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I'm looking for all (or most) theoretical semi-classical derivations of the maximal magnetic field intensity that there may be in the Universe. As an example, this paper evaluate a maximal field of about $3 \times 10^{12} \, \mathrm{teslas}$ :

https://arxiv.org/abs/1511.06679

but its calculations are buggy, especially on the first page, since there's a relativistic $\gamma$ factor missing in the first few formulae. When we take into account the $\gamma$ factor, it destroys the derivation of a maximal value for $B$ ! So this paper isn't satisfaying at all.

Please, I need semi-classical calculations with some rigor, using fully special relativity and some quantum bits only, without a full blown quantum electrodynamics calculation with Feynmann diagrams !

You may use the Heisenberg uncertainty principle, the quantification of angular momentum, Einstein/deBroglie relations, and even things about "vacuum polarization" or some other quantum tricks, but they all need to be properly justified to make the calculations convincing !

From the previous paper as a (buggy) example, is it possible to derive a theoretical maximal $B_{\text{max}}$ using classical electrodynamics and special relativity only (without quantum mechanics) ? I don't think it's possible, without at least general relativity which suggest that the field energy density cannot be larger than a certain limit without creating a black hole.


EDIT : Curiously, even in general relativity, it appears that there's no theoretical limit to the magnetic field strenght. The Melvin magnetic universe is an analytical solution to the Einstein field equation that is as close as possible to an uniform magnetic field. See for example this interesting paper from Kip Thorne :

http://authors.library.caltech.edu/5184/1/THOpr65a.pdf

The spacetime metric of Melvin's magnetic universe doesn't have any singularity, wathever the value of the field parameter, and there can be no gravitational collapse of the field under perturbations ! So apparently there's no maximal value of $B$ in classical general relativity, without matter!

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There is no maximum.

Consider a particle moving relativistically in a straight line. It can be derived from the Lienard-Wiechert potentials (full derivation in Griffiths 10.3 or at http://www.feynmanlectures.caltech.edu/II_26.html) that the electric field $\mathbf{E}(\mathbf{r},t)$ emitted by such a particle moving at velocity $\mathbf{v}$ is given by

$$\mathbf{E}(\mathbf{r},t)=\frac{q}{4\pi\epsilon_0}\frac{1-v^2/c^2}{(1-(v^2/c^2)\sin^2{\theta})^{3/2}}\frac{\mathbf{\hat{R}}}{R^2}$$

where $\mathbf{R}=\mathbf{r}-\mathbf{v}t$, and $\theta$ is the angle between $\mathbf{R}$ and $\mathbf{v}$. Suppose we stand to the side of this path and observe the electric field as the charge passes closest to us. Then $\theta=\pi/2$, so we substitute:

$$\mathbf{E}(\mathbf{r},t)=\frac{q}{4\pi\epsilon_0}\frac{1}{\sqrt{1-v^2/c^2}}\frac{\mathbf{\hat{R}}}{R^2}=\frac{q}{4\pi\epsilon_0}\gamma\frac{\mathbf{\hat{R}}}{R^2}$$

where $\gamma$ is the usual Lorentz factor. As the particle's velocity approaches $c$, $\gamma$ approaches infinity, so the electric field we observe goes to infinity without limit, even at a finite distance from the charge.

Similarly, the magnetic field around a charge relativistically moving in a straight line is given by

$$\mathbf{B}(\mathbf{r},t)=\frac{1}{c^2}(\mathbf{v}\times\mathbf{E})$$

and since the electric field goes to infinity without limit, the magnetic field must also go to infinity.

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  • $\begingroup$ This is classical only, and don't take quantum mechanics into account. $\endgroup$ – Cham Jul 25 '17 at 18:44
  • $\begingroup$ I suspect a flaw in this reasoning. It is uncomplete. If accelerated, the particle would emit radiation and lose energy. Classical electrodynamics also predicts radiation, which limits the magnetic field that could be produced and sustained. Also, other particles moving in your magnetic field would produce a counter magnetic field, and thus reduce it. $\endgroup$ – Cham Jul 25 '17 at 20:31
  • $\begingroup$ @Cham The particle is not accelerating. It's moving in a straight line with constant velocity. $\endgroup$ – probably_someone Jul 25 '17 at 20:34
  • $\begingroup$ Then you could reduce and even cancel your magnetic field by a simple change of reference frame ! In your case, you have $E^2 - B^2 > 0$ and $\vec{E} \cdot \vec{B} = 0$ (invariants). Your EM field is electric-like. Not a pure magnetic field. $\endgroup$ – Cham Jul 25 '17 at 20:37
  • $\begingroup$ @Cham Yes, you could. That's the whole point of relativity - magnetic fields are entirely a relativistic effect. $\endgroup$ – probably_someone Jul 25 '17 at 20:38
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Dimensional analysis suggest a few solutions :

We can use the following parameters : \begin{align}\tag{1} \mu_0 &\sim \frac{\mathrm{M} \mathrm{L}}{\mathrm{Q}^2}, &e &\sim \mathrm{Q}, &m_{e} &\sim M, &c &\sim \frac{\mathrm{L}}{\mathrm{T}}, &G &\sim \frac{\mathrm{L}^3}{\mathrm{M}\mathrm{T}^2}, &\hbar &\sim \frac{\mathrm{M}\mathrm{L}^2}{\mathrm{T}}. \end{align} Since the magnetic field strenght has dimensions $B \sim \mathrm{M}/\mathrm{Q}\mathrm{T}$, we can look out a function of the following form (rejecting all pure numbers). Without quantum mechanics: \begin{equation}\tag{2} B(\mu_0, e, m_{é}, c) = \mu_0^{\alpha} \: m_{e}^{\beta} \: c^{\gamma} \: e^{\delta}. \end{equation} The solution is unique : $\alpha = -1$, $\beta = 2$, $\gamma = 1$, $\delta = -3$, so \begin{equation}\tag{3} B_1(\mu_0, e, m_{é}, c) \approx \frac{m_e^2 \, c}{\mu_0 \, e^3} \approx 5 \times 10^{10} \, \mathrm{T}. \end{equation} This solution is interesting ! It corresponds to a magnetic field energy density $B_1^2/2\mu_0$ equivalent to $m_e c^2$ in a spherical volume of radius $e^2 / 4 \pi \varepsilon_0 m_e c^2$ (the electron's classical radius).

