Relative scalar: Difference between revisions

Revision as of 23:55, 15 August 2011

Revision as of 03:53, 12 October 2011

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: <math> J = \begin{vmatrix} \displaystyle \frac{\partial(x_1,\ldots,x_n)}{\partial(\bar{x}^1,\ldots,\bar{x}^n)} \end{vmatrix} , </math>

that is, the determinant of the Jacobian of the transformation.<ref name=lovelock>{{cite book| last=Lovelock| first=David| title=Tensors, Differential Forms, and Variational Principles|yeardate=April 1, 1989 | publisher=Dover | isbn=0486658406 | pages=400 | url=http://store.doverpublications.com/0486658406.html | coauthors=Rund | accessdate=19 April 2011 | page=103 | format=Paperback | chapter=4}}</ref> Relative scalars are an important special case of the more general concept of a [[relative tensor]].

==Ordinary scalar==

An '''ordinary scalar''' or '''absolute scalar'''<ref>{{cite book|last=Veblen|first=Oswald|title=Invariants of Quadratic Differential Forms|year=2004|publisher=Cambridge University Press|isbn=0521604842|pages=112}} (page 21)</ref> refers to the <math>w=0</math> case.

If <math>x^i</math> and <math>\bar{x}^j</math> refer to the same point <math>P</math> on the manifold, then we desire <math>\bar{f}(\bar{x}^j) = f(x^i)</math>. This equation can be interpreted two ways when <math>\bar{x}^j</math> are viewed as the "new coordinates" and <math>x^i</math> are viewed as the "original coordinates". The first is as <math>\bar{f}(\bar{x}^j) = f(x^i(\bar{x}^j))</math>, which "converts the function to the new coordinates". The second is as <math>f(x^i)=\bar{f}(\bar{x}^j(x^i))</math>, which "converts back to the original coordinates. Of course, "new" or "original" is is a relative concept.

There are many physical quantities that are represented by ordinary scalars, such as temperature and pressure.

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[[Category:Tensors]]

[[Category:Tensors in general relativity]]

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