In so-called 'natural unit', it is said that physical quantities are measured in the dimension of 'mass'. For example, $\text{[length]=[mass]}^{-1}$ and so on.

In quantum field theory, the dimension of coupling constant is very important because it determines renormalizability of the theory.

However, I do not see what exactly the mathematical meaning of 'physical dimension' is. For example, suppose we have self-interaction terms $g_1\cdot \phi\partial^\mu \phi \partial_\mu \phi$ and $g_2 \cdot \phi^4$, where $\phi$ is a real scalar field, $g_i$ are coupling constants and we assume $4$ dimensional spacetime.

Then, it is stated in standard physics books that the scalar field is of mass dimension $1$ and so $g_1$ must be of mass dimension $-1$ and $g_2$ is dimensionless. But, these numbers do not seem to play any 'mathematical' role.

To clarify my questions,

  1. What forbids me from proclaiming that $\phi$ is dimensionless instead of mass dimension $1$?

  2. What is the exact difference between a dimensionless coupling constant and a coupling constant of mass dimension $-1$?

These issues seem very fundamental but always confuse me. Could anyone please provide a precise answer?

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    $\begingroup$ The term is "dimensional analysis". Wikipedia has a section on mathematical formulation. Your specific questions might fit better on Physics SE but I am not qualified enough to say for sure. $\endgroup$
    – Wojowu
    Aug 25, 2021 at 14:04
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    $\begingroup$ Mathematically speaking, a quantity with a non-trivial physical dimension is acted non-trivially on by the multiplicative group of positive real numbers, which implicitly corresponds to rescaling a particular unit. This may or may not help your intuition on this point. $\endgroup$ Aug 25, 2021 at 14:32
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    $\begingroup$ mathoverflow.net/questions/52585 $\endgroup$ Aug 25, 2021 at 17:51
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    $\begingroup$ if you are even interested in some applications you can check Barenblatt's book "scaling, self similraity and intermediate asymptotics"and Goldenfeld's book on phase transitions and renormalization group $\endgroup$
    – Frappa
    Aug 26, 2021 at 13:45
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    $\begingroup$ You ask whether you have some choice in defining what has what units. In general you do, although consistency will generally require that if you change the definition of one quantity's units, you make changes to the units of other quantities as well. A paper that gets into this is Dicke, “Mach’s principle and invariance under transformation of units,” Phys Rev 125 (1962) 2163. For example, there are various inconsistent conventions as to what units the metric should have in general relativity. I have a treatment of this topic in sec. 5.11 of my GR book: lightandmatter.com/genrel $\endgroup$
    – user21349
    Aug 26, 2021 at 18:18

2 Answers 2


Mathematically, the concept of a physical dimension is expressed using one-dimensional vector spaces and their tensor products.

For example, consider mass. You can add masses together and you know how to multiply a mass by a real number. Thus, masses should form a one-dimensional real vector space $M$.

The same reasoning applies to other physical quantities, like length, time, temperature, etc. Denote the corresponding one-dimensional vector spaces by $L$, $T$, etc.

When you multiply (say) some mass $m∈M$ and some length $l∈L$, the result is $m⊗l∈M⊗L$. Here $M⊗L$ is another one-dimensional real vector space, which is capable of “storing” physical quantities of dimension mass times length.

Multiplicative inverses live in the dual space: if $m∈M$, then $m^{-1}∈M^*$, where $\def\Hom{\mathop{\rm Hom}} \def\R{{\bf R}} M^*=\Hom(M,\R)$. The element $m^{-1}$ is defined as the unique element in $M^*$ such that $m^{-1}(m)=1$, where $-(-)$ denotes the evaluation of a linear functional on $M$ on an element of $M$. Observe that $m ⊗ m^{-1} ∈ M⊗M^* ≅ \R$, where the latter canonical isomorphism sends $(f,m)$ to $f(m)$, so $m^{-1}$ is indeed the inverse of $m$.

Next, you can also define powers of physical quantities, i.e., $m^t$, where $m∈M$ is a mass and $t∈\R$ is a real number. This is done using the notion of a density from differential geometry. (The case $\def\C{{\bf C}} t\in\C$ works similarly, but with complex one-dimensional vector spaces.) In order to do this, we must make $M$ into an oriented vector space. For a one-dimensional vector space, this simply means that we declare one out of the two half-rays in $M∖\{0\}$ to be positive, and denote it by $M_{>0}$. This makes perfect sense for physical quantities like mass, length, temperature.

Once you have an orientation on $M$, you can define $\def\Dens{\mathop{\rm Dens}} \Dens_d(M)$ for $d∈\R$ as the one-dimensional (oriented) real vector space whose elements are equivalence classes of pairs $(a,m)$, where $a∈\R$, $m∈M_{>0}$. The equivalence relation is defined as follows: $(a,b⋅m)∼(a b^d,m)$ for any $b∈\R_{>0}$. The vector space operations are defined as follows: $0=(0,m)$ for some $m∈M_{>0}$, $-(a,m)=(-a,m)$, $(a,m)+(a',m)=(a+a',m)$, and $s(a,m)=(sa,m)$. It suffices to add pairs with the same second component $m$ because the equivalence relation allows you to change the second component arbitrarily.

Once we have defined $\Dens_d(M)$, given $m∈M_{>0}$ and $d∈\R$, we define $m^d∈\Dens_d(M)$ as the equivalence class of the pair $(1,m)$. It is easy to verify that all the usual laws of arithmetic, like $m^d m^e = m^{d+e}$, $m^d n^d = (mn)^d$, etc., are satisfied, provided that multiplication and reciprocals are interpreted as explained above.

