Relative scalar

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

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

on an n-dimensional manifold obeys the following equation

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

where

$J={\frac {\partial (x_{1},\ldots ,x_{n})}{\partial ({\bar {x}}^{1},\ldots ,{\bar {x}}^{n})}},$

that is, the determinate [could determinant have been meant here?] of the transformation.

Scalar

An ordinary scalar refers to the $w=0$ case.

If $x^{i}$ and ${\bar {x}}^{j}$ refer to the same point $P$ on the manifold, then we desire ${\bar {f}}({\bar {x}}^{j})=f(x^{i})$. This equation can be interpreted two ways when ${\bar {x}}^{j}$ are viewed as the "new coordinates" and $x^{i}$ are viewed as the "original coordinates". The first is as ${\bar {f}}({\bar {x}}^{j})=f(x^{i}({\bar {x}}^{j}))$, which "converts the function to the new coordinates". The second is as $f(x^{i})={\bar {f}}({\bar {x}}^{j}(x^{i}))$, 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.

Elementary example

Suppose the temperature in a room is given in terms of the function $f(x,y,z)=2x+y+5$ in Cartesian coordinates (x,y,z) and we desire the function in cylindrical coordinates (r,t,h). The two coordinate systems are related by the following sets of equations:

$r={\sqrt {x^{2}+y^{2}}}\,$

$t=\arctan(y/x)\,$

$h=z\,$

and

$x=r\cos(t)\,$

$t=r\sin(t)\,$

$z=h.\,$

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

Consider the point $P$ whose Cartesian coordinates are $(x,y,z)=(2,3,4)$ and whose corresponding value in the cylindrical system is $(r,t,h)=({\sqrt {13}},\arctan {(3/2)},4)$. A quick calculation shows that $f(2,3,4)=12$ and ${\bar {f}}({\sqrt {13}},\arctan {(3/2)},4)=12$ also. This equality would have held for any chosen point P. Thus, $f(x,y,z)$ is the "temperature function in the Cartesian coordinate system" and ${\bar {f}}(r,t,h)$ is the "temperature function in the cylindrical coordinate system".

One way to view these functions are 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. We could have been given ${\bar {f}}$ and wished to have derived the Cartesian temperature function $f$. This just flips the notion of "new" vs the "original" coordinate system.

Suppose that one wishes to integrate these functions over "the room", which we will denote by $D$. (Yes, integrating temperature is strange but that's partly what's to be shown.) Suppose the region $D$ is given in cylindrical coordinates as $r$ from $[0,2]$, $t$ from $[0,\pi /2]$ and $h$ from $[0,2]$ (that is, our "room" is a quarter slice of a cylinder of radius and height 2). The integral of $f$ over the region $D$ is

$\int _{0}^{2}\!\int _{-{\sqrt {2^{2}-x^{2}}}}^{\sqrt {2^{2}-x^{2}}}\!\int _{0}^{2}\!f(x,y,z)\,dz\,dy\,dx=64/3+20\pi$.

The value of the integral of ${\bar {f}}$ over the same region is

$\int _{0}^{2}\!\int _{0}^{\pi /2}\!\int _{0}^{2}\!{\bar {f}}(r,t,h)\,dh\,dt\,dr=12+10\pi$.

They are not equal! 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.

Scalar density

A scalar density refers to the $w=1$ case.

Generalization

Scalar densities are special cases of tensor densities.

See also

• Jacobian matrix and determinant