Relation between the Faraday 2–form and the Faraday tensor

by Stéphane Haussler

Warning

under construction

I have personally been very confused by the relationship between the Faraday tensor and the Faraday 2–form. In particular, I was confused about the components of the Faraday tensor \(F_{μν}\) and the Faraday 2–form, also noted \(F_{μν}\). In particular, I was confused about the components obtained in row/column representation, their symmetries, especially when lowering or raising indices. I somehow expected to always obtain the same components, which is the case for the fully covariant or fully contravariant case, but is not the case in mixed form. After all, 2–forms transform like tensors. Maybe this is obvious to most, but maybe helpful to few. Hence, in this page, I explore the relationship between the electromagnetic field tensor (Faraday tensor) and the Faraday 2–form that I have derived from Maxwell’s equations:

• Deriving the Faraday tensor from the 1865 Maxwell’s equations

• Deriving the Faraday 2–form from the 1865 Maxwell’s equations

In particular, we examine the effects of raising and lowering indices on both the Faraday 2-form, and the Faraday tensor. The core distinction lies in their mathematical structure. A tensor can be expressed in a tensor basis \(dx^μ ⊗ dx^ν\), and the overall tensor object \(T\) \(T=T_{μν} dx^μ ⊗ dx^ν\) can be symmetric, or antisymmetric, or combinations of both. Really, here we do nothing else than separate a rank 2 tensor into its symmetric and its antisymmetric part:

\[\begin{split}T_{μν} \: dx^μ ⊗ dx^ν =& \frac{1}{2} \: T_{μν}\left( dx^μ ⊗ dx^ν + dx^ν ⊗ dx^μ \right) + \\ & \frac{1}{2} \: T_{μν}\left( dx^μ ⊗ dx^ν - dx^ν ⊗ dx^μ \right)\end{split}\]

The tensor \(T\) is then antisymmetric if the tensor components \(T_{μν}\) are antisymmetric. In contrast, 2–forms are antisymmetric by nature and built from tensors by taking the antisymmetric part. The 2–form basis can be expressed combinations of tensor elements built to take the antisymmetric part \(T_{μν} \: dx^μ ⊗ dx^ν - dx^ν ⊗ dx^μ\).

In abstract index notation, the doubly covariant Faraday tensor is expressed without its tensor basis as:

\[F_{μν}\]

With explicit tensor basis and Einstein summation notation, this object is written as:

\[F_{μν} \: dx^μ ⊗ dx^ν\]

The Faraday 2–form is expressed as:

\[F^{♭♭} = \frac{1}{2} F_{μν} \: dx^μ ∧ dx^ν\]

Where we can rewrite the exterior product as a tensor product:

\[F^{♭♭} = \frac{1}{2} F_{μν} \: \left( dx^μ ⊗ dx^ν - dx^ν ⊗ dx^μ \right)\]

Since the components of the Faraday tensor are antisymmetric (\(F_{μν}=-F_{νμ}\)), we have:

\[\begin{split}F^{♭♭} &= \frac{1}{2} F_{μν} \: dx^μ ⊗ dx^ν - \frac{1}{2} F_{νμ} \: dx^ν ⊗ dx^μ \\ &= \frac{1}{2} F_{μν} \: dx^μ ⊗ dx^ν + \frac{1}{2} F_{μν} \: dx^μ ⊗ dx^ν \\ &= F_{μν} \: dx^μ ⊗ dx^ν\end{split}\]

The fundamental difference between the Faraday tensor and the Faraday 2-form is that the Faraday tensor innerhently has 4 ⨯ 4 = 16 components, of which only 6 are independent when taking into account the symmetries. The Faraday tensor is written with 16 components in row-column representation as:

Faraday tensor

\[\begin{split}\begin{bmatrix} & + \E^x \: dx ⊗ dt & + \E^y \: dy ⊗ dt & + \E^z \: dz ⊗ dt \\ - \E^x \: dt ⊗ dx & & - B^z \: dy ⊗ dx & + B^y \: dz ⊗ dx \\ - \E^y \: dt ⊗ dy & + B^z \: dx ⊗ dy & & - B^x \: dz ⊗ dy \\ - \E^z \: dt ⊗ dz & - B^y \: dx ⊗ dz & + B^x \: dy ⊗ dz & \\ \end{bmatrix}\end{split}\]

In contrast, the Faraday 2–form inherently has 6 independent components and can be written as a column of 6 elements:

Faraday 2–form

\[\begin{split}\left[ \begin{aligned} - & \E^x \: dt ∧ dx \\ - & \E^y \: dt ∧ dy \\ - & \E^z \: dt ∧ dz \\ & B^x \: dy ∧ dz \\ & B^y \: dz ∧ dx \\ & B^z \: dx ∧ dy \\ \end{aligned} \right]\end{split}\]

