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?
Best Answer
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
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.