Can you depict the electron's electromagnetic field?

Question

I've had a number of discussions with individuals about electromagnetism. A recurring issue concerns the distinction between field and force, and seems to be associated with the lack of unification in contemporary teaching. As we all know, Maxwell unified electricity and magnetism a hundred and fifty years ago, but IMHO it sometimes feels like it never happened.

I think my concern is best signalled by the lack of any depictions of the electromagnetic field. For example we can see this depiction of the electric field in Andrew Duffy's Physics 106 course, where we can also read "lines of force are also called field lines":

And we can see this GNUFDL picture of a magnetic field by user Stannered on Wikipedia:

But there aren't any depictions of the electromagnetic field. Why not? Note section 11.10 of Jackson's Classical Electrodynamics where he says "one should properly speak of the electromagnetic field Fμν rather than E or B separately". I'm sure most readers would agree with this. But then surely one should properly depict the electromagnetic field Fμν rather than E or B separately. So, can you? You can probably depict a gravitomagnetic field. But can you depict the electron's electromagnetic field? And if not, why not?

Answer 1

I don't think that nobody seriously thinks that unification of electricity and magnetism never happened or that it is unimportant. You just have to look at the standard model, and you'll find a term in the lagrangian (in units with $\hbar = c = \varepsilon_0 = 1 $): $$\mathcal{L}_{EM} = -\frac{1}{4}F^{\mu\nu}F_{\mu\nu}$$See? No electric or magnetic fields, but the electromagnetic tensor.

But you also have to be aware that the electric and magnetic fields can always be calculated from the electromagnetic tensor $$E^i = -F^{0i} \qquad B^i = -\frac{1}{2}\epsilon_{ijk}F^{jk}$$ The fact that you have a more powerful theory doesn't force you to use it in every occasion you can. For example, in this question you can solve the problem in an esasy way if you go to the reference frame where $\vec{B}=0$ or you can solve it in an involved way calculating the electromagnetic tensor in the original reference frame. Both are correct and you obtain the same results, but one of them is way easier. Do you prefer the complicated way because you have a better feeling of the physics of the problem? Perfect, but that doesn't disprove other equally valid approaches.

Now, to the visualization part. That is a bit tricky, because it involves both physics and perception and intuition. When you depict an vector field, let's say the velocity field in a fluid, you're conveying several things. First of all, of course, you're asigning three number to every point in space. But any collection of three numbers don't form a vector field: if you perform a rotation, the three numbers transform in a non-trivial way, they transform in the fundamental representation of the SO(3) group. If you depict them as arrows and rotate the figure, the arrows will rotate in the same way as the vector field does. So using arrows makes sense.

But you run in trouble when you try to depict things within a larger group, namely the Lorentz group SO(1,3). I can imagine rotating a diagram, but imaging boosting it is a little harder (at least for me). If you try to depict an antisymmetric second-order tensor like the electromagnetic tensor, you have to represent 6 independent degrees of freedom plus the way they transform under arbitrary Lorentz transforms (the example given by @CuriousOne in the comments doesn't seem too clear to me, because I don't know how color transforms into shape or orientation when rotating or boosting the figure). As you can see, this is a lot of information to put in a single diagram.

What it is usually done is forgetting about boosts (for depicting purposes only) and considering how the electromagnetic field transforms under the subgroup of rotations $SO(3) \subset SO(1,3) $. It isn't hard too see that three components transform as a vector and the other three as a pseudovector (yes, electric and magnetic fields again!). So the best way to depict the electromagnetic field, IMHO, is as two set of vector fields, keeping in mind that they are not independent, but they will mix together if you boost th whole thing.

I know that this is not the answer you're looking for, but sometimes we have to admit the limits to our perception and our ability to make nice sketches. But in the end, equations are the most reliable method to convey all the information, and they never lie!

ADDENDUM:

Reading your comments in the answers given to you, I think that the cause of your doubt is that you're mixing several concepts:

