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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.

Qmechanic
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Young Kindaichi
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3 Answers3

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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.

Qmechanic
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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.

ZeroTheHero
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  • For a second , Can you just leave $(q,p)$? Suppose it's not constructed, Lagrange had taken configuration space instead of $(q,\dot{q})$. What's the reason of this? I mean if both are independent then obvious choice should be later. – Young Kindaichi Nov 09 '20 at 05:06
  • I’m not sure I understand what you mean by “both are independent”. Of course Lagrangian mech. is in some sense precursor to Hamilton mech and in LMech you do use $(q,\dot q)$, but the formalism is less powerful because it does not allow for canonical transformations that can mix $q$ and $p$ and all these other things you miss out on. – ZeroTheHero Nov 09 '20 at 05:14
  • By independent means, I can independently specify the initial conditions. – Young Kindaichi Nov 09 '20 at 05:29
  • Hamiltonian developed his mechanics in 1833 while Lagrange's mechanics was developed in 1788. So at that time, we don't about canonical transformation and the benefits of using $(q,p)$. So why choose configuration space instead of $(q,\dot{q})$. – Young Kindaichi Nov 09 '20 at 05:33
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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.

AngusTheMan
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