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I would like to derive Euler's equations of rigid body motion from least action principle.

Suppose we are in free space so we have no gravity so Lagrangian is equal to kinetic energy.

$$ L = T = \int_M \rho(x)(x\cdot \Omega_B)^2 d^3x $$

where $\rho$ is density, $\Omega_B$ is angular velocity bivector in reference body frame and $M$ is the rigid body.

The action principle says(if I haven't messed up something) that

$$ 0 = \frac{d}{ds}\bigg|_{s=0}\int_0^T \int_M \rho(x)[x\cdot (\Omega_B(t)+s \delta \Omega(t))]^2 d^3x $$

for every $\delta \Omega$ with zero ends i.e. $\delta \Omega(0)=\delta \Omega(T) = 0$. I should get these equations:

$$ I(\Omega_B) - \Omega \times I(\Omega_B)=0 $$


$$ I(\Omega) = \int_M \rho(x) (x\wedge (x\cdot \Omega)) d^3x $$

But I can't see where I would get the time derivative of $\Omega_B$ from the action principle. Can someone tell me what I'm doing wrong? I guess there is some trouble with that $\delta \Omega$ can't be completely arbitrary.

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I think I know where is the problem. $\Omega$ is not state variable!!! It indicates only the rate of rotation. – Tom Aug 21 '13 at 11:16
up vote 3 down vote accepted

I) A Lagrangian variational principle for Euler's equations for a rigid body

$$ \tag{1} (DL)_i ~=~M_i, \qquad i\in\{1,2,3\}, $$

is e.g. explained in Ref. 1. Here the angular momentum $L_i$, $i\in\{1,2,3\}$, along the three principal axes of inertia is tied to the angular velocity $\omega_i$, $i\in\{1,2,3\}$, by the formula

$$\tag{2} L_i~:=~I_i \omega_i, \qquad i\in\{1,2,3\}, \qquad (\text{no sum over }i).$$

The covariant time-derivative $D$ of a vector $\eta_i$, $i\in\{1,2,3\}$, is defined as

$$\tag{3} (D\eta)_i ~:=~ \dot{\eta}_i+(\omega\times\eta)_i, \qquad i\in\{1,2,3\}. $$

The angular velocity vector $\omega$ plays the role of a non-Abelian gauge connection/potential.

II) To see the $so(3)$ Lie algebra, we map an infinitesimal rotation vector $\alpha$ into an antisymmetric real $3\times3$ matrix $r(\alpha)\in so(3)$ as

$$\tag{4} \alpha_i \quad\longrightarrow\quad r(\alpha)_{jk}~:=~\sum_{i=1}^3\alpha_i \varepsilon_{ijk}. $$

The $so(3)$ Lie-bracket is given by (minus) the vector cross product

$$\tag{5} [r(\alpha),r(\beta)]~=~r(\beta\times \alpha). $$

Similarly, for the corresponding $SO(3)$ Lie group, a finite rotation vector $\alpha$ maps into an orthogonal $3\times3$ rotation matrix $R(\alpha)\in SO(3)$ as explained in this Phys.SE post. Infinitesimally, for an infinitesimal rotation $|\delta\alpha| \ll 1$, the correspondence is

$$\tag{6} R(\delta\alpha)_{jk} ~=~\delta_{jk} + r(\delta\alpha)_{jk} + {\cal O}(\delta\alpha^2). $$

III) A finite non-Abelian gauge transformation $\omega\longrightarrow\omega^{\alpha}$ takes the form

$$\tag{7} r(\omega^{\alpha})~=~R(-\alpha) \left(\frac{d}{dt}-r(\omega)\right)R(\alpha), \qquad \alpha\in \mathbb{R}^3.$$

An infinitesimal non-Abelian gauge transformation $\delta$ takes the form

$$\tag{8} r(\delta\omega)~=~\frac{d}{dt}r(\delta\alpha)-[r(\omega),r(\delta\alpha)],$$

or equivalently

$$\tag{9} \delta\omega_i~=~(D\delta\alpha)_i, \qquad i\in\{1,2,3\}, $$

where $\delta\alpha$ denotes an infinitesimal rotation vector corresponding to an $so(3)$ Lie algebra element $r(\delta\alpha)$.

We call (7)-(9) gauge transformations for semantic reasons, because of their familiar form, but note that (most of) them are not unphysical/spurious transformations. We stress that the angular velocity $\omega$ is a physical variable.

IV) Finally we are ready to discuss the action principle. The finite rotation vector $\alpha(t)\in \mathbb{R}^3$ plays the role of independent dynamical variables for the action principle. One may think of virtual rotation paths $\alpha:[t_i,t_f]\to \mathbb{R}^3$ as a reparametrization of virtual angular velocity paths $\omega:[t_i,t_f]\to \mathbb{R}^3$. The action reads

$$\tag{10} S[\alpha,\omega]~=~\int_{t_i}^{t_f} \! dt ~L $$

with Lagrangian

$$\tag{11} L~=~\frac{1}{2} L^{\alpha}\cdot \omega^{\alpha} + M\cdot \alpha , $$


$$\tag{12} L^{\alpha}_i~:=~I_i \omega^{\alpha}_i, \qquad i\in\{1,2,3\}, \qquad (\text{no sum over }i).$$

The Lagrangian (11) consists of rotational kinetic energy plus a source term from the torque $M$. Here $\omega^{\alpha}$ is the actual angular velocity vector, while $\omega$ here (in contrast to above) is a fixed non-dynamical reference vector, which is not varied. It is sort of a gauge-fixing choice. Infinitesimal variation yields

$$ \tag{13}\delta L ~\stackrel{(11)}{=}~ L^{\alpha}\cdot \delta\omega^{\alpha} + M\cdot \delta\alpha ~\stackrel{(9)}{=}~ L^{\alpha}\cdot \left(\frac{d}{dt}\delta\alpha + (\omega^{\alpha}\times\delta\alpha)\right) + M\cdot \delta\alpha ,$$

which (after integration by part and appropriate boundary conditions) leads to Euler's equations (1) for the angular velocity vector $\omega^{\alpha}$.


  1. J.E. Marsden and T.S. Ratiu, Introduction to Mechanics and Symmetry, 1998.
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Correction to the last sentence in the answer (v2): Integration by part should be Integration by parts. – Qmechanic Oct 2 '14 at 21:00

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