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I’m trying to delve into lagrangian mechanics, mainly because i want to have a deeper understand of things lik Noether’s theorem.

What i don’t fully understand is what the lagrangian of a system actually is. I’ve heard it’s related to equations of motion for the system, and that’s it’s something like the kinetic energy minus the potential, but it’s full meaning i only have a limited grasp of.

As i understand it’s an important term in many areas of physics, so i’d like to have a better grasp of it, if anybody here can help.

So there was a relatively simple question i saw that i want to understand better, this is it:

Consider the Lagrangian $$L(q_1,q_2,\dot{q_1}, \dot{q_2})$$ of a conservative mechanical system described by generalised coordinates $$(q_1,q_2)$$. What condition must be satisfied for the momentum $p_1$ associated to $q_1$ to be conserved?

Now i understand that if you substitue in euler-lagrange equation, you get $$\frac{dL}{dq_1}=0$$

And apparently that implies the lagrangian L is independant of $q_1$. After thinking about it for a bit though, does this also imply spatial translational symmetry? I think it would since conservation of momentum is required for this question, but does this equation you end up with actually imply spatial translation symmetry because the lagrangian, which in some respects represents the path of motion, is independent of $q_1$?

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    $\begingroup$ The homogeneity of space leads to conservation of linear momentum of the system. and it will mean dL/dq1+dl/dq2 =0 $\endgroup$ – drvrm Aug 12 '18 at 5:45
  • $\begingroup$ @drvrm I know, i actually made a post about that second thing you thing you mentioned, but my question was if the lagrangian being independant of q1 had a connection to spatial translation. thank you though $\endgroup$ – Thatpotatoisaspy Aug 12 '18 at 6:00
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I should logically imply that. For momentum associated with q1 to be conserved the Lagrange should be independent of q1. Also translational symmetry implies momentum conservation. So it is like a=>b and b=>c so a=>c..

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    $\begingroup$ Ah, i thought so, thank you. I’m a bit new to this, so i was just kind of trying to figure it out on my own. $\endgroup$ – Thatpotatoisaspy Aug 12 '18 at 11:17

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