I Generalized Momentum is a linear functional of Velocity?

[chmodfree]

Generalized momentum is covariant while velocity is contravariant in coordinate transformation on configuration space, thus they are defined in the tangent bundle and cotangent bundle respectively.
Question: Is that means the momentum is a linear functional of velocity? If so, the way to construct this functional?

 

[jambaugh]

If the Lagrangian is quadratic in the coordinate velocities then it is linear at the point of the tangent bundle and can thus be expressed as the result of a linear mapping by a generalized inertia tensor, P_a = M_{ab}(Q,t)\dot{Q}^b

P_a = \frac{\partial \mathcal{L}(Q,\dot{Q},t)}{\partial \dot{Q}^a} = M_{ab}\dot{Q}^b with M not depending on any \dot{Q} if and only if the Lagrangian was quadratic in these velocities.

 

[chmodfree]

If the Lagrangian is quadratic in the coordinate velocities then it is linear at the point of the tangent bundle and can thus be expressed as the result of a linear mapping by a generalized inertia tensor, P_a = M_{ab}(Q,t)\dot{Q}^b

P_a = \frac{\partial \mathcal{L}(Q,\dot{Q},t)}{\partial \dot{Q}^a} = M_{ab}\dot{Q}^b with M not depending on any \dot{Q} if and only if the Lagrangian was quadratic in these velocities.

Oh I see, that means the contraction P_a\dot{Q}^a is a real.

 

[jambaugh]

Oh I see, that means the contraction P_a\dot{Q}^a is a real.

I don't know about real... I imagine someone might cook up a Lagrangian with complex inertia, but the contraction is a scalar.

 

[chmodfree]

I don't know about real... I imagine someone might cook up a Lagrangian with complex inertia, but the contraction is a scalar.

I mean it maps \dot{Q} from tangent bundle to the real number field as a functional... No problem now, thank you.