Why is it natural to make the electromagnetic field an antisymmetric, rank 2 tensor?

Question

As far as I know, the electromagnetic field strength tensor is defined to be the simplest object involving the electric and magnetic fields that transforms properly under Lorentz transformations. However, I don't get why such an object should be an antisymmetric rank $(2, 0)$ tensor, in a mathematically intuitive sense. (I guess you could say from a "geometric" perspective, but I'm not necessarily trying to visualize it, either.)

If we take the definition of the electric susceptibility tensor, for example, the fact that it is a $(1, 1)$ tensor is apparent from its definition. We assume that the polarization is linearly related to the electric field (as in $P^m = \epsilon_0\chi^m _n E^n$), and the most general transformation of this sort is a $(1, 1)$ tensor. Hence, the susceptibility tensor is naturally a rank $(1, 1)$ tensor. The symmetry of the tensor is also obvious, because we can always find principal axes along which the susceptibility is a scalar, and hence where the tensor is diagonal.

Such intuitiveness is not at all apparent for the electromagnetic tensor. There are a couple reasons I say this:

• Defining a tensor as a set of transforming components always leaves me completely unsatisfied. It's kind of like defining a vector as a tuple of components and a linear transformation as a matrix, which is equally ridiculous: many of the theorems of linear algebra are opaque and confusing when thinking about vectors as tuples, but are obviously true when thinking about vectors as linear objects (a simple example is the rank-nullity theorem). That's how the covariant formulation of electrodynamics feels to me, confusing and opaque, precisely because everything is in terms of components. It's for this reason that the argument "it's a rank $(2, 0)$ tensor because it works" fundamentally doesn't work for me.

  To summarize, I am not looking for explanations in terms of how the components transform.

• If we define a rank $(k, 0)$ tensor as a multilinear map \begin{align} T: V^* \times \cdots \times V^* \rightarrow \mathbb{R}, \end{align} the problems above all go away. There's just one problem: I have no idea how the electromagnetic field strength tensor is a multilinear map. The only two equations that I know of that directly involve the electromagnetic field tensor are $\partial_\mu F^{\mu \nu} = \mu_0 J^\nu$ and $K^\mu = qF^{\mu \nu} u_\mu$. The first one involves the tensor divergence, which doesn't seem to involve taking a covector as an input (although I'm really not experienced enough at tensor calculus to know). The second does take a covector, but it's the dual to the velocity. In that case, why isn't the field strength tensor a $(1, 1)$ tensor, to just take in... you know, the regular velocity? That might make a little bit more sense, but everything I've seen defines it either as a rank $(2, 0)$ tensor or as a rank $(0, 2)$ tensor. Why do they define it that way, what is there to gain in such a definition?

• There's also the question of antisymmetry in the tensor. What does that mean? Why should it be true? I'm attempting to study differential forms using Spivak's Calculus on Manifolds and any other resources I can find online, and antisymmetry for multilinear maps seems to be related to orientation. Why should the electromagnetic field strength tensor have anything to do with orientation (if it even does)?

So yeah, there's my main set of questions for the electromagnetic field strength tensor. I'm not looking for proofs of the tensor nature of the electromagnetic field; the proofs are actually the easiest part in this case. Essentially, all three of these issues boil down to the question of why an antisymmetric rank 2 tensor is the natural choice, not why it's the right one. If I were writing a covariant formulation of electrodynamics, why would it not only be right but obvious to choose such an object?

Answer 1

I think a very nice observation here is that if the force between field and current is $$ f^a = F^{a\mu} j_\mu $$ then in order for the force to be pure, i.e. rest-mass-preserving, it is sufficient that $F^{ab}$ be antisymmetric.

