Timeline for What is the physical significance of $ -\frac{\partial L}{\partial t} = \frac{\partial H}{\partial t} $?
Current License: CC BY-SA 4.0
11 events
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| Feb 23, 2021 at 21:15 | vote | accept | Roland Salz | ||
| Feb 22, 2021 at 22:01 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
Tried to make question more clear
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| Feb 22, 2021 at 22:00 | answer | added | Qmechanic♦ | timeline score: 5 | |
| Feb 20, 2021 at 10:34 | comment | added | Roland Salz | @ZeroTheHero, with the help of the first 2n equations we find $ \frac{dH}{dt} = \frac{\partial H}{\partial q_i} \dot{q}_i + \frac{\partial H}{\partial p_j} \dot{p}_j + \frac{\partial H}{\partial t} = \frac{\partial H}{\partial q_i} \frac{\partial H}{\partial p_i} - \frac{\partial H}{\partial p_j} \frac{\partial H}{\partial q_j} + \frac{\partial H}{\partial t} = \frac{\partial H}{\partial t} $. | |
| Feb 19, 2021 at 21:02 | comment | added | ZeroTheHero | $\partial H/\partial t \ne dH/dt$ in general. | |
| Feb 19, 2021 at 21:00 | comment | added | Roland Salz | @ZeroTheHero, I think we have $ -\frac{\partial L}{\partial t} = \frac{\partial H}{\partial t} = \frac{dH}{dt} $, so there is no difference. | |
| Feb 19, 2021 at 17:42 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
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| Feb 19, 2021 at 17:41 | comment | added | ZeroTheHero | The last should be $-\partial L/\partial t= dH/dt$, with full derivative of $H$. If the Lagrangian does not depend explicitly on $t$ then $H$ is conserved. | |
| Feb 19, 2021 at 17:36 | history | edited | Roland Salz | CC BY-SA 4.0 |
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| Feb 19, 2021 at 10:21 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
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| Feb 19, 2021 at 10:06 | history | asked | Roland Salz | CC BY-SA 4.0 |