Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

The Question

If the three vector electric and magnetic fields come from the four component four-potential, then is there a fourth component to the electric and magnetic field?

Related Question

I posted the following question: Express Maxwell's equations in terms of dipole field equations? . I remember along time ago I wrote Maxwell's equations down and I crossed the fields with the position vector and I was able to transform Maxwell's equations from a monopole charge source to something that appeared to be a dipole sourced field equations. When I did this it revealed a bizarre fourth component in the field equations which I think might be related to this question,

share|cite|improve this question

5 Answers 5

up vote 5 down vote accepted

To answer this question, you need to have a full geometric understanding of the Maxwell equations and what they represent.

Maxwell's equations are a garden-variety system of PDEs. In STA notation, it's simply

$$\nabla F = -J$$

We take it for granted that $F$ is a bivector, and thus has 6 components, and that $J$ is a vector, and thus has 4 components. But this equation describes up to eight separate equations. Why is that?

For an arbitrary bivector field $K$, the derivative $\nabla K$ can have both vector and trivector terms. That Maxwell's equations have only a vector source term is actually quite significant: this is part of the physical content of Maxwell's equations. We're saying the EM field is determined only by a vector current.

What would happen if there were a trivector current source term? It would be "magnetic" charge (magnetic monopoles) and an associated current. So right away we can appreciate what that source term would denote.

But wait, there's more! Let's say we had both electric and magnetic currents then. What kinds of fields could produce them?

As you've been trying to get at, these are the other two components of a field that could go into this differential equation. They are a scalar field and a pseudoscalar field. I'm not familiar with how these fields would manifest themselves, or what they would do.

So why don't we find out?

Let $\lambda$ be the scalar field. How would this affect Maxwell's equations with just a current source term?

Let $F = e_0 E + B$, where I've implicitly denoted that the magnetic field is a bivector. You can identify it as a vector instead and then consider $\epsilon_3 B$, but the net effect is pretty minimal.

Maxwell's equations then break down as

$$\nabla \cdot F = \partial_t E - e_0 \nabla_3 \cdot E + \nabla_3 \cdot B = -\rho e_0 - j$$


$$\nabla \wedge F = -e_0 \nabla_3 \wedge E - e_0 \partial_t B + \nabla_3 \wedge B = 0$$

Adding a scalar field $\lambda$ would affect only the vector part, with its gradient:

$$-J = \nabla \cdot F - e_0 \partial_t \lambda + \nabla_3 \lambda$$

So all in all, this would probably appear as some kind of extra current not associated with the motions of electric charges--or perhaps it would be indistinguishable from electric currents in some way, save that it permeates all space as a continuous function. It would look like there are some currents everywhere, in some sense. You can see why we don't even consider the existence of such a field. Unless it's very small, we would've detected it some time ago, since it interacts with the electric current source term.

An analysis of the magnetic pseudoscalar field would probably end up the same way.

So, is the Faraday tensor actually missing two extra components, a scalar field and a pseudoscalar field? I'd say no, but if you discover otherwise, you'll probably win a Nobel Prize. Good luck with that. As I said in the other question, do not be fooled into thinking that, just because $Fx$ has eight components, there are missing components of the Faraday field. There very likely aren't any such missing components. You can see this by considering what those components would do in the vanilla Maxwell equations, as I have done here.

Edit: some corrections on the relationship between this scalar field and gauge fixing.

This scalar field would remove the freedom to change $A$ through gauge transformations, as $\lambda$ would specify the divergence of $A$. Recall that gauge fixing relies upon the ability to perform the transformation,

$$A \mapsto A + \nabla \chi$$

For some scalar field $\chi$. This can be done because $\nabla \wedge (A + \nabla \chi) = \nabla \wedge A = F$, so the EM field is unchanged.

But if $\nabla \cdot A = \lambda$, then adding the gradient of a scalar field would change the value of $\lambda$ measurably, in all but the simplest cases:

$$\nabla \cdot (A + \nabla \chi) = \lambda + \nabla^2 \chi$$

Now, you would be restricted to gauge functions $\chi$ that are strictly harmonic. Harmonic fields are usually those that arise from some choice of boundary conditions--i.e. this would correspond to some choice of boundary condition, and the field contribution from currents would be unchanged. Of course, it strains imagination to consider how one would reasonably do this for gauge fixing. And if you found a transformation that preserves $\lambda$, it would not leave $F$ invariant in general.