Proof. Let $u^a$ be the 4-velocity of a charge $q$. Then $$ j^a = q u^a $$ and $$ f^\lambda u_\lambda = q F^{\lambda \mu} u_\mu u _\lambda. $$ But $u_\mu u _\lambda$ is symmetric so we have that the result is zero if $F^{ab}$ is antisymmetric: $$ \mbox{if } \;\; F^{ab} = -F^{ba} \;\; \mbox{ then } \;\; f^\lambda u_\lambda = 0 $$ and this is the property which guarantees that the force is pure.

(And for completeness, here is the proof of that: $$ f^a = \frac{d}{d\tau}( m u^a ) = \frac{dm}{d\tau} u^a + m \frac{du^a}{d\tau} $$ so $$ f^a u_a = \frac{dm}{d\tau} u^a u_a + m \frac{du^a}{d\tau} u_a = -c^2 \frac{dm}{d\tau} $$ using that 4-acceleration is orthogonal to 4-velocity (and I used signature $(-1,1,1,1)$).)

Answer 2

Forget everything you know about electromagnetism for a moment. Newton's second law (for constant mass) $m\ddot{x}=F$ tells us in Galilean-invariant terms how a scalar is linearly driven by a force, meaning its second time derivative appears on the LHS. Let's try the same thing for a Lorentz-invariant theory of a vector, only we'll call the driving RHS a current, and won't worry about a mass coefficient.

We can't have $\ddot{A}^\nu=j^\nu$; we can have $\square A^\nu=j^\nu$, or less concisely $\partial_\mu(\partial^\mu A^\nu)=j^\nu$. Come to think of it, we can more generally have $\partial_\mu(\partial^\mu A^\nu-k\partial^\nu A^\mu)=j^\nu$; the indices are still as nice. Since we can absorb a multiplicative constant into the current's definition based on your preferred physical units, there's no room to generalize. Now the only real question is what we want $k$ to be.

Since$$\partial_\nu j^\nu=\partial_\nu\partial_\mu(\partial^\mu A^\nu-k\partial^\nu A^\mu)\propto 1-k$$(because derivatives commute), the choice $k=1$ is unique in giving us a conserved current. What's more, the resulting equation in $A$ is the sourced part of Maxwell's equations. So the antisymmetry of $F^{\mu\nu}$ means the electromagnetic vector is driven by a conserved current. Noether's theorem says there must be a symmetry. You probably already know what it is: $\delta A_\mu=\partial_\mu\phi$ preserves $F$.

Yang-Mills theory is then just a nonlinear generalization of this.

Answer 3

The physical effects of the EM field depend only on the integral of the vector potential around closed loops (Wilson loops). If spacetime is simply connected then by Stokes' theorem you can write all such integrals in terms of integrals around infinitesimal closed loops, i.e., the curl of the vector potential. The field tensor is the curl.

The vector potential, in turn, represents the differential $U(1)$ rotation as you move from a spacetime point to a nearby point. Gauge transformations change the zero point of the $U(1)$ angle at each spacetime point, which changes $A$ but doesn't change the net phase rotation around closed loops.

The field is really a bivector or 2-form, not an antisymmetric tensor as such.

As a tensor, it probably isn't written in mixed form just because it isn't manifestly antisymmetric that way.

Answer 4

Heuristically, I guess one can argue for the naturalness from the standard 3-vector formulation in terms of $\vec E$ and $\vec B$, which arguably feels natural for nonrelativistic people. (The OP does not want explanations based on component transformations, but I feel this answer is more about physical intuition than about explicit Lorentz transformations.)

• First, it is clear that under a transformation to a moving system, $\vec E$ and $\vec B$ and transform into each other because, e.g., nonmoving charges become currents. (Since these transformations are boosts, it furthermore makes sense that $\vec E$ and $\vec B$ are mixed among time and space components.)
• Hence, we're looking for a tensor object that fits the six components of $\vec E$ and $\vec B$. Here, an antisymmetric rank-two tensor seems the easiest choice.
• The electric field is naturally a three-vector, and that fits with $E_i=F_{0i}$.
• The magnetic field also a three-vector, but it actually is an axial vector, and it appears with a cross product in the Lorentz force. Thus, it is in some sense natural already in three dimensions to think of it as an antisymmetric tensor $f_{ij}=\epsilon_{ijk}B_k$. (It also seems reasonably that in two dimensions, the magnetic field has one component, while the electric field has two.)

