# Comment on “Massive electrodynamics and the magnetic monopoles”

Michael Dunia<EMAIL_ADDRESS>Department of Physics,
California State University Fresno, Fresno, CA 93740-8031, USA  Timothy J.
Evans<EMAIL_ADDRESS>Department of Physics, California State
University Fresno, Fresno, CA 93740-8031, USA  Douglas Singleton
<EMAIL_ADDRESS>Department of Physics, California State University
Fresno, Fresno, CA 93740-8031, USA

###### Abstract

In this paper we correct previous work on magnetic charge plus a photon mass.
We show that contrary to previous claims this system has a very simple, closed
form solution which is the Dirac string potential multiplied by a exponential
decaying part. Interesting features of this solution are discussed namely: (i)
the Dirac string becomes a real feature of the solution; (ii) the breaking of
gauge symmetry via the photon mass leads to a breaking of the rotational
symmetry of the monopole’s magnetic field; (iii) the Dirac quantization
condition is potentially altered.

##  I Maxwell’s equations with magnetic charge and photon mass

Two common extensions of electrodynamics are to add a magnetic charge or a
photon mass into Maxwell’s equations. The former extension was consider by
Dirac dirac ; dirac1  through the introduction of a singular 3-vector
potential, which led to a quantization conditions between electric and
magnetic charge. The latter extension was proposed by Proca proca . Reference
joshi  considers both extensions together. The aim of this work is to correct
the analysis presented in joshi  and remark on some of the unusual features
that arise from the simple solution to this system that we find here.

Maxwell’s equations with both magnetic charge and photon mass can be written
in 4-vector forms as joshi

$\displaystyle\partial_{\nu}F^{\nu\mu}+m^{2}A^{\mu}=4\pi J^{\mu}_{(e)}$ (1)
$\displaystyle\partial_{\nu}{\cal F}^{\nu\mu}=4\pi J^{\mu}_{(m)}~{},$ (2)

where $J^{\mu}_{(e)}=(\rho_{e},{\bf J}_{e})$ and $J^{\mu}_{(m)}=(\rho_{m},{\bf
J}_{m})$ are the electric and magnetic 4-vector currents 111We use Gaussian
units and set $c=1$. Since we are using Gaussian units, we mostly use the
$2^{nd}$ edition of reference jackson  since this also uses Gaussian units.
The field strength tensor and its dual are defined via the 4-vector potential
$A^{\mu}=(\phi,{\bf A})$ as
$F^{\mu\nu}=\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}$ and ${\cal
F}^{\mu\nu}=\frac{1}{2}\epsilon^{\mu\nu\alpha\beta}F_{\alpha\beta}$. Finally,
$m^{2}A^{\mu}$ is the Proca photon mass term, with $m$ being the photon mass.
In general, due to symmetry, the divergence of the dual field strength tensor
is zero $\partial_{\nu}{\cal F}^{\nu\mu}=0$. Below we will discuss in what
sense one can have a magnetic charge source term in the right hand side of
(2).

Following reference joshi  we carry out our analysis of this system in
3-vector form. Writing out the 4-vector form of Maxwell’s equations from (1)
and (2) in 3-vector notation we have

$\displaystyle\nabla\cdot{\bf E}+m^{2}\phi=4\pi\rho_{e}$ (3)
$\displaystyle\nabla\times{\bf B}-\frac{\partial{\bf E}}{\partial t}+m^{2}{\bf
A}=4\pi{\bf J}_{e}$ (4) $\displaystyle\nabla\cdot{\bf B}=4\pi\rho_{m}$ (5)
$\displaystyle\nabla\times{\bf E}+\frac{\partial{\bf B}}{\partial t}=-4\pi{\bf
J}_{m}~{}.$ (6)

We start with the system of a magnetic charge, $g$, in the presence of a non-
zero photon mass, $m$. The magnetic charge density is
$\rho_{m}=g\delta^{3}({\bf r})$. We first assume there is no electric charge
(later we will consider both electric and magnetic charge together) we have
$\rho_{e}=0$ and ${\bf J}_{e}=0$. We fix the magnetic charge to be at rest so
${\bf J}_{m}=0$. For this set-up the fields are time-independent,
$\partial_{t}{\bf E}=\partial_{t}{\bf B}=0$. Since $\rho_{e}=0$ the scalar
potential is zero, $\phi=0$, which implies a zero electric field ${\bf E}=0$.
Using all these conditions the equations (3) (4) (5) (6) reduce to

$\displaystyle\nabla\times{\bf B}+m^{2}{\bf A}=0$ (7)
$\displaystyle\nabla\cdot{\bf B}=4\pi\rho_{m}=4\pi g\delta^{3}({\bf r})~{}.$
(8)

We first review the solution to (7) and (8) for the massless case when $m=0$.
In this case equation (8) is solved by the Dirac string potentials ${\bf
A}^{(0)}_{\pm}(r,\theta)=\frac{g}{r}\left(\frac{\pm
1-\cos\theta}{\sin\theta}\right){\bf{\hat{\varphi}}}$. The unit vector in the
$\varphi$-direction can be expanded as
${\bf{\hat{\varphi}}}=\frac{1}{\sin\theta}{\bf{\hat{z}}}\times{\bf{\hat{r}}}$
with ${\bf{\hat{r}}}=\frac{{\bf r}}{r}$ being the unit vector in the radial
direction. This vector potential is valid everywhere except at $r=0$ and along
$\theta=\pi$ or $\theta=0$ for the $+$ and $-$ signs respectively. It is also
possible (and useful) to write the Dirac string potential in cylindrical
coordinates as ${\bf A}^{(0)}_{\pm}(\rho,z)=\frac{g}{\rho}\left(\pm
1-\frac{z}{\sqrt{\rho^{2}+z^{2}}}\right){\bf{\hat{\varphi}}}$ where the
relationship of the $\rho$ and $z$ cylindrical coordinates to the spherical
polar coordinates is given via $\rho=r\sin\theta$ and $z=r\cos\theta$

Recent work shnir  adorno  heras  has emphasized the presence of an explicit
string piece to the magnetic field, in addition to the Coulomb part of the
magnetic field. In reference heras  a regularized version of the of the Dirac
string potential in cylindrical coordinates was considered

${\bf
A}^{(0)}_{\pm~{}\epsilon}(\rho,z)=\frac{g\Theta(\rho-\epsilon)}{\rho}\left(\pm
1-\frac{z}{\sqrt{\rho^{2}+z^{2}+\epsilon^{2}}}\right){\bf{\hat{\varphi}}}~{}.$
(9)

In (9) $\Theta(x)$ is the step function which equals $1$ when the argument is
positive and equals $0$ when the argument is negative, and $\epsilon$ is an
infinitesimal quantity which is taken to zero at the end. Taking the curl of
(9) and taking the limit $\epsilon\to 0$ at the end yields heras

${\bf B}^{(0)}=\lim_{\epsilon\to 0}\nabla\times{\bf
A}^{(0)}_{\pm~{}\epsilon}=g\frac{{\bf\hat{r}}}{r^{2}}\pm 4\pi
g\delta(x)\delta(y)\Theta(\mp z){\bf{\hat{z}}}$ (10)

Usually, only the first, Coulombic term, $g\frac{{\bf\hat{r}}}{r^{2}}$ is
written down, but a careful analysis (see shnir  adorno  and also heras  for a
recent, pedagogical and thorough exposition) shows the existence of the
second, string term, which is required to make sure that the divergence of a
curl is zero. In detail one can see that
$\nabla\cdot\left(g\frac{{\bf{\hat{r}}}}{r^{2}}\right)=4\pi g\delta^{3}({\bf
r})$ and also that $\nabla\cdot\left(\pm 4\pi g\delta(x)\delta(y)\Theta(\mp
z){\bf{\hat{z}}}\right)=-4\pi g\delta^{3}({\bf r})$, so that in total
$\nabla\cdot{\bf B}^{(0)}=\lim_{\epsilon\to 0}\nabla\cdot(\nabla\times{\bf
A}^{(0)}_{\pm~{}\epsilon})=0$. Thus it is only the first term in ${\bf
B}^{(0)}$ (i.e. $g\frac{{\bf{\hat{r}}}}{r^{2}}$) that gives the
$\delta$-function magnetic point source. The second term in ${\bf B}^{(0)}$
(i.e. $\pm 4\pi g\delta(x)\delta(y)\Theta(\mp z){\bf{\hat{z}}}$) does not have
a zero curl but rather gives $\nabla\times{\bf B}^{(0)}=\pm 4\pi g\Theta(\mp
z)\left[\delta(x)\delta^{\prime}(y){\bf\hat{x}}-\delta^{\prime}(x)\delta(y){\bf\hat{y}}\right]$,
where the primes indicate differentiation with respect to the argument, $x$ or
$y$. Thus $\nabla\times{\bf B}^{(0)}\neq 0$ in apparent violation of (7) with
$m=0$, but this non-zero curl for ${\bf B}$ represents an effective current
density associated with the Dirac string. By imposing the Dirac quantization
condition of $qg=n\frac{\hbar}{2}$ the string is made “invisible” for the most
part. Usually the effect of the string on the divergence and curl of ${\bf
B}^{(0)}$ is not discussed in detail. The recent works adorno  heras  have
pointed out these subtle issues connected with the Dirac string.

