# Why is it important that there is no variation of time $\delta t=0$ in the definition of virtual displacement?

In Goldstein's Classical mechanics I found a proposition that I don't understand:

Similarly, the arbitrary virtual displacement $$\delta \mathbf{r}_i$$ can be connected with the virtual displacement $$\delta q_i$$ by $$\delta \mathbf{r}_i = \sum_j \frac{\partial \mathbf{r}_i}{\partial q_j} \delta q_j. \tag{1.47}$$ Note that no variation of time, $$\delta t$$, is involved here, since a virtual displacement by definition considers only displacements of the coordinates. (Only then is the virtual displacement perpendicular to the force of constraint if the constraint itself is changing in time).

I don't understand the part within parenthesis. Why is it important that there is no variation of time $$\delta t=0$$ in the definition of virtual displacement?

My understanding is that the force of constraint is perpendicular to the surface, while the virtual displacement must be tangent to it (and this should be the case also if the constraints change with time).

I'm not asking for the definition of virtual displacement. I already know that virtual displacement by definition occurs at $$t=$$ const so that the differential of $$\mathbf{r}_i$$ does not vary with time.

• Look also at my answer here: physics.stackexchange.com/q/456771 Commented Dec 23, 2019 at 9:48
• I understand what a virtual displacement is but I don't understand this: 'Only then is the virtual displacement perpendicular to the force of constraint if the constraint itself is changing in time' Commented Dec 23, 2019 at 11:09

Example.

1. Consider a point particle constrained to move on a rotating rod $${\bf r}~=~q \begin{pmatrix} \cos (\omega t) \cr \sin(\omega t) \cr 0 \end{pmatrix} .\tag{1}$$ There are 2 holonomic constraints, and 1 generalized coordinate $$q$$. Infinitesimal displacements are of the form $$\delta {\bf r}~=~\delta q \begin{pmatrix} \cos (\omega t) \cr \sin(\omega t) \cr 0 \end{pmatrix}+ q\omega\delta t \begin{pmatrix} -\sin (\omega t) \cr \cos(\omega t) \cr 0 \end{pmatrix} .\tag{2}$$

2. The constraint force is (assumed to be) perpendicular to the rod, i.e. it belongs to the following 2-plane: $${\bf F}~\in {\rm span}_{\mathbb{R}}( \begin{pmatrix} -\sin (\omega t) \cr \cos(\omega t) \cr 0 \end{pmatrix},\begin{pmatrix} 0 \cr 0 \cr 1 \end{pmatrix} ).\tag{3}$$

3. The constraint force does no infinitesimal work iff the infinitesimal displacements (2) are perpendicular to the constraint force (3) iff we restrict to virtual displacements with $$\delta t~=~0.\tag{4}$$

4. Concerning virtual displacements, see also e.g. this, this & this Phys.SE posts.

• Why must "The constraint force belongs to the following 2-plane"? How is it significant? Commented Nov 6, 2022 at 7:17
• I updated the answer. Commented Nov 6, 2022 at 9:10
• How can there be two holonomic constraints in this system? The particle may move radially and rotate with the rod, so the only constraint here should be that there is no upward or downward motion, which leads to two generalized coordinates $r$ and $\theta$, right? Commented Sep 3, 2023 at 4:59
• Hi V Govind. Thanks for the feedback. Try to consider the particle position $(x^{\prime},0,0)$ from the rod's rest frame. Commented Sep 3, 2023 at 5:45