Taking these points together, the antisymmetric two-tensor is a natural and simple choice to embed the three-dimensional fields $\vec E$ and $\vec B$.

Answer 5

So a lot has changed in two years. I have a lot more to learn about electromagnetism and physics in general, but I feel that I have finally found an explanation that makes sense to me. It takes bits and pieces from other answers, so thanks everyone!

The short answer to the question? It's not. The natural representation of the EM field is a differential 2-form. The reason it's thought of as a tensor field is because it's computationally simpler, and because differential forms can be embedded in tensor algebra as linear objects.

So why am I convinced that this is true? It comes down to the fact that you can derive the entire theory of electromagnetism from three very reasonable assumptions (on top of special relativity). From these assumptions, not only do Maxwell's equations and the Lorentz force law pop right out, but the nature of the EM field becomes obvious. They are the following:

1. There exist two conserved electric and magnetic charge currents.

In the language of differential forms in spacetime, this means there are two 3-forms $J_e$ and $J_m$ such that $d J_e = d J_m = 0$ (where $d$ is the exterior derivative, mapping $k$-forms to $(k+1)$-forms). But by the Poincare lemma (and the convexity of spacetime), we know that these forms are exact and can be written as the differential of 2-forms, i.e. $J_e = dF_m$ and $J_m = dF_e$ where $F_m$ and $F_e$ are 2-forms. (I'll call these the primitives of $J_e$ and $J_m$.)

2. Electricity and magnetism are duals of one another, in the sense that $\star F_e = F_m$.

Here I've used the Hodge star dual to represent duality; if we work in terms of the components of the forms, it swaps the spacelike and timelike parts that make them up. This essentially says that the electric and magnetic fields are interchangeable. From this we get $d \star F_e = J_e$ and $dF_e = J_m$, which is the differential form version of Maxwell's equations with magnetic charge.

3. There exists an electric and a magnetic force, each of which is linearly related to the four-velocity $u$ of the test particle, and to each respective primitive.

It is natural to consider the four-velocity as a vector (being the tangent vector of a worldline), and the force to be a 1-form (being the differential of a Lagrangian). We therefore seek linear maps taking $V \times \Omega^2(V) \to T_p \Omega^1(V)$, and the most general such map is the contraction $\lrcorner$ (up to scaling). We are therefore essentially saying that $$ f_e \propto u \lrcorner F_e, \\ f_m \propto u \lrcorner \star F_e. $$ Calling each proportionality constant $q_e$ and $q_m$, and adding the forces, we conclude that $$ f = q_e u \lrcorner F_e + q_m u \lrcorner \star F_e, $$ which is the differential form version of the Lorentz force law with magnetic charge. You can set $J_m = 0$ wherever it appears, and you get the normal equations.

So, this is an elegant formulation of electrodynamics, but why does this naturally explain the features of the electromagnetic field strength?

1. Why is it a 2-form at all? Because it's essentially the integral of the current density, which is naturally a 3-form.

2. Why is it antisymmetric? The exterior algebra is isomorphic to the quotient of the tensor algebra by the symmetric ideal. More geometrically, as the integral of the current density 3-form (which is highly related to orientation e.g. through the orientation of the surface element), it has a close relation to orientation.

3. Why not a $(1, 1)$-tensor? This difference is a bit more subtle, but it comes down to the faulty assumption that force is a vector. In fact, since force is most naturally the differential of the Lagrangian, it should be considered to be a 1-form. In this sense, a map which takes a vector and outputs a 1-form is precisely a 2-form which, in tensor notation, is a $(0, 2)$-tensor.

