# Electromagnetic field tensor via tensor products?

Can the electromagnetic field tensor be thought of as the tensor product of two physically reasonable vectors? Are the vectors arbitrarily constructed so that their components simply give the values in the electromagnetic field tensor matrix? I'm not very comfortable with thinking of thinking about things like second rank tensors as anything other than tensor products of vectors, so showing how second rank tensor transformations like the e-m field tensor act under things like Lorentz transformations is like a game without any meaning to me at this point unfortunately, I'm thinking trying to interpret it in terms of tensor products of vectors makes things more intuitive (i.e. operators on simple vectors)

## 4 Answers

Any tensor has a number of equally valid perspectives on it, depending on how many, and which, indices you put upstairs and downstairs. Since you want to understand the electromagnetic field tensor in terms of operations on simple vectors, there's one "mixed" perspective which can help you get some intuition, and it is to think of this tensor as a linear transformation.

The Lorentz force is in fact the most general force (as allowed by special relativity) that is proportional to the particle's velocity. In relativistic notation, you can write it as $$\frac{dp^\mu}{d\tau}=qF^\mu_{\ \ \ \nu}u^\nu, \tag1$$ where $u^\nu=(\gamma,\gamma\vec\beta)$ is the particle's four-velocity, $p^\mu=mu^\mu$ its four-momentum, and $q$ is its electric charge.

Notice that I've used the form $F^\mu_{\ \ \ \nu}u^\nu$ for the contraction, instead of the equivalent form $F^{\mu\nu}u_\nu$ in terms of a two-tensor. The form in (1) takes a vector, $u_\nu$, (instead of a co-vector), and returns another vector. Thus, $F$ is in this perspective a linear transformation from 4D spacetime into itself.* These are hardly exotic objects!

On the other hand, how are we to make sense of all those weird tensor-y sounding properties? In particular, what does "antisymmetric" mean in this context? This arises from the fact that the (relativistic) norm of $u^\mu$ is always fixed to $u_\mu u^\mu\equiv1$, which in this formulation implies that $$0=m\frac{d}{d\tau}u_\mu u^\mu=2u_\mu m\frac{du^\mu}{d\tau}=q u_\mu F^\mu_{\ \ \ \nu}u^\nu.$$ This can be simplified by bringing back a more algebraic notation, using $F(u)$ for the four-vector with components $F^\mu_{\ \ \ \nu}u^\nu$, and using $u\cdot v=u_\mu v^\mu$ for the Lorentz inner product. In this sense, what we're asking is simply for $F(u)$ to be Lorentz-orthogonal to its input $u$: $$u\cdot F(u)=0.$$ Easy! To make this look more like an anti-symmetry condition, evaluate this equation changing $u$ for some other vector $v$ and for their sum $u+v$, and after simplifying you'll get $$u\cdot F(v)=-v\cdot F(u).$$

Casting things in this language also simplifies, enormously, what the transformation law under Lorentz transformations is for these tensors. If $p=qF(u)$ and $p$ and $u$ transform to $p'=\Lambda p$ and $u'=\Lambda u$, then for the transformed law, $\frac{dp'}{d\tau}=qF'(u')$ to hold, we need $$\frac{dp'}{d\tau}=\Lambda \frac{dp}{d\tau}=q\Lambda F(u)=q\Lambda F(\Lambda^{-1} u),$$ so the transformed linear transformation is simply $F'=\Lambda F \Lambda^{-1}$, and whatever else that comes down to in terms of components.

If you like this sort of perspective, I would heartily recommend The geometry of minkowski spacetime (G. L. Naber, Springer, 2012).

*Note, though, that this definition is only valid for future-pointing, timelike $u$'s, and needs to be extended by linearity to null and spacelike vectors. This is easy to do, though.

