the force between particle $$i$$ and $$j$$ can be derived from a potential i.e

$$\vec{F_{ij}}=-\vec{\nabla_i}V_{ij}$$

where $$\vec{\nabla_i}$$ is the gradient with respect to position vector $$\vec{r_i}$$

$$V$$ would usually only depend on $$\Vert\vec{r_i}-\vec{r_j}\Vert$$

My lecturer then defines the total energy of the $$n$$-body system as

$$E=\sum_i \frac{1}{2}m_i\vert\vec{\dot{r_i}}\vert^2 + \sum_{ij}V_{ij}$$

However, I am convinced that this is wrong since we’d be double counting potentials. My lecturer says we should indeed be double counting potential. In my opinion, the expression should read

$$E=\sum_i \frac{1}{2}m_i\vert\vec{\dot{r_i}}\vert^2 + \sum_{1\le{i}\lt{j}\le{n}}V_{ij}$$

For the second equation we also get:

$$\dot{E}=\sum_im_i\vec{\dot{r_i}}\bullet\vec{\ddot{r_i}} + \sum_{1\le{i}\lt{j}\le{n}}\vec{\nabla_i}V_{ij} \bullet \vec{\dot{r_i}} + \vec{\nabla_j}V_{ij} \bullet \vec{\dot{r_j}} = \sum_im_i\vec{\dot{r_i}}\bullet\vec{\ddot{r_i}} + \sum_{i}\sum_{j\neq{i}} \vec{\nabla_i}V_{ij} \bullet \vec{\dot{r_i}} = \sum_i \vec{\dot{r_i}} \bullet \Bigl(m_i \vec{\ddot{r_i}} + \sum_{j\neq{i}} \vec{\nabla_i}V_{ij} \Bigr) = \sum_i \vec{\dot{r_i}} \bullet \Bigl( m_i \vec{\ddot{r_i}} - \vec{F_{ij}^{tot}} \Bigr) = 0$$

So this seems reasonable to me. Is this actually correct?

We also have an example of the two-body problem in the notes where we only use $$V_{12}$$ as the potential, not $$V_{12}+V_{21}$$

• Yes. You are double counting. – Superfast Jellyfish Jan 21 at 14:02
• Yes, you should indeed present the case of $N=2$ to your lecturer to illustrate your point. – Ruslan Jan 21 at 15:32

I'll be using $$\vec{q}_i$$ instead of $$\vec{r}_i$$. Assuming $$V_{ij}$$ depends only on $$\vec{q_i}$$ and $$\vec{q_j}$$, one has \begin{align} \vec{F}_{net, \ i} &= \dot{\vec{p}_i} = -\frac{\partial E}{\partial \vec{q}_i} = \sum_{j,k = 1}^n -\frac{\partial V_{jk}}{\partial \vec{q}_i} \\&= \sum_{k = 1}^n -\frac{\partial V_{ik}}{\partial \vec{q}_i} -\frac{\partial V_{ki}}{\partial \vec{q}_i} = \sum_{k = 1}^n \vec{F}_{ik} - \frac{\partial V_{ki}}{\partial \vec{q}_i} \end{align} We haven't defined the latter term in this last expression yet. If we add in the reasonable requirement that $$V_{ij} = V_{ji}$$, then we find \begin{align} \vec{F}_{net, \ i} &= \sum_{k = 1}^n \vec{F}_{ik} - \frac{\partial V_{ki}}{\partial \vec{q}_i} \\&= \sum_{k = 1}^n \vec{F}_{ik} - \frac{\partial V_{ik}}{\partial \vec{q}_i} \\&= \sum_{k = 1}^n \vec{F}_{ik} + \vec{F}_{ik} \\&= 2 \sum_{k = 1}^n \vec{F}_{ik} \end{align} and we see that we have indeed double counted.