Timeline for Why can't we treat the Lagrangian as a function of the generalized positions and momenta and vary that? [duplicate]
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| May 15 at 18:01 | comment | added | Qmechanic♦ | See my Phys.SE answer here | |
| May 11 at 9:21 | comment | added | weirdmath | I still don't understand this. When we do Hamiltonian mechanics, we forget about how the p's initially came about and just declare them independent and vary them independently. I can in fact just write the integrand as pqdot - H(q, p, t) and vary the q's and p's independently. The main difference between this and what I'm attempting is that there, I would not be writing qdot = qdot(p, q) and instead just have it be varied automatically when q is varied. | |
| May 11 at 9:08 | comment | added | Confuse-ray30 | $q$ and $p$ are not independent if you just plug it back into the lagrangian, as seen by your equation for $p$. | |
| May 11 at 8:53 | history | edited | weirdmath | CC BY-SA 4.0 |
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| May 11 at 3:58 | history | closed |
hft Matt Hanson Hyperon |
Duplicate of Physical meaning of Legendre transformation | |
| May 11 at 0:39 | comment | added | weirdmath | Partially. I still want to know why writing $\dot q = \dot q(q, p)$ then writing $S = \int dt \, L(q, p)$ and varying $q$ and $p$ independently does not seem to work. | |
| May 10 at 23:57 | review | Close votes | |||
| May 11 at 4:05 | |||||
| May 10 at 23:37 | comment | added | hft | Does this answer your question? Physical meaning of Legendre transformation | |
| May 10 at 23:21 | comment | added | ACuriousMind♦ | Related: on the meaning of Legendre transformations and this answer of mine | |
| May 10 at 23:14 | history | edited | Níckolas Alves |
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| May 10 at 22:32 | history | edited | weirdmath | CC BY-SA 4.0 |
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| May 10 at 22:23 | history | asked | weirdmath | CC BY-SA 4.0 |