Relation between the Faraday 2–form and the Faraday tensor

by Stéphane Haussler

Warning

under construction

In this page, I relate the Faraday 2–form to the electormagnetic field tensor, and review the differences when rising or lowering and indices.

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:

\[F^μ{}_ν = F_{λν} η^{λμ}\]