The Born-Infeld model of nonlinear electrodynamics is described by the following Lagrangian $^1$

$${\cal {L_{\rm {BI}}}} = 4b^2\left( 1-\sqrt {1+\frac {F}{2b^2} } \right),$$

where $b$ is a maximum field strength and $F=F_{\mu\nu} F^{\mu\nu} = 2 \left(c^2\bf{B}^2 - \bf{E}^2\right)$. It was introduced in order to remove the divergence of the electron's self-energy in classical electrodynamics (Maxwell's theory) by introducing an upper bound of the electric field at the origin. After that in superstring theory, it was discovered that the effective Born-Infeld action describes the dynamics of D-branes and it is appeared the leading term in the low-energy effective action of the open string theory $^{2,3}$. Perhaps more importantly, as Heisenberg and Euler showed, Maxwell equations even in the vacuum have to be replaced by a nonlinear theory of electrodynamics in order to explain the vacuum polarization effects$^4$ and it is claimed that BI theory has also the same property.

But, how does this model classically explain the vacuum polarization effect? How does it relate to the results of QED in terms of the electric and magnetic permeability tensors of the vacuum?

$^1$ M. Born and L. Infeld, Foundations of the new field theory, Proc. Roy. Soc. Lond. A 144 (1934) 425

$^2$ E.S. Fradkin and A.A. Tseytlin, Non-linear electrodynamics from quantized strings, Phys. Lett. B 163 (1985) 123

$^3$ B. Zwiebach, A rst course in string theory, Cambridge university press (2004)

$^4$ Heisenberg, W., and H. Euler. "Consequences of Dirac theory of the positron." arXiv preprint physics/0605038 (2006)

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    $\begingroup$ I intended to ask this question today but finally I think I found the solution. Perhaps useful for some members. $\endgroup$
    – SG8
    Jun 10, 2021 at 17:37

1 Answer 1


The B-I Lagrangian is a possible UV completion of the Euler-Heisenberg Lagrangian, an effective theory of photons. You can expand the B-I Lagrangian for $F\ll b$ and obtain $$\mathcal{L}_\text{BI}=-b^2\left(\sqrt{-\det\left(g^{\mu\nu}+\frac{F^{\mu\nu}}{b}\right)}-1\right)\sim-\frac{1}{4}F^2-\frac{1}{32b^2}\left((F^2)^2+(\tilde{F}F)^2\right)$$ that can then be matched to the E-H Lagrangian.
The vacuum polarization effect isn't directly explained by the B-I Lagrangian. The reasoning is that the E-H Lagrangian (which we know to be an effective theory) can be used to explain vacuum polarization: one can expand its EoMs around a constant background and find that the fields $a_\mu=F_{\mu\nu}k^\nu$ and $\tilde{a}_\mu=\tilde{F}_{\mu\nu}k^\nu$ propagate independently, each with its own (vacuum) refractive index. One can then study their difference, which in a background of only magnetic fields is $$\Delta n\sim(c_2-c_1)\frac{B_\text{bg}^2}{\Lambda^4}$$ where $c_i/\Lambda^4$ are the coefficients of the E-H Lagrangian and $B_\text{bg}$ is the background magnetic field. Different UV completion of the E-H theory have different $c_i$, and one can measure their difference (and then compare them to the result of experiments). If the B-I theory is the correct UV-completion of the E-H, then one shouldn't observe vacuum polarization, because its $c_i$s are identical.

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    $\begingroup$ What does E-H stand for? $\endgroup$
    – Kosm
    Jun 29, 2022 at 18:01
  • $\begingroup$ @Kosm for Euler-Heisenberg. Thanks for the comment, I'll specify it in the answer. $\endgroup$ Jun 29, 2022 at 18:20

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