# Would this be a metric?

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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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 $c^2$.
@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