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The textbook I'm currently reading (Engineering Dynamics by Paley and Kasdin) states the internal work done on a system of $N$ particles (that is, the work done by all the internal forces in the system) is as follows:

$$\sum_{i=1}^N\sum_{j=1}^N\int_{C_i}\vec {F_{i, j}}\cdot d\vec{r_{i/o}}$$

where $C_i$ is the path that particle $i$ traverses and $\vec{F_{i,j}}$ is the internal force exerted by particle $j$ on particle $i$.

Now, it also claims that the expression above is equivalent to the following expression:

$$\frac{1}{2}\sum_{i=1}^N\sum_{j=1}^N\int_{C_i}\vec {F_{i, j}}\cdot d\vec{r_{i/j}}$$

My question is, how did the author get from the first expression to the other? The textbook skips the math but I still wanted to know how you get this result.

I'm pretty sure it uses the fact that

$$\sum_{i=1}^N\sum_{j=1}^N\int_{C_i}\vec {F_{i, j}}\cdot d\vec{r_{i/o}}=\sum_{i=1}^N\sum_{j=1}^N\int_{C_i}\vec {F_{j, i}}\cdot d\vec{r_{j/o}}$$

but I am not sure how to proceed from here.

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Here is an illustration that may help you see the concept:

internal work of two-body system

That square symbol was supposed to be dot product. Anyway, this shows two of the system's bodies undergoing a small displacement. They are shown at the same initial position but that doesn't matter, it just lets us see the vector relationships. The work is the sum of the dot product for the blue body and for the red body, but since the forces are equal and opposite by Newton's 3rd law, it works out to be a single dot product with the difference of the displacements, which is also the relative displacement of the two bodies! Notice how you only have to include that dot product once and it takes care of both bodies - that's why there's a factor of $\frac12$; otherwise you would be counting the pair twice, since the nested sum includes all body pairs in both orders.

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  • $\begingroup$ Thanks for the answer! In that case, what path would you integrate F•dr_1/2 over to get the total work? $\endgroup$ – roflmfao Aug 22 '20 at 21:17
  • $\begingroup$ @roflmfao The path is $\vec r_{1/2}(t)$ - the relative path between the two bodies - how their separation vector changes over time. You might say we're treating body #2 as the origin the whole time, but that's a little misleading - body #2 could be accelerating, so it's not an inertial reference frame - it's only the 3rd law that makes it work out that way. I think the moral is that, when you want to find an internal property of a system, it's generally going to be most easily expressed in terms of relative displacements, velocities, etc. between the parts of the system. $\endgroup$ – Adam Herbst Aug 23 '20 at 16:15
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As far as I understand from notation, $\vec{r}_{i/o}$ is the position vector of the $i$th particle and $\vec{r}_{j/o}$ is the position vector of the $j$th particle with respect to the reference frame $o$. Then, he writes the infinitesimal displacements as $\vec{dr}_{i/o}$ and write it in the integral which is the first equation. Then, he defines the vector $\vec{r}_{i/j}$ as the vector from $i$th particle to $j$th particle, or in other words $\vec{r}_{i/j} = \vec{r}_{j/o}- \vec{r}_{i/o}$. The infinitesimal displacement vector becomes $\vec{dr}_{i/j}$. Note that since we are talking about infinitesimal displacements, those two vectors are exactly the same in magnitude (infinitesimal) and pointing towards the same way. Therefore you can write the equation in second form.

Note: We may need to assume the space to be isotropic (which is) etc. to say that but it is classical physics after all and hey, I am a physicist so no problem.

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