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


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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    $\begingroup$ 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. $\endgroup$ – Ron Maimon Apr 2 '12 at 3:44
  • $\begingroup$ If you can, please explain what are the center of mass considerations that would make the statement correct. Thank you for the comment! $\endgroup$ – Webfarer Escape Apr 5 '12 at 13:19

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


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

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


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

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 $$ 2\sum_{i\neq j}\vec{F}_{ij}\cdot\delta\vec{r}_{j} ~\stackrel{(1)}{=}~ \sum_{i\neq j}(\vec{F}_{ij}-\vec{F}_{ji})\cdot\delta\vec{r}_{j} ~=~\sum_{i\neq j}\vec{F}_{ij}\cdot\delta\vec{r}_{j} -\sum_{i\neq j}\vec{F}_{ji}\cdot\delta\vec{r}_{j}$$ $$~\stackrel{i\leftrightarrow j}{=}~ \sum_{i\neq j}\vec{F}_{ij}\cdot \delta\vec{r}_{j} -\sum_{i\neq j}\vec{F}_{ij}\cdot \delta\vec{r}_{i} ~=~\sum_{i\neq j}\vec{F}_{ij}\cdot (\delta\vec{r}_{j}-\delta\vec{r}_{i})$$ $$~\stackrel{(6)}{=}~ \sum_{i\neq j}\vec{F}_{ij}\cdot \delta\vec{r}_{ij} ~\stackrel{(7)}{=}~0,\tag{8} $$

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.

  • $\begingroup$ 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. $\endgroup$ – Webfarer Escape Apr 5 '12 at 11:09
  • $\begingroup$ I updated the answer. Note that eq. numbers have changed. $\endgroup$ – Qmechanic Apr 5 '12 at 12:40
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    $\begingroup$ Thank you! This is one of the clearest answers I've ever received to a physics question of mine. $\endgroup$ – Webfarer Escape Apr 5 '12 at 13:04
  • $\begingroup$ @Qmechanic, I have a silly question. These internal forces, if satisfying the strong form of 3rd law, are supposed to be derivable from an internal potential that only depends on the separation between particles. In a rigid body, the separation is a constant and thus the potential too. Shouldn't this mean that the internal forces vanish? What am I missing? $\endgroup$ – GRrocks Dec 26 '18 at 13:13
  • $\begingroup$ An internal 2-body force that obeys the strong Newton's 3rd law may still depend on the angular orientation and may therefore have no potential. $\endgroup$ – Qmechanic Dec 26 '18 at 13:45

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