(Non-)holonomic geodesic i.e. constrained geodesics

Question

Fellows,

I have been working with classical mechanics during engineering classes to know Euler-Lagrange equation (EL) $\frac{d}{dt} \left(\frac{\partial L}{\partial \dot{q}}\right) - \frac{\partial L}{\partial q} = 0$. Its version is constrained by a set of equalities $A^\intercal \dot{q} = 0$ such that adds the term $A \lambda(t)$ at right-hand side, for lagrange multipliers $\lambda(t)$.

After learning some differential geometry, I learned the EL-equation is mechanics-equivalent to the covariant derivative of a geodesic curve $\nabla_\dot{q} \dot{q} = 0$. In a manifold, the constraints are such that for a given set of linearly independent vectors $\{a_i\}_m$, there are $m$ equalities which satisfy $\langle a_i, \dot{q}\rangle = 0$. The constrained geodesic takes format $\nabla_\dot{q} \dot{q} = \sum\limits_i a_i \lambda_i$. Einstein's notation is $a_i \lambda^i$.

My questions are:

1. How may I interpret both the EL equality and covariant derivative with right-hand term $a_i \lambda^i$?
2. In the case of EL-equation, the matricial format allows the definition of velocity vector $\dot{q}$ as linear combination $b_j p^j$ such that $\underbrace{\langle a_i, b_j\rangle}_{= 0} p^j = 0$, also known as the annihilator of a distribution. From the interpretation above, since covariant derivative $\nabla_\dot{q} \dot{q}$ is equal $a_i \lambda^i$, thus the following inner product with $n-m$ annihilator distribution vectors is equally null i.e. $\langle b_j, \nabla_\dot{q} \dot{q} \rangle = 0$. How do I obtain the Christoffel maps $\Gamma^k_{ij}$ of the constrained geodesic equation from here?

Surprised emoji :-O

Thanks for reading until here! :-)

Answer 1

The non-holonomic constraints you are considering are linear. When you start with a mechanical Lagrangian and add linear constrains, the following paper

Synge, J. L. Geodesics in non-holonomic geometry. Math. Ann. 99 (1928), no. 1, 738--751.

gives a positive answer. But it may not be the easiest read, and I suggest you look somewhere else first. In a more friendly style, you can find this result in e.g.

Lewis, Andrew D. Affine connections and distributions with applications to nonholonomic mechanics. Pacific Institute of Mathematical Sciences Workshop on Nonholonomic Constraints in Dynamics (Calgary, AB, 1997). Rep. Math. Phys. 42 (1998), no. 1-2, 135--164.

But, a word of caution: There is, in general, no way to interpret the Lagrange–D’Alembert equations as being geodesic equations for some Lagrangian. In other words: they need not be variational.

Answer 2

According to citation Lewis, quoted by @WishYouTheBest, in case velocity vector $\dot \gamma$ belongs to some distribution $B$ spanned by vectors $\{b_i\}$ i.e. $b_i p^i$, the covariant derivative $\nabla_{\dot \gamma} \dot \gamma$ belongs to its annihilator space of distribution $B$, $A$, given by vector set $\{a_j\}$ and such that $\langle b_i , a_j \rangle = 0$. Thus we conclude $\nabla_{\dot \gamma} \dot \gamma = a_j \lambda^j$. Therefore $\langle \nabla_{\dot \gamma} \dot \gamma, b_i \rangle = 0$. Since inner product $\langle u, v \rangle = g_{kl} u^k v^l$ and $(\nabla_{\dot \gamma} \dot \gamma)^k = \ddot{\gamma}^k + \Gamma^k_{ij} \dot{\gamma}^i \dot{\gamma}^j$ on Einstein's notation, we obtain equation $\dot{p}^k + I^k_{ij} p^i p^j = 0$ depending on Christoffel symbols $\Gamma^k_{ij}$ and vectors $\{b_i\}$ from there by substitution.