How will SR EM Lagrangian change if we find a magnetic charge? When we introduce electromagnetic field in Special Relativity, we add a term of
$$-\frac e c A_idx^i$$
into Lagrangian. When we then derive equations of motion, we get the magnetic field that is defined as $$\vec H=\nabla\times\vec A.$$
If we now take divergence of both sides of this definition, we automatically get
$$\nabla\cdot\vec H=0,$$
which is equivalent to inexistence of magnetic charges.
But suppose we've found a magnetic charge. What will change in our Lagrangian or in definition of electric and magnetic fields in this case to make $\nabla\cdot\vec H=\sigma$?
In this Phys.SE answer it's asserted that magnetic field would get an additional term "gradient of a scalar potential". Is this "a" scalar potential instead "the" $A^0$ potential?
 A: In the absence of magnetic monopoles, Maxwell's equations are
$$
\begin{align}
            \text d F &= 0   ,\\
  \text d{\star F} &= J_e ,
\end{align}
$$
where $J$ is the 4-current 3-form due to electric charges (assuming a metric with signature $(-,+,+,+)$). For cohomological reasons, from the first equation one can asserts that there exists a 1-form $A$ such that $F = \text d A$, and $A$ is the interpreted as the 4-potential $(\phi,\mathbf A)$ (up to the musical isomorphism between tangent and cotangent bundle to Minkowski spacetime). In the presence of magnetic monopoles (or charge, to even symmetrise terminology) the above equations would become
$$
\begin{align}
         \text d F &= J_m ,\\
  \text d{\star F} &= J_e ,
\end{align}
$$
where $J_m$ is the 4-current for magnetic charges. Therefore in this extended theory of electrodynamics both the Faraday tensor $F$ and its Hodge dual $\star F$ (sometimes also denoted by $G$) figure in constitutive equations.
Since $F$ is no longer a closed form, its expression must be modified by the introduction of a non-exact part, say $C$, so that
$$F = \text d A + C.$$
Since the equations are symmetric in $F$ and $\star F$ we can postulate there exist 1-forms $B$ and $D$ such that
$$\star F = \text d B + D,$$
and assume that $C$ depends on $B$, while $D$ depends on $A$. But since $\star\star = -1$ in special relativity, we conclude that
$$F = \text dA - \star\text dB,$$
which can be related to the Helmholtz decomposition into polar and axial part for twice differentiable vector fields.
The Lorentz force for a particle with electric charge $q_e$ and magnetic charge $q_m$ would be
$$K = \iota_u(q_e F + q_m G),$$
where $u$ is the particle's 4-velocity vector and $\iota$ denotes the interior product. The extra term can then be reproduced with a Lagrangian containing the extra term $B_\mu u^\mu$.

To make contact with the usual vector notation, observe that the Faraday tensor has the covariant matrix representation
$$F = \begin{bmatrix}0&-~\mathbf E^T\\\mathbf E&\star\mathbf H\end{bmatrix}$$
where $\star\mathbf H$ is the Hodge dual of the magnetic field $\mathbf H$, and can be thought as the linear map $(\star\mathbf H)\mathbf v = \mathbf v\times\mathbf H$ for any $\mathbf v\in\mathbb R^3$. 
Skew-symmetric tensors as the one above are then represented by a polar vector $\mathbf E$ and an axial vector $\mathbf H$, and can be denoted as $F=(\mathbf E,\mathbf H)$. Having defined this notation, the action of the Hodge dual is then $\star(\mathbf E,\mathbf H) = (\mathbf H,-\mathbf E)$ (up to a sign which I can't be bothered remembering). The exterior derivative of the 4-current $A$ is a tensor of the form above, and it turns out that
$$\text dA = \left(\nabla A^0+\frac{\partial\mathbf A}{\partial t},\nabla\times\mathbf A\right),$$
where the first  component is the polar part and the second one is the axial part. Hence with no magnetic charges we recover the electric and magnetic fields. Now for the extra potential $B=(B^0,\mathbf B)$ we have, using the rule for the Hodge dual discussed a few lines above,
$$\star\text dB = \left(\nabla\times\mathbf B, - \nabla B^0 - \frac{\partial\mathbf B}{\partial t}\right)$$
Remark Here $\mathbf B$ is an extra vector potential, not to be confused with the magnetic induction.
Reconstructing the Faraday tensor according to the prescription $F=\text dA - \star\text dB$ given above we then have, in terms of polar and axial parts
$$F = \left(\nabla A^0 + \frac{\partial\mathbf A}{\partial t} - \nabla\times\mathbf B, \nabla\times\mathbf A + \nabla B^0+\frac{\partial\mathbf B}{\partial t}\right),$$
whence
$$\mathbf E = \nabla A^0 + \frac{\partial\mathbf A}{\partial t} - \nabla\times\mathbf B$$
and
$$\mathbf H = \nabla\times\mathbf A + \nabla B^0 + \frac{\partial\mathbf B}{\partial t}.$$