We now move on to the massive case when $m\neq 0$ in (7). Reference joshi
gave a complex, not analytic form for the solution to equations (7) and (8).
Here we show that the proposed solution of joshi  is not correct; that in fact
these equations have a simple, closed form solution which is simply the Dirac
string potential multiplied by a Yukawa factor $e^{-mr}$. This mirrors the
Yukawa potential for electric charge which is just a Coulomb scalar potential
multiplied by $e^{-mr}$.

Our “guess” at a solution to equations (7) and (8) is to take ${\bf
A}^{(0)}_{\pm}$ and multiply it by $e^{-mr}$ giving evans

${\bf A}_{\pm}=e^{-mr}{\bf A}^{(0)}_{\pm}(r)=g\frac{e^{-mr}}{r}\left(\frac{\pm
1-\cos\theta}{\sin\theta}\right){\bf{\hat{\varphi}}}\\\ $ (11)

To obtain the magnetic field from (11) requires that we take the curl of the
vector potential in (11). This takes some care due to the singularity along
the $\pm z$-axis. Following appendix D of heras  we convert the vector
potential in (11) into cylindrical coordinates and regularize it via an
infinitesimal $\epsilon$ to give

${\bf A}_{\pm~{}\epsilon}=\frac{g\Theta(\rho-\epsilon)}{\rho}\left(\pm
1-\frac{z}{\sqrt{\rho^{2}+z^{2}+\epsilon^{2}}}\right)e^{-m\sqrt{\rho^{2}+z^{2}+\epsilon^{2}}}{\bf{\hat{\varphi}}}~{}.$
(12)

In the end we will take $\epsilon\to 0$. The magnetic field coming from (12)
is ${\bf B}_{\pm}=\nabla\times({\bf A}_{\pm~{}\epsilon}(r))$ which yields

$\displaystyle{\bf B}_{\pm}$ $\displaystyle=$ $\displaystyle
e^{-m\sqrt{\rho^{2}+z^{2}+\epsilon^{2}}}\nabla\times\left[\frac{g\Theta(\rho-\epsilon)}{\rho}\left(\pm
1-\frac{z}{\sqrt{\rho^{2}+z^{2}+\epsilon^{2}}}\right){\bf{\hat{\varphi}}}\right]$
(13) $\displaystyle+$
$\displaystyle\nabla\left(e^{-m\sqrt{\rho^{2}+z^{2}+\epsilon^{2}}}\right)\times\left[\frac{g\Theta(\rho-\epsilon)}{\rho}\left(\pm
1-\frac{z}{\sqrt{\rho^{2}+z^{2}+\epsilon^{2}}}\right){\bf{\hat{\varphi}}}\right].$

The first line of (13) is simply the result given in appendix D of heras
multiplied by a Yukawa exponential term. After taking the curl of the first
line in (13), taking the limit $\epsilon\to 0$ and converting back to
spherical polar coordinates the first line becomes
$g\frac{e^{-mr}}{r^{2}}{\bf{\hat{r}}}\pm 4\pi
ge^{-mr}\delta(x)\delta(y)\Theta(\mp z){\bf{\hat{z}}}$. The term in the second
line of (13), again after taking the limit $\epsilon\to 0$ and converting back
to spherical polar coordinates, is $mg\frac{e^{-mr}}{r}\left[\frac{\pm
1-\cos\theta}{\sin\theta}\right]{\bf{\hat{\theta}}}$, with
${\bf{\hat{\theta}}}=-\frac{1}{\sin\theta}({\bf{\hat{z}}}-\cos\theta{\bf{\hat{r}}})$.
Combining this results gives the total B-field as

$\displaystyle{\bf B}_{\pm}$ $\displaystyle=$ $\displaystyle
g\frac{e^{-mr}}{r^{2}}\left({\bf{\hat{r}}}+mr\left[\frac{\pm
1-\cos\theta}{\sin\theta}\right]{\bf{\hat{\theta}}}\right)\pm 4\pi
ge^{-m|z|}\delta(x)\delta(y)\Theta(\mp z){\bf{\hat{z}}}~{}$ (14)
$\displaystyle=$ $\displaystyle
g\frac{e^{-mr}}{r^{2}}\left({\bf{\hat{r}}}+mr\left[\frac{\pm
1-\cos\theta}{\sin\theta}\right]{\bf{\hat{\theta}}}\right)\pm\frac{2ge^{-m|z|}\delta(\rho)}{\rho}\Theta(\mp
z){\bf{\hat{z}}}~{}.$

The ${\bf\hat{r}}$ and ${\bf\hat{\theta}}$ components in ${\bf B}_{\pm}$ come
simply taking the curl of (11) naively, while the last, ${\bf\hat{z}}$ term is
the contribution of the string and arises from the regularization procedure of
appendix D of heras . In the second line of (14) we have used
$\delta(x)\delta(y)=\frac{\delta(\rho)}{2\pi\rho}$ to convert the string part
of the magnetic field into cylindrical coordinates. For this last term we have
used $e^{-mr}\delta(x)\delta(y)\to e^{-m|z|}\delta(x)\delta(y)$ through the
action of the two $\delta$-functions. It is straightforward to see that (11)
and (14) are exact solutions to the two magnetic equations (7) and (8) if the
$\pm 4\pi ge^{-m|z|}\delta(x)\delta(y)\Theta(\mp z){\bf{\hat{z}}}$ piece of
the magnetic field is neglected.

Next we want to verify the $\nabla\cdot{\bf B}_{\pm}=0$. Due to the
singularities in ${\bf B}_{\pm}$ from (14) it is easier to do this via the
integral form of the divergence theorem (i.e. $\int\nabla\cdot{\bf
B}_{\pm}d^{3}x=\int{\bf B}_{\pm}\cdot d{\bf a}=0$) rather than directly
calculating $\nabla\cdot{\bf B}_{\pm}$. Taking our surface to be a sphere of
radius $R$ the surface integral of the first term in (14) is

$\displaystyle\int g\frac{e^{-mr}}{r^{2}}{\bf{\hat{r}}}\cdot d{\bf
a}=g\int\frac{e^{-mR}}{R^{2}}{\bf{\hat{r}}}\cdot{\bf{\hat{r}}}R^{2}d\Omega=4\pi
ge^{-mR}~{},$ (15)

where the integration of the solid angle is $\int d\Omega=4\pi$. The second
term in (14) gives zero since ${\bf{\hat{\theta}}}\cdot d{\bf
a}\propto{\bf{\hat{\theta}}}\cdot{\bf\hat{r}}=0$. Next we look at the string
contribution i.e. the last term in (14). Let us look at the case when the
string is along the $-z$-axis for which the string part of the magnetic field
in (14) is $\frac{+2ge^{-m|z|}\delta(\rho)}{\rho}\Theta(-z){\bf{\hat{z}}}$.
This magnetic field will puncture the sphere along the negative $z$-axis at
$z=-R$ and due to the $\delta(\rho)$ part we only need to worry about a small
area patch, $\pi(\Delta\rho)^{2}$, located at $z=-R$ and parallel to the $xy$
plane so that $d{\bf a}\approx-{\bf\hat{z}}\rho d\varphi d\rho$. Since the
normal to the sphere is in the ${\bf\hat{r}}$ direction along the negative $z$
axis one has ${\bf\hat{r}}=-{\bf\hat{z}}$ at this location. With this
background the surface integral of the string part of the magnetic field along
the negative $z$-axis, ${\bf B}_{+}$, is

$\int\frac{2ge^{-mR}\delta(\rho)}{\rho}\Theta(-z){\bf{\hat{z}}}\cdot d{\bf
a}=2ge^{-mR}\int_{0}^{\Delta\rho}\int_{0}^{2\pi}\frac{\delta(\rho)}{\rho}{\bf\hat{z}}\cdot(-{\bf\hat{z}}\rho
d\rho d\varphi)=-4\pi ge^{-mR}~{}.$ (16)

Combining (15) and (16) shows that $\int{\bf B}_{+}\cdot d{\bf a}=0$ which by
the divergence theorem gives $\int\nabla\cdot{\bf B}_{+}d^{3}x=0$ and thus
implies $\nabla\cdot{\bf B}_{+}=0$. When the string term is neglected one has
$\nabla\cdot({\bf B}_{+}-{\bf B}_{string})=4\pi g\delta^{3}({\bf r})$ i.e. a
magnetic monopole. The same type of calculation carries through similarly for
the case when the string is along the positive $z$-axis i.e. for $-4\pi
ge^{-m|z|}\delta(x)\delta(y)\Theta(+z){\bf{\hat{z}}}=-\frac{2ge^{-m|z|}\delta(\rho)}{\rho}\Theta(+z){\bf{\hat{z}}}$.

