# motion in the body-fixed frame?

This is really basic, I'm sure: For rigid body motion, Euler's equations refer to $L_i$ and $\omega_i$ as measured in the fixed-body frame. But that frame is just that: fixed in the body. So how could such an observer ever measure non-zero $L$ or $\omega$?

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I fussed about this as well. My resolution: for these calculations the fixed-body frame is not to be considered as co-moving with the body, but rather a non-rotating frame that instantaneously aligns with the body.

The Euler angles translate between the body and the space frames. The Euler angles are indeed functions of time, and the fixed-body frame is as well, but angular velocity and momentum are measured with respect to a fixed "snapshot" of the body frame at a particular time.

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But doesn't that make the body frame inertial -- and isn't the whole point of the Euler equations that they take into account non-inertial "psuedo-torques"? An observer on a spinning object is certainly non-inertial. And in fact, the precession of a free symmetric top is observable by an observer in the body frame: earth's $\omega$ precesses in cone once every 300-400 days. –  gilonik Jul 25 '12 at 17:10
@gilonik, I think my answer was a bit too concise, sorry. The body frame is indeed non-inertial. However, to calculate angular velocity, one first establishes an inertial frame that coincides with the body frame at a particular time, and then determines the infinitesimal rotation of the body frame with respect to the inertial frame in a time dt. The angular velocity is that infinitesimal rotation / dt. A reference is Goldstein, Classical Mechanics, section 4-9. –  Art Brown Jul 25 '12 at 19:23
That helps, thanks. The resolution must be in the instantaneous/infinitesimal caveat. So maybe I just can't picture it. Here's the point: why isn't the body-fixed frame the "rotational rest frame"? I know it isn't: The free precession of the earth shows that we in the body-fixed frame observe an $\vec{L}$ even though we're rotating with the earth. That this $\vec{L}$ differs in magnitude (and direction?) from that observed in the space-fixed frame is fine; that it's non-zero is where I'm getting mixed up. –  gilonik Jul 25 '12 at 20:34
@gilonik, right, a body-fixed observer might think the body was not spinning (if the surroundings were ignored). Crucially, however, the body-fixed frame is not an inertial frame, so motion is complicated by Coriolis forces. That effect is described by (dG/dt)s=(dG/dt)b + wxG, relating the rate of change in a vector G in the space and body frames, with the angular momentum vector w as defined above. (In that sense, w quantifies the non-inertial-ness.) Setting G=L=Iw, one gets the Euler equations. It's only with that definition of w that one deduces the correct equations of motion. –  Art Brown Jul 25 '12 at 21:35

Your reference is probably referring to the angular momentum and velocity of the fixed body frame relative to some inertial frame.

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