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\usepackage {graphicx}
\newcommand{\qua}{(\gamma^{m}D_{m})^{2}}
\newcommand{\dc}{\gamma^{m}D_{m}}
\newcommand{\eff}{E_{eff(2+1)}}
\newcommand{\effp}{E_{eff(3+1)}}
\newcommand{\del}{\delta_{l,s}(k,r)}
\newcommand{\opa}{\frac{1}{2}tr\ln(\not\!\! D^{2}+m^{2}_{f})}

\begin{document}


\title{The strong magnetic field asymptotic behaviour
 for the fermion-induced  effective energy in the presence of a magnetic flux tube}
\author{Pavlos Pasipoularides  \footnote{paul@central.ntua.gr} \\
       Department of Physics, National Technical University of
       Athens \\ Zografou Campus, 157 80 Athens, Greece}
\date{ }
       \maketitle


\begin{abstract}
The strong magnetic field behaviour of the fermion-induced
effective energy in the presence of an inhomogeneous magnetic
field, has recently attract attention, see Refs.
\cite{1,2,3,5,2str}. In Ref. \cite{3}, we presented an asymptotic
formula for the fermion-induced effective energy in 3+1 dimensions
in the presence of a cylindrically symmetric inhomogeneous strong
magnetic field of finite magnetic flux $\Phi$. However, the proof
of the formula was presented in a condensed form and there are
some points which were not clearly explained. The aim of this work
is to present the complete proof in detail.
\end{abstract}

\section{Introduction}

Many authors have dealt with the numerical study of the
fermion-induced effective energy, in the presence of magnetic
fields of the form of a flux tube, see Refs. \cite{1,2,3,4,5}.

A similar topic, is the study of the gluon-induced effective
energy in the presence of a colored magnetic flux tube
\cite{9,10}. However, in this case there is an essential
difference: the effective energy has nonzero imaginary part and so
the system is unstable.

It is worth to mention the recently developed method in Refs.
\cite{4,6,7}, which is applicable even for magnetic fields which
do not exhibit a special kind of symmetry. For example, this
method can be applied to a system of two separated magnetic flux
tubes.

In Ref. \cite{3} we presented an asymptotic formula (Eq. (49)) for
the effective energy when the characteristic magnetic field
strength $B_{m}=\int\vec{B}(\vec{x})\cdot d\vec{S} /\pi d^{2}$ (or
the magnetic flux $\Phi$) tends to infinity and the spatial size
$d$ of the magnetic flux tube is kept fixed. However, the proof of
the formula was presented in a condensed form and there are some
points which were not clearly explained. The aim of this work is
to present the complete proof in detail.

In addition, in Sec. 4 we check if the derivative expansion
\cite{der4,der5} gives the correct asymptotic behaviour for the
effective energy (Eq. (23) below), when $B_{m}\rightarrow +\infty$
(the mass of the fermions $m_{f}$ and the range of the magnetic
flux tube $d$ are kept fixed).

\section{Effective energy in the presence of a magnetic flux tube}

The 3+1 dimensional fermion-induced effective energy in the
presence of a magnetic field is given by the equation:
\begin{eqnarray}
E_{eff(3+1)}=-\frac{1}{2iT}Tr\ln(\not\!\!D^{2}+m^{2}_{f})
\end{eqnarray}
where  $\not\!\!\!D=\gamma^{\mu}D_{\mu}$ $(\mu=0,1,2,3)$ and
$D_{\mu}=\partial_{\mu}-ieA_{\mu}$. The gamma matrices satisfy the
relationship $\{\gamma^{\mu},\gamma^{\nu}\}=2g^{\mu\nu}$ and $T$
is the total length of time.

We have shown in Ref. \cite{3} that a magnetic field with strength
independent of $z$ coordinate, directed towards $z$-axis, has
renormalized effective energy equal to \footnote{For the case of
massless fermions ($m_{f}=0$) the renormalized effective energy is
given in appendix A of Ref. \cite{3}.}
\begin{eqnarray}
E_{eff(3+1)}^{(ren)}=\frac{1}{4\pi}\sum_{\{n\}}(E_{\{n\}}+m_{f}^{2})\ln(\frac{E_{\{n\}}+m_{f}^{2}}{m_{f}^{2}})-\frac{1}{4\pi}\sum_{\{n\}}E_{\{n\}}
\end{eqnarray}
where $E_{\{n\}}$ are the eigenvalues of the planar operator
$(\gamma^{m}D_{m})^{2}$ $(m=1,2)$, and $\{n\}$ is a set of quantum
numbers. Note that $E_{\{n\}}\geq 0$ since the operator
$(\gamma^{m}D_{m})^{2}$ is positive definite as the square of a
hermitian operator. For the sake of simplicity we will drop the
index $(ren)$ in the rest of this paper.