Next we move to the curl of the magnetic field which is

$\displaystyle\nabla\times{\bf B}_{\pm}$ $\displaystyle=$
$\displaystyle-m^{2}g\frac{e^{-mr}}{r}\left(\frac{\pm
1-\cos\theta}{\sin\theta}\right){\bf{\hat{\varphi}}}\pm 4\pi g\Theta(\mp
z)e^{-m|z|}\left[\delta(x)\delta^{\prime}(y){\bf\hat{x}}-\delta^{\prime}(x)\delta(y){\bf\hat{y}}\right]$
(17) $\displaystyle=$ $\displaystyle-m^{2}g\frac{e^{-mr}}{r}\left(\frac{\pm
1-\cos\theta}{\sin\theta}\right){\bf{\hat{\varphi}}}\pm 2g\Theta(\mp
z)e^{-m|z|}\left[\frac{\delta(\rho)}{\rho^{2}}-\frac{\delta^{\prime}(\rho)}{\rho}\right]{\bf\hat{\varphi}}~{}.$

In the second line we have written the string contribution in cylindrical
coordinates. Combining the first ${\bf\hat{\varphi}}$ term in (17) with ${\bf
A}_{\pm}$ from (11) multiplied by $m^{2}$ we see that these terms cancel and
thus (7) is satisfied, except for the string term. As in the massless case, we
are left with the current density of the string

$\displaystyle{\bf J}_{string}$ $\displaystyle=$ $\displaystyle\pm g\Theta(\mp
z)e^{-m|z|}\left[\delta(x)\delta^{\prime}(y){\bf\hat{x}}-\delta^{\prime}(x)\delta(y){\bf\hat{y}}\right]$
(18) $\displaystyle=$ $\displaystyle\pm\frac{g}{2\pi}\Theta(\mp
z)e^{-m|z|}\left[\frac{\delta(\rho)}{\rho^{2}}-\frac{\delta^{\prime}(\rho)}{\rho}\right]{\bf\hat{\varphi}}~{}.$

This current density is generally not discussed directly, with the exception
of works like shnir  adorno  heras .

There are two points worth commenting on in relation to the above closed form
Yukawa-Dirac string solution:

  1. 1.

The potential for a magnetic charge with a massive photon is obtained in
exactly the same way as in the case of electric charge where one takes the
Coulomb potential, $\phi=\frac{q}{r}$, and multiplies it by $e^{-mr}$ to
obtain the Yukawa potential, $\phi=q\frac{e^{-mr}}{r}$.

  2. 2.

The string singularity along the negative/positive axis of ${\bf
A}^{(0)}_{\pm}$, for the massless photon case, is a gauge artifact, since
${\bf A}^{(0)}_{+}$ and ${\bf A}^{(0)}_{-}$ are related by a gauge
transformation. For the massive case ${\bf A}_{+}$ and ${\bf A}_{-}$ are no
longer related to one another by a gauge transformation.

Point number 2 has major consequences for the Dirac quantization condition
dirac ; dirac1  between a magnetic charge $g$ and electric charge $q$ which is

$qg=n\frac{\hbar}{2}~{}.$ (19)

One way of arriving at the condition in (19) is by requiring that the
Aharonov-Bohm (AB) phase ab  associated with a charge $q$ moving in the
background vector potential ${\bf A}_{\pm}^{(0)}$ be undetectable i.e. that
the AB phase be some integer multiple of $2\pi$ ryder ; heras . However, now
the the string singularity is not only a feature of the 3-vector potential,
but is also a feature of the magnetic field. Thus one can not require that the
effect of AB phase vanish, since now one has a physical string/solenoid.

There is another way of obtaining the quantization condition of (19) – the
fiber bundle approach of Wu and Yang wu-yang  where one defines the vector
potential in a way that it is non-singular in the region that it covers e.g.
the vector potential is defined as ${\bf A}^{(0)}_{+}$ over the northern
hemisphere ($0<\theta\leq\pi/2$) and ${\bf A}^{(0)}_{-}$ over the southern
hemisphere ($\pi/2<\theta\leq\pi$). The two vector potentials are related by
the gauge transformation ${\bf A}^{(0)}_{-}={\bf A}^{(0)}_{+}-\nabla(\lambda)$
with $\lambda=2g\varphi$. This implies that the wavefunctions over these two
hemispheres are related by
$\Psi_{+}=e^{iq\lambda}\Psi_{-}\to\Psi_{+}=e^{2iqg\varphi}\Psi_{-}$. Since the
wavefunction must be single value as $\varphi\to 2\pi$ this means the phase
should be some integer multiple of $2\pi$ i.e. $4\pi qg=2\pi n$ which
immediately yields the quantization condition from (19). This Wu-Yang fiber
bundle approach also fails in the case of a massive photon since, unlike ${\bf
A}^{(0)}_{+}$ and ${\bf A}^{(0)}_{-}$, the Yukawa-Dirac vector potentials,
${\bf A}_{+}$ and ${\bf A}_{-}$ from (11), are not related by a gauge
transformation.

##  II Modified Dirac quantization condition

One of the most interesting outcomes of Dirac’s formulation of magnetic charge
in terms of the string potential, ${\bf A}^{(0)}_{\pm}$, is the existence of a
quantization condition between electric and magnetic charges given in (19) for
the massless photon case. As shown in the preceding section, when the photon
is massive the Dirac quantization condition (19), can not be obtained using
the method of hiding the AB phase associated with the Dirac string or the Wu-
Yang fiber bundle method. There is a third approach to the Dirac quantization
condition saha ; saha1 ; wilson  which involves requiring the quantization of
the field angular momentum between a magnetic charge $+g$ and an electric
charge $+q$. This method might work. In the massless photon case the field
angular momentum of an electric charge-magnetic charge system is jackson

${\bf L}_{EM}^{(0)}=\frac{1}{4\pi}\int{\bf r}\times({\bf E}\times{\bf
B})d^{3}x=-qg{\bf{\hat{R}}}~{},$ (20)

where ${\bf{\hat{R}}}$ is the unit vector from the magnetic charge to the
electric charge. We have reserved the relative locations of the magnetic and
electric charges from that used in jackson  since we place the magnetic charge
at the origin (this is because in the massive photon case the magnetic field
is more complex and thus the calculations are easier with the magnetic charge
at the origin). Requiring that the magnitude of the angular momentum, ${\bf
L}_{EM}^{(0)}$, be some integer multiple of $\frac{\hbar}{2}$ immediately
gives (19).

The field angular momentum method of obtaining (19) has been used to study the
Dirac quantization condition in the case of a massive photon joshi . In
addition to reference joshi  there are a host of other works joshi2 ; cafaro ;
guimaraes ; goldhaber ; goldhaber2  which look the quantization condition when
the photon is massive. These works come to different and contradictory
conclusions, and in fact as noted in goldhaber  sometimes the same author came
to differing conclusions about the possibility of the viability of the Dirac
quantization condition in the presence of a photon mass (compare the
conclusions of references goldhaber  and goldhaber2  on this point). One of
the reasons for this is that the string potential and associated magnetic
field have, up until now, been incorrectly given. In the section above we have
corrected this problem and given the correct string potential (11) and
magnetic field (14). We can use the corrected potential and field to
investigate the question of what happens to the Dirac quantization condition
when the photon is massive.

