A few weeks ago, I exchanged a number of e-mails with someone about the Lanczos tensor and the Weyl-Lanczos equation. One of the things I derived is worth recording here for posterity.

The Lanczos tensor is an interesting animal. It can be thought of as the source of the Weyl curvature tensor, the traceless part of the Riemann curvature tensor. The Weyl tensor, together with the Ricci tensor, fully determine the Riemann tensor, i.e., the intrinsic curvature of a spacetime. Crudely put, whereas the Ricci tensor tells you how the volume of, say, a cloud of dust changes in response to gravity, the Weyl tensor tells you how that cloud of dust is distorted in response to the same gravitational field. (For instance, consider a cloud of dust in empty space falling towards the Earth. In empty space, the Ricci tensor is zero, so the volume of the cloud does not change. But its shape becomes distorted and elongated in response to tidal forces. This is described by the Weyl tensor.

Because the Ricci tensor is absent, the Weyl tensor fully describes gravitational fields in empty space. In a sense, the Weyl tensor is analogous to the electromagnetic field tensor that fully describes electromagnetic fields in empty space. The electromagnetic field tensor is sourced by the four-dimensional electromagnetic vector potential (meaning that the electromagnetic field tensor can be expressed using partial derivatives of the electromagnetic vector potential.) The Weyl tensor has a source in exactly the same sense, in the form of the Lanczos tensor.

The electromagnetic field does not uniquely determine the electromagnetic vector potential. This is basically how integrals vs. derivatives work. For instance, the derivative of the function $$y=x^2$$ is given by $$y’=2x$$. But the inverse operation is not unambiguous: $$\int 2x~ dx=x^2+C$$ where $$C$$ is an arbitrary integration constant. This is a recognition of the fact that the derivative of any function in the form $$y=x^2+C$$ is $$y’=2x$$ regardless of the value of $$C$$; so knowing only the derivative $$y’$$ does not fully determine the original function $$y$$.

In the case of electromagnetism, this freedom to choose the electromagnetic vector potential is referred to as the gauge freedom. The same gauge freedom exists for the Lanczos tensor.

Solutions for the Lanczos tensor for the simplest case of the Schwarzschild metric are provided in Wikipedia. A common characteristic of these solutions is that they yield a quantity that “blows up” at the event horizon. This runs contrary to accepted wisdom, namely that the event horizon is not in any way special; a freely falling space traveler would never know that he is crossing it.

But as it turns out, thanks to the gauge freedom of the Lanczos tensor, it is easy to construct a solution (an infinite family of solutions, as a matter of fact) that do not behave like this at the horizon.

Well, it was a fun thing to compute anyway.