Mathematical rules for dimensional analysis

Question

In dimensional analysis we treat the units as their own objects, in this case $1m$ is a unit multiplied by a number, and forms a 'quantity' is this explicitly the same object as the unit itself?

Having $1m=m$ implies that perhaps $-1m=-m$, I think it perhaps makes sense to make a distinction between the unit and the quantity, although $m$ could refer to a to a length, perhaps $'1m'$ is the length and we can show that $2m=2*(1m)$? We would never give a length of $'m'$, the issue with this is then $2(1*m)$ = $2*m$ which seems to be an issue based on the definition of multiplication?

If we can have an object just called $m$ and it is defined under addition (in the case it's defined as $1*m$) then we must have an additive inverse and $-m$ should exist, how do we deal with this issue? $-m$ should make no sense in the context of the unit.

How about taking $0*m$ is $0*m=0*s$? I believe this is not the case, as dimension needs to be preserved in multiplication, but I'm not sure.

It doesn't seem all the rules of algebra apply to units so it causes me confusion.

Answer 1

Here's one way to formalize things. As a warning, I would imagine most working physicists would consider this to be massive overkill, but it will provide a framework for answering your questions so we will press on. For the moment, let's consider only things with dimensions of mass, length, and time; the generalization to include other dimensions (e.g. temperature) is straightforward.

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First, define a physical dimension as an element of $\langle a,b,c\rangle\in \mathbb Q^3$ - that is, an ordered triple of rational numbers. We identify $\langle 1,0,0\rangle $ with the dimension of mass, $\langle 0,1,0\rangle$ with the dimension of length, and $\langle 0,0,1\rangle $ with the dimension of time. In (possibly) more familiar notation, we might write

$$\langle a,b,c\rangle \equiv [M]^a[L]^b [T]^c$$

The space of physical dimensions constitutes a (rational) vector space, which we call $\mathscr D$. I emphasize that these objects are not units like kg or m/s, but rather dimensions such as mass or length per time. The "addition" operation for dimensions (which corresponds to the multiplication of physical quantities, see $(2)$ below) will be denoted $\star$, and defined as $$\langle a,b,c\rangle \star \langle d,e,f\rangle := \langle a+d,b+e,c+f\rangle$$ The "scalar multiplication" operation for dimensions (which corresponds to the exponentiation of physical quantities, see $(3)$ below) is defined as $$\langle a,b,c\rangle \wedge d := \langle d\cdot a,d\cdot b,d\cdot c\rangle$$

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Next, a physical quantity is a pair $(x,D)$ where $x\in \mathbb R$ and $D\in \mathscr D$. Adopting the SI system of units, we may make the following definitions: $$\matrix{\mathrm{kg} \equiv \bigg(1,\langle 1,0,0\rangle\bigg)\\ \mathrm m \equiv \bigg(1,\langle0,1,0\rangle\bigg) \\ \mathrm s \equiv \bigg(1,\langle0,0,1\rangle\bigg)}$$

Physical quantities of the same dimension may be added together as follows: $$(x,D) + (y,D) := (x+y,D)\tag{1}$$ Additionally, physical quantities of arbitrary dimension may be multiplied together: $$(x,D_1) * (y,D_2) := (x\cdot y, D_1 \star D_2)\tag{2}$$ They can also be exponentiated: $$(x,D)^y = (x^y, D\wedge y )\tag{3}$$

What we usually call dimensionless quantities are really quantities of the form $(x,\mathbf 0)$. We would ordinarily denote such a thing simply by $x\in \mathbb R$.

Finally, we may define derived units for convenience - for example, $$ g \equiv 10^{-3} \mathrm kg \equiv \bigg(10^{-3}, \langle 1,0,0\rangle\bigg)$$ and so forth.

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Having defined all of these rules, we can provide your questions with unambiguous answers.

In dimensional analysis we treat the units as their own objects, in this case $1m$ is a unit multiplied by a number, and forms a 'quantity' is this explicitly the same object as the unit itself?

