Excuse my naive question and please let me explain it:

In everyday life we experience 3 spatial "dimensions" + time etc. Usually the 3 dimensions are represented by a coordinate system and mathematically as the vector space $R^3$. In constrast, the word "dimension" in dimensional analysis has a more objective realism which is based on physical measurements. For instance let $l$ be a transcendental number.

Then $1,l,l^2,l^3,\cdots$ will be linearly independent over the rational numbers, and hence can serve as a basis in a vector space. For instance, instead of saying the point in spatial dimension has coordinates $(x,y,z)$ we might as well write:

$$x+y \cdot L + z \cdot L^2$$

where $L$ represents a "transcendental number over the rationals" and is associated with meter or kilometer etc.

Take for instance the derived unit $s/t^k$ for $k=0,1,2$ where $s$ denotes length and $t$ denotes time. Then $s$=length, $s/t =$ velocity, $s/t^2=$ acceleration. If we view these as a basis of a vector space, then this vector space has 3 dimensions. But no one would count subjectively these as dimensions.

On the other hand the seven basic SI-units (or 6 if you do not want to count $mol$), could be seen as a transcendental numbers over the rationals or reals, and hence powers of those transcendental numbers could give a basis for vector spaces. Monomials of these transcendent numbers would correspond, as is done in dimensional analysis, to derived units. Adding and subtracting for instance $LT^{-1}+M$ would give a point in the "space of physical quantities" (Edit: and indicate the measured velocity and mass of an object). As a vector space over the reals this space has infinite dimension but has transcendence degree of $7$. Every derived / basic physical quantity measured by SI-Units would correspond to a point in this space / field of transcendence degree $6$ (or $7$ if you count $mol$ which I will not do):


Excuse my naive question: Is there any reason from physics to discard this point of view?

Edit: It seems that the main idea of this question can be implemented through the ring (Laurent polynomial ring) $L:=\mathbb{R}[T,T^{-1},L,L^{-1},M,M^{-1},I,I^{-1},\Theta, \Theta^{-1},J,J^{-1}]$

second Edit: I was asked to give at least one application of this idea, which I would like to do:


Let $k(x,y) = \frac{xy}{x^2+y^2-xy}$ for $x \neq 0,y \neq 0, x,y \in \mathbb{R}$ be a Jaccard-Similarity / positive definite kernel defined on $\mathbb{R}$.

We can define a similarity and positive definite kernel on the Laurent polynomial ring $L$ as :

$$K(x,y) := \frac{1}{N_x + N_y - N_{xy}} \sum_{X_i^{\alpha_i}=Y_j^{\beta_j}} k(a_i,b_j)$$

for $x = \sum_{i} a_i X^{\alpha_i},y = \sum_{j} b_j X^{\beta_j}$ and $X = (T , L, M, I, \Theta,J)$, $\alpha_i, \beta_j \in \mathbb{Z}^6$ and $X^{\alpha_i},X^{\beta_j}$ are multinomials, and $N_x = $ number of nonzero $a_i$, $N_y =$ number of nonzero $b_j$, $N_{xy} =$ number of $(X^{\alpha_i} = X^{\beta_j})$.

Since $k(ca,cb) = k(a,b)$ for all $a,b,c \neq 0$, we deduce that $K(c \cdot x,c \cdot y) = K(x,y)$ for all $x,y \in L$, $c \neq 0, c \in \mathbb{R}$. Hence this (or any other similarity and positive definite kernel $k$ with $k(ca,cb) = k(a,b)$. This is to make sure, that rescaling of physical units, does not change the similarity between objects.) gives us a possibility to measure the similarity / inner product of two physical objects $A,B$ each of which is defined through measurements $x = \sum_{i} a_i X^{\alpha_i}$ and $y = \sum_{j} b_j X^{\beta_j}$.

Since there are different possibilities to measure similarites / define positive definite kernels, there should be different possibilities to define equality / similarity between two physical objects $A$ and $B$.

Hence aposteriori $L$ is a Hilbert space by the Aaronszajn-Kolmogorov theorem.


