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 = \begin{vmatrix} \displaystyle \frac{\partial(x_1,\ldots,x_n)}{\partial(\bar{x}^1,\ldots,\bar{x}^n)} \end{vmatrix} ,

that is, the determinant of the Jacobian of the transformation. ^[1] 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^[2] 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:

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

t = \arctan(y/x) \,

h = z \,

and

x = r \cos(t) \,

y = r \sin(t) \,

z = h. \,

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

.^[3]

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

.^[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 $\bar{f}$ 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

,^[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 $f(x,y,z)= 2 x + y + 5$ 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

,^[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.

References

↑ Lovelock, David; Rund, Hanno (1 April 1989). "4". Tensors, Differential Forms, and Variational Principles (Paperback). Dover. p. 103. ISBN 0-486-65840-6. Retrieved 19 April 2011.
↑ Veblen, Oswald (2004). Invariants of Quadratic Differential Forms. Cambridge University Press. p. 21. ISBN 0-521-60484-2. Retrieved 3 October 2012.
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