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would a matrix $M$ with diagonal entries not necessarily equal 1, i.e. diag $M = (a,1,1,1)$ be a metric if $a \neq 1$ or $\neq 0$? I.e. in this case would this be like some sort of more general euclidean metric or just not a metric at all?

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up vote 6 down vote accepted

For $a<0$, it's a possible metric for spacetime. For $a>0$, it's a possible metric for a 4-dimensional Euclidean space. For $a=0$, it's degenerate, and in many cases it's not possible to work with a degenerate metric, e.g., the machinery of general relativity requires that the metric be nondegenerate. It doesn't matter whether $a$ has a particular nonzero value $a_1$ or another nonzero value $a_2$ with the same sign; under a change of coordinates, its value can change from one of these to another. For example, in relativity, if you switch between natural units and SI units, you'll pick up factors of $c^2$.

It doesn't matter whether the elements have absolute value 1. In SR, one generally makes this choice for convenience, because it's possible. In GR, you can't make the metric have constant components.

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Good answer. Generally, a metric has to be nondegenerate (as Ben notes), symmetric (that is, switching rows to columns and vice-versa for the matrix representation gives you the same metric), and bilinear (vector arguments passed to the metric can be subdivided linearly). Anything else following tensor rules is allowed. – B. Elliott May 14 '13 at 16:01
You can work consistently with null submanifolds in spacetime, and there is nothing wrong with the intrinsic metric, but you are stuck with a non-linked metric and a cometric, and if you want the full differential geometry machinery, you'll be stuck with some extrinsic geometry. – Jerry Schirmer May 14 '13 at 16:05
@JerrySchirmer: I don't necessarily disagree with your comment, but I would put it a different way. For example, in the limit $c\rightarrow\infty$, you can in some sense recover Newtonian physics from relativity. In this limit the metric becomes degenerate. The way I would express it is that Newtonian spacetime simply doesn't have a metric. What Newtonian physics has is a spatial metric. – Ben Crowell May 14 '13 at 19:00

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