• note that the most 'natural' placement of indices is $p_\mu$ and $F_{\mu\nu}$ – Christoph Nov 13 '13 at 18:05
• @Christoph I disagree. The most natural placement of indices on $F$ is, as I argue above, $F^\mu_{\ \ \ \nu}$. The placement of the index on $p$ can be up or down - it doesn't really matter; the dual is canonically isomorphic to the primal if there's an inner product. If you use $p_\mu$, of course, you should switch $F$ to $F_\mu^{\ \ \ \nu}$. However, the notation $F_{\mu\nu}$ forces a different vector/covector character on $u^\nu$ and $p_\mu=mu_\mu$ in the same equation, $p_\mu=qF_{\mu\nu}u^\nu$. While that's correct, I can't see how it's 'simpler' or more 'natural'. – Emilio Pisanty Nov 13 '13 at 18:14
• the lower indices of $F$ are natural because that's what you get from gauge theory - in general, it's a Lie-algebra-valued two-form (and becomes a regular one after contraction with a generalized charge, a co-adjoint orbit); to see that the lower index for $p$ is natural, look at the Legendre-transformation from tangent (ie velocity phase space) to cotangent (ie momentum phase space) when going from the Lagrangian to the Hamiltonian formulation; also, minimal coupling and contraction with velocities to get an energy suggest that momentum should be a covector – Christoph Nov 13 '13 at 19:45
• Sure, that's fair enough. Lagrangian mechanics does have $p$, and therefore $A$, as a covector, and this does make $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$ the most natural in that setting. (I'd encourage you to add an answer explaining the two-form perspective on $F$.) On the other hand, on a 'Newtonian' perspective where $p=mu$ - which is the usual port of entry into relativity -, having upper indices on $p$ and $F$ as a linear transformation makes the most sense. – Emilio Pisanty Nov 13 '13 at 19:52
• note that $F_{\mu\nu}$ can still be considered a linear transformation: ${\rm T}^*M\otimes{\rm T}^*M\cong{\rm Hom}({\rm T}M,{\rm T}^*M)$; if you accept the general premise (velocity as vector, momentum as covector), mass would be promoted from a simple number to a map ${\rm T}M\to{\rm T}^*M$ - in this case it's just a multiple of the metric, in the case of Lagrangian mechanics you'd use the fiber derivative; to fully geometrize Newtonian mechanics, you need additional structure - eq $(1)$ as-is is not well-defined from a differential-geometric perspective – Christoph Nov 13 '13 at 20:08

In a given reference frame, the electromagnetic field can be written as sum of electric and magnetic fields.

In the language of differential forms on a 4-dimensional space-time (and up to sign I might have messed up), this looks like $$F = E + \star B$$ where $$E = \sum_i E_i \,{\rm d}x^i\wedge{\rm d}t \\ B = \sum_i B_i \,{\rm d}x^i\wedge{\rm d}t$$ As mentioned in this answer, it cannot in general be written as a single exterior product. Also, a Lorentz boost will mix the components of $E$ and $B$, which is the reason why we came up with the single entity $F$ in the first place.

At each each point $x\in M$ of space-time, we also have $$F_x\in\bigwedge^2{\rm T}^*_xM\subset{\rm T}^*_xM\otimes{\rm T}^*_xM\cong\rm{Hom}({\rm T}_xM,{\rm T}^*_xM)$$ ie we can understand the electromagnetic field as a map $$F:{\rm T}M\to {\rm T}^*M$$ mapping a velocity vector to a covector, which is (up to charge and some fiddling with the geometry) just the Lorentz force.

If you're asking for a deeper reason why the electromagnetic field is a 2-form, you should look into gauge theory (in particular classical Yang-Mills theory). The electromagnetic field ends up being the curvature 2-form of a principal connection, which is the exterior covariant differential of the (in general Lie-algebra valued) connection 1-form (the vector potential).

First, the electromagnetic tensor is a bivector and so an exterior product is required.

Second, it can be shown that, in general, the electromagnetic tensor is the sum of two bivectors and thus, is not a simple bivector, i.e., is not the exterior product of two vectors.

Yes this is possible. It is the product of the gradient vector with the potential. Use the Octonions to form the product. The relevant Octonion multiplication table is defined by the following quaternionic triples:(Ijk), (iJk), (ijK), (IJK), (Iim), (Jjm), (Kkm). The lowercase are vectors and the uppercase are bivectors.

Note that m $\sqrt {-1}=l$

Say the vector potential is A and the four dimensional vector gradient operator is D, Using $i, j, k, l$ as the unit vectors. Then DA=-D•A+( DyAz-DzAy)I +(DzAx-DxAz) J +(DxAy-DyAx) K+$\sqrt {-1}${(DxAt+DtAx) l +(DyAt+DtAy) J+(DzAt+DtAz) K}. Using the lorentz gauge the term D•A is zero. The 6 Independent components of the electromagnetic upper index tensor are composed of the 3 real components of I J K and the 3 imaginary components of I J K.

• Can you give a more explicit answer? I have never heard of Octonions or quaternionic triples before. – Steven Mathey Apr 5 '15 at 0:38
• Steven details are coming in a few days when I get a chance to write out the equations. – Scott Apr 7 '15 at 0:14