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Relative scalar

## Summary

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

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

on an n-dimensional manifold obeys the following equation

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

where

${\displaystyle J=\left|{\dfrac {\partial (x_{1},\ldots ,x_{n})}{\partial ({\bar {x}}^{1},\ldots ,{\bar {x}}^{n})}}\right|,}$

that is, the determinant of the Jacobian of the transformation.[1] A scalar density refers to the ${\displaystyle w=1}$ 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 ${\displaystyle w=0}$  case.

If ${\displaystyle x^{i}}$  and ${\displaystyle {\bar {x}}^{j}}$  refer to the same point ${\displaystyle P}$  on the manifold, then we desire ${\displaystyle {\bar {f}}({\bar {x}}^{j})=f(x^{i})}$ . This equation can be interpreted two ways when ${\displaystyle {\bar {x}}^{j}}$  are viewed as the "new coordinates" and ${\displaystyle x^{i}}$  are viewed as the "original coordinates". The first is as ${\displaystyle {\bar {f}}({\bar {x}}^{j})=f(x^{i}({\bar {x}}^{j}))}$ , which "converts the function to the new coordinates". The second is as ${\displaystyle f(x^{i})={\bar {f}}({\bar {x}}^{j}(x^{i}))}$ , 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 ${\displaystyle f(x,y,z)=2x+y+5}$  in Cartesian coordinates ${\displaystyle (x,y,z)}$  and the function in cylindrical coordinates ${\displaystyle (r,t,h)}$  is desired. The two coordinate systems are related by the following sets of equations: {\displaystyle {\begin{aligned}r&={\sqrt {x^{2}+y^{2}}}\\t&=\arctan(y/x)\\h&=z\end{aligned}}}  and {\displaystyle {\begin{aligned}x&=r\cos(t)\\y&=r\sin(t)\\z&=h.\end{aligned}}}

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

Consider the point ${\displaystyle P}$  whose Cartesian coordinates are ${\displaystyle (x,y,z)=(2,3,4)}$  and whose corresponding value in the cylindrical system is ${\displaystyle (r,t,h)=({\sqrt {13}},\arctan {(3/2)},4)}$ . A quick calculation shows that ${\displaystyle f(2,3,4)=12}$  and ${\displaystyle {\bar {f}}({\sqrt {13}},\arctan {(3/2)},4)=12}$  also. This equality would have held for any chosen point ${\displaystyle P}$ . Thus, ${\displaystyle f(x,y,z)}$  is the "temperature function in the Cartesian coordinate system" and ${\displaystyle {\bar {f}}(r,t,h)}$  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 ${\displaystyle {\bar {f}}}$  and wished to have derived the Cartesian temperature function ${\displaystyle f}$ . 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 ${\displaystyle D}$ . (Yes, integrating temperature is strange but that's partly what's to be shown.) Suppose the region ${\displaystyle D}$  is given in cylindrical coordinates as ${\displaystyle r}$  from ${\displaystyle [0,2]}$ , ${\displaystyle t}$  from ${\displaystyle [0,\pi /2]}$  and ${\displaystyle h}$  from ${\displaystyle [0,2]}$  (that is, the "room" is a quarter slice of a cylinder of radius and height 2). The integral of ${\displaystyle f}$  over the region ${\displaystyle D}$  is[citation needed] ${\displaystyle \int _{0}^{2}\!\int _{0}^{\sqrt {2^{2}-x^{2}}}\!\int _{0}^{2}\!f(x,y,z)\,dz\,dy\,dx=16+10\pi .}$  The value of the integral of ${\displaystyle {\bar {f}}}$  over the same region is[citation needed] ${\displaystyle \int _{0}^{2}\!\int _{0}^{\pi /2}\!\int _{0}^{2}\!{\bar {f}}(r,t,h)\,dh\,dt\,dr=12+10\pi .}$  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 ${\displaystyle {\bar {f}}}$  included a factor of the Jacobian (which is just ${\displaystyle r}$ ), we get[citation needed] ${\displaystyle \int _{0}^{2}\!\int _{0}^{\pi /2}\!\int _{0}^{2}\!{\bar {f}}(r,t,h)r\,dh\,dt\,dr=16+10\pi ,}$  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 ${\displaystyle f(x,y,z)=2x+y+5}$  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 ${\displaystyle {\bar {f}}(r,t,h)=(2r\cos(t)+r\sin(t)+5)r}$ . This time ${\displaystyle f(2,3,4)=12}$  but ${\displaystyle {\bar {f}}({\sqrt {13}},\arctan {(3/2)},4)=12{\sqrt {29}}}$ . As before is integral (the total mass) in Cartesian coordinates is ${\displaystyle \int _{0}^{2}\!\int _{0}^{\sqrt {2^{2}-x^{2}}}\!\int _{0}^{2}\!f(x,y,z)\,dz\,dy\,dx=16+10\pi .}$  The value of the integral of ${\displaystyle {\bar {f}}}$  over the same region is ${\displaystyle \int _{0}^{2}\!\int _{0}^{\pi /2}\!\int _{0}^{2}\!{\bar {f}}(r,t,h)\,dh\,dt\,dr=16+10\pi .}$  They are equal. The integral of mass density gives total mass which is a coordinate-independent concept. Note that if the integral of ${\displaystyle {\bar {f}}}$  also included a factor of the Jacobian like before, we get[citation needed] ${\displaystyle \int _{0}^{2}\!\int _{0}^{\pi /2}\!\int _{0}^{2}\!{\bar {f}}(r,t,h)r\,dh\,dt\,dr=24+40\pi /3,}$  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.