If the magnetic field is cylindrically symmetric with finite
magnetic flux, the spectrum of the operator
$(\gamma^{m}D_{m})^{2}$ is continuous  \footnote{Besides
continuous spectrum, zero modes may exist according to the
Aharonov-Casher theorem \cite{12}.}. Thus we can set
$E_{\{n\}}=k^{2}$.

In order to perform the summation over $\{n\}$ that appears in Eq.
(2), we need a density of states, given by the equation:
\begin{equation}
\rho(k)=\sum_{l,s}\left(\rho_{l,s}^{(free)}(k)+\frac{1}{\pi}\frac{d\delta_{l,s}(k)}{dk}\right)=\sum_{l,s}\rho_{l,s}^{(free)}(k)+\frac{1}{\pi}\frac{d\Delta(k)}{dk}
\end{equation}
where $\sum_{l,s}\rho_{l,s}^{(free)}(k)$ is the density of states
for free space, and $\delta_{l,s}(k)$ is the phase shift which
corresponds to $l^{th}$ partial wave with momentum $k$ and spin
$s$. The function $\Delta(k)$ is defined by the equation:
\begin{equation}
\Delta(k)=\lim_{L\rightarrow+\infty}\sum_{s,l=-L}^{L}\delta_{l,s}(k)
\end{equation}
From Eqs. (2) and (3), if we drop the field independent term
$\sum_{l,s}\rho_{l,s}^{(free)}(k)$ and integrate by parts, we
obtain
\begin{equation}
 E_{eff(3+1)}=-\frac{1}{2\pi^{2}}\int_{0}^{+\infty}k\ln\left(\frac{k^{2}+m^{2}_{f}}{m_{f}^{2}}\right) (\Delta(k)-c) dk
\end{equation}
where $c=\lim_{k \rightarrow +\infty}\Delta(k)$. Our numerical
study, in Ref. \cite{3}, shows that $c=-\pi \phi^{2}$ ($\phi=e
\Phi/2 \pi$). This means that $c$ is independent of the special
form of the magnetic field we examine and depends only on the
total magnetic flux of the field $\Phi$. Also, in Ref. \cite{3},
we see that the function $\Delta(k)$, when $k\rightarrow +\infty$,
tends to $c$ fairly rapidly, and thus, the integral over $k$ in
Eq. (5) is convergent.

Eq. (5) is suitable for the numerical computation of the 3+1
dimensional fermion-induced effective energy in the presence of a
magnetic flux tube. The calculation of the phase shifts is
performed by solving an ordinary differential equation. For
details see Refs. \cite{3,phase}.

\section{The asymptotic behaviour for the effective
 energy when $B_{m}\rightarrow +\infty$ and $d$ and $m_{f}$ are kept fixed}

The function $\Delta(k)$ depends on three quantities: $k$,
$B_{m}=2\phi/d^{2}$ and $d$. From these we construct the following
two dimensionless quantities\footnote{We have assumed the
rescaling $B_{m}\rightarrow eB_{m}$.}: $k B_{m}^{-1/2}$ and
$B_{m}d^{2}$ (=$2\phi$). For dimensional reasons the function
$\Delta(k)$ can be put into the form
$\Delta(k)=G(kB_{m}^{-\frac{1}{2}},B_{m} d^{2})$. Setting
$\Delta(k)=G(kB_{m}^{-\frac{1}{2}},B_{m} d^{2})$ and making the
change of variable $y=kB_{m}^{-\frac{1}{2}}$ in Eq. (5) we obtain
\begin{eqnarray}
E_{eff(3+1)}&=&-\frac{B_{m}}{2\pi^{2}}\int_{0}^{+\infty}y\ln\left(\frac{y^{2}+m^{2}_{f}/B_{m}}{m^{2}_{f}/B_{m}}\right)\left(G(y,B_{m}
d^{2})-c\right)dy\\
&=&-\frac{B_{m}}{2\pi^{2}}\int_{0}^{+\infty}y\left(\ln(y^{2}+m^{2}_{f}/B_{m})-\ln(m^{2}_{f}/B_{m})\right)\left(G(y,B_{m}
d^{2})-c\right)dy
\end{eqnarray}

