The Minkowski so-called 'metric'

[jcsd]

I've just been thinking (prolly a bad idea): Lorentzian metrics aren't actually metrics at all are they? In fact they're not even pseudometrics, so what are they exactly and why do we call them metrics (Actually I can probably guess that as they perform the same role a mteric does and they are symmetric and obey the triangle inequality)?

 

[Fredrik]

A metric is just a just a symmetric non-degenerate tensor field of type (0,2). The metric of general relativity is not positive definite so it can't be called Riemannian, but it's still a metric.

 

[jcsd]

I know it's not postive definite and it's pseudo-Riemannian, what I am actually talking about is 'metric' in the most primitive mathematical sense. i.e. a set S forms a metric space when combined with a function [itex]d:S^2 \rightarrow R[/itex], known as the metric which obeys the following axioms for all [itex]x,y,z \in S[/itex].

[tex]d(x,y) = d(y,x)[/tex]
[tex]d(x,y) \geq 0[/tex]
[tex]d(x,y) = 0 \iff x=y[/tex]
[tex]d(x,z) + d(y,z) \geq d(x,y)[/tex]

 

[jcsd]

Ambitwistor (to give credit where it's due) has supplied me with the answer:

The conditions on the metric in a pseudo-Riemannian metric space are sufficiently relaxed that the metric may be of the form of the Minkowksi metric.

 

[pmb_phy]

I've just been thinking (prolly a bad idea): Lorentzian metrics aren't actually metrics at all are they? In fact they're not even pseudometrics, so what are they exactly and why do we call them metrics (Actually I can probably guess that as they perform the same role a mteric does and they are symmetric and obey the triangle inequality)?

The term "metric" means "to measure". The functions you gave are metrics in the sense that they provide some sort of measure. The metric *tensor* is an example of a mapping from vectors to scalars, e.g. ds^2 = g_ab dx^a dx^b where dx = vector and it gives a "measure" of the norm of a vector 'length' and the 'interval' between two points.

Pete

 

[jcsd]

The term "metric" means "to measure". The functions you gave are metrics in the sense that they provide some sort of measure. The metric *tensor* is an example of a mapping from vectors to scalars, e.g. ds^2 = g_ab dx^a dx^b where dx = vector and it gives a "measure" of the norm of a vector 'length' and the 'interval' between two points.

Pete

yes obviously it performs the same role as a metric does in a metric space and can intutively be thought as the distance. It was just that the Minkowski metirc does not meet the normal defintion of a metric, which worried me as after all when defining a metric for a vector space we usually require the set of vectors and the metric to form a metric space and we usually treat (for example) the set of all radius vectors in Minkowski space as a real vector space.

As I said Ambitwistor answerd this for me by saying that the conditions on the metric in a pseudo-Riemannian metric space were sufficently relaxed that the Minkowski metric is a suitable function to act as a metric in such a space.

There's no great distinction between a metric and a metric tensor , a metric tensor merely defines the metric for a vector space.