Timeline for Time dependence of the Lagrangian of a free particle?
Current License: CC BY-SA 3.0
8 events
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| Mar 3, 2015 at 21:55 | comment | added | Sofia | @stringcosmologyapplicant yes. Before imposing the condition of minimal action, the velocity is treated as an independent variable because at a given time $\tau$ between $t_1$ and $t_2$ and at an whatsoever place in space, $\vec r$, the velocity $\vec v$ can be 5m/s, 0.32m/s, -200m/s, God knows what and in which direction. It behaves as an independent variable. Only by imposing the condition of minimum action, will we find that it has to depend on time in a certain way. And the condition of minimum action leads to the Euler-Lagrange eqs., which lead us explicitly to the form of dependence. | |
| Mar 3, 2015 at 21:36 | comment | added | singularity | So it seems it doesn't make sense to talk about time dependence before applying the least action principle, since we do not really know the actual time dependence and the actual trajectory. | |
| Mar 3, 2015 at 21:27 | history | edited | Sofia | CC BY-SA 3.0 |
added 1 character in body
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| Mar 3, 2015 at 21:16 | history | edited | Sofia | CC BY-SA 3.0 |
adding some explanation that I gave in the comments
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| Mar 3, 2015 at 21:08 | comment | added | Sofia | @stringcosmologyapplicant Only when we learn what is the trajectory, by choosing the path that minimizes the action, we learn that the velocity depends on time in a certain way. But, as long as we don't define a trajectory, i.e. $\vec r(t), \vec {\dot r(t)}$ what we know about the velocity? Do we know on what it depends? | |
| Mar 3, 2015 at 21:06 | comment | added | Sofia | @stringcosmologyapplicant No, no. I am saying that before applying the least action principle and find the trajectory of the object, we have a Lagrangian dependent on velocity I the kinetic energy term, and position and (eventually) time in the potential energy. We have no idea how the velocity depends on time. There is a continuum of forms of dependence, because there is a continuum of forms of trajectories available. This is why, in the Lagrangian we take the velocity as a variable in itself. (I continue) | |
| Mar 3, 2015 at 20:45 | comment | added | singularity | Thanks for the response. I am sorry but your last sentence is a bit confusing to me. Are you saying that arbitrary trajectories may have time dependence of velocity but since we have chosen a Lagrangian which has time-independent velocity, the final equation of motion would have no relationship between time and velocity. | |
| Mar 3, 2015 at 20:31 | history | answered | Sofia | CC BY-SA 3.0 |