In the paper $D$-dimensional moments of inertia by Bender and Mead (1995), the authors show that the moment of inertia about an axis through its diameters is $\frac{D-1}{D+2}MR^2$, whereas that about an axis orthogonal to all the diameters is $\frac{D}{D+2}MR^2$. Is there therefore a $D$-dimensional analogue to the perpendicular axis theorem $I_x+I_y=I_z$?

Edit: I don’t mean something like an inequality as suggested in the comments, but more of an equality linking, say $D$ times of $\frac{D-1}{D+2}MR^2$ and $\frac{D}{D+2}MR^2$?

  • $\begingroup$ $ D\ge2$ thus the perpendicular axis theorem is correct $I_x+I_y\ge I_z,~$ $2D-1 >D$ $\endgroup$ – Eli Apr 23 at 6:54
  • $\begingroup$ @Eli I know that’s true but not what I’m looking for; I’m more curious as to whether there’s a stronger equality relating $\{I_i\}_1^D\}$ and $I_{D+1}$. $\endgroup$ – user107224 Apr 23 at 8:20
  • $\begingroup$ Might be worth noting that for the generalization to arbitrary $D$, angular momentum $L$ and angular velocity $W$ must be represented as antisymmetric matrices instead of as vectors. They are still related by an equation involving the moment of inertia tensor shown in Buzz's answer. $\endgroup$ – Chiral Anomaly Apr 25 at 17:06

The perpendicular axis theorem for two-dimensional mass distributions in three spatial dimensions generalizes to describe the moments of a mass distribution that is flat in one direction in $D$ dimensions. The moment of inertia for the axis perpendicular to the mass, $I_{D}$, is proportional to the sum of all the momenta of inertia for the other $D-1$ axes.

The tensor of inertia has general entries $${\bf I}_{jk}=\int dM\,(r^{2}\delta_{jk}-x_{j}x_{k}),$$ where $dM$ indicates an integral over the mass distribution and $r^{2}=\sum_{i=1}^{D} x_{i}^{2}$ is the distance squared from the origin. With the origin of the coordinates placed at the center of mass and the axes chosen to be the principal axes of the body, the tensor is represented by a diagonal a matrix. (This is just a consequence of the fact that the tensor ${\bf I}$ is manifestly symmetric, so it may be diagonalized by am orthogonal rotation.) The diagonal entries are the single-axis moments about the $D$ principal axes, such as $$I_{1}=\int dM\,\sum_{i=2}^{D}x_{i}^{2}.$$

The perpendicular axis theorem comes about when the mass distribution in $D=3$ dimensions is planar (that is, two dimensional). In that case, the direction perpendicular to the laminar mass distribution (call it the $x_{3}$-direction) is one of the principal axis directions. Since $x_{3}=0$ wherever there is mass, the three moments are given by $$I_{1}=\int dM_{2}\,x_{2}^{2}, \quad I_{2}=\int dM_{2}\,x_{1}^{2}, \quad I_{3}=\int dM_{2}\,\left(x_{1}^{2}+x_{2}^{2}\right),$$ where now $dM_{2}$ is the two-dimensional integration (in the $xy$-plane) over the in-plane mass distribution. [Formally, we may write the two- and three-dimensional integration measures as $dM=\delta(x_{3})\, dx_{3}\, dM_{2}$.] From the forms of these integrals, it is now trivial that $I_{1}+I_{2}=I_{3}$

If we have instead a $(D-1)$-dimensional mass distribution in $D$-dimensional space, we can orient $x_{D}$-direction to lie normal to the hyperplane of the mass distribution. Then the first $D-1$ principal moments about the center of mass are $$I_{j}=\int dM_{D\,-\,1}\,\left(-x_{j}^{2}+\sum_{i=1}^{D\,-\,1}x_{i}^{2}\right),$$ while the moment for the direction normal to the mass distribution is $$I_{D}=\int dM_{D\,-\,1}\,\sum_{i=1}^{D\,-\,1}x_{i}^{2}.$$ (Again, $dM_{D\,-\,1}$ denotes the integration over the $D-1$ dimensions of the mass distribution.) From the forms of the integrals for $I_{j}$ ($j<D$) and $I_{D}$, it is clear that $$\sum_{j=1}^{D\,-\,1}I_{j}=\sum_{j=1}^{D\,-\,1}\int dM_{D\,-\,1}\,\left(-x_{j}^{2}+\sum_{i=1}^{D\,-\,1}x_{i}^{2}\right)=-\sum_{j=1}^{D\,-\,1}\int dM_{D\,-\,1}\,x_{j}^{2}+\sum_{j=1}^{D\,-\,1}I_{D}\\=-I_{D}+(D-1)I_{D}=(D-2)I_{D}.$$ This is the final result, the generalization of the theorem to $$\sum_{j=1}^{D\,-\,1}I_{j}=(D-2)I_{D}.$$

If you have a mass distribution that is flat in more than one direction—so it has $D-2$ dimensions or less—you can generalize this result fairly simply, to show that the sum of the moments of inertia in the hyperplane of the mass distribution is proportional to the sum of the moments in the perpendicular directions. More precisely, if the mass distribution is restricted to a $d$-dimensional hyperplane of the $D$-dimensional space (with $0<d<D$), then $$\sum_{j=1}^{d}I_{j}=\frac{d-1}{D-d}\sum_{j=d\,+\,1}^{D}I_{j}.$$


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