Class of accuracy in electrical measurements: Difference between revisions

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en>Nedim Ardoğa
 
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In mathematics, a '''relative scalar''' (of weight ''w'') is a scalar-valued function whose transform under a coordinate transform,
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:<math>
\bar{x}^j = \bar{x}^j(x^i)
</math>
 
on an ''n''-dimensional manifold obeys the following equation
 
:<math>
\bar{f}(\bar{x}^j) = J^w f(x^i)
</math>
 
where
 
: <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 |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.
 
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 a relative concept.
 
There are many physical quantities that are represented by ordinary scalars, such as temperature and pressure.
 
===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> r = \sqrt{x^2 + y^2} \, </math>
:<math> t = \arctan(y/x) \, </math>
:<math> h = z \, </math>
and
:<math> x = r \cos(t) \, </math>
:<math> t = r \sin(t) \, </math>
:<math> z = h. \, </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.
 
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 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
:<math> \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
:<math> \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
:<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> \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> \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
:<math> \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.
 
==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.
 
==See also==
*[[Jacobian matrix and determinant]]
 
==References==
{{Reflist}}
 
[[Category:Tensors]]
[[Category:Tensors in general relativity]]

Latest revision as of 06:09, 15 April 2014

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