# Inertia tensor of rotated object

Suppose we have computed the inertia tensor of an object about its COM. Suppose the object is then rotated, is there a simple transformation that connects the new inertia tensor to the old one through the rotation matrix?

• The transformation rule is the same for any tensor with two indices. Have you learned how tensors transform? Commented May 19, 2021 at 3:35
• en.wikipedia.org/wiki/… Commented May 19, 2021 at 4:00
• Ah I see, you invert rotate your input vector, apply the inertia tensor and rotate back. I'm curious, is there an analytical proof this will produce the same result as recalculating the inertia tensor? Commented May 19, 2021 at 4:17
• That is one way to think about it, but it won’t work when you have a tensor with three or more indices, so I don’t recommend thinking that way. The analytical proof that you get the same result is that my second and third formulas are identical; they just use different notation. Commented May 19, 2021 at 5:10

Lets say $$I$$ is the moment of Inertia tensor before rotation (or any orthogonal transformation).

This means the angular momentum and angular velocity relates as :

$$\vec L = I \vec \omega$$

and the components of $$L$$ are given by :

$$L=I\omega$$ ...(1)

where $$L$$ and $$\omega$$ are $$3\times 1$$ matrices and $$I$$ is a $$3\times 3$$ matrix.

Now eq(1) would remain true even in the rotated frame. In rotated frame, the components of $$L$$ would be $$R\times L$$ and that of $$\omega$$ be $$R\times \omega$$.

so premultiplying eq(1) with the rotation matrix $$R$$ :

$$R\times L = R \times I\times \omega$$

$$R\times L = R \times I\times (R^T\times R) \times \omega$$

$$(R\times L) = R \times I\times R^T\times (R \times \omega)$$

$$L' = R \times I\times R^T\times \omega '$$ ...(2)

So given we know the rotation matrix R, Inertia tensor in a new coordinate system can be found out using (2)

If a 3D rotation $$R$$ rotates Cartesian vectors according to

$$V_i’=R_{ik}V_k$$

then it rotates Cartesian tensors with two indices (including the inertia tensor) according to

$$I_{ij}’=R_{ik}R_{jl}I_{kl},$$

and similarly for more indices. The way to remember this is “each index gets rotated” via contraction with the rotation matrix.

The index notation here uses the Einstein convention that a repeated index is summed over all possible values (in this case, from 1 to 3).

Index notation is more flexible than the matrix notation

$$\mathbf I’=\mathbf R\,\mathbf I\,\mathbf R^\top$$

because it works for tensors with any number of indices.

• Could you explain how this result is derived? Commented May 19, 2021 at 4:32
• It is saying that a tensor transforms like the outer product of two vectors. If you look at the inertia tensor, it involves the outer product $x_ix_j$. (It also involves $\delta_{ij}$ and that's a different story. It’s an invariant tensor because of the orthogonality relation for rotation matrices.) Commented May 19, 2021 at 4:35
• Like I wrote, is there an easy way to show it algebraically? The definition of the inertia tensor involves nonlinear terms. Commented May 19, 2021 at 4:45
• The definition of the inertia tensor involves nonlinear terms. Yes, like the $x_ix_j$ that I just mentioned. Each one of these transforms with a rotation matrix, because that is what vectors like $\mathbf r$ do. Comments are not the place to ask additional questions beyond what you already asked, which was only “is there a simple transformation that connects the new inertia tensor to the old one through the rotation matrix?” You did not ask for a derivation. If you need one, please post a new question. I have answered what you asked, and a bit more. Commented May 19, 2021 at 4:58
• So I found a proof basically take the definition of angular momentum about some vector for a particle, apply the transformation matrix to both the position and the rotation vectors, pull out the rotation matrix and you get H_r = RH = R I w, then you use the fact that the rotation matrix transpose is it's inverse and get H_r = RIR^tRw, Rw = w_r Commented May 19, 2021 at 6:03