The Wikipedia article on dimensional analysis says:

the dimensions form an abelian group under multiplication

This is used to justify the manipulation of ratios of incommensurable quantities. My question is, how do we know that the (physical) dimensions form an abelian group under multiplication? Is it just that at the outset we decided that abelian groups are easy to deal with, and so designed the dimensional system explicitly to form an abelian group, or is the abelian-groupness of physical dimensions derived from more basic properties?

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    $\begingroup$ There really isn't anything fancy going on. It's just that multiplication is commutative. I think that Wikipedia article used unnecessary terminology. $\endgroup$ Oct 4, 2019 at 2:43

1 Answer 1


It follows from what we know about dimensions that they form an abelian group under multiplication, in that sense. If I have a mass times a velocity, dimensionally that is equivalent to a velocity times a mass.

Now, you might say, what about dimensionful operators being multiplied that don't commute? While the operators themselves may not commute, the dimensions still do. For example, famously,

$$[x,p] = i\hbar,$$

and on the left hand side we are subtracting two quantities which while not equal, have the same dimension, as the right hand side. This is because in terms of the physical intuition, there is no difference between a distance times energy, or energy times distance (which is one way to think of the dimensions of $\hbar$).

If we have a set of dimensions, $[X_i]$ for $i=1,\dots,k$, for any $i$, $[X_i]^m [X_i]^n = [X_i]^{m+n}$. The abelian group in question is then generated by these $[X_i]$, and it is abelian, free and finitely generated. If you had say, a single dimension, this would obviously give you $\mathbb Z$.

  • $\begingroup$ Thank you for your response, I think this makes sense. Could you provide the name of the first formula please? I'm not a physicist myself and I'm not familiar with it. $\endgroup$
    – pelrow
    Oct 4, 2019 at 16:18
  • $\begingroup$ @pelrow $[x,p] = i\hbar$ is the canonical commutation relation, for the position operator $x$ and momentum operator $p$ in quantum theory. $\endgroup$
    – JamalS
    Oct 4, 2019 at 19:07

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