- #1
pellman
- 684
- 5
For a Lagrangian [itex]L(x^k,\dot{x}^k)[/itex] which is homogeneous in the [itex]\dot{x}^k[/itex] in the first degree, the usual Hamiltonian vanishes identically. Instead an alternative conjugate momenta is defined as
[itex]y_j=L\frac{\partial L}{\partial \dot{x}^j}[/itex]
which can then be inverted to give the velocities as a function of the position and momenta
[itex]\dot{x}^i=\phi^{i}(x^k,y_k)[/itex]
The Hamiltonian is then equal to the Lagrangian with the velocities replaced with this function
[itex]H(x^k,y_k)=L(x^k,\phi^{k}(x^l,y_l))[/itex]
We then find that
[itex]\dot{x}^i=H\frac{\partial H}{\partial y_i}[/itex]
which is one half of the Hamilton equations of motion. But what about [itex]\dot{y}_i[/itex]?
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.
[itex]y_j=L\frac{\partial L}{\partial \dot{x}^j}[/itex]
which can then be inverted to give the velocities as a function of the position and momenta
[itex]\dot{x}^i=\phi^{i}(x^k,y_k)[/itex]
The Hamiltonian is then equal to the Lagrangian with the velocities replaced with this function
[itex]H(x^k,y_k)=L(x^k,\phi^{k}(x^l,y_l))[/itex]
We then find that
[itex]\dot{x}^i=H\frac{\partial H}{\partial y_i}[/itex]
which is one half of the Hamilton equations of motion. But what about [itex]\dot{y}_i[/itex]?
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.