# Does the conservation of $\frac{\partial L}{\partial\dot{q}_i}$ necessarily require $q_i$ to be cyclic?

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.

• 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. Jul 6 '19 at 5:19
• @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. Jul 6 '19 at 5:37
• 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. Jul 6 '19 at 5:55
• 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. Jul 6 '19 at 9:55
• @knzhou What should be the result of integration? $C_1$ be independent of $\dot{q}$? Jul 7 '19 at 5:24

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 $$t,q,\dot{q}$$. $$\Box$$

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.