Deriving Hamiltonian for 2-DoF Aero-Elastic System

[thrillhouse86]

Hi all,

I am in a bit of a dilly of a pickle of a rhubarb of a jam with determining the Hamiltonian of a specific system. For background information it is an 2-DoF aero-elastic system where I am (temporarily) neglecting the aerodynamic lift and moment terms.

Being an intrinsically engineering problem, the full equations of motion that include the lift and moment terms have been formulated using Newtonian mechanics. I have temporarily ignored these nonconservative terms and derived a Lagrangian which, upon applying the Euler-Lagrange equations reproduces the equations of motion.

Now I have determined the Hamiltonian by defining the canonical momentum as:
p_{i} = dL/dq{dot}_i and applied the Legendre transform (sure enough it is H = T + V) and I now have the Hamiltonian in terms of my generalised position and velocities - the problem is that I cannot for the life of me write the hamiltonian in terms of the generalised position and momentum.

So I guess my question is two-fold

1. Is there a systematic way of determining the Hamiltonian in terms of the generalized position & momentum given the Hamiltonian in terms of the generalised position and velocity ? Or is the only way a character building algebraic excersize ?

2. Is it possible to have a conservative system which has a Lagrangian but does not have a Hamiltonian ?

Cheeers,
Thrillhouse86

 

[Bob S]

It's been a long time, but
1) Isn't momentum simply the mass times velocity?
2)Acording to my old text, "Thus for a system in which L = T - V and in which the transformation equations don't explicitly contain the time, H is equal to the total energy of the system." i.e., H = T + V.

 

[Dale]

Now I have determined the Hamiltonian by defining the canonical momentum as:
p_{i} = dL/dq{dot}_i and applied the Legendre transform (sure enough it is H = T + V) and I now have the Hamiltonian in terms of my generalised position and velocities - the problem is that I cannot for the life of me write the hamiltonian in terms of the generalised position and momentum.

Use the expressions for the canonical momenta (p_{i} = dL/dq{dot}_i), solve those for the generalized velocities and substitute into the Hamiltonian. Then you will have an expression in terms of generalized momenta rather than generalized velocities.

 

[thrillhouse86]

Thanks DaleSpam - it might be worth pointing out to anyone as dull as me that writing the relationship between the canonical momentum and the generalised velocities as a matrix equation gives one a systematic manner of writing the canonical momentum(s) in terms of the generalized velocities.

also thanks for trying to help me out Bob S - but I think what you are describing is the kinematic momentum - and this often, but is not necessarily the same as the canonical momentum. The classical example is to consider the canonical momentum for a charged particle in an electric and magnetic field - it turns out the canonical momentum is the classical momentum plus some term associated with charge and the electric field

 

[Dale]

2. Is it possible to have a conservative system which has a Lagrangian but does not have a Hamiltonian ?

I have thought about this a bit and come to the conclusion that no, it is not possible. However, it is probably possible that the expressions for the canonical momenta may not be invertible in all cases and therefore there may not always be a closed-form expression for the generalized velocities in terms of the generalized momenta. That would mean that the Hamiltonian (although it would certainly exist) could only be found numerically.

 

[Danger]

a bit of a dilly of a pickle of a rhubarb of a jam

Sounds like something that you'd find in Turbo's pantry.
Sorry... off-topic...