This question is relevant to Euler's angles and Euler's equations for a rigid body. Why aren't $\omega_1$, $\omega_2$ and $\omega_3 = 0$ in the body frame? How can we measure $\vec\omega$?

  • 1
    $\begingroup$ May I suggest that you change the title of the question to "Why is body frame angular velocity nonzero?" $\endgroup$ Commented Nov 29, 2013 at 22:51

3 Answers 3


Great question; I remember being so confused by this when I first took analytic mechanics.

The components of the angular velocity "in the body frame" aren't zero because when one writes these components, one isn't referring to measurements of the motions of the particles in the body frame (because, of course, the particles are stationary in this frame). Instead, one is referring to angular velocity as measured in an inertial frame but whose components have simply been written with respect to a time-varying basis that is rotating with the body.

In practice, we make measurements of the positions $\mathbf x_i(t)= (x_i(t),y_i(t),z_i(t))$ of the particles in an inertial frame. Then, we note that for a rigid body (let's consider pure rotation for simplicity), the position of each particle $i$ satisfies \begin{align} \mathbf x_i(t) = R(t) \mathbf x_i(0) \end{align} for some time-dependent rotation $R(t)$. Then we compute $\boldsymbol\omega(t) = (\omega^x(t),\omega^y(t),\omega^z(t))$ in the standard way in terms of $R(t)$. To see how this is done in detail, see, for example


Once we have $\boldsymbol\omega$, we can write its components with respect to any basis we like. If we write it in the standard ordered basis $\{\mathbf e_i\}$, then we'll just get $\omega_x(t)$ as its components. If we write it in some basis $\{\mathbf e_{i,B}(t)\}$ that is rotating with the body (like one that points along the principal axes of the body) then we get different components $\omega^i_B(t)$, and these are the body components.

Main Point Reiterated. Angular velocity is being measured with respect to an inertial frame, but its components can be taken with respect to any basis we wish such as one rotating with the body.

  • $\begingroup$ @Artemisia Sure thing. I can totally relate to that feeling. I once spent half an hour in office hours with my TA as he tried to explain this to me, but I didn't get it until grad school. I'd like to think his explanation just wasn't that great, but I must concede that maybe I just wasn't smart enough to understand back then. $\endgroup$ Commented Nov 29, 2013 at 20:25
  • $\begingroup$ Haha I see. I have another question... typing it out now :) $\endgroup$
    – Artemisia
    Commented Nov 29, 2013 at 20:37
  • $\begingroup$ @joshphysics physics.stackexchange.com/questions/595241/… please answer this. $\endgroup$ Commented Nov 20, 2020 at 17:26
  • $\begingroup$ @joshphysics when we say "measured" does it also mean that in principle an observer relative to the body frame could calculate $\omega$ in his reference system measuring the effects of the centrifugal and coriolis acceleration? thanks $\endgroup$
    – ebenezer
    Commented May 11 at 15:19

Because the angular velocity vector ω is “measured” in the inertial frame and is just expanded/written using the body frame basis. And we do this to get rid of diagonalization of the moment of inertia tensor.


Its because motion is relative. When you say a body is moving (or rotating), your statement must exactly involve two frames. You can never make a statement regarding a motion that involves only single frame. So you are always either saying there is motion in frame x with respect to frame y or vice versa. You never say that there is motion in frame x without reference to any other frame.

Now, assume you are standing in world frame and you say that body's frame is moving with respect to frame frame with velocity v. However, if you were sitting on the body then you wouldn't say that body has stopped moving but rather you would say that rest of the world is now moving with velocity -v with respect to you. Similarly, when body is rotating and if you are sitting on the body, the relative angular velocity between you and rest of the world is not zero but - $\omega$.

So the bottom line... When you say some vector quantity is x in world frame you are implicitly saying that other frame you are referring to is body frame. Similarly when you say some quantity is x in body frame, you are implicitly saying that other frame is world frame. The relative velocity of body frame wrt to body frame, of course would be 0 at all times.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.