# The Unapologetic Mathematician

## Minkowski Space

Before we push ahead with the Faraday field in hand, we need to properly define the Hodge star in our four-dimensional space, and we need a pseudo-Riemannian metric to do this. Before we were just using the standard $\mathbb{R}^3$, but now that we’re lumping in time we need to choose a four-dimensional metric.

And just to screw with you, it will have a different signature. If we have vectors $v_1=(x_1,y_1,z_1,t_1)$ and $v_2=(x_2,y_2,z_2,t_2)$ — with time here measured in the same units as space by using the speed of light as a conversion factor — then we calculate the metric as:

$\displaystyle g(v_1,v_2)=x_1x_2+y_1y_2+z_1z_2-t_1t_2$

In particular, if we stick the vector $v=(x,y,z,t)$ into the metric twice, like we do to calculate a squared-length when working with an inner product, we find:

$\displaystyle g(v,v)=x^2+y^2+z^2-t^2$

This looks like the Pythagorean theorem in two or three dimensions, but when we get to the time dimension we subtract $t^2$ instead of adding them! Four-dimensional real space equipped with a metric of this form is called “Minkowski space”. More specifically, it’s called 4-dimensional Minkowski space, or “(3+1)-dimensional” Minkowski space — three spatial dimensions and one temporal dimension. Higher-dimensional versions with $n-1$ “spatial” dimensions (with plusses in the metric) and one “temporal” dimension (with minuses) are also called Minkowski space. And, perversely enough, some physicists write it all backwards with one plus and $n-1$ minuses; this version is useful if you think of displacements in time as more fundamental — and thus more useful to call “positive” — than displacements in space.

What implications does this have on the coordinate expression of the Hodge star? It’s pretty much the same, except for the determinant part. You can think about it yourself, but the upshot is that we pick up an extra factor of $-1$ when the basic form going into the star involves $dt$.

So the rule is that for a basic form $\alpha$, the dual form $*\alpha$ consists of those component $1$-forms not involved in $\alpha$, ordered such that $\alpha\wedge(*\alpha)=\pm dx\wedge dy\wedge dz\wedge dt$, with a negative sign if and only if $dt$ is involved in $\alpha$. Let’s write it all out for easy reference:

$\displaystyle\begin{aligned}*1&=dx\wedge dy\wedge dz\wedge dt\\ *dx&=dy\wedge dz\wedge dt\\ *dy&=dz\wedge dx\wedge dt\\ *dz&=dx\wedge dy\wedge dt\\ *dt&=dx\wedge dy\wedge dz\\ *(dx\wedge dy)&=dz\wedge dt\\ *(dz\wedge dx)&=dy\wedge dt\\ *(dy\wedge dz)&=dx\wedge dt\\ *(dx\wedge dt)&=-dy\wedge dz\\ *(dy\wedge dt)&=-dz\wedge dx\\ *(dz\wedge dt)&=-dx\wedge dy\\ *(dx\wedge dy\wedge dz)&=dt\\ *(dx\wedge dy\wedge dt)&=dz\\ *(dz\wedge dx\wedge dt)&=dy\\ *(dy\wedge dz\wedge dt)&=dx\\ *(dx\wedge dy\wedge dz\wedge dt)&=-1\end{aligned}$

Note that the square of the Hodge star has the opposite sign from the Riemannian case; when $k$ is odd the double Hodge dual of a $k$-form is the original form back again, but when $k$ is even the double dual is the negative of the original form.

## 2 Comments	»

1. Hi John,

I just wanted to let you know that I’ve been loving this series. Connecting what you talked about before to something slightly more familiar has been really helpful in my understanding.

That’s all I wanted to say, but I wanted to say it before I forgot to.

Comment by | March 7, 2012 | Reply

2. [...] First off, I’m going to use a coordinate system where the speed of light is 1. That is, if my unit of time is seconds, my unit of distance is light-seconds. Mostly this helps keep annoying constants out of the way of the equations; physicists do this basically all the time. The other thing is that I’m going to work in four-dimensional spacetime, meaning we’ve got four coordinates: , , , and . We calculate dot products by writing . Yes, that minus sign is weird, but that’s just how spacetime works. [...]

Pingback by | July 17, 2012 | Reply

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