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,
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What forbids me from proclaiming that $\phi$ is dimensionless instead of mass dimension $1$?
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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?
Best Answer
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 $λ=1/\dim M$.