# What is the Planck scale magnetic field strength?

Using the constants $\mu_0$ (or $\varepsilon_0$), $c$, $\hbar$, $e$ and $G$, it is possible to define two quantities with units of magnetic field : \begin{align} B_1 &= \sqrt{\frac{\mu_0 c^7}{\hbar G^2}} \equiv \sqrt{\frac{c^5}{\varepsilon_0 \hbar G^2}} \approx 8 \times 10^{53} \, \mathrm{T}, \tag{1} \\[12pt] B_2 &= \frac{c^3}{G e} \approx 3 \times 10^{54} \, \mathrm{T}. \tag{2} \end{align} Which one is really the Planck magnetic field?

While $B_2$ is simpler, I suspect it should be $B_1$, because it doesn't use the electric charge unit. $e$ is not exactly as universal as $\mu_0$. $B_1$ uses the Planck constant, so it's consistent to call it a Planck "unit", while $B_2$ doesn't use that constant. Also, because of the square root, $B_1$ is a bit more of the same shape as the Planck length : \begin{equation}\tag{3} L_{P} \equiv \sqrt{\frac{\hbar G}{c^3}}. \end{equation} The Planck units are presented on wikipedia: https://en.wikipedia.org/wiki/Planck_units but it doesn't tell anything about the magnetic field.

We could also argue that $B_1$ is the answer because we can find it by equating the magnetic field energy density with the Planck density (dropping all the dimensionless constants) : \begin{equation} \frac{B_1^2}{2 \mu_0} = \frac{M_P c^2}{L_P^3}. \end{equation} But then, we could also find $B_2$ by equating the Planck cyclotron angular frequency with the Planck energy : \begin{equation} \hbar \omega_{\text{cyclotron}} \equiv \hbar \, \frac{e B_2}{2 M_P} = M_P c^2. \end{equation} Both methods are arbitrary.

So what is the Planck magnetic strength?

• The answer is neither; you can put in as many powers of $\alpha$ or $\alpha^{-1}$ as you want. In general, we don't allow powers of $\alpha$ in the other Planck units, because that brings specific information about electromagnetism into play, complicating things for no reason; why should it be related to quantum gravity? But this particular Planck unit you would like to form is explicitly related to electromagnetism. So we have to admit $\alpha$, and now it's totally arbitrary which power of it to choose. (That's also why lists of Planck units typically don't have a Planck magnetic field.) – knzhou Aug 31 at 7:30
• @knzhou, I agree with the $\alpha$ thing, but the "unit" should use the simplest expression possible. Since there doesn't seem to be a unique unit of magnetic field, it may be an indication that there is no theoretical maximal value in Nature. Magnetic fields could be as intense as we wish. – Cham Aug 31 at 12:21

Planck units are found simply by multiplying together powers of certain constants; one does not consider specific physical laws to get them, which is equivalent to motivating specific multiplicative constants. (We don't do it this way because setting each Planck unit to $1$, the ultimate goal of having Planck units, would be impossible on a law-based approach.)
Coulomb's constant $k_C=\frac{1}{4\pi\varepsilon_0}=\frac{\mu_0 c^2}{4\pi}$, which appears in an inverse-square law the same way $G$ does, is a Planck unit just like $G$. Thus in Planck units $\frac{\mu_0}{4\pi}=1$, so the Planck unit you want is $\frac{B_1}{\sqrt{4\pi}}$. It definitely isn't $B_2=\frac{c^3}{G\sqrt{\alpha}q_P}$, with Planck charge $q_P=\sqrt{4\pi\varepsilon_0 c\hbar}=\sqrt{\frac{c\hbar}{k_C}}$.
• Why favor the so called "Planck charge" $\sqrt{4 \pi \varepsilon_0 \hbar c}$, which is arbitrary, instead of $\sqrt{\hbar / \mu_0 c}$ or even $e$ ? In the case of electrodynamics, it feels so arbitrary, and ambiguous since $\varepsilon_0 \, \mu_0 \, c^2 \equiv 1$. – Cham Jul 29 '17 at 16:16
• @Cham The aim is to use $G$, its electrostatic counterpart $k_C$ and $c,\,\hbar,\,k_B$. Using $e$ makes no sense because, in Planck units, $e^2$ is the fine structure constant, an important dimensionless coupling parameter far from unity. – J.G. Jul 29 '17 at 16:23
• $k_C$ is from a classical macroscopic law. It's as arbitrary as using $\varepsilon_0$ or $\mu_0$ alone. And $e^2$ isn't the fine structure. It's $\alpha = k_C e^2 / \hbar c$. At the end of the wikipedia page, they do say that some authors are using $k_C$, and others are using other choices. – Cham Jul 29 '17 at 17:13
• @Cham I don't think you understand the way Planck units define an equivalence class. Since $\alpha/e^2=k_C/\hbar c$ is a Planck unit, it's "1 in Planck units". You can think of either $\alpha$ or $e$ as the parameter, but we don't use dimensionful parameters as Planck units. – J.G. Jul 30 '17 at 6:37