Wikipedia:Reference desk/Archives/Mathematics/2009 July 20

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July 20 edit

Magnitude of a vector in different units edit

Is there a definition, in the general case, for the magnitude of a vector whose components are in incommensurable units? NeonMerlin 00:12, 20 July 2009 (UTC)[reply]

I don't think so. What do you want one for? Algebraist 00:15, 20 July 2009 (UTC)[reply]
Dimensional analysis? Is Nondimensionalization of any help? 67.117.147.249 (talk) 03:37, 20 July 2009 (UTC)[reply]
NeonMerlin - what do you mean by "incommensurable units" ? Do you mean that the basis elements have different dimensions e.g. ex is in metres and ey is in seconds, so a typical member of the space might be "10 metres + 5 seconds" ? I don't think this makes sense as a physical vector space, because the dimension (in the physics sense) of a "vector" in this space is undefined. Gandalf61 (talk) 10:46, 20 July 2009 (UTC)[reply]

Certainly! See metric tensor. The usual formula for the length s of the vector (x,y)

 

generalizes to

 

when the components are in different units. The coefficients   and   are the squares of the magnitudes of the base vectors:

 
 

If the unit vectors of the coordinate system are not even orthogonal to one another, the formula reads

 

where

 

The formulas are simplified by renaming variables:

 
 
 

Here the superscript no longer means exponentiation, but rather indexing. The formula is further simplified by the Einstein convention of implying summation when the same index occurs once downstairs and once upstairs.

 

This is readily generalized to the three dimensional case of stereometry, and to the four dimensional case of relativity theory.

Bo Jacoby (talk) 08:26, 20 July 2009 (UTC).[reply]