We know that for a body with an axis of symmetry of order greater than 2 has two equal moments of inertia. But can the converse be proved? Namely, given only that two principal moments are equal, can it be deduced that an axis of symmetry of order greater than 2 exists for the body?

If the claim is false, is there a counterexample which happens to have two equal moments, but does not "look" symmetric?

The claim seems to be true to me, but beyond the observation that choice of two of the principal axes are arbitrary in a plane (due to degeneracy of the moment of inertia tensor), I do not know how to rigorously prove the existence of an axis of symmetry.

Are there standard ways to make (and prove) strong statements of symmetry in physics given a few calculated quantities?


2 Answers 2


It is not difficult to construct a counter-example in which all three principal moments of inertia are equal but the distribution of mass has no symmetry.

The easiest mass distribution is probably 6 point masses lying on the Oxyz axes with the centre of mass at O. The fact that the masses all lie on the axes ensures that these are the principal axes of inertia, because the products of inertia are zero.

There are 12 variables (6 masses, 6 distances from the origin) and 5 constraints. 3 constraints are of the form $m_1x_1=m_2x_2$. These ensure that the centre of mass is at the origin for each axis. 2 constraints are of the form $m_1x_1^2+m_2x_2^2=m_3y_1^2+m_4y_2^2$. These ensure that the moment of inertia is the same for all axes. The number of variables far exceeds the number of constraints, so it is easily possible to choose values satisfying all 5 constraints.

One solution is :

x-axis : masses of 20 & 10 units at distances of +1 & -2 units from O.
y-axis : masses of 15 & 5 units at distances of +1 & -3 units from O.
z-axis : masses of 6 & 4 units at distances of +2 & -3 units from O.

The masses are placed asymmetrically on each axis but the CM is at O. The moments of inertia about each axis are :
$I_x=15*(+1)^2+5*(-3)^2+6*(+2)^2+4*(-3)^2=120$ units $I_y=20*(+1)^2+10*(-2)^2+6*(+2)^2+4*(-3)^2=120$ units $I_z=20*(+1)^2+10*(-2)^2+15*(+1)^2+5*(-3)^2=120$ units

This counter-example shows that symmetry is not a necessary condition for the 3 principal moments of inertia to be equal.

  • $\begingroup$ +1 Nice answer. Why the factor of 2 in the last equation though? $\endgroup$
    – Diracology
    Sep 30, 2017 at 13:29
  • $\begingroup$ @Diracology Thanks. The equation was intended to show equal contributions from each axis, with the moment being the sum of moments for any 2 axes. I have revised my answer to make it clearer. $\endgroup$ Sep 30, 2017 at 17:39

It depends how sophisticated you are in defining "symmetry".

For example, consider a uniform square, with an hole in the shape of an equilateral triangle. The centroid of the triangle is at the center of the square but the orientation of the vertices of the triangle is arbitrary with respect to the sides of the square.

It is easy to show that plane figure has equal moments of inertia, but it doesn't obviously "look" symmetric.


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