Are there higher moments of vector quantities?

Question

The $n$-th moment of mass of a point mass $m$ with respect to a point located at $\boldsymbol r$ is $$ \mu_n(m) = m\boldsymbol r^n $$ So the 0th moment is the total mass $m$, the 1st moment is $m\boldsymbol r$, the 2nd is $I=mr^2$, the 3rd is $mr^2\boldsymbol r$, the 4th is $mr^4$, etc.

The 1st moment of a vector quantity $\boldsymbol F$, such as force, with respect to a point located at $\boldsymbol r$, is $$ \mu_1(\boldsymbol F) = \boldsymbol r\times\boldsymbol F $$ And I assume that its 0th moment is just the total force $\boldsymbol F$. So my question is: what would be the 2nd, 3rd, etc. moments look like? I haven't been able to find even a passing reference about it. The simplest explanation would be that they are all zero because $$ \mu_2(\boldsymbol F) = \boldsymbol r\times\boldsymbol r\times\boldsymbol F = \boldsymbol 0 $$ and so on. But I don't know whether this is indeed the correct expression for the $n$-th moment of a vector quantity. It may very well be that $$ \mu_2(\boldsymbol F) = r^2\boldsymbol F \neq \boldsymbol 0 $$ $$ \mu_3(\boldsymbol F) = r^2\boldsymbol r\times\boldsymbol F \neq \boldsymbol 0 $$ etc, and I'd be none the wiser. So is there a more general expression for the moment of a vector (tensor?) quantity?

Answer 1

So with the caveat that people might mean different things by the same term and so learning a fixed definition now might only lead to confusion later, the following is I think a reasonable definition of a $k^{th}$ moment:

0. We are only trying to define the moment of a quantity defined at a single point. Often people interpret moment as relating to some distribution, in which case you need to integrate this definition.

1. The $k^{th}$ moment of a rank $s$ tensor is a rank $s+k$ tensor. (If distinguishing contr- and co-variant ranks then taking a moment increases the contra-variant rank).

2. In components, taking the moment of a rank $s$ tensor $T^{i_1,\dots i_s}$ results in a quantity with one extra index $M_1(T)^{i_1,\dots,i_s, j}=T^{i_1,\dots i_s}r^j$.

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How this fits with the definitions in the OP:

Centre of mass: This is the first moment of the mass which is a scalar (rank $0$ tensor) $m$. The first moment is the centre of mass $R^i=mr^i$.

Moment of inertia: The full moment of inertia tensor is $I^{ij}=mr^ir^j$. Often we work about a given axis passing through the centre of mass (often chosen to be at the origin). This gives a single number, the result of contracting with a diagonal tensor whose entries are 1 for the perpendicular components and 0 for the parallel components.

Moment of force: The full tensor as I have defined it is $T^{ij}=F^i r^j$. You can get to a vector by contracting with the Levi-Civita symbol: $\tau_k = \epsilon_{ijk}F^ir^j = (\vec{F}\times \vec{r})_k$.

So in general what happens is that there is a rather high order tensor and maybe the physical quantities you care about are some contraction of it with some other tensor to reduce the rank to something manageable. When communicating with someone else, its therefore important to know exactly what they mean since there are many things you could contract with and thus many viable 'moments'.