Timeline for Advantages of Lagrangian Mechanics over Newtonian Mechanics
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Jun 21, 2018 at 20:16 | comment | added | docscience | @MariusMatutiae Your comment "non-conservative". Isn't that just because the scope of what is 'system' is limited, not considering perhaps other parts that describe where the energy has gone? Can't you take a 'non-conservative dissipative system' and make it conservative by expanding the scope of what the system is? | |
| Jun 21, 2018 at 20:11 | comment | added | docscience | The relaxation methods - like adding terms that take infinite rates of change to finite rates? Is there physics behind that (perhaps even rationalizations) or is it purely a math trick? | |
| May 8, 2016 at 15:30 | comment | added | Ian | Ah, I misunderstood the coin example. I thought you were referring to forces dependent on higher derivatives rather than forces not parallel to the velocity. In the attached article the dissipation function as defined does not couple the degrees of freedom. Would it be possible to introduce a generalized dissipation function that couples the Euler Lagrange equations such that the full complexity of Newton's equations can be treated? | |
| May 8, 2016 at 14:38 | comment | added | MariusMatutiae | @Ian Not really: dissipative forces are non-conservative, hence they cannot be derived from a potential (which is conservative), not even a velocity-dependent potential. Also, adding higher-order derivatives of position changes intrinsically the nature of the equations of motion, from a second-order one (Newton's) to a third order equation. This would mean there would be another independent solution of all problems known, which however has no physical meaning. A case such as this occurs, when one studies radiation reaction (see Jackson's Electrodynamics), and the new solution... | |
| May 8, 2016 at 4:25 | comment | added | Ian | You say that Lagrangian mechanics only includes velocity dependent dissipative forces. Is this correct? Could one not write a Lagrangian that depends on higher derivatives of position and then minimize the action to get generalized Euler-Lagrange equations that are equally general as Newton's equations? | |
| May 7, 2016 at 14:17 | vote | accept | Das | ||
| Aug 8, 2016 at 1:32 | |||||
| May 6, 2016 at 10:51 | history | answered | MariusMatutiae | CC BY-SA 3.0 |