17
$\begingroup$

Multiplication and division of units of measurement seems to be defined, e.g., $\mathrm{N}\times\mathrm{m}=\mathrm{N}\text{-}\mathrm{m}$, m/s = m/s (a rather unsatisfactory example---help).

Exponentiation of units seems also to be defined, e.g., $\mathrm{m}^3=\mathrm{m}\times\mathrm{m}\times\mathrm{m}$.

Addition and subtraction seem to be defined for some units of measurement, e.g., 1 m + 2 m = 3 m; but not for others: 273 K + 273 K is nonsensical. Length (m) is an extensive property; temperature (K) is an intensive property.

Addition of mixed units is defined---sort of---if we apply scaling.

Reduction to simplest terms is not well defined, e.g., $1 \mathrm{m} + 1 \mathrm{ft}$.

The above anecdotes are enough to suggest that units of measurement are not rational or real numbers.

But, what are they?

Addendum

After doing just a little reading on torsors, I think that in order to answer this question, we have to e specific about our model of the thing we are measuring as well as the unit of measurement.

people once thought only temperature differences could be measured - but then they discovered absolute zero. As soon as we pick units of temperature, temperatures are elements of an R-torsor. When absolute zero was discovered, this R-torsor was revealed to be R itself.

(Baez J [Downloaded 2025-09-17] Torsors Made Easy)

If energy is (still) a torsor and temperature is not (or isn't any longer), then energy and temperature are different kinds of objects.

Improve this question
$\endgroup$
6
  • 2
    $\begingroup$ Related: physics.stackexchange.com/q/98241/25301, physics.stackexchange.com/q/476965/25301, physics.stackexchange.com/q/337092/25301, probably others in the Linked/Related sections of those $\endgroup$ Commented Sep 6 at 15:19
  • 6
    $\begingroup$ You can indeed add a delta temperature to a "base" temperature, so 273 K + 273 K will equal 546 K. In addition, units are not mathematical objects. They are what "connects" mathematical equations to the real world. $\endgroup$ Commented Sep 6 at 15:35
  • $\begingroup$ I think your question is answered by physics.stackexchange.com/a/13087/174766. $\endgroup$ Commented Sep 6 at 16:41
  • 4
    $\begingroup$ The canonical (but quite advanced) blog post on this topic: terrytao.wordpress.com/2012/12/29/…. $\endgroup$ Commented Sep 6 at 17:44
  • 1
    $\begingroup$ Addition of temperatures can, in fact, be useful in some very limited circumstances, e.g. where you have several objects and you wish to know their average temperature. Computing the average involves addition. $\endgroup$ Commented Sep 7 at 0:55

6 Answers 6

Reset to default
13
$\begingroup$

The space of lengths is a one-dimensional ordered real vector space $L$.

We don't want to identify $L$ with the real numbers because the product of lengths is not a length and there is no distinguished length corresponding to the distinguished real number $1$.

We want $L$ to be ordered because there is an important physical difference between a positive length and a negative length; choosing an ordering comes down to deciding which lengths are declared positive.

The space of areas is another such vector space; explicitly it is the space $A=L\otimes_{\mathbb R}L$. The space of time intervals is another such space $T$. The space of velocities is another such space $V$; explicitly $V=L\otimes_{\mathbb R}T^{-1}$ (where $V^{-1}$ denotes the space dual to $V$).

Given a pair of lengths $l_1$ and $l_2$, there is an associated area, namely $l_1\otimes l_2$.

Given a length $l$ and a time interval $t$, there is an associated velocity, namely $l\otimes t^*$ where $t^*:T\rightarrow {\mathbb R}$ is the unique map taking $t$ to $1$.

Et cetera.

Improve this answer
$\endgroup$
6
  • $\begingroup$ With this structure is there any way to get units with fractional exponents? Eg $12 \ \mathrm{m^{3/2}}$ $\endgroup$ Commented Sep 6 at 21:02
  • 2
    $\begingroup$ @Dale: It would take some work, but I'm guessing you could do it, by introducing a one-dimensional vector space $M_2$ that you think of as "the space of square roots of masses" and adding some structure in the form of an isomorphism $M_2\otimes_{\mathbb R}M_2\rightarrow M$ (where $M$ is the space of masses). This sort of thing adds fractional exponents ``one denominator at a time'', and you might prefer a structure that incorporates all of them at once. I can imagine what that would look like, but would have to think about the details. $\endgroup$ Commented Sep 6 at 23:37
  • $\begingroup$ The space of velocities is not a linear vector space, as I realized while I was writing my answer to complement yours. The space of four-velocities is linear, though. $\endgroup$ Commented Sep 7 at 3:45
  • $\begingroup$ What is $\otimes_{\mathbb R}$? $\endgroup$ Commented Sep 8 at 3:28
  • 1
    $\begingroup$ @TheRizzler: If $V$ and $W$ are real vector spaces, then $V\otimes_{\mathbb R} W$ (which is called the tensor product of $V$ and $W$ over ${\mathbb R})$ is the universal recipient of a bilinear map from the product $V\times W$. It has a basis the various $v_i\otimes w_j$ where the $v_i$ form a basis of $V$ and the $W_j$ form a basis of $W$. $\endgroup$ Commented Sep 8 at 3:47
7
$\begingroup$

