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?

  • 1
    $\begingroup$ 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.) $\endgroup$ – knzhou Aug 31 '19 at 7:30
  • $\begingroup$ @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. $\endgroup$ – Cham Aug 31 '19 at 12:21
  • $\begingroup$ @knzhou, isn't your comment an answer? Please take a look at David Z's answer here: Answering in comments $\endgroup$ – Alfred Centauri May 14 '20 at 18:45

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 $q_P=\sqrt{4\pi\varepsilon_0 c\hbar}=\sqrt{\frac{c\hbar}{k_C}}$.

  • $\begingroup$ 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 $\varepsilon_0 \, \mu_0 \, c^2 \equiv 1$. $\endgroup$ – Cham Jul 29 '17 at 16:16
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    $\begingroup$ @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. $\endgroup$ – J.G. Jul 29 '17 at 16:23
  • $\begingroup$ $k_C$ 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. $\endgroup$ – Cham Jul 29 '17 at 17:13
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    $\begingroup$ @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. $\endgroup$ – J.G. Jul 30 '17 at 6:37
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    $\begingroup$ See the last paragraph of the History section on the Wikipedia page : "Planck did not adopt any electromagnetic units. However, since the non-rationalized gravitational constant, G, is set to 1, a natural extension of Planck units to a unit of electric charge is to also set the non-rationalized Coulomb constant, ke, to 1 as well. Another convention is to use the elementary charge as the basic unit of electric charge in Planck system. Such system is convenient for black hole physics. The two conventions for unit charge differ by a factor of the square root of the fine-structure constant." $\endgroup$ – Cham Jul 30 '17 at 12:55

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