• Field lines: In a vector field, field lines are a set of curves whose tangent vectors at every point form the vector field. Note that filed lines are a mathematical concept, even though in some special cases we can attach them some physical significance. Note also that field lines can be defined for vector fields, but $F_{\mu\nu}$ is a tensor field, not a vector field. So, you can't define filed lines for the electromagnetic tensor (but you can for electric and magnetic fields!)
• Force lines: A special case of field lines, where the vector field is the force felt by a test particle. In a purely electrostatic setting, filed lines and force lines are equal (up to a constant) $$\vec{F}(\vec{r}) = q \vec{E}(\vec{r})$$ so you can use them interchangeably. But when magnetic fields are present, the force in a test particle is the Lorentz force $$\vec{F}(\vec{r}, \vec{v}) = q\vec{E}(\vec{r}) + q\vec{v}\times\vec{B}(\vec{r})$$ or if you prefer a more unified fashion $$\frac{dp_\mu}{d\tau} = q F_{\mu\nu}\frac{dx^\nu}{d\tau} = q F_{\mu\nu} u^\nu$$The force now depends on the speed of the particle [to be exact, the rate of change of the four-momentum depends on the four-velocity of the particle]: you can no longer associate a force to every point of space in a univocal way, so you can't define force lines. In your second figure, a particle moving in the direction of the wire will feel a force perpendicular to its velocity and to the magnetic field, and therefore will be attracted/repelled to the wire. But a particle moving in the direction of the magnetic field, in the same point of space, won't feel any magnetic force at all.
• Depicting the electromagnetic field: The electromagnetic field is a bunch of numbers defined at every point that determine (following the equations given above) the dynamics of charged particles. They belong together because they are not independient, they mix when you rotate or boost your system. Anyway you try to depict them must reflect this mixing. This is the question that you posed (even though not what you meant, according to your comments), and this is what I tried to explain in the first version of my answer. Maybe you should ask yourself what is the real physical significance of the electromagnetic field, and what is the physical significance that you're trying to attach it.

Answer 2

This sketch might help ;)

It is an interactive picture of a charged particle in a plane, with the electric field lines and projections of the magnetic field lines also drawn. Here is what I thought about while making this visualization. The original question had depictions of an electric field due to a static charge and a magnetic field due to a current. My understanding of the situation (which I stole from Bosoneando haha) was that Coulomb's law and the Biot-Savart law are inadequate to describe situations where the light speed limit is important, and static pictures are not so good at conveying the subtleties of what happens with a field in that limit. Because position, velocity, and a particle's recent history all need to be taken into account to properly describe the classical electromagnetic field surrounding the particle, I felt an interactive visualization might best capture the physics of the situation.

Answer 3

I know of another geometric interpretation of the electromagnetic field that is invariant under Lorentz transformations and more general transformations of coordinates. It is not well known and requires you to think of two-dimensional surfaces in four-dimensional spacetime. But it can be visualized (as easily as any visualization of spacetime), is coordinate independent, and gives you all of the information about the field tensor using families of surfaces rather than numbers.

Integrable Surfaces

First let's consider the special case of a field where everywhere the electric field vector is perpendicular to the magnetic field vector. If we consider the four-vector $B^\mu$ with time component 0 and space components equal to the three-dimensional magnetic field vector, then it is annihilated by the electromagnetic field tensor $$F_{\mu\nu}B^\nu = 0.$$ Due to the antisymmetry of $F$ there must be another direction $u^\mu$ that is also annihilated. If we choose it to be orthogonal to $B$ it will be in the direction of the Poynting vector.

Now in some frame at some time slice consider the magnetic field. This three-dimensional space is foliated by all of the curves of the magnetic field lines. As time passes the magnetic field lines will change. And in fact we can consistently think of each field line like a string moving in time with four-velocity $u^\mu$, the fact that this works ultimately comes from the Frobenius theorem. So each magnetic field line traces out a two-dimensional sheet as it moves in four-dimensional spacetime. And the foliation of space-time by these sheets is coordinate independent.

This in itself doesn't fully specify the electromagnetic field, we need one more scalar that specifies the magnetic flux, i.e. how many of these magnetic field line sheets are intersecting a given spatial surface at a given point in spacetime. In fact if you understand differential forms the whole electromagnetic field tensor can be thought of as an area form for this flux. It is simple to depict this aspect in the usual way by making magnetic field line sheets denser in regions where the magnetic flux is higher.

What I have been describing is the case where the two directions annihilated by the field tensor are the magnetic field vector and a Poynting four-vector, which may be (1,0,0,0) if it's a static magnetic field. The two-dimensional sheets are tangent to these two directions. But what if we are instead dealing with a static electric field like for a point charge? There is no magnetic field but there are still two directions annihilated by the field tensor - the two spatial directions perpendicular to the electric field. The surfaces tangent to these are just the familiar equipotential surfaces. If you boost the point charge out of rest you will get a non-zero magnetic field which will be tangent to the surface, so the only difference from the previous case is that the other direction $u$ tangent to the surface is a spacelike rather than timelike. As before, this family of surfaces in spacetime tells you the `direction' of the field and the density of the surfaces tells you the 'magnitude.'