So the supposed existence of this function $\lambda$ would have profound consequences toward gauge fixing. It does not absolutely forbid it as I originally thought, but it puts serious constraints on fixing that we probably would've run into by now.

share|cite|improve this answer
Thank you for providing an excellent answer. It sounds like this extra scalar term that could appear in the electromagnetic field breaks charge conservation? Is it possible that this scalar term could be described in terms of the four potential? – linuxfreebird Mar 19 '14 at 8:06
+1. This answer makes me want to spend more time studying geometric algebra. – Kyle Kanos Mar 19 '14 at 12:43
@linuxfreebird: Yes, this scalar field would completely specify $\nabla A$, and then by the Minkowski analogue of the Helmholtz theorem, that would specify $A$ up to boundary conditions. The existence of such a scalar field would be manifest in breaking the ability to do gauge fixing. For instance, you ordinarily could use the Lorenz gauge by setting $\nabla (\nabla \cdot A) = 0$. But here, $\lambda = \nabla \cdot A$, and if $\lambda$ were to exist, you would have no a priori reason to set its gradient (and thus, its associated current terms) to zero. – Muphrid Mar 19 '14 at 13:37
I should say $\lambda$ would specify $\nabla \cdot A$, and that in addition to $F = \nabla \wedge A$ would completely specify $A$ up to boundary conditions. – Muphrid Mar 19 '14 at 13:47
So the fourth component of the electromagnetic field is the Lorentz gauge term? The Lorentz gauge term is set to zero for gauge fixing purposes and to satisfy the U(1) conditions? I think that makes sense. – linuxfreebird Mar 19 '14 at 13:48

Actually, the electric and magnetic fields from one combined tensor called the electromagnetic field tensor. This is a rank-2 tensor and takes the form* $$ F^{\mu\nu}=\left(\begin{array}{cccc} 0 & -E_x & -E_y & -E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0 \end{array}\right) $$ It has the following properties:

  1. It is anti-symmetric (so $F^{12}=-F^{21}$)
  2. It is traceless
  3. It has 16 elements, but only 6 distinct values
  4. When multiplied by its dual tensor ($G^{\mu\nu}$) it gives a Lorentz invariant value of $4\mathbf{B}\cdot\mathbf{E}$
  5. The inner product, $F_{\mu\nu}F^{\mu\nu}=2(B^2-E^2)$, is also a Lorentz invariant

You can also derive Maxwell's equations through the tensor by applying $\partial_\mu$ to it. Gauss's law and Ampere's law come from $$ \partial_\mu F^{\mu\nu}=4\pi J^\nu $$ where $J^\mu=\left(\rho,\,\mathbf{j}\right)$ is the four-current. The Dirac equation and Faraday's law come from applying the Bianchi identity to get $$ \partial_\gamma F_{\mu\nu} + \partial_\mu F_{\nu\gamma} + \partial_\nu F_{\gamma\mu}=0 $$ Or more concisely, $$\partial_{[\mu}F_{\nu\gamma]}=0$$

I am an astrophysicist, so I use cgs units; in SI, all electric fields have a factor of $1/c$.

share|cite|improve this answer
It's not missing components. This tensor is derived through $F^{\mu\nu}=\partial^\mu A^\nu-\partial^\nu A^\mu$. When $\mu=\nu$, then $F^{\mu\mu}=0$ by necessity. – Kyle Kanos Mar 15 '14 at 1:31
Well, time components are given by the $0$ index, so $F^{0\nu}$ and $F^{\mu0}$ return the electric field. It seems, then that the electric field is the temporal component. Chris White's answer here might also interest you. – Kyle Kanos Mar 15 '14 at 1:37
I finally was able to answer the following question:… Please investigate this question, because there is a fourth component in the field. – linuxfreebird Mar 18 '14 at 15:53
Shouldn't $J^\mu$ have a $\nu$ superscript instead? – Ruslan Mar 19 '14 at 11:06
@ruslan: yes it should, thanks for catching that, it has been fixed now. – Kyle Kanos Mar 19 '14 at 11:21

This is more an extended comment to address the comments to Kyle's answer

For example if there was a time component to the electric and magnetic field

In a relativistic context, the electric and magnetic field components are not components of separate, related vector fields but, rather, are components of a 2nd rank tensor field; the electric and magnetic fields are part of one geometric object, not two.

Indeed, a clue to this is found in the fact that the magnetic field is, in 3D, a pseudo-vector field rather than a vector field.

So, in fact, the question "what is the time component of the electric and magnetic field" actually presumes a falsehood; it presumes that the electric and magnetic fields are separate but related four-vectors.

But, they're not.

