1 Erratum to: Eur. Phys. J. C (2019) 79:550 https://doi.org/10.1140/epjc/s10052-019-7068-x

In this Erratum we correct some numerical coefficients for the low-energy effective action we obtained in our paper [1].

  1. 1.

    The Eq. (37) must be read as:

    $$\begin{aligned} \mathcal{L}_\mathrm {eff}= & {} \frac{g_t^2}{2}A_0A^0 -\frac{{{\tilde{\alpha }}}_2}{2} \left( \partial _0A_i\partial ^0A^i-\frac{6}{5}\partial _0A_i\partial ^iA^0\nonumber \right. \\&\left. -\frac{6}{5}\partial _iA_0\partial ^0A^i+\frac{36}{25}\partial _iA_0\partial ^iA^0 \right) \nonumber \\&-\frac{{{\tilde{\alpha }}}_5m^4}{2}(\partial _iA_j\partial ^iA^j-\partial _iA_j\partial ^jA^i)\nonumber \\&-\frac{9{{\tilde{\alpha }}}_2}{50}\partial _iA_0\partial ^iA^0-\frac{{{\tilde{\beta }}} m^4e^2}{4}\left( A_iA^i-\frac{a_ia^i}{e^2}\right) ^2,\nonumber \\ \end{aligned}$$
    (37)
  2. 2.

    The Eqs. (41–44) must be read as:

    $$\begin{aligned} A_0\rightarrow & {} \frac{5m{{\tilde{\alpha }}}_5^{1/4}}{6{{\tilde{\alpha }}}_2^{1/2}}A_0, \end{aligned}$$
    (41a)
    $$\begin{aligned} A_i\rightarrow & {} \frac{1}{m{{\tilde{\alpha }}}_5^{1/4}}A_i, \end{aligned}$$
    (41b)
    $$\begin{aligned} \partial _0\rightarrow & {} \frac{m^2{{\tilde{\alpha }}}_5^{1/4}}{{{\tilde{\alpha }}}_2^{1/2}}\partial _0, \end{aligned}$$
    (41c)
    $$\begin{aligned} \partial _i\rightarrow & {} \frac{1}{{{\tilde{\alpha }}}_5^{1/4}}\partial _i, \end{aligned}$$
    (41d)
$$\begin{aligned} \mathcal{L}_\mathrm {eff}= & {} -\frac{m^2}{4}F_{\mu \nu }F^{\mu \nu } -\frac{m^2}{8}(\partial _iA_0)^2 +\frac{g_t^2}{2}\frac{25m^2{{\tilde{\alpha }}}_5^{1/2}}{36{{\tilde{\alpha }}}_2}A_0A^0\nonumber \\&-\frac{{{\tilde{\beta }}} m^2e^2}{4{{\tilde{\alpha }}}_5}\left( A_iA^i-\frac{a_ia^i}{e^2}\right) ^2. \end{aligned}$$
(42)
$$\begin{aligned} \mathcal{L}_\mathrm {eff}= & {} -\frac{m^2}{4}F_{\mu \nu }F^{\mu \nu } -\frac{{{\tilde{\beta }}} m^2e^2}{4{{\tilde{\alpha }}}_5}\left( A_i A^i-\frac{a_i a^i}{e^2}\right) ^2, \end{aligned}$$
(43)
$$\begin{aligned} \mathcal{L}_{\mathrm {eff}}= & {} -\frac{m^2}{2}F_{0i}F^{0i}+\frac{m^2}{4}F_{ij}F^{ij}\nonumber \\&-\frac{{{\tilde{\beta }}} m^2e^2}{4|{{\tilde{\alpha }}}_5|}\left( A_i A^i-\frac{a_i a^i}{e^2}\right) ^2. \end{aligned}$$
(44)

None of physical conclusions of our paper are changed under these corrections.