With a massive photon the electric charge-magnetic charge system still carries
a field angular momentum, but with some changes. First, the momentum density
(the time-space component of the energy-momentum tensor, $T^{0i}$) is altered.
For the massless photon case $T^{0i}=({\bf E}\times{\bf B})^{i}$, while for
the massive photon case $T^{0i}=({\bf E}\times{\bf B})^{i}+m^{2}\phi{\bf
A}^{i}$ (see chapter 12 of Jackson’s $3^{rd}$ edition jackson3 , in particular
problem 12.16). Thus the angular momentum density becomes
$\frac{1}{4\pi}\left({\bf r}\times({\bf E}\times{\bf B}_{\pm})+m^{2}\phi{\bf
r}\times{\bf A}_{\pm}\right)$. To take account of the electric field and
potential of the electric charge $q$ we again assume that the charges are
fixed so that $\rho_{e}=q\delta^{3}({\bf r}-{\bf R})$ and
$\rho_{m}=g\delta^{3}({\bf r})$, where ${\bf R}$ is the location of $q$ and we
have assumed the magnetic charge is at the origin. Since the charges are at
rest the current densities are zero ${\bf J}_{e}={\bf J}_{m}=0$. This implies
the fields are time-independent, $\partial_{t}{\bf E}=\partial_{t}{\bf B}=0$.
Thus equations (7) and (8) for ${\bf B}$ and ${\bf A}$ are supplemented by
$\nabla\cdot{\bf E}+m^{2}\phi=4\pi\rho_{e}=4\pi q\delta^{3}({\bf r}-{\bf R})$
and $\nabla\times{\bf E}=0$ for ${\bf E}$ and $\phi$. The solution to these
two additional equations is

$\phi=q\frac{e^{-mr^{\prime}}}{r^{\prime}}~{}~{}~{}~{}~{}~{};~{}~{}~{}~{}~{}{\bf
E}=-\nabla\phi=q\left(1+mr^{\prime}\right)\frac{e^{-mr^{\prime}}}{{r^{\prime}}^{2}}{\bf{\hat{r}^{\prime}}}~{},$
(21)

where for convenience we have defined ${\bf r}^{\prime}={\bf r}-{\bf R}$. Note
that $\phi$ is of the usual Yukawa form.

We have not been able to work out the general case of the field angular
momentum with a massive photon, since the presence of the physical string
greatly complicates the calculation. There are two special cases where the
field angular momentum can be worked out exactly: (i) when the electric charge
and magnetic charge are both at the origin; (ii) when the displacement vector
between the electric charge and magnetic charge is lined up with the physical
Dirac string e.g. if both the displacement vector between the charges and the
physical string lie along the $z$-axis. In this work we make this assume
simply for ease of calculation (i.e. we have not been able to work out the
general case analytically). However, there may be some argument to constrain
the electric charge to be inside the current density ${\bf J}_{string}$. To
investigate this possibility one would need to not put the various electric
and magnetic source terms in by hand, but would need to make them dynamical by
associating them with some field coupled to the electromagnetic gauge field.
We are investigating this possibility, but finding such field interacting
solutions is difficult.

We begin with the first special case when both the electric charge and
magnetic charge are located at $r=0$ and we take the string to be along either
the $+z$ or $-z$ axis. Using the fields and potentials from (21) (14) and (11)
the field angular momentum density is

$\frac{1}{4\pi}\left({\bf r}\times({\bf E}\times{\bf B}_{\pm})+m^{2}\phi{\bf
r}\times{\bf
A}_{\pm}\right)=-\frac{qgm}{4\pi}(1+2mr)\frac{e^{-2mr}}{r^{2}}\left[\frac{\pm
1-\cos\theta}{\sin\theta}\right]{\bf{\hat{\theta}}}~{}.$ (22)

Recalling that
${\bf{\hat{\theta}}}=\cos\theta\cos\varphi{\bf{\hat{x}}}+\cos\theta\sin\varphi{\bf{\hat{y}}}-\sin\theta{\bf{\hat{z}}}$
the $d\varphi$ integration of the volume integral of ${\bf r}\times({\bf
E}\times{\bf B})+m^{2}\phi_{e}{\bf r}\times{\bf A}_{\pm}$ will give zero for
the ${\bf{\hat{x}}},{\bf{\hat{y}}}$ components and $2\pi$ for the
${\bf{\hat{z}}}$. Performing the $dr$ and $d\theta$ integrals (we change
$d\theta\to dx$ via the substitution $x=\cos\theta$) gives

$\displaystyle\frac{1}{4\pi}\int\left({\bf r}\times({\bf E}\times{\bf
B}_{\pm})+m^{2}\phi{\bf r}\times{\bf A}_{\pm}\right)d^{3}x$ $\displaystyle=$
$\displaystyle\frac{qgm}{2}\int_{0}^{\infty}(1+2mr)e^{-2mr}dr\int_{-1}^{1}(\pm
1-x)dx{\bf{\hat{z}}}$ (23) $\displaystyle=$ $\displaystyle\pm
qg{\bf{\hat{z}}}~{}.$

On the surface this result is exactly the same as in massless photon case.
However, in the usual massless photon case if the electric and magnetic charge
are both placed at $r=0$ one gets zero field angular momentum when the chagres
are placed at the same location. First, for the radial parts of the electric
and magnetic fields one has ${\bf E}^{(0)}_{r}\times{\bf
B}^{(0)}_{\pm~{}r}\propto{\bf{\hat{r}}}\times{\bf{\hat{r}}}=0$ which gives
${\bf r}\times({\bf E}^{(0)}_{r}\times{\bf B}^{(0)}_{\pm~{}r})=0$. Second,
there is also a contribution from the from the radial electric field and the
string part of the magnetic field as given by the $\pm 4\pi
g\delta(x)\delta(y)\Theta(\mp z){\bf{\hat{z}}}$ in (10). In this case the
radial part of the electric field and this string piece give ${\bf
E}^{(0)}_{r}\times{\bf
B}^{(0)}_{\pm~{}string}\propto\delta(x)\delta(y){\bf{\hat{r}}}\times{\bf{\hat{z}}}=-\delta(x)\delta(y)\sin\theta{\bf{\hat{\varphi}}}$,
where we have used the fact that
${\bf{\hat{z}}}={\bf{\hat{r}}}\cos\theta-{\bf{\hat{\theta}}}\sin\theta$,
${\bf{\hat{r}}}\times{\bf{\hat{r}}}=0$ and
${\bf{\hat{r}}}\times{\bf{\hat{\theta}}}={\bf{\hat{\varphi}}}$. Finally, using
all this one finds that ${\bf r}\times({\bf E}^{(0)}_{r}\times{\bf
B}^{(0)}_{\pm~{}string})\propto-\delta(x)\delta(y)\sin\theta~{}{\bf{\hat{r}}}\times{\bf{\hat{\varphi}}}=\delta(x)\delta(y)\sin\theta~{}{\bf{\hat{\theta}}}$.
Since $\sin\theta=\frac{\sqrt{x^{2}+y^{2}}}{r}$ then the two delta functions,
$\delta(x)\delta(y)$ will make $\sin\theta\to 0$ and thus the entire integrand
and integral go to zero. Thus the string piece of the magnetic field will not
contribute to the field angular momentum if the electric charge sits on the
same axis at the string.

Returning now to the massive photon case for the special case when both the
electric charge and magnetic charge are at the origin, we see that the field
angular momentum in (23) comes from the vector potential piece $m^{2}\phi{\bf
r}\times{\bf A}_{\pm}$, and from the ${\bf\hat{\theta}}$ part of the magnetic
field in ${\bf r}\times({\bf E}\times{\bf B}_{\pm})$. This field angular
momentum in (23) points along $+{\bf{\hat{z}}}$ if the singularity is along
the $-z$-axis, and points along $-{\bf{\hat{z}}}$ if the singularity is along
the $+z$-axis. In this special case one recovers the standard quantization
condition of (19) in agreement with references cafaro ; guimaraes ; goldhaber
and in disagreement with reference joshi ; joshi2 ; goldhaber2 .

