Why does the dimension of the electric charge depend on the number of spacetime dimensions?

Question

We can find via dimensional analysis that the dimension of the electric charge varies with the dimension of space-time $(D+1)$: $$[\text{charge}] = (\text{eV})^{(3-D)/2}.$$ It is dimensionless if there are three spatial dimensions ($D=3$). You can see below the way I did it. The question is: Why does this occur? What is the meaning of this?

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I'm using the Heaviside-Lorentz Units with Natural Units ($\hbar = c = 1$) so that all dimensions can be expressed in energy ($\text{eV}$).:

$$ \begin{align} [x] &= [t] = (\text{eV})^{-1}\\ [p] &= [E] = \text{eV} \end{align} $$

Using the action from Maxwell theory $$ S = - \frac{1}{4} \int d^{1+D} x F^{\mu \nu} F_{\mu \nu} $$ and making a dimensional analysis, $$ \begin{align} [S] &= [x]^{1+D} [F^{\mu \nu}]^2\Rightarrow\\ 1 &= (\text{eV})^{-D-1} [F^{\mu \nu}]^2\Rightarrow\\ [F^{\mu \nu}] &= (\text{eV})^{(D+1)/2} \end{align} $$ Then using the inhomogeneous equation $$ \partial_\nu F^{\mu \nu} = J^\mu $$ and making an dimensional analysis $$ \begin{align} [\partial_\nu] [F^{\mu \nu}] &= [J^\mu]\\ eV eV^{(D+1)/2} &= [J^\mu]\\ [J^\mu] = (\text{eV})^{(D+3)/2} \end{align} $$

Now lets look the charge dimension. The relation of the charge with the $(1+D)$ density current is

$$[J^\mu] = \frac{[\text{charge}]}{[x]^D}$$

So, $$ \begin{align} (\text{eV})^{(D+3)/2} &= (\text{eV})^{D} [\text{charge}]\\ [\text{charge}] &= (\text{eV})^{(3-D)/2} \end{align} $$

We get that only in $3+1$-dimensions the charge is dimensionless. If we are at $1+1$, and at $1+5$ charge is just like time.

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At another place I was told that it (dimension of charge) have relation with the renormalizability of QED. And that the fact that charge is dimensionless only in $3+1$ is related with Maxwell equations being conformally invariant at $3+1$.

Answer 1

Regularizing QED with dimensional regularization, we switch from $4\rightarrow d$ dimensions. The action can now be written as

$S = \int d^dx\mathcal{L}_d\ ,$

where the subscript $d$ empathizes the dimension. The action has to be dimensionless, so $\mathcal{L}_n$ has dimensions $m^d$. (I use $m$ for mass, or equivalent, energy dimension, s.t. $[x]=m^{-1}$.)

The QED Lagrangian is

$\mathcal{L}_{QED}= \bar\psi \left(i\gamma^\mu(\partial_\mu -ie_d QA_\mu)-m\right) \psi -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} \ .$

Dimensional analysis yields:

$[x]=m^{-1} \\ [\partial_{\mu}]=m^1\\ [F_{\mu\nu}]=m^{\frac{d}{2}}\quad \Rightarrow \quad [A_\mu]=m^{\frac{d}{2}-1}\\ [\psi]=m^{\frac{d-1}{2}} \quad \Rightarrow \quad [e_d]=m^{\frac{4-d}{2}} $

So the fact that the charge, i.e. the coupling of QED, is not dimensionless in $d\neq 4$ comes from the requirement of renormalizability of the theory.

The meaning of this is, that e.g. in a (renormalizable) $2+1$ theory, there is no scale invariance. (Actually, there is no scale invariance in QED either, since the coupling increases with increasing energy due to quantum corrections.)

There is no physical meaning to this, since we live in a world with $3+1$ (extended) dimensions.

Answer 2

Comment to the question (v6): It seems that OP is basically seeing the effect that Gauss's law forces Coulomb's law to be

$$\tag{1} F~=~ k_e \frac{Q_1Q_2}{r^{D-1}} $$

in $D$ spatial dimensions. If we choose Lorentz–Heaviside/CGS/Gaussian units with $c=1=\hbar$, then Coulomb's constant $k_e$ becomes dimensionless. It then follows from Coulomb's law (1) that the dimension of charge is

$$\tag{2} [Q] = [E]^{\frac{3-D}{2}}$$

when expressed in dimension $[E]$ of energy.