All Questions
50
questions
3
votes
1
answer
664
views
Hamilton-Jacobi formalism and on-shell actions
My question is essentially how to extract the canonical momentum out of an on-shell action.
The Hamilton-Jacobi formalism tells us that Hamilton's principal function is the on-shell action, which ...
- 649
8
votes
5
answers
716
views
Why can't we obtain a Hamiltonian from the Lagrangian by only substituting?
This question may sound a bit dumb. Why can't we obtain the Hamiltonian of a system simply by finding $\dot{q}$ in terms of $p$ and then evaluating the Lagrangian with $\dot{q} = \dot{q}(p)$? Wouldn't ...
- 590
0
votes
1
answer
3k
views
Generalized momentum conjugate and potential $U(q, \dot q)$
On Goldstein's "Classical Mechanics" (first ed.), I have read that
if $q_j$ is a cyclic coordinate, its generalized momentum conjugate $p_j$ is costant.
He obtained that starting from Lagrange's ...
- 1,133
0
votes
1
answer
191
views
Non-relativistic Kepler orbits
Consider the Newtonian gravitational potential at a distance of Sun:
$$\varphi \left ( r \right )~=~-\frac{GM}{r}.$$
I write the classical Lagrangian in spherical coordinates for a planet with mass $...
- 307
9
votes
2
answers
3k
views
How does the canonical momentum $p_i\equiv\frac{\partial L}{\partial\dot q_i}$ transform under a coordinates change $\mathbf q\to\mathbf Q$?
The canonical momentum is defined as
$$p_{i} = \frac {\partial L}{\partial \dot{q_{i}}}, $$
where $L$ is the Lagrangian.
So actually how does $p_{i}$ transform in one coordinate system $\textbf{q}$ to ...
- 91