SeReNiTy said:
The derivation is if the lagrangian contained velocity terms, derive what the conjugate momentum will be if the system is rotated by a angle in some direction.
Well I hope you know the lagrangian is given by:
[tex]L=T-V = \frac{1}{2} m (\dot{x}+\dot{y}+\dot{z}) - V(x,y,z)[/tex]
In cartesian coordinates. The conjugate momentum for a particular coordinate is given by:
[tex]\frac{\partial L}{\partial \dot{x}}= p_x[/tex]
Now if the system was rotated by a particular angle how would you then modify the original expression to deal with a rotation. If you consider the x,y,z coordinates to be a vector [tex]\mathbf{r}[/tex] rotated about any arbitrary unit vector [tex]\mathbf{\hat{a}}[/tex] by a small angle [tex]\theta[/tex].
Now all you have to do is findout how the x,y,z coordinates are related to [tex]\mathbf{\hat{a}}[/tex] through vector [tex]\mathbf{r}[/tex] and plug them into the lagrangian and see what the turn up. In some books they set [tex]\mathbf{\hat{a}}[/tex] parallel to the z-axis for ease of computation and normally call it [tex]\mathbf{\hat{k}}[/tex] instead.
Like I say I don't have the textbook but I assume this is something to do with conservation of angular momentum?
Need any more pointers just post.