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?

  • $\begingroup$ I think your first expression is wrong - shouldn't it be $A_\mu J^\nu \to q A_\mu U^\mu = q \gamma ({\bf A} \cdot {\bf v} - \phi)$ for point-particle sources? $\endgroup$
    – tparker
    Commented Jul 5, 2017 at 18:09
  • $\begingroup$ @tparker it's not wrong, but yours is neither. Mine is just in ${}+{}-{}-{}-{}$ metric signature, unlike yours. $\endgroup$
    – Ruslan
    Commented Jul 5, 2017 at 18:39
  • $\begingroup$ What? No, I wasn't talking about the sign. $\gamma({\bf A} \cdot {\bf v} - \phi)$ is not the same thing as ${\bf A} \cdot d\bf{x}$. What does "$d{\bf x}$" even mean in a Lagrangian? $\endgroup$
    – tparker
    Commented Jul 5, 2017 at 18:42
  • $\begingroup$ @tparker $x^i$ is the point on the world line of a particle. The expression $A_i dx^i$ expands as $A^0dt-\mathbf Ad\mathbf r$, where $d\mathbf r\equiv \mathbf v dt$. After you expand $dt$ in terms of $ds$ you'll get the $\gamma$. My notation is from Landau&Lifshitz vol. II $\S16$. $\endgroup$
    – Ruslan
    Commented Jul 5, 2017 at 18:51
  • $\begingroup$ Ah, got it. I was assuming the convention that Roman letters only run over spatial indices. Except I think in your notation it should be $$-\frac{e}{c} A_i \frac{dx^i}{d\tau},$$ where $\tau$ is the particle's proper time. Your expression doesn't have the units of a Lagrangian density. $\endgroup$
    – tparker
    Commented Jul 5, 2017 at 19:28

1 Answer 1


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 $$\mathbf H = \nabla\times\mathbf A + \nabla B^0 + \frac{\partial\mathbf B}{\partial t}.$$

  • 1
    $\begingroup$ Hmm, can I ask you to rewrite this in simpler terms, namely in components and without these scary stars, of which I have no idea (I've not studied differential geometry)?.. $\endgroup$
    – Ruslan
    Commented Feb 9, 2015 at 12:52
  • $\begingroup$ Thanks, this is much better. But how would the $-\frac e c A_i dx^i$ term in the Lagrangian change with this extra $B$ potential? $\endgroup$
    – Ruslan
    Commented Feb 9, 2015 at 14:36
  • $\begingroup$ Shouldn't your matrix have $\star \mathbf H$ not $\star\mathbf B$? $\endgroup$
    – Ryan Unger
    Commented Feb 9, 2015 at 14:37
  • $\begingroup$ @Ruslan please see the edits $\endgroup$
    – Phoenix87
    Commented Feb 9, 2015 at 15:02
  • 1
    $\begingroup$ @WetSavannaAnimalakaRodVance I think you are right about the signature. Although I usually prefer the other convention, I think the equation $dG = J$ is correct in the (-,+,+,+) signature. With this in mind, the matrix representation of F is the covariant one, i.e. that of $F_{\mu\nu}$. $\endgroup$
    – Phoenix87
    Commented Jul 22, 2017 at 8:26

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