Timeline for Is the Hessian of Hamilton's function positive-definite?
Current License: CC BY-SA 2.5
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 2, 2009 at 22:26 | vote | accept | Theo Johnson-Freyd | ||
| Nov 25, 2009 at 14:00 | comment | added | José Figueroa-O'Farrill | Yes, sorry -- I must have misunderstood. I thought your last question referred to a general lagrangian. If that is the case, I stand by what I said. | |
| Nov 25, 2009 at 6:11 | comment | added | Theo Johnson-Freyd | I certainly don't deny that "the Hessian depends on the second derivative of the potential". But you're too quick in your answer. As I said in the examples section, I can explicitly solve EOM when the potential V(q) is at most quadratic, with arbitrary sign, including when it is bounded below, and so in this case (and only this case) I can explicitly write the action and its Hessian. And unless I made a mistake, if V(q) is quadratic, then the Hessian is a perfect square depending on V''(q). | |
| Nov 25, 2009 at 3:12 | history | answered | José Figueroa-O'Farrill | CC BY-SA 2.5 |