# Is the Moment of Inertia tensor symmetric due to rotational invariance of space?

In standard classical mechanics' textbooks (for instance Morin) one finds that while calculating the angular momentum vector of a rigid body for a 3-D general case from the definition:

$$L=r \times p$$

The requirement to define a moment of inertia tensor, which encodes the full information of the mass distribution undergoing rotations, naturally arises.

We also find that the $$I_{i,j}$$ terms (for $$i\neq j$$) remain invariant under exchange of $$x$$ and $$y$$ since their product is involved.

This means that the moment of inertia is symmetric.

Now, from the derivation it wasn't clear to me why it comes out that way.

I am guessing since products of coordinates are involved in the non-diagonal terms it has to with rotational invariance of the space. I say this in the sense that rotating $$x$$ into $$y$$ coordinates leave the tensor invariant but I am not satisfied with my guess.

"Besides the mathematical derivation how can the symmetric property of the MOI tensor be explained on physical grounds? And if the matrix was not symmetric what would that imply for the space around us?"

• Angular momentum is $L=r \times m\,v$
– Eli
Commented Nov 18, 2021 at 13:26
• I must be drunk while writing that formula. Thanks for pointing it out!
– Lost
Commented Nov 18, 2021 at 15:53

The derivation of the mass moment of inertia tensor from the summation of particles has the form of a symmetric matrix.

Each particle of mass $$m_i$$ located at $$\boldsymbol{r}_i$$ contributes to the MMOI tensor by an amount that equals

$$\mathrm{I} = \sum_i m_i \left( (\boldsymbol{r}_i \cdot \boldsymbol{r}_i) \mathbf{1} - \boldsymbol{r}_i \odot \boldsymbol{r}_i \right)$$

where $$\cdot$$ is the vector inner product (obviously symmetric), and $$\odot$$ is the vector outer product (also symmetric when the arguments are identical).

In terms of components, of $$\boldsymbol{r}_i = \pmatrix{x\\y\\z}$$ then

$$\mathrm{I} = \sum_i m_i \left( \begin{bmatrix} x^2+y^2+z^2 & & \\ & x^2+y^2+z^2 & \\ & & x^2+y^2+z^2 \end{bmatrix} - \begin{bmatrix} x^2 & x y & x z \\ x y & y^2 & y z \\ x z & y z & z^2 \end{bmatrix} \right)$$

The above is an expansion of the following term

$$\boldsymbol{r}_i \times ( \boldsymbol{\omega} \times \boldsymbol{r}_i ) = \boldsymbol{\omega} ( \boldsymbol{r}_i \cdot \boldsymbol{r}_i ) - \boldsymbol{r}_i (\boldsymbol{r}_i \cdot \boldsymbol{\omega}) = \left( (\boldsymbol{r}_i^\top \cdot \boldsymbol{r}_i) \mathbf{1} - \boldsymbol{r}_i \boldsymbol{r}_i ^\top \right) \boldsymbol{\omega}$$

This symmetry is a result of the moment arm about the rotation axis being used twice. Once to establish the translational velocity of the particle (the $$\boldsymbol{\omega}\times \boldsymbol{r}_i$$ part), and once to take the moment of momentum (the $$\boldsymbol{r}_i \times m_i \boldsymbol{v}_i$$ part)

Alternatively the above can be also be expressed as

$$\mathrm{I} = \sum_i m_i \left( - [ \boldsymbol{r}_i \times] [\boldsymbol{r}_i \times] \right)$$

where each $$[\boldsymbol{r}_i \times]$$ is there anti-symmetric matrix

$$[\boldsymbol{r}_i \times] = \begin{bmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{bmatrix}$$

and the fact that the square of a skew-symmetric matrix is a symmetric matrix.

• If I remember correctly $v=\vec \omega \times \vec r$ is a a special case when the particle goes in a circle. Or is that formula general?
– Lost
Commented Nov 18, 2021 at 16:01
• "This symmetry is a result of the moment arm about the rotation axis being used twice". This seems correct. Would this be true even if the space was not rotationally invariant?
– Lost
Commented Nov 18, 2021 at 16:03
• @Lost - you define MMOI tensor by fixing the center of mass, and rotating about an axis through the COM. Then $\vec{v} = \vec{\omega} \times \vec{r}$ gives us the velocity of each particle on the body. For a rigid body, each particle does indeed go around in a circle. Commented Nov 18, 2021 at 22:15

starting with the kinetic energy (only rotation) \begin{align*} &T=\sum_{\rho=1}^{N}\,m_\rho\,\frac{\mathbf{v}_\rho^{2}}{2}\\ &\text{with}\\ &\mathbf v=\mathbf\omega\times\mathbf r\quad \Rightarrow\\ &T=\sum\,\frac{m }{2}\,(\mathbf\omega \times\mathbf r )^2= \sum\,\frac{m }{2}\,\left[\omega ^2\,r ^2 -\left(\mathbf\omega\cdot\mathbf r\right) ^2\right] \end{align*} with \begin{align*} T_h = &\left[\omega^2\,r^2 -\left(\mathbf\omega\cdot\mathbf r\right)^2\right]\\ =&\omega_i^2\,x^{i2}-\omega_i\,x^i\,x_k\,\omega^k\\ =&\omega_i\,\omega^k\,\delta_k^i\,x^{j2}-\omega_i\,\omega^k\,x^i\,x_k\,\\ =&\omega_i\,\omega^k\left(\delta_k^i\,x^{j2}-\,x^i\,x_k\right) \end{align*} hence \begin{align*} &T=\frac 12\,\omega_i\,\omega^k\underbrace{\left[\sum m \left(x^{j2}\,\delta_k^i-\,x^i\,x_k\right) \right]} _{\mathbf I_{ik}} \end{align*}

from here you can see that the inertia tensor

\begin{align*} \mathbf I_{ik}=\mathbf I_{ki} \end{align*}

is symmetric