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One has, let's say, the inertia matrix of an object about it's center of gravity: $I_p$ in the coordinate system ($i, j, k$). What is the inertia matrix in another coordinate system, say ($i_1,j_1, k_1$) with $$ \begin{bmatrix} i \\ j\\ k\end{bmatrix} = A \cdot \begin{bmatrix} i_1 \\ j_1 \\ k_1\end{bmatrix}$$

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  • $\begingroup$ PS. this is called a congruent transformation (en.wikipedia.org/wiki/Matrix_congruence) $\endgroup$ – John Alexiou Jul 12 '16 at 14:27
  • $\begingroup$ @ja72 Are you sure; in general its a similarity transformation, which means the same as a congruence in the orthonormal case $\endgroup$ – Selene Routley Jul 13 '16 at 8:06
  • $\begingroup$ When a matrix is transformed with ${\rm I}^\star = A \,{\rm I}\, A^\top$ as in this case it is called so. That is what I learned in college! $\endgroup$ – John Alexiou Jul 13 '16 at 13:07
  • $\begingroup$ This is not the case. I was interested in a general transformation $\endgroup$ – C Marius Jul 13 '16 at 13:41
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For what it's worth, here's how you work it out from the transformation of the angular momentum - I began this answer last night - but here is how Martin's Answer would play out with deducing the transformation laws from the assumption of vector transformation laws for the angular momentum and angular velocity instead of the assumption of energy's being a scalar and angular velocity a vector as in Martin's answer.

The definition of the inertia tensor is the linear homogeneous transformation characterizing a body that maps the angular velocity vector to the angular momentum:

$$\vec{L} = I\,\vec{\omega}\tag{1}$$

Vectors, by your definition of symbols, in this picture transform as $X \mapsto A\,X$. So we write $\vec{L} = A\,\vec{L}^\prime$, likewise $\vec{\omega} = A\,\vec{\omega}^\prime$, where the primed vectors are the components in the new co-ordinates. Substitution into (1) and fore-multiplying both sides by $A^{-1}$ (naturally your transformation is nonsingular) yields:

$$\vec{L}^\prime = A^{-1}\,I\,A\,\vec{\omega}^\prime$$

whence you can read off the transformed inertia matrix from the definition given in (1) and naturally this is the same as Martin's answer.

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  • $\begingroup$ I just realized that I restrict myself to orthogonal transformations by assuming that the scalar products remains invariant. So your answer is more fundamental and does not need any assumptions. However in the case $A^\mathrm T \neq A^{-1}$ one will see that the scalar products are also transformed. That's what you get if you only think about Lorentz transformations :-). $\endgroup$ – Martin Ueding Jul 13 '16 at 7:53
  • $\begingroup$ @MartinUeding Don't worry, I had to mess about with this quite a bit before I clarified things. I began with the integral definition of I and got myself into a right mess. The scalar products indeed gets transformed because you in general have a metric tensor which is the identity matrix in the orthonormal case but in general it's a symmetric billinear form $\langle X,\,Y\rangle = X^T \,G\,Y$. So $G$ transforms by $G\mapsto A^T\,G\,A$. $I$ is a mixed valence tensor (mapping vectors to vectors), which explains the difference in transformation law. $\endgroup$ – Selene Routley Jul 13 '16 at 8:04
  • $\begingroup$ So $L = I\omega$ and $L_1 = I_1\omega_1$ being the same vector but in different coordinate system one has $L_1 = A^{T}L$ and $\omega_1 = A^{T}\omega$ therefore $ A^{T}L = I_1A^{T}\omega = A^{T}I\omega$ hence $(I_1A^{T} - A^{T}I)\omega = 0$ for all $\omega$. So $I_1A^{T} - A^{T}I = O$ and from here $I_1 = A^{T}IA^{-T}$. Is this true in every reference frame ... ? I feel much more comfortable with math than physics ... reference frames ... $\endgroup$ – C Marius Jul 13 '16 at 8:21
  • $\begingroup$ @CMarius Yes, you've simply restated my argument, although now you've used $A^T$ instead of $A$. It has to be general to be consistent with the three postulates (1) $I$ is the linear homogeneous mapping defining AM in terms of $\vec{\omega}$, (2) AM transforms like a position vector and (3) $\omega$ transforms like a position vector. Being a mathematician, you'll understand if I say that the cleanest definitions of AM and $\omega$ are as two-forms, so we can cheat and make them behave like vectors in 3D if we take their Hodge stars. $\endgroup$ – Selene Routley Jul 13 '16 at 8:52
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You are given $\mathbf A$ which a matrix describing the linear transformation. This matrix will allow you to move from the new coordinates (subscript 1) to the old coordinates (no subscript).

