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I'm fine with Lagrangian mechanics without an integral of motion, but for some reason I keep seeing these integrals mentioned within the context of the Lagrangian with no real explanation as to their purpose. All I can gather so far is it results from when for seemingly no reason, someone decides to spontaneously set the equations of motion equal to each other after finding one that the $\frac{dL}{dt}$ term of equation for one coordinate is zero is zero, and then therefore that the $\frac{d}{dt} \frac{dL}{dt}$ term must also be zero, and I don't know why I should particularly care about when that is the case when all kinds of complicated and ambiguously convoluted equations of motion can be set to zero.

Then there is usually some mention of phase space and momentum being conserved when equations, however, the equations of motion often contain acceleration terms, which seems to contradict the idea that these integrals of motion are dependent solely on velocity and position and mass. From what I have sees, they should clearly be dependent on velocity, position, mass AND acceleration. I should be good to go for modeling a physical system if I have the acceleration of each coordinate, so what do I need this integral of motion for?

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    $\begingroup$ I'm not sure I understand. An integral of the motion is a conserved quantity - like momentum, angular momentum, energy, etc. Are you asserting that understanding the conserved quantities of a system is not worthwhile? $\endgroup$ – J. Murray Mar 1 '18 at 4:28
  • $\begingroup$ I'm not asserting anything, I'm asking what should be asserted. "An integral of motion is a conserved quantity" okay, so that implies it's not a coordinate or vector, but a number. Great, so what's important about this conserved quantity? How do you use integrals of motion for it? $\endgroup$ – user182023 Mar 1 '18 at 8:47
  • $\begingroup$ can you clarify "when that is the case when all kinds of complicated and ambiguously convoluted equations of motion can be set to zero."? A conserved quantity $Q$ is defined by $dQ/dt=0$. That's not terribly convoluted. The statement is that $\frac{\partial L}{\partial \dot{q}_i}$ is conserved when $\frac{\partial L}{\partial q_i}=0$. $\endgroup$ – ZeroTheHero Mar 1 '18 at 13:52

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