# 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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I have yet to find a physics book that doesn't make this really confusing. If one has a vector fixed in inertial space, its components as viewed in a moving frame are obtained by the dot product of the vector with the moving unit triad fixed to the body but moving relative to inertial space. While the inertial frame would measure its components as constants with time, the moving system would measure components that vary with time because the unit triad to which the components refer are moving relative to the fixed vector under investigation. The body rotates about an axis through an angle that can be described in the inertial frame, but all of its points are not moving if one is tied to and moving with the moving body frame. In fact, the only way one can deduce that he or she is tied to the body is via the coriolis force due to the rotational acceleration experienced. The body rotates through an axis and angle, each of which, in general, varies with time, relative to a secondary frame (often inertial) within which the motion of every point on the body can be observed as moving.

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