General Electromagnetic Fields

Now for a general field it is no longer true that the electric field is perpendicular to the magnetic field, and it is no longer true that spacetime is foliated by integrable magnetic field line sheets or equipotentials. But there is a simple modification. The electric field being perpendicular to the magnetic field is equivalent to saying the electromagnetic field can be written in terms of scalar fields $X$ and $Y$ as follows, $$ F_{\mu\nu} = \partial_\mu X \partial_\nu Y - \partial_\mu Y \partial_\nu X.\quad[1]$$

In general the electromagnetic field comes from the gauge potential $A_\mu$ $$ F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu.$$

And in general a four vector can be written in terms of four scalars via Darboux's theorem $$A_\mu = X_1 \partial_\mu Y_1 + X_2 \partial_\mu Y_2$$

So a general field tensor can be written in terms of two tensors, each with the form of equation 1. And we can associated a family of two dimensional sheets to each tensor.

Summary

The electromagnetic field can be represented by two families of two-dimensional surfaces in four-dimensional spacetime. If we pick an appropriate three-dimensional slice, we will see two families of one-dimensional field lines. These are not the electric and magnetic fields. If you add up the flux of both familes through a surface it will give you the integral of the electromagnetic field (as a two-form) over that surface. If the electric field is everywhere perpendicular to the magnetic field, only one family of surfaces is required and they are just the sheets traced out by magnetic field lines in spacetime (or equipotentials if the magnetic field vanishes).

Answer 4

I personally struggle with visualizing the electron's electromagnetic field $F_{\mu\nu}$ at a stroke because there are too many variables involved: six interrelated quantities at each point in space. However, as Bosoneando points out, those six quantities are trivially related to the usual electric and magnetic fields by \begin{align} E_i &= -F_{0i} & B_i &= \frac12 \epsilon_{ijk} F^{ij} \end{align} if you'll permit me a little sloppiness with raised vs. lowered indices. That means it's completely correct to visualize $F_{\mu\nu}$ as two vector fields in the volume surrounding the electron.

The electric field is given by the monopole term, as you point out. The magnetic field is not given by the circulate-about-the-wire field in your second image (v1 of your post): that is instead the field of a current, or the superposition of the field of many moving electrons. The magnetic field of a stationary electron is the field of a magnetic dipole. Unlike most dipole field illustrations, the electron's magnetic field has the "far-field" dipole shape at all length scales; the electron's magnetic dipole moment seems to be an intrinsic property of the particle, rather than a consequence of some internal structure.

A satisfying way to turn these field visualizations into computational tools that are more intuitive than "shut up and calculate" is to recall the energy density of an electromagnetic field, $$ U = \frac12\left( \epsilon_0 E^2 + \frac1{\mu_0}B^2 \right), $$ and remember that systems tend to evolve towards states with lower total energy. An electrodynamical system will therefore tend to minimize the volume of strong field. This, with your first diagram, makes it obvious why opposite charges are attracted to each other: an electric dipole field due to two separated charges has more nonzero field volume than the field due to charges which are superimposed. In a uniform magnetic field a dipole will experience a torque which tends to align the dipole with the external field, which produces a minimum in the combined magnetic field in the plane of the dipole. A magnetic dipole in a non-uniform magnetic field will still experience this torque, but also force in the direction of the gradient of the magnetic field strength: an aligned dipole wants to live where the magnetic field is at its strongest, in order to cancel as much magnetic field energy as possible, while an anti-aligned dipole wants to live where the magnetic field is weak. You have experienced this interaction with toy magnets: they like to repel each other, but they really like to flip around and attract each other.

Answer 5

You can certainly draw pictures which contain both the $\vec E$ and $\vec B$ field lines in different colors, and together they specify the electromagnetic field. The only problem is that, in this form, the vector fields themselves are not Lorentz-covariant so if you, say, assume that the electron's stationary E-field and zero B-field are the same in an boosted reference frame, then you will predict that moving charges never generate magnetic fields, which is very wrong.

But you see these sorts of illustrations for example on this google search.

The best alternative solution is probably depicting the 4-potential $A^\mu$, which also completely specifies the electromagnetic field: you depict $\phi$ with a color gradient and $\vec A$ with lines. Probably the reason why this is not done so much is that it has a similar problem: you have to know how the scalar field turns into the vector field under Lorentz transforms. Another reason is that, due to $\nabla\cdot \vec E = \rho$, there is a straightforward way to say that $\vec E$ "comes out of positive charges" and has continuous field lines otherwise. The analogous way of drawing field lines for $\vec A$ obeys the Lorentz gauge condition that $\nabla\cdot\vec A=-\dot\phi/c$, so $\vec A$ needs to come out of places where $\phi$ is decreasing, and into places where it is increasing.

Actually that picture probably looks pretty cool; the travelling electron now has a reduced $\phi$ due to its direct Lorentz contraction (does it also get a length contraction? It might appear "squished", too), and that part "bleeds out" into a cloud of $\vec A$ which points in the direction the electron is going. But it's not a common visualization by any means.