Timeline for Lagrangian $L' = L + \frac{df}{dt}$ gives the same equations of motion
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 4, 2015 at 20:05 | vote | accept | doetoe | ||
| May 3, 2015 at 15:08 | comment | added | gautam1168 | @doetoe Yes your question has got me thinking about these things. | |
| May 3, 2015 at 13:51 | answer | added | Qmechanic♦ | timeline score: 3 | |
| May 3, 2015 at 13:47 | answer | added | Valter Moretti | timeline score: 5 | |
| May 3, 2015 at 13:44 | comment | added | doetoe | @gautam1168 Note that your derivation looks plausible, but in fact you are assuming many things, for example that the time derivative of $f$ is not the time derivative of the function $f$ itself, but the time derivative of $f$ evaluated in the points of the curve whose variation you're computing. Also you're tacitly assuming that $f$ only depends on the generalized coordinate of the curve (and possibly an independent time parameter), but the derivation breaks down when it also depends on the velocity of the curve. | |
| May 3, 2015 at 13:41 | review | Close votes | |||
| May 3, 2015 at 16:47 | |||||
| May 3, 2015 at 13:36 | comment | added | doetoe | @gautam1168 Rather than emphasizing the independence in the second, the first is only meaningful if we assume an explicit dependence. But I don't mean to nitpick, I was genuinely confused. The comments of Noiralef and Qmechanics cleared up my doubts. | |
| May 3, 2015 at 13:28 | comment | added | gautam1168 | How is the second way emphasizing the independence of $q$ and $\dot q$? Does it even matter what the extra term is a function of? $$\delta \int \frac{df}{dt}dt=\delta\int df=\delta (f_2-f_1)=0$$ | |
| May 3, 2015 at 13:09 | comment | added | doetoe | @Qmechanic My question was mainly if it was still possible to consider $q$ and $\dot q$ as independent arguments. In the second form we could write $L'(a,b,c) = L(a,b,c) + \frac{\partial f}{\partial a}(a,c)b + \frac{\partial f}{\partial c}(a,c)$ without changing anything other than the names given to the arguments, in the first form that is not the case. | |
| May 3, 2015 at 12:51 | comment | added | Qmechanic♦ | Since $\frac{df}{dt}= \frac{\partial f}{\partial q}\dot{q} + \frac{\partial f}{\partial t}$ is an identity (via the chain rule), one is free to use any of the two forms one would like. They are equivalent. Related: physics.stackexchange.com/q/174137/2451 , physics.stackexchange.com/q/87628/2451 and links therein. | |
| May 3, 2015 at 12:24 | comment | added | Noiralef | I think the answer is yes. Probably its usually written in this form because the $\frac{df}{dt}$ is shorter, catchier and hence easier to remember. But I remember that in my second semester I didn't understand for quite a while how something could not be the time derivative of something else. Your way of writing it would have helped me a lot, I guess. | |
| May 3, 2015 at 12:16 | history | asked | doetoe | CC BY-SA 3.0 |