I'm studying rigid body motion on Landau but I'm having troubles to understand this proof of the fact that the angular velocity $\vec{\Omega}$ is constant for a rigid body.

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My doubt is about the two last equations in the last two lines of the text. If I use the two I get

$\vec{V'}=\vec{V}+(\vec{\Omega}-\vec{\Omega'})\times \vec{r'} +\vec{\Omega} \times\vec{a}$

How are (31.3) derived from this?

  • 1
    $\begingroup$ using $\boldsymbol\Omega'=\boldsymbol\Omega$? $\endgroup$ Apr 2, 2016 at 14:22
  • $\begingroup$ Isn't that the definition of a rigid body? A clump of particles sharing a common angular velocity such that velocities transform as $$\vec{v}_A = \vec{v}_B + \vec{\omega} \times (\vec{r}_A - \vec{r}_B)$$ $\endgroup$ Apr 2, 2016 at 14:41
  • $\begingroup$ @AccidentalFourierTransform But isn't $\vec{\Omega}=\vec{\Omega'}$ what is being proved? $\endgroup$
    – Sørën
    Apr 2, 2016 at 14:45

1 Answer 1


You have

$$ \boldsymbol{v} = \boldsymbol{V'} + \boldsymbol{\Omega'} \times \boldsymbol{r'} $$


$$ \boldsymbol{v} = \boldsymbol{V} + \boldsymbol{\Omega} \times (\boldsymbol{r'}+\boldsymbol{a}) $$

You collect the $\boldsymbol{r'}$ terms

$$ \boldsymbol{v} = \boldsymbol{V'} + \boldsymbol{\Omega'} \times \boldsymbol{r'} = \left( \boldsymbol{V}+ \boldsymbol{\Omega} \times \boldsymbol{a} \right) + \left( \boldsymbol{\Omega} \times \boldsymbol{r'} \right) $$

which is solved uniquely when

$$ \begin{align} \boldsymbol{\Omega'} & = \boldsymbol{\Omega} \\ \boldsymbol{V'} &= \boldsymbol{V}+ \boldsymbol{\Omega} \times \boldsymbol{a} \end{align} $$


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