What you have to do now is to transform the inertia matrix $\mathbf I$ into the new basis $\mathbf I_1$. Here is how to come up with the transformation rule: The rotational energy is a scalar (just one number), therefore it has to be the same in every coordinate system. This is $$ E = \frac12 \vec \omega^\mathrm T \mathbf I \vec \omega \,. $$ You know that $\mathbf A$ will bring you from the old coordinates to the new coordinates. You obtain the new coordinates from the old ones by using $\mathbf A^{-1}$. So $\mathbf A^{-1} \vec \omega$ will be the angular rotational velocity vector in the new coordinate system.

Use the above equation and insert $\mathbf A \mathbf A^{-1}$ into the energy equation above twice: between $\vec \omega^\mathrm T$ and $\mathbf I$ as well as $\mathbf I$ $\vec \omega$. Identify the angular velocity vectors in the new system and the new matrix in between.

What do you get for the matrix? Please do it yourself first before looking at the spoiler text below :-).


One can rewrite the rotational energy as follows:
$$ E = \frac12 \vec \omega^\mathrm T \mathbf I \vec \omega = \frac12 (\mathbf A \mathbf A^{-1} \vec \omega)^\mathrm T \mathbf I \mathbf A \mathbf A^{-1} \vec \omega = \frac12 \underbrace{\vec \omega^\mathrm T \mathbf A^{-1\,\mathrm T}}_{\vec \omega_1^\mathrm T} \underbrace{\mathbf A^\mathrm T \mathbf I \mathbf A}_{\mathbf I_1} \underbrace{\mathbf A^{-1} \vec \omega}_{\vec \omega_1} \,. $$

So one would rather say that $I_1 = \mathbf A^\mathrm T \mathbf I \mathbf A$ in the general case. However, a transformation which you use in cases like these are orthogonal transformations. Those are just rotations, so they are a physical symmetry of the system. Also the matrix $\mathbf I$ is symmetric, therefore it can be diagonalized with an orthogonal transformation. Therefore we may assume that $\mathbf A \in \mathrm{SO}(3)$ and therefore $\mathbf A^\mathrm T$ = $\mathbf A^{-1}$. Then we do get the solution that I wrote previously: $\mathbf I_1 = \mathbf A^{-1} \mathbf I \mathbf A $.

Going the other way will give $\mathbf I = \mathbf A \mathbf I_1 \mathbf A^{-1} $.


This differs from Rod's answer a bit because he directly obtains $\mathbf I_1 = \mathbf A^{-1} \mathbf I \mathbf A$ without having to assume $\mathbf A \in \mathrm{SO}(3)$. The reason for this is really subtle and will only become clear once you learn about special relativity with four-vectors and the metric tensor.

I have just assumed that the rotational energy will stay the same in every coordinate system. That is not always the case, that is only the case for transformations which leave the scalar product invariant. This is somewhat of a tautology, therefore I missed in the first go.

The invariance of scalar products depends on the invariance of the metric tensor. Only orthogonal transformations, those from SO(3), leave this metric tensor invariant. So in one has to restrict coordinate transformations right from the beginning to those.

Once you looked into special relativity, you will get that the group of symmetry operations for spacetime is something like $\mathrm{SO}(1, 3) \times \mathbb R^4$. So one directly restricts to special orthogonal matrices from the start, the scalar product is left invariant and one can do the argumentation that I did earlier.

Rod needs to assume way less in order to derive the result. So I think his answer is the cleaner one as I have made too strong assumptions.

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  • $\begingroup$ $I$ is the inertia matrix ? $\endgroup$ – C Marius Jul 11 '16 at 15:57
  • $\begingroup$ Yes. You called it $I_p$, I wanted to follow your notation with the subscript 1. Therefore I removed the $p$ in the subscript. $\endgroup$ – Martin Ueding Jul 11 '16 at 16:00
  • $\begingroup$ Thank you very much! I can not up vote your answer here on physics stack exchange, I do not have enough reputation yet. But: the answer is useful for my question, I have now to see if my question is useful for my problem ... I am trying to model a bike ... :) $\endgroup$ – C Marius Jul 11 '16 at 16:05
  • $\begingroup$ This trick of inserting 1 or a unit matrix happens all the time in physics. Then you regroup and redefine your terms. $\endgroup$ – Martin Ueding Jul 11 '16 at 16:06
  • $\begingroup$ Look: $E = \frac{1}{2}{\overrightarrow{\omega}}^{T}I\overrightarrow{\omega}$ is the energy in the first coordinate system, and $E_1 = \frac{1}{2}{\overrightarrow{\omega}_1}^{T}I_1\overrightarrow{\omega}_1$ is the energy in the second coordinate system. Of course $E_1 = E$ then writing $\omega = A^{T}\omega_1$ one obtains $$ \omega^{T}I\omega = \omega_1^{T} AA^{-1}I_1A^{-T}A^{T}\omega_1 = \omega^{T} A^{-1} I_1 A^{-T}\omega$$. Therefore $I = A^{-1} I_1 A^{-T}$ ... ?! $\endgroup$ – C Marius Jul 13 '16 at 7:20

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