I keep seeing the phrase "angular velocity with respect to the body frame", or "angular momentum with respect to the body frame" in context of rotation of rigid bodies, especially in the derivation of the Euler equations that is usually derived from:

$$(\frac d {dt}\vec L)_\text{inertial}=(\frac d {dt}\vec L)_\text{body}+\vec\omega\times\vec L_\text{body}$$

However, how can a rigid body have angular momentum or angular velocity with respect to its body frame, if the body frame is rotating with it too? I don't understand what is meant by $(\frac d {dt}\vec L)_\text{body}$, can someone please explain?

  • 1
    $\begingroup$ Most likely a duplicate. $\endgroup$
    – Kurt G.
    Sep 7 at 13:40
  • $\begingroup$ " angular velocity with respect to the body frame" means that the components of the angular velocity are given in body frame and not in inertial frame $\endgroup$
    – Eli
    Sep 7 at 15:18

1 Answer 1


The general rule you are referring to is

$$ \frac{\rm d}{{\rm d} t} \left\{ \blacklozenge \right\}_{\rm interial} = \frac{\rm d}{{\rm d} t} \left\{ \blacklozenge \right\}_{\rm body} + \vec{\omega} \times \left\{ \blacklozenge \right\}_{\rm body}$$

but the notation above is indeed confusing, since the frames of references do not apply to the quantities at hand, but to the derivative operator. As with any vector equation, all parts above must be in the same basis vector for it to have meaningful results, and so adding vectors of different basis vectors (inertial vs. body) is nonsensical.

This is why the most common form of the above is

$$ \underbrace{ \frac{\rm d}{{\rm d} t} \left\{ \blacklozenge \right\} }_{\text{change due to time and rotation}} = \underbrace{\frac{\partial }{{\partial} t} \left\{ \blacklozenge \right\}}_{\text{change due to time}} + \underbrace{ \vec{\omega} \times \left\{ \blacklozenge \right\} }_{\text{change due to rotation}}$$

All of the vector quantities must be expressed on the same basis vector, but the change considered for each derivative is different.

Take for example angular momentum about the center of mass on the inertial reference frame

$$ \vec{L}_C = {\rm I}_C \vec{\omega} $$

Since ${\rm I}_C$ must be on the inertial basis vector, the components of this tensor change in time due to the rotation only. This is expressed as

$$ \begin{gather} {\rm I}_C = {\rm R}\, \left\{ {\rm I}_C \right\}_{\rm body}\,{\rm R}^\intercal \\ \frac{\rm d}{{\rm d}t} \left\{ {\rm I}_C \right\}_{\rm body} = 0 \\ \end{gather}$$

where ${\rm R}$ is the change of basis rotation vector and its components change over time also due to the rotation only.

So applying the rule to angular momentum, we have

$$ { \frac{\rm d}{{\rm d} t} \left\{ {\rm I}_C \vec{\omega} \right\} } = {\frac{\partial }{{\partial} t} \left\{ {\rm I}_C \vec{\omega} \right\}} + {\vec{\omega} \times \left\{ {\rm I}_C \vec{\omega} \right\} }$$

and from the product rule

$$ \require{cancel} \frac{\rm d}{{\rm d} t} \left\{ {\rm I}_C \vec{\omega} \right\} = \left\{ \cancel{ \frac{\partial }{{\partial} t} {\rm I}_C } \right\}\vec{\omega} + {\rm I}_C \left\{ \frac{\partial }{{\partial} t} \vec{\omega} \right\} + \vec{\omega} \times \left\{ {\rm I}_C \vec{\omega} \right\} $$

or simplified and related to the law of motion $\vec{\tau}_C = \frac{\rm d}{{\rm d} t} \left\{ {\rm I}_C \vec{\omega} \right\}$

$$ \vec{\tau}_C = {\rm I}_C \dot{\vec{\omega}} + \vec{\omega} \times \left\{ {\rm I}_C \vec{\omega} \right\}$$

which is Euler's law of rotational motion in terms of angular velocity.

In summary, when you see a $\left\{ \blacklozenge \right\}_{\rm body}$ you can interpreted as the quantities if the body was not rotating, and the $\vec{\omega}\times$ part adds the effect of rotation back into the expression.

For example, lets say you have a particle on a rotating disk moving radially outwards at some time frame with speed $v$. At the same time frame the direction of motion is $\boldsymbol{e}$, expressed in the inertial basis vectors.

So at all times you have

$$ \boldsymbol{v} = v \boldsymbol{e} $$

What is the acceleration of the particle?

Apply the law above to get

$$\frac{\rm d}{{\rm d}t} \boldsymbol{v} = \frac{\partial}{\partial t}\left\{ v \boldsymbol{e} \right\} + \boldsymbol{\omega} \times \boldsymbol{v} $$

but when rotation is ignored (or from the point of view of the body) the direction vector does not change, so

$$\require{cancel} \frac{\partial}{\partial t}\left\{ v \boldsymbol{e} \right\} = \dot{v} \boldsymbol{e} + v \left\{ \cancel{ \frac{\partial }{\partial t} \boldsymbol{e} } \right\}$$

which means the acceleration of the particle is

$$ \boldsymbol{a} = \dot{v} \boldsymbol{e} + \boldsymbol{\omega} \times \boldsymbol{v} $$

where $\dot{v} \boldsymbol{e}$ is called the Eulerian or spatial acceleration and $\boldsymbol{\omega} \times \boldsymbol{v}$ is the Coriolis acceleration.

If you do the same with angular acceleratgion you will find the corriolis term to vanish to get

$$\boldsymbol{\alpha} = \boldsymbol{\dot{\omega}} $$


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