What is the connection between moments in probability theory and the moment of inertia?

Question

My question arises as the moment of inertia (MOI) has been described as a second moment. In my understanding if the MOI is indeed a second moment of a distribution of mass, this suggests the MOI could be interpreted as the variance in the distribution of mass about a particular axis and more interesting interpretations could be drawn from higher moments. However, I have not been able to find any definitive evidence that the MOI is indeed the second central moment of a distribution that describes the location of mass in a body.

(Q1) What is the connection between the second moment of the spatial distribution of mass and the MOI?

Moreover, a basic example leads to more questions than answers.

Example 1: Consider the following definition for the MOI, $I$, about an axis containing the center of mass, $$I = \int\limits_{m}r^2 \mathop{dm}\quad\quad(1)$$ where $r$ is the distance from the axis to the infinitesimal mass, $\mathop{dm}$.

For a thin disc of uniform density characterized by radius $R$ and height $H$ (with $H \ll R$), the MOI about the axis $z$ (taken to be the axis of rotational symmetry) follows from elementary manipulations of the differential mass element, $$\mathop{dm} = \rho \mathop{dV} = \frac{M}{\pi R^2 H} r\mathop{dr}\mathop{d\theta}\mathop{dz}.\quad\quad(2) $$ Then, the MOI of the disc about $z$, $I_z$ is simply, $$ I_z = \int\limits_{-H/2}^{H/2}\int\limits_{0}^{2\pi}\int\limits_{0}^{R} r^2 \frac{M}{\pi R^2 H} r\mathop{dr}\mathop{d\theta}\mathop{dz} = \frac{M}{2}R^2. \quad\quad(3) $$

In mathematics, the $n$th central moment, $\mu_{n}$, of a probability density function of a single variable is given by $$\mu_{n} = \int_{-\infty}^{\infty}(x - \mu_{1}^{'})^{n} f(x)\mathop{dx} \quad\quad(4) $$ where $\mu_{1}^{'}$ indicates the first raw moment of the distribution.

Returning to the example and assuming (A1) $I_z = \mu_{2}$ and (A2) $c = 0$ as the ''first moment (normalized by total mass) is the center of mass'' [1] and the center of mass (COM) intuitively lies at ($0$, $0$, $0$) in the (cartesian) principal axis frame, the probability density of the mass, $f(r)$, follows as $$ f(r) = \frac{2Mr}{R^2} \quad\quad(5) $$ where the factor of $2\pi H r$ associated with the differential volume element has been absorbed into $f$. Verifying the total mass is indeed the first moment gives, $$ M = \int_{0}^{R}f(r) \mathop{dr} = \frac{2M}{R^2}\left(\frac{R^2}{2}\right) = M, \quad\quad(6) $$ but verification of the COM via calculation uncovers an inconsistency with (A2), $$ \mu_{1}^{'} = \int_{0}^{R} r f(r) dr=\frac{2MR}{3} \neq 0. \quad\quad(7) $$ In my understanding, this result would indicate the average value of $r$ for the probability density defined earlier is $2MR/3$. I suspect this could be occurring because of the use of cylindrical coordinates, but I have a feeling that there is something else wrong (potentially with needing an expression for the $n$th moment of a multivariate distribution which would enable the use of cartesian coordinates with $f(x, y)$). After considering @Tristan's answer, I think Equation (7) more appropriately captures the average distance of mass from the $z$-axis rather than a component of the COM vector.

(Q1.1) Is the previous statement a correct interpretation of Equation (7)?

(Q1.2) It appears that the components of the MOI tensor and the COM vector are found by averaging about different combinations of coordinates. For example,

Example 2: The $x$ component of the COM vector is given by $$ x_{\text{COM}} = \iiint x' f(x',y',z')\mathop{dx'}\mathop{dy'}\mathop{dz'} $$ where $x'$ indicates the $x$-coordinate in the lab frame.

Is there a physically relevant quantity that would correspond to a marginalization about a particular axis? (In Example 1 it would appear that $f(r)$ is a marginalization about the $\theta$ and $z$ -axis and corresponds to the distribution of mass about the $z$-axis, though I suspect there could be something more interesting for higher order moments or averaging over other quantities.)

(Q2) What is the general form for the moments of a 3D mass distribution (e.g., $f(x, y, z)$ or $f(\mathbf{r})$) that includes the MOI about an arbitrary axis as special cases for the second moment?

(Q3) While David Z.'s comments on this post are close to what I am looking for, I have not been able to find another resource to verify his general expression and I am having considerable difficulty relating it to the basic example indicated here. Could you please point me to a reference on higher-order moments of multivariate distributions (potentially in the context of mass distributions)?

References

[1] https://en.wikipedia.org/wiki/Moment_(mathematics)

Answer 1

This is quite a big question, so I'll tackle questions (1) and (2) the best I can here. I will leave the other two for you to think about given this information.

