2
$\begingroup$

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 $V_{12}+V_{21}$

$\endgroup$
  • 1
    $\begingroup$ Yes. You are double counting. $\endgroup$ – Superfast Jellyfish Jan 21 at 14:02
  • 1
    $\begingroup$ Yes, you should indeed present the case of $N=2$ to your lecturer to illustrate your point. $\endgroup$ – Ruslan Jan 21 at 15:32
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thank you very much for your response! I'm not quite sure if I'm drawing the right conclusions from your answer: From my understanding, you have shown that if we double count potentials then we double count force and hence my approach is correct? $\endgroup$ – Leon231000 Jan 21 at 14:24
  • 1
    $\begingroup$ That's right. I should have made that clearer. The point is, by the end of the calculation it's clear we have double counted the forces, and that happened because we double counted the potentials. $\endgroup$ – Charles Hudgins Jan 21 at 14:36
  • $\begingroup$ Amazing! thank you so much! $\endgroup$ – Leon231000 Jan 21 at 14:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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