Using the power operation operations we just defined, we can now see that the equivalence class of $(a,m)$ is equal to $a⋅m^d$, where $m∈M_{>0}$, $m^d∈\Dens_d(M)_{>0}$, and $a⋅m^d∈\Dens_d(M)$. This makes the meaning of the equivalence relation clear.

In particular, for $d=-1$ we have a canonical isomorphism $\Dens_{-1}(M)→M^*$ that sends the equivalence class of $(1,m)$ to the element $m^{-1}∈M^*$ defined above, so the two notions of a reciprocal element coincide.

If you are dealing with temperature without knowing about the absolute zero, it can be modeled as a one-dimensional real affine space. That is, you can make sense of a linear combination $$a_1 t_1 + a_2 t_2 + a_3 t_3$$ of temperatures $t_1$, $t_2$, $t_3$ as long as $a_1+a_2+a_3=1$, and you don't need to know about the absolute zero to do this. The calculus of physical quantities can be extended to one-dimensional real affine spaces without much difficulty.

None of the above constructions make any noncanonical choices of physical units (such as a unit of mass, for example). Of course, if you do fix such a unit $μ∈M_{>0}$, you can construct an isomorphism $\R→\Dens_d(M)$ that sends $a∈\R$ to $aμ^d$, and the above calculus (including the power operations) is identified with the usual operations on real numbers.

In general relativity, we no longer have a single one-dimensional vector space for length. Instead, we have the tangent bundle, whose elements model (infinitesimal) displacements. Thus, physical quantities no longer live in a fixed one-dimensional vector space, but rather are sections of a one-dimensional vector bundle constructed from the tangent bundle. For example, the volume is an element of the total space of the line bundle of 1-densities $\Dens_1(T M)$, and the length is now given by the line-bundle of $λ$-densities $\Dens_λ(T M)$, where $λ=1/\dim M$.

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    $\begingroup$ @Wolfgang: If you are talking about kelvins as a unit of measurement for temperature, this means you know about the absolute zero and temperature is an element of a real vector space. In this case, multiplication by 2 is well-defined, corresponding, in physical terms, to doubling the average kinetic energy of particles in the system. For scales without absolute zero, multiplication by 2 is not defined, only affine combinations are defined, since temperatures in this case form a one-dimensional real affine space. $\endgroup$ Aug 26, 2021 at 0:12
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    $\begingroup$ @user347489: The metric $g$ is a bilinear form on the tangent bundle that takes values in $L⊗L$, i.e., has physical dimension length squared. If we have a curve $γ$, we can make sense of $g(γ'(t),γ'(t))^{1/2}∈L$, as described in the answer. This yields a function valued in $L$, which we can integrate over the domain of $γ$. The result is an element of $L$, which is the length of $γ$. $\endgroup$ Aug 26, 2021 at 6:47
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    $\begingroup$ @DeaneYang: I have not seen it presented in this specific form in the literature, definitely not with densities. I was thinking about densities at some point, wondering how to define the line of densities for a single vector space, as opposed to just a bundle (where you can simply take powers of transition functions). Once I figured this out, it became clear that for one-dimensional spaces this gives a coordinate-free way to take powers of their elements, such as physical quantities. $\endgroup$ Aug 26, 2021 at 18:21
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    $\begingroup$ @user3840170, fractional powers do appear. The square root of time in the heat equation, for example. $\endgroup$
    – Deane Yang
    Aug 26, 2021 at 18:56
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    $\begingroup$ @user3840170 There are absolutely fractional powers of quantities and units in physics! In addition to Deane Yang example above, in noise analysis one frequently refers to and plots the square root of the spectral density function of a random process, whose units are $\mathrm{V}/\sqrt{\mathrm{Hz}}$, $\mathrm{A}/\sqrt{\mathrm{Hz}}$ etc. $\endgroup$ Aug 27, 2021 at 10:02
  1. The action appears in an exponent, so it must be dimensionless. That then fixes the dimension of each term which appears in the action and "forbids you from proclaiming that $\phi$ is dimensionless".

To find the mass dimension of the field $\phi$ you can argue as follows: The action is the integral of the Lagrangian over $d$ space-time coordinates $x$, each of which has mass dimension $-1$, so the mass dimension of the Lagrangian is $d$. Hence the field $\phi$ must have mass dimension $d/2-1$ to ensure that the kinetic contribution $\propto (\partial\phi/\partial x)^2$ has mass dimension $d$.

  1. Whether or not a coupling constant has a dimension will depend on the number of space-time dimensions in which you work, there is no fundamental difference between the various numbers.

A term $g_2 \phi^4$ will have a coupling constant $g_2$ of mass dimension $4-d$. In 3+1 space-time dimensions $g_2$ is dimensionless. Similarly, the coupling constant $g_1$ in the term $g_1\cdot \phi\partial^\mu \phi \partial_\mu \phi$ must have mass dimension $1-d/2$, which equals $-1$ for $d=4$.

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    $\begingroup$ FWIW this sounds like a physicist's answer. Not that I'm complaining, since my background is in physics and this is very much like the answer I would have given in that context, but I did find Dmitri's answer more useful for understanding the deeper mathematical foundation of units. $\endgroup$
    – David Z
    Aug 28, 2021 at 0:34
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    $\begingroup$ true, I'm a physicist, but what other other answer than this can one give to question #1 of the OP: "What forbids me from proclaiming that $\phi$ is dimensionless instead of mass dimension 1" ? $\endgroup$ Aug 28, 2021 at 6:25
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    $\begingroup$ Fair enough :-) Sorry, my comment came out sounding a little more dismissive than I meant it. $\endgroup$
    – David Z
    Aug 28, 2021 at 6:53

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