We now move to the second special case – having the string direction align
with the displacement between the charges. We will take the magnetic charge to
be located at $r=0$ and we will first consider the case when the physical
string lies along the $-z$ axis i.e. we will use the ${\bf B}_{+}$ and ${\bf
A}_{+}$ fields from (14) and (11) respectively. The magnetic charge is placed
at the origin since it is the more complicated field. Next we place the
electric charge on the $\pm z$-axis a distance $R$ from the origin so that its
position is given by $\pm R{\bf\hat{z}}$. For $-R{\bf\hat{z}}$ the string will
run through the charge $q$, while for $+R{\bf\hat{z}}$ the charge $q$ will be
located outside the string. The electric potential and electric field for $q$
for these locations is given by (21) with ${\bf r^{\prime}}={\bf r}\mp
R{\bf\hat{z}}$ depending if the electric charge is at $+R{\bf\hat{z}}$ or
$-R{\bf\hat{z}}$. The magnitude of the position vector from the electric
charge is $r^{\prime}=\sqrt{r^{2}+R^{2}\mp 2rR\cos\theta}$. We present the
details of the calculation of the field angular momentum in the appendix. The
summary of these calculations from the appendix is

  1. i

String along $-z$-axis, electric charge at $r=-R{\bf\hat{z}}$: ${\bf
L}_{EM}={\bf L}_{EM}^{point}+{\bf
L}_{EM}^{A+B^{\theta}}=2qge^{-mR}{\bf\hat{z}}$

  2. ii

String along $-z$-axis, electric charge at $r=+R{\bf\hat{z}}$: ${\bf
L}_{EM}={\bf L}_{EM}^{point}+{\bf L}_{EM}^{A+B^{\theta}}=0$

  3. iii

String along $+z$-axis, electric charge at $r=-R{\bf\hat{z}}$: ${\bf
L}_{EM}={\bf L}_{EM}^{point}+{\bf L}_{EM}^{A+B^{\theta}}=0$

  4. iv

String along $+z$-axis, electric charge at $r=+R{\bf\hat{z}}$: ${\bf
L}_{EM}={\bf L}_{EM}^{point}+{\bf
L}_{EM}^{A+B^{\theta}}=-2qge^{-mR}{\bf\hat{z}}$

The usual result for the field angular momentum in the case of a massless
photon is given in (20) as ${\bf L}^{(0)}_{EM}=qg{\bf\hat{R}}$. The results
above – minus the factor $e^{-mR}$ – seem to indicate either twice this value
(for cases (i) and (iv)) or $0$ (for cases (ii) and (iii)). Cases (i) and (iv)
may be understood as the addition of the point and string contributions to
give twice the result of the massless case, $2qg$, with a multiplicative
$e^{-mR}$ factor to account for the photon mass. Cases (ii) and (iii) may be
understood as a cancellation of the point and string contributions. Note that
in the above summary of the results for ${\bf L}_{EM}$ we have included the
contribution of the magnetic field coming from the point/Yukawa part, the
vector potential part and the ${\bf\hat{\theta}}$ component, but we have
apparently neglected the contribution coming from the string part of the
magnetic field (e.g. the ${\bf\hat{z}}$ term in (14)). As explained in the
appendix the contribution from the string part of the magnetic field vanishes
when the electric charge sits on the same axis as the string.

If we now apply the requirement that ${\bf L}_{EM}$ from above must be some
integer multiple of $\hbar/2$ we would get conditions from cases (i) and (iv)
like

$2qge^{-mR}=n\frac{\hbar}{2}~{}.$ (24)

This quantization condition involves not only the charges, $q,g$, but also
involves mass and position, $m$ and $R$. To check more carefully whether the
heuristic arguments leading to (24) are correct, one should calculate the
commutators of the total angular momentum operators 222One should check that
$[L_{i},L_{j}]=i\hbar\epsilon_{ijk}L_{k}$ where
$L_{i}=L_{i}^{particle}+L_{i}^{field}$ to see if these work out and, if so,
under what conditions. The calculation of the commutators for the total
angular momentum (particle angular momentum plus field angular momentum) of
the electric charge-magnetic charge system for the massless photon case was
carried in fierz ; lipkin ; yang . If the particle carrying either the
electric charge or the magnetic charge has an intrinsic spin then this should
be included in the total angular momentum as ${\bf L}_{tot}={\bf
L}_{spin}+{\bf L}_{orbital}+{\bf L}_{field}$. It is this ${\bf L}_{tot}$ which
should satisfy the angular momentum commutation relationship.

A system with a field angular momentum that is closer to the present case of
electric charge-magnetic charge plus photon mass is the field angular momentum
of an electric charge, $q$, plus a magnetic dipole, ${\bf m}$, with a massless
photon. Such a system of electric charge-magnetic dipole was investigated in
singleton-plb  as a way of addressing the nucleon-spin puzzle. In singleton-98
it was shown that the total angular momentum of the electric charge-magnetic
dipole system exactly satisfied the standard commutator for angular momentum.
We are currently studying the general case of the field angular momentum of
the electric charge-magnetic charge system with a massive photon and the
viability (or not) of the Dirac quantization condition. One similarity between
the electric-charge magnetic dipole system with the electric-charge magnetic-
charge plus massive photon system is that both have a special direction – in
the first case it is the direction of the magnetic dipole, ${\bf m}$, and in
the second case it is the direction of the magnetic string.

##  III Summary and Conclusion

In this work we have given the Dirac string potential and magnetic field,
equations (11) and (14), for a electromagnetism with a magnetic charge plus
massive photon. This corrects the previous results for this system given in
reference joshi . The vector potential in (11), is similar to what occurs for
the Yukawa scalar potential for electric charge given in (21) – one takes the
$m=0$ potential (either Coulomb scalar potential or Dirac string potential)
and multiplies it by the Yukawa factor $e^{-mr}$. This was shown by explicit
calculations using the Maxwell equations with magnetic charge and photon mass
given by (3) (4) (5) (6).

Combining photon mass and magnetic charge turned the Dirac string from a gauge
artifact into a real, physical string, and spoiled the spherical symmetry of
the magnetic field associated with the magnetic charge due to the addition of
the ${\bf\hat{\theta}}$ dependent string part in (14). Thus adding a photon
mass not only spoils the gauge symmetry but also spoils the radial symmetry of
the magnetic field. This connection between gauge and spatial symmetries
warrants further investigation.

In section II we investigated the fate of the Dirac quantization condition in
the presence of a photon mass. Since the string singularity went from a gauge
artifact, in the massless photon case, to a real, physical feature in the
massive photon case, the approach to the Dirac quantization condition which
required the vanishing of the Aharonov-Bohm phase of the string ryder  no
longer worked. Also the fiber bundle approach wu-yang  to the Dirac
quantization condition was no longer viable. However the method of obtaining
the Dirac quantization condition through the quantization of the field angular
momentum of the electric charge - magnetic charge system saha ; saha1 ; wilson
still gave a potential path toward finding some form of the Dirac quantization
condition when $m\neq 0$. We were not able to obtain a closed form expression
for ${\bf L}_{EM}$ in the general case, but we were able to investigate two
special cases: (i) when the electric charge and magnetic charge where at the
same location and (ii) when the displacement vector between the charges was
aligned along the direction of the string. In the first case we did recover
the original Dirac quantization condition (19), and in the second case we
obtained a modified variant of the Dirac condition given in equation (24)
which involved both $m$ and $R$. We are currently working on using the correct
vector potential and magnetic fields from (11) and (14) to perform an analysis
similar to references fierz ; lipkin ; yang  by looking at how the commutation
relationship for the full angular momenta, ( i.e.
$[L_{i},L_{j}]=i\hbar\epsilon_{ijk}L_{k}$) plays out for a massive photon; to
see if this is possible or not, and if so what kind of condition needs to be
imposed on the electric charges, magnetic charges, and potentially the photon
mass and displacement between the charges.

In this work we have also included the magnetic field and current density
source connected with the Dirac string – see equations (14) and (17). These
fields and sources for the Dirac string are not generally emphasized, but
recent works shnir  adorno  heras  have pointed out the subtle, but crucial
importance of these quantities. In particular one may ask about the nature of
the source of the effective string current in (17), ${\bf J}_{string}=\pm
g\Theta(\mp
z)\left[\delta(x)\delta^{\prime}(y){\bf\hat{x}}-\delta^{\prime}(x)\delta(y){\bf\hat{y}}\right]$.
Note that this current density has a rotational character and thus might be
expected to contribute to the angular momentum of the system. Within the
context of the present work we can not address this question since we have put
the sources in by hand e.g. we had taken the magnetic point source to have the
form $\rho_{m}=g\delta^{3}({\bf r})$ without saying what fields give rise to
this source. To properly and consistently take into account the sources, we
should include an additional field interacting with the gauge field in the
manner of the t’ Hooft - Polyakov monopole thooft  polyakov  or the Prasad-
Sommerfield solution prasad . We are currently looking into this possibility
of modeling the inserted-by-hand magnetic sources with field sources.

