KleZMeR said:
No, I haven't. It asks: a Lagrange with kinetic energy in cartesian coordinates KE = [m/2]*(dx^2, dy^2, dz^2), and a potential in cartesian V = V(x^2 + y^2 , z) is invariant under some symmetry, find the corresponding invariant quantity. What I see is that the potential (V) is a function of some cylindrical shape. But I do not know what the "invariant quantities" are?
If you know the symmetry transformations, (the transformations which leave the Lagrangian invariant), you can easily deduce the corresponding constants of motion. The term “invariant quantities” is not accurate.
Consider your Lagrangian
L = \frac{ 1 }{ 2 } \left( ( \dot{ x }_{ 1 } )^{ 2 } + ( \dot{ x }_{ 2 } )^{ 2 } + ( \dot{ x }_{ 3 } )^{ 2 } \right) - V ( x_{ 1 }^{ 2 } + x_{ 2 }^{ 2 } , x_{ 3 } ) .
By inspection, we see that both KE and PE are invariant under rotations in the x_{ 1 } x_{ 2 }-plane. Therefore, the corresponding constant of motion is given by 3rd component of the angular momentum L_{ 3 } = x_{ 1 } p_{ 2 } - x_{ 2 } p_{ 1 }.
To see this, we need the explicit form of the transformations, i.e.,
\bar{ x }_{ 1 } = x_{ 1 } \cos ( \theta ) - x_{ 2 } \sin ( \theta ) ,
\bar{ x }_{ 2 } = x_{ 2 } \cos ( \theta ) + x_{ 1 } \sin ( \theta ) ,
and
\bar{ x }_{ 3 } = x_{ 3 } .
To find the constant of motion, it is sufficient to consider the infinitesimal version of the above transformations. We obtain this by setting \cos ( \theta ) = 1 and \sin ( \theta ) = \theta,
\bar{ x }_{ 1 } = x_{ 1 } - x_{ 2 } \theta , \ \ \Rightarrow \ \ \delta x_{ 1 } = - \theta x_{ 2 } , \ \ (1a)
\bar{ x }_{ 2 } = x_{ 2 } + x_{ 1 } \theta , \ \ \Rightarrow \ \ \delta x_{ 2 } = \theta x_{ 1 } , \ \ (1b)
and
\delta x_{ 3 } = 0 . \ \ \ \ \ (1c)
From these infinitesimal transformations, we find the following important relation
\frac{ d }{ d t } ( \delta x_{ i } ) = \delta ( \frac{ d }{ d t } x_{ i } ) \equiv \delta \dot{ x }_{ i } . \ \ \ \ (2)
The fact that the transformations leave the Lagrangian invariant (i.e., symmetry transformations) mean that
0 = \delta L = \sum_{ i = 1 }^{ 3 } \frac{ \partial L }{ \partial x_{ i } } \delta x_{ i } + \sum_{ i = 1 }^{ 3 } \frac{ \partial L }{ \partial \dot{ x }_{ i } } \delta \dot{ x }_{ i } . \ \ \ (3)
Now, if we use the Euler-Lagrange equation of motion in the first tern, and equation (2) in the second term, we find
\frac{ d }{ d t } \left( \sum \frac{ \partial L }{ \partial \dot{ x }_{ i } } \delta x_{ i } \right) = \frac{ d }{ d t } ( \sum p_{ i } \delta x_{ i } ) = 0 .
Or, using the transformations (1a)-(1c),
\frac{ d }{ d t } ( x_{ 1 } p_{ 2 } - x_{ 2 } p_{ 1 } ) \equiv \frac{ d L_{ 3 } }{ d t } = 0 .
Thus rotational invariance means that L_{ 3 } = x_{ 1 } p_{ 2 } - x_{ 2 } p_{ 1 } is time-independent on the physical trajectories (constant of motion corresponding to the invariance under rotations in the xy-plane).
Sam