Why not a $(q,\dot{q})$ space in Lagrangian Mechanics?

Question

We know that the Lagrangian $\mathcal{L}(q,\dot{q},t)$ which is function of generalized co-ordinate, generalized velocity and time. We consider the dynamics of particle is in configuration space. But As we know that $q$ and $\dot{q}$ are independent, Why not use both $q$ and $\dot{q}$ to construct a state space? The co-ordinates in the configuration space are not sufficient after all.

Answer 1

Yes, OP is right, although this is nothing new. Given a dynamical system [described by a first-order Lagrangian $L(q,v,t)$], the kinematic state $(q,v)$ of the system is a point in the tangent bundle $TQ$ of the configuration space $Q$, cf. e.g. my Phys.SE answer here.

Answer 2

The $(q,\dot q)$ space misses some "nice" features of the usual phase space, like Liouville invariants, the simplicity and symmetry of Hamilton's equations of motion, the invariance of form of H's equations of motions under (canonical) transformations.

There are in addition some deep geometrical reasons to use $(q,p)$ rather than $(q,\dot q)$, some of which are very closely related to the above.

Answer 3

As pointed out by other answers, the configuration space $Q$ is an $n$-dimensional space coordinated by $(q)$. The tangent bundle $TQ$ is a $2n$-dimensional space parametrised by $(q,v)$.

For a trajectory $\gamma(q)$ on $Q$, it happens that this can be 'lifted' to $TQ$ precisely when the velocity is the time derivative of $q$, i.e. $v=dq/dt$. When this occurs, $\gamma$ is said to be an integral curve. There are many potential vectors in the tangent space above a given point on $Q$; however, only integral curves represent the dynamics.