classical mechanics - Virtual displacement and generalized coordinates

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: $$\sum _{i=1} \frac{\partial \phi _j}{\partial \mathbf r _i} \cdot \dot {\mathbf r}_i=-\frac{\partial \phi _j}{\partial t}.\tag{2}$$

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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.

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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.

Answer

When I wrote this question some years ago, I was very confused about those "virtual displacements". Now I realize that analytical mechanics is one of those parts of physics where knowing the proper mathematical language, differential geometry in this case, can make your life incredibly easier.

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Virtual displacements and generalized coordinates.

Lagrangian mechanics takes place on a manifold $M$, which is embedded in $\mathbb R^{3N}$ via a (possibly non constant) mapping $\iota _t$. Virtual displacements are nothing but tangent vectors to $\iota _t (M)$. When $\iota _t=\iota _0$ is constant, virtual displacements also coincide with the velocity vectors of curves on $\iota _0 (M)$.

Generalized coordinates are the charts of the base manifold $M$; my above parametrization $\mathbf r _i(q_h;t)$ can be understood as a composition:

$$Q\times \mathbb R \mapsto M\times \mathbb R \mapsto \mathbb R ^{3N},$$

$$(q,t)\mapsto (x(q),t)\mapsto \iota _t(x(q))\equiv r(q,t).$$

For fixed $t_0$, $r(q,t_0)$ parametrizes $\iota _{t_0} (M)$ and therefore $\frac{\partial r}{\partial q _i}$ is a tangent vector to $\iota _{t_0} (M)$, that is, it is a virtual displacement.

To make contact with the OP, the embedding $\iota _t (M)$ is directly defined via cartesians equations:

$$\phi _j (r,t)=0\qquad r\in \mathbb R ^{3N}, j=1,2,\dots ,3N-d,$$ virtual displacements are orthogonal to the $3N-d$ gradients $\nabla \phi _j$, as in equation $(*)$ of the OP. [Here $r$ denotes the $N$-tuple $r=(\mathbf r_1,\dots, \mathbf r _N)\in \mathbb R ^{3N}$]

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There's actually no problem with what I said after my last equation in the OP. The first term vanishes because each $\frac{\partial r}{\partial q _i}$ is separately a tangent vector. The second term vanishes because $\frac{\partial r}{\partial t}$ is the actual velocity of a point that is stationary with respect to the base manifold $M$ (e.g., a material point stationary at $\theta =0$ of a rotating ring).