Relative scalar

In mathematics, a relative scalar (of weight w) is a scalar-valued function whose transform under a coordinate transform, [math]\displaystyle{ \bar{x}^j = \bar{x}^j(x^i) }[/math]

on an n-dimensional manifold obeys the following equation

[math]\displaystyle{ \bar{f}(\bar{x}^j) = J^w f(x^i) }[/math]

where

[math]\displaystyle{ J = \left| \dfrac{\partial(x_1,\ldots,x_n)}{\partial(\bar{x}^1,\ldots,\bar{x}^n)} \right| , }[/math]

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

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

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

Consider the point [math]\displaystyle{ P }[/math] whose Cartesian coordinates are [math]\displaystyle{ (x,y,z)=(2,3,4) }[/math] and whose corresponding value in the cylindrical system is [math]\displaystyle{ (r,t,h)=(\sqrt{13},\arctan{(3/2)},4) }[/math]. A quick calculation shows that [math]\displaystyle{ f(2,3,4)=12 }[/math] and [math]\displaystyle{ \bar{f}(\sqrt{13},\arctan{(3/2)},4)=12 }[/math] also. This equality would have held for any chosen point [math]\displaystyle{ P }[/math]. Thus, [math]\displaystyle{ f(x,y,z) }[/math] is the "temperature function in the Cartesian coordinate system" and [math]\displaystyle{ \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]\displaystyle{ \bar{f} }[/math] and wished to have derived the Cartesian temperature function [math]\displaystyle{ 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]\displaystyle{ D }[/math]. (Yes, integrating temperature is strange but that's partly what's to be shown.) Suppose the region [math]\displaystyle{ D }[/math] is given in cylindrical coordinates as [math]\displaystyle{ r }[/math] from [math]\displaystyle{ [0,2] }[/math], [math]\displaystyle{ t }[/math] from [math]\displaystyle{ [0,\pi/2] }[/math] and [math]\displaystyle{ h }[/math] from [math]\displaystyle{ [0,2] }[/math] (that is, the "room" is a quarter slice of a cylinder of radius and height 2). The integral of [math]\displaystyle{ f }[/math] over the region [math]\displaystyle{ D }[/math] is [math]\displaystyle{ \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]\displaystyle{ \bar{f} }[/math] over the same region is [math]\displaystyle{ \int_0^2 \! \int_{0}^{\pi/2} \! \int_0^2 \! \bar{f}(r,t,h) \, dh \, dt \, dr = 12 + 10 \pi. }[/math] 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]\displaystyle{ \bar{f} }[/math] included a factor of the Jacobian (which is just [math]\displaystyle{ r }[/math]), we get [math]\displaystyle{ \int_0^2 \! \int_{0}^{\pi/2} \! \int_0^2 \! \bar{f}(r,t,h) r \, dh \, dt \, dr = 16 + 10 \pi, }[/math] 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]\displaystyle{ 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]\displaystyle{ \bar{f}(r,t,h)= (2 r \cos(t)+ r \sin(t) + 5) r }[/math]. This time [math]\displaystyle{ f(2,3,4)=12 }[/math] but [math]\displaystyle{ \bar{f}(\sqrt{13},\arctan{(3/2)},4)=12\sqrt{29} }[/math]. As before is integral (the total mass) in Cartesian coordinates is [math]\displaystyle{ \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]\displaystyle{ \bar{f} }[/math] over the same region is [math]\displaystyle{ \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]\displaystyle{ \bar{f} }[/math] also included a factor of the Jacobian like before, we get [math]\displaystyle{ \int_0^2 \! \int_{0}^{\pi/2} \! \int_0^2 \! \bar{f}(r,t,h) r \, dh \, dt \, dr = 24 + 40 \pi / 3 , }[/math] 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