From the Faraday 2–form to the doubly covariant Faraday tensor

We can decompose the exterior product into tensor products:

\[\begin{split}F^{♭♭} = \left[ \begin{alignedat}{2} - & \E^x \: ( dt ⊗ dx &-& dx ⊗ dt ) \\ - & \E^y \: ( dt ⊗ dy &-& dy ⊗ dt ) \\ - & \E^z \: ( dt ⊗ dz &-& dz ⊗ dt ) \\ & B^x \: ( dy ⊗ dz &-& dz ⊗ dy ) \\ & B^y \: ( dz ⊗ dx &-& dx ⊗ dz ) \\ & B^z \: ( dx ⊗ dy &-& dy ⊗ dx ) \\ \end{alignedat} \right]\end{split}\]

Rerorganizing the terms with a row/column representation, we get the Faraday tensor:

\[\begin{split}F^{♭♭} = \begin{bmatrix} & + \E^x \: dx ⊗ dt & + \E^y \: dy ⊗ dt & + \E^z \: dz ⊗ dt \\ - \E^x \: dt ⊗ dx & & - B^z \: dy ⊗ dx & + B^y \: dz ⊗ dx \\ - \E^y \: dt ⊗ dy & + B^z \: dx ⊗ dy & & - B^x \: dz ⊗ dy \\ - \E^z \: dt ⊗ dz & - B^y \: dx ⊗ dz & + B^x \: dy ⊗ dz & \\ \end{bmatrix}\end{split}\]

Difference between tensor basis and bivector basis

It turns out that the row-column representation in the double covariant tensor basis \(dx^μ ⊗ dx^ν\) and the 2–form basis \(dx^μ ∧ dx^ν\) is the same. The 2–form basis is related to the double covariant tensor basis through:

\[dx^μ ∧ dx^ν = dx^μ ⊗ dx^ν - dx^ν ⊗ dx^μ\]

Hence we can also write:

\[\begin{split}F^{♭♭} = \left[ \begin{aligned} - & \E^x \: dt ∧ dx \\ - & \E^y \: dt ∧ dy \\ - & \E^z \: dt ∧ dz \\ & B^x \: dy ∧ dz \\ & B^y \: dz ∧ dx \\ & B^z \: dx ∧ dy \\ \end{aligned} \right] = \frac{1}{2} \left[ \begin{alignedat}{3} - & \E^x \: (dt ∧ dx &+& dt ∧ dx) \\ - & \E^y \: (dt ∧ dy &+& dt ∧ dy) \\ - & \E^z \: (dt ∧ dz &+& dt ∧ dz) \\ & B^x \: (dy ∧ dz &+& dy ∧ dz) \\ & B^y \: (dz ∧ dx &+& dz ∧ dx) \\ & B^z \: (dx ∧ dy &+& dx ∧ dy) \\ \end{alignedat} \right] = \frac{1}{2} \left[ \begin{alignedat}{3} - & \E^x \: (dt ∧ dx &-& dx ∧ dt) \\ - & \E^y \: (dt ∧ dy &-& dy ∧ dt) \\ - & \E^z \: (dt ∧ dz &-& dz ∧ dt) \\ & B^x \: (dy ∧ dz &-& dz ∧ dy) \\ & B^y \: (dz ∧ dx &-& dx ∧ dz) \\ & B^z \: (dx ∧ dy &-& dy ∧ dx) \\ \end{alignedat} \right]\end{split}\]

Which can then be rearanged into a row-column form:

\[\begin{split}F^{♭♭} = \frac{1}{2} \begin{bmatrix} & + \E^x \: dx ∧ dt & + \E^y \: dy ∧ dt & + \E^z \: dz ∧ dt \\ - \E^x \: dt ∧ dx & & - B^z \: dy ∧ dx & + B^y \: dz ∧ dx \\ - \E^y \: dt ∧ dy & + B^z \: dx ∧ dy & & - B^x \: dz ∧ dy \\ - \E^z \: dt ∧ dz & - B^y \: dx ∧ dz & + B^x \: dy ∧ dz & \\ \end{bmatrix}\end{split}\]

The row-column repesentation with implicit basis is then exactly the same up to a factor \(\frac{1}{2}\):

\[\begin{split}\begin{bmatrix} & + \E^x & + \E^y & + \E^z \\ - \E^x & & - B^z & + B^y \\ - \E^y & + B^z & & - B^x \\ - \E^z & - B^y & + B^x & \\ \end{bmatrix}\end{split}\]

I am laying this here because it personally cost me a lot of confusion, since I expected the same behavoir when lowering or raising the indices, which is not the case because we are dealing with different basis.

Calulating the mixed Faraday tensor, we have:

\[F^μ{}_ν = F_{λν} η^{λμ}\]