# What are the principal axes?

1. Apr 5, 2007

### quasar987

The book was vague on this subject and as a result, everyone is in disagreement on this.

All my friends think that the directions of the principal axes are the eigenvectors of the tensor of inertia wrt some Oxyz axes.

I think that the directions of the principal axes are obtained by applying on the Oxyz axes the matrix R that diagonalizes the tensor of inertia. (i.e. the matrix whose lines are the eigenvectors of I)

The T.A. thinks we're both right.

2. Apr 5, 2007

### neurocomp2003

Principal Axes are the coordinate Axes of the rotation R that diagonalizes I. (defined in Fowles and Cassiday)

"the matrix whose lines are the eigenvectors of I" are just the standard axes.

3. Apr 5, 2007

### quasar987

This is pretty much the definition I have in my book. But does that mean that the principal axes are the eigenvectors of I or the standard axes acted on by R?

4. Apr 5, 2007

### robphy

Last edited by a moderator: May 2, 2017
5. Apr 5, 2007

### quasar987

I've read all your source robphy and none seem to support my position.

Why do you say both are correct?

And how could both be correct? I have before me a concrete problem and clearly the eigenvectors and the Oxyz system acted on by R give different directions.

6. Apr 5, 2007

### robphy

The moment of inertia is a 3-dimensional positive-definite symmetric matrix.

So, its eigenvalues are real and its eigenvectors are mutually orthogonal [or can be chosen to be]. (These eigenvectors are the principal axes, and the eigenvalues are the principal moments.)

So, there is a rotation that will orient the xyz-triad along the triad of mutually-orthogonal eigenvectors. If I am not mistaken, that rotation will diagonalize the moment of inertia matrix. It amounts to having chosen the axes to study the object along its principal axes.

Maybe you should post the concrete problem.

7. Apr 5, 2007

### quasar987

I will, because this is too mysterious.

There is a square of uniform mass M and sides 'a' in the Oxy plane with corners at (0,0), (a,0), (0,a) and (a,a).

Everyone agrees that the inertia tensor is

$$I=M\left(\begin{array}{ccc}a^2/3&-a^2/4&0\\-a^2/4&a^2/3&0\\0&0&2a^2/3\end{array}\right)$$

Everyone also agrees that the eigenvectors are

$$\omega^{(1)}=\left(\begin{array}{c}0 \\ 0\\ 1 \end{array}\right)$$
$$\omega^{(2)}=\left(\begin{array}{c}-1 \\ 1\\ 0 \end{array}\right)$$
$$\omega^{(3)}=\left(\begin{array}{c}1 \\ 1\\ 0 \end{array}\right)$$

However, this means that the matrix R which diagonalizes I is

$$R=\left(\begin{array}{ccc}0&0&1\\-1&1&0\\1&1&0\end{array}\right)$$

And now if I apply R on, say, $$\hat{x}$$, I get

$$R\hat{x}=\left(\begin{array}{ccc}0&0&1\\-1&1&0\\1&1&0\end{array}\right)\left(\begin{array}{c}1 \\ 0\\ 0 \end{array}\right)=\left(\begin{array}{c}0 \\ -1\\ 1 \end{array}\right)$$

which is not equal to any of the eigenvectors.

8. Apr 5, 2007

### marcusl

No, you wrote down the transpose of R. The eigenvectors are the columns of R, not the rows.

9. Apr 5, 2007

### quasar987

Ok, it's an error in the book then because it is clearly indicated how to construct R from the eigenvectors and he lines up the eigenvectors (as opposed to "rows up").

Thanks for pointing that out marcusl.

10. Apr 6, 2007

### lpfr

I apologize. I just want to add the principal axes are not only eingevalues of something. They are also the axis of bigger moment of inertia, the axis of least moment of inertia and the third, perpendicular to previous ones.

11. Apr 6, 2007

### marcusl

12. Apr 6, 2007

### robphy

Yeah, me too... I guess.

13. Apr 6, 2007

### marcusl

Rob, you gave all the answers. I just caught the error...