dEdt said:
My class is starting to cover E&M in Lorentz covariant form, and obviously the subject of tensors came up. The problem is that my prof defines tensors in terms of coordinates, which is ugly and against the spirit of relativity. Is there a way of doing tensors coordinate-free in a physics friendly fashion?
One approach is to use multivectors and wedge products.
What are bivectors? Oriented planes, the same way vectors can be taken to represent oriented lines with magnitudes. If you have two basis vectors ##e_x, e_y##, then the bivector ##e_x \wedge e_y## represents the unit bivector of the xy-plane. In 3+1 spacetime, there are six such planes: tx, ty, tz, xy, yz, and zx.
6 components...sound familiar? It should. These are exactly the components of the electromagnetic field tensor. What we call ##F_{\mu \nu}## is a bivector (field). We can extract the components of ##F## the way you would with vectors:
$$A_\mu = A \cdot e_\mu \\
F_{\mu \nu} = F \cdot (e_\nu \wedge e_\mu)$$
(We've slightly extended the dot product's definition, but this is the general idea.)
The use of the wedge makes a lot of expressions simpler that otherwise, in index notation, would require explicit use of antisymmetric summations that are harder to geometrically interpret. We know, for instance, that the EM field obeys
$$\partial_\alpha F_{\beta \gamma} + \partial_\beta F_{\gamma \alpha} + \partial_\gamma F_{\alpha \beta} = 0$$
But using the wedge product and the vector derivative $\nabla = e^\mu \partial_\mu$, we get a more easily condensed result,
$$\nabla \wedge F = 0$$
This is a 4D analogue to curl. The other equation is ##\nabla \cdot F = -\mu_0 j## (constants and sign depending on exact conventions). This capture's Maxwell's equations outside of material media.
The use of bivectors and wedges makes the process of converting from traditional EM fields to expressions that are covariant straightforward. Consider, for example, the EM bivector for a stationary point charge at the origin.
$$F = -Q \mu_0 c \frac{e_t \wedge r}{4\pi|r|^3}$$
We identify ##e_t## as the four-velocity of the point-charge, ##u##. ##r##, the position vector in three-space, must be entirely orthogonal to ##e_0##, so if four-position is ##s##, then ##r = -e_0 \cdot (e_0 \wedge s)##. Again, we replace ##e_0## by ##u## to get
$$F = -\mu_0 Qc\frac{u \wedge s}{4\pi |u \cdot (u \wedge s)|^3}$$
In this way, one can compute the EM bivector for any uniformly moving point charge given the four-velocity $u$, and in whatever coordinate system we choose.