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 |last1=Lovelock |first1=David |last2=Rund |first2=Hanno |authorlink2=Hanno Rund |title=Tensors, Differential Forms, and Variational Principles |date=1 April 1989 | publisher=Dover | isbn=0-486-65840-6 | url=http://store.doverpublications.com/0486658406.html | accessdate=19 April 2011 | format=Paperback | chapter=4 | page=103}}</ref> A '''scalar density''' refers to the <math>w=1</math> case.

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===Weight 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> yx &= 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.

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Suppose that one wishes to ''integrate'' these functions over "the room", which will be 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}}

:<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>.<ref>http://www.wolframalpha.com/input/?i=int_0^2+int_0^%28sqrt%282^2-x^2%29%29+int_0^2+%282+x+%2B+y+%2B+5%29+dz+dy+dx</ref>

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>.<ref>http://www.wolframalpha.com/input/?i=int_0^2+int_0^%28pi%2F2%29+int_0^2+%282+r+cos%28t%29+%2B+r+cos%28t%29+%2B+5%29+dh+dt+dr</ref>

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 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>,<ref>http://www.wolframalpha.com/input/?i=int_0^2+int_0^%28pi%2F2%29+int_0^2+%282+r+cos%28t%29+%2B+r+cos%28t%29+%2B+5%29+r^2+dh+dt+dr</ref>

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>),

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.

we get

:<math> \int_0^2 \! \int_{0}^{\pi/2} \! \int_0^2 \! \bar{f}(r,t,h) r \, dh \, dt \, dr = 16 + 10 \pi</math>,<ref>http://www.wolframalpha.com/input/?i=int_0^2+int_0^%28pi%2F2%29+int_0^2+%282+r+cos%28t%29+%2B+r+cos%28t%29+%2B+5%29+r^2+dh+dt+dr</ref>

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.

===Weight 1 example===

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

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>,<ref>http://www.wolframalpha.com/input/?i=int_0^2+int_0^%28pi%2F2%29+int_0^2+%282+r+cos%28t%29+%2B+r+cos%28t%29+%2B+5%29+r^2+dh+dt+dr</ref>

which is not equal to the previous case.