- #1

Boorglar

- 210

- 10

While studying the equations of motion for a free rigid body, I decided to work out the Lagrangian in terms of the Euler angles as generalized coordinates. I got the following expression for the kinetic energy (rotational, ignoring the center of mass movement), which is also the Lagrangian:

[tex]L = \frac{1}{2}\left(I_1ω_1^2 + I_2ω_2^2 + I_3ω_3^2\right) = \frac{1}{2}\left[I_1\left(\dot{\phi}\sinθ\sinψ + \dot{θ}\cosψ\right)^2 + I_2\left(\dot{\phi}\sinθ\cosψ-\dot{θ}\sinψ\right)^2 + I_3\left(\dot{ψ}+\dot{\phi}\cosθ\right)^2\right][/tex]

The [itex]ω_i[/itex] are the components of the angular velocity in the body frame, and can be written in terms of the euler angles ([itex]\phi,\ θ,\ ψ[/itex] for precession, nutation and spin respectively of the body frame). The [itex]I_i[/itex] are the principal moments of inertia.

Now Noether's Theorem implies that if the Lagrangian is rotationally symmetric, angular momentum is conserved. But in this case, the Lagrangian does not seem to be rotationally symmetric. Indeed, a rotation would correspond to a change in one of the Euler angles. And yet [itex]\frac{∂L}{∂θ},\ \frac{∂L}{∂ψ} ≠ 0[/itex].

Does this mean that not every conservation law arises from symmetries of the Lagrangian, or have I misunderstood what the rotational symmetries should be (i.e. a change in one Euler angle is not actually a rotation)?