lagrangian formalism - Is it valid to replace the equations of motion inside a symmetry?

For example, this symmetry:

$$\delta q^{i}=\epsilon(q^{i}-2\dot{q}^{i}t)$$

it's derivative is:

$$\delta\dot{q}^{i}=-\epsilon(\dot{q}^i +2\ddot{q}^i t)$$

There appears $\ddot{q}^{i}$ in this expression, so I am tempted to replace the equations of motion there, but don't know if that is valid. I know you can't do that kind of stuff sometimes.

What I am trying to show is that the EOM of the Lagrangian $L=\frac{1}{2}\dot{q}_{i}\dot{q}^{i}-V(q_{i}q^{i})$ are invariant (or maybe not) under this symmetry.

Answer

Yes it is valid. The problem can be viewed this way. There is a manifold, a fiber bundle (a jet bundle actually) $A$, with basis $\mathbb R$ where the coordinate $t$ ranges and fibers $Q_t$ locally covered by coordinates $q^1,\ldots, q^n, \dot{q}^1, \ldots, \dot{q}^n$.

The solutions of E.-L. equations arising from a Lagrangian $L=L(t, q, \dot{q})$ are nothing but the integral curves $$\mathbb R \ni t \mapsto \gamma(t) = (t, q(t), \dot{q}(t)) \in A$$ of the dynamical vector field with local form

$$Z = \frac{\partial}{\partial t} + \sum_k \dot{q}^k \frac{\partial}{\partial q^k} + \sum_k A^k(q, \dot{q})\frac{\partial}{\partial \dot{q}^k}$$

Imposing that $\gamma$ is an integral curve of $Z$ we have

$$1=1, \quad \frac{dq^k}{dt} = \dot{q}^k(t)\:, \quad \frac{d\dot{q}^k}{dt} = A^k(t, q(t), \dot{q}(t))$$

the last requirement is nothing but the Euler-Lagrange differential equation system written into its normal form, separating on the left-hand side the derivatives of highest order from the other derivatives.

A one-parameter group of dynamical symmetries first of all admits a generator $X$. This is a vector field on $A$ whose integral lines are nothing but the evolution of the points of $A$ subjected to the symmetry.

A dynamical symmetry, by definition, moves solutions of the E.-L. solutions into solutions of E.-L- solutions. It is possible to prove that this is equivalent to $$[X,Z]=0 \tag{1}$$ where the bracket is the standard Lie bracket of vector fields. What is the natural structure of $X$ used in classical mechanics? I mean the one leading to the standard formulation of Noether theorem. Here is

$$X = 0\frac{\partial}{\partial t} + \sum_k X^k(t,q, \dot{q}) \frac{\partial}{\partial q^k} + \sum_k Z(X^k)(t, q, \dot{q})\frac{\partial}{\partial \dot{q}^k}\tag{2}$$

The factor $0$ means that the fixed time $t$ fibers are fixed under the symmetry. The functions $X^k(t,q)$ are arbitrary and $$Z(X^j) := \frac{\partial X^j}{\partial t} + \sum_k \dot{q}^k \frac{\partial X^j}{\partial q^k} + \sum_k A^k(q, \dot{q})\frac{\partial X^j}{\partial \dot{q}^k}\tag{3}$$ This is exactly you are doing written into another language.

The term $A^k$ disappears if $X$ is only function of $t$ and $q$. In this case one says that the symmetry is geometric.

To be complete I conclude saying that, if the Lagrangian of the system is $L$ and $Z$ is constructed out of it, and $X$ has the form (2), the invariance condition
$$X(L)=0 \tag{4}$$

implies both (1), so that we have a dynamical symmetry, and $$Z(N)=0\tag{5}$$ where $$N(t,q,\dot{q}) = \sum_{k=1}^n X^k \frac{\partial L}{\partial \dot{q}^k}\:.$$ The identity (5) just says that $N$ is a conserved quantity along the motion of the system.