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

$$ J = \left| \dfrac{\partial(x_1,\ldots,x_n)}{\partial(\bar{x}^1,\ldots,\bar{x}^n)} \right|, $$

that is, the determinant of the Jacobian of the transformation. A scalar density refers to the $$w=1$$ 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 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 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 $$f(x,y,z) = 2 x + y + 5$$ in Cartesian coordinates $$(x,y,z)$$ and the function in cylindrical coordinates $$(r,t,h)$$ is desired. The two coordinate systems are related by the following sets of equations: $$ \begin{align} r &= \sqrt{x^2 + y^2} \\ t &= \arctan(y/x) \\ h &= z \end{align} $$ and $$ \begin{align} x &= r \cos(t) \\ y &= r \sin(t) \\ z &= h. \end{align} $$

Using $$\bar{f}(\bar{x}^j) = f(x^i(\bar{x}^j))$$ allows one to derive $$\bar{f}(r,t,h)= 2 r \cos(t)+ r \sin(t) + 5$$ 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 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 $$\bar{f}$$ 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 will be denoted 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, the "room" is a quarter slice of a cylinder of radius and height 2). The integral of $$f$$ over the region $$D$$ is $$ \int_0^2 \! \int_{0}^\sqrt{2^2-x^2} \! \int_0^2 \! f(x,y,z) \, dz \, dy \, dx = 16 + 10 \pi.$$ The value of the integral of $$\bar{f}$$ 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. 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 $$\bar{f}$$ included a factor of the Jacobian (which is just $$r$$), we get $$ \int_0^2 \! \int_{0}^{\pi/2} \! \int_0^2 \! \bar{f}(r,t,h) r \, dh \, dt \, dr = 16 + 10 \pi,$$ 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 $$f(x,y,z) = 2 x + y + 5$$ 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 $$\bar{f}(r,t,h)= (2 r \cos(t)+ r \sin(t) + 5) r$$. This time $$f(2,3,4)=12$$ but $$\bar{f}(\sqrt{13},\arctan{(3/2)},4)=12\sqrt{29}$$. As before is integral (the total mass) in Cartesian coordinates is $$ \int_0^2 \! \int_{0}^\sqrt{2^2-x^2} \! \int_0^2 \! f(x,y,z) \, dz \, dy \, dx = 16 + 10 \pi.$$ The value of the integral of $$\bar{f}$$ over the same region is $$ \int_0^2 \! \int_{0}^{\pi/2} \! \int_0^2 \! \bar{f}(r,t,h) \, dh \, dt \, dr = 16 + 10 \pi.$$ They are equal. The integral of mass density gives total mass which is a coordinate-independent concept. Note that if the integral of $$\bar{f}$$ also included a factor of the Jacobian like before, we get $$ \int_0^2 \! \int_{0}^{\pi/2} \! \int_0^2 \! \bar{f}(r,t,h) r \, dh \, dt \, dr = 24 + 40 \pi / 3 ,$$ 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.