How do constraints work in Lagrangian systems?
I have a question about the discussion of constrained Lagrangian systems in the book Mathematical Aspects of Classical and Celestial Mechanics by Arnold et al. (section 1.2.5).
The Lagrangian system is defined on a manifold $M$. The constraints pick out a submanifold $S$ of the phase space $TM$, and are of the form $$f_1(q,\dot{q},t)=\cdots=f_n(q,\dot{q},t)=0,$$ with the requirement that $f_{1\dot{q}}'$, ..., $f_{n\dot{q}}'$ be linearly independent at each point. The virtual velocities of the state $(q,\dot{q})$ at time $t$ are the set of vectors $\xi\in T_qM$ satisfying $$f_{1\dot{q}}'\cdot \xi=\cdots=f_{n\dot{q}}'\cdot \xi=0.$$
I'm confused by these definitions. What if we want to impose a constraint $f$ that only depends on $q$? Then $f_{\dot{q}}'=0$, and all vectors at every point would be virtual velocities. But doesn't that mean there are no constraints on the system?
1 answer
As apparently seems par for this book, the theorem statement you're paraphrasing involves vaguely defined or completely undefined terms. For example, it talks about "admissible (smooth) paths" but this is the first use of the term "admissible" in the text.^{[1]}
Anyway, the situation you describe violates the assumptions that the covectors are linearly independent. Even in the case where you only have one, the set $\{0\}$ is a linearly dependent set.
More practically, constraints of the sort you describe, i.e. that are functions of the (generalized) positions (and possibly time) alone, are called holonomic. If you think about, what it means to say that $f(\mathbf q, t) = 0$, an acceptable constraint of the form the authors describe arises. The constraint $f(\mathbf q, t) = 0$ states that at each point $\mathbf q$, $f$ is a conserved quantity. This then gives rise to a continuity equation expressing this conservation, namely $\dot{\mathbf q} \cdot \nabla_\mathbf q f + \frac{\partial f}{\partial t} = 0$. That is, the amount of $f$ flowing into a point $\mathbf q$ has to exactly match the amount that $f$ decreases with time. We can then define $g(\mathbf q, \dot{\mathbf q}, t) = \dot{\mathbf q} \cdot \nabla_\mathbf q f + \frac{\partial f}{\partial t}$ and use this as a constraint. It will typically give rise to a nonzero gradient with respect to $\dot{\mathbf q}$.
For the case where $f$ is not timevarying. The continuity equation states that the only allowed tangent vectors are those normal to the gradient, i.e. those in the directions of constant $f$. In this case, the $\mathbf v$ such that $g(\mathbf q, \mathbf v) = 0$ pick out a (hyper)plane through the origin which, if translated to $\mathbf q$, would be tangent to the (hyper)surface $f(\mathbf q) = 0$ at $\mathbf q$. This is exactly the tangent space we'd want in this situation. The gradient of $g(\mathbf q, \mathbf v)$ with respect to $\mathbf v$ is just $\nabla_\mathbf q f$ again. The virtual velocity vectors are again just the vectors of this tangent (hyper)plane.
If $f$ is timevarying, then at a fixed point and time, $\mathbf q$ and $t$, the $\mathbf v$ such that $g(\mathbf q, \mathbf v, t) = 0$ will again pick out a (hyper)plane, but this time it will not be a (hyper)plane through the origin (unless $\frac{\partial f}{\partial t} = 0$). However, $\nabla_\mathbf v g(\mathbf q, \mathbf v, t)$ will still be $\nabla_\mathbf q f$ and thus the virtual velocities will be the same as if $f$ wasn't timevarying but instead was frozen at its current shape.

The notation $f'_q$ is bizarre. The authors like to use $'$ to indicate differentiation with respect to whatever they feel like at the moment, but this is clearly a problem for multiparameter functions. So instead of explicitly writing $\frac{\partial f}{\partial q}$ or following other authors and writing $f_q$, they use $f'_q$ presumably to (help) distinguish subscripts that are indexes from subscripts that mean differentiation. Unfortunately, they don't seem to articulate this convention anywhere, they also use $'$ for just distinct but related variables, and they (presumably) typo the definition of "Lagrangian derivative" as $(L'_{{\dot q}_i})^\cdot  L_{q_i}$ where $(L'_{{\dot q}_i})^\cdot  L'_{q_i}$ was presumably intended. ↩︎
1 comment
Surely is phase space the tangent bundle? Ī̲ expected the cotangent bundle of the configuration space to play this rôle. — Incnis Mrsi 30 days ago