I understand there're already similar posts, however I still like to check my understanding and resolve some confusion points. Any help would be appreciated!

Given $\mathbf{F} = \dot{\mathbf{p}}$ by the Newton's second law, the D'Alembert's principle states that:

$$(\mathbf{F} - \dot{\mathbf{p}}) \cdot \delta \mathbf{r} = 0\tag{1} $$

The virtual displacement is understood as a tangent vector on the configuration manifold (source1, source2). I understand that the line integral of work can be defined as the sum of dot product of the force field and the tangent vectors along the curve (source). Does it mean $(\mathbf{F} - \dot{\mathbf{p}}) \cdot \delta \mathbf{r}$ is basically something of the same nature as the integrand of the line integral? That is to say, $(\mathbf{F} - \dot{\mathbf{p}}) \cdot \delta \mathbf{r}$ is the infinitestimal summand of the line integral $$\oint (\mathbf{F} - \dot{\mathbf{p}}) \cdot d\mathbf{r}.\tag{2} $$ (My language might be sloppy here, if it's confusing, please let me know!)

Now assuming d'Alembert's principle can be related to somewhat integral form, I wonder how does a single virtual displacement $\delta \mathbf{r}$ is related to the variation in the entire path? Is there always a variation corresponding to an arbitrary single point virtual displacement at a particular time? Wikipedia only explains how the virtual displacement is derived from the variation.

I hope my question makes sense. Basically what I found is that although people talk about the d'Alembert's principle being the differential principle and Least Action principle being the integral principle (Goldstein chapter 2.1 par. 1), I don't really find the symmetry easily. Specifically the Least Action principle seems to have a very intuitive interpretation (nudging the entire path a little), whereas it seems harder to visualise what d'Alembert's principle means, i.e. what is a single virtual displacement at a particular time? I'm trying to connect that to the concept of variation that's more intuitive to me.

Please let me know if my direction is correct. As a self learner, it's a bit hard to verify my understanding sometimes!


2 Answers 2


For me, the D'Alembert principle is simply a mathematical consequence of the Newton's second law.

$\mathbf F = m\mathbf a \implies \mathbf F.\delta \mathbf r = m\mathbf a.\delta \mathbf r$, because if 2 vectors are equal, their dot products with the same vector are also equal.

From that principle, valid for classical mechanics, the Euler-Lagrange equation can be derived when the force is conservative. And from EL equation, the priciple of least action can also be derived. That last part is shown here.

  1. Concerning OP's eq. (1), it's important not to fall into the trap of believing that d'Alembert's principle is a trivial consequence of Newton's 2nd law, cf. e.g. this Phys.SE post.

  2. Secondly, d'Alembert's principle is a statement about infinitesimal virtual displacements, i.e. oversimplied/morally it's a statement about that ${\bf F}_j^{(a)} - \dot{\bf p}_j$ is normal to the constrained submanifold. It is not supposed to be integrated into a contour integral (2) per se.

  3. One may show that d'Alembert's principle implies Lagrange equations, cf. e.g. Ref. 1 and my Phys.SE answer here.

  4. Usually if all generalized forces have generalized potentials, then Lagrange equations can be promoted to Euler-Lagrange equations, i.e. there exists a stationary action principle.


  1. H. Goldstein, Classical Mechanics, Chapter 1.
  • $\begingroup$ Thanks for the answer, and correcting my typo! 1. Yes I kind of understand that and Goldstein mentioned it as well that we're making the assumption that constraint forces do no work. 2. Thanks for explaining that it's not supposed to be integrated. Does that mean d'Alembert's principle basically is a statement about constraint force do no work? (or F - dot{p} is normal to the tangent space in your work). 3. If that's the case, does it mean d'Alembert's principle is no more than an assumption? 4. What does Euler-Lagrange equations differ from the Lagrange equations? $\endgroup$
    – Rui Liu
    Jul 17, 2021 at 20:55
  • $\begingroup$ The answers to these questions can be found in the hyperlinks. $\endgroup$
    – Qmechanic
    Jul 18, 2021 at 18:39

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