The meaning of this kernel is to compare two physical objects. For instance $A = 10 m/s + 1 kg$, $B = 9 m/s + 2kg$, $C = 1 m/s^2+10kg$. Then $K(A,B) = 226/276 = 0.8278$, $K(A,C) = 10/91= 0.10989$, $K(B,C) = 5/21 = 0.2381$. Hence $A$ is most similar to $B$, $B$ is most similar to $A$, $C$ is most similar to $B$, and $A,C$ are the most dissimilar physical objects in this list.

  • 1
    $\begingroup$ If $l$ is a transcendental number, then $\{\,1,l\,\}$ is not linearly independent over the reals. So I stopped reading. $\endgroup$ Nov 30, 2021 at 9:29
  • 2
    $\begingroup$ @GerryMyerson: I am sorry you did not read the full question, and also sorry for the inconvenience. I meant transcendental like $x$ in $\mathbb{R}(x)$, or like in transcendence degree: en.wikipedia.org/wiki/Transcendence_degree $\endgroup$ Nov 30, 2021 at 9:34
  • $\begingroup$ $x$ is not a number. $\endgroup$ Nov 30, 2021 at 9:40
  • 1
    $\begingroup$ @GerryMyerson: Ok, then one could take $\mathbb{Q}$ instead of $\mathbb{R}$ and replace $x$ by transcendental numbers. I will correct the question. $\endgroup$ Nov 30, 2021 at 9:43
  • 2
    $\begingroup$ There are no physical reasons to consider a quantity like $LT^{-1}+M$. On the contrary, in physics, there are many speculations based on playing with dimensions. It is how the Planck time and the Planck length are introduced : by supposing that an unknown theory would produce as a byproduct the Planck time and the Planck length which, because of their values, are interpreted as the minimal time and length if the spacetime was discrete (which does not prove of course that the spacetime is discrete). Since physics under the Planck length is unknown, it is just speculation. $\endgroup$ Dec 1, 2021 at 8:59

1 Answer 1


I don't think there are many meaningful situations where physical quantities of different dimension are added together (in fact, this is widely regarded as a taboo, and precisely the sort of mistake that dimensional analysis is supposed to prevent us from doing), so I don't think viewing the space of physical quantities as a field is the most fecund or useful description.

Instead, let me offer the following alternative point of view, which I think better matches what physics does and why it does it, and which is, at any rate, how I think of physical dimensions:

Consider the “group of homogeneities$G = (\mathbb{R}_+^\times)^k$ which is the product of $k$ copies of the positive real numbers, where $k$ is the number of “base units” (so $k=7$ in the SI system, but this depends on the sort of theory we wish to study: economists might have reasons to consider currencies or commodities as different units, whereas relativists will not consider time and length to be different units): the point is that $G$ acts on physical quantities by multiplying each one of the $k$ base quantities by the corresponding positive real number (e.g., $g=(2,3,1,\ldots)$ might multiply lengths by $2$, times by $3$ and preserve masses).

The crucial physics point is that: insofar as the choice of units is arbitrary, $G$ acts as a group of physics-preserving symmetries (e.g., multiplying all lengths by $2$ and times by $3$ while preserving masses should keep physics the same). So we seek quantities that are covariant under $G$. This is the whole point of dimensional analysis.

Now the group $G^*$ of characters of $G$, i.e., the group of continuous morphisms $\lambda\colon G\to \mathbb{R}_+^\times$ is the group $\mathbb{Z}^k$ with $(d_1,\ldots,d_k) \in \mathbb{Z}^k$ sending $g := (g_1,\ldots,g_k) \in G$ to $g_1^{d_1}\cdots g_k^{d_k}$. Really we should forget about $k$ and care about the group $G$ which might not have a clear “basis of dimensions” (e.g., is it electrical charge or electrical current which is the “base unit”? this is meaningless) and might even incorporate other transformations than positive-real-homogeneities (e.g., changes of sign on certain quantities might be physics-preserving).

So I propose that: a “physical dimension” is an element $\lambda$ of $G^*$ and a quantity of that dimension is an element of the corresponding representation $V_\lambda$ of $G$ (which is $1$-dimensional because $G$ is abelian), i.e., $G$ acts on $V_\lambda$ by $g \mapsto v \mapsto g^\lambda\cdot v$.

