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Suppose a rigid body is invariant under a rotation around an axis $\mathsf{A}$ by a given angle $0 \leq \alpha_0 < 2\pi$ (and also every multiple of $\alpha_0$).

Is it true that in this case the axis $\mathsf{A}$ is a principal axis of the rigid body?

If so, how to prove it? Do you have any references for a proof?

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I think that using Poinsot's ellipsoid (inertia ellipsoid) would serve as intuitive proof which with the aid of some formalism would legitimize it. –  qoqosz Jun 6 '12 at 15:03
    
@qoqosz Could you provide some details for your idea? –  martin Jun 7 '12 at 14:33
    
the moments of inertia around axes perpendicular to A must be all the same (they form a circle if $\alpha_0 < \pi$) - so it's a symmetrical top but I'm not sure if this leads to something. –  qoqosz Jun 8 '12 at 17:20
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3 Answers

up vote 3 down vote accepted

Your statement is true.

Proof:

Let $\rho$ be the mass density of the rigid body.

Remember that the tensor of inertia $I$ is given by:

$$ \vec{v}^t I \vec{w} = \int d^3b\, \rho(\vec{b}) (\vec{v} \cdot \vec{w} - (\vec{v}\cdot \vec{b})(\vec{w}\cdot \vec{b})) $$

for all $\vec{v},\vec{w} \in \mathbb{R}^3$.

Now take an orthogonal matrix $O$ which represents a rotation around an axis $\mathsf{A}$ with direction vector $\vec{n}$, i.e. $O\vec{n} = \vec{n}$.

Your invariance means that $\rho$ is invariant under $O$, i.e. $\rho(O \vec{x}) = \rho(\vec{x})$ for all $\vec{x}$.

Next show that the inertia tensor commutes with $O$: $IO = OI$: $$ \begin{align*} \vec{v}^t I O \vec{w} &= \int d^3b\, \rho(\vec{b}) (\vec{v} \cdot O \vec{w} - (\vec{v}\cdot \vec{b})(O \vec{w}\cdot \vec{b})) \\ &= \int d^3b\, \rho(\vec{b}) (O^t \vec{v} \cdot \vec{w} - (\vec{v}\cdot \vec{b})(\vec{w}\cdot O^t \vec{b})) \\ &= \int d^3b\, \rho(O^t \vec{b}) (O^t \vec{v}\cdot \vec{w} - (O^t \vec{v}\cdot O^t \vec{b})(\vec{w}\cdot O^t \vec{b})) \\ &= \int d^3b\, \rho(\vec{b}) (O^t \vec{v}\cdot \vec{w} - (O^t \vec{v}\cdot \vec{b})(\vec{w}\cdot \vec{b})) \\ &= \int d^3b\, \rho(\vec{b}) ((\vec{v}^t O)^t\cdot \vec{w} - ((\vec{v}^t O)^t \cdot \vec{b})(\vec{w}\cdot \vec{b})) \\ &= \vec{v}^t O I \vec{w} \end{align*} $$

for all $\vec{v},\vec{w} \in \mathbb{R}^3$

Then one sees that $I\vec{n}$ is again an eigenvector of $O$ because $O(I\vec{n}) = IO\vec{n} = I\vec{n}$.

Now $I$ has only one real eigenvalue $1$ with eigenspace $\mathbb{R}\vec{n}$. This implies that there is a unique $\lambda$ with $I\vec{n} = \lambda \vec{n}$. Thus $\vec{n}$ is an eigenvector of $I$ i.e. $\vec{n}$ points along a principal axis of $I$.

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As a starting point, Wikipedia says on this issue:

The principal axes are often aligned with the object's symmetry axes. If a rigid body has an axis of symmetry of order $m$, meaning it is symmetrical under rotations of 360°/$m$ about the given axis, that axis is a principal axis. When , the rigid body is a symmetrical top. If a rigid body has at least two symmetry axes that are not parallel or perpendicular to each other, it is a spherical top, for example, a cube or any other Platonic solid.

But it gives no canonical reference.

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Thanks. Do you have a proof? –  martin Jun 6 '12 at 13:31
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At least as phrased ("by a given angle $\alpha_0$"), axis $\mathsf{A}$ can be trivially shown not to necessarily be a principal axis. If $\alpha_0$ is a multiple of $2\pi$, every rigid body is invariant under a rotation about any axis.

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Thanks, I added this constraint to the question. –  martin Jun 6 '12 at 14:37
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