# Are basis vectors imaginary in special relativity?

It sounds absurd, but it seems to come naturally:

I know that $g_{\mu \nu }=e_\mu \cdot e_\nu$ (source)

and that the diagonal elements of $\eta_{\mu \nu}$ are typically $\pm(1, -1, -1, -1)$.

This seems to very clearly suggest that the basis vectors corresponding to negative metric components could be $(0, i, 0, 0)$ or similar.

If these are not the basis vectors in SR, I would like to ask:

1. What are the basis vectors, then? (of course they're variable, but what would a Cartesian basis look like?)

2) If $g_{\mu \nu }=e_\mu \cdot e_\nu$ does not hold for those vectors, why not?

• You probably know that in the old books, they had $(-ict, x, y, z)$. Then 50 years ago they dropped the imaginary bit in front of time.
– user108787
Oct 31, 2016 at 22:44
• Oct 31, 2016 at 23:41
• More on complex numbers in SR. Nov 1, 2016 at 11:38

The basis vectors are exactly what you'd expect, $$(1, 0, 0, 0), \quad (0, 1, 0, 0), \quad (0, 0, 1, 0), \quad (0, 0, 0, 1).$$ However, the inner product, i.e. the way we combine two vectors into a number, is not the same as the usual dot product. Using your notation, we're changing the definition of $\cdot$, not the definition of the $e_{\mu}$.

You are correct that it's possible to continue working with the dot product formally if we define some of the basis vectors to have imaginary components. That's how it was done in the past, but it's a bad idea: time and lengths just aren't complex numbers. They're perfectly real, so moving to a complex vector space doesn't make physical sense. (Moreover, the dot product itself is unnatural in a complex vector space, where the Hermitian inner product fits better.)

• I think I understand, but let me ask: before using a metric with negative values on the diagonal, I was under the impression that if you are in an orthonormal basis, and you are in flat space, your inner product will just be the dot product. From what you are saying, in SR, we have redefined the "usual" inner product, so this is no longer the case. So even though the basis vectors are cartesian, I still need to use the metric to find their inner product. Is this correct? Oct 31, 2016 at 23:34
• @doublefelix Yes, you need the metric. Basically, the equation $g_{\mu\nu} = e_\mu \cdot e_\nu$ is a fake. You won't use the right-hand side to determine the left-hand side, because the $\cdot$ is different. You use the left to determine the right. Oct 31, 2016 at 23:40
• Also, if you're defining an orthonormal basis to be a set of orthogonal vectors all with length $1$, then the fact that negative signs exist in the metric tells you that orthonormal bases, under that definition, don't exist in SR. Oct 31, 2016 at 23:41
• I agree that the equation seems useless now. But suppose that I want to change to some non-orthonormal coordinates. Can I then find the (new) components of my metric using an analogous form of that equation, namely the same one but with the dot product redefined as the SR inner product? What I am asking is: If the new basis vectors' components in cartesian coordinates are $(e'_{\mu})_{\alpha}$. Then is $\eta'_{\mu \nu}=(e'_{\mu})_{\alpha} \eta_{\alpha \beta} (e'_{\nu})_{\beta}$ ? Sorry for the many questions. I have learned a lot already. Nov 1, 2016 at 0:01
• @doublefelix Yup, that works! (The more common way to arrive at your equation is to simply say, the metric is a rank 2 tensor, and just write down the result. But this is a nice way of picturing it.) Nov 1, 2016 at 0:03

There are several approaches to this question. A long time ago people did use the hack of making some of the basis vectors complex in an attempt to make things look 'more natural'. But this doesn't actually solve any problems worth solving: in fact it makes them worse.

If you wanted to treat the metric as corresponding to an inner product, then it should actually do so. In particular, for a vector space over any field we need $\langle \vec{u},\vec{u}\rangle \ge 0$ with equality only when $\vec{u}=\vec{0}$. This is just as true for a vector space over $\mathbb{C}$ as it is for one over $\mathbb{R}$.

But this simply is not true for the metric of relativity: the whole structure of the thing rests on their being null vectors: vectors which are not zero but whose 'lengths' are zero. There is no way out of this problem other than to realise that the metric in relativity does not correspond to an inner product, because it is not positive definite.

That being the case it is far simpler to abandon any hacky approach involving complex numbers, and to accept the metric as what it is: not actually a metric at all, but something which has all the properties of a metric except positive-definiteness.

(An earlier version of this called it a 'pseudometric', but as pointed out by John Davis in a comment, it is not even that: a pseudometric is nonnegative, but can be zero for distinct vectors.)

• Very important that you pointed out that the "metric" in Minkowski space is not really a metric it is a pseudometric. Oct 31, 2016 at 23:56
• It is not even a pseudometric, though the exact classification is not really that important as the properties of Minkowski space are well-studied in themselves. Nov 1, 2016 at 0:01
• The problem with trying to find a classification for the Minkowski metric that is a weaker version of a metric space, is that it divides separations into two types: spacelike and timelike (along with degenerate null separations). This isn't really similar enough and a little too specific to fit comfortably into some kind of hierarchy of metric space-like algebraic structures. Nov 1, 2016 at 2:14
• The reason though the term "metric" is still used is because of the Minkwoski inner product, that is in many ways similar to a metric-inducing inner product. I would therefore call it a "Minkowski metric" or perhaps a "pseudo-Riemannian metric" Nov 1, 2016 at 2:18
• Pseudo-Riemannian is an established term in the literature, as is Lorentzian [geometry]. Nov 1, 2016 at 16:18