Usually, when people talk about Lagrangians they are talking about a function of configuration space variables $q_i$ and their time derivatives $\dot q_i$. This is a function $L = L(q_i, \dot q_i,t)$. However, there is another notion of the Lagrangian, the "Hamiltonian Lagrangian", which is also a function of $p_i$. $$ L_H(q_i, \dot q_i, p_i,t) = p_i \dot q_i - H(q_i, p_i,t) $$ Noether's theorem tells us that for every "off-shell" symmetry of of $L$ (these are transformations that change $L$ by a total time derivative) we have a conserved quantity. However, there are some symmetries that can only be seen using the Hamiltonian Lagrangian $L_H(q_i, \dot q_i, p_i,t)$, and not with the configuration space Lagrangian $L(q_i, \dot q_i,t)$. A famous example is the hidden $SO(4)$ symmetry in the Kepler problem with gives rise to the conservation of the Laplace Runge Lenz vector. This symmetry apparently "mixes up" the $q$'s and $p$'s in such a way that the configuration space Lagrangian cannot capture.

So we might say that symmetries with mix up $q_i$ and $p_i$ are not apparent in the configuration space Lagrangian and call it a day. However, this answer does not make much sense after closer inspection. For example, the time translational symmetry, generated by the conserved quantity $H$, certainly "mixes up $q$ and $p$" and yet is present in the configuration space Lagrangian in terms of the symmetry $q_i \to q_i + \varepsilon \dot q_i$.

Furthermore, consider this: all "off shell" symmetries are symmetries of the equations of motion. (The converse is not true.) This is just because if a path is stationary, then the path subjected to the symmetry transformation will also be stationary, because the action of the path and all nearby variations will be unchanged. Therefore, even the hidden $SO(4)$ symmetry must represent a symmetry of the equations of motion (probably has to do with changing the eccentricity of the orbit) and therefore ought to be expressible as a symmetry of the configuration space Lagrangian.

So, all that said, why are some symmetries invisible to the configuration space Lagrangian? Is there a robust criterion for determining if a symmetry is invisible in this way?

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    $\begingroup$ Laplace Runge Lenz arises because the Hamilton-Jacobi equation (which is not Hamiltonian mechanics it's it's own formulation) for the action in the Kepler problem is superintegrable en.wikipedia.org/wiki/… and it can be formulated in a strictly Lagrangian setting using a transformation involving just position and velocity, see aapt.scitation.org/doi/10.1119/1.1986202 $\endgroup$
    – bolbteppa
    Jan 24, 2019 at 19:54

1 Answer 1

  1. If the Legendre transformation is regular, then there is a bijective correspondence between quasisymmetries of the Hamiltonian action and quasisymmetries of the corresponding Lagrangian action, cf. this Phys.SE post.

  2. Of course a Lagrangian quasisymmetry may involve (higher) time-derivatives of the $q^i$ variables, and this seems to be why some authors call it "subtle/hidden/invisible" in configuration space, but it's still there.

  3. Example: The Laplace-Runge-Lenz vector conservation discussed in this Phys.SE post.

  • $\begingroup$ So in other words, you are saying I have it backwards. If the Legendre transformation is "regular" (which I assumes means you have a bijection between $\dot q_i$ and $p_i$ then you are saying all $L_H$ symmetries descend into $L$ symmetries. However, you could have $L$ symmetries (depending on higher derivatives of $q_i$ which cannot be seen in $L_H$, correct? $\endgroup$ Jan 24, 2019 at 20:34
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    $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Jan 24, 2019 at 21:08

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