Relative scalar: Difference between revisions

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In mathematics, a '''relative scalar''' (of weight ''w'') is a [[scalar-valued function]] whose transform under a coordinate transform,

:<math display="block">

\bar{x}^j = \bar{x}^j(x^i)

</math>

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on an ''n''-dimensional manifold obeys the following equation

:<math display="block">

\bar{f}(\bar{x}^j) = J^w f(x^i)

</math>

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where

: <math display="block"> J = \begin{vmatrix}left| \displaystyle \fracdfrac{\partial(x_1,\ldots,x_n)}{\partial(\bar{x}^1,\ldots,\bar{x}^n)} \end{vmatrix}right| , </math>

that is, the determinant of the [[Jacobian matrix and determinant|Jacobian]] of the transformation.<ref name=lovelock>{{cite book| last|last1=Lovelock| first|first1=David |last2=Rund |first2=Hanno |authorlink2=Hanno Rund |title=Tensors, Differential Forms, and Variational Principles |yeardate=April 1, April 1989 | publisher=Dover | isbn=0486658406 | pages=4000-486-65840-6 | url=http://store.doverpublications.com/0486658406.html | coauthors=Rund | accessdate=19 April 2011 | page=103 | format=Paperback | chapter=4 | page=103}}</ref> A '''scalar density''' refers to the <math>w=1</math> case.

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 |authorlink=Oswald Veblen |title=Invariants of Quadratic Differential Forms |url=http://www.cambridge.org/us/knowledge/isbn/item1156775/?site_locale=en_US |accessdate=3 October 2012 |year=2004 |publisher=[[Cambridge University Press]] |isbn=05216048420-521-60484-2 |pagespage=11221}} (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.

===ElementaryWeight 0 example===

Suppose the temperature in a room is given in terms of the function <math>f(x,y,z) = 2 x + y + 5</math> in Cartesian coordinates <math>(x,y,z)</math> and the function in cylindrical coordinates <math>(r,t,h)</math> is desired. The two coordinate systems are related by the following sets of equations:

<math display="block"> \begin{align}

:<math> r &= \sqrt{x^2 + y^2} \, </math>\

:<math> t &= \arctan(y/x) \, </math>\

:<math> h = z \, </math>

h &= z

\end{align} </math>

and

<math display="block"> \begin{align}

:<math> x = r \cos(t) \, </math>

:<math> tx &= r \sincos(t) \, </math>\

y &= r \sin(t) \\

:<math> z = h. \, </math>

z &= h.

\end{align} </math>

Using <math>\bar{f}(\bar{x}^j) = f(x^i(\bar{x}^j))</math> allows one to derive <math>\bar{f}(r,t,h)= 2 r \cos(t)+ r \sin(t) + 5</math> as the transformed function.

Consider the point <math>P</math> whose Cartesian coordinates are <math>(x,y,z)=(2,3,4)</math> and whose corresponding value in the cylindrical system is <math>(r,t,h)=(\sqrt{13},\arctan{(3/2)},4)</math>. A quick calculation shows that <math>f(2,3,4)=12</math> and <math>\bar{f}(\sqrt{13},\arctan{(3/2)},4)=12</math> also. This equality would have held for any chosen point <math>P</math>. Thus, <math>f(x,y,z)</math> is the "temperature function in the Cartesian coordinate system" and <math>\bar{f}(r,t,h)</math> is the "temperature function in the cylindrical coordinate system".

One way to view these functions is as representations of the "parent" function that takes a point of the manifold as an argument and gives the temperature.

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The problem could have been reversed. One could have been given <math>\bar{f}</math> and wished to have derived the Cartesian temperature function <math>f</math>. This just flips the notion of "new" vs the "original" coordinate system.

Suppose that one wishes to ''integrate'' these functions over "the room", which will be denoted by <math>D</math>. (Yes, integrating temperature is strange{{Says who|date=July 2011}} but that's partly what's to be shown.) Suppose the region <math>D</math> is given in cylindrical coordinates as <math>r</math> from <math>[0,2]</math>, <math>t</math> from <math>[0,\pi/2]</math> and <math>h</math> from <math>[0,2]</math> (that is, the "room" is a quarter slice of a cylinder of radius and height 2).

The integral of <math>f</math> over the region <math>D</math> is{{cn|date=August 2022}}

:<math display=block> \int_0^2 \! \int_{-\sqrt{2^2-x^2}0}^\sqrt{2^2-x^2} \! \int_0^2 \! f(x,y,z) \, dz \, dy \, dx = 64/316 + 2010 \pi.</math>.

The value of the integral of <math>\bar{f}</math> over the same region is{{cn|date=August 2022}}

:<math display=block> \int_0^2 \! \int_{0}^{\pi/2} \! \int_0^2 \! \bar{f}(r,t,h) \, dh \, dt \, dr = 12 + 10 \pi.</math>.

They are not equal. The integral of antemperature ordinaryis scalarnot dependsindependent onof the coordinate system used. ThisIt coordinateis dependencenon-physical tendsin tothat removesense, anyhence physical"strange". meaningNote fromthat if the integral of an<math>\bar{f}</math> ordinaryincluded scalar.a factor of the Jacobian (which is just <math>r</math>), we get{{cn|date=August 2022}}

<math display=block> \int_0^2 \! \int_{0}^{\pi/2} \! \int_0^2 \! \bar{f}(r,t,h) r \, dh \, dt \, dr = 16 + 10 \pi,</math>

which ''is'' equal to the original integral but it is not however the integral of ''temperature'' because temperature is a relative scalar of weight 0, not a relative scalar of weight 1.

==Scalar=Weight density1 example===

If we had said <math>f(x,y,z) = 2 x + y + 5</math> was representing mass density, however, then its transformed value

A '''scalar density''' refers to the <math>w=1</math> case.

should include the Jacobian factor that takes into account the geometric distortion of the coordinate

system. The transformed function is now <math>\bar{f}(r,t,h)= (2 r \cos(t)+ r \sin(t) + 5) r</math>. This time

<math>f(2,3,4)=12</math> but <math>\bar{f}(\sqrt{13},\arctan{(3/2)},4)=12\sqrt{29}</math>. As before

is integral (the total mass) in Cartesian coordinates is

<math display=block> \int_0^2 \! \int_{0}^\sqrt{2^2-x^2} \! \int_0^2 \! f(x,y,z) \, dz \, dy \, dx = 16 + 10 \pi.</math>

The value of the integral of <math>\bar{f}</math> over the same region is

<math display=block> \int_0^2 \! \int_{0}^{\pi/2} \! \int_0^2 \! \bar{f}(r,t,h) \, dh \, dt \, dr = 16 + 10 \pi.</math>

They are equal. The integral of mass ''density'' gives total mass which is a coordinate-independent concept.

Note that if the integral of <math>\bar{f}</math> also included a factor of the Jacobian like before, we get{{cn|date=August 2022}}

<math display=block> \int_0^2 \! \int_{0}^{\pi/2} \! \int_0^2 \! \bar{f}(r,t,h) r \, dh \, dt \, dr = 24 + 40 \pi / 3 ,</math>

which is not equal to the previous case.

==Other cases==

Weights other than 0 and 1 do not arise as often. It can be shown the determinant of a type (0,2) tensor is a relative scalar of weight 2.

==Generalization==

Relative scalars are special cases of [[relative tensor]]s.

==See also==

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[[Category:Tensors]]

[[Category:Tensors in general relativity]]

[[Category:Scalars]]

{{math-stub}}