I'm having some trouble understanding virtual work and displacement, especially a particular section of Goldstein. I'll use an example to explain my difficulty, but I realize this might be the product of misunderstanding of virtual displacement.

Consider a pendulum whose length $\ell(t)$ varies as a function of time. It has two degrees of freedom: one along the rod, and one along its angular position; we can describe this system with generalized/polar-plane coordinates $r$ and $\phi$ and thus the constraint is holonomic.

As it swings, both $\phi(t)$ and $\ell(t)$ change and thus the displacement $d\mathbf{r}$ is not perpendicular to the constraint force of the rod. However, the virtual displacement $\delta \mathbf{r}$ points along the direction of angular motion, i.e., perpendicular to the rod at all times. Therefore we can dissect the displacement as

$$ d\mathbf{r} = \delta\mathbf{r} + \Delta\mathbf{r},$$

where $\Delta \mathbf{r}$ is a displacement along the direction of the rod, such that $||\Delta \mathbf{r}|| = \Delta \ell$ in that instant of time ($dt = 0$). We saw this decomposition also in this physics.SE answer on virtual displacement.

Now in Goldstein (section 1.4, p. 17) we have the claim $$ \sum_{i} \mathbf{F}_i \cdot \delta\mathbf{r}_i = 0,$$ where $\mathbf{F}_i = \mathbf{F}_{a,i} + \mathbf{f}_i$ is the total force, $\mathbf{F}_{a,i}$ is the applied force, and $\mathbf{f}_i$ is the constraint force.

Thus we could expand our sum for every $i$th particle to see that

$$ \sum_i \left[ \left(\mathbf{F}_{a,i} \cdot \delta\mathbf{r}_i\right) + \left(\mathbf{f}_i \cdot \delta\mathbf{r}_i\right)\right] = 0.$$

And this is zero by Goldstein's claim that the system in equilibrium. It is easy for me to see that $\mathbf{f}_i \cdot \Delta\mathbf{r}_i$ is nonzero because those vectors are parallel, and likewise that $\mathbf{f}_i \cdot \delta\mathbf{r}_i = 0$ because these vectors are orthogonal.

However, it is not clear to me (at least geometrically) why the other scalar product involving $\mathbf{F}_{a,i}$ is zero. Is this just a glorified statement of energy conservation? How does the original sum (a single term in my example) evaluate to zero for the pendulum?


1 Answer 1


Reading through the section you mention it is clear that Goldstein is using two assumptions to derive your last equation.

(1) The particles are in equilibrium, so that the total force each is zero. Literally, $\mathbf{F}_i = 0$ for all $i$. So your second equation is trivially true.

(2) Next, he restricts the derivation to systems for which (his italics) "the net virtual work of forces of constraint is zero." In other words, $\mathbf{f}_i \cdot \delta\mathbf{r}_i=0$ for all $i$.

As a consequence of these assumptions, you find that the virtual work of the applied forces is also zero, which is mathematically evident from your post.

This isn't a glorified statement of the conservation of energy. It is a simple mathematical fact that the dot product of a zero vector and any other vector is zero. You are going at the derivation in the wrong direction where you are trying to derive Godlstein's Equation 1.40 from his equation 1.43, when in fact you need to be working the other way.

Further, your pendulum example is probably not a good example in this case because the bob is not in equilibrium. If it is oscillating then there is a net force on it. It isn't clear from your description of the system what is going on, but if you can draw a free body diagram, enumerate the forces and find that $\mathbf{F} \ne 0$, then you are violating the assumptions that go in to the derivation.

  • $\begingroup$ Is Goldstein not trying to generalize the principle of virtual work to systems where $\mathbf{F}_i \neq 0$? $\endgroup$
    – zh1
    Aug 30, 2018 at 19:41
  • $\begingroup$ He does later by adding in a "reversed effective force" which is defined such that $\mathbf{F}_i-\mathbf{p}_i=0$. But the mathematics are unchanged. You start with the assumption that sum of the forces, $\mathbf{F}_i$, and the reversed effective force, $\mathbf{p}_i$ is zero and then you assume that the forces of constraint do no work and then you find his equation 1.45, after which he argues for why you can can drop the superscript $(a)$. $\endgroup$ Aug 30, 2018 at 19:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.