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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6 Answers
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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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$\begingroup$ 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. $\endgroup$– gilonikCommented Jul 25, 2012 at 17:10
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$\begingroup$ @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. $\endgroup$ Commented Jul 25, 2012 at 19:23
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$\begingroup$ 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. $\endgroup$– gilonikCommented Jul 25, 2012 at 20:34
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$\begingroup$ @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. $\endgroup$ Commented Jul 25, 2012 at 21:35
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The frame is only instantaneously aligned with the body frame. The measuring frame is not moving, but the body frame is. So the motion and momentum measure non zero because is it only the alignment that is used and not the motion for measuring. The equations of motion are still on an inertial frame, just not aligned with the world coordinate system.
The way I realized this is by considering forces also. Forces are always described in inertial frames, and can be rotated at any orientation. For the vector equations of motion to work all quantities must be on the same coordinate frame, so velocities, accelerations, momenta, forces and moments all have to be on an inertial frame all the time.
answered Nov 28, 2015 at 1:47
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As shown in physics mechanics textbooks, using a passive rotation, any vector $\vec A$ can be considered as the same vector in both the inertial and a rotating frame, and $$ (1) {d\vec A \over dt} = {d^*\vec A \over dt} + \vec \omega \times \vec A$$ where ${d \over dt}$ is in the inertial frame and ${d^* \over dt}$ is the derivative of the same vector expressed using coordinates of a frame rotating at angular velocity $\vec \omega$ with respect to the inertial frame.
Relationship (1) applies to a given vector $\vec A$, and for $\vec A$ taken to the angular momentum in the space frame, $\vec L$, ${d^*\vec L \over dt}$ in (1) is the derivative of the angular momentum in the space frame expressed in body frame coordinates. The angular momentum in the body frame, call it $\vec L^*$, is zero. In general $\vec L \ne \vec L^*$, and $({d \vec L \over dt}) \ne ({d^* \vec L^* \over dt})$. If we apply (1) to $\vec L$, we have ${d\vec L \over dt} = {d^*\vec L \over dt} + \vec \omega \times \vec L$. If we apply (1) to $\vec L^*$, we have ${d\vec L^* \over dt} = {d^*\vec L^* \over dt} + \vec \omega \times \vec L^* = \vec \omega \times \vec L^*$.
In summary, relationship (1) applies to the same vector in both coordinate systems, and the angular momentum vector in the rotating system is in general not the same as the angular momentum vector in the inertial system.
Consider a rigid body in rotation about a fixed point. $\vec L = \bf I \vec \omega$ is the angular momentum of the body in an inertial frame (the space axes) where $\bf I$ is the inertia tensor and $\vec \omega$ is the angular velocity of the body about the point of rotation. Let $\vec M$ be the total external torque in the inertial frame. $$(2) \vec M = {d(\bf I \vec \omega) \over dt}$$ In the space frame $\bf I$ is not constant.
The non-inertial body frame fixed in the body rotates with the body (and with respect to the inertial space frame). In the body frame $\bf I$ is constant- call is $\bf I_B$, so we would like to apply relationship (1) to simplify relationship (2) using $\bf I_B$.
We have $$(3) \vec M = {d(\bf I \vec \omega) \over dt} ={\bf I_Bd^*( \vec \omega) \over dt} + \vec \omega \times \vec (\bf I_B \vec \omega)$$ considering the * frame to be rotating body axes where $\bf I_B$ is constant in that frame. Taking the body axes to be the principal axes in relationship (3) yields Euler's equations of motion.
The right-hand side of relationship (3) expresses the rate of change of the angular momentum in the space frame in terms of body frame coordinates. The right-hand side is not the rate of change of the angular momentum in the rotating body frame; with respect to the rotating frame the body is fixed (not rotating) and the angular momentum is zero.
To consider translation as well as rotation is more complicated but for the angular momentum evaluated with respect to the moving center of mass, relationship (1) is still valid.
answered Jul 19, 2022 at 3:30
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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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You must remember that the 'body-frame' (which is the frame 'inside' the rotating body) is a frame and that OTHER bodies (which may, for example, be rotating about the same point with a different speed) will appear in the 'body-frame' to have an angular speed less than if observed in an inertial (stationary) frame.
To help visualise, if you stick your arm out to the side (parallel to your chest) and start spinning, whilst looking ONLY at your fist, your fist would appear stationary to a blurry background to YOU. Now if you continue spinning but additionally move your arm, YOU would observe your fist moving at a slow speed, whereas an outside person would observe the fist increase its speed further.
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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.