I think the right answer is that "a unit" (or better, "a dimension," as defined below) is a point in a seven-parameter space whose coordinates are rational numbers. Each "point" in this space (e.g. $\rm J\ mol^{-1}\ K^{-1}$) corresponds to a type of physical quantity (e.g. specific heat capacity), and the rules for manipulating these quantities must be discovered by experiment. Some of these quantities, as WillO writes in another answer, are nice linear vector spaces, like one-dimensional distances. But others aren't, like one-dimensional speeds.

In the International System of Units, the definition of a unit is:

The value of a quantity is generally expressed as the product of a number and a unit. The unit is a particular example of the quantity concerned which is used as a reference, and the number is the ratio of the value of the quantity to the unit.

For a particular quantity different units may be used. For example, the value of the speed v of a particle may be expressed as v = 25 m/s or v = 90 km/h, where metre per second and kilometre per hour are alternative units for the same value of the quantity speed.

Before stating the result of a measurement, it is essential that the quantity being presented is adequately described. This may be simple, as in the case of the length of a particular steel rod, but can become more complex when higher accuracy is required and where additional parameters, such as temperature, need to be specified.

The standard then defines the seven famous reference quantities:

  • the second is the duration required for a certain number of microwave oscillations to occur from cesium atoms mistreated in a particular way
  • the meter is the length traveled by light in a certain small fraction of a second
  • the ampere is the current associated with a certain number of elementary charges per second through an electrical device
  • the kilogram is the mass of an object which interacts in a certain way with a Kibble balance or other such inertia-measuring device
  • the kelvin is the change in temperature which results in a particular change in thermal energy
  • the mole is a particular size of a collection of things (think "a dozen" or "a baker's dozen" or "a gross," but a somewhat larger collection)
  • the candela is the brightness of light with a particular wavelength

Note that, unlike in previous versions of SI, the cesium clock standard is the only particular reference quantity which is definitive. We've gotten rid of the meter rod, the prototype kilogram, and the triple point of undeuterated water:

Instead of each definition specifying a particular condition or physical state, which sets a fundamental limit to the accuracy of realization, a user is now free to choose any convenient equation of physics that links the defining constants to the quantity intended to be measured.

We then define the dimension of a quantity by the coordinates of its reference value in a seven-dimensional vector space:

$$ \text{dim}\ Q = \mathsf T^\alpha \mathsf L^\beta \mathsf I^\gamma \mathsf M^\delta \mathsf \Theta^\epsilon \mathsf N^\zeta \mathsf J^\eta, $$

where $\alpha$ is the exponent associated with time, $\beta$ the exponent associated with length, et cetera. For most types of quantities, the exponents are small integers or rational numbers.

As WillO writes in another answer, quantities with some dimensions effectively occupy their own one-dimensional vector space. Lengths (dimension $\mathsf L$) are closed under addition and obey scalar multiplication; multiplying two lengths gives you a quantity in a different vector space $\mathsf L^2$, which we call an area.

Addition and subtraction seem to be defined for some units ... but not for others: 273 K + 273 K is nonsensical.

I don't know that this is the case. The existence of the Celsius scale shows that it certainly makes sense to talk about the difference between two temperatures. Temperature is weird because it's fundamentally a derivative, and it shows up in the denominator of its own definition, too. Frequency is similarly weird, and also involves a denominator. If I play a 440 Hz tone and "add" a 445 Hz tone, I'll hear a 5 Hz "beat frequency," but I generally won't hear anything at 885 Hz.

On the other hand, adding two velocities like

$$ \vec u \neq \vec v_1 + \vec v_2 $$

has been discovered by experiment to be wrong; a more correct rule is

$$ \vec u = \frac{ \vec v_1 + \vec v_2 }{ 1 + {\vec v_1 \cdot \vec v_2}/c^2} $$

which arises from treating velocities as a space of "four-vectors" as discussed in your favorite book on special relativity. For things like baseballs and airplanes, the (wrong) linear addition formula for velocity introduces errors starting in about the fourteenth significant figure, so sometimes we forget about its wrongness.

