A Question about Virtual Work related to Newton's Third Law In describing d'Alembert's principle, the lecture note I was provided with states that the total force $\mathbb F_l$ acting on a particle can be taken as,
$$\mathbb F_l=F_l+\sum_mf_{ml}+C_l,$$
where $F_l$ is the sum of the applied forces on $l^{th}$ particle, $f_{ml}$ being the internal force on $l^{th}$ particle due to an $m^{th}$ particle, and $C_l$ denoting the constraint forces. However considering the law of action and reaction it further states that  $f_{ml}+f_{lm}=0$, which I have no trouble understanding. But considering a virtual displacement $\delta \bf{r}_l$ on $l^{th}$ particle, in the next line it concludes that the virtual work $\delta W$ done should be,
$$\delta W~=~\sum_{l=1}^N(F_l+C_l)\centerdot\delta\textbf{r}_l,$$
ignoring $f_{ml}$ terms. But if we take those terms into account shouldn't it be
$$\delta W~=~\sum_{l=1}^N(F_l+C_l)\centerdot\delta\textbf{r}_l+\sum_l\sum_mf_{ml}\centerdot\delta\textbf{r}_l.$$

In other words, I don't see how comes
$$\sum_l\sum_mf_{ml}\centerdot\delta\textbf{r}_l~\stackrel?=~0.$$

For the purpose of illustrating my problem, consider a system of two particles for which one can expand the above double summation and write
$$f_{11}\centerdot\delta\textbf{r}_1+f_{21}\centerdot\delta\textbf{r}_1+f_{12}\centerdot\delta\textbf{r}_2+f_{22}\centerdot\delta\textbf{r}_2.$$
How can this add up to zero in general? Even if I assume $f_{11}=0$ and $f_{22}=0$, I am left with
$$f_{21}\centerdot\delta\textbf{r}_1+f_{12}\centerdot\delta\textbf{r}_2.$$
Do I have to assume $\delta\textbf{r}_1=\delta\textbf{r}_2$ i.e. that the virtual displacements of the particles correspond to merely a displacement of the system? Or have I missed out on something?
 A: I) Let us first recall Newton's third law.

*

*Definition. The weak Newton's third law says that mutual forces of action and reaction are equal and opposite between two particles at position $\vec{r}_i$ and $\vec{r}_j$,
$$ \vec{F}_{ij}+\vec{F}_{ji}~=~\vec{0}.\tag{1}$$


*Definition. The strong Newton's Third law says besides eq. $(1)$ that the forces are also collinear,
$$\vec{F}_{ij} ~\parallel ~\vec{r}_{ij},\tag{2}$$
i.e. parallel to the difference in positions
$$\vec{r}_{ij}~:=~\vec{r}_j-\vec{r}_i.\tag{3}$$
II) The strong Newton's third law is by itself not enough to ensure that the double sum

$$\sum_{i\neq j}\vec{F}_{ij}\cdot\delta\vec{r}_{j} ~\stackrel{?}{=}~0\tag{4}$$

vanishes. We need an extra assumption, e.g. rigidity. If all the distances $|\vec{r}_{ij}|$ are constrained/fixed (imagine e.g. a rigid body made out of the particles), then all virtual displacements $\delta\vec{r}_i$  must satisfy
$$ 0~=~\delta|\vec{r}_{ij}|^2~=~2\vec{r}_{ij}\cdot \delta\vec{r}_{ij}, \tag{5}$$
where
$$\delta\vec{r}_{ij}~\stackrel{(3)}{=}~\delta(\vec{r}_j-\vec{r}_i)~=~\delta\vec{r}_j-\delta\vec{r}_i. \tag{6}$$
Collinearity $(2)$ and rigidity $(5)$ then imply that
$$ 0~\stackrel{(2)+(5)}{=}~\vec{F}_{ij}\cdot \delta\vec{r}_{ij}.\tag{7}$$
Then the double sum $(4)$ vanishes
$$\begin{align} 
2\sum_{i\neq j}\vec{F}_{ij}\cdot\delta\vec{r}_{j}
~\stackrel{(1)}{=}~&
\sum_{i\neq j}(\vec{F}_{ij}-\vec{F}_{ji})\cdot\delta\vec{r}_{j}\cr
~=~&\sum_{i\neq j}\vec{F}_{ij}\cdot\delta\vec{r}_{j}
-\sum_{i\neq j}\vec{F}_{ji}\cdot\delta\vec{r}_{j}\cr
~\stackrel{i\leftrightarrow j}{=}~&
\sum_{i\neq j}\vec{F}_{ij}\cdot \delta\vec{r}_{j}
-\sum_{i\neq j}\vec{F}_{ij}\cdot \delta\vec{r}_{i}\cr
~=~&\sum_{i\neq j}\vec{F}_{ij}\cdot (\delta\vec{r}_{j}-\delta\vec{r}_{i})\cr
~\stackrel{(6)}{=}~&
\sum_{i\neq j}\vec{F}_{ij}\cdot \delta\vec{r}_{ij}\cr
~\stackrel{(7)}{=}~&0,\end{align} \tag{8}
$$
as we wanted to prove. In the third equality of eq. $(8)$, we renamed the two summation variables $i\leftrightarrow j$ in the second term.
