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Say we have a system with a given Lagrangian $L(q_i, \dot q_i, t)$.

A continuous transformation $ Q_i= q_i +εK_i $ is a symmetry when L doesn't change in first order with respect to $ε$. In the same manner, it's a symmetry of action when the action $ S= \int_{t_1}^{t_2} Ldt $ doesn't change in first order.

Is there, given the Lagrangian, a specific method for finding all the symmetries of this Lagrangian/action? If we suppose a random transformation $ Q= q +εK $ and then demand that it's a symmetry how do we proceed? Do we know what/how many symmetries to expect? Thanks in advance :)

EDIT If it makes things easier, suppose the special case of a body in a homogenous and vertical gravitational field with $L={1 \over 2} m {\dot {\vec r}}^2 -m \vec g \vec r$. How do we find the symmetries of this lagrangian and of its action? I'd prefer an answer to the general case, but solving this example would give me an insight as well.

Arbiter
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    Given a lagrangian you do not demand that a given transformation is a symmetry. You only verify whether it is or not. There is no general procedure in finding symmetries of a lagrangian. Moreover, physicists normally do the opposite way, they construct lagrangians that satisfy given symmetries. – Diracology Apr 27 '17 at 17:44
  • Possible duplicates: https://physics.stackexchange.com/q/295077/2451 , https://physics.stackexchange.com/q/206355/2451 and links therein. – Qmechanic Apr 27 '17 at 18:34
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    So, notice that your specific Lagrangian only depends on the vector r, and the scalar distance $r^2$. And your gr is a dot product that picks the z (vertical) axis component of r, call it z. Think about how you can move that figure and still have it look the same. For instance, moving up and down on the z axis changes the vertical distance, so it is not the same. A transformation to move along the z axis is not a symmetry then. Try others. Physicists mostly do it by thinking thru it like the example. – Bob Bee Apr 28 '17 at 01:43

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