In the strong magnetic field case ($B_{m}/m_{f}^{2}>>1$) we
obtain:
\begin{eqnarray}
E_{eff(3+1)}\rightarrow
&-&\frac{B_{m}}{\pi^{2}}\int_{0}^{+\infty}y\ln y\left(G(y,B_{m}
d^{2})-c\right)dy \nonumber \\
&-&\frac{B_{m}}{2\pi^{2}}\ln(B_{m}/m^{2}_{f})\int_{0}^{+\infty}y\left(G(y,B_{m}
d^{2})-c\right)dy
\end{eqnarray}
There are many ways to achieve a large ratio $B_{m}/m_{f}^{2}$ by
changing the three independent variables $B_{m}$, $m_{f}$ and $d$.
However, the most interesting case is when $B_{m}\rightarrow
+\infty$ and $m_{f}$ and $d$ are kept fixed (see also Refs.
\cite{1,2,3,4}). This means that we keep the spatial size of the
magnetic field configuration fixed and we increase the
characteristic magnetic field strength $B_{m}$ or the magnetic
flux $\phi=B_{m}d^{2}/2$.

From Eq. (8) we see that the asymptotic behaviour of the effective
energy when $B_{m}\rightarrow +\infty$ ($m_{f}$ and $d$ are kept
fixed) depends on the asymptotic behaviour of the integrals:
\begin{eqnarray}
I_{1}(B_{m}d^{2})&=&\int_{0}^{+\infty}y\left(G(y,B_{m}
d^{2})-c\right)dy \\ I_{2}(B_{m}d^{2})&= & \int_{0}^{+\infty}y\ln
y\left(G(y,B_{m} d^{2})-c\right)dy
\end{eqnarray}
In the case of the integral of Eq. (9) it was possible to find an
analytical expression:
\begin{equation}
\int_{0}^{+\infty} y(G(y,B_{m}d^{2})-c)dy=\frac{1}{12B_{m}}\int
d^{2}\vec{x} B^{2}(\vec{x})
\end{equation}
In order to give a proof for the above equation we will use Eq.
(13) in Ref. \cite{3}
\begin{equation}
Tr(\gamma^{m}D_{m})^{2}=\sum_{\{n\}}E_{\{n\}}=-\frac{1}{6\pi}\int
d^{2}\vec{x} B^{2}(\vec{x})\;
\end{equation}
By using Eq. (3) for the density of states we obtain
\begin{eqnarray}
\sum_{\{n\}}E_{\{n\}}=\int_{0}^{+\infty}
k^{2}\rho(k)dk=\int_{0}^{+\infty}
k^{2}\left(\sum_{l,s}\rho_{l,s}^{(free)}(k)+\frac{1}{\pi}\frac{d\Delta(k)}{dk}\right)dk
\end{eqnarray}
If we drop the field independent term
$\sum_{l,s}\rho_{l,s}^{(free)}(k)$ and set again
$\Delta(k)=G(kB_{m}^{-\frac{1}{2}},B_{m} d^{2})$ and
$y=kB_{m}^{-1/2}$ we find
\begin{eqnarray}
\sum_{\{n\}}E_{\{n\}}&=&\int_{0}^{+\infty}
k^{2}\frac{1}{\pi}\frac{d\Delta(k)}{dk}dk \nonumber \\
&=&-\frac{2}{\pi}\int_{0}^{+\infty} k(\Delta(k)-c)dk \nonumber \\
&=&-\frac{2}{\pi}B_{m}\int_{0}^{+\infty}
y(G(kB_{m}^{-1/2},B_{m}d^{2})-c)dy
\end{eqnarray}
Eq. (11) is a straightforward result of Eqs. (14) and (12).

For convenience we write
\begin{equation}
B(\vec{x})=B_{m}\tilde{B}(\vec{x}/d)
\end{equation}
where the magnetic field $\tilde{B}(\vec{x})$ has a characteristic
magnetic field strength $\tilde{B}_{m}=1$ and range $d=1$, Eq.
(11) can then be written in the form
\begin{equation}
\int_{0}^{+\infty}
y(G(y,B_{m}d^{2})-c)dy=\frac{B_{m}d^{2}}{12}\int d^{2}\vec{x}
\tilde{B}^{2}(\vec{x})
\end{equation}
From the above relationship it is obvious that the integral
$\int_{0}^{+\infty} y(G(y,B_{m}d^{2})-c)dy$ is proportional to
$B_{m}d^{2}$.

