# Mathematical rules for dimensional analysis

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.

• Can we have −m as −1∗m?? Why to stop here ? Maybe you want at least $[0^\circ..360^\circ]~m^\circ$ "degree-meters" for describing exact position vector ? So the answer is no, unit is a measuring type definition and should not depend on measurement direction. However quantity which it measures,- can depend on direction. Sep 6, 2022 at 9:02
• Worth reading if you want to do a deep dive into the algebra of dimensional analysis: terrytao.wordpress.com/2012/12/29/… Sep 7, 2022 at 23:37
• 5m can be treated the same as 5x. x is a variable that can be positive or negative. Having a coefficient of 5 in front of that x does not change the character of the variable. Accordingly, "m" is a unit of length all on its own, regardless of the coefficient in front of it. Sep 7, 2022 at 23:38

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.

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 $$\langle a,b,c\rangle \wedge d := \langle d\cdot a,d\cdot b,d\cdot c\rangle$$

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.

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 $$0\ \mathrm m \equiv \bigg(0, \langle 0,1,0\rangle\bigg) \neq \bigg(0, \langle 0,0,1\rangle \bigg) \equiv 0\ \mathrm s$$

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

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

• Awesome answer. Do you have any references for further reading? Sep 8, 2022 at 0:29
• @Shaktyai Andrew linked this blog post by Terry Tao in a comment to the question; while my approach does not precisely align with either formalization which he describes explicitly, it is a phenomenal discussion. Sep 8, 2022 at 1:28
• @Shaktyai And thank you for the kind comment :) Sep 8, 2022 at 1:41
• Brilliant answer, so it is as I thought, I just find it strange that we can have $-m$ but I guess it's just one of those things Sep 8, 2022 at 8:42
• I guess when you try to formalize such a thing, you get answers that are required mathematically, but don't make physical sense Sep 8, 2022 at 9:03

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.

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.
• can we consider $1$ the number that returns the unit as an answer in multiplication then? Where as $2$ would give us something which is mathematically double in magnitude? I'm really asking if mathematically speaking they are definitely the same object. Sep 5, 2022 at 16:02
• yes, they are the same. I don't see any reason (besides philosophy arguments) that makes $1 m \ne m$, like I don't see any reason that makes $1 x \ne x$. Sep 5, 2022 at 16:04
• Because $x$ is a number, the fact that they are the same is well defined, it's best not to assume anything as they are units but we treat them like variables. Sep 5, 2022 at 16:06
• $x$ could be more than a pure number. Think at the second law of Dynamics, $m \mathbf{a} = \matfbf{F}$, no symbol represents a pure number here Sep 5, 2022 at 16:09
• @user37577 You can think of m as being an unspecified constant (a number), and then you can scale that (resize it s times by multiplying it by a scalar s, to get sm: 0.5 m, 2 m, 100 m, ...). So, if I wanted to know how long is 2 m, you could give me something that you know represents the length m - like some stick (some standard), and I can use any measuring device using any units whatsoever to measure the stick, and then multiply whatever I got by 2 to get the correct length. That's what 2 m means. Sep 5, 2022 at 17:19