Timeline for How is a Hamiltonian constructed from a Lagrangian with a Legendre transform
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| S Dec 19, 2017 at 23:00 | history | suggested | tommy1996q | CC BY-SA 3.0 |
Bigger brackets
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| Dec 19, 2017 at 15:52 | review | Suggested edits | |||
| S Dec 19, 2017 at 23:00 | |||||
| Feb 22, 2017 at 10:16 | comment | added | lalala | @user30750 If the derivative matrix of the Langrangian w.r.t. to velocities cannot be inverted (aka has eigenvector with eigenvalue 0) these have to be taken as primary constraints. | |
| Oct 8, 2013 at 16:19 | comment | added | Trimok | The OP Lagrangian here was very very special, because it depends only on $q$, so we have $p=0$. Now, passing to the hamiltonian $H(p,q)$, we have to keep this information $(p=0)$, if we want to recover the usual Euler-Lagrange equations $(2)$ (see also the beginning of the answer), from the hamiltonian relations $(1)$. I suppose a more rigorous answer would be based on the Legendre transformation | |
| Oct 8, 2013 at 16:12 | history | edited | Trimok | CC BY-SA 3.0 |
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| Oct 8, 2013 at 15:56 | comment | added | user30750 | Trimok Why we have to add constraints with Hamiltonian representation to be equivalent the Lagrangian representation? What is/are physically or mathematically reason/s? Thank you very much for the answers. | |
| Jun 12, 2013 at 9:48 | vote | accept | 0range | ||
| Jun 12, 2013 at 9:35 | history | answered | Trimok | CC BY-SA 3.0 |