This $V_\lambda$ is $1$-dimensional, so it is $\mathbb{R}$ as an $\mathbb{R}$-vector space, but it is so in a non-canonical way, and a unit of measure for the physical dimension $\lambda$ is a basis for $V_\lambda$ (probably a positive basis since $V_\lambda$ still has a natural orientation).

Taking the product of two physical quantities of dimensions $\lambda$ and $\mu$ consists of taking the tensor product $V_\lambda \otimes V_\mu \buildrel\sim\over\to V_{\lambda+\mu}$, while the inverse of a physical quantity $\lambda$ should be seen as living in the dual space $V_\lambda^\vee \buildrel\sim\over\to V_{-\lambda}$. The sum, on the other hand, is meaningful only insofar as it compatible under the action of $\lambda$, i.e., adding elements of the same $V_\lambda$. (Of course, you are welcome to consider $V_\lambda \oplus V_{\lambda'}$, in some cases it might be meaningful, e.g., adding different currencies in economics, but the direct sum is then canonically split.)

What you propose to do, by taking arbitrary rational combinations of the base units and considering them as transcendental quantities, would be germane if the physically relevant acting group were the full Cremona group of automorphisms of rational fractions. I don't think we can make any physical sense of such an action, so this point of view is bound to remain very artificial.

PS: I'd like to add the following remark showing that the group of symmetries (so, in my description, $k$) might very well depend on the kind of theory being considered: Americans like to measure heights in feet and plane distances in miles, which makes some kind of sense in a situation where one might apply different homotheties on both; the “fundamental slope” of $5280\,\mathrm{ft}/\mathrm{mi}$ breaks this symmetry, exactly in the same way the “fundamental speed” known as the speed of light of $299\,792\,458\,\mathrm{m}/\mathrm{s}$ breaks the possibility of applying different homotheties on times and lengths: there is no fundamental difference between $5280\,\mathrm{ft}/\mathrm{mi}$ and $299\,792\,458\,\mathrm{m}/\mathrm{s}$ from the dimensional point of view: they are conversion factors breaking the possibility of choosing two units independently which, from the pre-breaking point of view, appears as a fundamental constant of nature (which can be measured experimentally: if the foot and the mile had been defined separately using a reference height and a reference plane length, the former would be an experimental quantity until redefinition of one of the units, exactly like the speed of light in the SI system).

  • 1
    $\begingroup$ Thank you for your detailed insight in this question. I will need some time to think about it and absorb your ideas. (+1) $\endgroup$ Nov 30, 2021 at 9:42
  • $\begingroup$ You could add $LT^{-1}+M$ to indicate the velocity and the mass of an object that you measure, etc. $\endgroup$ Nov 30, 2021 at 9:57
  • 1
    $\begingroup$ @stackExchangeUser May I gently suggest that "velocity and mass" might be better considered as a vector (velocity,mass) rather than the sum velocity+mass. So for example if a mass of 10kg is moving with a velocity of 20m/s, it seems more natural to record this as (20,10) or perhaps (20m/s,10kg), rather than as 30 (being 20+10=30). The problems are (a) is the 30 from a 10kg mass moving at 20m/s, or a 5kg mass moving at 25m/s, or...? (b) this will have terrible trouble with changing units, e.g., kilograms to grams (30 becomes 10020 ?!). $\endgroup$ Dec 1, 2021 at 8:28
  • $\begingroup$ @ZachTeitler: My proposal is to use $m/s$ a $LT^{-1}$ which are just variables and hence transcendent over $\mathbb{R}$ and so distinct physical (derived or basic) units represent different vector dimensions, since they are linear independent over $\mathbb{R}$. I am not suggesting to add $10+20=30$ but to see $20 m/s + 10kg$ as a formal sum such as is done with multivariate Laurent polynomials. $\endgroup$ Dec 1, 2021 at 8:38
  • $\begingroup$ I will be interested if you can give one example where something is gained by doing so. $\endgroup$
    – mme
    Dec 1, 2021 at 8:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.