Euclidian space versus Minkowski spacetime

[michel123456]

Splitted from a tooo long thread:

 

Summary:

 

(...)

 

Michael123456 (...) can't seem to grasp the difference between Minkowski space with the Minkowski metric as opposed to Euclidean space with the Euclidean metric.

(...)

 

That is correct.

What happens to Minkowski spacetime when one dimension is taken away?

 

From the 4 dimensions x,y,z & t, eliminate z for example.

The resulting reduced spacetime x,y,t looks quite a lot with Euclidian space, or do I miss something?

Edited February 28, 2011 by michel123456

[timo]

You're probably missing the keyword "metric" and its meaning.

[ajb]

What happens to Minkowski spacetime when one dimension is taken away?

 

From the 4 dimensions x,y,z & t, eliminate z for example.

The resulting reduced spacetime x,y,t looks quite a lot with Euclidian space, or do I miss something?

 

 

As a topological space it is just [math]\mathbb{R}^{3}[/math]. However, you have the metric to take care of.

 

Euclidean space (3 dimensional) has a metric [math]diag(1,1,1)[/math] (in any Euclidean coordinates). The 1+2 dimensional Minkowski space has a metric [math]diag(-1,1,1)[/math] (in any "inertial" coordinates).

 

The subtle, but quite important difference is the minus sign.

[michel123456]

Does that mean it is mirrored?

[ajb]

Does that mean it is mirrored?

 

What do you mean by mirrored?

[michel123456]

enantiomorphic

 

Like your left & right hand.

[ajb]

possess no improper rotations?

 

Have a look at the Wikipedia entry on the Euclidean group.

[michel123456]

Funny conversation.

 

Reflection?

 

Lets take an even more reduced spacetime: only x,t

 

I suppose the negative is induced by time.

What does that mean from a geometric point of vue? Do euclidian laws of geometry change in such a spacetime?

Edited February 28, 2011 by michel123456