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The Minkowski so-called 'metric'

  1. Nov 12, 2004 #1

    jcsd

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    I've just been thinking (prolly a bad idea): Lorentzian metrics aren't actually metrics at all are they? Infact 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)?
     
    Last edited: Nov 12, 2004
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  3. Nov 12, 2004 #2

    Fredrik

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    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.
     
  4. Nov 12, 2004 #3

    jcsd

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    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]
     
    Last edited: Nov 12, 2004
  5. Nov 12, 2004 #4

    jcsd

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    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.
     
    Last edited: Nov 12, 2004
  6. Nov 16, 2004 #5
    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
     
  7. Nov 16, 2004 #6

    jcsd

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    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.
     
    Last edited: Nov 17, 2004
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