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The kinetic energy of a fixed, rotating rigid body is $$ T =\frac{1}{2}\mathbf{\omega}\mathbf{I}\mathbf{\omega}=\frac{1}{2}I_{xx}\omega_x^2 +\frac{1}{2}I_{yy}\omega_y^2 + \frac{1}{2}I_{zz}\omega_z^2 + I_{xy}\omega_x\omega_y + I_{yz}\omega_y\omega_z + I_{zx}\omega_z\omega_x, $$ and when the reference frame is rotated to coincide with the principal axis frame, the inertia tensor is diagonalised and the kinetic energy becomes $$ T = \frac{1}{2}I_x'\omega_x'^2 + \frac{1}{2}I_y'\omega_y'^2 + \frac{1}{2}I_z'\omega_z'^2. $$qftqft This is the form of kinetic energy in the Lagrangian formulation when used to solve the motion of e.g a spinning top.

But how is this legal? The reference frame used to describe $\mathbf{\omega'}$ and $\mathbf{I}'$ is non-inertial. Surely newton's laws and hence the euler-lagrange equations won't work here due to fictitious forces?

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There is no problem here because in usual rigid-body mechanics, the angular velocity is simply being decomposed with respect to the principal axes (which, in general, vary with time as the body rotates). The physics is not actually being done in the body frame because the angular velocity and momentum of the body in the body-fixed frame are, by definition, zero. The angular velocity and momentum vectors used are still those of the inertial laboratory frame. This is valid because tensors are basis-independent.

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But how is this legal? The reference frame used to describe $\mathbf{\omega'}$ and $\mathbf{I}'$ is non-inertial.

This is not what we do when we switch to the principle axis frame. The principle axis frame is inertial and non-rotating; we have simply rotated it some number of degrees in one or two directions so that off-diagonal terms vanish.

In other words, we are not switching to a rotating coordinate system, we are just choosing new axes such that they align with the principle axes of rotation. This amounts to, at most, one rotation in 3D space, since the equations we are dealing with have no specific reference to a location on any of the coordinate axes (if they did, we’d just add a translation in too to get the coordinate system origin onto whatever point is necessary).

Surely newton's laws and hence the euler-lagrange equations won't work here due to fictitious forces?

For Newton, yes, but I don’t believe I’m mistaken in saying that you can add terms to the Lagrangian to account for rotation, even if you were in a rotating reference frame, so as to re-validate the Euler-Lagrange equations.

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  • $\begingroup$ So if a body is rotating, the principal axes don't rotate with it? $\endgroup$ Commented Nov 4 at 17:52
  • $\begingroup$ @jeffreygorwinkle If they did rotate with it, you would find that the angular momenta would not be conserved in time as those are the axes about which the object rotates. So unless some torque is acting on the body, its principal axes should not rotate. $\endgroup$ Commented Nov 4 at 18:00
  • $\begingroup$ Principal axes are material axes. If the body rotates, they rotate as well: direction and vectors of a Cartesian basis change, the components of the tensor of inertia w.r.t. that material referenxe frame don't change. $\endgroup$ Commented Nov 4 at 23:59
  • $\begingroup$ Then you're right in recalling that Newton's laws (in their common expression) holds w.r.t. an inertial reference frame, but you're wrong if you think that you can't use any non-inertial reference frame to express vector (and tensor) physical quantities. Just remember that the equations in physics are vector equations, and don't depend on the reference frame or coordinates used to write them in components $\endgroup$ Commented Nov 5 at 0:04

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