Minkowski metric is negative?

Question

From what I understand about metrics, they have to be positive. However the Minkowski 'metric' can be negative for time-like separation of events. Is the Minkowski metric not actually a real metric?

Answer 1

Part of the confusion in your question is due to the ambiguous use of the word "metric": it may mean either a "distance function", or a "metric tensor".

Indeed, on a Riemannian manifold the metric tensor $g$ is positive definite at every point and gives birth to a distance function $d$ by the formula

$$d(x,y) = \inf _c \int \limits _a ^b \sqrt {g(\dot c(t), \dot c(t) )} \ \Bbb d t \ ,$$

where the infimum is taken among all the differentiable curves $c$ with $c(a) = x$ and $c(b) = y$.

A Minkowskian manifold, on the other hand, is a special kind of pseudo-Riemannian (or semi-Riemannian) manifold. This time the metric tensor $g$ need not be positive at every point, anymore (all the other conditions from metric tensors on Riemannian manifolds must still be verified, though). Consequently, you can't use the same formula as above to construct a distance, because you may end up with imaginary numbers (square roots of negative numbers). In order to avoid this, you just add a modulus function under the square root, and obtain a pseudo-distance (or semi-distance):

$$d(x,y) = \inf _c \int \limits _a ^b \sqrt { |g(\dot c(t), \dot c(t) )|} \ \Bbb d t \ .$$

This $d$ has all the usual properties of a distance function, save for one: if $d(x,y) = 0$ it does no longer follow that $x=y$. This has an interesting consequence: on a Riemannian manifold, it can be proven that the topology of the manifold coincides with the topology given by the distance; this is no longer true on Hausdorff pseudo-Riemannian manifolds (the underlying topology is Hausdorff, while the topology given by the pseudo-distance can be shown no to be so).