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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?

Qmechanic
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Webfarer Escape
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1 Answers1

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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.

Qmechanic
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    Thank you! This is one of the clearest answers I've ever received to a physics question of mine. – Webfarer Escape Apr 05 '12 at 13:04
  • @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? – GRrocks Dec 26 '18 at 13:13
  • 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. – Qmechanic Dec 26 '18 at 13:45
  • @GRrocks. I'm reviving this because I'm reading along to Taylor's book and I had a similar thought when he said that internal interaction energies are constant (by definition) in rigid bodies, and thus we might suspect that taking the negative of the gradient would lead to a force that vanishes. But we need to be clear about the gradient used. The force on a particle is the gradient with respect to its coordinates of its potential function, this is not constant (ie. $U(|\mathbf{r_1}-\mathbf{r_2}|$ is not a constant with respect to $\mathbf{r_1}$ but rather to $|\mathbf{r_1}-\mathbf{r_2}|$). – EE18 Apr 29 '20 at 17:27
  • @Qmechanic is this what you were getting at? – EE18 Apr 29 '20 at 17:28
  • @1729_SR Yes, that's exactly it. The potential is a constant in the separation, not individual positions. – GRrocks Apr 30 '20 at 06:07