There are a couple of places in the SI where quantities which aren't directly comparable wind up with the same dimension. For example, torque ($\rm N\,m$) and energy ($\rm J$) both have dimension $\mathsf L^2 \mathsf T^{-2} \mathsf M^1$, but they don't really measure the same thing --- this is probably something to do with the radian being officially dimensionless. You might make the same remark about the hertz (for periodic phenomena, which usually get radians in their descriptions) versus the becquerel (for stochastic phenomena).

Improve this answer
$\endgroup$
9
  • 8
    $\begingroup$ The mathematical operation of adding two velocities $u+v$ is never wrong. Of course you use it in the numerator of your correct equation. Rather, the formula for how velocities transform in different reference frames, which could be anything $f(u,v)$, happens not to be direct addition in general. But that has nothing to do with whether m/s can be added. $\endgroup$ Commented Sep 7 at 4:26
  • $\begingroup$ "the cesium clock standard is the only particular reference quantity which is definitive": I don’t understand this comment. The definition of the second will be changed in the future (a not-so-distant one) to refer to optical transitions of other atomic species, which allow the implementation of more stable and accurate clocks. There is a roadmap for this change. $\endgroup$ Commented Sep 7 at 7:08
  • 2
    $\begingroup$ @RC_23 My intent was to distinguish between the algebraic manipulation you describe in your answer, which the BIPM's SI handbook calls "the quantity calculus" or "the algebra of quantities," versus the physical rules for combining quantities, such as the asker's observation that temperatures are intensive. Obeying the "algebra of quantities" for velocities is necessary but not sufficient for adding them together. $\endgroup$ Commented Sep 7 at 12:07
  • $\begingroup$ @MassimoOrtolano When the definition of the second is changed in the future, then different frequency standards will define the second. As you say, the current definition of the second does still reference one particular experiment. $\endgroup$ Commented Sep 7 at 12:20
  • 1
    $\begingroup$ @MassimoOrtolano Perhaps I should have said "defining" instead. The BIPM brochure is a little clearer that the cesium standard is the definition for now, but Appendix 2 discusses the moves towards higher-stability standards. $\endgroup$ Commented Sep 7 at 19:06
2
$\begingroup$

Essentially, physical units are modeled as elements of torsors under a group. This is a more precise model that resolves the issue of a fixed zero of vector spaces. Formally, for set-theoretic purposes, for a group $G$, a $G$-torsor is a set $X$ equipped with a free and transitive right (or left) action of $G$, which is for $\cdot : X \times G \rightarrow X$ a right group action, that

  • Free Action: The action has no fixed points. For any $x \in X$, if $x \cdot g = x$, then $g$ must be the identity element $e \in G$.
  • Transitive Action: For any two elements $x,y \in X$, there exists a unique group element $g \in G$, such that $y = x \cdot g$.

For a more formal discussion, see https://ncatlab.org/nlab/show/torsor.

For our purposes here in modeling physical units, this is usually for $G = \mathbb{R}^\times_{>0}$ the multiplicative group of positive real numbers. One example of a $G$-torsor $X$ in this case is the set of all physical length values, an element $x \in X$ a number paired with a unit. The group $\mathbb{R}^\times_{>0}$ acts on $X$ by scalar multiplication, representing the changing of scale or converting between units.

Torsors are essentially just vector spaces regarded up to a choice of origin, analogous to affine spaces. There is then also the notion of an associated vector space, the choice of an origin, corresponding to a choice of standard unit.

For a more detailed discussion, see https://ncatlab.org/nlab/show/physical+unit.

Improve this answer
$\endgroup$
5
  • $\begingroup$ This and other answers lead me to believe that I am not qualified to mark a "correct" answer. I am "accepting" this answer because it seems to be leading me down a path toward a deeper understanding. $\endgroup$ Commented Sep 9 at 14:35
  • $\begingroup$ I don't think I have the prerequisite knowledge to understand this answer well. I have skimmed through a text on Abstract Algebra, and the unfamiliar terms here seem to match up with the terms introduced there. Would an introductory text on Abstract Algebra be a good place to begin gaining the background needed to understand your answer? $\endgroup$ Commented Sep 9 at 14:38
  • $\begingroup$ @AnaNimbus Well, the mathematics used here is quite minimal, just torsors and vector spaces, so essentially just basic group theory and linear algebra would work. Abstract algebra does help, especially if you are going down the line of abstraction $\endgroup$ Commented Sep 9 at 14:43
  • $\begingroup$ Shouldn't the origin be the identity dimension (ie $m^0$), and hence be independent of the standard unit chosen? Ah, no, you these members are elements of $m^1$ and $m^2$ etc. So this doesn't really cover how elements of $m^1$ naturally multiply to form elements of $m^2$. $\endgroup$ Commented Sep 9 at 18:51
  • $\begingroup$ I really like this answer. I think it comes closest to answering the question. But I think that units have additional structure beyond the torsor. The torsor only defines multiplication of a torsor with a real to get another torsor and division of two torsors to get a real. So with only the torsor structure you could say $30 s=0.5 min$ and $(1 min)/(1 s)=60$. But you need additional structure to handle addition within a torsor and the creation of new torsors through multiplication. E.g. $1min+30s=90s$ and $1min*30s=1800s^2$ $\endgroup$ Commented Sep 10 at 17:17
2
$\begingroup$