Answer 6

Maxwell’s equations sit in exact analogy to the equations of fluid flow. (Maxwell, Riemann and others even tried to describe the field as flows, rotations of the aether.) Antisymmetric tensors in 3D correspond to rotations, in 4D to rotations and flows. (While symmetric tensors describe compressions,extensions,shears.)

Answer 7

It's related to gauge symmetry and an Utiyama type theorem.

Contemporary understanding of the electromagnetic field is that its potentials have a gauge transformation, and that the dynamics for the field (and all other fields that interact with it) have symmetry with respect to this transform. In particular, if the potentials are denoted component-wise as $A_μ$, for $μ = 0,1,2,3$, then the transform is given by $ΔA_μ = ∂_μ χ$ for some function $χ$ of the coordinates $\left(x^μ: μ = 0,1,2,3\right)$, where $∂_μ$ denotes the derivative operator $∂/∂x^μ$.

So, now consider the following question: what is the most general dynamics given by a gauge-invariant first order Lagrangian density $𝔏$; that is: a gauge-invariant density $𝔏\left(A_μ, ∂_μ A_ν\right)$ that is a function only of the potentials and their first order derivatives? Define the following differential coefficients: $$𝔉^μ = \frac{∂𝔏}{∂A_μ}, \hspace 1em 𝔓^{μν} = \frac{∂𝔏}{∂\left(∂_μA_ν\right)}.$$ Then a variation in $𝔏$ induced by a variation $ΔA_μ$ in $A_μ$ along with the induced variation in its first order derivatives $Δ\left(∂_μ A_ν\right) = ∂_μ\left(ΔA_ν\right)$ yields the following variation in the Lagrangian density: $$\begin{align} Δ𝔏 &= 𝔉^μ ΔA_μ + 𝔓^{μν} Δ\left(∂_μ A_ν\right) \\ &= 𝔉^μ ΔA_μ + 𝔓^{μν} ∂_μ\left(ΔA_ν\right), \end{align}$$ where I'm using the summation convention. Applying this to the gauge transformation, we have: $$Δ𝔏 = 𝔉^μ ∂_μ χ + 𝔓^{μν} ∂_μ ∂_ν χ.$$ Noting that $∂_μ ∂_ν χ = ∂_ν ∂_μ χ$, we can just as well write the second term as: $$Δ𝔏 = 𝔉^μ ∂_μ χ + \frac 1 2 \left(𝔓^{μν} + 𝔓^{νμ}\right) ∂_μ ∂_ν χ.$$

The requirement that $Δ𝔏 = 0$, for all gauge transformations by (second order continuously differentiable) functions $χ$ means that the coefficients of $∂_μ χ$ and $∂_μ ∂_ν χ$ should separately be 0, or that: $$𝔉^μ = 0, \hspace 1em 𝔓^{μν} + 𝔓^{νμ} = 0.$$ Thus $𝔓^{μν} = -𝔓^{νμ}$. These are the conditions required for the differential coefficients of a Lagrangian density that has invariance under gauge transformation.

Substituting this back into the variational of the Lagrangian, we find that: $$\begin{align} Δ𝔏 &= 𝔓^{μν} Δ\left(∂_μ A_ν\right) \\ &= \frac 1 2 𝔓^{μν} Δ\left(∂_μ A_ν - ∂_ν A_μ\right), \end{align}$$ where in the second line, we use the anti-symmetry of $𝔓^{μν}$. Therefore, $𝔏\left(A_μ, ∂_μ A_ν\right)$ must be independent of $A_μ$ and have dependence on the gradients $∂_μ A_ν$ only through the combinations $F_{μν} = ∂_μ A_ν - ∂_ν A_μ$, thereby reducing to a function of the form $𝔏\left(F_{μν}\right)$.

Thus, the dynamics given by the Lagrangian density $𝔏$ may only involve the anti-symmetric tensor $F_{μν} = -F_{νμ}$, instead of the gradients outright.