# A simple metric explanation

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• Jun 27th 2006, 03:42 AM
topsquark
A simple metric explanation
(This post has the theme to "Mission: Impossible" running in the background.)

I am currently having an e-conversation with someone who thinks Special Relativity is incorrect. I have no problem discussing this with this person, though I support Special Relativity based on the evidence, however it seems his Math background is a bit lacking.

The problem is this. I worked out a metric for his theory so I could compare it with the Minkowski metric. I then pointed out to him that his metric is variable and depends on the relative velocity of the reference frame with respect to his aether frame. His response? What's a metric?

I've apparently spent too much time doing "Physics Math." I can't come up with a simple description for what a metric is. Could someone please give me an explanation of what a metric is that would be understood by someone who (apparently) knows nothing beyond vector addition and intro Calculus?

Thank you!

-Dan
• Jun 27th 2006, 05:34 AM
ThePerfectHacker
Quote:

Originally Posted by topsquark
Could someone please give me an explanation of what a metric is that would be understood by someone who (apparently) knows nothing beyond vector addition and intro Calculus?

Given a non-empty set $\displaystyle X$.
Such as function such as,
$\displaystyle d:X\times X\to \mathbb{R}^+$
which satisfies.
$\displaystyle d(x,y)\doublearrow x=y$
$\displaystyle d(x,y)=d(y,x)$
$\displaystyle d(x,y)\leq d(x,z)+d(z,y)$
Underthese condition we have a metric space $\displaystyle d$ on set $\displaystyle X$ denoted by $\displaystyle (X,d)$.
---
The intuitive, concept is to capture the idea of distance for any sets. Cuz,
1)It is measured in real numbers (positive) like any distance.
2)Makes no difference which point is first and second.
3)Triangular inequality for distance.

I do not see any more basic way to explain it rather than saying its is distance defined for any sets.

(I hate these type of people that argue with known concepts that they are incorrect and yet themselves never fully studied them!)
• Jun 27th 2006, 07:25 AM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
Given a non-empty set $\displaystyle X$.
Such as function such as,
$\displaystyle d:X\times X\to \mathbb{R}^+$
which satisfies.
$\displaystyle d(x,y)\doublearrow x=y$
$\displaystyle d(x,y)=d(y,x)$
$\displaystyle d(x,y)\leq d(x,z)+d(z,y)$
Underthese condition we have a metric space $\displaystyle d$ on set $\displaystyle X$ denoted by $\displaystyle (X,d)$.
---
The intuitive, concept is to capture the idea of distance for any sets. Cuz,
1)It is measured in real numbers (positive) like any distance.
2)Makes no difference which point is first and second.
3)Triangular inequality for distance.

I do not see any more basic way to explain it rather than saying its is distance defined for any sets.

(I hate these type of people that argue with known concepts that they are incorrect and yet themselves never fully studied them!)

But this does not cover the metric of SR

RonL
• Jun 27th 2006, 12:46 PM
topsquark
Quote:

Originally Posted by CaptainBlack
But this does not cover the metric of SR

RonL

You're both right. ThePerfectHacker has explained (in Mathspeak) what a metric is, and CaptainBlack pointed out that the Minkowski metric is indefinite, so we can only construct what I believe is called the "size" of a vector ( $\displaystyle \sqrt{|x \cdot x|}$ ), not a length. (TPH: You described an "inner product" which is a type of metric.)

And yeah, I don't think he really knows that much about SR. I'm about to give up on this guy. (sigh) I don't want to say anything bad about him as he's put an awful lot of effort into his theory. My problem is he's sort of focussing on the "paradoxes" which I've never really studied. I'm trying to work with him on other applications, but then I start running into the "Math wall." Which, frankly, I can't teach that well over e-mails!

(If only SR and GR were intuitive, then I could do this easily!)

Ah well. Enough griping. Back to work! :)

-Dan