## References

  * (1) P.A.M. Dirac, Proc. Roy. Soc. A 133, 60-72 (1931).
  * (2) P.A.M. Dirac, Phys. Rev. 74, 817-830 (1948).
  * (3) A. Proca, J.Phys.Radium 7, 347 (1936).
  * (4) A. Yu. Ignatiev and G. C. Joshi, Phys. Rev. D 53, 984-992 (1996).
  * (5) J.D. Jackson, Classical Electrodynamics $2^{nd}$ edition, (John Wiley & Sons, 1975).
  * (6) Y.M. Shnir, Magnetic Monopoles (Springer, Berlin 2005.
  * (7) T.C. Adorno, D.M. Gitman, and A.E. Shabad, Proc. of the Steklov Inst. of Math. 309, 1 (2020).
  * (8) R. Heras, Contemp. Phys. 59, 331 (2018).
  * (9) T. Evans and D. Singleton, Int. J. Mod. Phys. A 33, 1850064 (2018).
  * (10) Y. Aharonov and D. Bohm, Phys. Rev. 115, 484 (1959).
  * (11) L.H. Ryder, Quantum Theory of Fields 2nd edition (Cambridge University Press 1996).
  * (12) T.T. Wu and C,N. Yang, Phys. Rev. D 12, 3845 (1975).
  * (13) M.N. Saha, Ind. J. Phys. 10, 145-151 (1936).
  * (14) M.N. Saha, Phys. Rev. 75, 1968 (1949).
  * (15) H.A. Wilson, Phys. Rev. 75, 309 (1949).
  * (16) A.Yu. Ignatiev and G. C. Joshi, Mod.Phys.Lett. A 11, 2735 (1996)
  * (17) C. Cafaro, S. Capozziello, Ch. Corda, and S.A. Ali, Adv.High Energy Phys., 69835 (2007).
  * (18) M.S. Guimaraes, R. Rougemont, C. Wotzasek, and C.A.D. Zarro, Phys.Lett.B 723, 422 (2013).
  * (19) A. S. Goldhaber and R. Heras, “Dirac Quantization Condition Holds with Nonzero Photon Mass”, e-Print: 1710.03321
  * (20) A. S. Goldhaber and M. M. Nieto, Rev.Mod.Phys. 43, 277 (1971).
  * (21) J.D. Jackson, Classical Electrodynamics $3^{rd}$ Edition, (John Wiley & Sons, 1999).
  * (22) M. Fierz, Helv. Phys. Acta 17, 27 (1944).
  * (23) H.J. Lipkin, W.I. Weisberger, and M. Peshkin, Ann. Phys. 53, 203 (1969).
  * (24) C.N. Yang, Annals New York Academy of Sciences 294, 86 - 97 (1977).
  * (25) D. Singleton, Phys. Lett. B 427, 155 (1998).
  * (26) D. Singleton, Am.J.Phys. 66, 697 (1998)
  * (27) G. ’t Hooft, Nucl.Phys.B 79, 276 (1974).
  * (28) A. M. Polyakov, JETP Lett. 20, 194 (1974).
  * (29) M.K. Prasad and C. M. Sommerfield, Phys.Rev.Lett. 35, 760 (1975).

##  Appendix A Field angular momentum for displaced electric charge

In this appendix we present the details of the calculation of the field
angular momentum in the special case where the magnetic charge and electric
charge are at different points, but with the displacement vector between the
charges aligned with physical string. We start with the magnetic field and
vector potential with the string along the $-z$ axis given by (14) and (11) as

${\bf
B}_{+}=g\frac{e^{-mr}}{r^{2}}\left({\bf{\hat{r}}}+mr\left[\frac{1-\cos\theta}{\sin\theta}\right]{\bf{\hat{\theta}}}\right)~{}~{}~{}~{};~{}~{}~{}~{}{\bf
A}_{+}=g\frac{e^{-mr}}{r}\left(\frac{1-\cos\theta}{\sin\theta}\right){\bf{\hat{\varphi}}}~{}.$
(25)

We take the electric field and potential for a charge located on the $z$-axis
at $-R$ (inside the string) given by

${\bf E}=q(1+mr^{\prime})\frac{e^{-mr^{\prime}}}{(r^{\prime})^{3}}{\bf
r^{\prime}}~{}~{}~{}~{};~{}~{}~{}~{}\phi=q\frac{e^{-mr^{\prime}}}{r^{\prime}}~{},$
(26)

where ${\bf r^{\prime}}={\bf r}+R{\bf\hat{z}}$ and
$r^{\prime}=\sqrt{r^{2}+R^{2}+2rR\cos\theta}$. The case when the charge sits
at $+R{\bf\hat{z}}$ (outside the string) has a parallel calculation but with
the changes ${\bf r^{\prime}}={\bf r}-R{\bf\hat{z}}$ and
$r^{\prime}=\sqrt{r^{2}+R^{2}-2rR\cos\theta}$. We find that for the former
case the field angular momentum of the string-part adds to that of the point-
part, while in the latter case the the string-part cancels that of the point-
part. In the next two subsections we calculate the contribution to the field
angular momentum of the point-part and string-part of the fields in (25) and
(26).

###  A.1 Point-part

Here we calculate the point-part contribution to the field angular momentum
density from the radial part of the magnetic field ${\bf
B}^{point}_{+}=g\frac{e^{-mr}}{r^{2}}{\bf{\hat{r}}}$. The double cross product
from ${\bf r}\times({\bf E}\times{\bf B}^{point}_{+})$ works out as ${\bf
r}\times({\bf r^{\prime}}\times{\bf\hat{r}})=R\sin\theta{\bf
r}\times{\bf\hat{\varphi}}=-rR\sin\theta{\bf\hat{\theta}}$. The field angular
momentum for this point-part is now

${\bf
L}_{EM}^{point}=-\frac{Rqg}{4\pi}\int\frac{\sin\theta(1+mr^{\prime})e^{-mr^{\prime}}e^{-mr}}{r{r^{\prime}}^{3}}{\bf\hat{\theta}}d^{3}x~{}.$
(27)

We can perform the integration over $d\varphi$ since the only $\varphi$
dependence is in ${\bf\hat{\theta}}$. This yields
$\int_{0}^{2\pi}{\bf\hat{\theta}}d\varphi=-2\pi\sin\theta{\bf\hat{z}}$.
Collecting terms and making the standard change of variables $x=\cos\theta$ we
obtain

${\bf
L}_{EM}^{point}=\frac{Rqg}{2}{\bf\hat{z}}\int_{0}^{\infty}\int_{-1}^{1}\frac{(1-x^{2})(1+mr^{\prime})e^{-mr^{\prime}}e^{-mr}}{r{r^{\prime}}^{3}}~{}dx~{}r^{2}dr~{}.$
(28)

Using Mathematica the $dx$ integration yields (taking into account
$r^{\prime}=\sqrt{r^{2}+R^{2}+2rRx}$)

$\displaystyle{\bf
L}_{EM}^{point}=\frac{qg}{R^{2}m^{3}}{\bf\hat{z}}\int_{0}^{\infty}\frac{e^{-mr}}{r^{2}}\bigg{(}$
$\displaystyle e^{-m|r-R|}(-1+m^{2}rR-m|r-R|)$ (29)
$\displaystyle+e^{-m(r+R)}(1+m^{2}rR+m(r+R))\bigg{)}~{}dr~{}.$

Due to the presence of the $|r-R|$ term, the $dr$ integration has to be broken
up into the range $0\leq r\leq R$ (when $|r-R|=R-r$) and $R\leq r\leq\infty$
(when $|r-R|=r-R$). Using Mathematica the $dr$ integration yields

${\bf
L}_{EM}^{point}=\frac{qge^{-mR}}{m^{2}R^{2}}\left(-2mR+(1+mR)(\gamma+\ln(2mR))+e^{2mR}(-1+mR)Ei(-2mR)\right){\bf\hat{z}}~{}.$
(30)

In (30) $\gamma\approx 0.577216$ is the Euler–-Mascheroni constant and
$Ei(x)=-\int_{-x}^{\infty}\frac{e^{-t}}{t}dt$ is the exponential integral
function.