(1) What is the connection between the second moment of the spatial distribution of mass and the MOI?

It seems you have the clear understanding that the moment of inertia (MOI) is directly connected to the second moment of the spatial distribution but seem to have problems in your results due to improper assumptions. The inconsistency, at least that I can see, is coming from the fact that the center of mass should be at $r = 0$, but your overall calculation shows $\mu' = \frac{3}{2MR} \neq 0$.This is likely arising from your use of cylindrical coordinates, and as we know, in cylindrical coordinates the radius r is not directly indicative of the center of mass in the z-axis. So, you need to consider both the radial and vertical distributions.

For example, a thin disc in three dimensions, with a uniform density across its volume, we get that $\rho = \frac{\pi R^2 H}{M}$. This then provides a new mass element dm that can be identified as $m = \rho dV = \rho (2\pi rH) dr$.The center of mass for a thin disc should then be at $(r_{cm}, \theta_{cm}, z_{cm})$ or also noted as $r_{cm} = 0, \theta_{cm}$ and $z_{cm} = \frac{H}{2}$.We can now fully correct for this calculation by doing the following: $\mu_1' = \int_0^R \int_0^H r \left( \frac{R^2}{2M} r \right) \, dr \, dz$.

(2) What is the general form for the moments of a 3D mass distribution (e.g., f(x,y,z) or f(r)) that includes the MOI about an arbitrary axis as special cases for the second moment?

To conceptualize moments in 3D, we can simply extend everything to the $\mathbb{R}^3$ plane. We can denote the mass distribution in 3D by the following: $\mu_n = \iiint_V (r - \mu)^n f(x, y, z) \, dV$. The first moment (COM) can be defined via the three linking functions

1. $\mu_x = \frac{1}{M} \iiint_V x f(x, y, z) \, dV$.
2. $\mu_y = \frac{1}{M} \iiint_V y f(x, y, z) \, dV$.
3. $\mu_z = \frac{1}{M} \iiint_V z f(x, y, z) \, dV$.

M being the total mass given by $M = \iiint_V f(x, y, z) \, dV$.

And so for any arbitrary axis specified by a unit vector $\mathbf{u} = (u_x, u_y, u_z)$, the moment of inertia can be generalized as $I_u = \iiint_V \left[ r^2 - (r \cdot u)^2 \right] f(x, y, z) \, dV$.

Answer 2

Indeed @Tristan is right. If you have a coordinate $r \in [0, \infty)$, you will be hard pressed to find a distribution where $\langle r \rangle = 0$ that is not $\delta(r)$.

I think a good way to look at the moments of a mass distribution is just the standard multipole expansion (see: https://en.wikipedia.org/wiki/Multipole_expansion).

Basically, the spherical harmonics form a set of basis functions:

$$r^l Y_l^m(\theta, \phi)$$

(note, you will usually see a radial weighting $1/r^l$ or even a sum...that's used for fields formed by charge distributions...but we're talking about the distribution itself).

So at zeroth order, we're looking at:

$$ r^l Y_l^m \propto 1 $$

which is just the total mass.

At $l=1$, if you look at $rY_1^m$,those are linear combinations of:

$$ (\hat x, \hat y, \hat z) $$

so it is literally the lever arm (center-of-mass), and is zero with with well chosen coordinates.

At $l=2$, you looking at the second moment projected onto the 5 shapes that are quadrupoles.

I cannot find a nice picture of real spherical harmonics in cartesian coordinates, but they are for $m=(-2,-1,0,1,2)$:

$$(xy, yz, 3z^2-r, xz, x^2-y^2) $$

The $m=\pm 1$ are a misaligned $z$ axis (an axis tilt) and can be diagonalized away.

The $m=0$ term represents prolate or oblateness, depending on sign.

From $m =\pm 2$ you just get an equation for a conic section...that just describes the cross-section in the $x-y$ plane: if it is (on average) a circle, those are zero. If it's an ellipse, they are not. That is: it describes the azimuthal asymmetry at the lowest non trivial order... something like a tidal bulge on a planet.

Of course, real shapes are much more complicated, but all the MOI cares about is the projection onto these quadrupole shapes.

Finally, back to your point about a moment of a distribution, it is that, but it's a dyadic product, not a square:

$$ \langle 3\vec r \vec r\rangle $$

which has 6 DoF ($l=0$ and $l=2$), so we subtract the scalar ($l=0$) part off:

$$ I \propto \langle 3\vec r \vec r-r^2\delta \rangle $$

Now really finally, higher order (even) moments just describe the shape more accurately, reproducing it exactly in the limit $l\rightarrow\infty$ (see for example, the EGM, which is the shape of the earth out to $l$ in the thousands).