Physical units are just convenient choices of physical quantity such that we may express any other physical quantity by combining them via multiplication and exponentiation. They are very much like a basis for a vector space in that way. In the formalism presented here, $1\ \mathrm m$ and $\mathrm m$ are the same thing because $$1\ \mathrm m \equiv 1 * \bigg(1,\langle 0,1,0\rangle\bigg) = \bigg(1\cdot 1,\langle 0,1,0\rangle\bigg) = \bigg(1,\langle 0,1,0\rangle\bigg) \equiv \mathrm m$$

Having $1m=m$ implies that perhaps $-1m=-m$

This is also true for the same reason, though we would never write the latter expression because it could easily lead to confusion.

We would never give a length of "$m$"

True, but that's just a matter of convention and natural language.

If we can have an object just called $m$ and it is defined under addition (in the case it's defined as $1*m$) then we must have an additive inverse and $-m$ should exist, how do we deal with this issue? $-m$ should make no sense in the context of the unit.

I don't understand your objection. $-\mathrm m$ is a perfectly reasonable physical quantity (it might be a coordinate, for example), but to avoid awkwardness and misunderstandings we would pretty much always write that as $-1\ \mathrm m$.

How about taking $0*m$ is $0∗m=0∗s$? I believe this is not the case, as dimension needs to be preserved in multiplication, but I'm not sure.

I would agree. Two physical quantities $(x,D_1)$ and $(y,D_2)$ are equal if $x=y$ and $D_1=D_2$. In particular, $0\ \mathrm m \neq 0\ \mathrm s$ because $$0\ \mathrm m \equiv \bigg(0, \langle 0,1,0\rangle\bigg) \neq \bigg(0, \langle 0,0,1\rangle \bigg) \equiv 0\ \mathrm s$$

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I'll conclude by demonstrating that $0.5$ m/s = $50$ cm/s using my formalism. Note that

$$0.5\ \mathrm{m/s} \equiv \bigg(0.5,\langle 0,1,-1\rangle\bigg)$$ Next, $1\ \mathrm{cm} = 10^{-2}\ \mathrm{m} \equiv \big(10^{-2},\langle 0,1,0\rangle\big)$ and so $1 \ \mathrm{cm/s} = \big(10^{-2},\langle 0,1,-1\rangle\big)$. Therefore, we have that

$$50 \ \mathrm{cm/s} \equiv 50*\bigg(10^{-2},\langle 0,1,-1\rangle\bigg) = \bigg(0.5,\langle 0,1,-1\rangle\bigg) \equiv 0.5\ \mathrm{m/s}$$

Answer 2

It doesn't seem all the rules of algebra apply to units so it causes me confusion.

Indeed, units of length form a positive space, not a vector space. The space is closed under addition and the multiplication with positive numbers. See "An Algebraic Approach to Physical Scales" for much more information.

Answer 3

The measure of a physical quantity can be interpreted as the multiplication of a pure number, here $1$, and a unit of measurement, here $m$, that has the same physical dimension of the measured physical quantities. A $m$ indicates a length (and thus a physical quantity) of $1\,m$, like $2 \, m$ means twice that length.

When you write the length of $1 \, m$, I think that you can write $m$ implying that it's the same as $1 \, m$, but I wouldn't do it, while I'd explicitly write the pure number as well.

In dimensional analysis we don't care about the pure numbers, but only about the physical dimensions, each of which can have several different units of measurement: as an example a the physical dimension length $L$ can have several different units of measurement, like meters ($m$), millimetres ($mm$), inches ($in$), feet ($ft$),...

As an example, the dimensional analysis of the Newton's second principle of dynamics for a system with constant mass reads

$m \mathbf{a} = \mathbf{F}$,

and you need to check that $[m \mathbf{a}] = [\mathbf{F}]$, i.e.

$M \dfrac{L}{t^2} = F$,

given that:

• $M$ is the physical dimension of mass;
• $L/t^2$ is the physical dimension of acceleration, being $L$ the one for length, $t$ for time;
• $F$ is the physical dimension of force.