###  A.2 ${\bf A}$ and $B^{\theta}$ parts

Here we calculate the contribution to the field angular momentum density from
those parts of the vector potential and magnetic field that are proportional
to $\left(\frac{1-\cos\theta}{\sin\theta}\right)$. From (25) one can see that
there are two contributions: ${\bf
B}^{\theta}_{+}=gm\frac{e^{-mr}}{r}\left[\frac{1-\cos\theta}{\sin\theta}\right]{\bf{\hat{\theta}}}$
and ${\bf
A}_{+}=g\frac{e^{-mr}}{r}\left(\frac{1-\cos\theta}{\sin\theta}\right){\bf{\hat{\varphi}}}$.
We will calculate the latter term first.

The contribution to the field angular momentum density from ${\bf A}_{+}$ is
$\frac{m^{2}}{4\pi}\phi({\bf r}\times{\bf A}_{+})$. The cross product yields
${\bf\hat{r}}\times{\bf\hat{\varphi}}=-{\bf\hat{\theta}}$. As previously the
$d\varphi$ integration gives
$\int_{0}^{2\pi}{\bf\hat{\theta}}d\varphi=-2\pi\sin\theta{\bf\hat{z}}$. Using
$\phi$ and ${\bf A}_{+}$ from (26) and (25) respectively we find

${\bf
L}_{EM}^{A}=\frac{m^{2}qg}{2}{\bf\hat{z}}\int_{0}^{\infty}r^{2}dr\int_{-1}^{1}dx\frac{e^{-mr}e^{-mr^{\prime}}(1-x)}{r^{\prime}}~{}.$
(31)

We have made the change of variables $x=\cos\theta$. Using Mathematica to
carry out the $dx$ integration yields

$\displaystyle{\bf
L}_{EM}^{A}=\frac{qg}{2mR^{2}}{\bf\hat{z}}\int_{0}^{\infty}dre^{-mr}\bigg{(}$
$\displaystyle-e^{-m|r-R|}(1+m|r-R|-2m^{2}rR)$ (32)
$\displaystyle+e^{-m(r+R)}(1+m(r+R))\bigg{)}~{}.$

Due to the $|r-R|$ term one again needs to split the $dr$ integration up into
two ranges – $0\leq r\leq R$ and $R\leq r\leq\infty$. Using Mathematica to do
the $dr$ integration yields

${\bf L}_{EM}^{A}=\frac{1}{4}qge^{-mR}(1+2mR){\bf\hat{z}}~{}.$ (33)

Now we move on to ${\bf B}^{\theta}_{+}$. We first compute the double cross
product ${\bf r}\times({\bf r^{\prime}}\times{\bf\hat{\theta}})={\bf
r}\times(r+r\cos\theta){\bf\hat{\varphi}}=-r(r+R\cos\theta){\bf\hat{\theta}}$.
As before the only $\varphi$ dependence comes from ${\bf\hat{\theta}}$ so we
can perform the $d\varphi$ integration
$\int^{2\pi}_{0}{\bf\hat{\theta}}d\varphi=-2\pi\sin\theta{\bf\hat{z}}$.
Collecting terms and making the change of variable $x=\cos\theta$ we find

${\bf
L}_{EM}^{B^{\theta}}=\frac{mqg}{2}{\bf\hat{z}}\int_{0}^{\infty}r^{2}dr\int_{-1}^{1}dx\frac{e^{-mr}e^{-mr^{\prime}}(1-x)(1+mr^{\prime})(r+Rx)}{(r^{\prime})^{3}}~{}.$
(34)

Carrying out the $dx$ integration via Mathematica yields

$\displaystyle{\bf L}_{EM}^{B^{\theta}}$ $\displaystyle=$
$\displaystyle\frac{qg}{2m^{2}R^{2}}{\bf\hat{z}}\int_{0}^{\infty}r^{2}dr\frac{e^{-mr}}{r^{3}}\bigg{(}-e^{-m|r-R|}\Big{[}2m|r-R|+2+m^{2}r(r-3R)\pm
2m^{3}r^{2}R\Big{]}$ (35) $\displaystyle+$ $\displaystyle
e^{-m(r+R)}\Big{[}2m(r+R)+2+m^{2}r(r+R)\Big{]}\bigg{)}~{}.$

The $+2m^{3}r^{2}R$ term in the first line is for the case when $r<R$ and the
$-2m^{3}r^{2}R$ term in the first line is for the case when $r>R$. As
previously the $dr$ integration has to be split into $r<R$ and $r>R$ ranges.
Using Mathematica  the $dr$ integration of (35) gives

$\displaystyle{\bf L}_{EM}^{B^{\theta}}=\frac{qge^{-mR}}{4m^{2}R^{2}}$
$\displaystyle\Big{(}$ $\displaystyle
mR(8+7mR-2m^{2}R^{2})-4(1+mR)(\gamma+\ln(2mR))$ (36) $\displaystyle-$
$\displaystyle 4e^{2mR}(-1+mR)Ei(-2mR)\Big{)}{\bf\hat{z}}~{}.$

As before $\gamma$ is the Euler-–Mascheroni constant and $Ei(x)$ is the
exponential integral function. Adding together the two parts from (36) and
(33) gives the total string contribution from the case when the string is
along the $-z$-axis and the electric charge is embedded in the string at
$r=-R{\bf\hat{z}}$ as

$\displaystyle{\bf L}_{EM}^{A+B^{\theta}}={\bf L}_{EM}^{A}+{\bf
L}_{EM}^{B^{\theta}}$ $\displaystyle=$
$\displaystyle\frac{qge^{-mR}}{m^{2}R^{2}}\Big{(}2mR(1+mR)-(1+mR)(\gamma+\ln(2mR))-e^{2mR}(-1+mR)Ei(-2mR)\Big{)}{\bf\hat{z}}~{}.$

###  A.3 Total field angular momentum

In this subsection we obtain the total field angular momentum by combining the
results from the previous two subsections for the four different cases: (i)
string along the $-z$-axis and electric charge located at $r=-R{\bf\hat{z}}$;
(ii) string along the $-z$-axis and electric charge located at
$r=+R{\bf\hat{z}}$ ; (iii) string along the $+z$-axis and electric charge
located at $r=-R{\bf\hat{z}}$; (iv) string along the $+z$-axis and electric
charge located at $r=+R{\bf\hat{z}}$. Now in the above calculations we have
apparently not included the contribution coming from the string part of the
magnetic field (i.e. the $\pm 4\pi ge^{-m|z|}\delta(x)\delta(y)\Theta(\mp
z){\bf{\hat{z}}}$ term in (14)). However, it is straightforward to check that
the field angular momentum from this part of the magnetic field is zero when
the electric charge is on the same axis as the string, as is the case when the
photon is massless (see the calculation after equation (23)).

Case (i) is the case that we explicitly calculated in the above two
subsections. Adding together the point part from (30) and string part from
(A.2) gives

${\bf L}_{EM}={\bf L}_{EM}^{point}+{\bf
L}_{EM}^{A+B^{\theta}}=2qge^{-mR}{\bf\hat{z}}$ (38)

Case (iv) can be obtained from case (i) by simply exchanging the positive and
negative $z$-axis. This switch changes the signs of (30) and (A.2). Adding
these sign changed terms together the total field angular momentum becomes
${\bf L}_{EM}={\bf L}_{EM}^{point}+{\bf
L}_{EM}^{A+B^{\theta}}=-2qge^{-mR}{\bf\hat{z}}$ i.e. the same magnitude but
opposite direction from case (i).

