# 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?

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You haven't missed something--- this is incorrect as stated. There are several ways to make it a correct statement, depending on center of mass considerations, or rigidity. You should give the context of the derivation. – Ron Maimon Apr 2 '12 at 3:44
If you can, please explain what are the center of mass considerations that would make the statement correct. Thank you for the comment! – Hiran Apr 5 '12 at 13:19

I) Let us first recall Newton's third law. 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 $\vec{r}_j$,

$$\vec{f}_{ij}+\vec{f}_{ji}~=~\vec{0}.\qquad\qquad(1)$$

The strong Newton's Third law says besides eq. $(1)$ that the forces are also collinear,

$$\vec{f}_{ij} ~\parallel ~\vec{r}_{ij},\qquad\qquad(2)$$

i.e. parallel to the difference in positions

$$\vec{r}_{ij}~:=~\vec{r}_j-\vec{r}_i.\qquad\qquad(3)$$

II) The strong Newton's third law is by itself not enough to ensure that the double sum

$$\sum_{i\neq j}\vec{f}_{ij}\cdot\delta\vec{r}_{j} \qquad\qquad(4)$$

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

$$0~=~\delta|\vec{r}_{ij}|^2~=~2\vec{r}_{ij}\cdot \delta\vec{r}_{ij}, \qquad\qquad(5)$$

where

$$\delta\vec{r}_{ij}~\stackrel{(3)}{=}~\delta(\vec{r}_j-\vec{r}_i)~=~\delta\vec{r}_j-\delta\vec{r}_i. \qquad\qquad(6)$$

Equations $(2)$ and $(5)$ then imply that

$$0~=~\vec{f}_{ij}\cdot \delta\vec{r}_{ij}.\qquad\qquad(7)$$

Then the double sum $(4)$ vanishes $$2\sum_{i\neq j}\vec{f}_{ij}\cdot\delta\vec{r}_{j} ~\stackrel{(1)}{=}~ \sum_{i\neq j}(\vec{f}_{ij}-\vec{f}_{ji})\cdot\delta\vec{r}_{j} ~=~\sum_{i\neq j}\vec{f}_{ij}\cdot\delta\vec{r}_{j} -\sum_{i\neq j}\vec{f}_{ji}\cdot\delta\vec{r}_{j}$$ $$~\stackrel{i\leftrightarrow j}{=}~ \sum_{i\neq j}\vec{f}_{ij}\cdot \delta\vec{r}_{j} -\sum_{i\neq j}\vec{f}_{ij}\cdot \delta\vec{r}_{i} ~=~\sum_{i\neq j}\vec{f}_{ij}\cdot (\delta\vec{r}_{j}-\delta\vec{r}_{i})$$ $$~\stackrel{(6)}{=}~ \sum_{i\neq j}\vec{f}_{ij}\cdot \delta\vec{r}_{ij} ~\stackrel{(7)}{=}~0,\qquad\qquad(8)$$

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.

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Can you please elaborate on how the double sum with $\delta\vec{r}_j$ in it (i.e. $\sum_{i\neq{j}}\vec{f}_{ij}\centerdot\delta\vec{r}_j$), can be taken to be equivalent to that with $\delta\vec{r}_{ij}$ in it (eq. 6)? Thank you for your answer and for your constructive edits to both my question and to your answer. – Hiran Apr 5 '12 at 11:09
I updated the answer. Note that eq. numbers have changed. – Qmechanic Apr 5 '12 at 12:40
Thank you! This is one of the clearest answers I've ever received to a physics question of mine. – Hiran Apr 5 '12 at 13:04