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?


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 $$P^{ab} = I^{abcd}~\omega_{cd},\\ I^{abcd} = \frac12~\int_{\mathbb R^N} d^Nr~\rho(r)\left(r^a~g^{bc}~r^d - r^b~g^{ac}~r^d\right).$$

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.