A unit is mathematically the same as a variable.

$$1a+1a = 2a $$

$$2a \cdot 3a = 6a^2 $$

$$4a/2b = 2~a/b$$

Replace $a$ and $b$ with any unit (m, s, K, ft, lb, fathom, acre) and all the same algebraic rules apply.

And you can certainly add Kelvin together.

Saying something is "15 m" long means its length $L$ divided by the length of a meter is the raw number 15.

$$ \frac L {\rm meter} = 15$$

$$L = 15~ \rm m$$

Improve this answer
$\endgroup$
8
  • 5
    $\begingroup$ No it's not. 15m + 15m^2 does not make any sense, if you consider them units of length and area. Or 12s + 15kg. You can perfectly well have a formula in n unknowns, but for units it's undefined. $\endgroup$ Commented Sep 7 at 13:56
  • $\begingroup$ I'm not sure what you mean. $x$ and $x^2$ are not commensurate and cannot be added algebraically, which is the same for meters and square meters. Of course you can compute $x+x^2$ by inputting a pure number $(2)+(2)^2= 6$, but unit variables are never substituted by pure numbers in normal usage $\endgroup$ Commented Sep 7 at 16:02
  • $\begingroup$ @infinitezero 15 m + 15 m² may be semantically meaningless, but there's no harm in it being syntactically valid. 5/0 is also syntactically valid but semantically meaningless, and no one really complains about that. $\endgroup$ Commented Sep 7 at 16:19
  • 2
    $\begingroup$ I am with @infinitezero on this. For a mathematical structure to represent units then $15 m + 15 m^2$ should be undefined. The operation $+$ should not work with incommensurate units $\endgroup$ Commented Sep 8 at 0:39
  • $\begingroup$ if we make an analogy with tensor algebra $$ T(V)=\bigoplus _{k=0}^{\infty }T^{k}V =K\oplus V \oplus (V\otimes V) \oplus (V\otimes V\otimes V)\oplus \cdots,$$ :-) $\endgroup$ Commented Sep 8 at 8:16
0
$\begingroup$

Units are not mathematical objects, but nouns used to identify the connection of (real or conceptual) objects to an abstraction (i.e. a number). Math is useful because sometimes a mathematical measure is analogous to a reality.

Improve this answer
$\endgroup$
0
$\begingroup$

With integer exponents, dimension-attached values form a formal multinomial, possibly a multinomial power series.

Unit conversion then looks like "modding out" by the conversion; ie setting $(m-3.28084 ft)$ to zero (which implicitly sets $(m-3.28084 ft)^2$ to zero, which is $$(m^2 -2 * 3.28084 m ft + 3.28084^2 ft^2) = (m^2 -2 * 3.28084 ft(3.28084 ft) + 3.28084^2 ft^2) = (m^2 - 3.28084^2 ft^2)$$ as expected (once you "mod out" that 3.28084 feet is 1 meter, squaring the "mod out" also zeros out the proper number of square feet equals a square meter).

Adding in non-integral exponents, I don't know the name for that space; but polynomials with non-integer exponents aren't that weird. The math to handle formally converting between $m^{1.5}$ and $ft^{1.5}$ is going to be amusing, but it will just work out to $1 m^{1.5} = 3.28084^{1.5} ft^{1.5}$.

Of interest is that scalar values in this space correspond to dimensionless units in physics; they don't depend on your choice of variables ($m$ or $ft$ or whatever).

Physics, in practice, tends to only use a few different variables. And equations that aren't a single coefficient of some multinomial variable product (possibly empty) are treated as non-physical.

The mathematical foundation of this doesn't care about that, however. You can reformulate the math and get values whose "variable dimension" is $m^2 + m^-1$; these values care about the ratio between the different dimensions (a different ratio leads to a different vector in the variable-space), making choice of dimension semi-physical.

Improve this answer
$\endgroup$

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.