Class of accuracy in electrical measurements: 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> | |||
\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]] |
Revision as of 13:48, 15 December 2013
In mathematics, a relative scalar (of weight w) is a scalar-valued function whose transform under a coordinate transform,
on an n-dimensional manifold obeys the following equation
where
that is, the determinant of the Jacobian of the transformation. [1] A scalar density refers to the 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[2] refers to the case.
If and refer to the same point on the manifold, then we desire . This equation can be interpreted two ways when are viewed as the "new coordinates" and are viewed as the "original coordinates". The first is as , which "converts the function to the new coordinates". The second is as , 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 in Cartesian coordinates and the function in cylindrical coordinates is desired. The two coordinate systems are related by the following sets of equations:
and
Using allows one to derive as the transformed function.
Consider the point whose Cartesian coordinates are and whose corresponding value in the cylindrical system is . A quick calculation shows that and also. This equality would have held for any chosen point . Thus, is the "temperature function in the Cartesian coordinate system" and 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 and wished to have derived the Cartesian temperature function . 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 . (Yes, integrating temperature is strange but that's partly what's to be shown.) Suppose the region is given in cylindrical coordinates as from , from and from (that is, the "room" is a quarter slice of a cylinder of radius and height 2). The integral of over the region is
- .[3]
The value of the integral of over the same region is
- .[4]
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 included a factor of the Jacobian (which is just ), we get
- ,[5]
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 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 . This time but . As before is integral (the total mass) in Cartesian coordinates is
The value of the integral of over the same region is
They are equal. The integral of mass density gives total mass which is a coordinate-independent concept. Note that if the integral of also included a factor of the Jacobian like before, we get
- ,[6]
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
References
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