Relative scalar: Difference between revisions

Revision as of 17:08, 20 April 2011

Latest revision as of 15:33, 9 May 2023

(33 intermediate revisions by 16 users not shown)

Line 1:

AIn mathematics, a '''relative scalar''' (of weight <math>&nbsp;''w</math>'') is a [[scalar-valued function]] whose transform under a coordinate transform,

:<math>

<math display="block">

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

</math>

on an <math>''n</math>''-dimensional manifold obeys the following equation

:<math>

<math display="block">

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

</math>

where <math>J=\frac{\partial(x_1,\ldots,x_n)}{\partial(\bar{x}^1,\ldots,\bar{x}^n)}</math>, that is, the determinate of the transformation.<ref name=lovelock>{{cite book|last=Lovelock|first=David|title=Tensors, Differential Forms, and Variational Principles|year=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>

where

==Scalar==

An ordinary scalar refers to the <math>w=0</math> case.

<math display="block"> J = \left| \dfrac{\partial(x_1,\ldots,x_n)}{\partial(\bar{x}^1,\ldots,\bar{x}^n)} \right| , </math>

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.

where <math>J=\frac{\partial(x_1,\ldots,x_n)}{\partial(\bar{x}^1,\ldots,\bar{x}^n)}</math>, that is, the determinatedeterminant of the [[Jacobian matrix and determinant|Jacobian]] of the transformation.<ref name=lovelock>{{cite book |lastlast1=Lovelock |firstfirst1=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=0-521-60484-2 |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 we desire 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> r display="block"> \sqrtbegin{x^2 + y^2align}</math>

r &= \sqrt{x^2 + y^2} \\

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

:<math> h = z</math>

h &= z

\end{align} </math>

and

:<math> x display= r"block"> \cosbegin{(t)align}</math>

:<math> tx &= r \sin{cos(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 useone 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 areis as representations of the "parent" function that takes a point of the manifold as an argument and gives the temperature.

The problem could have been reversed. WeOne 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 we will denotebe denoted by <math>D</math>. (Yes, integrating temperature is strange 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}}

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, our "room" is a quarter slice of a cylinder of radius and height 2).

:<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 integral of <math>f</math> over the region <math>D</math> is

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

:<math> \int_0^2 \! \int_{-\sqrt{2^2-x^2}}^\sqrt{2^2-x^2} \! \int_0^2 \! f(x,y,z) \, dz \, dy \, dx = 64/3 + 20 \pi</math>.

:<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>.

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

They are not equal. The integral of temperature is not independent of the coordinate system used. It is non-physical in that sense, hence "strange". Note that if the integral of <math>\bar{f}</math> included a factor of the Jacobian (which is just <math>r</math>), we get{{cn|date=August 2022}}

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

<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>

They are not equal! Clearly, the integral of a ordinary scalar depends on the coordinate system used. This coordinate dependence tends to remove any physical meaning from the integral of an ordinary scalar.

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.

==GeneralizationOther 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.

Scalar densities are special cases of [[Tensor density|tensor densities]].

==See also==

Line 56 ⟶ 82:

[[Category:Tensors]]

[[Category:Tensors in general relativity]]

[[Category:Scalars]]

{{math-stub}}