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Oct 11, 2023 at 16:39 comment added NinjaDarth Actually, the $-\frac{1}{4μ_0}$ is $-¼ε_0c\sqrt{|g|}$. There is a $\sqrt{|g|}$ factor there, and for all the other terms in the action integral. In addition, the dimensions for the components $F_{μν}$, $g_{μν}$, $∂_μ$, $d^4x$, etc. depend on what type of coordinates you're using (e.g. $[∂_μ] = 1$ if $x^μ$ is an angular coordinate). In particular, $d^4x$ can have all sorts of different dimensions (e.g. $TL$ for spherical coordinates and $x^0 = t$). Instead, it's $\sqrt{|g|}d^4x$ that has the consistent dimension ($L^4$), where L denotes length, T time.
Apr 3, 2023 at 21:28 comment added Ghoster I think you can do that, but I don’t recall it being done that way in textbooks.
Apr 3, 2023 at 6:42 comment added Gere Yes, you would replace $F'=eF$ accordingly. That leaves you with $\mu_0e^2$ which depends only on a dimensionless $\alpha$ (and $\hbar=c=1$). Hence, you have removed all constants and are left with only $\alpha$. Effectively you are using different units for $A$ / charge and hence the space of solutions is unaffected. If you don't want $\alpha$ in front of $F^2$, you could rescale $A$ again to have $\sqrt{\alpha}$ in front of $A$. But am I right, that you can simply remove all constants and be left with dimensionless $\alpha$ only?
Apr 2, 2023 at 20:36 comment added Ghoster I don’t think so. If you scale $A$, you have to consider what happens to $F$. And what would be the point of suppressing the quantity that perturbation theory expands in powers of?
Apr 2, 2023 at 17:27 comment added Gere Can I absorb $e$ into $A$ like $A'=eA$ and replace $\mu_0$ by $\alpha$ to get rid of $e$ completely?
Mar 19, 2023 at 17:32 comment added Ghoster And then $e$ isn't 1 anymore? It isn’t 1 in any of these four unit systems. If it were, you wouldn’t see it written in $D_\mu$.
Mar 19, 2023 at 17:29 comment added Ghoster My recommendation is that, now that you see that QED can be done in any unit system, forget about all of them except the simple particle physics units, where everything is measured in terms of energy.
Mar 19, 2023 at 17:26 comment added Ghoster This makes Coulomb’s Law be $F=q_1q_2/r^2$, so there is no silly $1/4\pi\epsilon_0$ as in SI. Similarly, there is no need for a $\mu_0$ in the force between two currents.
Mar 19, 2023 at 17:24 comment added Ghoster This article explains how charge can be defined in terms of mass, length, and time. A unit of charge defined this way is such that if you put it on each of two point particles separated by one unit of distance you get a repulsion of one unit of force.
Mar 19, 2023 at 17:16 comment added Ghoster Do you know which book would mention $\mu_0$? Wikipedia’s article on the covariant formulation of classical electromagnetism has it in $\mathscr L$.
Mar 19, 2023 at 8:07 comment added Gere I don't understand how $\mu_0$ can get dropped in other units though. For that you would need to fix units such that $\varepsilon_0=1$? And then $e$ isn't 1 anymore? Maybe I'm lost. Do you know a reference discussing how to make $\mu_0=1$?
Mar 19, 2023 at 8:03 comment added Gere Wow, thanks! That's more extensive than I wished for. My main stumbling block was $\mu_0$ which did not appear in the references I found. I don't have all books at hand. Do you know which book would mention $\mu_0$?
Mar 19, 2023 at 8:01 vote accept Gere
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