The metric induced by the p-norm:

$d((x_1,\dotsc,x_n),(y_1,\dotsc,y_n)) = \left(\sum_{i=1}^n |x_i-y_i|^p\right)^{1/p}$

is often called the Minkowski distance.

There is also Minkowski space, which as I understand is a bit like Euclidean 4-space. And there is the Minkowski metric tensor defined for it.

Is there a relationship between Minkowski distance and the Minkowski metric tensor? If not, why is the metric induced by the p-norm called Minkowski distance? Does anybody have a reference for this name?

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    $\begingroup$ Isn't the relationship just Hermann Minkowski? $\endgroup$ – martini Apr 24 '12 at 10:27
  • $\begingroup$ @martini Well, that's what I assumed, but what has Minkowski got to do with the metric which I mentioned? Basically I'm looking for a better source for the name and origin of this metric than Wikipedia. I thought the origin might have been from the metric tensor, but apparently not. $\endgroup$ – Borbus Apr 24 '12 at 15:43
  • $\begingroup$ @Borbus, why did you said you think is like the euclidean 4-space? isn't it more like euclidean n-space? $\endgroup$ – user286485 Jan 1 '17 at 18:11

No. The Minkowski metric tensor is actually based on a very different metric: $$d((x_1,...x_n),(y_1,...,y_n))=\sum_{i=1}^{n-1}{(x_i-y_i)^2}- (x_n-y_n)^2$$ Which in fact is not really a metric, since $d$ can be smaller than zero (if you want to be formal - this is a pseudo-Riemannian manifold)

  • $\begingroup$ Ok, that's what I thought the metric tensor looked like, or it can be +--- apparently (where time is the first component). It's still a mystery to me why the metric I mentioned is named after Minkowski in that case. $\endgroup$ – Borbus Apr 24 '12 at 15:44
  • $\begingroup$ As @martini pointed out, they are both named after the same person - that's it. $\endgroup$ – nbubis Apr 24 '12 at 16:44
  • $\begingroup$ Not quite. The correct formula is $d((x_1,...x_n),(y_1,...,y_n))=\sqrt{ |\sum_{i=1}^{n-1}{(x_i-y_i)^2}- (x_n-y_n)^2|}$ and it is a pseudo-distance. By abuse, though, it is still called "metric". $\endgroup$ – Alex M. May 10 '17 at 15:07

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