Since a rank 2 tensor has two indices, we can properly speak of the time-time component, the time-space components, and the space-space components of the tensor but not the time or the space component(s).

share|cite|improve this answer
I finally was able to answer the following question:… Please investigate this question, because there is a fourth component in the field. – linuxfreebird Mar 18 '14 at 15:54

The Faraday $2$-tensor $F$ describing the electromagnetic field strength is antisymmetric, so there is no room in $1\!+\!3$ dimensions for any extra components besides the $3$ electric and $3$ magnetic ones. This is straightforward, but why is it antisymmetric?

A hint towards the nature of the electromagnetic field can be found by expressing Maxwell's equations in terms of both the Faraday tensor $F$ and the Maxwell tensor $H$: $$\begin{eqnarray*}\partial_{[i}F_{jk]} = 0\text{,}&\quad&\partial_{[i}H_{jk]} = J_{ijk} =_\text{def}\frac{1}{3!}\epsilon_{ijkl}J^l\text{.}\end{eqnarray*}$$ Physically, $F \Leftrightarrow (\mathbf{E},\mathbf{B})$ is the field strength and $H \Leftrightarrow (\mathbf{D},\mathbf{H})$ is the electromagnetic excitation. The former is gives Gauss's law for magnetism and Faraday's law of induction, while latter gives the Gauss's and Ampère-Maxwell laws.

What's notable about this form is that none of Maxwell's equations care about the metric of spacetime at all. Rather, the metric appears as part of the Hodge star in a separate law linking the Maxwell and Faraday tensors: $$H \propto \star F\text{,}$$ a good way to interpret which is that it is a particularly simple constitutive relation giving the dielectric and magnetic properties of spacetime. For example, both the Born-Infeld nonlinear electrodynamics can be described by an alternative constitutive relation while keeping the above Maxwell's equations the same, and so can the effects of first-order QED vacuum corrections derived by Heisenberg and Euler.

If the three vector electric and magnetic fields come from the four component four-potential, then is there a fourth component to the electric and magnetic field?

Now let's flip the above observation into an argument. Electromagnetism is not gravity, so while we might need the metric at some point to turn it into a fully predictive theory, we should be able to put the equations describing the electromagnetic field itself into a form independent of both the metric and the connection. Therefore, the equations of electromagnetism should still make sense even if spacetime had neither a metric nor a connection. What's left besides topology is differential structure.

Conclusion: Electromagnetism must be describable by differential forms, and differential forms correspond to covariant antisymmetric tensors. In $n$ dimensions, the number of independent components of a $k$-form is $C(n,k)$. Thus, if we know that the electric and magnetic fields intermix in transformations across frames and so must be parts of the same tensor/$k$-form, the total number of components for $n=4$ must be one of $\{1,4,6\}$.

Having $4$ for both the electric and magnetic field would make $8$, so that's a no-go.

By the way, if we also already know that the field strength $2$-form has a potential, $F = \mathrm{d}A$, then $A$ must be a $1$-form and have four independent components... but we shouldn't have expected all that apparent freedom to be physical in the first place, because $A\mapsto A+\chi$ for any $1$-form $\chi$ with $\mathrm{d}\chi = 0$ produces the same $F$. Note that $F = \mathrm{d}A$ implies that $\mathrm{d}F = 0$, which is Gauss's law for magnetism and Faraday's law of induction.

share|cite|improve this answer
I finally was able to answer the following question:… Please investigate this question, because there is a fourth component in the field. – linuxfreebird Mar 18 '14 at 15:54

The answers provided by Alfred Centauri and Kyle Kanos contain factual statements, but there is more hidden geometry in the problem. It is true that the electric and magnetic field vectors are not true vectors, but rather pseudo-vectors as stated by Alfred Centauri that belong to rank 2 field tensors as stated by Kyle Kanos. However, Kyle Kanos did mention the fields come from the exterior calculation $\partial^\mu A^{\nu} - \partial^{\nu}A^{\mu}$, which suggests that the EM-fields can be understood as possible bivectors of four-dimensional space Using the remapping of a hodge dual, one might be able to remap the bivector components to four components of a four-vector of space-time as a way to condense the notation. This would imply a possible forth component to the EM-field.

share|cite|improve this answer
The bivector link includes a discussion on the E&M field tensor, showing that the time-like component is the electric field, just as I had stated. – Kyle Kanos Mar 15 '14 at 2:00
The hodge dual in 4 dimensions just takes bivectors to bivectors. In this case, it just gives you the dual EM field tensor. – Muphrid Mar 15 '14 at 6:02
I posted an additional question related to this one concerning the two electromagnetic tensors:… – linuxfreebird Mar 15 '14 at 14:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.