# What are Euler's equations

1. Jul 24, 2014

### Greg Bernhardt

Definition/Summary

Euler's three equations relate the net torque on a rotating rigid body to the moment of inertia tensor and the angular momentum.

They are valid only in a frame of reference fixed in the body along three (perpendicular) principal axes, and therefore rotating with it.

They are valid only if the origin of the axes is at the centre of mass, or is stationary.

They are the three components along those axes of a single tensor equation.

Equations

$$\tau_1\ =\ I_1\,\frac{d\omega_1}{dt} + (I_3\ -\ I_2)\omega_2\omega_3$$

$$\tau_2\ =\ I_2\,\frac{d\omega_2}{dt} + (I_1\ -\ I_3)\omega_3\omega_1$$

$$\tau_3\ =\ I_3\,\frac{d\omega_3}{dt} + (I_2\ -\ I_1)\omega_1\omega_2$$

Extended explanation

Frame of reference fixed in space:

The rate of change of the angular momentum vector of a rigid body about any point equals the net torque vector about that point (the moment of all the external forces acting on that body):

$$\mathbf{\tau}_{net}\ =\ \frac{d\mathbf{L}}{dt}$$

This is the rotational version of Newton's second law.

The angular momentum vector of a rigid body about a point equals the moment of inertia tensor about that point "times" the angular momentum vector ($\mathbf{L}\ =\ \tilde{I}\,\mathbf{\omega}$), and so the net torque vector about that point is:

$$\mathbf{\tau}_{net}\ =\ \frac{d\mathbf{L}}{dt}\ =\ \frac{d}{dt}\left(\tilde{I}\,\mathbf{\omega}\right)\ =\ \tilde{I}\,\frac{d}{dt}\left(\mathbf{\omega}\right)\ +\ \frac{d\tilde{I}}{dt}\left(\mathbf{\omega}\right)$$

This only applies if that point is stationary or is the centre of mass or has a velocity v parallel to the velocity of the centre of mass: in any other case, there is an additional term, v x mvc.o.m..

Although the moment of inertia tensor, $\tilde{I}$, is fixed in the body, it is not fixed in space, and so $d\tilde{I}/dt$ is not zero (unless ω lies along a principal axis).

Frame of reference fixed in the body:

However, if we change to a frame of reference fixed in the body, then $d\tilde{I}/dt$ is zero, although there is an added "cross" term:

$$\mathbf{\tau}_{net}\ =\ \frac{d\mathbf{L}}{dt}\ +\ \mathbf{\omega}\times \mathbf{L}\ =\ \frac{d}{dt}\left(\tilde{I}\,\mathbf{\omega}\right)\ \ +\ \ \mathbf{\omega}\times \left(\tilde{I}\,\mathbf{\omega}\right)\ =\ \tilde{I}\,\frac{d}{dt}\left(\mathbf{\omega}\right)\ \ +\ \ \mathbf{\omega}\times \left(\tilde{I}\,\mathbf{\omega}\right)$$

which, expressed relative to three perpendicular axes fixed in the body along principal axes with moments of inertia $I_1\ I_2\ \text{and}\ I_3$, gives the three Euler's equations:

$$\tau_1\ =\ I_1\,\frac{d\omega_1}{dt} + (I_3\ -\ I_2)\omega_2\omega_3$$

$$\tau_2\ =\ I_2\,\frac{d\omega_2}{dt} + (I_1\ -\ I_3)\omega_3\omega_1$$

$$\tau_3\ =\ I_3\,\frac{d\omega_3}{dt} + (I_2\ -\ I_1)\omega_1\omega_2$$

Comparison:

Euler's equations give the same net torque as the fixed-frame equations.

The only reason why Euler's equations may be preferred to the fixed-frame equations is that the moment of inertia tensor is changing in the latter, but only the angular momentum vector is changing in the former …

and a non-zero derivative of a tensor is a lot nastier to work with than a non-zero derivative of a vector!

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