I could not find an analytical expression, like that of Eq. (16),
for the integral $I_{2}(B_{m}d^{2})$ of Eq. (10). However, a
numerical study of this integral shows that its asymptotic
behaviour, for $B_{m}\rightarrow +\infty$ ,is the same with that
of integral $\int_{0}^{+\infty} y(G(y,B_{m}d^{2})-c)dy$,
independently of the particular shape of the magnetic field. Thus
\begin{equation} \int_{0}^{+\infty}y \ln
y \left(G(y,B_{m}d^{2})-c \right) dy \rightarrow c1\: B_{m}d^{2}
\end{equation}
where the constant $c1$ is independent of $B_{m}d^{2}$, and its
value depends only on the special form of the magnetic field
configuration.
\begin{figure}[h]
\begin{center}
\includegraphics[scale=0.5,angle=-90]{grstrong.ps}
\end{center}
\caption { $-(1/B_{m}d^{2}) \int_{0}^{\infty}y \ln y
(G(y,B_{m}d^{2})-c) dy $ as a function of $B_{m}d^{2}$ for the
magnetic fields $B_{1}$, $B_{2}$, $B_{3}$, $B_{4}$.} \label{1}
\end{figure}

In Fig. \ref{1} we present our numerical calculations for four
magnetic field configurations with quite different kinds of
inhomogeneity.
\begin{eqnarray}
B_{1}(r)&=&B_{m}\: \frac{1}{(r^{2}/d^{2}+1)^{2}}\\
B_{2}(r)&=&B_{m}\: \exp(-\frac{r^{2}}{d^{2}})\\ B_{3}(r)&=&B_{m}\:
\theta(d-r) \\ B_{4}(r)&=&B_{m}\:3(1-\frac{r}{d})\: \theta(d-r)
\end{eqnarray}
We expect that our numerical work, presented in Fig. \ref{1},
convinces the reader that the asymptotic behaviour of
$I_{2}(B_{m}d^{2})$, as given by Eq. (17), does not depend on the
kind of inhomogeneity of the magnetic flux tube.

Thus, when $B_{m}\rightarrow +\infty$ and $m_{f}$ and $d$ are kept
fixed, from Eqs. (8),(16) and (17) we obtain
\begin{eqnarray}
E_{eff(3+1)}\rightarrow
-\frac{c1\:B_{m}^{2}d^{2}}{\pi^{2}}-\frac{B_{m}^{2}d^{2}}{24\pi^{2}}\ln(B_{m}/m^{2}_{f})\int
d^{2}\vec{x} \tilde{B}^{2}(\vec{x})
\end{eqnarray}
We see that the logarithmic term dominates, so we can write
\begin{eqnarray}
E_{eff(3+1)}^{asympt}=-\frac{B_{m}^{2}d^{2}}{24\pi^{2}}\ln(B_{m}/m^{2}_{f})\int
d^{2}\vec{x} \tilde{B}^{2}(\vec{x})
\end{eqnarray}
If, instead of the characteristic magnetic field strength
$B_{m}=\Phi/\pi d^{2}$ we use the parameter $\phi=e\Phi/2 \pi$ we
see that the fermion-induced effective energy, for $\phi
\rightarrow +\infty$, is proportional to $-\phi^{2}\ln\phi$.
\begin{eqnarray}
E_{eff(3+1)}^{asympt} \sim -\phi^{2}\ln\phi
\end{eqnarray}
This asymptotic behaviour is true, independently of the particular
shape of the magnetic flux tube, as it was shown in the above
numerical study (See Fig. \ref{1}).

Note that the above asymptotic formula, of Eq. (23), is the
correct and for other cases of large ratio $B_{m}/m_{f}^{2}$. For
example we note two: a) for $m_{f}\rightarrow 0$ when $B_{m}$ and
$d$ are kept fixed and b) for $d\rightarrow 0$ when $m_{f}$ and
$\phi=B_{m} d^{2}/2$ are kept fixed. For these two cases, the
asymptotic formula, of Eq. (23), is obtained straightforwardly
from Eqs. (8) and (16), and thus knowledge of the asymptotic
behaviour of the integral $I_{2}(B_{m}d^{2})$ is not necessary.

