units - Is $0,mathrm{m}$ dimensionless?

Is $0 \, \mathrm m = 0 \, \mathrm s = 0 \,\mathrm {kg} = 0$? How do we define $[0 \, \mathrm m]$?

I once was given an assignment where I was asked to deduce and write down some physical quantity. It turned out that this quantity was $0$ in one unit or another, so I decided to drop the physical units because I figured that it didn't matter anyhow.

But I got to thinking:

If $0 \, \mathrm m = 0$, then we could add units of different dimensions (an operation not normally defined) like so: $5 + 0 \,\mathrm m = 5$.

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  Does $[ \cdot ]$ map $0 \, \mathrm m$ to both $0$ and $L$?

  
  


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  Is there an established regime of thinking about physical quantities as an abstract algebra? Perhaps as a vector space where formal sums such as $5 \, \mathrm m + 10 \, \mathrm s$ are allowed?

  
  


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  Would it be improper to drop the units of measurement of $0 \,\mathrm m$ like this in an academic paper?

  
  
  

Answer

This is really not that big of a deal. In essence, though, the dimensions of zero are ill-defined, and $[0]$ (i.e. the dimensions of zero) is (a) not meaningfully defined, and (b) never used in practice.

Let me start by making one thing clear:

Would it be improper to drop the units of measurement of $0 \,\mathrm m$ like this in an academic paper?

Yes, this is perfectly OK, and it is standard practice.

The dimensionality map $[\,·\,]$ has multiple different conventions, but they all work in much the same way. The key fact is that physical quantities form a vector space under multiplication, with exponentiation (over the field $\mathbb Q$) taking the role of scalar multiplication. (This is the essential reason why dimensional analysis often boils down to systems of linear equations, by the way.) The different base dimensions - mass, length, time, electric charge, etc. - are assumed to be algebraically independent and to span the space, and the dimensionality map $[\,·\,]$ reads out the 'coordinates' of a given physical quantity in terms of some pre-chosen canonical basis.

This only works, however, if you exclude zero from the game. The quantity $1\,\mathrm m$ has a multiplicative inverse, but $0\,\mathrm m$ does not, so if you included it it would break the vector space axioms. This is in general OK - you're not forced to keep those properties - but it does preclude you from using the tools built on that vector space, most notably the dimensionality map. Thus $[0]$ doesn't map to anything.

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Since you explicitly asked for it, here's one way you can formalize what I said above. (For another nice analysis, see this nice blog post by Terry Tao.)

• A positive physical quantity consists of a 8-tuple $(x,m,l,t,\theta,c,q,i) \in\mathbb R^\times\times \mathbb Q^7$, where $\mathbb R^\times=(0,\infty)$ is the real multiplicative group. This is usually displayed in the form $$ x\,\mathrm{kg}^m\mathrm{m}^l\mathrm s^t\mathrm K^\theta\mathrm A^c \mathrm{mol}^q \mathrm{cd}^i.$$


• The multiplication of two physical quantities $p=(x,m,l,t,\theta,c,q,i)$ and $p'=(x',m',l',t',\theta',c',q',i')$ is defined as $$pp'=(xx',m+m',l+l',t+t',\theta+\theta',c+c', q+q',i+i').$$ The multiplicative identity is $1=(1,0,0,0,0,0,0,0)$, and the multiplicative inverse of $p$ is $1/p=(1/x,-m,-l,-t,-\theta,-c,-q,-i)$.


• The exponentiation of a physical quantity $p$ to an exponent $r\in\mathbb Q$ is defined as $$p^r=(x^r,rm,rl,rt,r\theta,rc,rq,ri).$$

You can then easily check that these two operations satisfy the vector space axioms. The above construction is, in fact, a specific instantiation of the abstract vector space $\mathcal Q$ of physical quantities, but it suffices to take one specific example to show that this works.

As an aside, the choice for $\mathbb Q$ as the scalar field is because (a) it's essential for the vector space structure, and (b) it's still somewhat reasonable to have things like $\mathrm {m}^{-3/2}$ (e.g. the units of a wavefunction). On the other hand, things like $\mathrm {m}^{\pi}$ cannot be made to make sense.

The dimensionality map $[\,·\,]$ is, first and foremost, an equivalence relation, that of commesurability. That is, we say that for $p,p'\in\mathcal Q$, $$[p]=[p']\Leftrightarrow p/p'=(x,0,0,0,0,0,0,0)\text{ for some }x\in\mathbb R^\times.$$ This is in fact all you really need to do dimensional analysis, as I argued here, but it's still useful to go on for a bit.

The really useful vector space, if you want to do dimensional analysis, is the vector space of physical dimensions: the space of physical quantities once we forget about their numerical value. This is the quotient space of $\mathcal Q$ over the commesurability equivalence relation: $$\mathcal D=\mathcal Q/[\,·\,]=\{[p]\,:\,p\in\mathcal Q\}.$$ (Here I've abused notation slightly to make $[p]$ the equivalence class of $p$, i.e. the set of all physical quantities commensurable to $p$.) The vector space of physical dimensions, $\mathcal D$, has the same operations as in $\mathcal Q$:

• $[p][p']=[pp']$, and


• $[p]^r=[p^r]$.

It is easy to check that these definitions do not depend on the specific representatives $p$ and $p'$, so the operations are well-defined.

Dimensional analysis takes place in $\mathcal D$. From the definitions above, you can prove that the seven base units give rise to a basis $\{[1\,\mathrm {kg}], [1\,\mathrm m],\ldots,[1\,\mathrm{cd}]\}$ for $\mathcal D$. More physically, though,

• the seven base units are algebraically independent, which means that they cannot be expressed as multiples of each other, and


• they are enough to capture the dimensions of all physical quantities.

These are the key physical requirements on a set of base units for the abstract space $\mathcal Q$.

After this, you're all set, really. And it should be clear that there's no way to make zero fit into this scheme at all.