Timeline for Noether Theorem and Energy conservation in classical mechanics
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 27, 2015 at 14:15 | comment | added | Arturo don Juan | @pppqqq and Mark Eichenlaub, thank you. Both of your comments completely cleared up my confusion. | |
| Mar 26, 2015 at 23:26 | comment | added | pppqqq | @ArturoDonJuan the problem here is in the interpretation of "time translation invariance". The OP assumed that this meant $\frac{\text d}{\text{d}t} L=0$, $\text d / \text d t$ being the total derivative of the lagrangian along the classical path (see equation (1-2-3) of the OP). This is not correct. Instead, it is the partial derivative $\partial L /\partial t$ which is zero. Moreover, this is not a property of the path, as $\text d /\text{d} t = 0$ is, but is a property of the lagrangian. | |
| Mar 26, 2015 at 23:17 | history | edited | pppqqq | CC BY-SA 3.0 | deleted 17 characters in body |
| Mar 19, 2015 at 22:28 | comment | added | Mark Eichenlaub | I find it hard to understand your question. The numerical value of the Lagrangian is not constant. It is not constant for infinitesimal periods. It is not constant for finite periods. The function form of the Lagrangian is usually constant. So if the Lagrangian is $\frac{1}{2}mv^2 - mgh$, it stays that, and doesn't become $\frac{1}{3}mv^2 - mgh$ at some later time or anything like that. The actual value of the Lagrangian of course changes because $v$ and $h$ change. | |
| Mar 19, 2015 at 21:32 | comment | added | Arturo don Juan | Did you mean that if the motion itself (i.e. $q$ and $\dot{q}$) is held fixed, then the Lagrangian is invariant with finite/infinitesimal time displacement? | |
| Mar 19, 2015 at 21:29 | comment | added | Arturo don Juan | @MarkEichenlaub and also pppqqq, you say that infinitesimal time displacement does not necessarily preserve the value of the Lagrangian (Mark's comment that I gave a +1 was helpful). But then how does 'finite' time displacement differ from this? You are saying that in an infinitesimal time frame it may not be invariant, but in a finite time frame it is? Can't you just 'choose' your finite time frame to be infinitesimal? I must have a bad misunderstanding of what you are saying. | |
| Jan 19, 2014 at 21:34 | comment | added | pppqqq | @JakobH I've tried to elaborate on the “my problem is seeing how this notion emerges naturally from time-translation invariance” theme. | |
| Jan 19, 2014 at 21:33 | history | edited | pppqqq | CC BY-SA 3.0 | added 2030 characters in body |
| Jan 19, 2014 at 19:58 | comment | added | Mark Eichenlaub | pppqqq Is correct. Your procedure begins by assuming that for a given trajectory, the value of the Lagrangian is constant. I.E., you throw a ball through the air and over the course of its flight kinetic-potential energy stays constant. This is wrong, so of course the result you get is wrong. | |
| Jan 19, 2014 at 19:18 | history | edited | pppqqq | CC BY-SA 3.0 | added 774 characters in body |
| Jan 19, 2014 at 19:05 | comment | added | jak | thanks for your reply. I have no problem prooving that $\dot H = 0$, but my problem is seeing how this notion emerges naturally from time-translation invariance | |
| Jan 19, 2014 at 18:57 | comment | added | pppqqq | The difference is that you are taking $q$ and $\dot q$ as functions of time, instead of simply a vector of $\mathbb R ^{2n}$ (or, better a tangent vector to $\mathbb R ^n$). I'm not sure about what does the technical term “infinitesimal time displacement” mean, since you can see that the two conditions I've wrote are indeed equivalent. However, if we take “$\partial _t L=0$“ as a definition of “time translational symmetry“, the proof of $\dot H = 0$ is really a simple calculation. | |
| Jan 19, 2014 at 18:43 | comment | added | jak | Unfortunately I can't see how this is different from what i did. I worked with infinitesimal transformations and therefore neglecting quadratic terms. Can you give me another tip how the term -L should emerge? | |
| Jan 19, 2014 at 18:28 | history | answered | pppqqq | CC BY-SA 3.0 |