\section{Derivative expansion for $B_{m}\rightarrow +\infty$ when $d$ and $m_{f}$ are kept fixed}

An approximate way for the computation of the fermion-induced
effective energy is the derivative expansion \cite{der4,der5}.
This method gives accurate results for smooth magnetic field
configurations. Note that the derivative expansion fails for the
magnetic fields of Eqs. (20) and (21), as the magnetic field of
Eq. (20) is discontinuous and the magnetic field of Eq. (21) has
discontinuous first derivative.

Our aim is to check if the derivative expansion succeeds in giving
the correct asymptotic behaviour for the effective energy in the
strong magnetic field regime ($B_{m}\rightarrow +\infty$, $d$ and
$m_{f}$ are kept fixed).

If we keep the first two terms of the derivative expansion, the
effective energy, in the case of 3+1 dimensions, is given by the
equation
\begin{equation}
E_{eff(3+1)}^{(der)}=E^{(0)}_{(3+1)}[B]+E^{(1)}_{(3+1)}[B,(\partial
B)^{2}]
\end{equation}
where
\begin{eqnarray}
 E_{(3+1)}^{(0)}=\int d^{2}\vec{x}\frac{B^{2}(\vec{x})}{8\pi^{2}}\int_{0}^{+\infty}\frac{1}{s^{2}}(\coth(s)-\frac{1}{s}-\frac{s}{3})
 \: e^{-s \: m_{f}^{2}/B(\vec{x})} ds
\end{eqnarray}
\begin{eqnarray}
E_{(3+1)}^{(1)} =(\frac{1}{8\pi})^{2} \int d^{2}\vec{x}
\frac{(\nabla B)^{2}}{B(\vec{x})}\int_{0}^{+\infty}\frac{1}{s}
 (\frac{d}{ds})^{3}[s\coth(s)] \:e^{-s \:m_{f}^{2}/B(\vec{x})} ds
\end{eqnarray}

For $B_{m}\rightarrow +\infty$ from Eqs. (26) and (27) we obtain
\begin{eqnarray}
 E_{(3+1)}^{(0)}\rightarrow -\frac{B_{m}^{2}d^{2}}{24\pi^{2}}\ln(B_{m}/m^{2}_{f})\int
d^{2}\vec{x} \tilde{B}^{2}(\vec{x})
\end{eqnarray}
\begin{eqnarray}
E_{(3+1)}^{(1)} \rightarrow B_{m}(\frac{1}{8\pi})^{2} 24
\zeta'(-2) \int d^{2}\vec{x} \frac{(\nabla
\tilde{B})^{2}}{\tilde{B}(\vec{x})}
\end{eqnarray}
Eq. (29) can be derived straight from Eq. (27) if we assume that
$B_{m}>>m_{f}^{2}$ (for details see Ref. \cite{der5}). In order to
derive Eq. (28) from Eq. (26) we have used a similar way with that
presented in Ref. \cite{strong}.

Thus Eq. (28), which gives the asymptotic behaviour of the
derivative expansion for large $B_{m}$, is in agreement with our
asymptotic formula of Eq. (23). This means, that the derivative
expansion is a good approximation for the effective energy, when
$B_{m}\rightarrow +\infty$ ($d$ and $m_{f}$ are kept fixed)

\section{Conclusions}

We emphasize that a step in the derivation of the asymptotic
behavior of integral $I_{2}(B_{m}d^{2})$, of Eq. (10), was
obtained by performing numerical calculations and not with an
analytic method. We computed numerically $I_{2}(B_{m}d^{2})$ for
four magnetic field configurations with quite different kinds of
inhomogeneity (see Eqs. (18), (19), (20) and (21)). As we see in
Fig. \ref{1}, $I_{2}(B_{m}d^{2})$ is proportional to $B_{m}d^{2}$
(for $B_{m}d^{2}>>1$) independently of the specific form of the
magnetic field configurations. Consequence of this result is that
and the final formula, which states that $E_{eff(3+1)}^{asympt}
\sim -\phi^{2}\ln\phi$ for $\phi\rightarrow +\infty$, is valid
independently of the specific form of the magnetic field.

Finally we showed, comparing with our asymptotic formula of Eq.
(23), that the leading term of the derivative expansion gives the
correct asymptotic behaviour for the fermion-induced effective
energy in the strong magnetic field regime even in case (like that
of the magnetic fields of Eqs. (20) and $(21)$) when the
derivative expansion is invalid mainly where the next to the
leading term diverges.

\section{Acknowledgements}
I am grateful to Professor G. Tiktopoulos for numerous valuable
discussions, as well as Professors K. Farakos and G. Koutsoumbas
for useful comments and suggestions.

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\end{document}

