Virtual Displacement – Virtual Displacement and Generalized Coordinates in Classical Mechanics

classical-mechanicsconstrained-dynamicsdifferential-geometrylagrangian-formalismmathematical physics

I have a doubt regarding the expression of a virtual displacement using generalized coordinates. I will state the definitions I'm taking and the problem.

The system is composed by $n$ points with positions $\mathbf r _i$ and subject to $3n-d$ constraints of the form: $$\phi _j (\mathbf r _1, \mathbf r _2,…,\mathbf r _n,t)=0\qquad (1\leq j \leq 3n-d), \tag{1}$$
that, deriving with respect to time, gives: $$\sum _{i=1} \frac{\partial \phi _j}{\partial \mathbf r _i} \cdot \dot {\mathbf r}_i=-\frac{\partial \phi _j}{\partial t}.\tag{2}$$


According to my notes, a set of possible velocities $(\mathbf v_1,\mathbf v_2,…,\mathbf v_n)$ is one that satisfies the above system of $j$ equations (with $v_i$ in the place of $\dot r _i$), while a set of virtual velocities is one that satisfies the homogeneous system
$$\sum _{i=1} \frac{\partial \phi _j}{\partial \mathbf r _i} \cdot \dot {\mathbf r}_i=0.\tag{*}$$
Finally, a virtual displacement is given by the product of a virtual velocity by a quantity $\delta t$, with the dimensions of time.


I have the following problem. Suppose that I have a parametrization of the configuration space at time $t$ in the form: $$\mathbf r _i = \mathbf r _i (q_1,\dots ,q_d;t).$$ That is: $$\phi _j(\{\mathbf r _i (q_1,\dots,q_d;t)\},t)=0$$ for all $q=(q_1,\dots,q_d)\in Q$ and $t\in [t_1,t_2]$.
Now, according to my notes, if such a parametrization is given, the general form of a virtual displacement is: $$\delta \boldsymbol r _i =\sum _h \frac{\partial \mathbf r _i}{\partial q _h}\delta q _h.$$

Let $q(t)$ be a curve in the coordinate's space. By taking the total derivative of both sides of the precedent equation, I obtain: $$\sum _i \frac{\partial \phi _j}{\partial \mathbf r _i}\cdot (\sum _h \frac{\partial \mathbf r_i}{\partial q _h} \dot q _h)+\sum _i \frac{\partial \phi _j}{\partial \mathbf r _i}\cdot \frac{\partial \mathbf r _i}{\partial t} +\frac{\partial \phi _j}{\partial t}=0.$$ But the first term is zero because it is the product of the gradients $\nabla _{\mathbf r _i}\phi _j$ with the virtual velocities $\mathbf v _i$. But, in this case, it looks like that the second+third terms should be zero.

I suspect that there's an error, I don't see why the second+term should always give $0$ and I would like a proof check of what I wrote above.

Best Answer

I see your question can be expressed in words as "when the virtual displacements/velocities agree with the allowed ones?" that's, as you said,

$ \frac{\partial \mathbf{x}_{i}}{\partial t} = 0 $

that is to say that the position vector $r$ is expressed in terms of $q_{k}$'s only and doesn't contain $t$ explicitly, so as the constraints. i.e. the system is scleronomic .

example of this system is the pendulum with inextensible string, you will find that virtual displacements and velocities are the same as the allowed ones, and the last term you'r asking about vanishes.

for another case, think about the same pendulum but with extensible string, say $l = 0.2 t$ .

"the virtual displacement is not always the allowed one, the same for the virtual velocity"

I hope my answer helps you and I think you'll find "Greenwood- Classical Dynamics" useful for you.

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