Case (ii) requires that one repeat the calculations of the above two
subsections that lead to (30) (33) and (36) but with ${\bf r}^{\prime}={\bf
r}-R{\bf\hat{z}}$ and $r^{\prime}=\sqrt{r^{2}+R^{2}-2rR\cos\theta}$. We do not
give the details explicitly but the results are

$\displaystyle{\bf L}_{EM}^{point}$ $\displaystyle=$
$\displaystyle\frac{qge^{-mR}}{m^{2}R^{2}}\left(2mR-(1+mR)(\gamma+\ln(2mR))-e^{2mR}(-1+mR)Ei(-2mR)\right){\bf\hat{z}}~{},$
$\displaystyle{\bf L}_{EM}^{A}$ $\displaystyle=$
$\displaystyle\frac{1}{4}qge^{-mR}{\bf\hat{z}}~{},$ (39) $\displaystyle{\bf
L}_{EM}^{B^{\theta}}$ $\displaystyle=$
$\displaystyle\frac{qge^{-mR}}{4m^{2}R^{2}}\Big{(}-mR(8+mR)+4(1+mR)(\gamma+\ln(2mR))$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}+4e^{2mR}(-1+mR)Ei(-2mR)\Big{)}{\bf\hat{z}}~{}.$

Adding up all the terms in (A.3) gives

${\bf L}_{EM}={\bf L}_{EM}^{point}+{\bf L}_{EM}^{A}+{\bf
L}_{EM}^{B^{\theta}}=0~{},$ (40)

In this case the different contributions to the field angular momentum cancel.

Case (iii) can be obtained from case(ii) simply by flipping the $z$-axis
around exchanging the positive and negative $z$-axis. This changes the overall
sign in front of each term in (A.3). When adding these together one again
obtains the result that the field angular momentum is zero, ${\bf L}_{EM}={\bf
L}_{EM}^{point}+{\bf L}_{EM}^{A}+{\bf L}_{EM}^{B^{\theta}}=0$.

###  A.4 Limits and checks of the field angular momentum results

We now want to check the above result, and come to a physical understanding of
the field angular momentum by taking the $m\to 0$ and/or $R\to 0$ limits. In
looking at the expression for the point contribution, ${\bf L}^{point}_{EM}$
(see (30) and (A.3)), and the two string contributions, ${\bf
L}_{EM}^{string-A}$ and ${\bf L}_{EM}^{string-B}$ (see (33), (36) and (A.3))
one sees that all of these terms depend on $mR$. Thus the two limits $m\to 0$
and $R\to 0$ are intertwined with one another. Defining the new variable
$x=mR$ we can write the point contribution as

${\bf
L}_{EM}^{point}=\pm\frac{qge^{-x}}{x^{2}}\left(-2x+(1+x)(\gamma+\ln(2x))+e^{2x}(-1+x)Ei(-2x)\right){\bf\hat{z}}~{},$
(41)

where the $+$ sign is for (30) and the $-$ sign is for (A.3). Using
Mathematica and taking $x\to 0$ one finds

$\lim_{x\to 0}{\bf L}_{EM}^{point}=\pm qg{\bf\hat{z}}~{}~{}~{}~{}~{}{\rm
Limit~{}~{}}x,m\to 0~{}~{}{\rm but}~{}~{}R\neq 0.$ (42)

One must be careful since the above limit only applies for $m\to 0$, but not
$R\to 0$. For $R\to 0$ the vector direction of both the electric and magnetic
field are solely in the ${\bf\hat{r}}$ direction. Thus $\lim_{R\to 0}{\bf
E}^{point}\times{\bf B}^{point}\propto{\bf\hat{r}}\times{\bf\hat{r}}=0$ and
${\bf L}_{EM}^{point}=0$. This is in agreement with the massless case where
${\bf L}_{EM}^{(0)}=0$ when the electric and magnetic charges sit on top of
one another. In summary the limit in (42) does agree with the $m=0$ case if
the electric charge and magnetic charge are displaced from one another. When
$R=0$ from the outset, going back to the beginning of the calculation in the
appendix and using the fact that ${\bf\hat{r}}\times{\bf\hat{r}}=0$ leads to
the consistent result ${\bf L}_{EM}^{point}=0$.

Next we look at the contribution from the vector potential for cases (i) and
(ii) when the string is along the $-z$-axis. When the charge lies at
$-R{\bf\hat{z}}$ equation (33) gives ${\bf
L}_{EM}^{A}=\frac{1}{4}qge^{-x}(1+2x){\bf\hat{z}}$; when the charge lies at
$+R{\bf\hat{z}}$ equation (A.3) gives ${\bf
L}_{EM}^{A}=\frac{1}{4}qge^{-x}{\bf\hat{z}}$. In either case (i) or (ii) one
finds that $\lim_{x\to 0}{\bf L}_{EM}^{A}=\frac{1}{4}qg{\bf\hat{z}}$ i.e. this
piece of the field angular momentum is along the $+z$-direction. For cases
(iii) and (iv) when the string is placed along the $+z$-axis the limit $x\to
0$ will simply have the opposite sign from cases (i) and (ii). In summary

$\lim_{x\to 0}{\bf L}_{EM}^{A}=\pm\frac{1}{4}qg{\bf\hat{z}}~{}~{}~{}~{}~{}{\rm
Limit~{}~{}}x,R\to 0~{}~{}{\rm but}~{}~{}m\neq 0~{},$ (43)

with the $+$ sign being for the string along the $-z$-axis and the $-$ sign
being for the string along the $+z$-axis. For $x\to 0$ but $m=0$ one finds
${\bf L}_{EM}^{A}=0$, since this term is missing from the outset – see (31).

Finally, we tackle the limit of the $\theta$ part of the magnetic field, ${\bf
L}_{EM}^{B^{\theta}}$. For cases(i) and (ii) (given in (36) and (A.3)) taking
the limit $x\to 0$ but $m\neq 0$ one finds that $\lim_{x\to 0}{\bf
L}_{EM}^{point-B}=\frac{3}{4}qg{\bf\hat{z}}$. Running through cases (iii) and
(iv) one finds that $\lim_{x\to 0}{\bf
L}_{EM}^{point-B}=-\frac{3}{4}qg{\bf\hat{z}}$. Summarizing this contribution
in the limit gives

$\lim_{x\to 0}{\bf
L}_{EM}^{string-B}=\pm\frac{3}{4}qg{\bf\hat{z}}~{}~{}~{}~{}~{}{\rm
Limit~{}~{}}x,R\to 0~{}~{}{\rm but}~{}~{}m\neq 0~{}.$ (44)

As with the contribution ${\bf L}_{EM}^{A}$ the $+$ sign is for the string
along the $-z$-axis and the $-$ sign is for the string along the $+z$-axis.
For $x\to 0$ but $m=0$ ${\bf L}_{EM}^{B^{\theta}}=0$ since this term is
missing from the outset – see (34).

These limits confirm previous results. If $m=0$ to begin with and then ${\bf
L}_{EM}^{A}={\bf L}_{EM}^{B^{\theta}}=0$ while from (42) ${\bf
L}_{EM}^{point}=\pm qg{\bf\hat{z}}$ which agrees with ${\bf
L}_{EM}^{(0)}=qg{\bf\hat{R}}$ from (20). If $R=0$ but $m\neq 0$ then ${\bf
L}_{EM}^{point}=0$ while ${\bf L}_{EM}^{A}=\pm\frac{1}{4}qg{\bf\hat{z}}$ and
${\bf L}_{EM}^{B^{\theta}}=\pm\frac{3}{4}qg{\bf\hat{z}}$ from (43) and (44).
Adding these together gives ${\bf L}^{A+B^{\theta}}_{EM}=\pm qg{\bf\hat{z}}$
which agrees with the result in (23). In this case all the field angular
momentum comes from the ${\bf A}$ and ${\bf B}^{\theta}$ parts.

Finally, these limits also support the four cases when the electric charge is
displaced from the magnetic charge, as enumerated below equation (23). For
case (i) both the point and ${\bf A}$ plus ${\bf B}^{\theta}$ contributions
point in the $+z$ direction and have magnitudes $qg$ so that the results add
for the two parts to give $2qg{\bf\hat{z}}$ times an $e^{-mR}$ factor. For
case (ii) one changes the location of the electric charge which reverses the
direction of ${\bf L}_{EM}^{point}$ but not the direction of ${\bf
L}_{EM}^{A}$ or ${\bf L}_{EM}^{B^{\theta}}$, so that now the point and ${\bf
A}$ plus ${\bf B}^{\theta}$ contributions cancel giving ${\bf L}_{EM}=0$. Case
(iii) is simply case (ii) but with both the string and electric charge flipped
along the $z$-axis relative to case (ii). Thus for case (iii) the string and
point charges again cancel giving ${\bf L}_{EM}=0$. Finally, case (iv) is the
same as case (i) but with both the string and point contributions flipped
relative to the $z$-axis. The string and point charges again add giving ${\bf
L}_{EM}=-2qg{\bf\hat{z}}$ times an $e^{-mR}$ factor.