The meaning of the time derivative of potential energy

Question

Here's the link: http://www.feynmanlectures.caltech.edu/I_13.html

\begin{equation} \label{Eq:I:13:16} \frac{d}{dt}\sum_i\biggl(\sum_j-\frac{Gm_im_j\mathbf r_{ij}}{r_{ij}^3}\biggr)\cdot \mathbf v_i= \sum_{\text{pairs}}\biggl[ \frac{Gm_im_j\mathbf{r}_{ij}}{r_{ij}^3}\cdot \mathbf{v}_i+ \frac{Gm_jm_i\mathbf{r}_{ji}}{r_{ji}^3}\cdot\mathbf{v}_j\biggr] \tag{13.16} \end{equation}

Look at equation 13.16 and the explanation below it, which I am copying here:

In Eq. (13.16), on the other hand, $\sum_{\text{pairs}}$ means that given values of $i$ and $j$ occur only once. Thus the particle pair 1 and 3 contributes only one term to the sum. To keep track of this, we might agree to let $i$ range over all values 1, 2, 3, …, and for each $i$, let $j$ range only over values greater than $i$. Thus if $i=3$, $j$ could only have values 4, 5, 6, … But we notice that for each $i,j$ value there are two contributions to the sum, one involving $\mathbf{v}_i$, and the other $\mathbf{v}_j$, and that these terms have the same appearance as those of Eq. (13.15), where all values of $i$ and $j$ (except $i=j$) are included in the sum. Therefore, by matching the terms one by one, we see that Eqs. (13.16) and (13.15) are precisely the same, but of opposite sign, so that the time derivative of the kinetic plus potential energy is indeed zero.

Now what I don't get is how do the terms of equation 13.16 and equation 13.15 cancel out exactly? There is a $\mathbf{v}_j$ term in 13.16 and there is no $\mathbf{v}_j$ term in 13.15. So how can they be cancelled out to become zero?

Answer 1

It's a shorthand sort of reasoning.

So this is a useful exercise in general. What was actually discovered is that you have this difference between two terms $D_{ij}$ and that difference satisfies the "antisymmetry condition",$$D_{ij} = -D_{ji}.$$ Consider first a sum over all of both indices, $$S = \sum_{i=0}^N\sum_{j=0}^N D_{ij}.$$You can rewrite this for free as two half-sums,$$S = \frac12 \sum_{i=0}^N\sum_{j=0}^N D_{ij} + \frac12 \sum_{i=0}^N\sum_{j=0}^N D_{ij},$$and then in any one of those sums, $i$ and $j$ are just abstract symbols and the answer must not change if we were to swap those symbols physically $i \leftrightarrow j$, giving $$S = \frac12 \sum_{i=0}^N\sum_{j=0}^N D_{ij} + \frac12 \sum_{j=0}^N\sum_{i=0}^N D_{ji},$$and by the distributive property of multiplication over addition we can rearrange these two summation signs on the right and distribute the factor of 1/2 over it too, so the summation is just $$S = \sum_{i=0}^N\sum_{j=0}^N\left(\frac12 D_{ij} + \frac12 D_{ji}\right),$$ at which point we recognize that we know the answer: if $D_{ij} = -D_{ji}$ then by straightforward rearrangement of that equation, $D_{ij} + D_{ji} = 0.$ So $S = \sum_{ij} 0 = 0.$

Well what you've got is not a sum over all $i,j$, it is the sum over a set $P$ of $(i,j)$ pairs. We can write this sum with the "is in" symbol $\in$ as $$S = \sum_{(i, j) \in P} D_{ij}.$$But in your mind go through that argument again: it is not hard to see the condition on $P$ which makes this happen:

Call the summation domain $P$ symmetric if whenever some pair $(i, j) \in P,$ then also it turns out that $(j, i)\in P.$ Any sum of an antisymmetric matrix over a symmetric domain is zero.

What Feynman is observing is this "all $(i, j)$ with $i\ne j$" domain is symmetric and hence the sum must be zero. He just does the rewriting in his head; he sees $$S = \sum_{(i,j)\in P} \left(A_{ij} - A_{ji}\right),$$and he immediately realizes that since $P$ is symmetric he can exchange $i\leftrightarrow j$ on the right hand side to get $\sum_{ij}\left( A_{ij} - A_{ij}\right) = \sum 0 = 0.$