Saturday, June 15, 2019

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




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.



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.




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 $r=(\mathbf r_1,\dots, \mathbf r _N)\in \mathbb R ^{3N}$]




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).


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