Constrain forces and virtual displacements

[ralqs]

I've read conflicting descriptions of the relationship between constraint forces and virtual displacements that I'd like to clarify.

Suppose $\vec{C}_i$ is the constraint force on the ith particle, and $\delta \vec{r}_i$ is its virtual displacement. Is it the case that $\vec{C}_i\cdot\vec{r}_i=0$? Or can one only say that $\sum_i \vec{C}_i\cdot\vec{r}_i=0$?

The first would be true, as I see it, if the virtual displacements were always perpendicular to the constraint forces, even if the constraints were moving...or does the movement of the constraint even matter if time is kept constant?

And if the first isn't true, how would one prove the second formula?

This evidently doesn't matter for the derivation of the Euler-Lagrange equations, but I would still like to clear this confusion up.

 

[Valerie Park]

The first statement is true: \vec{C}_i\cdot\vec{r}_i=0. This is because the constraint force is always perpendicular to the virtual displacement, regardless of the motion of the constraint. To prove the second statement, one can use the fact that the total force on the system is equal to the time derivative of the momentum, and thus the sum of the constraint forces will be equal to the rate of change of the momentum. This can be expressed mathematically as \sum_i \vec{C}_i\cdot\vec{r}_i=\frac{d}{dt}\sum_i\vec{p}_i, which implies that \sum_i \vec{C}_i\cdot\vec{r}_i=0.