Can the Moment of Inertia Tensor be diagonal in a reference frame where the Center of Mass of the system is not on some of the axes?

  • $\begingroup$ It's not too hard to show (using the parallel-axis theorem for the inertia tensor) that if the CM inertia tensor is diagonal with your axes aligned in a particular way, then the inertia tensor with respect to a set of translated axes is diagonal if and only if the CM lies on one of the coordinate axes. But that's a special case of what you're asking here, which is why this is a comment rather than an answer. $\endgroup$ Commented Jul 14, 2022 at 13:16
  • $\begingroup$ The center of mass is a point , it doesn’t lie on some axes? $\endgroup$
    – Eli
    Commented Jul 14, 2022 at 14:02
  • $\begingroup$ @Eli: That point can lie on a coordinate axis. Its coordinates might be $(3,0,0)$, for example, meaning that it lies on the $x$-axis. $\endgroup$ Commented Jul 14, 2022 at 14:08
  • $\begingroup$ @Michael Agree , but this axis has nothing to do with the inertia tensor coordinates system $\endgroup$
    – Eli
    Commented Jul 14, 2022 at 14:13
  • $\begingroup$ I mean the axes of the reference frame, a generic orthonormal base of R³ $\endgroup$ Commented Jul 14, 2022 at 14:18

2 Answers 2


Yes, we can have a system whose CM is not on a coordinate axis which also has a diagonal inertia tensor. As an example, consider a system consisting of four point masses $m$ at the points $(1,1, 1)$, $(1,1, -1)$, $(-1,1, 1)$, and $(1,-1,1)$. Then the CM of the system lies at $$ (x_{CM}, y_{CM}, z_{CM}) = \left( \frac{1}{2}, \frac{1}{2}, \frac{1}{2} \right), $$ which does not lie on any of the coordinate axes or even in any of the coordinate planes.

Meanwhile, the product of inertia $I_{xy}$ is $$ I_{xy} = - \sum_i m_i x_i y_i = - m \left[ (1)(1) + (1)(1) + (1) (-1) + (-1)(1) \right] = 0. $$ The products of inertia $I_{xz}$ and $I_{yz}$ also vanish by a similar logic. Thus, we have a diagonal inertia tensor.

  • $\begingroup$ I have confirmed the above is true with CAD,. $\endgroup$
    – JAlex
    Commented Jul 14, 2022 at 21:23

Given the general inertia tensor about the center of mass as

$$ {\rm I}_C = \begin{bmatrix} I_{xx} & I_{xy} & I_{xz} \\ I_{xy} & I_{yy} & I_{yz} \\ I_{xz} & I_{yz} & I_{zz} \end{bmatrix} $$

and the parallel axis theorem in 3D given the center of mass at a position $(x,y,z)$ from the reference frame origin, the MMOI about the origin is

$$ {\rm I}_0 = {\rm I}_C + m \begin{bmatrix} y^2+z^2 & - x y & x z \\ -x y & x^2+z^2 & -y x \\ -x z & - y z & x^2+y^2 \end{bmatrix} $$

You are asking if the above can be diagonal.

So for example you are looking for cases where

$$ \begin{aligned} I_{xy} - m\, x y & = 0 \\ I_{xz} - m\, x z & = 0 \\ I_{yz} - m\, y z & = 0 \end{aligned} $$

If the the center of mass coordinates are not on the axes, then $x \neq 0$, $y \neq 0$, and $z \neq 0$.

The solution of the above is

$$ \begin{aligned} x & = \sqrt{ \frac{I_{x y} I_{x z} }{m\, I_{y z}}} \\ y & = \sqrt{ \frac{I_{x y} I_{y z} }{m\, I_{x z}}} \\ z & = \sqrt{ \frac{I_{x z} I_{y z} }{m\, I_{x y}}} \\ \end{aligned} $$

A second solution exists also with $(-x,-y,-z)$.

Now from the three cross terms $I_{xy}$, $I_{xz}$, $I_{yz}$ not all of them will have the same sign. The problem here, is that not all will be positive. In fact, most of the time one the cross terms is negative. This means their product, and the thing under the square root will be negative, and as a result the location of the reference frame will be a complex number (or a solution won't be available).

So you can construct a theoretical set of cross terms to diagonalize the MMOI tensor, but I think in real life this will never happen. At best, only one of the cross terms will be eliminated.

As an example, I used CAD to construct and arbitrary shape. If the above is correct, then I should be able to find a solution for $(x,y,z)$.


The values I get from CAD are

Quantity Symbol Value Units
Mass $m$ $2.4543256$ ${\rm kg}$
Center of mass $\boldsymbol{cg}$ $\pmatrix{ -25.108283 \\ -3.8535717 \\ -59.530438}$ ${\rm mm}$
MMOI Cross terms $I_{xy}$ $511.959$ ${\rm kg\,mm^2}$
$I_{xz}$ $1314.109$ ${\rm kg\,mm^2}$
$I_{yz}$ $-1503.307$ ${\rm kg\,mm^2}$

The problem now is that all the terms under the square root are negative, and the reference location should be

$$ \boldsymbol{r} = \pmatrix{ 13.5034 {\rm i} \\15.44755 {\rm i} \\ 39.65117 {\rm i} } \; {\rm mm}$$

When calculating the MMOI from the reference point, I get only one zero non-diagonal term

$${\bf I}_r = \begin{bmatrix} 4771.346 & 1023.918 & 2628.218 \\ 1023.918 & 5063.45 & 0 \\ 2628.218 & 0 & 1318.564 \end{bmatrix}\;{\rm kg \,mm^2} $$

  • $\begingroup$ I think the answer is real life is no actually. See above edits for a bit more details. $\endgroup$
    – JAlex
    Commented Jul 14, 2022 at 21:03

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