What is the definition of the moment of inertia tensor?

Question

I can find volume integrals for the moment of inertia in 2D and 3D, but is there a definition that works in an arbitrary number of (spatial!) dimensions?

Answer 1

I actually have no real idea, so this answer is kind of just me trying to talk through what the answer might have to look like.

What describes an orientation in $N$ dimensions?

You have an $N$-dimensional rigid object in $N$ dimensional space, and fixing $N$ points on it should in principle fix it in the space. You can take one of those to be the center of mass, later relaxing that; the other $N-1$ points can be characterized as vectors from the center of mass to their destination. Since it's a rigid body each of these vectors will not change in length but only in angle. Even then if you think about adding them as constraints, each one has a fixed angle with the constraints you've already added, so one starts with $N-1$ degrees of freedom for the first, then $N-2$ for the second, on down to $1$ for the last. So a rigid body in $N$ dimensions presumably has an orientation with $N(N-1)/2$ dimensions. In 2D this predicts that orientation is one-dimensional (just an angle), in 3D this predicts that it is three-dimensional (the vector $\vec \omega$), presumably in 4D it is 6-dimensional.

If that's correct then probably the right way to describe orientation in an arbitrary number of spatial dimensions is with an antisymmetric [0, 2] tensor $\omega_{ab}$. In 3D we have that a point at position $\vec r$ from the center of mass has velocity $\vec v = \vec \omega \times \vec r$ around that center of mass; in fact $\bullet\mapsto\vec \omega \times \bullet$ is precisely such an antisymmetric tensor, suggesting the analogue that $$v^a = g^{ab}~\omega_{bc}~r^c$$ in $N$ dimensions. (Of course your metric tensor $g = \operatorname{diag}(1,1,\dots 1)$ in Cartesian coordinates so you do not have to maintain the Einstein summation convention as such if you want to be sloppy, but I will endeavor to not be sloppy here.)

What about angular momentum?

The angular momentum meanwhile is I think something more than just an antisymmetric tensor, as $\vec L = \vec r \times \vec p$ would seem to generalize to $\epsilon_{ab\dots yz} ~r^y ~p^z$ which gets complicated because the orientation tensor $\epsilon$ has valence $[0,N]$ in $N$ dimensions, so in 2D it's just $\epsilon_{ab}$ but in 3D its $\epsilon_{abc}$ and in 4D it becomes $\epsilon_{abcd}$.

This suggests that the true moment-of-inertia tensor, which must generate a $[0, N-2]$-valence angular momentum tensor out of a $[0, 2]$-valence orientation, is a $[2, N-2]$-valence tensor, and that would make it very hard to give a perfect formula. But most of its components appear to simply be shuffled around by the orientation tensor so maybe that's inessential and there's just a $[4, 0]$-valence tensor sitting underneath there? If so one would expect to define a "pre-angular-momentum" tensor $P^{ab} = \frac12\big(r^a~p^b-r^b~p^a\big)$ or so, expecting that any symmetric part gets destroyed by the orientation tensor anyway when we form $L_{a\dots x} = \epsilon_{a\dots xyz}~P^{yz}$.

How would that answer your question?

If all of that hand-waving reasoning is right, then putting those two expressions together with some mass density $\rho$ and momentum density $p = \rho~v$, one would find that $$P^{ab} = \int_{\mathbb R^N} d^Nr~\rho(r)~\frac12\left(r^a~g^{bc}~\omega_{cd}~r^d - r^b~g^{ac}~\omega_{cd}~r^d\right),$$ and thus that there is a $[4,0]$-valence moment of inertia tensor that looks like $$P^{ab} = I^{abcd}~\omega_{cd},\\ I^{abcd} = \frac12~\int_{\mathbb R^N} d^Nr~\rho(r)\left(r^a~g^{bc}~r^d - r^b~g^{ac}~r^d\right).$$