- #1
pellman
- 683
- 6
For a Lagrangian L(x^k,\dot{x}^k) which is homogeneous in the \dot{x}^k in the first degree, the usual Hamiltonian vanishes identically. Instead an alternative conjugate momenta is defined as
y_j=L\frac{\partial L}{\partial \dot{x}^j}
which can then be inverted to give the velocities as a function of the position and momenta
\dot{x}^i=\phi^{i}(x^k,y_k)
The Hamiltonian is then equal to the Lagrangian with the velocities replaced with this function
H(x^k,y_k)=L(x^k,\phi^{k}(x^l,y_l))
We then find that
\dot{x}^i=H\frac{\partial H}{\partial y_i}
which is one half of the Hamilton equations of motion. But what about \dot{y}_i?
I am following Hanno Rund The Hamilton-Jacobi equation in the Calculus of Variations. But Rund moves on from this point to the H-J equation, leaving me wondering about this question.
y_j=L\frac{\partial L}{\partial \dot{x}^j}
which can then be inverted to give the velocities as a function of the position and momenta
\dot{x}^i=\phi^{i}(x^k,y_k)
The Hamiltonian is then equal to the Lagrangian with the velocities replaced with this function
H(x^k,y_k)=L(x^k,\phi^{k}(x^l,y_l))
We then find that
\dot{x}^i=H\frac{\partial H}{\partial y_i}
which is one half of the Hamilton equations of motion. But what about \dot{y}_i?
I am following Hanno Rund The Hamilton-Jacobi equation in the Calculus of Variations. But Rund moves on from this point to the H-J equation, leaving me wondering about this question.