If we add the Planck constant : $\hbar \sim \mathrm{M} \mathrm{L}^2 / \mathrm{T}$, then there's an infinity of solution from dimensional analysis.

If we use the set $\mu_0$, $c$, $e$ and $G$ (without $m_e$, which is weird), then there is a unique Planck-like solution : \begin{equation}\tag{4} B_2(\mu_0, e, c, G) \approx \frac{c^3}{G e} \approx 2.5 \times 10^{54} \, \mathrm{T}. \end{equation} Using the set $\mu_0$, $c$, $m_e$ and $G$ is weirder without the electron's charge. There's also only one solution in this case : \begin{equation}\tag{5} B_3(\mu_0, m_e, c, G) \approx \frac{c^4}{m_e} \, \sqrt{\frac{\mu_0}{G^3}} \approx 2 \times 10^{76} \, \mathrm{T}. \end{equation}

If we use the set $\mu_0$, $e$, $m_e$ and $\hbar$ (without relativity), then there is a another interesting solution : \begin{equation}\tag{6} B_4(\mu_0, e, m_e, \hbar) \approx \frac{\hbar \, m_e^2}{\mu_0^2 \, e^5} \approx 5 \times 10^{11} \, \mathrm{T}. \end{equation} The ratio of (3) on (6) is the fine structure constant $\alpha \equiv e^2/4\pi \varepsilon_0 \hbar c \approx 1/137$ (up to a $4 \pi$ factor): \begin{equation}\tag{7} B_1 \equiv B_4 \, 4 \pi \alpha. \end{equation} So this naturally brings a question : using the full set $\mu_0$, $e$, $m_e$, $c$, $\hbar$ (without gravity), are all the solutions of the following form ? \begin{equation}\tag{8} B_5 \approx \frac{m_e^2\, c}{\mu_0 \, e^3} \, f(\alpha) = B_1 \, f(\alpha), \end{equation} where $f(\alpha)$ is an arbitrary function of the fine structure constant ? I think the answer is yes ! Notice that if $f(\alpha) \approx 4 \pi \alpha$, then \begin{equation}\tag{9} B_5(e, m_e, c, \hbar) \approx \frac{m_e^2 \, c^2}{e \hbar} \sim 4 \times 10^{9} \, \mathrm{T}, \end{equation} This is the Schwinger's limit : https://en.wikipedia.org/wiki/Schwinger_limit It can be found by equating the cyclotron energy to the rest mass energy $\hbar \omega_e = m_e c^2$.

Also, I think there should not be any maximal magnetic field in the classical limit, i.e $B_{\text{max}} \rightarrow \infty$ when $c \rightarrow \infty$ or $\hbar \rightarrow 0$. Expression (8) fullfill this constraint while (6) should be rejected because of $\hbar$ at the numerator, but there's an ambiguity with $\alpha$, depending if we consider $\mu_0$ or $\varepsilon_0$ as independant of $c$ : \begin{equation}\tag{10} \alpha = \frac{e^2}{4 \pi \varepsilon_0 \hbar c} = \frac{\mu_0 \, e^2 \, c}{4 \pi \hbar}. \end{equation}

Finally, using the classical set $\mu_0$, $m_e$, $e$ and $G$ without relativity and quantum mechanics gives a puzzling value : \begin{equation}\tag{11} B_6(\mu_0, e, m_e, G) = \frac{m_e^3}{e^4} \, \sqrt{\frac{G}{\mu_0^3}} \sim 7 \times 10^{-12} \, \mathrm{T}. \end{equation} This tiny magnetic field can be found by equating the magnetic field energy density with the gravity's pressure inside a sphere of radius $e^2/4\pi \varepsilon_0 m c^2$.

Any comments on this ?

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    $\begingroup$ Dimensional analysis is good for determining the proportionality relations between different quantities, but it still leaves a constant of proportionality undetermined. This constant of proportionality could be arbitrarily big or small, so unless you assume this constant is of order 1 (and there's no good reason to do so), dimensional analysis tells you absolutely nothing here. $\endgroup$ – probably_someone Jul 25 '17 at 20:17
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    $\begingroup$ I partially agree. The unknown pure number cannot be that big or small, if it's not exactly 0. If there's really a limit imposed by the theory, the pure number should not be very different than 1 (by one or 2 orders of magnitude, say). It would be extremelly strange to get a pure number from the theory like $10^{-7}$. $\endgroup$ – Cham Jul 25 '17 at 20:24
  • $\begingroup$ Why would it be strange? In other areas of physics, the proportionality constants are extremely far from being of order 1. The proportionality constant between energy and frequency is of order $10^{-34}$, for example. Coulomb's constant is $10^9$, and Newton's $G$ is $10^{-11}$. $\endgroup$ – probably_someone Jul 25 '17 at 20:30
  • $\begingroup$ These are NOT pure numbers ! They have units ! $\endgroup$ – Cham Jul 25 '17 at 20:36

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