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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?

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    $\begingroup$ This feels like a semantic question - as you've noted, there are quantities that are non-zero that make sense to call by this name. Whether anyone does will depend on if those quantities ever come up in applications. $\endgroup$
    – jacob1729
    Jan 4 at 13:24
  • $\begingroup$ So there is no general definition of moment? $\endgroup$ Jan 4 at 13:33
  • $\begingroup$ I don't think so - I could write an answer giving what I think it probably a sufficiently general definition to cover all cases - but it would for instance differ from yours since what I want to call the "first moment" of a scalar would be a vector (the centre of mass coordinate). $\endgroup$
    – jacob1729
    Jan 4 at 13:48
  • $\begingroup$ Yes, my first moment should be a vector quantity, I'll correct that now. I'd be interested in your answer if you don't mind writing it. $\endgroup$ Jan 4 at 14:17
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    $\begingroup$ Note that the moment of inertia is already a tensor, and the rules for combining a tensor and a vector give rise to a rich family of behaviors in rotating systems which seem relevant to your interests here. $\endgroup$
    – rob
    Jan 4 at 14:36
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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:

  1. 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.

  2. 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).

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


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'.

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  • $\begingroup$ Thank you! This is exactly what I was looking for. Makes so much sense. The only thing I'd add is that the rank s tensor in point 2 could also have lower indices to make explicit the remark in parentheses from point 1. $\endgroup$ Jan 4 at 15:46
  • $\begingroup$ Also, I think the indices added need not be contravariant. I.e. the moment of inertia may be expressed as $ I_{ij} = m r_i r_j $ depending on where you want the indices to end up. $\endgroup$ Jan 4 at 16:40
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    $\begingroup$ Sure, but position is normally upper. Its kind of pointless to distinguish in this case though since in the cases where the distintion is relevant, its also not the case that position forms a vector. But yeah there's nothing stopping you from lowering indices in this definition. $\endgroup$
    – jacob1729
    Jan 4 at 16:50

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