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What are the assumptions behind the Lagrangian derivation of energy? I understand that we're searching for a function $L$ that describes a set of physics so that solving the energy minimization problem

$$\begin{array}{rcl} \arg\min\limits_{q} && \int_{t_1}^{t_2} L(q,\dot{q},t) dt\\ \textrm{st}&&q(t_1)=q_1\\ &&q(t_2)=q_2 \end{array}$$

determines the path of an a particle $q:[t_1,t_2]\rightarrow\mathbb{R}^3$ from time $t_1$ to time $t_2$. Eventually, we find that $L(q,\dot{q},t)=\frac{1}{2}m \dot{q}^2$. What's not clear to me are the assumptions behind the setup for the energy minimization problem. It looks like to me that there's an assumption about the lack of forces like gravity. Except, I know that this derivation can be used to eventually derive that $F=m\ddot{q}$, so we don't yet have a concept of force. Also, I know there's a corollary that shows that the eventual solution $q$ is such that $\dot{q}=c$ or $\ddot{q}=0$. I'm sure that makes sense in relation to the problem setup, which is what I'm trying to clarify. Finally, I do know that we need to assume

  1. Time and space are homogenous
  2. Space is isotropic
  3. Galilean invariance

Anyway, I'm looking for the other assumptions and I'm pretty sure it has to do with lack of other forces or some kind of related concept.

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  • $\begingroup$ All that is required is Newtonian mechanics; you always get the same equations of motion. The only forces that can be ignored are the forces of constraint, which is because they do no work. The easiest problems are ones with conservative forces (i.e., expressible with potentials); otherwise you must introduce generalized forces, and the equations are messier. $\endgroup$ – Peter Diehr Apr 25 '16 at 2:41
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    $\begingroup$ You can do Lagrangians with inhomogeneous and non-isotropic space and time, you just don't get energy and momentum conservation. $\endgroup$ – CuriousOne Apr 25 '16 at 5:46
  • $\begingroup$ Small quibble: you're extremising the action, which has the units of angular momentum, not energy. $\endgroup$ – J.G. Aug 7 '17 at 22:50
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  1. Firstly, given a differentiable Lagrangian $L(q,\dot{q},t)$, we can always form the Lagrangian energy function $$\tag{1} h ~:=~\sum_ip_i \dot{q}^i-L ,\qquad p_i ~:=~\frac{\partial L }{\partial \dot{q}^i }. $$

  2. Secondly, make the assumption that

$$\tag{2} \text{The Lagrangian } L=L(q,\dot{q}) \text{ has no }{\it explicit} \text{ time dependence.} $$

  1. Then Noether's theorem implies that the energy function (1) is conserved on-shell. For details, see e.g. this Phys.SE post.
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  • $\begingroup$ Thanks for the answer. Though, I'm not sure this quite answers things. To me, point number 2 is covered by homogeneity in space in time. In the derivation toward $\frac{1}{2}m\dot{q}^2$, we eventually assume that $L$ is only a function of $\dot{q}$. As far as point 1, it's not clear to me how this relates to the formulation above, which is sufficient for deriving $\frac{1}{2}m\dot{q}^2$. Really, I'm looking for some kind of missing assumption that precludes any sort of outside force or related idea that leads to the simple, plain, energy derivation $\frac{1}{2}m\dot{q}^2$. $\endgroup$ – wyer33 Apr 25 '16 at 19:04
  • $\begingroup$ For the derivation of the Lagrangian for a free non-relativistic point particle, see also e.g. this Phys.SE post. $\endgroup$ – Qmechanic Apr 25 '16 at 19:11
  • $\begingroup$ Given the above setup, I can prove the result I want. What's missing above as well as in your link are a complete list of the assumptions behind the derivation. The problem I have isn't in the proof; it's in the setup. $\endgroup$ – wyer33 Apr 25 '16 at 19:18

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