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If a generalized coordinate $q_i$ is cyclic, the conjugate momentum $p_i=\frac{\partial L}{\partial\dot{q}_i}$ is conserved.

Is the converse also true? To state more explicitly, if a conjugate momentum $$p_i=\frac{\partial L}{\partial\dot{q}_i}=C_1\tag{1}$$ is conserved, will $q_i$ be necessarily cyclic? If we integrate $(1)$, we get $$L=C_1(q_i,\dot{q}_i)\dot{q}_i+C_2(q_i) q_i\tag{2}$$ From $(2)$, it is evident that the conservation of $p_i$ does not necessarily imply $q_i$ is cyclic. $q_i$ is cyclic only if $C_2=0$ which is only a special case.

Assuming my little observation is correct what is an example (perhaps a physical one) of such a situation i.e., a conserved $p_i$ with a non-cyclic $q_i$? I cannot immediately think of one.

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    $\begingroup$ I do not understand what "integrate" means. If $C_2\neq 0$, then EL equations produce $d p_i/dt = C_2$, so that it is not constant in time. However, using the existence an uniqueness theorem for EL equations it arises that $p_i$ is a constant of motion if and only if $q_i$ is cyclic. $\endgroup$ Jul 6 '19 at 5:19
  • $\begingroup$ @ValterMoretti From Eq. (1), if a time-derivative is taken, then $\frac{dp_i}{dt}=0$. Eq. (2), is obtained by simply integrating Eq. (1) w.r.t $\dot{q}_i$. I don't see where I did a mistake. $\endgroup$ Jul 6 '19 at 5:37
  • $\begingroup$ Sorry I do not understand. 'Constant' here means along the time evolution when solving EL equations. It has noting to do with that type of integration. $\endgroup$ Jul 6 '19 at 5:55
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    $\begingroup$ The way you “integrated” that equation doesn’t make sense. You can check that it gives an incorrect answer in even the simplest of examples. $\endgroup$
    – knzhou
    Jul 6 '19 at 9:55
  • $\begingroup$ @knzhou What should be the result of integration? $C_1$ be independent of $\dot{q}$? $\endgroup$ Jul 7 '19 at 5:24
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THEOREM. Assume that $L(t,q, \dot{q})$ is jointly $C^2$ in the considered coordinate patch and the Hessian matrix of coefficients $\frac{\partial^2 L}{\partial q^r \partial q^s}$ is everywhere non-singular.

Then, $p_k$ is a constant of motion (it is constant along every solution of EL equations) if and only if $q^k$ is cyclic ($\frac{\partial L}{\partial q^k}(t,q,\dot{q})=0$ for every choice of $t,q,\dot{q}$).

PROOF. If $L(t,q, \dot{q})$ is jointly $C^2$ and the Hessian matrix of coefficients $\frac{\partial^2 L}{\partial q^r \partial q^s}$ is non-singular, then for every choice of initial conditions $(t_0, q(t_0), \dot{q}(t_0))$ there is a local solution of EL equations satisfying those initial conditions (requiring for instance $C^3$ this solution turns out to be maximal and unique). Let us pass to the main statement.

If $\frac{\partial L}{\partial q^k}=0$ for every $(t, q, \dot{q})$, then every solution $t \mapsto (t, q(t), \dot{q}(t))=:\gamma(t)$ of EL equations $$\frac{d}{dt}\left( \frac{\partial L}{\partial \dot{q}}|_{\gamma(t)}\right)= \frac{\partial L}{\partial q^k}|_{\gamma(t)}\:, \quad \frac{dq^k}{dt}|_{\gamma(t)} = \dot{q}(t)$$ satisfies $$\frac{dp_k|_\gamma(t)}{dt}=\frac{d}{dt}\left( \frac{\partial L}{\partial \dot{q}}|_{\gamma(t)}\right)=0$$ so that $p_k$ is a constant of motion.

Vice versa, if for every solution $\gamma$ it holds $$\frac{d}{dt}\left( \frac{\partial L}{\partial \dot{q}}|_{\gamma(t)}\right)=0\:,$$ form EL equations we also have that $$\frac{\partial L}{\partial q^k}|_{\gamma(t)}=0\:.$$ To conclude, fix a kinetical state $(t,q,\dot{q})$. We know that there is a solution of EL equations which admits that state as initial conditions. Therefore, evaluating $p_k$ along that solution at time $t$, $$\frac{\partial L}{\partial q^k}(t,q,\dot{q})\left(=\frac{dp_k|_\gamma(t)}{dt}\right)=0\:,$$ for every choice of $t,q,\dot{q}$. $\Box$

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  1. The title question (v2) fails e.g. for static Lagrangians $L(q)$ independent of $\dot{q}$.

  2. More generally, the title question essentially asks about the possible existence of an inverse Noether theorem, see e.g. this Phys.SE post.

  3. One cannot naively integrate on-shell equations $\frac{\partial L}{\partial \dot{q}^j}~\approx~c_j$ to deduce an off-shell Lagrangian $L$ (even if $L$ is known to exists).

  4. Still not convinced? Try to work out what happens in the case of a free non-relativistic particle.

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